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2. Assign five college students to three villages in a certain town. If each village must have at least one student, then the number of different assignment schemes is $\qquad$ . | 2. 150 .
According to the number of college students allocated to each village, the only two types that meet the requirements are $1,1,3$ and $1, 2, 2$. Therefore, the number of different allocation schemes is
$$
C_{3}^{1} C_{5}^{3} C_{2}^{1}+C_{3}^{1} C_{5}^{2} C_{3}^{2}=60+90=150
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a_{n}=2^{n}, b_{n}=5 n-1\left(n \in \mathbf{Z}_{+}\right)$,
$$
S=\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{a_{2015}}\right\} \text {. }
$$
Then the number of elements in the set $S$ is | 5.504.
Since the set $\left\{b_{1}, b_{2}, \cdots, b_{2015}\right\}$ contains $2^{2015}$ elements, forming an arithmetic sequence with a common difference of 5, and each term has a remainder of 4 when divided by 5, we only need to consider the number of terms in the set $\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\}$ that have a remainder of 4 when divided by 5.
For $k \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
a_{4 k}=2^{4 k} \equiv 1(\bmod 5), \\
a_{4 k-1}=2^{4(k-1)} \times 2^{3} \equiv 2^{3} \equiv 3(\bmod 5), \\
a_{4 k-2}=2^{4(k-1)} \times 2^{2} \equiv 2^{2} \equiv 4(\bmod 5), \\
a_{4 k-3}=2^{4(k-1)} \times 2 \equiv 2(\bmod 5) . \\
\text { Let } 4 k-2 \leqslant 2015 \Rightarrow k \leqslant 504+\frac{1}{4}
\end{array}
$$
Therefore, the number of elements in set $S$ is 504. | 504 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $P(1,4,5)$ is a fixed point in the rectangular coordinate system $O-x y z$, a plane is drawn through $P$ intersecting the positive half-axes of the three coordinate axes at points $A$, $B$, and $C$ respectively. Then the minimum value of the volume $V$ of all such tetrahedrons $O-A B C$ is $\qquad$ | 4. 90 .
Let the plane equation be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where the positive numbers $a$, $b$, and $c$ are the intercepts of the plane on the $x$-axis, $y$-axis, and $z$-axis, respectively.
Given that point $P$ lies within plane $ABC$, we have $\frac{1}{a}+\frac{4}{b}+\frac{5}{c}=1$.
From $1=\frac{1}{a}+\frac{4}{b}+\frac{5}{c} \geqslant 3 \sqrt[3]{\frac{1}{a} \times \frac{4}{b} \times \frac{5}{c}}$ $\Rightarrow \frac{1}{6} a b c \geqslant 90$.
When $(a, b, c)=(3,12,15)$, $V$ reaches its minimum value of 90. | 90 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) From the 2015 positive integers 1, 2, $\cdots$, 2015, select $k$ numbers such that the sum of any two different numbers is not a multiple of 50. Find the maximum value of $k$.
untranslated part:
在 1,2 , $\cdots, 2015$ 这 2015 个正整数中选出 $k$ 个数,使得其中任意两个不同的数之和均不为 50 的倍数. 求 $k$ 的最大值.
translated part:
From the 2015 positive integers 1, 2, $\cdots$, 2015, select $k$ numbers such that the sum of any two different numbers is not a multiple of 50. Find the maximum value of $k$. | 10. Classify $1 \sim 2015$ by their remainders when divided by 50.
Let $A_{i}$ represent the set of numbers from $1 \sim 2015$ that have a remainder of $i$ when divided by 50, where $i=0,1, \cdots, 49$. Then $A_{1}, A_{2}, \cdots, A_{15}$ each contain 41 numbers, and $A_{0}, A_{16}, A_{17}, \cdots, A_{49}$ each contain 40 numbers.
According to the problem, at most one number can be selected from $A_{0}$ and $A_{25}$; and no numbers can be selected simultaneously from $A_{i}$ and $A_{50-i}$ (for $i=1,2, \cdots, 24$).
Notice that, when $1 \leqslant i \leqslant 15$, $A_{i}$ contains more numbers than $A_{50-i}$.
Therefore, all numbers from $A_{1}, A_{2}, \cdots, A_{15}$ can be selected.
For $16 \leqslant i \leqslant 24$, all numbers from $A_{i}$ (or $A_{50-i}$) can be selected.
In summary, the maximum value of $k$ is
$$
15 \times 41 + 9 \times 40 + 1 + 1 = 977 \text{. }
$$ | 977 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Real numbers $x, y, a$ satisfy $x+y=a+1$ and $xy=a^{2}-7a+16$. Then the maximum value of $x^{2}+y^{2}$ is $\qquad$ | 5. 32 .
Notice that,
$$
\begin{array}{l}
x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=(a+1)^{2}-2\left(a^{2}-7 a+16\right) \\
=-a^{2}+16 a-31=-(a-8)^{2}+33 . \\
\text { Also, }(x-y)^{2}=(a+1)^{2}-4\left(a^{2}-7 a+16\right) \geqslant 0 \\
\Rightarrow-3 a^{2}+30 a-63 \geqslant 0 \Rightarrow 3 \leqslant a \leqslant 7 .
\end{array}
$$
Thus, on the closed interval $[3,7]$, the function
$$
f(a)=-(a-8)^{2}+33
$$
is increasing.
Therefore, its maximum value is obtained at $a=7$, i.e., the maximum value of $x^{2}+y^{2}$ is 32. | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a triangle with sides as three consecutive natural numbers, the largest angle is twice the smallest angle. Then the perimeter of the triangle is $\qquad$ | II, 7.15.
Let the three sides of $\triangle ABC$ be $AB=n+1, BC=n, AC=n-1$, and the angle opposite to side $AC$ be $\theta$, and the angle opposite to side $AB$ be $2\theta$.
According to the Law of Sines and the Law of Cosines, we have
$$
\begin{array}{l}
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \frac{n+1}{n-1}=2 \cos \theta, \\
\cos \theta=\frac{(n+1)^{2}+n^{2}-(n-1)^{2}}{2 n(n+1)} \\
=\frac{n+4}{2(n+1)} .
\end{array}
$$
From equations (1) and (2), we get
$$
\frac{n+1}{n-1}=\frac{n+4}{n+1} \Rightarrow n=5 \text {. }
$$
Therefore, the perimeter of the triangle is 15. | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the minimum value of the function
$$
f(x)=|x-1|+|x-2|+\cdots+|x-10|
$$
. ${ }^{[1]}$ | 【Analysis】By the geometric meaning of absolute value, $\sum_{i=1}^{n}\left|x-a_{i}\right|$ represents the sum of distances from the point corresponding to $x$ on the number line to the points corresponding to $a_{i}(i=1,2, \cdots, n)$. It is easy to know that when the point corresponding to $x$ is in the middle of the points corresponding to $a_{i}$ (when $n=2 k+1$ $(k \in \mathbf{N})$) or between the middle two points (when $n=2 k$ $\left(k \in \mathbf{Z}_{+}\right)$), the sum of distances is minimized.
When $x \in[5,6]$, $f(x)$ achieves its minimum value of 25.
By generalizing Example 1, it is not difficult to obtain
Conclusion 1 Suppose $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. For the function
$$
f(x)=\sum_{i=1}^{n}\left|x-a_{i}\right| \text {, }
$$
(1) When $n=2 k+1(k \in \mathbf{N})$, the point where $f(x)$ achieves its minimum value is $a_{k+1}$, and
$$
f(x)_{\min }=\sum_{i=k+2}^{n} a_{i}-\sum_{i=1}^{k} a_{i} ;
$$
(2) When $n=2 k\left(k \in \mathbf{Z}_{+}\right)$, every point in the interval $\left[a_{k}, a_{k+1}\right]$
is a point where $f(x)$ achieves its minimum value, and
$$
f(x)_{\min }=\sum_{i=k+1}^{n} a_{i}-\sum_{i=1}^{k} a_{i} .
$$ | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given a moving large circle $\odot O$ that is externally tangent to a fixed small circle $\odot O_{1}$ with radius 3 at point $P, AB$ is the external common tangent of the two circles, with $A, B$ being the points of tangency. A line $l$ parallel to $AB$ is tangent to $\odot O_{1}$ at point $C$ and intersects $\odot O$ at points $D, E$. Then $C D \cdot C E=$ | 4.36.
As shown in Figure 5, connect $AP$, $PB$, $O_{1}C$, $BO_{1}$, $PC$, and draw the common tangent line of the two circles through point $P$, intersecting $AB$ at point $Q$.
It is easy to prove that $\angle APB=90^{\circ}$, points $B$, $O_{1}$, and $C$ are collinear, $\angle BPC=90^{\circ}$, and points $A$, $P$, and $C$ are collinear.
By the secant theorem and the projection theorem, we get
$$
CD \cdot CE = AC \cdot PC = BC^2 = 36 .
$$ | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let real numbers $x_{1}, x_{2}, \cdots, x_{1999}$ satisfy the condition $\sum_{i=1}^{1990}\left|x_{i}-x_{i+1}\right|=1991$.
And $y_{k}=\frac{1}{k} \sum_{i=1}^{k} x_{i}(k=1,2, \cdots, 1991)$. Try to find the maximum value of $\sum_{i=1}^{1990}\left|y_{i}-y_{i+1}\right|$. ${ }^{[3]}$ | 【Analysis】For $k=1,2, \cdots, 1990$, we have
$$
\begin{array}{l}
\left|y_{k}-y_{k+1}\right|=\left|\frac{1}{k} \sum_{i=1}^{k} x_{i}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}\right| \\
=\left|\frac{1}{k(k+1)}\left(\sum_{i=1}^{k} x_{i}-k x_{k+1}\right)\right| \\
\leqslant \frac{1}{k(k+1)} \sum_{i=1}^{k} i\left|x_{i}-x_{i+1}\right| . \\
\text { Hence } \sum_{i=1}^{190}\left|y_{i}-y_{i+1}\right| \\
\leqslant \sum_{i=1}^{1990}\left[i\left|x_{i}-x_{i+1}\right| \sum_{k=i}^{1990} \frac{1}{k(k+1)}\right] \\
=\left(\sum_{i=1}^{1990}\left|x_{k}-x_{k+1}\right|\right)\left(1-\frac{k}{1991}\right) \\
\leqslant 1991\left(1-\frac{1}{1991}\right)=1990 .
\end{array}
$$
The maximum value 1990 is achieved when $x=1991, x_{2}=x_{3}=\cdots=x_{1991}=0$. | 1990 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let positive integers $a_{1}, a_{2}, \cdots, a_{31}, b_{1}, b_{2}, \cdots, b_{31}$ satisfy
$$
\begin{array}{l}
\text { (1) } a_{1}<a_{2}<\cdots<a_{31} \leqslant 2015, \\
b_{1}<b_{2}<\cdots<b_{31} \leqslant 2015 ; \\
\text { (2) } a_{1}+a_{2}+\cdots+a_{31}=b_{1}+b_{2}+\cdots+b_{31} \text {. } \\
\text { Find } S=\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\cdots+\left|a_{31}-b_{31}\right|
\end{array}
$$
the maximum value.
(Supplied by He Yijie) | 1. Define the sets
$$
\begin{array}{l}
A=\left\{m \mid a_{m}>b_{m}, 1 \leqslant m \leqslant 31\right\}, \\
B=\left\{n \mid a_{n}<b_{n}, 1 \leqslant n \leqslant 31\right\} . \\
\text { Let } S_{1}=\sum_{m \in A}\left(a_{m}-b_{m}\right), S_{2}=\sum_{m \in B}\left(b_{n}-a_{n}\right) .
\end{array}
$$
Then $S=S_{1}+S_{2}$.
From condition (2), we know
$$
\begin{array}{l}
S_{1}-S_{2}=\sum_{m \in A \cup B}\left(a_{m}-b_{m}\right)=0 \\
\Rightarrow S_{1}=S_{2}=\frac{S}{2} . \\
\text { When } A=\varnothing, S=2 S_{1}=0 .
\end{array}
$$
Assume $A \neq \varnothing$, then $B \neq \varnothing$. In this case, $|A|, |B|$ are positive integers, and $|A|+|B| \leqslant 31$.
$$
\begin{array}{l}
\text { Let } u=a_{k}-b_{k}=\max _{m \in A}\left\{a_{m}-b_{m}\right\}, \\
v=b_{l}-a_{l}=\max _{n \in B}\left\{b_{n}-a_{n}\right\} .
\end{array}
$$
We will prove: $u+v \leqslant 1984$.
Without loss of generality, assume $1 \leqslant k<l \leqslant 31$.
Then $u+v=a_{k}-b_{k}+b_{l}-a_{l}$
$$
=b_{31}-\left(b_{31}-b_{l}\right)-b_{k}-\left(a_{l}-a_{k}\right) \text {. }
$$
From condition (1), we have
$$
\begin{array}{l}
b_{31} \leqslant 2015, b_{31}-b_{l} \geqslant 31-l, \\
b_{k} \geqslant k, a_{l}-a_{k} \geqslant l-k .
\end{array}
$$
Thus, $u+v \leqslant 2015-(31-l)-k-(l-k)$ $=1984$.
Clearly, $S_{1} \leqslant u|A|, S_{2} \leqslant v|B|$, hence,
$$
\begin{array}{l}
1984 \geqslant u+v \geqslant \frac{S_{1}}{|A|}+\frac{S_{2}}{|B|} \geqslant \frac{S_{1}}{|A|}+\frac{S_{2}}{31-|A|} \\
=\frac{S}{2} \cdot \frac{31}{|A|(31-|A|)} \geqslant \frac{31 S}{2 \times 15 \times 16} \\
\Rightarrow S \leqslant \frac{2 \times 15 \times 16}{31} \times 1984=30720 .
\end{array}
$$
On the other hand, if we take
$$
\begin{array}{l}
\left(a_{1}, a_{2}, \cdots, a_{16}, a_{17}, a_{18}, \cdots, a_{31}\right) \\
=(1,2, \cdots, 16,2001,2002, \cdots, 2015), \\
\left(b_{1}, b_{2}, \cdots, b_{31}\right)=(961,962, \cdots, 991),
\end{array}
$$
then conditions (1) and (2) are satisfied. In this case,
$$
S=2 S_{1}=2 \times 16 \times 960=30720 \text {. }
$$
In conclusion, the maximum value of $S$ is 30720. | 30720 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. In the coordinate plane, points with both coordinates as integers are called integer points, and triangles with all three vertices as integer points are called integer point triangles. Find the number of integer point right triangles $OAB$ with a right angle at the origin $O$ and with $I(2015,7 \times 2015)$ as the incenter. | 14. Let point $A$ be in the first quadrant.
Let $\angle x O I=\alpha$.
Then $\tan \alpha=7$,
$k_{O A}=\tan \left(\alpha-\frac{\pi}{4}\right)=\frac{\tan \alpha-1}{1+\tan \alpha}=\frac{3}{4}$
$\Rightarrow k_{O B}=-\frac{4}{3}$.
Since $A$ and $B$ are integer points, set
$$
A\left(4 t_{1}, 3 t_{1}\right), B\left(-3 t_{2}, 4 t_{2}\right)\left(t_{1} 、 t_{2} \in \mathbf{Z}_{+}\right) \text {. }
$$
Thus, $|O A|=5 t_{1},|O B|=5 t_{2}$.
Let the inradius of $\triangle O A B$ be $r$. Then
$r=\frac{\sqrt{2}}{2}|O I|=\frac{\sqrt{2}}{2} \times 5 \sqrt{2} \times 2015=5 \times 2015$.
Also, $r=\frac{|O A|+|O B|-|A B|}{2}$
$\Rightarrow|A B|=|O A|+|O B|-2 r$
$\Rightarrow|A B|^{2}=(|O A|+|O B|-2 r)^{2}=|O A|^{2}+|O B|^{2}$
$\Rightarrow\left(5 t_{1}+5 t_{2}-2 \times 5 \times 2015\right)^{2}=25 t_{1}^{2}+25 t_{2}^{2}$
$\Rightarrow\left(t_{1}+t_{2}-2 \times 2015\right)^{2}=t_{1}^{2}+t_{2}^{2}$.
Set $t_{1}=x+2015, t_{2}=y+2015$. Then
$(x+y)^{2}=(x+2015)^{2}+(y+2015)^{2}$
$\Rightarrow x y=2015 x+2015 y+2015^{2}$
$\Rightarrow(x-2015)(y-2015)=2 \times 2015^{2}$
$=2 \times 5^{2} \times 13^{2} \times 31^{2}$.
Since $|O A|>2 r,|O B|>2 r$, we know
$x-2015, y-2015 \in \mathbf{Z}_{+}$.
Note that, $2 \times 5^{2} \times 13^{2} \times 31^{2}$ has $2 \times 3 \times 3 \times 3=54$ positive divisors.
Thus, there are 54 pairs of $(x, y)$ that satisfy the conditions.
Therefore, there are 54 triangles that satisfy the conditions. | 54 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. For any positive integer $m$, the set
$$
\{m, m+1, m+2, \cdots, m+99\}
$$
any $n(n \geqslant 3)$-element subset of it, always contains three elements that are pairwise coprime. Find the minimum value of $n$. | 15. Consider the set $\{1,2, \cdots, 100\}$ with $m=1$, and its 67-element subset, whose elements are even numbers and odd numbers divisible by 3, i.e.,
$$
P=\{2,4, \cdots, 100,3,9, \cdots, 99\} .
$$
Clearly, there do not exist three pairwise coprime elements in set $P$.
Thus, $n \leqslant 67$ does not meet the requirement.
We now prove that $n=68$ meets the requirement.
First, we prove a lemma.
Lemma For any positive integer $m$, in any five-element subset of the set
$$
\{m, m+1, \cdots, m+5\}
$$
there are always three elements that are pairwise coprime.
Proof Let $A$ be a five-element subset of the set $\{m, m+1, \cdots, m+5\}$.
Notice that among $m, m+1, \cdots, m+5$, there are three odd and three even numbers, and exactly one of them is a multiple of 5.
Thus, if set $A$ contains three odd numbers, these three odd numbers must be pairwise coprime, and the conclusion holds.
If set $A$ contains two odd and three even numbers, since at most one of the three even numbers is a multiple of 3, and at most one is a multiple of 5, there must be one even number that is neither a multiple of 3 nor 5. This number is coprime with the two odd numbers, and the conclusion holds.
Returning to the original problem.
For any positive integer $m$, partition the set
$$
\{m, m+1, \cdots, m+99\}
$$
into the following 17 sets:
$$
A_{i}=\{m+6(i-1)+j \mid j=0,1, \cdots, 5\}
$$
$(i=1,2, \cdots, 16)$,
$$
A_{17}=\{m+96, m+97, m+98, m+99\} .
$$
Let $M$ be a 68-element subset of the set $\{m, m+1, \cdots, m+99\}$.
(1) If set $M$ has four elements from set $A_{17}$, since when $m$ is odd, $m+96, m+97, m+98$ are pairwise coprime; and when $m$ is even, $m+97, m+98, m+99$ are pairwise coprime, therefore, set $M$ contains at least three elements that are pairwise coprime.
(2) If set $M$ has at most three elements from set $A_{17}$, then set $M$ has at least 65 elements from sets $A_{1}, A_{2}, \cdots, A_{16}$.
By the pigeonhole principle, set $M$ must have at least five elements from one of these sets, say $A_{1}$. By the lemma, there exist three pairwise coprime elements among them. Therefore, set $M$ always contains three pairwise coprime elements.
Thus, $n=68$ meets the requirement, i.e., for any positive integer $m$, any 68-element subset of the set $\{m, m+1, \cdots, m+99\}$ contains three elements that are pairwise coprime.
In conclusion, the minimum value of $n$ is 68.
