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11. Given that $a$ and $b$ are real numbers, satisfying:
$$
\sqrt[3]{a}-\sqrt[3]{b}=12, \quad a b=\left(\frac{a+b+8}{6}\right)^{3} \text {. }
$$
Then $a-b=$ $\qquad$ (Proposed by Thailand) | 11. 468 .
Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}$. Then $x-y=12, 6xy=x^{3}+y^{3}+8$.
Thus, $x^{3}+y^{3}+2^{3}-3 \times 2xy=0$
$$
\Rightarrow(x+y+2)\left(x^{2}+y^{2}+2^{2}-xy-2x-2y\right)=0 \text {. }
$$
Therefore, $x+y+2=0$, or
$$
x^{2}-xy+y^{2}-2x-2y+4=0 \text {. }
$$
Equation (1) can be rewritten as
$$
\frac{1}{2}(x-y)^{2}+\frac{1}{2}(x-2)^{2}+\frac{1}{2}(y-2)^{2}=0 \text {. }
$$
Thus, $x=y=2$, but this does not satisfy $x-y=12$.
Hence, $x+y+2=0$.
Therefore, $x=5, y=-7$.
Thus, $a=125, b=-343, a-b=468$. | 468 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the equation $\left|x^{2}-2 a x+b\right|=8$ has exactly three real roots, and they are the side lengths of a right triangle. Find the value of $a+b$.
(Bulgaria) | II. 1. Note that, for the equation
$$
x^{2}-2 a x+b-8=0
$$
the discriminant $\Delta_{1}=4\left(a^{2}-b+8\right)$; for the equation
$$
x^{2}-2 a x+b+8=0
$$
the discriminant $\Delta_{2}=4\left(a^{2}-b-8\right)$.
According to the problem, the original equation has exactly three real roots, so one of the discriminants is 0, and the other is greater than 0.
Since $\Delta_{1}>\Delta_{2}$, then $a^{2}-b-8=0$.
Thus, the repeated root of equation (2) is $a$, and the two roots of equation (1) are $a+4$ and $a-4$.
Therefore, $a^{2}+(a-4)^{2}=(a+4)^{2}$
$$
\begin{array}{l}
\Rightarrow a=16 \Rightarrow b=a^{2}-8=248 \\
\Rightarrow a+b=264 .
\end{array}
$$ | 264 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Color the six vertices of a regular pentagonal pyramid using at most six different colors, such that the two endpoints of any edge are colored differently. If one coloring method can be obtained from another by rotation, they are considered the same method. How many different coloring methods are there?
(Romania) | 2. Since the problem does not specify the number of colors to choose from, we will calculate the coloring methods based on the actual number of colors used.
If fewer than four colors are used, the bottom face can have at most two colors. Since 5 is an odd number, the colors of the bottom face vertices cannot be alternated. In this case, there are no valid coloring methods.
If six colors are actually used, without considering rotational equivalence, there are $6!=720$ methods. After removing rotations, there are 720 $\div 5=144$ methods.
If five colors are actually used, the coloring of the bottom face vertices must be in the form $A B A C D$. By rotation, the color $B$ can be moved to a fixed vertex, so the number of coloring methods is the permutation of 5 colors, which is $5!=120$.
If four colors are actually used, the coloring of the bottom face vertices must be in the form $A B A B C$. By rotation, the color $C$ can be moved to a fixed vertex, so the number of coloring methods is the permutation of 4 colors, which is $4!=24$.
In summary, the total number of coloring methods is
$144+120+24=288$. | 288 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. For what positive integer $k$ does $\frac{20^{k}+18^{k}}{k!}$ achieve its maximum value?
(Hong Kong, China, Contributed) | 5.19.
Let $A_{k}=\frac{20^{k}+18^{k}}{k!}$. Then
$A_{k+1}=\frac{20^{k+1}+18^{k+1}}{(k+1)!}$,
$\frac{A_{k+1}}{A_{k}}=\frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}$.
Notice that,
$$
\begin{array}{l}
\frac{A_{k+1}}{A_{k}}>1 \Leftrightarrow \frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}>1 \\
\Leftrightarrow 20^{k+1}+18^{k+1}>(k+1)\left(20^{k}+18^{k}\right) \\
\Leftrightarrow 20^{k}(19-k)+18^{k}(17-k)>0 .
\end{array}
$$
Upon inspection, when $k=1,2, \cdots, 18$, equation (1) holds; when $k=19,20, \cdots$, equation (1) does not hold.
Therefore, when $k=19$, $A_{k}$ reaches its maximum value. | 19 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In the square in Figure 1, 10000 points are randomly thrown. Then the estimated number of points that fall into the shaded area (the equation of curve $C$ is $x^{2}-y=0$) is ( ).
(A) 5000
(B) 6667
(C) 7500
(D) 7854 | 5. B.
According to the problem, the area of the blank region is
$$
\int_{0}^{1} x^{2} \mathrm{~d} x=\left.\frac{1}{3} x^{3}\right|_{0} ^{1}=\frac{1}{3} \text {. }
$$
Therefore, the area of the shaded part is $\frac{2}{3}$, and the ratio of the area of the shaded part to the area of the blank part is $2: 1$.
Thus, the number of points falling into the shaded part is
$$
10000 \times \frac{2}{3} \approx 6667 .
$$ | 6667 | Geometry | MCQ | Yes | Yes | cn_contest | false |
16. Nine consecutive positive integers are arranged in an increasing sequence $a_{1}, a_{2}, \cdots, a_{9}$. If $a_{1}+a_{3}+a_{5}+a_{7}+a_{9}$ is a perfect square, and $a_{2}+a_{4}+a_{6}+a_{8}$ is a perfect cube, then the minimum value of $a_{1}+a_{2}+\cdots+a_{9}$ is $\qquad$. | 16. 18000 .
Let these nine numbers be $a-4, a-3, \cdots, a+4$. Then $5 a=m^{2}, 4 a=n^{3}, S=9 a$. Thus, $a=\frac{m^{2}}{5}=\frac{n^{3}}{4} \Rightarrow 4 m^{2}=5 n^{3}$.
Clearly, $m$ and $n$ are both multiples of 10.
Let $m=10 m_{1}, n=10 n_{1}$, then $4 m_{1}^{2}=50 n_{1}^{3}$.
Take $m_{1}=10, n_{1}=2$. At this point,
$$
\begin{array}{l}
n=20, m=100 \\
\Rightarrow a=2000, S=9 a=18000 .
\end{array}
$$ | 18000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Mr. Wanda often forgets the numbers he should remember, such as his friends' phone numbers, the password of the safe, etc. For this reason, the manufacturer specially designed a password lock for his office safe, with keys labeled $0 \sim 9$. It is known that the password of the safe is a three-digit number, but as long as two of the digits in the entered three-digit number are correct, the safe will open. Mr. Wanda has forgotten the password he set again. How many times does he need to try at least to ensure that the safe will definitely open? | Four, 50 times.
First, introduce several notations. Let
$E=\{1,3,5,7,9\}, F=\{0,2,4,6,8\}$
$\Phi=E \cup F=\{0,1, \cdots, 9\}$,
$\Omega=\{x y z \mid x, y, z \in \Phi\}$ (the set of all three-digit codes),
$\Omega_{11}=\{x a b \in \Omega \mid x \in \Phi, a, b \in E\}$,
$\Omega_{12}=\{a y b \in \Omega \mid y \in \Phi, a, b \in E\}$,
$\Omega_{13}=\{a b z \in \Omega \mid z \in \Phi, a, b \in E\}$,
$\Omega_{1}=\Omega_{11} \cup \Omega_{12} \cup \Omega_{13}$ (the set of all three-digit codes with at least two odd numbers),
$\Omega_{2}=\{c d w, c w d, w c d \in \Omega \mid w \in \Phi, c, d \in F\}$ (the set of all three-digit codes with at least two even numbers).
Then $\Omega=\Omega_{1} \cup \Omega_{2}$.
(1) Existence.
Prove that a subset $X$ of $\Omega$ with 50 elements can be constructed such that any three-digit code in $\Omega$ can be opened by a three-digit code in $X$.
Select 25 numbers $x_{a b}, y_{a b}, z_{a b} \in \Phi$, $a, b \in E$, and construct three subsets of $\Omega$ each with 25 elements:
$X_{11}=\left\{x_{a b} a b \in \Omega \mid a, b \in E\right\}$,
$X_{12}=\left\{a y_{a b} b \in \Omega \mid a, b \in E\right\}$,
$X_{13}=\left\{a b z_{a b} \in \Omega \mid a, b \in E\right\}$.
Then for $i=1,2,3$, any three-digit code in the set $\Omega_{1 i}$ can be opened by a three-digit code in the set $X_{1 i}$. Thus, any three-digit code in the set $\Omega_{1}$ can be opened by a three-digit code in the set
$X_{1}=X_{11} \cup X_{12} \cup X_{13}$.
Next, take special $X_{11}, X_{12}, X_{13}$ such that $X_{11} = X_{12} = X_{13}$, thus $X_{1} = X_{11} = X_{12} = X_{13}$ is a set containing 25 elements, for example:
$X_{1}=X_{11}=X_{12}=X_{13}$
$=\left\{\begin{array}{lllll}111 & 133 & 155 & 177 & 199 \\ 319 & 331 & 353 & 375 & 397 \\ 517 & 539 & 551 & 573 & 595 \\ 715 & 737 & 759 & 771 & 793 \\ 913 & 935 & 957 & 979 & 991\end{array}\right\}$
And any three-digit code in $\Omega_{1}$ can be opened by one of these 25 three-digit codes in $X_{1}$.
Similarly, a set $X_{2}$ of 25 three-digit codes composed entirely of even numbers can be constructed such that any three-digit code in $\Omega_{2}$ can be opened by a three-digit code in $X_{2}$.
Let $X=X_{1} \cup X_{2}$. By the construction of $X_{1}$ and $X_{2}$, we know $|X| = 50$, and any three-digit code in $\Omega = \Omega_{1} \cup \Omega_{2}$ can be opened by a three-digit code in $X$.
(2) 50 is the minimum.
We will prove by contradiction that if a subset $X$ of $\Omega$ satisfies that any three-digit code in $\Omega$ can be opened by a three-digit code in $X$, then $|X| \geqslant 50$.
Assume $w=|X| \leqslant 49$, and let
$X=\left\{\left(a_{1} b_{1} c_{1}\right),\left(a_{2} b_{2} c_{2}\right), \cdots,\left(a_{w} b_{w} c_{w}\right)\right\}$.
Consider the three sequences
$a_{1}, a_{2}, \cdots, a_{w} ; b_{1}, b_{2}, \cdots, b_{w} ; c_{1}, c_{2}, \cdots, c_{w}$. Then in these three sequences, at least two sequences must contain all the digits $0,1, \cdots, 9$. Otherwise, suppose the digit $a (a \in \Phi)$ does not appear in $a_{1}, a_{2}, \cdots, a_{w}$, and the digit $b (b \in \Phi)$ does not appear in $b_{1}, b_{2}, \cdots, b_{w}$. Then, the three-digit code $a b 0 \in \Omega$ cannot be opened by any three-digit code in $X$, which is a contradiction.
Without loss of generality, assume the set of different digits appearing in $a_{1}, a_{2}, \cdots, a_{w}$ is $A$, and let $x \in A$ appear the fewest times, which is $m_{1}$ times.
