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Example 5 Let the set of all integer points (points with integer coordinates) in the plane be denoted as $S$. It is known that for any $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ in $S$, there exists another point $P$ in $S$ such that the segments $A_{i} P(i=1,2, \cdots, n)$ do not contain any points from $S$ internally. Find the maximum possible value of $n$.
|
The maximum value of the required $n$ is 3.
On one hand, when $n \geqslant 4$, take 4 points $A(0,0), B(1,0), C(0,1)$, $D(1,1)$ in $S$ (the remaining $n-4$ points are chosen arbitrarily), then for any other point $P$ in $S$, analyzing the parity of the coordinates of $P$, there are only 4 cases: (odd, odd), (odd, even), (even, odd), and (even, even). Therefore, among $A, B, C, D$, there is one point such that the midpoint of the line segment connecting it to $P$ is a point in $S$, hence $n \leqslant 3$.
On the other hand, we prove: for any three points $A, B, C$ in $S$, there exists another point $P$ in $S$ such that the interiors of $A P, B P, C P$ do not contain any integer points.
Note that for integer points $X\left(x_{1}, y_{1}\right)$ and $Y\left(x_{2}, y_{2}\right)$ on the plane, the necessary and sufficient condition for the line segment $X Y$ to have no integer points in its interior is $\left(x_{1}-x_{2}, y_{1}-y_{2}\right)=1$. Therefore, we can assume $A\left(a_{1}, a_{2}\right), B\left(b_{1}, b_{2}\right), C(0,0)$.
To find a point $P(x, y)$ such that $A P, B P, C P$ have no integer points in their interiors, $x, y$ need to satisfy the following conditions:
$$\left\{\begin{array}{l}
x \equiv 1(\bmod y) \\
x-a_{1} \equiv 1\left(\bmod y-a_{2}\right) \\
x-b_{1} \equiv 1\left(\bmod y-b_{2}\right)
\end{array}\right.$$
Using the Chinese Remainder Theorem, if there exists $y \in \mathbf{Z}$ such that $y, y-a_{2}, y-b_{2}$ are pairwise coprime, then an $x$ satisfying (2) exists, thus finding a point $P$ that meets the requirements.
Take $y=m\left[a_{2}, b_{2}\right]+1, m \in \mathbf{Z}, m$ to be determined, then $\left(y, y-a_{2}\right)=\left(y, a_{2}\right)=\left(1, a_{2}\right)=1$, similarly $\left(y, y-b_{2}\right)=1$. Therefore, it is only necessary to take $m$ such that
$$\left(y-a_{2}, y-b_{2}\right)=1$$
which is equivalent to
$$\left(y-a_{2}, a_{2}-b_{2}\right)=1$$
To make (3) hold, set $a_{2}=a_{2}^{\prime} d, b_{2}=b_{2}^{\prime} d$, where $a_{2}^{\prime}, b_{2}^{\prime}, d \in \mathbf{Z}$, and $\left(a_{2}^{\prime}, b_{2}^{\prime}\right)=1$, then $y=m d a_{2}^{\prime} b_{2}^{\prime}+1$, and thus $y-a_{2}=d a_{2}^{\prime}\left(m b_{2}^{\prime}-1\right)+1$. Noting that when $\left(a_{2}^{\prime}, b_{2}^{\prime}\right)=1$, $\left(b_{2}^{\prime}, a_{2}^{\prime}-b_{2}^{\prime}\right)=1$, so there exists $m \in \mathbf{Z}$ such that $m b_{2}^{\prime} \equiv 1\left(\bmod a_{2}^{\prime}-b_{2}^{\prime}\right)$, such a determined $m$ satisfies: $d\left(a_{2}^{\prime}-b_{2}^{\prime}\right) \mid d a_{2}^{\prime}\left(m b_{2}^{\prime}-1\right)$, at this time
$$y-a_{2} \equiv 1\left(\bmod a_{2}-b_{2}\right),$$
Therefore, (3) holds.