(Provided by Chen Deyan) | 68 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. After removing all perfect squares from the sequence of positive integers $\{1,2, \cdots\}$, the remaining numbers form a sequence $\left\{a_{n}\right\}$ in their original order. Then $a_{2015}=$ $\qquad$ . | 6.2060.
Let $k^{2}2014
\end{array}\right. \\
\Rightarrow k=45, a_{2015}=2060 .
\end{array}
$ | 2060 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$, and
$$
|b-c| \cos \frac{A}{2}=8,(b+c) \sin \frac{A}{2}=15 \text {. }
$$
Then $a=$ | 2. 17 .
From the cosine theorem, we get
$$
\begin{aligned}
a^{2}= & b^{2}+c^{2}-2 b c \cos A \\
= & \left(b^{2}+c^{2}\right)\left(\sin ^{2} \frac{A}{2}+\cos ^{2} \frac{A}{2}\right)- \\
& 2 b c\left(\cos ^{2} \frac{A}{2}-\sin ^{2} \frac{A}{2}\right) \\
= & (b+c)^{2} \sin ^{2} \frac{A}{2}+(b-c)^{2} \cos ^{2} \frac{A}{2} \\
= & 15^{2}+8^{2}=17^{2} .
\end{aligned}
$$
Therefore, $a=17$. | 17 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $\left\{a_{n}\right\}$ be a monotonically increasing sequence of positive integers, satisfying
$$
a_{n+2}=3 a_{n+1}-a_{n}, a_{6}=280 \text {. }
$$
Then $a_{7}=$ | 7.733.
From the problem, we have
$$
\begin{array}{l}
a_{3}=3 a_{2}-a_{1}, \\
a_{4}=3 a_{3}-a_{2}=8 a_{2}-3 a_{1}, \\
a_{5}=3 a_{4}-a_{3}=21 a_{2}-8 a_{1}, \\
a_{6}=3 a_{5}-a_{4}=55 a_{2}-21 a_{1}=280=5 \times 7 \times 8, \\
a_{7}=3 a_{6}-a_{5}=144 a_{2}-55 a_{1} .
\end{array}
$$
Since $(21,55)=1$, and $a_{2}>a_{1}$ are both positive integers, it follows that $5\left|a_{1}, 7\right| a_{2}$.
Let $a_{1}=5 b, a_{2}=7 c\left(b, c \in \mathbf{Z}_{+}\right)$, we get $11 c-3 b=8$.
It is easy to verify that $a_{1}=5, a_{2}=7$.
Then $a_{7}=144 a_{2}-55 a_{1}=733$. | 733 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given positive integers $a$, $b$, $c$, $d$ satisfy $a^{2}=c(d+29)$, $b^{2}=c(d-29)$. Then the value of $d$ is $\qquad$. | $=, 1.421$
Let $(a, b)=d^{\prime}, a=d^{\prime} a_{1}, b=d^{\prime} b_{1},\left(a_{1}, b_{1}\right)=1$.
From the given conditions, dividing the two equations yields
$$
\begin{array}{l}
\frac{a^{2}}{b^{2}}=\frac{d+29}{d-29} \Rightarrow \frac{a_{1}^{2}}{b_{1}^{2}}=\frac{d+29}{d-29} \\
\Rightarrow\left\{\begin{array}{l}
d+29=a_{1}^{2} t, \\
d-29=b_{1}^{2} t
\end{array}\left(t \in \mathbf{Z}_{+}\right)\right. \\
\Rightarrow 2 \times 29=\left(a_{1}+b_{1}\right)\left(a_{1}-b_{1}\right) t .
\end{array}
$$
Since $a_{1}+b_{1}$ and $a_{1}-b_{1}$ have the same parity,
$$
t=2, a_{1}+b_{1}=29, a_{1}-b_{1}=1 \text {. }
$$
Thus, $a_{1}=15, b_{1}=14, d=421$. | 421 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $a$ and $b$ are positive integers, the fraction $\frac{a}{b}$, when converted to a decimal, contains the consecutive digits $\overline{2015}$. Then the minimum value of $b$ is . $\qquad$ | 5. 129.
Given that multiplying a fraction by an appropriate power of 10 can result in the first four digits after the decimal point being $\overline{2015}$, we only need to find the minimum value of $b$ for the fraction $\frac{a}{b}=0.2015 \cdots$.
On one hand,
$\frac{a}{b}-\frac{1}{5}=0.2015 \cdots-0.2=0.0015$
$$
125 \text {. }
$$
We will test the values below.
When $b=126$,
$\frac{25}{126}=0.1984 \cdots \neq 0.2015 \cdots$;
When $b=127$, $\frac{26}{127}=0.2047 \cdots$;
When $b=128$, $\frac{26}{128}=0.2031 \cdots$;
When $b=129$, $\frac{26}{129}=0.2015 \cdots$.
Therefore, the minimum value of $b$ is 129. | 129 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the sequence $\left\{a_{n}\right\}$ :
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, a_{3}=5, \\
a_{n}=a_{n-1}-a_{n-2}+a_{n-3}(n=4,5, \cdots) .
\end{array}
$$
Then the sum of the first 2015 terms of this sequence $S_{2015}=$ | -、1.6045.
From the given information, we have $a_{4}=3$, and the sequence $\left\{a_{n}\right\}$ has a period of 4. Therefore, $S_{2015}$
$$
\begin{array}{l}
=503\left(a_{1}+a_{2}+a_{3}+a_{4}\right)+a_{1}+a_{2}+a_{3} \\
=503(1+3+5+3)+1+3+5=6045 .
\end{array}
$$ | 6045 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $f(x)=a x^{5}+b x^{3}+c x+10$, and $f(3)$ $=3$. Then $f(-3)=$ $\qquad$ | 5. 17 .
Notice that, for any $x$, we have $f(x)+f(-x)=20$.
Therefore, $f(-3)=17$. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Try to find the largest integer $k$, such that for each positive integer $n$, we have
$$
1980^{k} \left\lvert\, \frac{(1980 n)!}{(n!)^{1980}} .\right.
$$ | Notice,
$$
\begin{array}{l}
1980=2^{2} \times 3^{2} \times 5 \times 11, \\
v_{11}(1980!)=\sum_{i \geqslant 1}\left[\frac{1980}{11^{i}}\right]=197 . \\
\text { Since }\left[\frac{m}{5^{i}}\right] \geqslant\left[2 \times \frac{m}{11^{i}}\right] \geqslant 2\left[\frac{m}{11^{i}}\right], \text { hence } \\
v_{5}(m!) \geqslant 2 v_{11}(m!) .
\end{array}
$$
Since $\left[\frac{m}{5^{i}}\right] \geqslant\left[2 \times \frac{m}{11^{i}}\right] \geqslant 2\left[\frac{m}{11^{i}}\right]$, hence
similarly, $v_{3}(m!) \geqslant 3 v_{11}(m!)$,
$$
v_{2}(m!) \geqslant 5 v_{11}(m!) \text {. }
$$
Then the maximum $k$ satisfying $1980^{k} \mid 1980!$ is 197.
For any prime $p$,
$$
\begin{array}{l}
v_{p}\left(\frac{(1980 n)!}{(n!)^{1980}}\right) \\
=v_{p}((1980 n)!)-1980 v_{p}(n!) \\
=\frac{1980 S_{p}(n)-S_{p}(1980 n)}{p-1} .
\end{array}
$$
Using $S_{p}(m n) \leqslant S_{p}(m) S_{p}(n)$, we get
$$
\begin{array}{l}
v_{p}\left(\frac{(1980 n)!}{(n!)^{1980}}\right) \geqslant \frac{1980-S_{p}(1980)}{p-1} S_{p}(n) \\
\geqslant v_{p}(1980!) .
\end{array}
$$
Therefore, for any positive integer $n$, we have
$$
1980^{197} \left\lvert\, \frac{(1980 n)!}{(n!)^{1980}}\right. \text {. }
$$
In summary, the maximum $k$ is 197. | 197 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Polynomial
$$
p(x)=x^{3}-224 x^{2}+2016 x-d
$$
has three roots that form a geometric progression. Then the value of $d$ is $\qquad$ | 2. 729 .
Let the three roots of the polynomial be $a$, $b$, and $c$, and $b^{2}=a c$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a+b+c=224, \\
a b+b c+c a=2016, \\
a b c=d .
\end{array}\right.
$$
Then $b=\frac{b(a+b+c)}{a+b+c}=\frac{a b+b c+a c}{a+b+c}=9$.
Thus, $d=a b c=b^{3}=729$. | 729 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $a, b, c, d$ are prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. [1] | Let
\[
\begin{array}{l}
a b c d=x+(x+1)+\cdots+(x+34) \\
=5 \times 7(x+17) .
\end{array}
\]
Assume \( a=5, b=7, c \leqslant d, c d=x+17 \).
Thus, \( d_{\text {min }}=5 \).
If \( d=5 \), then \( c_{\text {min }}=5 \), at this point,
\[
x=8, a+b+c+d=22 \text {; }
\]
If \( d=7 \), then \( c_{\text {min }}=3 \), at this point,
\[
x=4, a+b+c+d=22 \text {; }
\]
If \( d \geqslant 11 \), then
\[
a+b+c+d \geqslant 5+7+2+11=25 \text {. }
\]
In summary, the minimum value of \( a+b+c+d \) is 22. | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given positive integer $n=a b c<10000, a, b, c$ are all prime numbers, and $2 a+3 b=c, 4 a+c+1=4 b$. Find the value of $n$.
| From the given equations, we have
$$
b=6 a+1, c=20 a+3 \text{. }
$$
Then $a(6 a+1)(20 a+3)<10000$
$\Rightarrow 12 a^{3}<10000 \Rightarrow$ prime $a<5$.
Combining $b=6 a+1, c=20 a+3$ being primes, we can determine
$$
a=2, b=13, c=43, n=1118 \text{. }
$$ | 1118 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $S=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{980100}}$. Find the greatest positive integer $[S]$ that does not exceed the real number $S$. | Notice that, $\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}$.
Therefore, $\frac{1}{\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$,
$2(\sqrt{n+1}-\sqrt{n})=\frac{2}{\sqrt{n+1+\sqrt{n}}}<\frac{1}{\sqrt{n}}$.
Thus, $1977<2(\sqrt{980101}-\sqrt{2})$
$$
<S<2(\sqrt{980100}-1)=1978 \text {. }
$$
Hence, $[S]=1977$. | 1977 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 For integer pairs $(a, b)(0<a<b<1000)$, a set $S \subseteq\{1,2, \cdots, 2003\}$ is called a "jump set" of the pair $(a, b)$: if for any element pair $\left(s_{1}, s_{2}\right)$, $s_{1} 、 s_{2} \in$ $S,\left|s_{1}-s_{2}\right| \notin\{a, b\}$.
Let $f(a, b)$ be the maximum number of elements in a jump set of the pair $(a, b)$. Find the maximum and minimum values of $f(a, b)$. | 【Analysis】This method is quite typical, the minimum value is obtained using the greedy idea, while the maximum value is derived using the pigeonhole principle. Unfortunately, during the exam, most candidates only answered one part correctly.
For the minimum value, take $a=1, b=2$, then
$$
\begin{array}{l}
\{1,2,3\},\{4,5,6\}, \cdots, \\
\{1999,2000,2001\},\{2002,2003\}
\end{array}
$$
each set can contribute at most one element.
Thus, $f(1,2) \leqslant 668$.
Next, we prove that the minimum value is indeed 668, which requires proving that for any $(a, b), f(a, b) \geqslant 668$.
This proof process is similar to Example 4.
Since adding $x$ to the jump set $S$ prevents $x+a$ and $x+b$ from being added, after the first 1 is taken, 1, $1+a$, and $1+b$ are removed. Each time, "greedily" take the smallest number $x$ that can be taken, and remove $x$, $x+a$, and $x+b$. This means that taking one element can prevent at most three elements from being taken, thus,
$$
f(a, b) \geqslant\left\lceil\frac{2003}{3}\right\rceil=668 \text {, }
$$
where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.
In summary, $f_{\text {min }}=668$. | 668 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Let positive integers $a_{1}, a_{2}, \cdots, a_{31}, b_{1}, b_{2}$, $\cdots, b_{31}$ satisfy
$$
\begin{array}{l}
\text { (1) } a_{1}<a_{2}<\cdots<a_{31} \leqslant 2015, \\
b_{1}<b_{2}<\cdots<b_{31} \leqslant 2015 ;
\end{array}
$$
$$
\text { (2) } a_{1}+a_{2}+\cdots+a_{31}=b_{1}+b_{2}+\cdots+b_{31} \text {. }
$$
Find the maximum value of $S=\sum_{i=1}^{31}\left|a_{i}-b_{i}\right|$. | 【Detailed Estimation】It is known that, $a_{31} \leqslant 2015, a_{30} \leqslant 2014, \cdots \cdots$
Thus, for all $1 \leqslant i \leqslant 31$, we have $a_{i} \leqslant 1984+i$.
Similarly, $b_{i} \leqslant 1984+i$.
Also, $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots \cdots$
Thus, for all $1 \leqslant i \leqslant 31$, we have $a_{i} \geqslant i$.
Similarly, $b_{i} \geqslant i$.
Furthermore, $a_{i+1} \geqslant a_{i}+1$,
$a_{i+2} \geqslant a_{i+1}+1 \geqslant a_{i}+2, \cdots \cdots$.
Thus, for all $1 \leqslant i < j$. Then
$$
\begin{array}{l}
\left(a_{i}-b_{i}\right)+\left(b_{j}-a_{j}\right)=a_{i}-\left(b_{i}-b_{j}\right)-a_{j} \\
\leqslant(1984+i)-(i-j)-j=1984 . \\
\text { Hence } \frac{31 A}{240} \leqslant 1984 \\
\Rightarrow A \leqslant 64 \times 240=15360 \\
\Rightarrow S=2 A \leqslant 30720 .
\end{array}
$$
Finally, starting from the condition for equality, we can take
$$
p=15, q=16 \text {. }
$$
Also note that,
$$
15360=1024 \times 15=960 \times 16 \text {. }
$$
We can construct
$$
\begin{aligned}
& \left(a_{1}, a_{2}, \cdots, a_{31}\right) \\
& =(1,2, \cdots, 15,2000,2001, \cdots, 2015), \\
& \left(b_{1}, b_{2}, \cdots, b_{31}\right)=(1025,1026, \cdots, 1055), \\
& \text { at this time, } S=1024 \times 15+960 \times 16=30720, \\
\text { and } & a_{1}+a_{2}+\cdots+a_{31} \\
& =b_{1}+b_{2}+\cdots+b_{31}=32240,
\end{aligned}
$$
which meets the requirements.
In summary, the maximum value of $S$ is 30720. | 30720 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the set $M=\{1,2, \cdots, 12\}$, and the three-element set $A=$ $\{a, b, c\}$ satisfies $A \subset M$, and $a+b+c$ is a perfect square. Then the number of sets $A$ is $\qquad$. | ,- 1.26 .
From $6 \leqslant a+b+c \leqslant 33$, we know that the square numbers within this range are $9, 16, 25$. Let's assume $a<b<c$.
If $a+b+c=9$, then the possible values for $c$ are 6, 5, 4, in which case,
$$
A=\{1,2,6\},\{1,3,5\},\{2,3,4\} \text {; }
$$
If $a+b+c=16$, then $7 \leqslant c \leqslant 12$, we can have
$$
\begin{aligned}
A= & \{1,3,12\},\{1,4,11\},\{2,3,11\}, \\
& \{1,5,10\},\{2,4,10\},\{1,6,9\}, \\
& \{2,5,9\},\{3,4,9\},\{1,7,8\}, \\
& \{2,6,8\},\{3,5,8\},\{3,6,7\}, \\
& \{4,5,7\} ;
\end{aligned}
$$
If $a+b+c=25$, then $10 \leqslant c \leqslant 12$, in this case,
$$
\begin{aligned}
A= & \{2,11,12\},\{3,10,12\},\{4,9,12\}, \\
& \{4,10,11\},\{5,8,12\},\{5,9,11\}, \\
& \{6,7,12\},\{6,8,11\},\{6,9,10\}, \\
& \{7,8,10\} .
\end{aligned}
$$
In total, we have 26 sets $A$. | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The volume of a rectangular prism is 8 cubic centimeters, and the total surface area is 32 square centimeters. If the length, width, and height form a geometric sequence, then the sum of all the edges of this rectangular prism is $\qquad$ | 4.32 cm.
Let the common ratio be $q$, and the length, width, and height be $q a, a, \frac{a}{q}$, respectively. Then $a^{3}=8 \Rightarrow a=2$.
$$
\begin{array}{l}
\text { Also } 2\left(q a \cdot a+a \cdot \frac{a}{q}+q a \cdot \frac{a}{q}\right)=32 \\
\Rightarrow q+\frac{1}{q}+1=4 .
\end{array}
$$
Therefore, the sum of the 12 edges is
$$
4\left(q a+\frac{a}{q}+a\right)=8\left(q+\frac{1}{q}+1\right)=32 \text {. }
$$ | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) If the subset $A$ of the set $M=\{1,2, \cdots, 200\}$ consists of elements each of which can be expressed as the sum of the squares of two natural numbers (allowing the same number), find the maximum number of elements in the set $A$.
| 11. Notice that, the squares not exceeding 200 are $0^{2}, 1^{2}, 2^{2}, \cdots, 14^{2}$.
First, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form $k^{2}+0^{2}$, and there are 14 such numbers; while the sum of each pair of numbers in $1^{2}$, $2^{2}, \cdots, 10^{2}$ (allowing the same number) is in the set $M$, and there are $\mathrm{C}_{10}^{2}+10=55$ such numbers, among which, numbers of the form $x^{2}+x^{2}$ are 10, and numbers of the form $x^{2}+y^{2}(x \neq y)$ are $\mathrm{C}_{10}^{2}$.
Second, numbers of the form $11^{2}+x^{2}(x=1,2, \cdots, 8)$ are 8,
numbers of the form $12^{2}+x^{2}(x=1,2, \cdots, 7)$ are 7, numbers of the form $13^{2}+x^{2}(x=1,2, \cdots, 5)$ are 5, and numbers of the form $14^{2}+x^{2}(x=1,2)$ are 2, totaling 22.
Consider the repeated cases.
Notice that, if
$$
\begin{array}{l}
x=a^{2}+b^{2}, y=c^{2}+d^{2}(a \neq b, c \neq d), \\
\text { then } x y=(a c+b d)^{2}+(a d-b c)^{2} \\
=(a c-b d)^{2}+(a d+b c)^{2} .
\end{array}
$$
Numbers not exceeding 40 that can be expressed as the sum of squares of two different positive integers are
$$
5 、 10 、 13 、 17 、 20 、 25,26,29 、 34,37 、 40 \text {, }
$$
The product of each number in this group with 5, and $13^{2}$, are all in the set $M$, and can each be expressed as the sum of squares in two ways, hence each is counted twice, totaling 12 repeated counts (10, $13 、 17 、 20$ with 10 are already included in the above product group).
Therefore, the maximum number of elements in set $A$ is $14+55+22-12=79$. | 79 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. In the plane, $m$ points have no three points collinear, and their convex hull is an $n$-sided polygon. By appropriately connecting lines, a grid region composed of triangles can be obtained. Let the number of non-overlapping triangles be $f(m, n)$. Then $f(2016,30)=$ $\qquad$ | One, 1.4000.
Since no three points are collinear, we have
$$
f(m, n)=(n-2)+2(m-n)=2 m-n-2 \text {. }
$$
Therefore, $f(2016,30)=2 \times 2016-30-2=4000$. | 4000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the set $T=\{1,2, \cdots, 2010\}$, for each non-empty subset of $T$, calculate the reciprocal of the product of all its elements. Then the sum of all such reciprocals is $\qquad$ | 6. 2010 .