Similarly, the set of different digits appearing in $b_{1}, b_{2}, \cdots, b_{w}$ is $B$, and let $y \in B$ appear the fewest times, which is $m_{2}$ times; the set of different digits appearing in $c_{1}, c_{2}, \cdots, c_{w}$ is $C$, and let $z \in C$ appear the fewest times, which is $m_{3}$ times. Since at least two sequences contain all the digits $0,1, \cdots, 9$, then
$m \leqslant\left[\frac{49}{10}\right]=4$.
Consider the subset $Y$ of $X$ consisting of elements with the last digit $z$, say
$Y=\left\{\left(a_{1} b_{1} z\right),\left(a_{2} b_{2} z\right), \cdots,\left(a_{m} b_{m} z\right)\right\}$
Let $U=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$,
$V=\left\{b_{1}, b_{2}, \cdots, b_{m}\right\}$
and $U$ contains $s$ different digits, $V$ contains $t$ different digits, then
$1 \leqslant s \leqslant m, 1 \leqslant t \leqslant m$.
Note: $0,1, \cdots, 9$ have $10-s$ digits not appearing in $U$, and $10-t$ digits not appearing in $V$.
Construct the set
$Z=\{a b z \in \Omega \mid a \notin U, b \notin V\}$.
Therefore, $|Z|=(10-s)(10-t)$. For each $a b z \in Z$, there exists at least one three-digit code $\alpha \beta \gamma \in X$ that can open $a b z$.
By the construction of $Z$, we know $\alpha=a, \beta=b, \gamma \neq z$. Let
$Z^{\prime}=\{a b c \in X \mid a \notin U, b \notin V, c \neq z\}$.
Then $\left|Z^{\prime}\right| \geqslant(10-s)(10-t)$.
Construct the subset of $X$
$W=\{(a b c) \mid(a b c) \in X, a \in U\}$
Since $U$ contains $s$ different digits, and each digit appears at least $m$ times in $X$, then $|W| \geqslant m s$.
By the construction of sets $Z^{\prime}$ and $W$, it is clear that $W \cap Z^{\prime}=\varnothing$.
Then $49 \geqslant w=|X| \geqslant m s+(10-s)(10-t)$
Thus $49 \geqslant 100-10(s+t)+s t+m s$
$=2(5-s)(5-t)+50+s(m-t) \geqslant 50$,
which is a contradiction. | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a regular 2019-gon, then, the maximum number of diagonals such that any two of them are either perpendicular or do not intersect except at endpoints. | 7.2016 .
On one hand, no two diagonals of the regular 2019-gon $\Gamma$ are perpendicular.
Assume there exist two diagonals of the regular 2019-gon $\Gamma$ that are perpendicular, say $AB \perp CD$, and $A, B, C, D$ are four vertices of $\Gamma$. Let $E$ be a point on the perpendicular bisector of $AB$, and let $E'$ be a point on the circumcircle of $\Gamma$ such that $EE'$ is its diameter.
Since $EC = E'D$, and $C, D, E$ are vertices of the regular 2019-gon $\Gamma$, it follows that $E'$ is also a vertex of $\Gamma$.
However, there cannot be two vertices of $\Gamma$ that are symmetric with respect to the center, so the assumption is false.
On the other hand, these non-intersecting diagonals can divide the regular 2019-gon $\Gamma$ into at most 2017 triangles, thus, at most 2016 diagonals can satisfy the condition. Furthermore, by selecting all the diagonals from a single vertex of $\Gamma$, the condition is met.
In summary, at most 2016 diagonals can satisfy the condition. | 2016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ with the general terms $a_{n}=2^{n}, b_{n}=5 n-2$. Then the sum of all elements in the set
$$
\left\{a_{1}, a_{2}, \cdots, a_{2019}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{2019}\right\}
$$
is $\qquad$ | 8.2184 .
Let the elements of $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ be arranged in ascending order to form the sequence $\left\{c_{n}\right\}$.
Then the problem is reduced to finding the number of solutions to the equation
$$
2^{n}=5 m-2\left(n, m \in \mathbf{Z}_{+}, n, m \leqslant 2019\right)
$$
Clearly, $m$ must be even.
Thus, we need to find the number of solutions to the equation
$$
2^{n}=10 p-2\left(n, p \in \mathbf{Z}_{+}, n \leqslant 2019, p \leqslant 1009\right)
$$
which is equivalent to finding the number of solutions to the equation
$$
2^{n-1}=5 p-1\left(n, p \in \mathbf{Z}_{+}, n \leqslant 2019, p \leqslant 1009\right)
$$
Clearly, $n=3, p=1$ satisfies the requirement.
If $n$ is even, then $4^{\frac{n}{2}}=5 m-2$, but
$$
(5-1)^{\frac{n}{2}} \equiv(-1)^{\frac{n}{2}}(\bmod 5),
$$
which is a contradiction.
Therefore, $n$ must be odd, so we need to find the number of solutions to the equation
$$
4^{q}=5 p-1\left(q, p \in \mathbf{Z}_{+}, q \leqslant 1009, p \leqslant 1009\right)
$$
By $4^{q}=(5-1)^{q} \equiv(-1)^{q}(\bmod 5)$, we know that $q$ must be odd.
Thus, $\left\{c_{n}\right\}$ is a geometric sequence with the first term 8 and common ratio 16.
$$
\text { Therefore, }\left\{a_{1}, a_{2}, \cdots, a_{2019}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{2019}\right\}
$$
has 3 elements, which are $2^{3}, 2^{7}, 2^{11}$.
Thus, the sum of the required elements is 2184. | 2184 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Periodical cicadas are insects with very long larval periods and brief adult lives. For each species of periodical cicada with larval period of 17 years, there is a similar species with a larval period of 13 years. If both the 17 -year and 13 -year species emerged in a particular location in 1900, when will they next both emerge in that location? | 5. 2121 | 2121 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
14. Find the least common multiple of each of the following pairs of integers
a) 8,12
d) 111,303
b) 14,15
e) 256,5040
c) 28,35
f) 343,999 . | 14. a) 24 b) 210 c) 140 d) 11211 e) 80640 f) 342657 | 140 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10. How many ways can change be made for one dollar using
a) dimes and quarters
b) nickels. dimes, and quarters
c) pennies, nickels, dimes, and quarters? | 10. a) 3
b) 29
c) 242 | 242 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
1. Find the values of the following sums
a) $\sum_{j=1}^{10} 2$
c) $\sum_{j=1}^{10} j^{2}$
b) $\sum_{j=1}^{10} j$
d) $\sum_{j=1}^{10} 2^{j}$. | 1. a) 20 b) 55 c) 385 d) 2046 | 2046 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
2. Find the values of the following products
a) $\prod_{j=1}^{5} 2$
c) $\prod_{j=1}^{5} j^{2}$
b) $\prod_{j=1}^{5} j$
d) $\prod_{j=1}^{5} 2^{j}$. | 2. a) 32 b) 120 c) 14400 d) 32768 | 14400 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
3. What is the ciphertext that is produced when the RSA cipher with key $(e, n)=(3,2669)$ is used to encipher the message BEST WISHES? | 3. 12151224147100230116 | 12151224147100230116 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Find $n$ ! for $n$ equal to each of the first ten positive integers. | 3. $1,2,6,24,120,720,5040,40320,362880,3628800$ | 3628800 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
12. Let $n=2^{t} \cdot p_{1}^{t} p_{2}^{t} \cdots p_{m}^{t}$ be the prime-power factorization of $n$. Let $a$ be an integer relatively prime to $n$. Let $r_{1}, r_{2}, \ldots, r_{m}$ be primitive roots of $p_{1}^{t_{1}}, p_{2}^{t_{2}}, \ldots, p_{m}^{t}$, respectively, and let $\gamma_{1}=\operatorname{ind}_{r_{1}} a\left(\bmod p_{1}^{t_{1}}\right), \quad \gamma_{2}=\operatorname{ind}_{r_{2}} a\left(\bmod p_{2}^{t_{2}}\right)$, $\ldots, \gamma_{m}=\operatorname{ind}_{r_{-}} a\left(\bmod p_{m}^{t}\right)$. If $t_{0} \leqslant 2$, let $r_{0}$ be a primitive root of $2^{t}$, and let $\gamma_{0}=\operatorname{ind}_{r_{0}} a\left(\bmod 2^{t}\right)$. If $t_{0} \geqslant 3$, let $(\alpha, \beta)$ be the index system of $a$ modulo $2^{k}$, so that $a \equiv(-1)^{\alpha} 5^{\beta}\left(\bmod 2^{k}\right)$. Define the index system of $a$ modulo $n$ to be $\left(\gamma_{0}, \gamma_{1}, \gamma_{2}, \ldots, \gamma_{m}\right)$ if $t_{0} \leqslant 2$ and $\left(\alpha, \beta, \gamma_{1}, \gamma_{2}, \ldots, \gamma_{m}\right)$ if $t_{0} \geqslant 3$.
a) Show that if $n$ is a positive integer, then every integer has a unique index system modulo $n$.
b) Find the index systems of 17 and $41(\bmod 120)$ (in your computations, use 2 as a primitive root of the prime factor 5 of 120 ).
c) Develop rules for the index systems modulo $n$ of products and powers analogous to those for indices.
d) Use an index system modulo 60 to find the solutions of $11 x^{7} \equiv 43(\bmod 60)$ | 12. b) $(0,0,1,1),(0,0,1,4)$
d) $x \equiv 17(\bmod 60)$ | 17 | Number Theory | proof | Yes | Yes | number_theory | false |
1. Find $\lambda(n)$, the minimal universal exponent of $n$, for the following values of $n$
a) 100
e) $2^{4} \cdot 3^{3} \cdot 5^{2} \cdot 7$
b) 144
f) $2^{5} \cdot 3^{2} \cdot 5^{2} \cdot 7^{3} \cdot 11^{2} \cdot 13 \cdot 17 \cdot 19$
c) 222
g) 10 !
d) 884
h) 20 !. | a) 20
b) 12
c) 36
d) 48
e) 180
f) 388080
g) 8640
h) 125411328000 | 48 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1. Find the sequence of two-digit pseudo-random numbers generated using the middle-square method, taking 69 as the seed. | 1. $69,76,77,92,46,11,12,14,19,36,29,84,5,25,62,84,5,25,62, \ldots$ | 62 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the number of prime numbers not exceeding 100. | We take $N=100$, the prime numbers not exceeding $\sqrt{100}=10$ are $2,3,5,7$, so by formula (20) we get
$$\begin{aligned}
\pi(100)= & 4-1+100-\left\{\left[\frac{100}{2}\right]+\left[\frac{100}{3}\right]+\left[\frac{100}{5}\right]+\left[\frac{100}{7}\right]\right\} \\
& +\left\{\left[\frac{100}{2 \cdot 3}\right]+\left[\frac{100}{2 \cdot 5}\right]+\left[\frac{100}{2 \cdot 7}\right]+\left[\frac{100}{3 \cdot 5}\right]+\left[\frac{100}{3 \cdot 7}\right]+\left[\frac{100}{5 \cdot 7}\right]\right\} \\
& -\left\{\left[\frac{100}{2 \cdot 3 \cdot 5}\right]+\left[\frac{100}{2 \cdot 3 \cdot 7}\right]+\left[\frac{100}{2 \cdot 5 \cdot 7}\right]+\left[\frac{100}{3 \cdot 5 \cdot 7}\right]\right\} \\
& +\left[\frac{100}{2 \cdot 3 \cdot 5 \cdot 7}\right] \\
= & 4-1+100-117+45-6+0=25
\end{aligned}$$
This is consistent with the result in Section 2 of Chapter 1. | 25 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 (i) Find the number of integers between 1 and 500 that are not divisible by any of 5, 6, 8;
(ii) Find the number of integers between 1 and 500 that are divisible by 5, or 6, or 8. | Consider the sequence $A$ composed of integers from 1 to 500. Property $P_{1}$: divisible by 5, property $P_{2}$: divisible by 6, property $P_{3}$: divisible by 8. Thus, (i) is to find the number of $B^{(0)}$, and (ii) is to find the number of $A^{(0)}-B^{(0)}$.