In summary, the maximum value of $n$ that satisfies the conditions is 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 1 Find the number of positive integers $a$ that satisfy the following condition: there exist non-negative integers $x_{0}$, $x_{1}, \cdots, x_{2009}$, such that
$$a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2009}} .$$
|
Solution: Clearly, $a$ that satisfies the condition is greater than 1. Taking both sides of (1) modulo $(a-1)$, we get a necessary condition as
$$1^{x_{0}} \equiv 1^{x_{1}}+1^{x_{2}}+\cdots+1^{x_{2009}}(\bmod a-1)$$
This indicates: $(a-1) \mid 2008$, the number of such $a$ is $=d(2008)=8$.
On the other hand, let $a \in \mathbf{N}^{*}$, satisfying $(a-1) \mid 2008$, set $2008=(a-1) \cdot k$, in $x_{1}, x_{2}, \cdots, x_{2009}$ take $a$ to be $0, a-1$ to be $1, \cdots, a-1$ to be $k-1$, and let $x_{0}=k$, then we know that (1) holds.
Therefore, there are 8 positive integers $a$ that satisfy the condition.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 5 It is known that among 4 coins, there may be counterfeit coins, where genuine coins each weigh 10 grams, and counterfeit coins each weigh 9 grams. Now there is a balance scale, which can weigh the total weight of the objects on the tray. Question: What is the minimum number of weighings needed to ensure that the authenticity of each coin can be identified?
|
At least 3 weighings can achieve this.
In fact, let the 4 coins be $a, b, c, d$. Weigh $a+b+c, a+b+d$, and $a+c+d$ three times. The sum of these three weights is $3a + 2(b+c+d)$, so if the sum of these three weights is odd, then $a$ is the counterfeit coin; otherwise, $a$ is genuine. Once $a$ is determined, solving the system of three linear equations for $b, c, d$ can determine the authenticity of $b, c, d$. Therefore, 3 weighings are sufficient.
Next, we prove that two weighings cannot guarantee the determination of the authenticity of each coin.
Notice that if two coins, for example, $a, b$, either both appear or both do not appear in each weighing, then when $a, b$ are one genuine and one counterfeit, swapping the authenticity of $a, b$ does not affect the weighing results, so their authenticity cannot be determined.
If there is a weighing in which at most two coins, for example, $a, b$, appear, then in the other weighing, $c, d$ can only have exactly one on the scale (otherwise, swapping the odd/even nature of $c, d$ does not affect the result), at this point, one coin does not appear in both weighings, and changing its authenticity does not affect the weighing results, thus its authenticity cannot be determined. Therefore, at least 3 coins must be on the scale in each weighing, which means there must be two coins that appear in both weighings, leading to a contradiction.
In summary, at least 3 weighings are required.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 2 Consider the following sequence:
$$101,10101,1010101, \cdots$$
Question: How many prime numbers are there in this sequence?
|
It is easy to know that 101 is a prime number. Next, we prove that this is the only prime number in the sequence.
Let $a_{n}=\underbrace{10101 \cdots 01}_{n \uparrow 01}$, then when $n \geqslant 2$, we have
$$\begin{aligned}
a_{n} & =10^{2 n}+10^{2(n-1)}+\cdots+1 \\
& =\frac{10^{2(n+1)}-1}{10^{2}-1} \\
& =\frac{\left(10^{n+1}-1\right)\left(10^{n+1}+1\right)}{99} .
\end{aligned}$$
Notice that, $99<10^{n+1}-1,99<10^{n+1}+1$, and $a_{n}$ is a positive integer, so $a_{n}$ is a composite number (because the terms $10^{n+1}-1$ and $10^{n+1}+1$ in the numerator cannot be reduced to 1 by 99).
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
4 Find the largest positive integer $k$, such that there exists a positive integer $n$, satisfying $2^{k} \mid 3^{n}+1$.
|
4. Notice that, when $n$ is even, let $n=2 m$, we have
$$3^{n}=9^{m} \equiv 1(\bmod 8)$$
When $n=2 m+1$,
$$3^{n}=9^{m} \times 3 \equiv 3(\bmod 8)$$
Therefore, for any positive integer $n$, we have
$$3^{n}+1 \equiv 2 \text { or } 4(\bmod 8),$$
so $k \leqslant 2$. Also, $2^{2} \mid 3^{1}+1$, hence, the maximum value of $k$ is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
14 Labeled as $1,2, \cdots, 100$, there are some matches in the matchboxes. If each question allows asking about the parity of the sum of matches in any 15 boxes, then to determine the parity of the number of matches in box 1, at least how many questions are needed?