Notice that, in the expansion of the product
$$
(1+1)\left(1+\frac{1}{2}\right) \cdots\left(1+\frac{1}{2010}\right)
$$
the $2^{2010}$ terms are all the reciprocals of positive integers, and each is precisely the reciprocal of the product of the numbers in one of the $2^{2010}$ subsets of set $T$. Removing the term 1, which corresponds to the empty set, the sum of the reciprocals is
$$
2 \times \frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{2011}{2010}-1=2010 .
$$ | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For integers $n \geqslant 3$, let
$$
f(n)=\log _{2} 3 \times \log _{3} 4 \times \cdots \times \log _{n-1} n \text {. }
$$
Then $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)=$ $\qquad$ | 2. 54
Notice,
$$
\begin{array}{l}
f(n)=\log _{2} 3 \times \frac{\log _{2} 4}{\log _{2} 3} \times \cdots \times \frac{\log _{2} n}{\log _{2}(n-1)} \\
=\log _{2} n .
\end{array}
$$
Thus, $f\left(2^{k}\right)=k$.
Therefore, $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)$
$$
=2+3+\cdots+10=54 \text {. }
$$ | 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of the white balls is. $\qquad$ | 3. 126.
First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls.
Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by swapping the positions of the red and white balls, we get an arrangement where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls, and vice versa.
Therefore, the number of arrangements where the sum of the numbers on the red balls is greater than the sum of the numbers on the white balls is equal to the number of arrangements where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls. Thus, the number of arrangements we are looking for is $\frac{1}{2} \mathrm{C}_{10}^{5}=126$. | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. (14 points) A flower shop purchases a certain number of roses from a farm at a price of 5 yuan per stem every day, and then sells them at a price of 10 yuan per stem. If they are not sold on the same day, the remaining roses are treated as garbage.
(1) If the flower shop purchases 16 stems of roses one day, find the function expression for the profit $y$ (unit: yuan) of that day in terms of the demand $n$ (unit: stems, $n \in \mathbf{N}$).
(2) The flower shop recorded the daily demand for roses (unit: stems) over 100 days, and organized the data as shown in Table 1.
Table 1
\begin{tabular}{|c|l|l|l|l|l|l|l|}
\hline Daily demand $n$ & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline Frequency & 10 & 20 & 16 & 16 & 15 & 13 & 10 \\
\hline
\end{tabular}
The frequencies of each demand over 100 days are used as the probabilities of each demand.
(i) If the flower shop purchases 16 stems of roses one day, and $X$ represents the profit (unit: yuan) of that day, find the distribution of $X$, its expected value, and variance.
(ii) If the flower shop plans to purchase 16 or 17 stems of roses one day, do you think they should purchase 16 or 17 stems? Please explain your reasoning. | 12. (1) When the daily demand $n \geqslant 16$, the profit $y=80$. When the daily demand $n<16$, the profit $y=10 n-80$. Therefore, the function expression of $y$ with respect to the positive integer $n$ is
$$
y=\left\{\begin{array}{ll}
10 n-80, & n<16 ; \\
80, & n \geqslant 16 .
\end{array}\right.
$$
(2) (i) The possible values of $X$ are $60, 70, 80$, and
$$
\begin{array}{l}
P(X=60)=0.1, P(X=70)=0.2, \\
P(X=80)=0.7 .
\end{array}
$$
Thus, the distribution of $X$ is as shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|}
\hline$X$ & 60 & 70 & 80 \\
\hline$P$ & 0.1 & 0.2 & 0.7 \\
\hline
\end{tabular}
The expected value of $X$ is
$$
\mathrm{E}(X)=60 \times 0.1+70 \times 0.2+80 \times 0.7=76 \text {; }
$$
The variance of $X$ is
$$
\begin{array}{l}
\mathrm{D}(X)=(60-76)^{2} \times 0.1+(70-76)^{2} \times 0.2+ \\
\quad(80-76)^{2} \times 0.7 \\
=44 .
\end{array}
$$
(ii) Method 1 The flower shop should purchase 16 roses per day. If the flower shop purchases 17 roses per day, let $Y$ represent the profit for the day (unit: yuan), then the distribution of $Y$ is as shown in Table 3.
Table 3
\begin{tabular}{|c|c|c|c|c|}
\hline$Y$ & 55 & 65 & 75 & 85 \\
\hline$P$ & 0.1 & 0.2 & 0.16 & 0.54 \\
\hline
\end{tabular}
The expected value of $Y$ is
$$
\begin{array}{l}
\mathrm{E}(Y)=55 \times 0.1+65 \times 0.2+ \\
75 \times 0.16+85 \times 0.54 \\
=76.4 ;
\end{array}
$$
The variance of $Y$ is
$$
\begin{aligned}
\mathrm{D}(Y)= & (55-76.4)^{2} \times 0.1+ \\
& (65-76.4)^{2} \times 0.2+ \\
& (75-76.4)^{2} \times 0.16+ \\
& (85-76.4)^{2} \times 0.54 \\
= & 112.04 .
\end{aligned}
$$
Thus, $\mathrm{D}(X)<\mathrm{D}(Y)$, meaning the profit fluctuation is smaller when purchasing 16 roses.
Additionally, although $\mathrm{E}(X)<\mathrm{E}(Y)$, the difference is not significant.
Therefore, the flower shop should purchase 16 roses per day.
Method 2 The flower shop should purchase 17 roses per day.
From Method 1, we know that $\mathrm{E}(X)<\mathrm{E}(Y)$, meaning the average profit is higher when purchasing 17 roses compared to purchasing 16 roses.
Therefore, the flower shop should purchase 17 roses per day. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Problem 5: In an $n \times n$ grid, 101 cells are colored blue. It is known that there is a unique way to cut the grid along the grid lines into some rectangles, such that each rectangle contains exactly one blue cell. Find the minimum possible value of $n$.
(2015, Bulgarian Mathematical Olympiad) | The minimum possible value of $n$ is 101.
First, prove the more general conclusion below.
Lemma Given an $n \times n$ grid $P$ with $m$ cells colored blue. A "good" partition of the grid $P$ is defined as: if the grid is divided along the grid lines into $m$ rectangles, and each rectangle contains exactly one blue cell. Then the grid $P$ has a unique good partition if and only if the $m$ blue cells form a rectangle.
Proof On the one hand, if the $m$ blue cells form a $m_{1} \times m_{2}\left(m_{1} 、 m_{2} \geqslant 1\right)$ rectangle $A$, cut the four corner cells of $A$ to form rectangles with the four corner cells of the grid $P$; the cells on the four sides of $A$ form $1 \times s(s \geqslant 1)$ rectangles with the cells at the top and bottom (left and right) ends of their respective rows (columns); each cell inside the rectangle $A$ forms a rectangle by itself. This constitutes a good partition $S$ of the grid $P$, and it is easy to see that the grid $P$ has only this unique good partition.
On the other hand, consider the grid $P$ has a unique good partition $S$. Call a grid line a "partition line" if it can divide the grid $P$ into two rectangles, and each rectangle contains at least one blue cell.
We now prove: every rectangle $A$ containing at least one blue cell has a good partition.
We use mathematical induction on the number of blue cells $k$ in the rectangle $A$ to prove conclusion (1).
If $k=1$, then by the definition of a good partition, conclusion (1) is obviously true.
Assume conclusion (1) holds for $k \leqslant t-1$.
Consider the case $k=t$. In this case, there exists a partition line that divides the rectangle $A$ into two rectangles $A_{1} 、 A_{2}$, and each rectangle contains at least one blue cell. By the induction hypothesis, $A_{1} 、 A_{2}$ each have a good partition. Therefore, the rectangle $A$ has a good partition, and conclusion (1) holds.
By conclusion (1) and the definition of a partition line, every partition line $l$ divides the grid $P$ into two rectangles, each of which has a good partition. Since $S$ is the unique good partition of the grid $P$, the partition $S$ includes all partition lines in the grid $P$.
Let $l_{1}, l_{2}, \cdots, l_{p}$ be all the vertical partition lines from left to right, and $m_{1}, m_{2}, \cdots, m_{q}$ be all the horizontal partition lines from bottom to top. Then they divide the grid $P$ into some rectangles, and each rectangle contains at most one blue cell. Since $S$ divides the grid $P$ using these partition lines, it results in $m$ rectangles, each containing exactly one blue cell.
Let $l_{0}$ be the vertical grid line closest to $l_{1}$ on the left with no blue cells to its left, $l_{p+1}$ be the vertical grid line closest to $l_{p}$ on the right with no blue cells to its right, and $m_{0} 、 m_{q+1}$ be similarly defined horizontal grid lines.
From the above discussion, the distance between $l_{i}$ and $l_{i+1}$
$d\left(l_{i}, l_{i+1}\right)=1(i=0,1, \cdots, p)$;
otherwise, a new partition line can be added between two partition lines.
Similarly, $d\left(m_{j}, m_{j+1}\right)=1(j=0,1, \cdots, q)$.
Therefore, the blue cells form a rectangle with $l_{0} 、 l_{p+1} 、 m_{0} 、 m_{q+1}$ as boundary lines.
The lemma is proved.
By the lemma, since the grid $P$ has a unique good partition, the 101 blue cells must form a rectangle. Therefore, it can only be a $1 \times 101$ rectangle.
Hence, the minimum value of $n$ is 101. | 101 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given
$$
5 x+16 y+33 z \geqslant 136\left(x 、 y 、 z \in \mathbf{R}_{+}\right) \text {. }
$$
then the minimum value of $x^{3}+y^{3}+z^{3}+x^{2}+y^{2}+z^{2}$ is | 3. 50 .
Notice that,
$$
\begin{array}{l}
x^{3}+x^{2}-5 x+3=(x+3)(x-1)^{2} \geqslant 0 \\
\Rightarrow x^{3}+x^{2} \geqslant 5 x-3, \\
y^{3}+y^{2}-16 y+20=(y+5)(y-2)^{2} \geqslant 0 \\
\Rightarrow y^{3}+y^{2} \geqslant 16 y-20, \\
z^{3}+z^{2}-33 z+63=(z+7)(z-3)^{2} \geqslant 0 \\
\Rightarrow z^{3}+z^{2} \geqslant 33 z-63 .
\end{array}
$$
Therefore, $x^{3}+y^{3}+z^{3}+x^{2}+y^{2}+z^{2}$
$$
\geqslant 5 x+16 y+33 z-86=50 \text {. }
$$
The minimum value 50 is achieved when and only when $x=1, y=2, z=3$. | 50 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
For any real number $x$, $[x]$ represents the greatest integer not exceeding $x$, and $\{x\}$ represents the fractional part of $x$. Then
$$
\begin{array}{l}
\left\{\frac{2014}{2015}\right\}+\left\{\frac{2014^{2}}{2015}\right\}+\cdots+\left\{\frac{2014^{2014}}{2015}\right\} \\
=
\end{array}
$$ | $-, 1.1007$.
Notice that,
$$
\frac{2014^{n}}{2015}=\frac{(2015-1)^{n}}{2015}=M+\frac{(-1)^{n}}{2015} \text {, }
$$
where $M$ is an integer.
Then $\left\{\frac{2014^{n}}{2015}\right\}=\left\{\begin{array}{ll}\frac{1}{2015}, & n \text { is even; } \\ \frac{2014}{2015}, & n \text { is odd. }\end{array}\right.$
Therefore, the original expression $=1007$. | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a_{1}, a_{2}, \cdots, a_{9}$ as any permutation of $1,2, \cdots, 9$. Then the minimum value of $a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9}$ is $\qquad$ | 2. 214.
Since $a_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, 9)$, by the AM-GM inequality, we have
$$
\begin{array}{l}
a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9} \\
\geqslant 3 \sqrt[3]{a_{1} a_{2} \cdots a_{9}}=3 \sqrt[3]{9!} \\
=3 \sqrt[3]{(2 \times 5 \times 7) \times(1 \times 8 \times 9) \times(3 \times 4 \times 6)} \\
=3 \sqrt[3]{70 \times 72 \times 72} \\
>3 \times 71=213 . \\
\text { Also, } 214=70+72+72 \\
=(2 \times 5 \times 7)+(1 \times 8 \times 9)+(3 \times 4 \times 6),
\end{array}
$$
Therefore, its minimum value is 214. | 214 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$.
Then the number of such $n$ is $\qquad$. | From $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right] \leqslant \frac{n}{3}+\frac{n}{6}=\frac{n}{2}$, knowing that equality holds, we find that $n$ is a common multiple of $3$ and $6$, i.e., a multiple of $6$. Therefore, the number of such $n$ is $\left[\frac{2006}{6}\right]=334$. | 334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
23. For any real numbers $a$, $b$, $c$, define an operation ※ with the following properties:
(1) $a ※(b ※ c)=(a ※ b) \cdot c$,
(2) $a ※ a=1$,
where “$\cdot$” denotes multiplication.
If the solution to the equation $2016 \times(6 ※ x)=100$ is $x=\frac{p}{q}(p, q$ are positive integers, $(p, q)=1)$, then the value of $p+q$ is $(\quad)$.
(A) 109
(B) 201
(C) 301
(D) 3049
(E) 33601 | 23. A.
$$
\begin{array}{l}
\text { Given } 2016 ※(6 ※ x)=(2016 ※ 6) \cdot x \\
\Rightarrow(2016 ※ 6) \cdot 6=\frac{600}{x} \\
\Rightarrow 2016 ※(6 ※ 6)=\frac{600}{x} .
\end{array}
$$
Since $a ※ a=1$, we have,
$$
\begin{array}{l}
6 ※ 6=2016 ※ 2016=1 . \\
\text { Therefore } 2016 ※(2016 ※ 2016)=\frac{600}{x} \\
\Rightarrow(2016 ※ 2016) \cdot 2016=\frac{600}{x} \\
\Rightarrow 1 \cdot 2016=\frac{600}{x} \Rightarrow x=\frac{25}{84} \\
\Rightarrow p+q=25+84=109 .
\end{array}
$$ | 109 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given positive integers $a, b, c (a < b < c)$ form a geometric sequence, and
$$
\log _{2016} a+\log _{2016} b+\log _{2016} c=3 .
$$
Then the maximum value of $a+b+c$ is $\qquad$ | One, 1.4066 273.
From the given, we know
$$
\begin{array}{l}
b^{2}=a c, a b c=2016^{3} \\
\Rightarrow b=2016, a c=2016^{2} .
\end{array}
$$
Since $a$, $b$, and $c$ are positive integers, when $a=1$, $c=2016^{2}$, $a+b+c$ reaches its maximum value
$$
2016^{2}+2017=4066273 .
$$ | 4066273 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
492 Select five subsets $A_{1}, A_{2}, \cdots, A_{5}$ from the set $\{1,2, \cdots, 1000\}$ such that $\left|A_{i}\right|=500(i=1,2$, $\cdots, 5)$. Find the maximum value of the number of common elements in any three of these subsets. | Let $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ be five 1000-dimensional vectors, and let $\alpha_{i}(j)$ denote the $j$-th element of $\alpha_{i}$. Then $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ satisfy
$$
\alpha_{i}(j)=\left\{\begin{array}{l}
1, j \in A_{i} ; \\
0, j \notin A_{i} .
\end{array}\right.
$$
Let the maximum number of common elements in any three subsets be $r$.
For any three vectors
$$
\begin{array}{l}
\boldsymbol{\beta}_{1}=\left(b_{1}, b_{2}, \cdots, b_{t}\right), \boldsymbol{\gamma}_{1}=\left(c_{1}, c_{2}, \cdots, c_{t}\right), \\
\boldsymbol{\delta}_{1}=\left(d_{1}, d_{2}, \cdots, d_{t}\right),
\end{array}
$$
define the ternary inner product operation
$$
\left(\boldsymbol{\beta}_{1}, \boldsymbol{\gamma}_{1}, \boldsymbol{\delta}_{1}\right)=b_{1} c_{1} d_{1}+b_{2} c_{2} d_{2}+\cdots+b_{t} c_{t} d_{t} .
$$
Clearly, this operation is linear and satisfies the commutative law.
By the problem, $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right)=500$.
For $1 \leqslant i<j<k \leqslant 5,\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right)$ equals the number of common elements in the sets $A_{i}, A_{j}, A_{k}$.
Let $\boldsymbol{\alpha}=\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{4}+\boldsymbol{\alpha}_{5}$. Then
$(\boldsymbol{\alpha}, \boldsymbol{\alpha}, \boldsymbol{\alpha})$
$$
\begin{aligned}
= & \sum_{i=1}^{5}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right)+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+ \\
& 6 \sum_{i<j<k}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right) \\
= & 2500+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+6 \sum_{i<j<k}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right) \\
\leqslant & 2500+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+60 r .
\end{aligned}
$$
Also, $\boldsymbol{\alpha}$ is a 1000-dimensional vector, and
$$
\boldsymbol{\alpha}=\left(a_{1}, a_{2}, \cdots, a_{1000}\right)\left(a_{i} \in\{0,1, \cdots, 5\}\right),
$$
satisfying $\sum_{i=1}^{1000} a_{i}=2500$.
At this point, $(\boldsymbol{\alpha}, \boldsymbol{\alpha}, \boldsymbol{\alpha})=\sum_{i=1}^{1000} a_{i}^{3}$, and $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)$ represents the number of common elements in the sets $A_{i}$ and $A_{j}$, so $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)=\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)$, where $(\boldsymbol{a}, \boldsymbol{b})$ denotes the usual inner product of vectors $\boldsymbol{a}$ and $\boldsymbol{b}$.
$$
\begin{array}{l}
\text { Hence } \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)=\sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right) \\
=(\boldsymbol{\alpha}, \boldsymbol{\alpha})-\sum_{i=1}^{5}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right) \\
=\sum_{i=1}^{1000} a_{i}^{2}-2500 .
\end{array}
$$
At this point,
$$
\begin{array}{l}
60 r \geqslant \sum_{i=1}^{1000} a_{i}^{3}-3\left(\sum_{i=1}^{1000} a_{i}^{2}-2500\right)-2500 \\
=\sum_{i=1}^{1000}\left(a_{i}^{3}-3 a_{i}^{2}\right)+5000 . \\
\text { Also, }\left(a_{i}-2\right)\left(a_{i}-3\right)\left(a_{i}+2\right) \geqslant 0 \\
\Leftrightarrow a_{i}^{3}-3 a_{i}^{2}-4 a_{i}+12 \geqslant 0 .
\end{array}
$$
Thus, for any $1 \leqslant i \leqslant 1000$,
$$
a_{i}^{3}-3 a_{i}^{2} \geqslant 4 a_{i}-12 \text {. }
$$
Therefore, $60 r \geqslant \sum_{i=1}^{1000}\left(4 a_{i}-12\right)+5000$
$$
=4 \times 2500-12000+5000=3000 \text {. }
$$
Thus, $r \geqslant 50$.
Below is a construction for $r=50$.
Divide the 1000 elements into 20 sets, each containing 50 elements. These 20 sets are divided into two groups, denoted as
$$
\begin{array}{l}
B_{123}, B_{124}, B_{125}, B_{134}, B_{135}, B_{145}, B_{234}, B_{235}, \\
B_{245}, B_{345} ; \\
C_{12}, C_{13}, C_{14}, C_{15}, C_{23}, C_{24}, C_{25}, C_{34}, C_{35}, C_{45} .
\end{array}
$$
Define $A_{i}=\{$ the union of all sets whose indices contain $i\}$.
At this point, for any $1 \leqslant i<j<k \leqslant 5$,
$$
A_{i} \cap A_{j} \cap A_{k}=B_{i j k},
$$
which has exactly 50 elements.