The number of $A(1)$ is: 100; the number of $A(2)$ is: 83; the number of $A(3)$ is: 62. $A(1,2)$ consists of integers divisible by 30, with a count of 16; $A(1,3)$ consists of integers divisible by 40, with a count of 12; $A(2,3)$ consists of integers divisible by 24, with a count of 20. $A(1,2,3)$ consists of integers divisible by 120, with a count of 4. Therefore,
$$\begin{array}{c}
B^{(0)}=500-(100+83+62)+(16+20+12)-(4)=299 \\
A^{(0)}-B^{(0)}=500-299=201
\end{array}$$ | 299 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
20. Find how many zeros are at the end of the decimal expression of 120!.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note above is not part of the translation but is provided to clarify the instruction. The actual translation is above it. | 20. Find the highest power of 10 that divides 120!, which is also the highest power of 5 that divides 120!. Therefore, there are 28 zeros. | 28 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
16. There are five sailors and a monkey on a small island. During the day, they collected some coconuts as food. At night, one of the sailors woke up and decided to take his share of the coconuts. He divided the coconuts into five equal parts, with one left over, so he gave the extra one to the monkey and hid his share, then went back to sleep. After a while, the second sailor woke up and did the same thing. When he divided the remaining coconuts into five equal parts, there was also one left over, which he gave to the monkey, and then he hid his share and went back to sleep. The other three sailors also did the same thing in turn. The next morning, they woke up and pretended nothing had happened, dividing the remaining coconuts into five equal parts, with none left over. How many coconuts were there originally, and how many coconuts did each of them get in total? | 16. This pile of coconuts has at least 3121. The number of coconuts each of the five people received in turn is $828, 703, 603$, 523, 459, and the monkey ate 5. | 3121 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find the last two digits of $3^{406}$ when written as a decimal number. | Solve: This is to find the smallest non-negative remainder $b$ when $3^{106}$ is divided by 100, i.e., $b$ satisfies
$$3^{406} \equiv b(\bmod 100), \quad 0 \leqslant b \leqslant 99$$
Notice that $100=4 \cdot 25, (4,25)=1$. Clearly, $3^{2} \equiv 1(\bmod 4), 3^{4} \equiv 1(\bmod 5)$. Note that 4 is the smallest power, by Example 5 in Chapter 1, §4, we know that for $3^{d} \equiv 1(\bmod 25)$ to hold, $4 \mid d$ must be true. Therefore, calculate:
$$\begin{array}{c}
3^{4} \equiv 81 \equiv 6(\bmod 25), \quad 3^{8} \equiv 36 \equiv 11(\bmod 25) \\
3^{12} \equiv 66 \equiv-9(\bmod 25), \quad 3^{16} \equiv-54 \equiv-4(\bmod 25) \\
3^{20} \equiv-24 \equiv 1(\bmod 25)
\end{array}$$
From this and $3^{20} \equiv 1(\bmod 4)$, by Property IX, we derive $3^{20} \equiv 1(\bmod 100), 3^{400} \equiv 1(\bmod 100)$. Therefore, $3^{406} \equiv 3^{400} \cdot 3^{6} \equiv 3^{6} \equiv 29(\bmod 100)$. So, the unit digit is 9, and the tens digit is 2.
If we do not use the property $4 \mid d$, we have to calculate the remainder $b_{j}$ of $3^{j}$ modulo 25 (to facilitate the calculation of $b_{j}$, the absolute smallest remainder should be taken), as shown in the table below:
\begin{tabular}{|c|cccccccccc|}
\hline$j$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$b_{j}$ & 3 & 9 & 2 & 6 & -7 & 4 & 12 & 11 & 8 & -1 \\
\hline
\end{tabular}
From the $3^{10} \equiv-1(\bmod 25)$ obtained here, we conclude $3^{20} \equiv 1(\bmod 25)$. | 29 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Let $m>n \geqslant 1$. Find the smallest $m+n$ such that
$$\text {1000| } 1978^{m}-1978^{n} \text {. }$$ | Solve: Using the congruence symbol, the problem is to find the smallest $m+n$ such that
$$1978^{m}-1978^{n} \equiv 0(\bmod 1000)$$
is satisfied. First, let's discuss what conditions $m, n$ must meet for the above equation to hold. Let $k=m-n$. Equation (9) becomes
$$2^{n} \cdot 989^{n}\left(1978^{k}-1\right) \equiv 0\left(\bmod 2^{3} \cdot 5^{3}\right)$$
By Property VII and Property IX, it is equivalent to
$$\left\{\begin{array}{l}
2^{n} \equiv 0\left(\bmod 2^{3}\right) \\
1978^{k}-1 \equiv 0\left(\bmod 5^{3}\right)
\end{array}\right.$$
From (10), we know $n \geqslant 3$. Next, we find the $k$ that satisfies (11). First, find the smallest $l$ such that
$$1978^{t}-1 \equiv 0(\bmod 5)$$
holds, denoted as $d_{1}$. Since
$$1978 \equiv 3(\bmod 5)$$
we have $d_{1}=4$. Next, find the smallest $h$ such that
$$1978^{h}-1 \equiv 0\left(\bmod 5^{2}\right)$$
holds, denoted as $d_{2}$. From Chapter 1, §4, Example 5, we know $4 \mid d_{2}$. Noting that
$$1978 \equiv 3\left(\bmod 5^{2}\right)$$
From the calculation in Example 2, we have $d_{2}=20$. Finally, find the smallest $k$ such that
$$1978^{k}-1 \equiv 0\left(\bmod 5^{3}\right)$$
holds, denoted as $d_{3}$. From Chapter 1, §4, Example 5, we know $20 \mid d_{3}$. Noting that
$$\begin{array}{l}
1978 \equiv-22\left(\bmod 5^{3}\right) \\
\begin{aligned}
(-22)^{20} & \equiv(25-3)^{20} \equiv 3^{20} \equiv(243)^{4} \\
& \equiv 7^{4} \equiv(50-1)^{2} \equiv 26\left(\bmod 5^{3}\right)
\end{aligned}
\end{array}$$
Through calculation, we get
$$\begin{array}{c}
1978^{20} \equiv 26\left(\bmod 5^{3}\right), \quad 1978^{40} \equiv(25+1)^{2} \equiv 51\left(\bmod 5^{3}\right) \\
1978^{60} \equiv(25+1)(50+1) \equiv 76\left(\bmod 5^{3}\right) \\
1978^{80} \equiv(50+1)^{2} \equiv 101\left(\bmod 5^{3}\right) \\
1978^{100} \equiv(100+1)(25+1) \equiv 1\left(\bmod 5^{3}\right)
\end{array}$$
Therefore, $d_{3}=100$. So, from Chapter 1, §4, Example 5, we know $100 \mid k$, and the smallest $k=100$. Thus, the necessary and sufficient conditions for (10) and (11), i.e., (9) to hold are
$$n \geqslant 3, \quad 100 \mid m-n$$
Therefore, the smallest $m+n=(m-n)+2 n=106$. If the congruence concept and its properties are not used to solve this problem, it would be very complicated. | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
7. Use the previous problem to solve (i) $3 x \equiv 1(\bmod 125)$; (ii) $5 x \equiv 1(\bmod 243)$. | 7. (i) $3 x \equiv 1\left(\bmod 5^{3}\right), 3 \cdot 2 \equiv 1(\bmod 5), 1-(1-3 \cdot 2)^{3}=126$, so $x=42$ is a solution to the original congruence equation. (ii) $5 x \equiv 1\left(\bmod 3^{5}\right) .5 \cdot(-1) \equiv 1(\bmod 3) .1-(1-5 \cdot(-1))^{5}=$ $-7775 . x=-1555$ is a solution. | -1555 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Solve the system of congruences
$$\left\{\begin{array}{l}
x \equiv 1(\bmod 3), \\
x \equiv-1(\bmod 5), \\
x \equiv 2(\bmod 7), \\
x \equiv-2(\bmod 11)
\end{array}\right.$$ | Let $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, satisfying the conditions of Theorem 1. In this case, $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$. We will find $M_{j}^{-1}$. Since $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$, we have
$$1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$$
Therefore, we can take $M_{1}^{-1}=1$. From $M_{2} \equiv(-2) \cdot(2) \cdot 1 \equiv 1(\bmod 5)$, we have
$$1 \equiv M_{2} M_{2}^{-1} \equiv M_{2}^{-1}(\bmod 5)$$
Therefore, we can take $M_{2}^{-1}=1$. From $M_{3} \equiv 3 \cdot 5 \cdot 4 \equiv 4(\bmod 7)$, we have
$$1 \equiv M_{3} M_{3}^{-1} \equiv 4 M_{3}^{-1}(\bmod 7)$$
Therefore, we can take $M_{3}^{-1}=2$. From $M_{4} \equiv 3 \cdot 5 \cdot 7 \equiv 4 \cdot 7 \equiv 6(\bmod 11)$, we have
$$1 \equiv M_{4} M_{4}^{-1} \equiv 6 M_{4}^{-1}(\bmod 11)$$
Therefore, we can take $M_{4}^{-1}=2$. Thus, by Theorem 1, the solution to the system of congruences is
$$\begin{aligned}
x \equiv & (5 \cdot 7 \cdot 11) \cdot 1 \cdot 1+(3 \cdot 7 \cdot 11) \cdot 1 \cdot(-1)+(3 \cdot 5 \cdot 11) \cdot 2 \cdot 2 \\
& +(3 \cdot 5 \cdot 7) \cdot 2 \cdot(-2)(\bmod 3 \cdot 5 \cdot 7 \cdot 11),
\end{aligned}$$
i.e., $\square$
$$x \equiv 385-231+660-420 \equiv 394(\bmod 1155)$$ | 394 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
15. Find all positive integer solutions \(a, b, c\) that satisfy \((a, b, c)=10,[a, b, c]=100\). | 15. $(a / 10, b / 10, c / 10)=1,[a / 10, b / 10, c / 10]=10 . a / 10, b / 10, c / 10$ can only take the values $1,2,5,10$ and satisfy the above two conditions. There are three possible scenarios: (i) $a / 10=b / 10=c / 10$, which is impossible; (ii) Two of the three are equal, and the other is uniquely determined by the conditions, they are $\{10,10,1\}$, $\{5,5,2\},\{2,2,5\},\{1,1,10\}$, and permutations, giving a total of $3 \cdot 4=12$ solutions; (iii) All three are distinct, any three distinct values can be chosen, they are $\{10,5,2\},\{10,5,1\},\{10,2,1\},\{5,2,1\}$ and permutations, giving a total of $6 \cdot 4=24$ solutions. Multiplying each number by 10 gives the solutions to the original problem, with a total of 36 solutions. | 36 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
7. Let $1975 \leqslant n \leqslant 1985$, ask which of these $n$ have primitive roots. | 7. Only $n=1979$ is a prime number. | 1979 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find the greatest common divisor of 6731 and 2809 | Solve:
\begin{aligned}
6731 & =2809 \times 2+1113 \\
2809 & =1113 \times 2+583 \\
1113 & =583 \times 1+530 \\
583 & =530+53 \\
530 & =53 \times 10+0
\end{aligned}
So $(6731,2809)=53$. For convenience in writing, the series of calculations above can be abbreviated as follows:
2 \begin{tabular}{|r|r|r}
6731 & 2809 & 2 \\
5618 & 2226 \\
\hline 1113 & 583 & 1 \\
583 & 530 \\
\hline 530 & 53 & \\
530 & $(0)$ & \\
\hline 0 & 53 &.