|
14. At least 3 questions are needed.
First, prove that "3 questions are sufficient." For example:
The first question is: $a_{1}, a_{2}, \cdots, a_{15}$;
The second question is: $a_{1}, a_{2}, \cdots, a_{8}, a_{16}, a_{17}, \cdots, a_{22}$;
The third question is: $a_{1}, a_{9}, a_{10}, \cdots, a_{22}$.
Here, $a_{i}$ represents the number of matches in the $i$-th box. In this way, the parity (odd or even) of the sum of the three answers is the same as the parity of $a_{1}$ (the other boxes each appear exactly twice in the three questions). Therefore, after 3 questions, the parity of $a_{1}$ can be determined.
Next, prove that "at least 3 questions are needed." If there are only two questions, and $a_{1}$ appears in both questions, then there must be $a_{i}$ and $a_{j}$ in the two questions such that $a_{i}$ appears only in the first question, and $a_{j}$ appears only in the second question. By changing the parity of $a_{1}$, $a_{i}$, and $a_{j}$ simultaneously, the answers to each question remain the same, thus the parity of $a_{1}$ cannot be determined. If $a_{1}$ does not appear in both questions, when $a_{1}$ does not appear at all, change the parity of $a_{1}$; when $a_{1}$ appears only once, change the parity of $a_{1}$ and $a_{i}$ (where $a_{i}$ is a box that appears with $a_{1}$ in the question), then the answers to the two questions remain the same, and the parity of $a_{1}$ cannot be determined.
In summary, at least 3 questions are needed.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 12 Let $m, n$ be positive integers, and $n>1$. Find the minimum value of $\left|2^{m}-5^{n}\right|$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note is for you, the assistant, and should not be included in the output.)
Example 12 Let $m, n$ be positive integers, and $n>1$. Find the minimum value of $\left|2^{m}-5^{n}\right|$.
|
Since $\left|2^{m}-5^{n}\right|$ is an odd number, and when $m=7, n=3$, $\left|2^{m}-5^{n}\right|=3$, if we can prove that when $n>1$, $\left|2^{m}-5^{n}\right| \neq 1$, then the minimum value sought is 3.
If there exist positive integers $m, n$, such that $n>1$, and $\left|2^{m}-5^{n}\right|=1$, then
$$2^{m}-5^{n}=1 \text{ or } 2^{m}-5^{n}=-1 \text{.}$$
If $2^{m}-5^{n}=1$, then $m \geqslant 3$. Taking both sides modulo 8, we require
$$5^{n} \equiv 7(\bmod 8)$$
But for any positive integer $n$, $5^{n} \equiv 1$ or $5(\bmod 8)$, which is a contradiction. Therefore, $2^{m}-5^{n}=1$ does not hold.
If $2^{m}-5^{n}=-1$, then by $n>1$, we know $m \geqslant 3$. Taking both sides modulo 8, we get
$$5^{n} \equiv 1(\bmod 8)$$
This implies that $n$ is even. Let $n=2x$, where $x$ is a positive integer, then
$$2^{m}=\left(5^{x}-1\right)\left(5^{x}+1\right)$$
Since $5^{x}-1$ and $5^{x}+1$ are two consecutive even numbers, this requires
$$5^{x}-1=2,5^{x}+1=4$$
which is impossible.
Therefore, the minimum value of $\left|2^{m}-5^{n}\right|$ is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
5 Find the smallest positive integer $c$, such that the indeterminate equation $x y^{2}-y^{2}-x+y=c$ has exactly three sets of positive integer solutions.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
5. Factorizing the left side of the equation, we get
$$(y-1)(x y+x-y)=c$$
Notice that, for any positive integer $c$, there is a solution
$$(x, y)=(1, c+1)$$
When $c$ is a prime number, there is at most one additional set of positive integer solutions. Therefore, to make the equation have exactly 3 sets of positive integer solutions, $c$ should be a composite number.