(Chen Kai, Xueersi Peiyou Beijing Branch, 100086) | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. For a finite set $A$ consisting of positive integers, if $A$ is divided into two non-empty disjoint subsets $A_{1}$ and $A_{2}$, and the least common multiple (LCM) of the elements in $A_{1}$ equals the greatest common divisor (GCD) of the elements in $A_{2}$, then such a partition is called "good". Find the minimum value of the positive integer $n$ such that there exists a set of $n$ positive integers with exactly 2015 good partitions. | 3. 3024.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\left(a_{1}<a_{2}<\cdots<a_{n}\right)$.
For any non-empty finite set of positive integers $B$, let $\operatorname{lcm} B$ and $\operatorname{gcd} B$ denote the least common multiple and greatest common divisor of the elements in $B$, respectively.
Consider any good partition $\left(A_{1}, A_{2}\right)$ of $A$. By definition, there exists a positive integer $d$ such that
$\operatorname{lcm} A_{1}=d=\operatorname{gcd} A_{2}$.
For any $a_{i} \in A_{1}$ and $a_{j} \in A_{2}$, we have
$a_{i} \leqslant d \leqslant a_{j}$.
Thus, there exists a positive integer $k(1 \leqslant k<n)$ such that
$A_{1}=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$,
$A_{2}=\left\{a_{k+1}, a_{k+2}, \cdots, a_{n}\right\}$.
Therefore, each good partition is determined by an element $a_{k}$ $(1 \leqslant k<n)$, which we call a "separator".
Define $l_{k}=\operatorname{lcm}\left(a_{1}, a_{2}, \cdots, a_{k}\right)$,
$g_{k}=\operatorname{gcd}\left(a_{k+1}, a_{k+2}, \cdots, a_{n}\right)$,
where $1 \leqslant k \leqslant n-1$. Then $a_{k}$ is a separator if and only if $l_{k}=g_{k}$.
To prove some properties of separators, we need the following lemma.
Lemma If $a_{k-1}$ and $a_{k}$ $(2 \leqslant k \leqslant n-1)$ are both separators, then $g_{k-1}=g_{k}=a_{k}$.
Proof Assume $a_{k-1}$ and $a_{k}$ are both separators.
$$
\begin{array}{c}
\text { By } l_{k-1}=g_{k-1} \Rightarrow l_{k-1} \mid a_{k} \\
\Rightarrow g_{k}=l_{k}=\operatorname{lcm}\left(l_{k-1}, a_{k}\right)=a_{k}, \\
g_{k-1}=\operatorname{gcd}\left(a_{k}, g_{k}\right)=a_{k} .
\end{array}
$$
The lemma is proved.
Property 1 For each $k=2,3, \cdots, n-2$, at least one of $a_{k-1}$, $a_{k}$, and $a_{k+1}$ is not a separator.
Proof of Property 1 By contradiction.
Assume $a_{k-1}$, $a_{k}$, and $a_{k+1}$ are all separators.
By the lemma, we have $a_{k+1}=g_{k}=a_{k}$, which is a contradiction.
Property 2 $a_{1}$ and $a_{2}$ cannot both be separators, and $a_{n-2}$ and $a_{n-1}$ cannot both be separators.
Proof of Property 2 Assume $a_{1}$ and $a_{2}$ are both separators.
By the lemma, we have $a_{2}=g_{1}=l_{1}=\operatorname{lcm}\left(a_{1}\right)=a_{1}$, which is a contradiction.
Similarly, assume $a_{n-2}$ and $a_{n-1}$ are both separators. By the lemma, we have $a_{n-1}=g_{n-1}=\operatorname{gcd}\left(a_{n}\right)=a_{n}$, which is a contradiction.
Properties 1 and 2 are proved.
Suppose a set $A$ with $n$ elements has exactly 2015 good partitions. Clearly, $n \geqslant 5$.
By Property 2, at most one element in $\left\{a_{1}, a_{2}\right\}$ and $\left\{a_{n-2}, a_{n-1}\right\}$ can be a separator.
Let $\lfloor x\rfloor$ denote the greatest integer not exceeding the real number $x$, and $\lceil x\rceil$ denote the smallest integer not less than the real number $x$.
By Property 1, at least $\left\lfloor\frac{n-5}{3}\right\rfloor$ elements in $\left\{a_{3}, a_{4}, \cdots, a_{n-3}\right\}$ are not separators. Therefore, $A$ has at most
$$
(n-1)-2-\left\lfloor\frac{n-5}{3}\right\rfloor=\left\lceil\frac{2(n-2)}{3}\right\rceil
$$
separators.
$$
\text { Therefore, }\left\lceil\frac{2(n-2)}{3}\right\rceil \geqslant 2015 \Rightarrow n \geqslant 3024 \text {. }
$$
Finally, we prove that there exists a set $A$ with 3024 elements that has exactly 2015 separators.
In fact,
$$
A=\left\{2 \times 6^{i}, 3 \times 6^{i}, 6^{i+1} \mid 10 \leqslant i \leqslant 1007\right\}
$$
satisfies $|A|=3024$, and $3 \times 6^{i} (0 \leqslant i \leqslant 1007)$ and $6^{i} (1 \leqslant i \leqslant 1007)$ are all separators.
In summary, the minimum value of $n$ is 3024. | 3024 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $m$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=m, a_{1}=\varphi(m), \\
a_{2}=\varphi^{(2)}(m)=\varphi(\varphi(m)), \cdots, \\
a_{n}=\varphi^{(n)}(m)=\varphi\left(\varphi^{(n-1)}(m)\right),
\end{array}
$$
where $\varphi(m)$ is the Euler's totient function.
If for any non-negative integer $k$, we have $a_{k+1} \mid a_{k}$, find the largest positive integer $m$ not exceeding 2016.
(Weng Shiyou, problem contributor) | 3. It is known that if $a \mid b$, then $\varphi(a) \mid \varphi(b)$.
Moreover, when $m>2$, $\varphi(m)$ is even.
Therefore, it is only necessary to find integers $m$ such that $a_{1} \mid a_{0}$, which implies $a_{k+1} \mid a_{k}$.
If $m$ is an odd number greater than 2, it is impossible for $a_{1} \mid m$.
Let $m=2^{\alpha} \prod_{k=1}^{s} p_{k}^{\alpha_{k}}$, where $p_{1}, p_{2}, \cdots, p_{s}$ are all odd primes. Then
$$
\varphi(m)=m\left(1-\frac{1}{2}\right) \prod_{k=1}^{s}\left(1-\frac{1}{p_{k}}\right) .
$$
Thus, only when $s=1$ and $p_{1}=3$, we have $a_{1} \mid a_{0}$.
Therefore, $m$ must be of the form $2^{\alpha} \times 3^{\beta}$.
Hence, the largest positive integer value not exceeding 2016 is
$$
m=2^{3} \times 3^{5}=1944 \text {. }
$$ | 1944 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the set
$$
I=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{i} \in\{1,2, \cdots, 11\}\right\} \text {, }
$$
$A$ is a subset of $I$ and satisfies: for any
$$
\left(x_{1}, x_{2}, x_{3}, x_{4}\right) 、\left(y_{1}, y_{2}, y_{3}, y_{4}\right) \in A \text {, }
$$
there exist $i 、 j(1 \leqslant i<j \leqslant 4)$ such that
$$
\left(x_{i}-x_{j}\right)\left(y_{i}-y_{j}\right)<0 \text {. }
$$
Determine the maximum value of $|A|$, where $|A|$ denotes the number of elements in the set $A$. (Provided by Zhang Limin) | 8. First, consider the set satisfying $x_{1}+x_{2}+x_{3}+x_{4}=24$
$$
\begin{aligned}
A= & \left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{i} \in\{1,2, \cdots, 11\},\right. \\
& i=1,2,3,4\},
\end{aligned}
$$
The number of its elements is
$$
\begin{array}{l}
\mathrm{C}_{23}^{3}-4\left(\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{12}^{2}\right) \\
=\mathrm{C}_{23}^{3}-4 \mathrm{C}_{13}^{3}=891,
\end{array}
$$
Any two different vectors in this set satisfy the problem's conditions.
Next, consider the vector set
$$
\begin{array}{c}
B=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{1}, x_{2} \in\{1,2, \cdots,\right. \\
\left.11\}, x_{3}=x_{4}=a \in\{1,2, \cdots, 11\}\right\},
\end{array}
$$
For each vector $\left(x_{1}, x_{2}, a, a\right)$, define the vector group as follows:
If $a=1$, the vector group is
$$
\begin{array}{l}
\left(x_{1}, x_{2}, 1,1\right),\left(x_{1}, x_{2}, 1,2\right), \cdots, \\
\left(x_{1}, x_{2}, 1,11\right),\left(x_{1}, x_{2}, 2,11\right), \cdots, \\
\left(x_{1}, x_{2}, 11,11\right) ;
\end{array}
$$
If $a=11$, the vector group is
$$
\begin{array}{l}
\left(x_{1}, x_{2}, 1,1\right),\left(x_{1}, x_{2}, 2,1\right), \cdots, \\
\left(x_{1}, x_{2}, 11,1\right),\left(x_{1}, x_{2}, 11,2\right), \cdots, \\
\left(x_{1}, x_{2}, 11,11\right) ;
\end{array}
$$
If $a$ is odd and greater than 1 and less than 11, the vector group is
$$
\begin{array}{l}
\left(x_{1}, x_{2}, 1,1\right), \cdots,\left(x_{1}, x_{2}, 1, a-1\right), \\
\left(x_{1}, x_{2}, 1, a\right),\left(x_{1}, x_{2}, 2, a\right), \cdots, \\
\left(x_{1}, x_{2}, a-1, a\right),\left(x_{1}, x_{2}, a, a\right) \\
\left(x_{1}, x_{2}, a, a+1\right), \cdots,\left(x_{1}, x_{2}, a, 11\right), \\
\left(x_{1}, x_{2}, a+1,11\right), \cdots,\left(x_{1}, x_{2}, 11,11\right) ;
\end{array}
$$
If $a$ is even, the vector group is
$$
\begin{array}{l}
\left(x_{1}, x_{2}, 1,1\right), \cdots,\left(x_{1}, x_{2}, a-1,1\right), \\
\left(x_{1}, x_{2}, a, 1\right),\left(x_{1}, x_{2}, a, 2\right), \cdots, \\
\left(x_{1}, x_{2}, a-1, a\right),\left(x_{1}, x_{2}, a, a\right), \\
\left(x_{1}, x_{2}, a+1, a\right), \cdots,\left(x_{1}, x_{2}, 11, a\right), \\
\left(x_{1}, x_{2}, 11, a+1\right),\left(x_{1}, x_{2}, 11,11\right) .
\end{array}
$$
For the vector set
$$
\begin{array}{c}
C=\left\{\left(a, a, x_{3}, x_{4}\right) \mid x_{3}, x_{4} \in\{1,2, \cdots,\right. \\
\left.11\}, x_{1}=x_{2}=a \in\{1,2, \cdots, 11\}\right\}
\end{array}
$$
Similarly, define the vector group.
For each vector in set $A$, it belongs to exactly one vector group, and different vectors belong to different vector groups, determining these 891 vector groups. Any 892 vectors must have two belonging to the same vector group, hence no longer possessing the property set by the problem.
Therefore, the maximum value sought is 891. | 891 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ and $b$ are integers, $\frac{127}{a}-\frac{16}{b}=1$. Then the maximum value of $b$ is $\qquad$ . | $$
\text { Two, 1.2 } 016 .
$$
The original equation is transformed into $b=\frac{16 a}{127-a}=\frac{127 \times 16}{127-a}-16$. When and only when $127-a=1$, $b$ reaches its maximum value, at this point, $b=127 \times 16-16=2016$. | 2016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 If a natural number $K$ can be expressed as
$$
\begin{aligned}
K & =V(a, b, c) \\
& =a^{3}+b^{3}+c^{3}-3 a b c \quad (a, b, c \in \mathbf{N})
\end{aligned}
$$
then $K$ is called a "Water Cube Number".
Now, arrange all different Water Cube Numbers in ascending order to form a sequence $\left\{K_{n}\right\}$, called the Water Cube Number sequence, and let $K_{0}=0$.
(1) Find $n$ such that $K_{n}=2016$;
(2) Determine $K_{2016}$ and $S_{2016}=\sum^{2016} K_{n}$. | 【Analysis】Let $M=\{3 x \mid x \in \mathbf{N},(3, x)=1\}$.
Lemma A natural number $k$ is a water cube number if and only if $k \in \mathbf{N} \backslash M$.
Proof Note that,
$V(a, b, c)=a^{3}+b^{3}+c^{3}-3 a b c$
$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$=(a+b+c)^{3}-3(a+b+c)(a b+b c+c a)$.
Then when $3 \mid (a+b+c)$, $3^{2} \mid V(a, b, c)$;
while when $3 \nmid (a+b+c)$, $3 \nmid V(a, b, c)$.
Therefore, each number in the set $M$ is not a water cube number.
Next, we prove: each natural number outside the set $M$ is a water cube number.
When $k \in \mathbf{N} \backslash M$:
If $k=0$, then $V(1,1,1)=0$;
If $k>0$, then
$$
k \in\{3 n+1,3 n+2,9(n+1) \mid k \in \mathbf{N}\},
$$
For each natural number $n$, we have
$$
\begin{array}{l}
V(n, n, n+1)=3 n+1, \\
V(n, n+1, n+1)=3 n+2, \\
V(n, n+1, n+2)=9(n+1),
\end{array}
$$
Hence $k$ is a water cube number.
The lemma is proved.
According to the lemma, we know that a positive integer $k$ is not a water cube number if and only if $k$ has the form $9 n+3$ or $9 n+6$, where $n$ is any natural number, i.e., among every nine consecutive positive integers, exactly seven are water cube numbers, and the other two are not water cube numbers.
Now list all positive integers in sequence, and divide them into segments of nine numbers each, from left to right:
$$
\begin{array}{l}
1,2, \cdots, 9 ; 10,11, \cdots, 18 ; \cdots ; \\
9 m-8,9 m-7, \cdots, 9 m ; \cdots .
\end{array}
$$
In the first $m$ segments, each segment has exactly seven water cube numbers, and $9 m$ is the largest water cube number among them, then
$$
K_{7 m}=9 m \text {. }
$$
(1) According to equation (1),
$$
\begin{array}{l}
2016=9 \times 224=K_{7 \times 224}=K_{1568} \\
\Rightarrow n=1568 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence } S_{7 m}=\sum_{n=1}^{7 m} K_{n} \\
=\sum_{n=1}^{9 m} n-\sum_{n=0}^{m-1}(9 n+3)-\sum_{n=0}^{m-1}(9 n+6) \\
=\sum_{n=1}^{9 m} n-\sum_{n=0}^{m-1}(18 n+9) \\
=\frac{9 m(9 m+1)}{2}-9 \sum_{n=0}^{m-1}(2 n+1) \\
=\frac{9 m(9 m+1)}{2}-9 m^{2} \\
=\frac{9 m(7 m+1)}{2} .
\end{array}
$$
(2) Since 2016 is a multiple of both 7 and 9, then $m=288$.
Hence $K_{2016}=K_{7 \times 288}=9 \times 288=2592$,
$$
S_{2016}=\frac{9 m(7 m+1)}{2}=2614032 .
$$ | 2614032 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$, is called a "fractal sequence", and its construction method is as follows:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;
Then copy all the previous items $1, 1, 2$, and add the successor number 3 of 2, to get $a_{4}=1, a_{5}=1, a_{6}=2, a_{7}=3$;
Next, copy all the previous items $1,1,2, 1, 1, 2, 3$, and add the successor number 4 of 3, to get the first 15 items as $1,1,2,1,1,2,3,1,1,2,1,1,2,3,4$; and so on.
Try to find $a_{2000}$ and the sum of the first 2000 terms of the sequence $S_{2000}$. (2000, Nanchang City High School Mathematics Competition) | Solving the construction method of the sequence $\left\{a_{n}\right\}$, we easily know
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$
Generally, $a_{2^{n-1}}=n$, which means the number $n$ first appears at the $2^{n}-1$ term, and if
$$
m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right),
$$
then $a_{m}=a_{k}$.
Given $2000=2^{10}-1+977$,
$977=2^{9}-1+466,466=2^{8}-1+211$,
$211=2^{7}-1+84,84=2^{6}-1+21$,
$21=2^{4}-1+6$,
thus $a_{2000}=a_{977}=a_{466}=a_{211}$
$$
=a_{84}=a_{21}=a_{6}=2 \text {. }
$$
To find $S_{2000}$, we first calculate $S_{2^{n}-1}$.
From the construction method of the sequence $\left\{a_{n}\right\}$, we know that in the first $2^{n}-1$ terms of the sequence, there is exactly 1 $n$, 2 $n-1$s, $2^{2}$ $n-2$s, $\cdots, 2^{k}$ $n-k$s, $\cdots, 2^{n-1}$ 1s.
Thus, $S_{2^{n-1}}=n+2(n-1)+2^{2}(n-2)+$
$$
\begin{array}{c}
\cdots+2^{n-2} \times 2+2^{n-1} \times 1, \\
2 S_{2^{n-1}}=2 n+2^{2}(n-1)+2^{3}(n-2)+ \\
\cdots+2^{n-1} \times 2+2^{n} .
\end{array}
$$
Subtracting the above two equations, we get
$$
\begin{array}{l}
S_{2^{n}-1}=-n+\left(2+2^{2}+\cdots+2^{n-1}+2^{n}\right) \\
=2^{n+1}-(n+2) .
\end{array}
$$
When $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$,
$$
\begin{array}{l}
S_{m}=S_{2^{n}-1}+\sum_{i=1}^{k} a_{\left(2^{n}-1\right)+i} \\
=S_{2^{n}-1}+\left(a_{1}+a_{2}+\cdots+a_{k}\right)=S_{2^{n}-1}+S_{k} .
\end{array}
$$
Thus, $S_{2000}=S_{2^{10}-1}+S_{977}, S_{977}=S_{2^{9}-1}+S_{466}$,
$$
\begin{array}{l}
S_{466}=S_{2^{8}-1}+S_{211}, S_{211}=S_{2^{7}-1}+S_{84}, \\
S_{84}=S_{2^{6}-1}+S_{21}, S_{21}=S_{2^{4}-1}+S_{6}, S_{6}=8 .
\end{array}
$$
Substituting into equation (1), we get
$$
\begin{aligned}
S_{2000}= & \left(2^{11}-12\right)+\left(2^{10}-11\right)+\left(2^{9}-10\right)+ \\
& \left(2^{8}-9\right)+\left(2^{7}-8\right)+\left(2^{5}-6\right)+6 \\
= & 3950
\end{aligned}
$$ | 3950 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the function be
$$
f(x)=\sin ^{4} \frac{k x}{10}+\cos ^{4} \frac{k x}{10}\left(k \in \mathbf{Z}_{+}\right) .
$$
If for any real number $a$, we have
$$
\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\} \text {, }
$$
then the minimum value of $k$ is $\qquad$ | 6. 16 .
From the given conditions, we have
$$
\begin{array}{l}
f(x)=\left(\sin ^{2} \frac{k x}{10}+\cos ^{2} \frac{k x}{10}\right)^{2}-2 \sin ^{2} \frac{k x}{10} \cdot \cos ^{2} \frac{k x}{10} \\
=1-\frac{1}{2} \sin ^{2} \frac{k x}{5}=\frac{1}{4} \cos \frac{2 k x}{5}+\frac{3}{4},
\end{array}
$$
The function $f(x)$ reaches its maximum value if and only if $x=\frac{5 m \pi}{k}(m \in \mathbf{Z})$.
According to the conditions, we know that any open interval of length 1, $(a, a+1)$, contains at least one maximum point. Therefore,
$$
\frac{5 \pi}{k}5 \pi \text {. }
$$
Conversely, when $k>5 \pi$, any open interval $(a, a+1)$ contains a complete period of $f(x)$, at which point, $\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\}$.
In summary, the minimum value of $k$ is $[5 \pi]+1=16$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four distinct numbers from $1, 2, \cdots, 100$, satisfying
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \\
=\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2} .