\end{tabular}
Let $n \geqslant 3$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}$ be positive integers. Here we introduce a method to find the greatest common divisor of these $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$: We first find the greatest common divisor of $a_{1}$ and $a_{2}$, and if the greatest common divisor of $a_{1}$ and $a_{2}$ is $b_{1}$. Then we find the greatest common divisor of $b_{1}$ and $a_{3}$, and if the greatest common divisor of $b_{1}$ and $a_{3}$ is $b_{2}$, then $b_{2}$ is the greatest common divisor of $a_{1}, a_{2}$, and $a_{3}$. When $n \geqslant 4$, we find the greatest common divisor of $b_{2}$ and $a_{4}$, and if the greatest common divisor of $b_{2}$ and $a_{4}$ is $b_{3}$, then $b_{3}$ is the greatest common divisor of $a_{1}, a_{2}, a_{3}$, and $a_{4}$. When $n \geqslant 5$, we find the greatest common divisor of $b_{3}$ and $a_{5}$, $\cdots$. | 53 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find the least common multiple of 108, 28, and 42. | Since
$$108=2^{2} \times 3^{3}, 28=2^{2} \times 7, 42=2 \times 3 \times 7$$
we get
$$\{108,28,42\}=2^{2} \times 3^{3} \times 7=756$$ | 756 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 16 Han Xin Counts the Soldiers: There is a troop of soldiers. If they form a column of five, there is one person left over at the end. If they form a column of six, there are five people left over at the end. If they form a column of seven, there are four people left over at the end. If they form a column of eleven, there are ten people left over at the end. Find the number of soldiers. | Let $x$ be the number of soldiers we are looking for. According to the problem,
In Theorem 1, take $m_{1}=5, m_{2}=6, m_{3}=7, m_{4}=11, b_{1}=1$, $b_{2}=5, b_{3}=4, b_{4}=10$. Then we have
$$\begin{array}{l}
M=5 \times 6 \times 7 \times 11=2310 \\
M_{1}=\frac{2310}{5}=462 \\
M_{2}=\frac{2310}{6}=385 \\
M_{3}=\frac{2310}{7}=330 \\
M_{4}=\frac{2310}{11}=210
\end{array}$$
Let $M_{1}^{\prime}$ be a positive integer that satisfies $M_{1}^{\prime} M_{1} \equiv 1(\bmod 5)$, then $1 \equiv$ $M_{1}^{\prime} M_{1} \equiv 462 M_{1}^{\prime} \equiv 2 M_{1}^{\prime}(\bmod 5)$, so we get $M_{1}^{\prime}=3$. Let $M_{2}^{\prime}$ be a positive integer that satisfies $M_{2}^{\prime} M_{2} \equiv 1(\bmod 6)$, then $1 \equiv M_{2}^{\prime} M_{2} \equiv$ $385 M_{2}^{\prime} \equiv M_{2}^{\prime}(\bmod 6)$, so we get $M_{2}^{\prime}=1$. Let $M_{3}^{\prime}$ be a positive integer that satisfies $M_{3}^{\prime} M_{3} \equiv 1(\bmod 7)$, then $1 \equiv M_{3}^{\prime} M_{3} \equiv 330 M_{3}^{\prime} \equiv$ $M_{3}^{\prime}(\bmod 7)$, so we get $M_{3}^{\prime}=1$. Let $M_{4}^{\prime}$ be a positive integer that satisfies $M_{4}^{\prime} M_{4} \equiv 1 \quad(\bmod 11)$, then $1 \equiv M_{4}^{\prime} M_{4} \equiv 210 M_{4}^{\prime} \equiv M_{4}^{\prime} (\bmod$ 11). Therefore, by (46) we get
$$\begin{aligned}
x & \equiv 3 \times 462+5 \times 385+4 \times 330+10 \times 210 \\
& \equiv 6731 \equiv 2111(\bmod 2310)
\end{aligned}$$
Thus, we have
$$x=2111+2310 k, \quad k=0,1,2, \cdots$$
$$\begin{array}{l}
x \equiv 1(\bmod 5), x \equiv 5(\bmod 6), \\
x \equiv 4 \quad(\bmod 7), \quad x \equiv 10 \quad(\bmod 11).
\end{array}$$ | 2111 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Solve the following systems of congruences:
(i) $\left\{\begin{array}{l}x \equiv 3 \quad(\bmod 7) \\ x \equiv 5 \quad(\bmod 11)\end{array}\right.$
(ii) $\left\{\begin{array}{ll}x \equiv 2 & (\bmod 11), \\ x \equiv 5 & (\bmod 7), \\ x \equiv 4 & (\bmod 5) .\end{array}\right.$
(iii) $\left\{\begin{array}{ll}x \equiv 1 & (\bmod 7) \\ 3 x \equiv 4 & (\bmod 5) \\ 8 x \equiv 4 & (\bmod 9)\end{array}\right.$ | 5.
(i) Solution: By the Chinese Remainder Theorem,
Given
Given
$$\begin{array}{c}
b_{1}=3, \quad b_{2}=5, \quad m_{1}=7, \quad m_{2}=11, \\
m=m_{1} \cdot m_{2}=7 \times 11=77, \\
M_{1}=\frac{77}{7}=11, \quad M_{2}=\frac{77}{11}=7 . \\
11 M_{1}^{\prime}=1 \quad(\bmod 7), \text { so } M_{1}^{\prime}=2 . \\
7 M_{2}^{\prime}=1 \quad(\bmod 11), \text { so } M_{2}^{\prime}=8 .
\end{array}$$
Therefore, the solution is
$$\begin{aligned}
x & \equiv 3 \times 11 \times 2+5 \times 7 \times 8 \\
& \equiv 346 \equiv 38(\bmod 77)
\end{aligned}$$
(ii) Solution: The moduli are pairwise coprime, so the Chinese Remainder Theorem can be used.
$$\begin{array}{c}
b_{1}=2, \quad b_{2}=5, \quad b_{3}=4 \\
m_{1}=11, \quad m_{2}=7, \quad m_{3}=5 \\
m=m_{1} \cdot m_{2} \cdot m_{3}=11 \times 7 \times 5=385 \\
M_{1}=\frac{385}{11}=35, \quad M_{2}=\frac{385}{7}=55 \\
M_{3}=\frac{385}{5}=77
\end{array}$$
By
That is
$$35 M_{1}^{\prime} \equiv 1 \quad(\bmod 11)$$
So
$$\begin{array}{c}
(11 \times 3+2) M_{1}^{\prime} \equiv 1 \quad(\bmod 11) \\
2 M_{1}^{\prime} \equiv 1(\bmod 11)
\end{array}$$
We get
$$M_{1}^{\prime}=-5$$
Similarly, by
$$55 M_{2}^{\prime} \equiv 1(\bmod 7) ， \quad \text { i.e., } 6 M_{2}^{\prime} \equiv 1(\bmod 7) ，$$
We get
$$M_{2}^{\prime}=-1$$
By
$$77 M_{3}^{\prime} \equiv 1(\bmod 5) ， \text { i.e., } 2 M_{3}^{\prime} \equiv 1(\bmod 5) \text { ， }$$
We get
$$M_{3}^{\prime}=3$$
By the Chinese Remainder Theorem, the solution is
$$\begin{aligned}
x \equiv & 2 \times 35 \times(-5)+5 \times 55 \times(-1) \\
& +4 \times 77 \times 3 \equiv 299 \quad(\bmod 385)
\end{aligned}$$
(iii) Solution: From the second and third equations, we get
$$x \equiv 3(\bmod 5), \quad x \equiv 5(\bmod 9)$$
Combining the first equation with the above two equations, we can use the Chinese Remainder Theorem to solve.
$$\begin{array}{c}
b_{1}=1, \quad b_{2}=3, \quad b_{3}=5 \\
m_{1}=7, \quad m_{2}=5, \quad m_{3}=9 \\
m=m_{1} \cdot m_{2} \cdot m_{3}=7 \times 5 \times 9=315
\end{array}$$
$$M_{1}=\frac{315}{7}=45, \quad M_{2}=\frac{315}{5}=63, \quad M_{3}=\frac{315}{9}=35$$
By
$45 M_{1}^{\prime} \equiv 1 \quad(\bmod 7)$ , we get $M_{1}^{\prime}=-2$ .
By $\quad 63 M_{2}^{\prime} \equiv 1(\bmod 5)$ , we get $M_{2}^{\prime}=2$ .
By $\quad 35 M_{3}^{\prime} \equiv 1(\bmod 9)$ , we get $M_{3}^{\prime}=-1$ .
Therefore,
$$\begin{aligned}
x \equiv & 1 \times 45 \times(-2)+3 \times 63 \times 2 \\
& +5 \times 35 \times(-1) \equiv 113(\bmod 315)
\end{aligned}$$ | 299 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6. Solve the following problems: (Yang Hui: "Xu Gu Zhai Qi Suan Fa" (1275))
(i) When divided by 7, the remainder is 1; when divided by 8, the remainder is 2; when divided by 9, the remainder is 4. What is the original number?
(ii) When divided by 2, the remainder is 1; when divided by 5, the remainder is 2; when divided by 7, the remainder is 3; when divided by 9, the remainder is 5. What is the original number?
(iii) When divided by 11, the remainder is 3; when divided by 72, the remainder is 2; when divided by 13, the remainder is 1. What is the original number? | 6.
(i) Solution: Let the number be $x$, then according to the problem, we have
Here
By
By
By
$$\left.\begin{array}{c}
\left\{\begin{array}{l}
x \equiv 1 \quad(\bmod 7), \\
x \equiv 2 \quad(\bmod 8), \\
x \equiv 4 \quad(\bmod 9) .