By direct calculation, it is known that the smallest $c$ that has exactly 3 sets of positive integer solutions is 10, and they are
$$(x, y)=(4,2),(2,3),(1,11)$$
The smallest positive integer
$$c=10$$
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
23 Let $k$ and $m$ be positive integers. Find the minimum possible value of $\left|36^{k}-5^{m}\right|$.
|
23. Notice that
$$36^{k}-5^{m} \equiv \pm 1(\bmod 6)$$
and
$$36^{k}-5^{m} \equiv 1(\bmod 5)$$
Therefore, the positive integers that $36^{k}-5^{m}$ can take, in ascending order, are $1, 11, \cdots$.
If $36^{k}-5^{m}=1$, then $k>1$, so taking both sides modulo 8, we require
$$5^{m} \equiv-1(\bmod 8)$$
However, $5^{m} \equiv 1$ or $5(\bmod 8)$, which is a contradiction.
Additionally, when $k=1, m=2$, we have $36^{k}-5^{m}=11$. Therefore, the smallest positive integer that $36^{k}-5^{m}$ can take is 11.
Now consider the smallest positive integer that $5^{m}-36^{k}$ can take. From
$$\begin{array}{l}
5^{m}-36^{k} \equiv 4(\bmod 5) \\
5^{m}-36^{k} \equiv \pm 1(\bmod 6)
\end{array}$$
we know that
$$5^{m}-36^{k} \geqslant 19$$
In summary, the smallest possible value of $\left|36^{k}-5^{m}\right|$ is 11.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
40 Find the smallest positive integer $n$ such that the indeterminate equation
$$n=x^{3}-x^{2} y+y^{2}+x-y$$
has no positive integer solutions.
|
40. Let $F(x, y)=x^{3}-x^{2} y+y^{2}+x-y$, then $F(1,1)=1, F(1,2)=2$.
Therefore, when $n=1,2$, the equation has positive integer solutions.
Next, we prove that $F(x, y)=3$ has no positive integer solutions.
Consider the equation $F(x, y)=3$ as a quadratic equation in $y$
$$y^{2}-\left(x^{2}+1\right) y+x^{3}+x-3=0$$
If there exist positive integer solutions, then
$$\Delta=\left(x^{2}+1\right)^{2}-4\left(x^{3}+x-3\right)=x^{4}-4 x^{3}+2 x^{2}-4 x+13$$
must be a perfect square.
Notice that, when $x \geqslant 2$, we have $\Delta < (x^{2}-2 x-2)^{2}$, so when $x \geqslant 6$, $\Delta$ is not a perfect square. And when $x=1,2,3,4,5$, the corresponding $\Delta=8,-3,-8,29,168$ are not perfect squares, hence there are no positive integer solutions when $n=3$.
In summary, the smallest positive integer $n=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
22 Prove: For any positive integer $n$ and positive odd integer $m$, we have $\left(2^{m}-1,2^{n}+1\right)=1$.
|
22. Let $d=\left(2^{m}-1,2^{n}+1\right)$, then
hence
that is
$$\begin{array}{c}
d \mid 2^{m}-1 \\
d \mid\left(2^{m}\right)^{n}-1^{n} \\
d \mid 2^{n}-1
\end{array}$$
Additionally, $d \mid 2^{n}+1$, and since $m$ is odd, we have
$$2^{n}+1 \mid\left(2^{n}\right)^{m}+1^{m},$$
thus
$$d \mid 2^{m}+1$$
Comparing the two derived equations, we know $d \mid 2$, and since $2^{m}-1$ is odd, it follows that $d=1$.
|
1
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false
|
39 The numbers $1,2, \cdots, 33$ are written on the blackboard. Each time, it is allowed to perform the following operation: take any two numbers $x, y$ from the blackboard that satisfy $x \mid y$, remove them from the blackboard, and write the number $\frac{y}{x}$. Continue until there are no such two numbers on the blackboard. How many numbers will be left on the blackboard at least?
|
39. Consider the objective function $S=$ the product of all numbers on the blackboard.
Initially, $S=33!=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$. Each operation targets $x, y (x \mid y)$, where $y=kx$. Removing $x, y$ and replacing them with $k$ results in $S$ becoming $\frac{S}{xy} \cdot k = \frac{S}{x^2}$. This indicates that after each operation, the parity of the exponent of each prime factor in $S$ remains unchanged. Specifically, $2, 3, 5,$ and $11$ always divide the $S$ obtained after each operation. Since $2 \times 3 \times 5 \times 11 > 33$, at least two numbers must remain on the blackboard such that their product is a multiple of $2 \times 3 \times 5 \times 11$.