\end{array}
$$
Then the number of such ordered quadruples $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ is | 8. 40 .
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \\
\geqslant\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2},
\end{array}
$$
Equality holds if and only if
$$
\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\frac{a_{3}}{a_{4}},
$$
which means \(a_{1}, a_{2}, a_{3}, a_{4}\) form a geometric sequence.
Thus, the problem is equivalent to counting the number of geometric sequences \(\{a_{1}, a_{2}, a_{3}, a_{4}\}\) such that
$$
\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\} \subseteq\{1,2, \cdots, 100\}.
$$
Let the common ratio \(q \neq 1\) and \(q \in \mathbf{Q}\). Denote \(q=\frac{n}{m}\), where \(m\) and \(n\) are coprime positive integers, and \(m \neq n\).
First, consider the case \(n > m\).
In this case, \(a_{4}=a_{1}\left(\frac{n}{m}\right)^{3}=\frac{a_{1} n^{3}}{m^{3}}\).
Since \(m^{3}\) and \(n^{3}\) are coprime, \(l=\frac{a_{1}}{m^{3}} \in \mathbf{Z}_{+}\). Accordingly, \(a_{1}, a_{2}, a_{3}, a_{4}\) are \(m^{3} l, m^{2} n l, m n^{2} l, n^{3} l\), all of which are positive integers. This indicates that for any given \(q=\frac{n}{m}>1\), the number of sequences \(\{a_{1}, a_{2}, a_{3}, a_{4}\}\) that satisfy the conditions and have \(q\) as the common ratio is the number of positive integers \(l\) that satisfy the inequality \(n^{3} l \leqslant 100\), which is \(\left[\frac{100}{n^{3}}\right]\).
Since \(5^{3} > 100\), we only need to consider
$$
q=2,3, \frac{3}{2}, 4, \frac{4}{3}
$$
The total number of such geometric sequences is
$$
\begin{array}{l}
{\left[\frac{100}{8}\right]+\left[\frac{100}{27}\right]+\left[\frac{100}{27}\right]+\left[\frac{100}{64}\right]+\left[\frac{100}{64}\right]} \\
=12+3+3+1+1=20 .
\end{array}
$$
When \(n < m\), by symmetry, there are also 20 sequences \(\{a_{1}, a_{2}, a_{3}, a_{4}\}\) that satisfy the conditions.
In summary, there are 40 ordered tuples \(\left(a_{1}, a_{2}, a_{3}, a_{4}\right)\) that satisfy the conditions. | 40 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. | Three, using these ten points as vertices and the connected line segments as edges, we get a simple graph $G$ of order 10.
The following proves: The number of edges in graph $G$ does not exceed 15.
Let the vertices of graph $G$ be $v_{1}, v_{2}, \cdots, v_{10}$, with a total of $k$ edges, and use $\operatorname{deg}\left(v_{i}\right)$ to denote the degree of vertex $v_{i}$.
If $\operatorname{deg}\left(v_{i}\right) \leqslant 3$ for $i=1,2, \cdots, 10$, then $k=\frac{1}{2} \sum_{i=1}^{10} \operatorname{deg}\left(v_{i}\right) \leqslant \frac{1}{2} \times 10 \times 3=15$.
Assume there exists a vertex $v_{i}$ such that $\operatorname{deg}\left(v_{i}\right) \geqslant 4$. Without loss of generality, let $\operatorname{deg}\left(v_{1}\right)=n \geqslant 4$, and $v_{1}$ is adjacent to $v_{2}, v_{3}, \cdots, v_{n+1}$. Thus, there are no edges between $v_{2}, v_{3}, \cdots, v_{n+1}$, otherwise, a triangle would be formed. Therefore, there are exactly $n$ edges between $v_{1}, v_{2}, \cdots, v_{n+1}$.
For each $j(n+2 \leqslant j \leqslant 10)$, $v_{j}$ can be adjacent to at most one of $v_{2}, v_{3}$, $\cdots, v_{n+1}$ (otherwise, if $v_{j}$ is adjacent to $v_{s}$ and $v_{t}(2 \leqslant s<t \leqslant n+1)$, then $v_{1}, v_{s}, v_{j}, v_{t}$ would correspond to the four vertices of a spatial quadrilateral, which contradicts the given condition). Therefore, the number of edges between $v_{2}, v_{3}, \cdots, v_{n+1}$ and $v_{n+2}, v_{n+3}$, $\cdots, v_{10}$ is at most
$$
10-(n+1)=9-n \text {. }
$$
Among the $9-n$ vertices $v_{n+2}, v_{n+3}, \cdots, v_{n}$, since there are no triangles, by Turán's theorem, there are at most $\left[\frac{(9-n)^{2}}{4}\right]$ edges. Therefore, the number of edges in graph $G$ is
$$
\begin{array}{l}
k \leqslant n+(9-n)+\left[\frac{(9-n)^{2}}{4}\right] \\
=9+\left[\frac{(9-n)^{2}}{4}\right] \leqslant 9+\left[\frac{25}{4}\right]=15 .
\end{array}
$$
The graph in Figure 8 has 15 edges and meets the requirements. In summary, the maximum number of edges is 15. | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Several pairwise non-overlapping isosceles right triangles with leg lengths of 1 are placed on a $100 \times 100$ grid paper. It is known that the hypotenuse of any right triangle is the diagonal of some unit square; each side of a unit square is the leg of a unique right triangle. A unit square whose diagonals are not the hypotenuse of any right triangle is called a "blank square." Find the maximum number of blank squares. | 3. For a general $2 n \times 2 n$ grid paper, the maximum number of empty cells is $n(n-1)$.
In fact, the $2 n \times 2 n$ grid paper is enclosed by $2 n+1$ horizontal lines and $2 n+1$ vertical lines:
$$
\begin{array}{l}
\{(x, y) \mid x=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\}, \\
\{(x, y) \mid y=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\} .
\end{array}
$$
For an isosceles right triangle, if its right-angle vertex is at the bottom-left (top-left, bottom-right, top-right) corner of its cell, the isosceles triangle is called bottom-left (top-left, bottom-right, top-right). Bottom-left (top-left) and bottom-right (top-right) triangles are collectively referred to as bottom (top) triangles.
Let $u_{k}(0 \leqslant k \leqslant 2 n)$ denote the number of bottom right triangles with one leg on the line $y=k$, and $d_{k}(0 \leqslant k \leqslant 2 n)$ denote the number of top right triangles with one leg on the line $y=k$.
$$
\begin{array}{l}
\text { Then } u_{k}+d_{k}=2 n, u_{0}=d_{2 n}=2 n, \\
u_{k}+d_{k+1}=2 n+1 .
\end{array}
$$
Solving this, we get $u_{k}=2 n-k, d_{k}=k$. There are $u_{k}$ bottom triangles and $d_{k+1}$ top triangles between the lines $y=k$ and $y=k+1$. Therefore, the number of empty cells in this row is no more than
$$
2 n-\max \left\{u_{k}, d_{k+1}\right\} \text {. }
$$
Thus, the total number of empty cells is no more than
$$
2[0+1+\cdots+(n-1)]=n(n-1) \text {. }
$$
Below is an example with $n(n-1)$ empty cells.
The set of coordinates for the right-angle vertices of bottom-left triangles:
$$
\begin{array}{l}
\{(x, y) \mid x 、 y \in \mathbf{Z}, x 、 y \geqslant 0, x+y \leqslant n-1\} \cup \\
\{(x+k-1, n+k-x) \mid x 、 k=1,2, \cdots, n\},
\end{array}
$$
The set of coordinates for the right-angle vertices of top-right triangles:
$$
\begin{array}{l}
\{(2 n-x, 2 n-y) \mid x 、 y \in \mathbf{Z}, x 、 y \geqslant 0, \\
x+y \leqslant n-1\} \cup \\
\{(x+k-1, n+k-x) \mid x 、 k=1,2, \cdots, n\},
\end{array}
$$
The set of coordinates for the right-angle vertices of top-left triangles:
$$
\begin{array}{l}
\{(x, y) \mid x 、 y \in \mathbf{Z}, x=k, n+1+k \leqslant y \leqslant \\
2 n, k=0,1, \cdots, n-1\},
\end{array}
$$
The set of coordinates for the right-angle vertices of bottom-right triangles:
$$
\begin{array}{l}
\{(2 n-x, 2 n-y) \mid x, y \in \mathbf{Z}, x=k, \\
n+1+k \leqslant y \leqslant 2 n, k=0,1, \cdots, n-1\} .
\end{array}
$$
Thus, the required result is 2450. | 2450 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set
$$
S=\{1,2, \cdots, 12\}, A=\left\{a_{1}, a_{2}, a_{3}\right\}
$$
satisfy $a_{1}<a_{2}<a_{3}, a_{3}-a_{2} \leqslant 5, A \subseteq S$. Then the number of sets $A$ that satisfy the conditions is | 5. 185 .
Notice that, the number of all three-element subsets of set $S$ is
$$
\mathrm{C}_{12}^{3}=220 \text {. }
$$
The number of subsets satisfying $1 \leqslant a_{1}<a_{2}<a_{3}-5 \leqslant 7$ is $\mathrm{C}_{7}^{3}=35$, so the number of sets $A$ that meet the conditions of the problem is $220-35=185$. | 185 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If a positive integer can be expressed as the difference of cubes of two consecutive odd numbers, then the positive integer is called a "harmonious number" (for example, $2=1^{3}-(-1)^{3}, 26=3^{3}-1^{3}, 2, 26$ are both harmonious numbers). Among the positive integers not exceeding 2016, the sum of all harmonious numbers is $(\quad$.
(A) 6858
(B) 6860
(C) 9260
(D) 9262 | 3. B.
Notice that,
$$
(2 k+1)^{3}-(2 k-1)^{3}=2\left(12 k^{2}+1\right) \text {. }
$$
From $2\left(12 k^{2}+1\right) \leqslant 2016 \Rightarrow|k|<10$.
Taking $k=0,1, \cdots, 9$, we get all the harmonious numbers not exceeding 2016, and their sum is
$$
\begin{array}{l}
{\left[1^{3}-(-1)^{3}\right]+\left(3^{3}-1^{3}\right)+\left(5^{3}-6^{3}\right)+} \\
\cdots+\left(19^{3}-17^{3}\right) \\
=19^{3}+1=6860 .
\end{array}
$$ | 6860 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
a+b+c=5, a^{2}+b^{2}+c^{2}=15, \\
a^{3}+b^{3}+c^{3}=47 . \\
\text { Find }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
\end{array}
$$ | Given $a+b+c=5, a^{2}+b^{2}+c^{2}=15$, we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
Then $47-3 a b c=5(15-5)=50$
$$
\begin{aligned}
& \Rightarrow a b c=-1 . \\
& \text { And } a^{2}+a b+b^{2} \\
& =(a+b)(a+b+c)-(a b+b c+c a) \\
= & (5-c)-5=5(4-c) .
\end{aligned}
$$
Similarly,
$$
\begin{array}{l}
b^{2}+b c+c^{2}=5(4-a), \\
c^{2}+c a+a^{2}=5(4-b) . \\
\text { Therefore }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \\
=125(4-a)(4-b)(4-c) \\
=125[64-16 \times 5+4 \times 5-(-1)] \\
=625 .
\end{array}
$$ | 625 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=0,5 a_{n+1}=4 a_{n}+3 \sqrt{1-a_{n}^{2}}(n \in \mathbf{N}) \text {. }
$$
Let $S_{n}=\sum_{i=0}^{n} a_{i}$. Then $S_{51}=$ $\qquad$ | 2. 48 .
Let $a_{n}=\sin \alpha_{n}$.
Suppose $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$.
Then $30^{\circ}0 ; \\
\sin \left(\alpha_{n}-\theta\right), \cos \alpha_{n}<0 .
\end{array}\right.
$
Thus, $a_{n}=\left\{\begin{array}{l}\sin n \theta, n=0,1 ; \\ \sin 2 \theta, n=2 k\left(k \in \mathbf{Z}_{+}\right) ; \\ \sin 3 \theta, n=2 k+1\left(k \in \mathbf{Z}_{+}\right) .\end{array}\right.$
Therefore, when $n=0$, $S_{n}=0$;
when $n=1$, $S_{n}=\frac{3}{5}$.
Hence, $S_{51}=\frac{3}{5}+25 \sin 2 \theta+25 \sin 3 \theta=48$. | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. A drawer contains red and blue socks, with a total number not exceeding 2016. If two socks are randomly drawn, the probability that they are the same color is $\frac{1}{2}$. Then the maximum number of red socks in the drawer is . $\qquad$ | 5.990.
Let $x$ and $y$ be the number of red and blue socks in the drawer, respectively. Then
$$
\frac{x y}{\mathrm{C}_{x+y}^{2}}=\frac{1}{2} \Rightarrow (x-y)^{2}=x+y.
$$
Thus, the total number of socks is a perfect square.
Let $n=x-y$, i.e., $n^{2}=x+y$. Therefore, $x=\frac{n^{2}+n}{2}$.
Since $x+y \leqslant 2016$, we have $n \leqslant 44, x \leqslant 990$.
Hence, the maximum number of red socks in the drawer is 990. | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the set $I=\{1,2, \cdots, n\}(n \geqslant 3)$. If two non-empty proper subsets $A$ and $B$ of $I$ satisfy $A \cap B=\varnothing, A \cup$ $B=I$, then $A$ and $B$ are called a partition of $I$. If for any partition $A, B$ of the set $I$, there exist two numbers in $A$ or $B$ such that their sum is a perfect square, then $n$ is at least | 6. 15 .
When $n=14$, take
$A=\{1,2,4,6,9,11,13\}$,
$B=\{3,5,7,8,10,12,14\}$.
It is easy to see that the sum of any two numbers in $A$ and $B$ is not a perfect square.
Thus, $n=14$ does not meet the requirement.
Therefore, $n<14$ also does not meet the requirement.
Now consider $n=15$.
Use proof by contradiction. Assume that the sum of any two numbers in $A$ and $B$ is not a perfect square.
Without loss of generality, let $1 \in A$.
$$
\begin{array}{l}
\text { By } 1+3=2^{2}, 1+8=3^{2}, 1+15=4^{2} \\
\Rightarrow\{3,8,15\} \subseteq B ; \\
\text { By } 3+6=3^{2}, 3+13=4^{2}, 15+10=5^{2} \\
\Rightarrow\{6,13,10\} \subseteq A .
\end{array}
$$
But $6+10=4^{2}$, which contradicts the assumption that the sum of any two numbers in $A$ is not a perfect square. Therefore, $n=15$ meets the requirement. Thus, the minimum value of $n$ is 15. | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the set
$$
\{[x]+[2 x]+[3 x] \mid x \in \mathbf{R}\} \cap\{1,2, \cdots, 100\}
$$
has elements. | 4. 67 .
Let $f(x)=[x]+[2 x]+[3 x]$.
Then $f(x+1)=f(x)+6$.
When $0 \leqslant x<1$, all possible values of $f(x)$ are 0, 1, 2, 3.
Therefore, the range of $f(x)$ is
$$
S=\{6 k, 6 k+1,6 k+2,6 k+3 । k \in \mathbf{Z}\} \text {. }
$$
Thus, $S \cap\{1,2, \cdots, 100\}$ has $4 \times 17-1=67$ elements. | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the sequence $\left\{\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the number of rational terms in the first 2016 terms of the sequence $\left\{S_{n}\right\}$ is | - 1.43.
$$
\begin{array}{l}
\text { Given } \frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1} \\
\Rightarrow S_{n}=1-\frac{\sqrt{n+1}}{n+1} .
\end{array}
$$
Since $44<\sqrt{2016}<45$, the number of values for which $S$ is rational is 43. | 43 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 One evening, 21 people made $n$ phone calls to each other. It is known that among them, there are $m$ ($m$ is an odd number) people $a_{1}, a_{2}$, $\cdots, a_{m}$ such that $a_{i}$ and $a_{i+1}\left(i=1,2, \cdots, m ; a_{m+1}=a_{1}\right)$ made phone calls. If no three people among these 21 people made phone calls to each other, find the maximum value of $n$.
【Analysis】Construct a graph theory model to characterize the relationships between the elements in the problem. From the conditions given in the problem, we know that there exists an odd cycle in the graph. Since the number of points is finite, by the extreme principle, we know that there must exist the shortest odd cycle $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. Furthermore, in the shortest odd cycle, any two points $A_{i}$ and $A_{j}(j \neq i \pm 1)$ must not be connected, thus the internal structure of the odd cycle is fully understood. At this point, starting from the shortest odd cycle $C$, the entire graph can be divided into two parts: one part is the odd cycle $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$, and the other part is all the points except $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. We only need to understand the structure of each part and the connection situation between the two parts to grasp the overall structure. | Solve: Represent 21 people with 21 points.
If two people communicate by phone, connect the corresponding two points with an edge; otherwise, do not connect an edge, resulting in a graph $G$. It is known that graph $G$ contains an odd cycle, and there are no triangles in graph $G$. The task is to find the maximum number of edges $n$ in graph $G$.
Let the length of the shortest odd cycle $C$ in the graph be $2k+1$.
Since there are no triangles in graph $G$, we have $k \geqslant 2$.
Let $C$ be $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. Then, for any $i, j (1 \leqslant i, j \leqslant 2 k+1)$, $A_{i}$ and $A_{j} (j \neq i \pm 1)$ are not connected $\left(A_{0}=A_{2 k+2}=A_{1}\right)$ (otherwise, there would be a smaller odd cycle, contradicting the assumption). Apart from $A_{1}, A_{2}, \cdots, A_{2 k+1}$,
the remaining
$$
21-(2 k+1)=2(10-k)
$$
points have no triangles.
By Turán's theorem, the maximum number of edges among these $2(10-k)$ points is $\frac{1}{4}(20-2 k)^{2}=(10-k)^{2}$. Additionally, each of these $2(10-k)$ points cannot be connected to two adjacent vertices in $C$ (otherwise, there would be a triangle, contradicting the assumption). Therefore, each of these $2(10-k)$ points can be connected to at most $k$ vertices in $C$. Hence, the number of edges $n$ in the graph satisfies
$$
\begin{array}{l}
n \leqslant 2 k+1+(10-k)^{2}+2(10-k) k \\
=102-(k-1)^{2} \leqslant 102-(2-1)^{2}=101 .
\end{array}
$$
Suppose the 21 vertices of $G$ are $A_{1}, A_{2}, \cdots, A_{21}$,
$$
\begin{array}{l}
X=\left\{A_{1}, A_{2}, \cdots, A_{5}\right\}, \\
Y=\left\{A_{6}, A_{8}, A_{10}, \cdots, A_{20}\right\}, \\
Z=\left\{A_{7}, A_{9}, A_{11}, \cdots, A_{21}\right\} .
\end{array}
$$
In $X$, $A_{i}$ is connected to $A_{i+1} (i=1,2,3,4)$ and $A_{5}$ is connected to $A_{1}$. In $Y$, each point is connected to each point in $Z$, and each point in $Y$ is connected to two points in $X$, $A_{1}$ and $A_{3}$. Each point in $Z$ is connected to two points in $X$, $A_{2}$ and $A_{4}$. No other pairs of points are connected. Then the number of edges in graph $G$ is
$$
n=5+8 \times 8+2 \times 8 \times 2=101 \text {, }
$$
and graph $G$ contains an odd cycle $A_{1} A_{2} \cdots A_{5} A_{1}$, but no triangles.