\end{array}\right. \\
b_{1}=1, \quad b_{2}=2, \quad b_{3}=4, \\
m_{1}=7, \quad m_{2}=8, \quad m_{3}=9, \\
m=7 \times 8 \times 9=504,
\end{array}\right\}$$
Therefore,
$$\begin{array}{c}
x \equiv 1 \times 72 \times 4+2 \times 63 \times(-1)+4 \times 56 \\
\times(-4) \equiv-734 \equiv 274(\bmod 504)
\end{array}$$
(ii) Solution: Let the number be $x$, according to the problem, we have
$$\left\{\begin{array}{l}
x \equiv 1 \quad(\bmod 2) \\
x \equiv 2 \quad(\bmod 5) \\
x \equiv 3 \quad(\bmod 7) \\
x \equiv 5 \quad(\bmod 9)
\end{array}\right.$$
Here
$$\begin{array}{c}
b_{1}=1, \quad b_{2}=2, \quad b_{3}=3, \quad b_{4}=5 \\
m_{1}=2, \quad m_{2}=5, \quad m_{3}=7, \quad m_{4}=9 \\
m=m_{1} \cdot m_{2} \cdot m_{3} \cdot m_{4}=2 \times 5 \times 7 \times 9=630 \\
M_{1}=\frac{630}{2}=315, \quad M_{2}=\frac{630}{5}=126 \\
M_{3}=\frac{630}{7}=90, \quad M_{4}=\frac{630}{9}=70
\end{array}$$
By
By
By
By
Therefore,
$$\begin{array}{l}
315 M_{1}^{\prime} \equiv 1 \quad(\bmod 2), \text { get } M_{1}^{\prime}=1 \\
126 M_{2}^{\prime} \equiv 1 \quad(\bmod 5), \text { get } M_{2}^{\prime}=1 \\
90 M_{3}^{\prime} \equiv 1 \quad(\bmod 7), \\
70 M_{4}^{\prime} \equiv 1 \quad(\bmod 9), M_{3}^{\prime}=-1 \\
x=315+2 \times 126+3 \times 90 \times(-1)+5 \\
\times 70 \times 4 \equiv 1697 \equiv 437(\bmod 630)
\end{array}$$
(iii) Solution: Let the number be $x$, according to the problem, we have
$$\left\{\begin{array}{ll}
x \equiv 3 & (\bmod 11) \\
x \equiv 2 & (\bmod 72) \\
x \equiv 1 & (\bmod 13)
\end{array}\right.$$
Here
$$\begin{array}{c}
b_{1}=3, \quad b_{2}=2, \quad b_{3}=1, \\
m_{1}=11, \quad m_{2}=72, \quad m_{3}=13, \\
m=m_{1} \cdot m_{2} \cdot m_{3}=11 \times 72 \times 13=10296, \\
M_{1}=\frac{10296}{11}=936, \quad M_{2}=\frac{10296}{72}=143, \\
M_{3}=\frac{10296}{13}=792 . \\
936 M_{1}^{\prime} \equiv 1 \quad(\bmod 11), \text { get } M_{1}^{\prime}=1 . \\
143 M_{2}^{\prime} \equiv 1 \quad(\bmod 72), \text { get } M_{2}^{\prime}=-1 . \\
792 M_{3}^{\prime} \equiv 1 \quad(\bmod 13), \text { get } M_{3}^{\prime}=-1 .
\end{array}$$
Therefore,
$$\begin{array}{l}
x \equiv 3 \times 936+2 \times 143 \times(-1)+1 \times 792 \\
\times(-1) \equiv 1730(\bmod 10296)
\end{array}$$ | 274 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
9. Try to solve:
(i) $\left\{\begin{array}{l}x \equiv 2 \quad(\bmod 7), \\ x \equiv 5 \quad(\bmod 9), \\ x \equiv 11 \quad(\bmod 15) .\end{array}\right.$
(ii) There is a number whose total is unknown; when reduced by multiples of five, there is no remainder; when reduced by multiples of seven hundred and fifteen, there is a remainder of ten; when reduced by multiples of two hundred and forty-seven, there is a remainder of one hundred and forty; when reduced by multiples of three hundred and ninety-one, there is a remainder of two hundred and forty-five; when reduced by multiples of one hundred and eighty-seven, there is a remainder of one hundred and nine. What is the total number? (Huang Zongxian: "General Solution of the Method of Seeking One," answer: 10020.) | 9.
(i) Solution: Since $(7,9)=1,(7,15)=1,(9,15)=3,11-5=6$, and $3 \mid(11-5)$, by problem 8(i, ii), we know the system of congruences has a solution and is equivalent to
$$\left\{\begin{array}{l}
x \equiv 2 \quad(\bmod 7) \\
x \equiv 5 \quad(\bmod 9) \\
x \equiv 11 \equiv 1 \quad(\bmod 5)
\end{array}\right.$$
Solving the above using the Chinese Remainder Theorem:
$$\begin{array}{c}
b_{1}=2, \quad b_{2}=5, \quad b_{3}=1 \\
m_{1}=7, \quad m_{2}=9, \quad m_{3}=5 \\
m=m_{1} \cdot m_{2} \cdot m_{3}=7 \times 9 \times 5=315
\end{array}$$
$$M_{1}=\frac{315}{7}=45, \quad M_{2}=\frac{315}{9}=35, \quad M_{3}=\frac{315}{5}=63$$
From $45 M_{1}^{\prime} \equiv 1(\bmod 7)$, we get $M_{1}^{\prime}=-2$.
From $35 M_{2}^{\prime} \equiv 1(\bmod 9)$, we get $M_{2}^{\prime}=-1$.
From $63 M_{3}^{\prime} \equiv 1(\bmod 5)$, we get $M_{3}^{\prime}=2$.
Thus, the solution is
$$\begin{array}{c}
x \equiv 2 \times 45 \times(-2)+5 \times 35 \times(-1)+1 \times 63 \\
\times 2 \equiv-229 \equiv 86(\bmod 315) .
\end{array}$$
(ii) Solution: Let the total number be $x$. According to the problem, we have
$$\left\{\begin{array}{l}
x \equiv 0 \quad(\bmod 5) \\
x \equiv 10 \quad(\bmod 715) \\
x \equiv 140 \quad(\bmod 247) \\
x \equiv 245 \quad(\bmod 391) \\
x \equiv 109 \quad(\bmod 187)
\end{array}\right.$$
First, check if there is a solution. Since $715=5 \times 11 \times 13, 247=13 \times 19, 391=17 \times 23, 187=11 \times 17$, we have $(5,715)=5$, and $5 \mid(10-0)$; $(715,247)=13$, and $13 \mid(140-10)$; $(715,187)=11$, and $11 \mid(109-10)$; $(391,187)=17$, and $17 \mid(245-109)$; thus, by problem 8(i), the system of congruences has a solution. By problem 8(ii), appropriately choose the moduli so that the original system of congruences is equivalent to
$$\left\{\begin{array}{l}
x \equiv 0 \quad(\bmod 5) \\
x \equiv 10(\bmod 11 \times 13) \\
x \equiv 140 \equiv 7(\bmod 19) \\
x \equiv 245 \equiv 15(\bmod 23) \\
x \equiv 109 \equiv 7(\bmod 17)
\end{array}\right.$$
The moduli in the above system of congruences are pairwise coprime, so we can use the Chinese Remainder Theorem to solve it.
From $\square$
$$\begin{array}{c}
1062347 M_{1}^{\prime} \equiv 2 M_{1}^{\prime} \equiv 1 \quad(\bmod 5), \text { we get } M_{1}^{\prime}=3 . \\
37145 M_{2}^{\prime} \equiv 108 M_{2}^{\prime} \equiv 1 \quad(\bmod 11 \times 13) \\
\text { and } 11 \times 13 \equiv 143
\end{array}$$
Since
we have
$$\begin{array}{l}
143=108+35, \quad 108=3 \times 35+3, \\
35=11 \times 3+2, \quad 3=2+1,
\end{array}$$
$$\begin{aligned}
1 & =3-2=3-(35-11 \times 3)=12 \times 3-35 \\
& =12 \times(108-3 \times 35)-35 \\
& =12 \times 108-37 \times 35 \\
& =12 \times 108-37 \times(143-108) \\
& =49 \times 108-37 \times 143 .
\end{aligned}$$
Thus, $\square$
$108 \times 49 \equiv 1(\bmod 143)$, we get $M_{2}^{\prime}=49$.
From $279565 M_{3}^{\prime} \equiv 18 M_{3}^{\prime} \equiv 1(\bmod 19)$, we get $M_{3}^{\prime}=-1$.
From $230945 M_{4}^{\prime} \equiv 2 M_{4}^{\prime} \equiv 1(\bmod 23)$, we get $M_{4}^{\prime}=-11$.
From $312455 M_{5}^{\prime} \equiv 12 M_{5}^{\prime} \equiv 1(\bmod 17)$, we get $M_{5}^{\prime}=-7$.
Finally, the solution is
$$\begin{aligned}
x \equiv 10 & \times 37145 \times 49+7 \times 279565 \times(-1) \\
& +15 \times 230945 \times(-11)+7 \times 312455 \\
& \times(-7) \equiv 18201050-1956955-38105925 \\
& -15310295 \equiv-37172125 \\
\equiv & 10020(\bmod 5311735) .
\end{aligned}$$
Answer: The smallest total number is 10020.
$$\begin{array}{l}
b_{1}=0, \quad b_{2}=10, \quad b_{3}=7, \quad b_{4}=15, \quad b_{5}=7, \\
m_{1}=5, \quad m_{2}=11 \times 13, \quad m_{3}=19, \quad m_{4}=23, \\
m_{5}=17, \quad m=5 \times 11 \times 13 \times 19 \times 23 \times 17 \\
=5311735, \\
M_{1}=\frac{5311735}{5}=1062347, \\
M_{2}=\frac{5311735}{11 \times 13}=37145, \\
M_{3}=\frac{5311735}{19}=279565, \\
M_{4}=\frac{5311735}{23}=230945, \\
M_{5}=\frac{5311735}{17}=312455.
\end{array}$$ | 10020 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10. The distance between Port A and Port B does not exceed 5000 kilometers. Today, three ships depart from Port A to Port B at midnight simultaneously. Assuming the three ships sail at a constant speed for 24 hours a day, the first ship arrives at midnight several days later, the second ship arrives at 18:00 several days after that, and the third ship arrives at 8:00 a few days after the second ship. If the first ship travels 300 kilometers per day, the second ship travels 240 kilometers per day, and the third ship travels 180 kilometers per day, what is the actual distance between Port A and Port B in kilometers, and how long did each of the three ships travel? | 10. Solution: Let the distance between ports A and B be $x$ kilometers. The distance the second ship travels in 18 hours is $240 \times \frac{18}{24}=180$ kilometers, and the distance the third ship travels in 8 hours is $180 \times \frac{8}{24}=60$ kilometers. According to the problem, we have
$$\left\{\begin{array}{l}
x \equiv 0(\bmod 300) \\
x \equiv 180(\bmod 240) \\
x \equiv 60(\bmod 180)
\end{array}\right.$$
Since
$$\begin{array}{l}
(300,240)=60, \text { and } 60 \mid(180-0) ; \\
(300,180)=60, \text { and } 60 \mid(60-0) ; \\
(240,180)=60, \text { and } 60 \mid(180-60) ;
\end{array}$$
the system of congruences has a solution. Because
$$\begin{aligned}
300 & =2^{2} \times 3 \times 5^{2} \\
240 & =2^{4} \times 3 \times 5 \\
180 & =2^{2} \times 3^{2} \times 5
\end{aligned}$$
the original system of congruences is equivalent to
$$\left\{\begin{array}{l}
x \equiv 0 \quad\left(\bmod 5^{2}\right) \\
x \equiv 180 \equiv 4 \quad\left(\bmod 2^{4}\right) \\
x \equiv 60 \equiv 6 \quad\left(\bmod 3^{2}\right)
\end{array}\right.$$
We solve this using the Chinese Remainder Theorem. Here,
$$\begin{array}{c}
b_{1}=0, \quad b_{2}=4, \quad b_{3}=6 \\
m_{1}=5^{2}, \quad m_{2}=2^{4}, \quad m_{3}=3^{2} \\
m=5^{2} \times 2^{4} \times 3^{2}=3600 \\
M_{1}=\frac{3600}{5^{2}}=144, \quad M_{2}=\frac{3600}{2^{4}}=225 \\
M_{3}=\frac{3600}{9}=400
\end{array}$$
From $144 M_{1}^{\prime} \equiv 1\left(\bmod 5^{2}\right)$, i.e., $19 M_{1}^{\prime} \equiv 1\left(\bmod 5^{2}\right)$, we get $M_{1}^{\prime}=4$. From $225 M_{2}^{\prime} \equiv 1\left(\bmod 2^{4}\right)$, i.e., $M_{2}^{\prime} \equiv 1\left(\bmod 2^{4}\right)$, we get $M_{2}^{\prime}=1$. From $400 M_{3}^{\prime} \equiv 1\left(\bmod 3^{2}\right)$, i.e., $4 M_{3}^{\prime} \equiv 1(\bmod 9)$, we get $M_{3}^{\prime}=-2$. Therefore,
$$\begin{aligned}
x & \equiv 4 \times 225 \times 1+6 \times 400 \times(-2) \\
& \equiv-3900 \equiv 3300(\bmod 3600)
\end{aligned}$$
Since the distance between ports A and B does not exceed 5000 kilometers, the actual distance is 3300 kilometers.