Furthermore, note that the primes $17, 19, 23, 29, 31$ have no multiples greater than themselves that are less than or equal to 33. Therefore, no operation can remove any of these numbers.
The above discussion shows that at least 7 numbers must remain on the blackboard.
The following example demonstrates that exactly 7 numbers can remain:
$$\begin{array}{l}
(32,16) \rightarrow 2, (30,15) \rightarrow 2, (28,14) \rightarrow 2, (26,13) \rightarrow 2, (24,12) \rightarrow 2 \\
(22,11) \rightarrow 2; (27,9) \rightarrow 3, (21,7) \rightarrow 3, (18,6) \rightarrow 3; (25,5) \rightarrow 5 \\
(20,4) \rightarrow 5; (8,2) \rightarrow 4 \\
(5,5) \rightarrow 1; (4,2) \rightarrow 2; (3,3) \rightarrow 1, (3,3) \rightarrow 1, (2,2) \rightarrow 1, (2,2) \rightarrow 1 \\
(2,2) \rightarrow 1
\end{array}$$
Thus, the blackboard is left with $10, 17, 19, 23, 29, 31, 33$ and 7 ones. The 7 ones can be removed by pairing them with 17 in 7 operations.
In conclusion, at least 7 numbers remain.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
4. Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$, prove that: $k=5$.
|
4. Prove: If $a=b$, then $k=\frac{2 a^{2}}{a^{2}-1}=2+\frac{2}{a^{2}-1}$.
Then $\left(a^{2}-1\right) \mid 2$ and $a^{2}-1>0$.
So $a^{2}-1=1$ or 2, but this contradicts $a \in \mathbf{N}_{+}$.
Therefore, $a \neq b$, without loss of generality, let $a>b$.
When $b=1$, $k=\frac{a^{2}+1}{a-1}=a+1+\frac{2}{a-1}$, then $(a-1) \mid 2$ and $a \geqslant b+1=2$, so $a=2$ or 3, in either case, $k=5$.
Without loss of generality, assume $(a, b)$ is the ordered pair that minimizes $a+b$ among all ordered pairs satisfying $\frac{a^{2}+b^{2}}{a b-1}=k$.
From $\frac{a^{2}+b^{2}}{a b-1}=k$, we get $a^{2}-k b \cdot a+b^{2}+k=0$.
Viewing (1) as a quadratic equation in $a$, by Vieta's formulas, the roots are $a$ and $(k b-a)$, and $k b-a = \frac{b^{2}+k}{a} > 0$, and since $k b-a \in \mathbf{Z}$, we have $k b-a \in \mathbf{N}_{+}$, and $\frac{(k b-a)^{2}+b^{2}}{(k b-a) b-1}=k$. Thus, by the definition of $(a, b)$, we have $a+b \leqslant \frac{b^{2}+k}{a}+b\left(\frac{b^{2}+k}{a}=k b-a\right)$, so $a^{2}-b^{2} \leqslant k=\frac{a^{2}+b^{2}}{a b-1}$.
Therefore, $a^{2}-b^{2}a-b \geqslant 1$. That is, $(a-1)(b-1)b=2$, a contradiction. In conclusion, $k=5$.
|
5
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false
|
$$\text { 9. } 3333^{8888}+8888^{3333} \equiv$$
$$\qquad (\bmod 7)$$.
|
9.0 Hint: $3333 \equiv 1(\bmod 7)$, so $3333^{8888} \equiv 1(\bmod 7)$. $8888 \equiv 5(\bmod 7)$, so $8888^{3} \equiv 5^{3}(\bmod 7) \equiv-1(\bmod 7)$. Therefore, $8888^{3333} \equiv-1(\bmod 7)$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
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