In conclusion, the maximum value of $n$ is 101. | 101 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the sum of the first $n$ terms denoted as $S_{n}$. If $S_{6}=26, a_{7}=2$, then the maximum value of $n S_{n}$ is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 338 .
$$
\begin{array}{l}
\text { Given } S_{6}=26, a_{7}=2 \\
\Rightarrow a_{1}=6, d=-\frac{2}{3} \\
\Rightarrow S_{n}=-\frac{1}{3} n^{2}+\frac{19}{3} n \\
\Rightarrow n S_{n}=-\frac{1}{3} n^{3}+\frac{19}{3} n^{2} . \\
\text { Let } f(x)=-\frac{1}{3} x^{3}+\frac{19}{3} x^{2}(x \in \mathbf{R}) .
\end{array}
$$
Then $f^{\prime}(x)=-x^{2}+\frac{38}{3} x$.
By $f^{\prime}(x)=0 \Rightarrow x=0$ or $\frac{38}{3}$.
Thus, $f(n)\left(n \in \mathbf{Z}_{+}\right)$ is increasing in the interval $[1,12]$ and decreasing in the interval $[13,+\infty)$.
Notice that, $f(12)=336, f(13)=338$. Therefore, the maximum value is 338. | 338 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
9. (16 points) When $x \in [1,2017]$, find
$$
f(x)=\sum_{i=1}^{2017} i|x-i|
$$
the minimum value. | When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2016)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2017} i(i-x) \\
=\left(k^{2}+k-2017 \times 1009\right) x+ \\
\frac{2017 \times 2018 \times 4035}{6}-\frac{k(k+1)(2 k+1)}{3}
\end{array}
$$
is a linear function, and its minimum value is attained at the endpoints of the interval $[k, k+1]$.
Thus, the minimum term of the sequence $\{f(k)\}(1 \leqslant k \leqslant 2016)$ is the desired value.
Notice that,
$$
\begin{array}{l}
f(k+1)>f(k) \\
\Leftrightarrow k^{2}+k-2017 \times 1009>0 \\
\Leftrightarrow k \geqslant 1427 . \\
\text { Hence } f(x)_{\min }=f(1427)=801730806 .
\end{array}
$$ | 801730806 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (40 points) Given a $2016 \times 2016$ grid. Find the smallest positive integer $M$, such that it is possible to draw $M$ rectangles (with their sides on the grid lines) in the grid, and each small square in the grid has its sides included in the sides of one of the $M$ rectangles. | Second, $M=2017$.
First, we prove: All grid lines of a $2016 \times 2016$ square grid can be covered by the edges of 2017 rectangles.
In fact, all horizontal grid lines can be covered by the edges of 1008 $1 \times 2016$ rectangles and one $2016 \times 2016$ rectangle; all vertical grid lines can be covered by the edges of 1008 $2016 \times 1$ rectangles and the aforementioned $2016 \times 2016$ rectangle.
Thus, $M \leqslant 2017$.
Next, assume that the grid lines of a $2016 \times 2016$ square grid $A$ are covered by the edges of $M$ rectangles as required by the problem.
We will prove: $M \geqslant 2017$.
We call the top (bottom) edge of a rectangle the "head" ("foot").
For each rectangle in the grid $A$, we label it with a pair $(a, b)$: if the head of the rectangle is at the top of the grid $A$, we set $a=1$, otherwise $a=0$; if the foot of the rectangle is at the bottom of the grid $A$, we set $b=1$, otherwise $b=0$.
Among the $M$ rectangles, let the number of rectangles labeled $(1,1)$, $(1,0)$, $(0,1)$, and $(0,0)$ be $x$, $y$, $z$, and $w$, respectively.
There are 2015 horizontal grid lines inside the grid $A$, so $y+z+2w \geqslant 2015$;
There are 2017 vertical grid lines intersecting with the top of the grid $A$, so $2x+2y \geqslant 2017 \Rightarrow x+y \geqslant 1009$.
Similarly, $x+z \geqslant 1009$.
Thus, $2x+y+z \geqslant 2018$
$$
\begin{array}{l}
\Rightarrow 2(x+y+z+w) \geqslant 4033 \\
\Rightarrow M=x+y+z+w \geqslant 2017 .
\end{array}
$$ | 2017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. A certain unit distributes a year-end bonus of 1 million yuan, with first prize at 15,000 yuan per person, second prize at 10,000 yuan per person, and third prize at 5,000 yuan per person. If the difference in the number of people between third prize and first prize is no less than 93 but less than 96, then the total number of people who won awards in the unit is $\qquad$ . | 3. 147.
Let the number of first prize winners be $x$, the number of second prize winners be $y$, and the number of third prize winners be $z$.
Then $1.5 x + y + 0.5 z = 100$
$$
\begin{array}{l}
\Rightarrow (x + y + z) + 0.5 x - 0.5 z = 100 \\
\Rightarrow x + y + z = 100 + 0.5(z - x). \\
\text{Given } 93 \leqslant z - x < 96 \\
\Rightarrow 46.5 \leqslant 0.5(z - x) < 48 \\
\Rightarrow 146.5 \leqslant x + y + z < 148.
\end{array}
$$
Since the number of people is a positive integer, hence $x + y + z = 147$. | 147 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the curve
$$
(x-20)^{2}+(y-16)^{2}=r^{2} \text { and } y=\sin x
$$
have exactly one common point $P\left(x_{0}, y_{0}\right)$. Then
$$
\frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=
$$
$\qquad$ | 2. 20 .
From the given information, there is a common tangent line at point $P\left(x_{0}, y_{0}\right)$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\sin x_{0}-16}{x_{0}-20} \cos x_{0}=-1 \\
\Rightarrow \frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=20 .
\end{array}
$$ | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (20 points) Given sets $A$ and $B$ are both sets of positive integers, and $|A|=20,|B|=16$. Set $A$ satisfies the following condition: if $a, b, m, n \in A$, and $a+b=$ $m+n$, then $\{a, b\}=\{m, n\}$. Define
$$
A+B=\{a+b \mid a \in A, b \in B\} \text {. }
$$
Determine the minimum value of $|A+B|$. | 14. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{20}\right\}$,
$$
\begin{array}{l}
B=\left\{b_{1}, b_{2}, \cdots, b_{16}\right\}, \\
C_{j}=\left\{a_{i}+b_{j} \mid i=1,2, \cdots, 20\right\},
\end{array}
$$
where $j=1,2, \cdots, 16$.
Thus, $A+B=\bigcup_{j=1}^{16} C_{j}$.
We now prove: $\left|C_{m} \cap C_{n}\right| \leqslant 1(m \neq n)$.
In fact, suppose there exist $C_{m}, C_{n}(m \neq n), \left|C_{m} \cap C_{n}\right| \geqslant 2$. Then there exist $k_{1}, k_{2}, l_{1}, l_{2}$ such that
$$
\text { and } \begin{array}{l}
a_{k_{1}}+b_{m}, a_{k_{2}}+b_{m}, a_{l_{1}}+b_{n}, a_{l_{2}}+b_{n} \in b_{m}=a_{k_{2}}+b_{n}, a_{l_{1}}+b_{m}=a_{l_{2}}+b_{n}, \\
\\
a_{k_{1}} \neq a_{k_{2}}, a_{l_{1}} \neq a_{l_{2}} .
\end{array}
$$
Then $a_{k_{1}}-a_{k_{2}}=a_{l_{1}}-a_{l_{2}}$
$$
\Rightarrow a_{k_{1}}+a_{l_{2}}=a_{l_{1}}+a_{k_{2}},
$$
which implies $\left\{a_{k_{1}}, a_{l_{1}}\right\}=\left\{a_{k_{2}}, a_{l_{2}}\right\}$.
By conclusion (1), we know $a_{k_{1}}=a_{l_{1}}, a_{k_{2}}=a_{l_{2}}$, which is impossible, otherwise,
$$
\begin{array}{l}
a_{k_{1}}+b_{m} \in C_{n}, a_{l_{2}}+b_{n} \in C_{m} \\
\Rightarrow a_{k_{1}}=b_{n}, a_{b_{2}}=b_{m},
\end{array}
$$
which means $C_{m} \cap C_{n}$ can have at most one element, a contradiction.
Thus, $\left|C_{m} \cap C_{n}\right| \leqslant 1(m \neq n)$.
By the principle of inclusion-exclusion,
$$
\begin{array}{l}
\quad|A+B|=\left|\bigcup_{k=1}^{16} C_{k}\right| \\
\geqslant \sum_{k=1}^{16}\left|C_{k}\right|-\sum_{1 \leqslant m<n \leqslant 16}\left|C_{m} \cap C_{n}\right| \\
=320-120=200 . \\
\text { Let } A=\left\{2,2^{2}, \cdots, 2^{20}\right\}, \\
B=\left\{2,2^{2}, \cdots, 2^{16}\right\} .
\end{array}
$$
It is easy to verify that if $m \neq n \neq k$, then
$$
C_{m} \cap C_{n}=\left\{2^{m}+2^{n}\right\}, C_{m} \cap C_{n} \cap C_{k}=\varnothing .
$$
Therefore, $|A+B|=200$.
In conclusion, the minimum value of $|A+B|$ is 200. | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. If the sum of 12 distinct positive integers is 2016, then the maximum value of the greatest common divisor of these positive integers is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. 24 .
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d\left(\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1\right)$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then, $2016=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$$
S \geqslant 1+2+\cdots+12=78 \text {. }
$$
Note that, $S \mid 2016$, and 78 is not a factor of 2016.
Also, $2016=2^{5} \times 3^{2} \times 7$, its smallest positive factor greater than 78 is
$$
2^{2} \times 3 \times 7=84,2016=84 \times 24 .
$$
Thus, $d \leqslant 24$, and $d=24$ can be achieved, by setting
$$
\left(a_{1}, a_{2}, \cdots, a_{11}, a_{12}\right)=(1,2, \cdots, 11,18) .
$$ | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $m$ and $n$ be positive integers, and satisfy $24 m=n^{4}$. Then the minimum value of $m$ is $\qquad$ | 10. 54 .
$$
\begin{array}{l}
\text { Given } n^{4}=24 m=2^{3} \times 3 m \text {, we know that } \\
m_{\min }=2 \times 3^{3}=54 \text {. }
\end{array}
$$ | 54 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let four complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ satisfy
$$
\begin{array}{l}
\left|z_{1}-z_{2}\right|=1,\left|z_{3}-z_{4}\right|=2, \\
\left|z_{1}-z_{4}\right|=3,\left|z_{2}-z_{3}\right|=4, \\
z=\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right) .
\end{array}
$$
Then the maximum value of $|z|$ is | 5.14 .
Notice,
$$
\begin{array}{l}
|z|=\left|\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right)\right| \\
=\left|z_{1} z_{2}-z_{1} z_{4}-z_{3} z_{2}+z_{3} z_{4}\right| \\
=\left|\left(z_{1}-z_{2}\right)\left(z_{3}-z_{4}\right)+\left(z_{1}-z_{4}\right)\left(z_{2}-z_{3}\right)\right| \\
\leqslant\left|\left(z_{1}-z_{2}\right)\left(z_{3}-z_{4}\right)\right|+\left|\left(z_{1}-z_{4}\right)\left(z_{2}-z_{3}\right)\right| \\
=1 \times 2+3 \times 4=14 .
\end{array}
$$
Therefore, $|z|_{\max }=14$. | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Find the number of all positive integer solutions $(x, y, z)$ to the equation $\arctan \frac{1}{x}+\arctan \frac{1}{y}+\arctan \frac{1}{z}=\frac{\pi}{4}$. | 10. By symmetry, let $x \leqslant y \leqslant z$.
Taking the tangent of both sides of the given equation, we get
$\frac{\frac{1}{y}+\frac{1}{z}}{1-\frac{1}{y z}}=\frac{1-\frac{1}{x}}{1+\frac{1}{x}}$
$\Rightarrow \frac{y+z}{y z-1}=\frac{x-1}{x+1}=1-\frac{2}{x+1}$.
If $x \geqslant 5$, then
$1-\frac{2}{x+1} \geqslant 1-\frac{2}{5+1}=\frac{2}{3} \Rightarrow \frac{y+z}{y z-1} \geqslant \frac{2}{3}$.
However, when $z \geqslant y \geqslant x \geqslant 5$,
$$
\frac{y+z}{y z-1} \leqslant \frac{5+5}{25-1}=\frac{5}{12}<\frac{2}{3},
$$
which is a contradiction. Therefore, $x=2,3,4$.
When $x=2$,
$$
\begin{array}{l}
y z-1=3(y+z) \\
\Rightarrow(y-3)(z-3)=10 \\
\Rightarrow(y, z)=(4,13),(5,8) .
\end{array}
$$
When $x=3$,
$$
\begin{array}{l}
y z-1=2(y+z) \\
\Rightarrow(y-2)(z-2)=5 \\
\Rightarrow(y, z)=(3,7) .
\end{array}
$$
When $x=4$,
$$
\begin{array}{l}
3 y z-3=5(y+z) \\
\Rightarrow 3=5\left(\frac{1}{y}+\frac{1}{z}\right)+\frac{3}{y z} \leqslant 5\left(\frac{1}{4}+\frac{1}{4}\right)+\frac{3}{16},
\end{array}
$$
which is a contradiction.
Thus, the ordered triples $(x, y, z)$ that satisfy $z \geqslant y \geqslant x$ are
$$
(x, y, z)=(2,4,13),(2,5,8),(3,3,7) \text {. }
$$
By permuting the order, we can obtain $6+6+3=15$ solutions.
In conclusion, the number of ordered positive integer solutions to the equation is 15. | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $f(x)$ be a function defined on $\mathbf{R}$, if $f(0)$ $=1008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant 2(x+1), \\
f(x+12)-f(x) \geqslant 6(x+5) .
\end{array}
$$
Then $\frac{f(2016)}{2016}=$ $\qquad$ . | 9. 504 .
From the conditions, we have
$$
\begin{array}{l}
f(x+12)-f(x) \\
=(f(x+12)-f(x+8))+ \\
\quad(f(x+8)-f(x+4))+(f(x+4)-f(x)) \\
\leqslant 2((x+8)+1)+2((x+4)+1)+2(x+1) \\
=6 x+30=6(x+5) . \\
\text { Also, } f(x+12)-f(x) \geqslant 6(x+5), \text { thus, } \\
f(x+12)-f(x)=6(x+5) .
\end{array}
$$
Then, $f(2016)$
$$
\begin{array}{l}
=\sum_{k=0}^{167}(f(12 k+12)-f(12 k))+f(0) \\
=6 \sum_{k=0}^{167}(12 k+5)+1008 \\
=6 \times \frac{(2009+5) \times 168}{2}+1008 \\
=1008 \times 1008 .
\end{array}
$$
$$
\text { Therefore, } \frac{f(2016)}{2016}=\frac{1008}{2}=504 \text {. }
$$ | 504 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. It is known that Team A and Team B each have several people. If 90 people are transferred from Team A to Team B, then the total number of people in Team B will be twice that of Team A; if some people are transferred from Team B to Team A, then the total number of people in Team A will be 6 times that of Team B. Then, the original minimum number of people in Team A is. | 8. 153.
Let the original number of people in team A and team B be $a$ and $b$ respectively.
Then $2(a-90)=b+90$.
Suppose $c$ people are transferred from team B to team A. Then $a+c=6(b-c)$.
From equations (1) and (2), eliminating $b$ and simplifying, we get
$$
\begin{array}{l}
11 a-7 c=1620 \\
\Rightarrow c=\frac{11 a-1620}{7}=a-232+\frac{4(a+1)}{7} \text {. }
\end{array}
$$
Since $a$ and $c$ are positive integers, thus,
$$
c=\frac{11 a-1620}{7} \geqslant 1 \Rightarrow a \geqslant 148 \text {. }
$$
Also, $7 \mid 4(a+1),(4,7)=1$, hence, $7 \mid (a+1)$.
Therefore, the minimum value of $a$ is 153.
Thus, the original number of people in team A is at least 153. | 153 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
518 On a circle, initially write 1 and 2 at opposite positions. Each operation involves writing the sum of two adjacent numbers between them, for example, the first operation writes two 3s, the second operation writes two 4s and two 5s. After each operation, the sum of all numbers becomes three times the previous sum. After performing the operation sufficiently many times, find the sum of the number of 2015s and 2016s written. | Observe the pattern, and conjecture that after a sufficient number of operations, the $n$ numbers written equal $\varphi(n)$. After each operation, the property that adjacent numbers are coprime remains unchanged, and the new number written each time is the sum of its two neighbors. Therefore, the number we are looking for should be
$$
\varphi(2015)+\varphi(2016)=2016.
$$
Since the number written each time is the sum of the two numbers on its sides, observing the numbers on the circle in a clockwise direction, it is only necessary to prove that for any positive integer $n$ and any positive integer $k$ coprime with $n$, $n$ is written as the sum of its left neighbor $k$ and right neighbor $n-k$ exactly once.
We will prove this using the second principle of mathematical induction.
When $n=1,2,3$, it is obviously true.
When $n>3$, since $k$ and $n-k$ are coprime, they are not equal.
If $k>n-k$, according to the induction hypothesis, $k$ is written exactly once as the sum of its left neighbor $2k-n$ and right neighbor $n-k$;
If $k<n-k$, according to the induction hypothesis, $n-k$ is written exactly once as the sum of its left neighbor $k$ and right neighbor $n-2k$.
Therefore, by the second principle of mathematical induction, the conclusion holds. | 2016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Define the sequence $\left\{a_{n}\right\}: a_{n}$ is the last digit of $1+2+\cdots+n$, and $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then $S_{2016}=$ $\qquad$ . | 6.7 066 .
From the problem, we know
$$
\begin{array}{l}
\frac{(n+20)(n+20+1)}{2}=\frac{n^{2}+41 n+420}{2} \\
=\frac{n(n+1)}{2}+20 n+210 .
\end{array}
$$
Then $\frac{(n+20)(n+21)}{2}$ and $\frac{n(n+1)}{2}$ have the same last digit, i.e., $a_{n+20}=a_{n}$.
$$
\begin{array}{l}
\text { Therefore, } S_{2016}=S_{16}+100 S_{20} . \\
\text { Also, } S_{20}=a_{1}+a_{2}+\cdots+a_{20} \\
=1+3+6+0+5+1+8+6+5+5+ \\
\\
6+8+1+5+0+6+3+1+0+0 \\
=70, \\
S_{16}=66,
\end{array}
$$
Thus, $S_{2016}=S_{16}+100 S_{20}=7066$. | 7066 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. From five positive integers $a, b, c, d, e$, any four are taken to find their sum, resulting in the set of sums $\{44,45,46,47\}$, then $a+b+c+d+e=$ $\qquad$ . | 2. 57 .
From five positive integers, if we take any four to find their sum, there are five possible ways, which should result in five sum values. Since the set $\{44,45, 46,47\}$ contains only four elements, there must be two sum values that are equal. Therefore,
$$
\begin{array}{l}
44+44+45+46+47 \\
\leqslant 4(a+b+c+d+e) \\
\leqslant 44+45+46+47+47 \\
\Rightarrow 226 \leqslant 4(a+b+c+d+e) \leqslant 229 .
\end{array}
$$
Since $a, b, c, d, e$ are integers, we get
$$
\begin{array}{l}
4(a+b+c+d+e)=228 \\
\Rightarrow a+b+c+d+e=57
\end{array}
$$ | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the numbers $2, 3, 4, 6, 8, 9, 12, 15$ in a row so that the greatest common divisor of any two adjacent numbers is greater than 1. The total number of possible arrangements is ( ) .
(A) 720
(B) 1014
(C) 576
(D) 1296 | 4. D.
First, divide the eight numbers into three groups:
I $(2,4,8)$, II $(3,9,15)$, III $(6,12)$.
Since the numbers in group I and group II have no common factors, the arrangement that satisfies the condition must be:
(1) After removing the numbers 6 and 12, the remaining numbers are divided into three parts and arranged (in sequence) as I, II, I or II, I, II. In this case, there are $2 \times 3! \times 3! \times 2 = 144$ ways to arrange these six numbers. And 6 and 12 can be placed at the intersection of different parts in two different ways, totaling $2 \times 144 = 288$ ways.