Also,
$$\begin{array}{l}
\frac{3300}{300}=11 \\
\frac{3300}{240}=13 \frac{18}{24} \\
\frac{3300}{180}=18 \frac{6}{18}
\end{array}$$
Answer: The distance between ports A and B is 3300 kilometers. The first ship takes 11 days, the second ship takes 13 days and 18 hours, and the third ship takes 18 days and 8 hours. | 3300 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 9 Find the least common multiple of 24871 and 3468.
The text above is translated into English, preserving the original text's line breaks and format. | Since
thus $(24871,3468)=17$. By Lemma 10 we have
$$\{24871,3468\}=\frac{24871 \times 3468}{17}=5073684 .$$ | 5073684 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 10 Find the least common multiple of 513, 135, and 3114. | Since
$1\left|\begin{array}{r|r}513 & 135 \\ 405 & 108 \\ \hline 108 & 27 \\ 108 & \\ \hline 0 & 27\end{array}\right|^{3}$
Therefore, $(513,135)=27$, by Lemma 10 we have
$$\{513,135\}=\frac{513 \times 135}{27}=2565 .$$
Since
\begin{tabular}{|c|c|c|}
\hline 1 & 2565 & \multirow{2}{*}{\begin{tabular}{l}
3114 \\
2565
\end{tabular}} \\
\hline & 2196 & \\
\hline 1 & 369 & 549 \\
\hline & 360 & 369 \\
\hline 20 & 9 & 180 \\
\hline & & 180 \\
\hline & & 0 \\
\hline
\end{tabular}
Therefore, $(2565,3114)=9$, by Lemma 10 we have
$$\{2565,3114\}=\frac{2565 \times 3114}{9}=887490$$
Therefore,
$$\{513,135,3114\}=887490 .$$ | 887490 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 11 Find the least common multiple of $8127, 11352, 21672$ and 27090. | Since
\begin{tabular}{|c|c|c|}
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
8127 \\
6450
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{r}
11352 \\
8127
\end{tabular}
\end{tabular} \\
\hline 1 & 1677 & 3225 \\
\hline & 1548 & 1677 \\
\hline 12 & 129 & 1548 \\
\hline & & 1548 \\
\hline & 129 & 0 \\
\hline
\end{tabular}
Therefore, \((8127,11352)=129\), and by Lemma 10 we have
\[
\{8127,11352\}=\frac{8127 \times 11352}{129}=715176
\]
Since
\[
\begin{array}{|r|l|}
715176 & 21672 \\
715176 & \\
\hline 0 & 21672
\end{array}
\]
Therefore, \((715176,21672)=21672\), and by Lemma 10 we have
\[
\{715176,21672\}=\frac{715176 \times 21672}{21672}=715176
\]
Since
\[
2\left|\begin{array}{r|r|}715176 & 27090 \\ 704340 & 21672 \\ \hline 10836 & 5418 \\ 10836 & \\ \hline 0 & 5418\end{array}\right|
\]
Therefore, \((715176,27090)=5418\), and by Lemma 10 we have
\[
\{715176,27090\}=\frac{715176 \times 27090}{5418}=3575880
\]
Thus, we have
\[
\{8127,11352,21672,27090\}=3575880.
\] | 3575880 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 12 A steel plate, 1丈 3尺 5寸 long and 1丈 5寸 wide. Now it is to be cut into equally sized squares, the squares must be as large as possible, and no steel plate should be left over. Find the side length of the square.
Note: In traditional Chinese units, 1丈 = 10尺, and 1尺 = 10寸. | Solution: Since we want the largest square, we need to find the largest side length of the square. To find the largest side length of the square, we need to find the greatest common divisor (GCD) of 135 inches and 105 inches. Since
$$135=3^{3} \times 5, \quad 105=3 \times 5 \times 7$$
we have $(135,105)=15$.
Answer: The side length of the square is 1 foot 5 inches. | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
9. Use the Euclidean algorithm to find the greatest common divisor:
(i) 51425, 13310.
(ii) $353430, 530145, 165186$.
(iii) $81719, 52003, 33649, 30107$. | 9
(i) Solution: Since
1 \begin{tabular}{r|r|r}
51425 & 13310 \\
39930 & 11495 \\
\hline 11495 & 1815 & 6 \\
10890 & 1815 \\
\hline 605 & 0
\end{tabular}
Therefore, $(51425,13310)=605$.
(ii) Solution: Since
We get
$$(353430,530145)=176715$$
Also,
$14 \left\lvert\,$\begin{tabular}{r|r|}
176715 & 165186 \\
165186 & 161406 \\
\hline 11529 & 3780 \\
11340 & 3780 \\
\hline 189 & 0
\end{tabular}\right.,
We get
$(176715,165186)=189$.
Therefore, $(353430,530145,165186)=189$.
(iii) Solution: Since
$1\left|\begin{array}{r|r|r}81719 & 52003 \\ 52003 & 29716 \\ \hline 29716 & 22287 \\ 22287 & 1 \\ \hline 7429 & 0\end{array}\right|$
We get $(81719,52003)=7429$
Since
\begin{tabular}{|c|c|c|}
\hline & 33649 & 7429 \\
\hline & 29716 & 3933 \\
\hline 1 & 3933 & 3496 \\
\hline & 3496 & 3496 \\
\hline 8 & 437 & 0 \\
\hline
\end{tabular}
We get
$(33649,7429)=437$.
Since
\begin{tabular}{|c|c|c|c|}
\hline & 30107 & 437 & 68 \\
\hline & 29716 & 391 & \\
\hline 1 & 391 & 46 & 8 \\
\hline & 368 & 46 & \\
\hline 2 & 23 & 0 & \\
\hline
\end{tabular}
We get
$(30107,437)=23$.
Therefore, $(81719,52003,33649,30107)=23$. | 23 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12. Use the properties of the greatest common divisor from the previous question to find the greatest common divisor of the following:
(i) $216,64,1000$.
(ii) $24000,36000,144000$. | 12.
(i) Solution: The property of the greatest common divisor (GCD) from the previous problem can obviously be extended to more than two numbers, so
$$\begin{array}{c}
(216,64,1000)=\left(6^{3}, 4^{3}, 10^{3}\right) \\
=(6,4,10)^{3}=2^{3}=8
\end{array}$$
(ii) Solution:
$$\begin{array}{l}
(24000,36000,144000) \\
=1000 \times(24,36,144) \\
=1000 \times 12 \times(2,3,12) \\
=1000 \times 12=12000
\end{array}$$ | 12000 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
14. There is a rectangular room that is 5.25 meters long and 3.25 meters wide. Now, square tiles are used to cover the floor, and it is required to exactly cover the entire room. What is the maximum side length of the square tiles used? | 14. Solution: To pave the entire room with square tiles, the maximum side length of the tiles should be the greatest common divisor (GCD) of the room's length and width. To eliminate decimals, we use centimeters as the unit of length, with the room's length being 525 cm and the width 325 cm. White text
$$\begin{array}{l}
(525,325)=5 \times(105,65) \\
\quad=5 \times 5 \times(21,13)=5 \times 5=25
\end{array}$$
So $(525,325)=25$.
Answer: The maximum side length of the tiles is 25 cm. | 25 | Geometry | math-word-problem | Yes | Yes | number_theory | false |
18. Venus and Earth are at a certain position relative to the Sun at a certain moment. It is known that Venus orbits the Sun in 225 days, and Earth orbits the Sun in 365 days. How many days at least will it take for both planets to return to their original positions simultaneously? | 18. Solution: The time required for both planets to return to their original positions must be a common multiple of the time each planet takes to orbit the sun once. To find the least amount of time for them to return to their original positions, we need to find the least common multiple (LCM) of their orbital periods. Since
$$\begin{array}{l}
225=3^{2} \times 5^{2} \\
365=5 \times 73
\end{array}$$
the LCM of $\{225,365\}=3^{2} \times 5^{2} \times 73=16425$.
Answer: The two planets will simultaneously return to their original positions after at least 16425 days. | 16425 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
19. Design a packaging box with a square base to transport four different sizes of chess sets, where the base of each chess box is also square, with side lengths of 21 cm, 12 cm, 14 cm, and 10.5 cm, respectively. To ensure that the packaging box can completely cover the base regardless of which size of chess set it is transporting, what is the minimum side length of the packaging box's base in centimeters? | 19. Solution: To ensure that each type of chess box can completely cover the bottom of the packaging box, the side length of the bottom of the packaging box should be a common multiple of the side lengths of the bottom of each chess box. Therefore, the smallest side length of the box bottom is the least common multiple of the side lengths of the bottoms of the various chess boxes. We use millimeters as the unit of length to ensure that all side lengths are integers. Using the method of prime factorization, we get
$$\begin{array}{l}
210=2 \times 3 \times 5 \times 7 \\
120=2^{3} \times 3 \times 5 \\
140=2^{2} \times 5 \times 7 \\
105=3 \times 5 \times 7
\end{array}$$
So $\{210,120,140,105\}=2^{3} \times 3 \times 5 \times 7=840$.
Answer: The minimum side length of the box bottom is 840 millimeters, which is 84 centimeters. | 84 | Geometry | math-word-problem | Yes | Yes | number_theory | false |
20. In the performance of a group gymnastics, it is required that when the formation changes to 10 rows, 15 rows, 18 rows, and 24 rows, the formation can always form a rectangle. How many people are needed at minimum for the group gymnastics performance? | 20. Solution: Since the formation needs to be a rectangle, the number of people must be a multiple of the number of rows. Finding the minimum number of people is essentially finding the least common multiple (LCM) of the row numbers. Using the method of prime factorization, we get
$$\begin{array}{ll}
10=2 \times 5, & 15=3 \times 5 \\
18=2 \times 3^{2}, & 24=2^{3} \times 3
\end{array}$$
Therefore, $\{10,15,18,24\}=2^{3} \times 3^{2} \times 5=360$.
Answer: The minimum number of people required is 360. | 360 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Convert the following decimal numbers to octal numbers:
(i) 420,
(ii) 2640. | 3 .
(i) Solution: From $\frac{420}{8}=52+\frac{4}{8}$, we get $b_{0}=4$. From $\frac{52}{8}=6+\frac{4}{8}$, we get $b_{1}=4, b_{2}=6$, which means $420=(644)_{8}$.
(ii) Solution: From $\frac{2640}{8}=330$, we get $b_{0}=0$. From $\frac{330}{8}=41+\frac{2}{8}$, we get $b_{1}=2$. From $\frac{41}{8}=5+\frac{1}{8}$, we get $b_{2}=1, b_{3}=5$, which means $2640=(5120)_{8}$. | 644 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 Find the greatest common divisor of $1008, 1260, 882$ and 1134. | Solve: Decompose these four numbers into prime factors
$$\begin{array}{l}
1008=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7=2^{4} \times 3^{2} \times 7 \\
1260=2 \times 2 \times 3 \times 3 \times 5 \times 7=2^{2} \times 3^{2} \times 5 \times 7 \\
882=2 \times 3^{2} \times 7^{2} \\
1134=2 \times 3^{4} \times 7
\end{array}$$
Therefore, $(1008,1260,882,1134)=2 \times 3^{2} \times 7=126$. | 126 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Prove that the remainder of $\left(12371^{16}+34\right)^{23+7+c}$ divided by 111 is equal to 70, where $c$ is any non-negative integer. | Example 2's proof: From $(70,111)=1$ and Example 8's $\left(12371^{56}+\right.$ $34)^{28} \equiv 70(\bmod 111)$, we get $\left(12371^{56}+34,111\right)=1$. Since $111=3 \times 37$ and Lemma 14, we have $\varphi(111)=2 \times 36=72$. Since $c$ is a positive integer and by Theorem 1, we have $\left(12371^{56}+\right.$ $34)^{72 c} \equiv 1(\bmod 111)$. Therefore, we obtain
$$\left(12371^{56}+34\right)^{72 c+28} \equiv 70(\bmod 111)$$ | 70 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 26 Find $\sigma(450)=$ ? | Since $45!=2 \times 3 \times 5^{2}$, by Lemma 10 we have
$$\begin{aligned}
\sigma(450) & =\frac{2^{2}-1}{2-1} \cdot \frac{3^{3}-1}{3-1} \cdot \frac{5^{3}-1}{5-1} \\
& =3 \times 13 \times 31=1209
\end{aligned}$$ | 1209 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 27 Find $\sigma_{2}(28)=$ ? | Since the factors of 28 are $1,2,4,7,14,28$, we have
$$\sigma_{2}(28)=1+2^{2}+4^{2}+7^{2}+14^{2}+28^{2}=1050$$ | 1050 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
side 28 find $\sigma_{3}(62)=$ ? | Since the factors of 62 are $1,2,31,62$, we have
$$\sigma_{3}(62)=1+2^{3}+31^{3}+62^{3}=268128$$ | 268128 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Find the remainder when $\left(12371^{55}+34\right)^{28}$ is divided by 111. | From $12371=111^{2}+50$, we get $12371 \equiv 50(\bmod 111)$. By Lemma 3, we have
$$12371^{56} \equiv 50^{56}(\bmod 111)$$
We also have $(50)^{28}=(125000)^{9}(50), 125000 \equiv 14(\bmod 111)$, so by Lemma 3, we get
$$(50)^{28} \equiv(14)^{9}(50)(\bmod 111)$$
Furthermore, $14^{3} \equiv 80(\bmod 111),(80)^{3} \equiv 68(\bmod 111) ，(68)(50) \equiv$ $70(\bmod 111)$, which gives us
$$(50)^{28} \equiv 70(\bmod 111)$$
By Lemma 3, we have $(50)^{*} \equiv 70^{2}(\bmod 111)$. We also have $70^{2} \equiv 16$ $(\bmod 111)$, so by (20) we get $12371^{56} \equiv 16(\bmod 111)$. By Lemma 3, we have $\left(12371^{56}+34\right)^{28} \equiv 50^{28}(\bmod 111)$. By (22), we get $\left(12371^{56}+34\right)^{28} \equiv 70(\bmod 111)$. Therefore, the remainder when $\left(12371^{56}+34\right)^{28}$ is divided by 111 is 70. | 70 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8. (i) Let $N=9450$, find $\varphi(N)$.
(ii) Find the sum of all positive integers not greater than 9450 and coprime with 9450. | 8. (i) Solution: Given $9450=2 \cdot 3^{3} \cdot 5^{2} \cdot 7$, and by Lemma 14, we have
$$\begin{aligned}
\varphi(N) & =9450\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \\
& =\frac{2 \cdot 3^{3} \cdot 5^{2} \cdot 7 \cdot 2 \cdot 4 \cdot 6}{2 \cdot 3 \cdot 5 \cdot 7}=2160
\end{aligned}$$
(ii) Solution: Let the sum of all positive integers not greater than 9450 and coprime with 9450 be $S$. By Problem 5, we have
$$\begin{aligned}
S & =\frac{1}{2} \cdot 9450 \cdot \varphi(9450) \\
& =\frac{1}{2} \times 9450 \times 2160 \\
& =10206000
\end{aligned}$$ | 10206000 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12. Find positive integers $\boldsymbol{n}$ and $m, n>m \geqslant 1$, such that the last three digits of $1978^{n}$ and $1978^{m}$ are equal, and make $n+m$ as small as possible. (20th International Mathematical Olympiad Problem) | 12. Solution: $1978^{n}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right)$
$$=2^{m} \cdot 989^{m}\left(1978^{n-m}-1\right)$$
Since the last three digits of $1978^{n}$ and $1978^{m}$ are the same, the last three digits of $1978^{n}-1978^{m}$ are all 0. Therefore, $1978^{n}-1978^{m}$ is divisible by 1000. And $1000=2^{3} \cdot 5^{3}$. Thus,
$$2^{3} \cdot 5^{3} \mid 2^{m} \cdot 989^{m}\left(1978^{n-m}-1\right)$$
Since $989^{m}$ and $1978^{n-m}-1$ are both odd, $2^{3} \mid 2^{m} . . m$ must be at least 3.
$$\text { Also, }\left(5^{3}, 2^{m} \cdot 989^{m}\right)=1 \text {, so } 5^{3} \mid\left(1978^{n \cdots m}-1\right) \text {, }$$
i.e.,
$$1978^{n-m} \equiv 1(\bmod 125)$$
The problem now is to find the smallest positive integer $n-m$ that satisfies the above congruence. At this point, taking $m=3, n+m=(n-m)+2 m$ will also be the smallest.
Since $\varphi(125)=5^{2} \cdot 4=100,(1978,125)=1$, by Theorem 1 we get
$$1978^{100} \equiv 1(\bmod 125)$$
We can prove that $(n-m) \mid 100$. Because otherwise:
$100=(n-m) q+r, q$ is an integer, $r$ is a positive integer,
and $0<r<n-m$, then
$$1978^{100}=1.978^{(n-m) q} \cdot 1978^{r} \equiv 1978^{r}(\bmod 125)$$
But
$$1978^{100} \equiv 1(\bmod 125)$$
so
$$1978^{r} \equiv 1(\bmod 125)$$
However, $r<n-m$, which contradicts the assumption that $n-m$ is the smallest positive integer that satisfies the congruence. Therefore, $(n-m) \mid 100$.
Since $125 | (1978^{n-m}-1)$, the last digit of $1978^{n-m}$ must be 1 or 6. It is easy to verify that the last digit of $1978^{n-m}$ is 6 only when $4 \mid(n-m)$. So $n-m$ is a multiple of 4, a divisor of 100, and can only be one of 4, 20, 100.
Since
$$\begin{aligned}
1978^{4} & =(125 \times 15+103)^{4} \equiv 103^{4}(\bmod 125) \\
103^{2} & =\left(3+4 \cdot 5^{2}\right)^{2} \equiv 3^{2}+2 \cdot 3 \cdot 4 \cdot 5^{2} \\
& \equiv 609 \equiv-16(\bmod 125) \\
103^{4} & \equiv(-16)^{2} \equiv 6(\bmod 125)
\end{aligned}$$
So
$$1978^{4} \equiv 1(\bmod 125)$$
And
$$1978^{20}=\left(1978^{4}\right)^{5} \equiv 6^{5} \equiv 1(\bmod 125)$$
Therefore, the smallest value of $n-m$ is 100. Now taking $m=3$, we have $n=103, n+m=106$. | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
11. If $3^{k} \mid 1000!$ and $3^{k+1} \nmid 1000!$, find $k$. | 11. Solution: From the result of the previous problem, we have.
$$\begin{aligned}
k= & {\left[\frac{1000}{3}\right]+\left[\frac{1000}{9}\right]+\left[\frac{1000}{27}\right]+\left[\frac{1000}{81}\right] } \\
& +\left[\frac{1000}{243}\right]+\left[\frac{1000}{729}\right] \\
= & 333+111+37+12+4+1=498
\end{aligned}$$ | 498 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Given $p=29$, try to find a primitive root of $p^{2}=841$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let's verify that 14 is a primitive root of 29. Since
$$\varphi(p)=28=(4)(7)=(2)(14)$$
To verify that 14 is indeed a primitive root of \( p = 29 \), it is not necessary to check for all \( m (1 \leqslant m < 28) \) that
$$14^{m} \equiv 1 \pmod{29}$$
but only the following conditions need to be verified (see Theorem 1 in the previous section):
$$14^{2} \equiv 1, 14^{4} \equiv 1, 14^{7} \equiv 1, 14^{14} \equiv 1 \pmod{29}$$
Calculations yield:
$$\begin{array}{c}
14^{2} \equiv 22, \quad 14^{4} \equiv 22^{2} \equiv (-7)^{2} \equiv 22 \\
14^{7} \equiv (14)^{4}(14)^{2}(14) \equiv (20)(22)(14) \equiv 12 \\
14^{14} \equiv (12)^{2} \equiv -1 \pmod{29}
\end{array}$$
Thus, the conditions in (21) are indeed satisfied, and 14 must be a primitive root modulo 29.
Further calculations yield:
$$\begin{aligned}
(14)^{28} & = (196)^{14} = (38416)^{7} \equiv (571)^{7} = (571)(326041)^{3} \\
& \equiv (571)(574)^{3} = (327754)(329476) \equiv (605)(645) \\
& = 390225 \equiv 1 \pmod{29^2}
\end{aligned}$$
Therefore, 14 is not a primitive root modulo \( 29^2 \). To find a primitive root of \( 29^2 \) from 14, we consider \( 14 + 29 = 43 \), since
$$43 \equiv 14 \pmod{29}$$
43 is still a primitive root modulo 29. However,
$$\begin{aligned}
(43)^{28} & = (3418801)^{7} \equiv (136)^{7} = (136)(18496)^{3} \\
& \equiv (136)(-6)^{3} = -29376 \equiv 59 \equiv 1 \pmod{29^2}
\end{aligned}$$
Thus, by Theorem 8, 43 is a primitive root modulo \( 29^2 \).
The method used in the above example is generally applicable. That is, if \( g \) is a primitive root modulo \( p \geqslant 3 \) and
$$g^{p-1} \equiv 1 \pmod{p^2}$$
then \( g + p \) must be a primitive root modulo \( p^2 \).
Since it is known that a primitive root modulo a prime \( p \) always exists, the above result implies that a primitive root modulo \( p^2 \) also always exists for an odd prime \( p \geqslant 3 \).
Surprisingly, for any odd prime power \( p^s, s \geqslant 2 \), condition (17) is also a necessary and sufficient condition for a primitive root \( g \) of \( p \) to remain a primitive root of \( p^s \). When \( g \) does not satisfy (17), taking \( g + p \) will yield a primitive root of \( p^s \). This is the result stated below. | 43 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Theorem 9 We have $g(8) \leqslant 42273$. | First, by applying the method of comparing coefficients or direct expansion, it is easy to verify that the following identity holds:
$$\begin{array}{l}
5040\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{4} \\
\quad=6 \sum(2 a)^{8}+60 \sum(a \pm b)^{8} \\
\quad+\sum(2 a \pm b \pm c)^{8}+6 \sum(a \pm b \pm c+d)^{8}
\end{array}$$
where the sum $\sum(2 a)^{8}$ represents the sum over each element in the set $\{a, b, c, d\}$, and the other sums have similar meanings. It is evident that the right-hand side of (31) consists of
$$6\binom{4}{1}+60\binom{4}{2}(2)+\binom{4}{1}\binom{3}{2}\left(2^{2}\right)+6\binom{4}{4}\left(2^{3}\right)=840$$
eighth powers of integers.
For any non-negative integer $n$, there exist integers $q \geqslant 0$ and $r$, $0 \leqslant r \leqslant 5039$, such that $n=5040 q+r$ holds.