(2) After removing the numbers 6 and 12, the remaining numbers are divided into two parts and arranged as I, II or II, I. In this case, there are $2 \times 3! \times 3!$ ways to arrange these six numbers. If 6 and 12 are both at the boundary between group I and group II, there are two ways to arrange them. Otherwise, only one is at the boundary, and the other has six possible positions, totaling $2 \times 3! \times 3! \times (2 + 2 \times 6) = 1008$.
Therefore, the total number of arrangements is $1008 + 288 = 1296$. | 1296 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
3. If three numbers are taken simultaneously from the 14 integers $1,2, \cdots, 14$, such that the absolute difference between any two numbers is not less than 3, then the number of different ways to choose is $\qquad$ | 3. 120 .
Let the three integers taken out be $x, y, z (x<y<z)$.
$$
\begin{array}{l}
\text { Let } a=x, b=y-x-2, \\
c=z-y-2, d=15-z .
\end{array}
$$
Thus, $a, b, c, d \geqslant 1$.
If $a, b, c, d$ are determined, then $x, y, z$ are uniquely determined.
Since $a+b+c+d=11$, it is equivalent to dividing 11 identical balls into four piles.
Using the stars and bars method, we get $C_{10}^{3}=120$. | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given ten points in space, where no four points lie on the same plane. Some points are connected by line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$
(2016, National High School Mathematics Joint Competition) | Proof Let $v$ be a vertex in graph $G$, and $N_{i}(v)$ denote the set of points at a distance $i$ from $v$. For example,
$$
N_{0}(v)=\{v\},
$$
$N_{1}(v)=\{u \mid u$ is adjacent to $v\}$,
$N_{2}(v)=\{w \mid w$ is adjacent to $u$, not adjacent to $v$, and $u$ is adjacent to $v\}$.
First, we prove a lemma.
Lemma Let $G=(V, E)$ be a graph with girth $g(G) \geqslant 5$, average degree, number of vertices, and number of edges be $d$, $n$, and $e$ respectively. Then there exists a vertex $v$ such that
$$
\left|N_{2}(v)\right| \geqslant d^{2}-\operatorname{deg} v.
$$
Proof Assume the conclusion does not hold. Then for any vertex $v$, we have
$$
\left|N_{2}(v)\right|<d^{2}-\operatorname{deg} v.
$$
Since $g(G) \geqslant 5$, for any vertex $v$, if $w \in N_{2}(v)$, then $w$ is adjacent to exactly one vertex in $N_{1}(v)$. Because $w \in N_{2}(v)$, there is at least one vertex in $N_{1}(v)$ that is adjacent to it. If there were two vertices adjacent to $w$, say $u_{1}$ and $u_{2}$, then $v, u_{1}, w, u_{2}$ would form a cycle of length 4, contradicting $g(G) \geqslant 5$.
Therefore, for any vertex $v$, we have
$\left|N_{2}(v)\right|=\sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)$.
Thus, $\sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)<d^{2}-\operatorname{deg} v$.
Summing over all $v$ gives
$\sum_{v \in V} \sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)<\sum_{v \in V}\left(d^{2}-\operatorname{deg} v\right) = n d^{2}-2 e$.
The left-hand side is $\sum_{v \in V}(\operatorname{deg} v)(\operatorname{deg} v-1)$, so
$$
\sum_{v \in V} \operatorname{deg}^{2} v < n d^{2} = n\left(\frac{\sum_{v \in V} \operatorname{deg} v}{n}\right)^{2}.
$$
Let $\boldsymbol{a}=(1,1, \cdots, 1)$,
$\boldsymbol{b}=\left(\operatorname{deg} v_{1}, \operatorname{deg} v_{2}, \cdots, \operatorname{deg} v_{n}\right)$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
|\boldsymbol{a} \cdot \boldsymbol{b}|^{2}=\left(\sum_{v \in V} \operatorname{deg} v\right)^{2} \\
\leqslant|\boldsymbol{a}|^{2}|\boldsymbol{b}|^{2}=n \sum_{v \in V} \operatorname{deg}^{2} v \\
\Rightarrow \sum_{v \in V} \operatorname{deg}^{2} v \geqslant n\left(\frac{\sum_{v \in V} \operatorname{deg} v}{n}\right)^{2},
\end{array}
$$
which is a contradiction.
The lemma is proved.
Since for any vertex $v$, we have
$$
n \geqslant\left|N_{0}(v)\right|+\left|N_{1}(v)\right|+\left|N_{2}(v)\right|,
$$
by the lemma, we get
$n \geqslant 1+\operatorname{deg} v+d^{2}-\operatorname{deg} v=1+d^{2}$
$\Rightarrow d \leqslant \sqrt{n-1}$.
Since $d=\frac{1}{|V|} \sum_{v \in V} \operatorname{deg} v=\frac{2 e}{n}$, we have
$e \leqslant \frac{n \sqrt{n-1}}{2}$.
Another proof: Construct a simple 10-vertex graph $G$ with the ten points as vertices and the connecting line segments as edges.
Since graph $G$ contains no triangles or quadrilaterals, $g(G) \geqslant 5$.
By the theorem, the number of edges is at most
$\frac{10 \times \sqrt{10-1}}{2}=15$.
Additionally, the Petersen graph satisfies the conditions, so the maximum number of edges is 15. | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Arrange the numbers in the set $\left\{2^{x}+2^{y} \mid x 、 y \in \mathbf{N}, x<y\right\}$ in ascending order. Then the 60th number is $\qquad$ (answer with a number).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3.2064.
It is known that the number of combinations $(x, y)$ satisfying $0 \leqslant x<y \leqslant n$ is $\mathrm{C}_{n+1}^{2}$.
Notice that, $\mathrm{C}_{11}^{2}=55<60<66=\mathrm{C}_{12}^{2}$.
Therefore, the 60th number satisfies $y=11, x=4$, which means the 60th number is $2^{11}+2^{4}=2064$. | 2064 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the maximum value of $n$ such that there exists an arithmetic sequence $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\sum_{i=1}^{n}\left|a_{i}\right|=\sum_{i=1}^{n}\left|a_{i}+1\right|=\sum_{i=1}^{n}\left|a_{i}-2\right|=507 .
$$
$(2005$, China Southeast Mathematical Olympiad) | 【Analysis】Let's set $a_{i}=a-i d(d>0, i=1,2$, $\cdots, n)$. Then the given system of equations becomes
$$
\left\{\begin{array}{l}
\sum_{i=1}^{n}|a-i d|=507, \\
\sum_{i=1}^{n}|a+1-i d|=507, \\
\sum_{i=1}^{n}|a-2-i d|=507 .
\end{array}\right.
$$
Thus, the absolute value sum function $f(x)=\sum_{i=1}^{n}|x-i d|$ has three "equal value points":
$$
f(a-2)=f(a)=f(a+1)=507,
$$
It must be that $n=2 k\left(k \in \mathbf{Z}_{+}\right)$, and
$$
\begin{array}{l}
\left\{\begin{array}{l}
a-2, a, a+1 \in[k d,(k+1) d], \\
f(k d)=507
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
d \geqslant 3, \\
k^{2} d=507
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
d \geqslant 3, \\
k^{2}=\frac{507}{d} \leqslant \frac{507}{3}=169
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
d \geqslant 3, \\
k \leqslant 13 .
\end{array}\right.
\end{array}
$$
Therefore, $n \leqslant 26$, with equality holding when $d=3, k=13$.
Thus, when $a-2=k d=39$, i.e., $a=41$, the corresponding arithmetic sequence is $\{41-3 k\}(k=1,2, \cdots, 26)$. Hence, $n_{\max }=26$. | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 There are 68 pairs of non-zero integers on the blackboard. For a positive integer $k$, at most one of the pairs $(k, k)$ and $(-k, -k)$ appears on the blackboard. A student erases some of these 136 numbers so that the sum of any two erased numbers is not 0. It is stipulated that if at least one number from a pair among the 68 pairs is erased, the student scores one point. Find the maximum score the student can achieve.
Note: These 68 pairs can include some identical pairs. ${ }^{[3]}$
(2010, USA Mathematical Olympiad) | Given that $(j, j)$ and $(-j, -j)$ can appear at most as one pair, we can assume that if $(j, j)$ appears, then $j > 0$ (otherwise, replace $j$ with $-j$). For a positive integer $k$, all $k$ or $-k$ can be deleted from the blackboard, but not both. For each $k > 0$, delete $k$ with probability $p$ and $-k$ with probability $1-p$.
Thus, the probability of scoring for each pair is at least $\min \left\{p, 1-p^{2}\right\}$. Therefore, the expected score for 68 pairs is $68 \min \left\{p, 1-p^{2}\right\}$.
Solving the equation $p = 1 - p^{2}$, we get $p = \frac{\sqrt{5} - 1}{2}$.
Therefore, the expected total score is at least
\[ 68 p = 68 \times \frac{\sqrt{5} - 1}{2} > 42. \]
Thus, there must exist a way to achieve a score of 43.
Finally, we provide an example to show that 44 points may not be feasible.
For $1 \leqslant i \leqslant 8$, each pair $(i, i)$ appears 5 times; for $1 \leqslant i \neq j \leqslant 8$, each pair $(-i, -j)$ appears 1 time: a total of $40 + 28 = 68$ pairs.
Suppose $k$ pairs of numbers from 1 to 8 are deleted. Then the first type of pairs can score $5k$ points; and among the second type of pairs, at least $\mathrm{C}_{k}^{2}$ pairs cannot be taken, so the score is $28 - \mathrm{C}_{k}^{2}$. The total score is $5k + 28 - \mathrm{C}_{k}^{2}$, and by completing the square, the maximum value is 43 points. | 43 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) A function calculator has a display screen and two operation keys. If the first operation key is pressed once, the number on the display screen will change to $\left[\frac{x}{2}\right]$ (where $[x]$ represents the greatest integer not exceeding the real number $x$); if the second operation key is pressed once, the number on the display screen will change to $4x+1$. Pressing either operation key once is called one operation. The number on the display screen is currently 1. Answer the following questions:
(1) Can the number 2000 appear on the display screen after a finite number of operations? Explain your reasoning.
(2) How many integers less than 2000 can appear on the display screen after a finite number of operations? | Three, (1) Impossible.
Convert the number to binary. Then pressing the first operation key means removing the last digit of the number on the display; pressing the second operation key means appending 01 to the number on the display.
When the initial number is 1, after performing the above two operations, the resulting number does not have two 1s adjacent. However, 2000 in binary is $(11111010000)_2$, which has two 1s adjacent. Therefore, the number 2000 cannot appear on the display.
(2) If the second operation is performed first, followed by the first operation, it is equivalent to appending a 0 to the original number. Hence, any binary number without two 1s adjacent can be obtained after a finite number of operations.
Thus, the number of integers less than 2000 that can appear on the display is equivalent to the number of natural numbers not greater than $(10101010101)_2$ and without two 1s adjacent.
Moreover, the number of natural numbers with $k(0 \leqslant k \leqslant 6)$ 1s and not greater than $(10101010101)_2$ is $\mathrm{C}_{12-k}^{k}$. Therefore, the number of natural numbers that meet the condition is
$$
1+11+45+84+70+21+1=233 .
$$ | 233 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) There are $12k$ people attending a meeting, each of whom has shaken hands with exactly $3k+6$ others, and for any two of them, the number of people who have shaken hands with both is the same. How many people attended the meeting? Prove your conclusion. | Three, abstracting a person as a point and two people shaking hands as a line connecting two points, then $A$ shaking hands with $B$ and $C$ corresponds to $\angle BAC$ in the graph. Since each person has shaken hands with $3k+6$ people, there are $12k \mathrm{C}_{3k+6}^{2}$ such angles in the graph. The number of pairs of people in the graph is $\mathrm{C}_{12k}^{2}$, so the number of common handshakes between any two people is
$$
d=\frac{12k \mathrm{C}_{3k+6}^{2}}{\mathrm{C}_{12k}^{2}}=\frac{3(k+2)(3k+5)}{12k-1} \in \mathbf{Z}_{+} \text{. }
$$
Noting that $(3,12k-1)=1$,
$$
\begin{aligned}
p & =\frac{(k+2)(3k+5)}{12k-1}=k+1+\frac{11-9k^{2}}{12k-1} \\
& =k+1-11+\frac{-3k(3k-44)}{12k-1} \in \mathbf{Z}_{+},
\end{aligned}
$$
and $(3k, 12k-1)=1$.
Thus, $(12k-1) \mid (3k-44)$.
Then $3k-44 \geqslant 12k-1$ or $44-3k \geqslant 12k-1$.
Solving, we get $k \leqslant -\frac{43}{9}$ (discard) or $k \leqslant 3$.
Upon inspection, only when $k=3$, $d \in \mathbf{Z}_{+}$. At this time, $d=6$.
In summary, there are 36 people attending the meeting, and the number of common handshakes between any two people is 6. | 36 | Combinatorics | proof | Yes | Yes | cn_contest | false |
2. A four-digit number divided by 433 has a quotient of $a$ and a remainder of $r$ $(a 、 r \in \mathbf{N})$. Then the maximum value of $a+r$ is $\qquad$ . | 2.454.
Let the four-digit number be $433 a+r(0 \leqslant r \leqslant 432)$. Since $433 \times 24=10392>9999$, then $a \leqslant 23$.
When $a=23$, $433 \times 23+40=9999$. At this point, $a+r=23+40=63$. When $a=22$, $433 \times 22+432=9958$. At this point, $a+r=22+432=454$. In summary, the maximum value of $a+r$ is 454. | 454 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
\begin{array}{c}
S=\left[\frac{1}{1}\right]+\left[\frac{2}{1}\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\left[\frac{3}{2}\right]+ \\
{\left[\frac{4}{2}\right]+\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{3}{3}\right]+\left[\frac{4}{3}\right]+} \\
{\left[\frac{5}{3}\right]+\left[\frac{6}{3}\right]+\cdots}
\end{array}
$$
up to 2016 terms, where, for a segment with denominator $k$, there are $2 k$ terms $\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]$, and only the last segment may have fewer than $2 k$ terms. Then the value of $S$ is | 6.1078.
$$
2+4+\cdots+2 \times 44=1980 \text {. }
$$
For any integer \( k \) satisfying \( 1 \leqslant k \leqslant 44 \), the sum includes \( 2k \) terms with the denominator \( k \): \(\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]\), whose sum is \( k+2 \).
Also, \( 2016-1980=36 \), so the sum includes 36 terms with the denominator 45: \(\left[\frac{1}{45}\right],\left[\frac{2}{45}\right], \cdots,\left[\frac{36}{45}\right]\), whose sum is zero.
$$
\text { Therefore, } S=\sum_{k=1}^{44}(k+2)=1078 \text {. }
$$ | 1078 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. If real numbers $a, b, c$ make the quadratic function $f(x) = a x^{2} + b x + c$ such that when $0 \leqslant x \leqslant 1$, always $|f(x)| \leqslant 1$. Then the maximum value of $|a| + |b| + |c|$ is $\qquad$ | 7. 17.
Take $x=0, \frac{1}{2}, 1$.
From the problem, we have
$$
\begin{array}{l}
|c| \leqslant 1,|a+2 b+4 c| \leqslant 4, \\
|a+b+c| \leqslant 1 .
\end{array}
$$
Let $m=a+2 b+4 c, n=a+b+c$.
Then $a=-m+2 n+2 c, b=m-n-3 c$
$$
\begin{aligned}
\Rightarrow & |a| \leqslant|m|+2|n|+2|c| \leqslant 8, \\
& |b| \leqslant|m|+|n|+3|c| \leqslant 8 \\
\Rightarrow & |a|+|b|+|c| \leqslant 17 .
\end{aligned}
$$
It is easy to verify,
$$
f(x)=8\left(x-\frac{1}{2}\right)^{2}-1=8 x^{2}-8 x+1 \text {, }
$$
When $0 \leqslant x \leqslant 1$, it always holds that $|f(x)| \leqslant 1$.
Therefore, the maximum value sought is 17. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. | Let the ten points be $A_{1}, A_{2}, \cdots, A_{10}$. Using these ten points as vertices and the line segments connecting them as edges, we obtain a simple graph $G$ of order 10. Let the degree of point $A_{i} (i=1,2, \cdots, 10)$ be $d_{i}$. Then the total number of edges in graph $G$ is $\frac{1}{2} \sum_{i=1}^{10} d_{i}$.
In graph $G$, we call the figure formed by two line segments sharing a common endpoint an angle. Thus, there are $\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}$ different angles.
By the given condition, for any angle $\angle A_{i} A_{j} A_{k}$ in graph $G$, $A_{i}$ and $A_{k}$ are not adjacent. For any two angles $\angle A_{i_{1}} A_{j_{1}} A_{k_{1}}$ and $\angle A_{i_{2}} A_{j_{2}} A_{k_{2}}$, we have
$\left\{A_{i_{1}}, A_{k_{1}}\right\} \neq\left\{A_{i_{2}}, A_{k_{2}}\right\}$.
Thus, the complement graph of $G$ has at least $\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}$ edges.
Therefore, the total number of edges $S$ satisfies
$$
\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}=S \leqslant \mathrm{C}_{10}^{2}-\frac{1}{2} \sum_{i=1}^{10} d_{i} \text {, }
$$
which implies $\mathrm{C}_{10}^{2} \geqslant \sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}+\frac{1}{2} \sum_{i=1}^{10} d_{i}=\frac{1}{2} \sum_{i=1}^{10} d_{i}^{2}$.
Also, $\frac{1}{2} \sum_{i=1}^{10} d_{i}^{2} \geqslant \frac{1}{2} \times \frac{1}{10}\left(\sum_{i=1}^{10} d_{i}\right)^{2}$, so
$$
\mathrm{C}_{10}^{2} \geqslant \frac{1}{2} \times \frac{1}{10}\left(\sum_{i=1}^{10} d_{i}\right)^{2} \Rightarrow \frac{1}{2} \sum_{i=1}^{10} d_{i} \leqslant 15 \text {. }
$$
A specific construction is shown in Figure 1. | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the sum of the digits of the natural number $x$ be $S(x)$. Then the solution set of the equation $x+S(x)+S(S(x))+S(S(S(x)))=2016$ is $\qquad$ | $-1 .\{1980\}$.
It is easy to see that $x<2016$.
Note that, the sum of the digits of natural numbers less than 2016 is at most 28, for example, $S(1999)=28$, which indicates,
$$
S(x) \leqslant 28 \text {. }
$$
Furthermore, $S(S(x)) \leqslant S(19)=10$.
Finally, $S(S(S(x))) \leqslant 9$.
From the equation we get
$$
\begin{array}{l}
x=2016-S(x)-S(S(x))-S(S(S(x))) \\
\geqslant 2016-28-10-9=1969 .
\end{array}
$$
Thus, $x \in\{1969,1970, \cdots, 2015\}$.
Also, $x, S(x), S(S(x)), S(S(S(x)))$ have the same remainder when divided by 9, and 2016 has a remainder of 0 when divided by 9, so each number has a remainder of 0 when divided by 9.
Therefore, $x$ can only be $1971, 1980, 1989, 1998, 2007$.
Upon inspection, only $1980+18+9+9=2016$. Therefore, the solution set of the equation is $\{1980\}$. | 1980 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set $A=\left[\frac{7}{8}\right]+\left[\frac{7^{2}}{8}\right]+\cdots+\left[\frac{7^{2016}}{8}\right]$. Then the remainder when $A$ is divided by 50 is $\qquad$ | 3. 42.
Since $\frac{7^{2 k-1}}{8}$ and $\frac{7^{2 k}}{8}$ are not integers, and
$$
\frac{7^{2 k-1}}{8}+\frac{7^{2 k}}{8}=7^{2 k-1},
$$
for any $k \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
{\left[\frac{7^{2 k-1}}{8}\right]+\left[\frac{7^{2 k}}{8}\right]=7^{2 k-1}-1} \\
\equiv 7(-1)^{k-1}-1(\bmod 50) .
\end{array}
$$
Thus, $A \equiv 7(1-1+1-1+\cdots+1-1)-1008$ $\equiv 42(\bmod 50)$. | 42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Rectangle $R$ is divided into 2016 small rectangles, with each small rectangle's sides parallel to the sides of rectangle $R$. The vertices of the small rectangles are called "nodes". For a line segment on the side of a small rectangle, if both endpoints are nodes and its interior does not contain any other nodes, then this line segment is called a "basic segment". Considering all possible divisions, find the maximum and minimum number of basic segments. ${ }^{[4]}$ | Consider a graph $G$ with all nodes as vertices and basic segments as edges. Let the number of vertices in graph $G$ be $v$, and the number of edges be $e$. Treat the external region of rectangle $R$ as one face (region). Then, the total number of faces in graph $G$ is $f=2017$.
By Euler's formula, we have $v+f-e=2$.
Thus, $e=v+2015$.
Note that, the points with degree 2 in graph $G$ are exactly 4, the rest of the points have a degree of 3 or 4. Points with degree 2 are the vertices of one rectangle, points with degree 3 are the vertices of two rectangles, and points with degree 4 are the vertices of four rectangles.
On one hand, the total number of vertices of all small rectangles is $4 \times 2016$.
On the other hand, the total number of vertices of all small rectangles is no less than $2(v-4)+4=2 v-4$.
Thus, $2 v-4 \leqslant 4 \times 2016 \Rightarrow v \leqslant 4034$.
Therefore, from equation (1), we get $e \leqslant 6049$.
When the rectangle $R$ is divided into $1 \times 2016$ small rectangles, the above equality holds.
Thus, the maximum value of $e$ is $e_{\text {max }}=6049$.
Next, consider the minimum value of $e$.
Suppose the rectangle $R$ is divided using $a$ horizontal lines and $b$ vertical lines, excluding the boundaries.
Since $a$ horizontal lines and $b$ vertical lines can divide the area into at most $(a+1)(b+1)$ regions, we have
$$
(a+1)(b+1) \geqslant 2016 \text {. }
$$
Let the number of points with degree 3 be $x$, and the number of points with degree 4 be $y$. Then,
$$
v=4+x+y, 4+2 x+4 y=4 \times 2016 .
$$
Each horizontal and vertical line has its endpoints as the vertices of two rectangles, thus they are all points with degree 3, and these points are distinct.
Hence, $x \geqslant 2 a+2 b$.
Combining with equation (2), we get
$$
\begin{array}{l}
x \geqslant 2(a+1)+2(b+1)-4 \\
\geqslant 4 \sqrt{(a+1)(b+1)}-4 \\
\geqslant 4 \sqrt{2016}-4>175.59 \\
\Rightarrow x \geqslant 176 .
\end{array}
$$
Thus, from equations (3) and (4), we get
$$
v=2019+\frac{1}{2} x \geqslant 2017 \text {. }
$$
From equation (1), we immediately get $e \geqslant 4122$, where the equality holds in the $42 \times 48$ division.
Thus, $e_{\min }=4122$. | 4122 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the length of intervals $(m, n)$, $[m, n)$, $(m, n]$, and $[m, n]$ to be $n-m$ (where $n, m \in \mathbf{R}$, and $n > m$). Then the sum of the lengths of the intervals of real numbers $x$ that satisfy
$$
\frac{1}{x-20}+\frac{1}{x-17} \geqslant \frac{1}{512}
$$
is $\qquad$ . | $-1.1024$.
Let $a=20, b=17, c=\frac{1}{512}$.
Then $a>b>c>0$.
The original inequality is equivalent to $\frac{2 x-(a+b)}{(x-a)(x-b)} \geqslant c$.
When $x>a$ or $x0, f(a)=b-a<0$.
Let the two real roots of $f(x)=0$ be $x_{1}$ and $x_{2}$ $\left(x_{1}<x_{2}\right)$. Then the interval of $x$ that satisfies $f(x) \leqslant 0$ is $\left(a, x_{2}\right]$, with the length of the interval being $x_{2}-a$.
Similarly, when $b<x<a$, the interval of $x$ that satisfies $f(x) \geqslant 0$ is $\left(b, x_{1}\right]$, with the length of the interval being $x_{1}-b$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=\frac{a c+b c+2}{c}=a+b+\frac{2}{c} \text {. }
$$
Then the sum of the lengths of the intervals of $x$ that satisfy the conditions is
$$
\begin{array}{l}
x_{2}-a+x_{1}-b=a+b+\frac{2}{c}-a-b \\
=\frac{2}{c}=1024 .
\end{array}
$$ | 1024 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the function
$$
\begin{aligned}
f(x)= & a \tan ^{2017} x+b x^{2017}+ \\
& c \ln \left(x+\sqrt{x^{2}+1}\right)+20,
\end{aligned}
$$
where $a$, $b$, and $c$ are real numbers. If $f\left(\ln \log _{5} 21\right)=17$, then $f\left(\ln \log _{21} 5\right)=$ $\qquad$ | 3. 23 .
Let $g(x)$
$$
=a \tan ^{2017} x+b x^{2017}+c \ln \left(x+\sqrt{x^{2}+1}\right) \text {. }
$$
Then $g(-x)=-g(x)$
$$
\begin{array}{l}
\Rightarrow f(-x)-20=-(f(x)-20) \\
\Rightarrow f(-x)=40-f(x) .
\end{array}
$$
Therefore, $f\left(\ln \log _{21} 5\right)=f\left(-\ln \log _{5} 21\right)$
$$
=40-f\left(\ln \log _{5} 21\right)=23 \text {. }
$$ | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. For any positive integer $n$, define
$$
S(n)=\left[\frac{n}{10^{[\lg n]}}\right]+10\left(n-10^{[\lg n]}\left[\frac{n}{10^{[\lg n]}}\right]\right) \text {. }
$$
Then among the positive integers $1,2, \cdots, 5000$, the number of positive integers $n$ that satisfy $S(S(n))=n$ is $\qquad$ . | 7. 135.
Let $t=10^{[18 n]}$, then
$$
S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right)
$$
Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$, and $\left[\frac{n}{t}\right]$ is the leading digit of $n$.
We will discuss the cases separately.
(1) If $n$ is a one-digit number, then all such $n$ satisfy the condition, and there are 9 such $n$;
(2) If $n$ is a two-digit number, let $n=\overline{x y}$, then
$$
S(n)=\overline{y x}, S(S(n))=\overline{x y} \text {, }
$$
There are 81 such $n$, where $x, y \neq 0$;
(3) If $n$ is a three-digit number, let $n=\overline{x y z}$, then
$$
S(n)=\overline{y z x}, S(S(n))=\overline{z x y} \text {, }
$$
Thus, $x=y=z \neq 0$, and there are 9 such $n$;
(4) If $n$ is a four-digit number, let $n=\overline{x y z w}$, then
$$
S(S(n))=\overline{z w x y} \text {, }
$$
Thus, $w=y, z=x$, and there are 36 such $n$, where $x=1,2,3,4, y \neq 0$.
In summary, the number of positive integers $n$ that satisfy the condition is 135. | 135 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given the ellipse $\Gamma: \frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, a line passing through the left focus $F(-2,0)$ of the ellipse $\Gamma$ with a slope of $k_{1}\left(k_{1} \notin\{0\right.$, $\infty\})$ intersects the ellipse $\Gamma$ at points $A$ and $B$. Let point $R(1,0)$, and extend $A R$ and $B R$ to intersect the ellipse $\Gamma$ at points $C$ and $D$ respectively. The slope of line $C D$ is $k_{2}$. Write $\frac{k_{1}^{2}}{k_{2}^{2}}$ as a reduced fraction $\frac{a}{b}$ (where $a$ and $b$ are coprime positive integers). Then $a^{2}+b=$ | 9. 305 .
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, $C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$,
$l_{A R}: x=\frac{x_{1}-1}{y_{1}} y+1$.
Substitute into the equation of the ellipse $\Gamma$, eliminate $x$ to get
$$
\frac{5-x_{1}}{y_{1}^{2}} y^{2}+\frac{x_{1}-1}{y_{1}} y-4=0 \text {. }
$$
By Vieta's formulas, we have
$$
y_{1} y_{3}=-\frac{4 y_{1}^{2}}{5-x_{1}}\left(y_{1} \neq 0\right) \Rightarrow y_{3}=\frac{4 y_{1}}{x_{1}-5} \text {. }
$$
Substitute into the equation of the line $A R$ to get $x_{3}=\frac{5 x_{1}-9}{x_{1}-5}$.
Similarly, $x_{4}=\frac{5 x_{2}-9}{x_{2}-5}, y_{4}=\frac{4 y_{2}}{x_{2}-5}$.
Thus, $k_{2}=\frac{y_{3}-y_{4}}{x_{3}-x_{4}}$
$$
=\frac{4\left(y_{1} x_{2}-5 y_{1}-y_{2} x_{1}+5 y_{2}\right)}{16\left(x_{2}-x_{1}\right)} \text {. }
$$
Since points $A, F, B$ are collinear, we have
$$
\begin{array}{l}
\frac{y_{1}}{x_{1}+2}=\frac{y_{2}}{x_{2}+2} \\
\Rightarrow y_{1} x_{2}-y_{2} x_{1}=2\left(y_{2}-y_{1}\right) \\
\Rightarrow k_{2}=\frac{7}{4} \cdot \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7}{4} k_{1} \\
\Rightarrow \frac{a}{b}=\frac{k_{1}^{2}}{k_{2}^{2}}=\frac{16}{49} \Rightarrow a=16, b=49 \\
\Rightarrow a^{2}+b=305 .
\end{array}
$$ | 305 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the maximum value of the positive integer $r$ such that: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. | Three, first explain $r \leqslant 200$.
Take $k \in\{1,2, \cdots, 10\}$. Let
$$
A_{k}=\{100 k-99,100 k-98, \cdots, 100 k\} \text {. }
$$
Consider the set
$$
\begin{array}{l}
A_{1} \cup A_{5} \cup A_{6} \cup A_{7} \cup A_{9}, A_{1} \cup A_{2} \cup A_{7} \cup A_{8} \cup A_{10}, \\
A_{2} \cup A_{3} \cup A_{6} \cup A_{8} \cup A_{9}, A_{3} \cup A_{4} \cup A_{7} \cup A_{9} \cup A_{10}, \\
A_{4} \cup A_{5} \cup A_{6} \cup A_{8} \cup A_{10},
\end{array}
$$
It can be seen that these satisfy the problem's conditions and each set has 200 elements. Thus, $r \leqslant 200$.
Define $a_{i j}=\left\{\begin{array}{l}1, i \in A_{j} ; \\ 0, \text { otherwise, }\end{array} m_{i}=\sum_{j=1}^{5} a_{i j}\right.$, where $i=1,2, \cdots, 1000 ; j=1,2, \cdots, 5$.
$$
\begin{array}{l}
\text { Then } \sum_{i=1}^{1000} m_{i}=2500 . \\
\text { By } \sum_{1 \leqslant i<j \leqslant 5}\left|A_{i} \cap A_{j}\right|=\sum_{i=1}^{1000} \mathrm{C}_{m_{i}}^{2} \\
=\frac{1}{2}\left(\sum_{i=1}^{1000} m_{i}^{2}-\sum_{i=1}^{1000} m_{i}\right), \\
\sum_{i=1}^{1000} m_{i}^{2} \geqslant \frac{1}{1000}\left(\sum_{i=1}^{1000} m_{i}\right)^{2},
\end{array}
$$
we know that $\sum_{i=1}^{1000} m_{i}^{2}$ takes its minimum value when $m_{i}$ are as close to each other as possible.
Suppose there are $x$ 2's and $y$ 3's.
Then $\left\{\begin{array}{l}x+y=1000, \\ 2 x+3 y=2500\end{array} \Rightarrow x=y=500\right.$.
Thus, $\sum_{i=1}^{1000} m_{i}^{2} \geqslant 500 \times 2^{2}+500 \times 3^{2}=6500$
$\Rightarrow \sum_{1 \leqslant i<j \leqslant 5}\left|A_{i} \cap A_{j}\right| \geqslant 2000$.
Therefore, there must exist $1 \leqslant i<j \leqslant 5$, such that
$\left|A_{i} \cap A_{j}\right| \geqslant \frac{2000}{C_{5}^{2}}=200$.
Hence, $r \geqslant 200$. | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the number of integers in the set $\left\{\left.\frac{2015[a, b]}{a+b} \right\rvert\, a 、 b \in \mathbf{Z}_{+}\right\}$. | Let $d=(a, b)$, and $a=A d, b=B d$, where $A$ and $B$ are coprime positive integers.
Since $[a, b]=A B(a, b)$, then
$(a+b)|2015[a, b] \Leftrightarrow(A+B)| 2015 A B$.
Because $(A, B)=1$, we have
$(A+B, A)=1,(A+B, B)=1$,
$(A+B, A B)=1$.
Thus, $(A+B) \mid 2015$.
For a fixed divisor $k$ of 2015 greater than 1, and $k$ is odd, let the number of positive integers less than $k$ and coprime to $k$ be $\varphi(k)$. Suppose these numbers are
$$
1=c_{1}<c_{2}<\cdots<c_{\varphi(k)} \leq k-1
$$
and
$$
c_{1}+c_{2}+\cdots+c_{\varphi(k)}=\frac{\varphi(k)(k-1)}{2} \quad \text{and} \quad \varphi(1)} \varphi(k)=n-1$.
To prove that in the set $\left\{\frac{1}{n}, \frac{2}{n}, \cdots, \frac{n-1}{n}\right\}$, when the fractions are simplified to their lowest terms, the denominators can only be the divisors $k$ of $n$ that are greater than or equal to 2, and the number of fractions with denominator $k$ is $\varphi(k)$, the total number of fractions in lowest terms is
$$
\sum_{\substack{k \mid n \\ k>1}} \varphi(k)=n-1 .
$$
The lemma is proved.
By the lemma, the number of integers in the original set is
$$
\sum_{\substack{k \mid 2015 \\ k>1}} \frac{\varphi(k)}{2}=\frac{2015-1}{2}=1007 \text {. }
$$
[Note] For problems involving the greatest common divisor or the least common multiple, setting $d=(a, b)$, and $a=A d, b=B d$ (where $A$ and $B$ are coprime integers) is a common transformation. Using this transformation can simplify the problem and pave the way for solving it. | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Red, blue, green, and white four dice, each die's six faces have numbers $1, 2, 3, 4, 5, 6$. Simultaneously roll these four dice so that the product of the numbers facing up on the four dice equals 36, there are $\qquad$ possible ways. | 4. 48 .
$$
\begin{array}{l}
36=6 \times 6 \times 1 \times 1=6 \times 3 \times 2 \times 1 \\
=4 \times 3 \times 3 \times 1=3 \times 3 \times 2 \times 2 .
\end{array}
$$
For each of the above cases, there are respectively
$$
\begin{array}{l}
\frac{4!}{(2!)(2!)}=6,4!=24, \\
\frac{4!}{2!}=12, \frac{4!}{(2!)(2!)}=6
\end{array}
$$
possibilities.
In total, there are 48 possibilities. | 48 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given natural numbers $a, b, c$ whose sum is $S$, satisfying $a+b=1014, c-b=497, a>b$. Then the maximum value of $S$ is ( ).
(A) 1511
(B) 2015
(C) 22017
(D) 2018 | $\begin{array}{l}\text { I. 1.C. } \\ \text { Given } S=a+b+c=1014+b+497 \text {, and } a>b \\ \Rightarrow 1014=a+b \geqslant b+1+b \\ \Rightarrow b \leqslant 506.6 \Rightarrow b_{\max }=506 \\ \Rightarrow S_{\text {max }}=1014+506+497=2017 .\end{array}$ | 2017 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Four. (50 points) Let $A=\{0,1, \cdots, 2016\}$. If a surjective function $f: \mathbf{N} \rightarrow A$ satisfies: for any $i \in \mathbf{N}$,
$$
f(i+2017)=f(i),
$$
then $f$ is called a "harmonious function".
$$
\begin{array}{l}
\text { Let } f^{(1)}(x)=f(x), \\
f^{(k+1)}(x)=f\left(f^{(k)}(x)\right)\left(k \in \mathbf{N}_{+}\right) .
\end{array}
$$
Suppose the "harmonious function" $f$ satisfies the condition: there exists a positive integer $M$, such that
(1) When $m<M$, if $i, j \in \mathbf{N}$, $i \equiv j+1(\bmod 2017)$,
then $f^{(m)}(i)-f^{(m)}(j) \not \equiv \pm 1(\bmod 2017)$;
(2) If $i, j \in \mathbf{N}, i \equiv j+1(\bmod 2017)$, then $f^{(M)}(i)-f^{(M)}(j) \equiv \pm 1(\bmod 2017)$.
Find the maximum possible value of $M$. | On the one hand, note that 2017 is a prime number.
Let $g$ be a primitive root modulo 2017, then the half-order of $g$ modulo 2017 is 1008.
$$
\text{Let } f(i) \equiv g(i-1)+1(\bmod 2017) \text{.}
$$
Since $(g, 2017)=1$, $g(i-1)+1$ runs through a complete residue system modulo 2017.
Thus, the mapping $f: \mathbf{N} \rightarrow A$ is surjective.
Also, $f(i+2017) \equiv f(i)(\bmod 2017)$, i.e., $f(i+2017)=f(i)$,
Hence, the defined $f$ is a "harmonious mapping".
According to the definition of $f$,
$$
f^{(k)}(i) \equiv g^{k}(i-1)+1(\bmod 2017) \text{.}
$$
At this point, from $i \equiv j+1(\bmod 2017)$, we get
$$
\begin{array}{l}
\left(g^{M}(i-1)+1\right)-\left(g^{M}(j-1)+1\right) \\
\equiv \pm 1(\bmod 2017) \\
\Leftrightarrow g^{M}(i-j) \equiv \pm 1(\bmod 2017) \\
\Leftrightarrow g^{M} \equiv \pm 1(\bmod 2017) .
\end{array}
$$
Noting that the half-order of $g$ modulo 2017 is 1008.
Thus, 1008 divides $M$.
Therefore, the required $M_{\max } \geqslant 1008$.
On the other hand, construct a convex 2017-gon $A_{0} A_{1} \cdots A_{2016}$, denoted as graph $G$. Connect lines according to the following rule: if $1 \leqslant m<M$, $i \in A$, $f^{m}(i)=a$, $f^{m}(i+1)=b$, then connect the segment $A_{a} A_{b}$.
Clearly, the connected segments are diagonals of graph $G$, and the connected segments are not repeated. Otherwise, if there exist two identical lines, i.e., there exist $i, j \in A$, and $1 \leqslant q<p<M$, such that $f^{(p)}(i)=f^{(q)}(j)$, and
$$
f^{(p)}(i+1)=f^{(q)}(j+1)
$$
or $f^{(p)}(i+1)=f^{(q)}(j-1)$.
$$
\text{Then }\left\{\begin{array}{l}
f^{(p-q)}(i)=j, \\
f^{(p-q)}(i+1) \equiv j+1 \text{ or } j-1(\bmod 2017) .
\end{array}\right.
$$
Noting that $0<p-q<M$, this contradicts the given conditions. Hence, the connected diagonals are not repeated.
Since a total of $2017(M-1)$ segments are connected, and the convex 2017-gon $G$ has $2017 \times 2017$ segments, thus,
$$
\begin{array}{l}
2017(M-1) \leqslant 2017 \times 1007 \\
\Rightarrow M \leqslant 1008 .
\end{array}
$$
In summary, the maximum value of $M$ is 1008. | 1008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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