First, consider $r$. Since $0 \leqslant r \leqslant 5039 < 3^{8}$, if $r$ can be expressed as the sum of eighth powers of non-negative integers, it can only be expressed as the sum of several $2^{8}$ and several $1^{8}$. Let
$$r=2^{8} k+1, k \geqslant 0, 0 \leqslant 1 \leqslant 2^{8}-1=255,$$
then it is clear that $k=\left[\frac{r}{2^{8}}\right] \leqslant\left[\frac{5039}{2^{8}}\right]=19$. $l=r-\left(2^{8}\right)(19) \leqslant 5039-(256)(19)=175.1$ is $r=(19) 2+1$ can be expressed as the sum of at most $19+175=194$ eighth powers of non-negative integers.
(2) If $0 \leqslant r<\left(2^{8}\right)(19)$. In this case, it is clear that $k \leqslant 18$ and $r=2^{8} k+l$ can be expressed as the sum of at most $k+l \leqslant 18+255-273$ eighth powers of non-negative integers. The fourth power of each $5040 q=5040 x_{1}^{+}+\cdots+5040 x_{i+1}^{+}$.
By Lagrange's theorem, each $x_{i}, 1 \leqslant i \leqslant g(4)$, can be expressed as the sum of four squares of non-negative integers $x_{i}=y_{i 1}^{2}+y_{i 2}^{2}+y_{13}^{2}+y_{i 4}^{2}$. Using the identity (31), for each $i (1 \leqslant i \leqslant g(4))$, the number $5040 x_{i}^{4}=5040\left(y_{i 1}^{2}+y_{i 2}^{2}+y_{i 3}^{2}+y_{i 4}^{2}\right)+$ can be expressed as the sum of 840 eighth powers of non-negative integers. Therefore, $5040 q$ can be expressed as the sum of $g(4) \times 840$ eighth powers of non-negative integers. Thus, $n=5040 q+r$ can be expressed as the sum of at most
$$g(4)(840)+273 \leqslant(50)(840)+273=42273$$
eighth powers of non-negative integers, completing the proof of Theorem 9. | 42273 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 1 Find the number of integers from 1 to 1000 that are not divisible by 5, nor by 6 and 8. | We use the notation $\operatorname{LCM}\left\{a_{1}, \cdots, a_{n}\right\}$ to represent the least common multiple of $n$ integers $a_{1}, \cdots, a_{n}$. Let $S$ be the set consisting of the natural numbers from 1 to 1000. Property $P_{1}$ is "an integer is divisible by 5", property $P_{2}$ is "an integer is divisible by 6", and property $P_{3}$ is "an integer is divisible by 8". $A_{i}(i=1,2,3)$ is the subset of $S$ consisting of integers with property $P_{i}$. Note that
$$\begin{array}{l}
\left|A_{1}\right|=\left[\frac{1000}{5}\right]=200, \\
\left|A_{2}\right|=\left[\frac{1000}{6}\right]=166, \\
\left|A_{3}\right|=\left[\frac{1000}{8}\right]=125,
\end{array}$$
Since $\operatorname{LCM}\{5,6\}=30$, we have
$$\left|A_{1} \cap A_{2}\right|=\left[\frac{1000}{30}\right]=33 \text {, }$$
Since $\operatorname{LCM}\{5,8\}=40$, we have
$$\left|A_{1} \cap A_{3}\right|=\left[\frac{1000}{40}\right]=25$$
Since $\operatorname{LCM}\{6,8\}=24$, we have
$$\left|A_{2} \cap A_{3}\right|=\left[\frac{1000}{24}\right]=41$$
Since $\operatorname{LCM}\{5,6,8\}=120$, we have
$$\left|A_{1} \cap A_{2} \cap A_{3}\right|=\left[\frac{1000}{120}\right]=8$$
Thus, by Theorem 1, the number of integers in $S$ that are not divisible by 5, 6, or 8 is
$$\begin{aligned}
\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}\right|= & |S|-\left|A_{1}\right|-\left|A_{2}\right|-\left|A_{3}\right|+\left|A_{1} \cap A_{2}\right|+\left|A_{1} \cap A_{3}\right| \\
& +\left|A_{2} \cap A_{3}\right|-\left|A_{2} \cap A_{2} \cap A_{3}\right| \\
= & 1000-200-166-125+33+25+41-8 \\
= & 600
\end{aligned}$$ | 600 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1. Find the number of positive integers among the first $10^{5}$ that are not divisible by $7, 11, 13$. | 1. Solution: According to Theorem 1 of this chapter, the number of integers sought is
$$\begin{array}{c}
10^{5}-\left[\frac{10^{5}}{7}\right]-\left[\frac{10^{5}}{11}\right]-\left[\frac{10^{5}}{13}\right]+\left[\frac{10^{5}}{7 \times 11}\right] \\
+\left[\frac{10^{5}}{7 \times 13}\right]+\left[\frac{10^{5}}{11 \times 13}\right]-\left[\frac{10^{5}}{7 \times 11 \times 13}\right]=10^{5} \\
-14285-9090-7692+1298+1098+699-99=71929
\end{array}$$ | 71929 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. A school organized three extracurricular activity groups in mathematics, Chinese, and foreign language. Each group meets twice a week, with no overlapping schedules. Each student can freely join one group, or two groups, or all three groups simultaneously. A total of 1200 students participate in the extracurricular groups, with 550 students joining the mathematics group, 460 students joining the Chinese group, and 350 students joining the foreign language group. Among them, 100 students participate in both the mathematics and foreign language groups, 120 students participate in both the mathematics and Chinese groups, and 140 students participate in all three groups. How many students participate in both the Chinese and foreign language groups? | 2. Solution: Since all 1200 students have joined at least one extracurricular group, the number of students who did not join any group is 0. We use $A_{1}, A_{2}, A_{3}$ to represent the sets of students who joined the math group, the Chinese group, and the English group, respectively. Thus, by the problem statement, we have
$$\left|A_{1}\right|=550,\left|A_{2}\right|=460,\left|A_{3}\right|=350$$
We also use $A_{12}$ to represent the set of students who joined both the math and Chinese groups, $A_{13}$ to represent the set of students who joined both the math and English groups, and $A_{23}$ to represent the set of students who joined both the Chinese and English groups, then we have
$$\left|A_{13}\right|=100, \quad\left|A_{12}\right|=120$$
We use $A_{123}$ to represent the set of students who joined all three groups, then
$$\left|A_{123}\right|=140$$
Noting the initial explanation, by Theorem 1 of this chapter, we get
$$\begin{array}{c}
0=1200-\left|A_{1}\right|-\left|A_{2}\right|-\left|A_{3}\right|+\left|A_{12}\right| \\
+\left|A_{13}\right|+\left|A_{23}\right|-\left|A_{123}\right|,
\end{array}$$
Thus, we have
$$\begin{aligned}
\left|A_{23}\right| & =-1200+550+460+350-100-120+140 \\
& =80
\end{aligned}$$
That is, the number of students who joined both the Chinese and English groups is 80. | 80 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
2. Find the smallest positive integer $a$, such that there exists a positive odd integer $n$, satisfying
$$2001 \mid\left(55^{n}+a \cdot 32^{n}\right)$$ | 2. From $2001=3 \times 23 \times 29$ and the conditions, we have
$$\left\{\begin{array}{ll}
a \equiv 1 & (\bmod 3) \\
a \equiv 1 & (\bmod 29), \\
a \equiv-1 & (\bmod 23)
\end{array}\right.$$
From the first two equations, we can set $a=3 \times 29 \times k+1$, substituting into the last equation gives
$$k \equiv 5(\bmod 23),$$
Thus, we get $a \geqslant 3 \times 29 \times 5+1=436$. When $a=436$, $2001 \mid(55+436 \times 32)$. Therefore, the smallest value is 436. | 436 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Find the smallest prime $p$ such that there do not exist $a, b \in \mathbf{N}$, satisfying
$$\left|3^{a}-2^{b}\right|=p$$ | 3. Notice that, $2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, 7=2^{3}-3^{0}, 11=3^{3}-$ $2^{4}, 13=2^{4}-3^{1}, 17=3^{4}-2^{6}, 19=3^{3}-2^{3}, 23=3^{3}-2^{2}, 29=2^{5}-3^{1}, 31=$ $2^{5}-3^{0}, 37=2^{6}-3^{3}$. Therefore, the required $p \geqslant 41$.
On the other hand, if $\left|3^{a}-2^{b}\right|=41$, there are two cases.
Case one: $3^{a}-2^{b}=41$, taking modulo 3 on both sides, we know $b$ is even; taking modulo 4 on both sides, we know $a$ is even. Let $a=2 m, b=2 n$, then $\left(3^{m}-2^{n}\right)\left(3^{m}+2^{n}\right)=41$, which gives $\left(3^{m}-2^{n}, 3^{m}+\right.$ $\left.2^{n}\right)=(1,41)$, leading to
$$2 \times 3^{m}=\left(3^{m}-2^{n}\right)+\left(3^{m}+2^{n}\right)=42$$
i.e., $3^{m}=21$, a contradiction.
Case two: $2^{b}-3^{a}=41$, in this case $b \geqslant 3$, taking modulo 8 on both sides leads to a contradiction.
In summary, the smallest prime $p=41$. | 41 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find the smallest positive integer $n$, such that the indeterminate equation
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$$
has integer solutions $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. | Note that for any $x \in \mathbf{Z}$, if $x$ is even, then $x^{4} \equiv 0(\bmod 16)$; if $x$ is odd, then $x^{2} \equiv 1(\bmod 8)$, and in this case, $x^{4} \equiv 1(\bmod 16)$.
The above discussion shows that $x_{i}^{4} \equiv 0$ or $1(\bmod 16)$, thus the remainder of $x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}$ modulo 16 lies between 0 and $n$. Therefore, when $n \leqslant 14$, equation (9) cannot hold (since $1599 \equiv 15(\bmod 16)$).
Notice that $5^{4}+12 \times 3^{4}+1^{4}+1^{4}=1599$, so when $n=15$, equation (9) has an integer solution $(5, \underbrace{3, \cdots, 3}_{12 \uparrow}, 1,1)$.
In summary, the smallest positive integer $n=15$. | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
7. Let $a, b, c, d$ all be prime numbers, satisfying: $a>3 b>6 c>12 d$, and
$$a^{2}-b^{2}+c^{2}-d^{2}=1749$$
Find the value of $a^{2}+b^{2}+c^{2}+d^{2}$. | 7. From the conditions, we know that $a, b, c$ are all odd numbers. If $d$ is odd, then $a^{2}-b^{2}+c^{2}-d^{2}$ is even, which is a contradiction. Therefore, $d$ is even, and thus $d=2$. Consequently, $a^{2}-b^{2}+c^{2}=1753$. From the conditions, we also know that $a \geqslant 3 b+2, b \geqslant 2 c+1, c \geqslant 5$, hence
$$\begin{aligned}
1753 & \geqslant(3 b+2)^{2}-b^{2}+c^{2}=8 b^{2}+12 b+4-c^{2} \\
& \geqslant 8(2 c+1)^{2}+12 b+4=33 c^{2}+32 c+12 b+12 \\
& \geqslant 33 c^{2}+160+132+12
\end{aligned}$$
Therefore, $c^{2}3 b$, we know $a-b>2 b \geqslant 22$. Also, $a-b$ and $a+b$ are both even, and $a, b$ are both odd, so $a-b=32$, and $a+b=54$. Thus, $a=43, b=11$, and consequently, $a^{2}+b^{2}+c^{2}+d^{2}=1999$. | 1999 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 Let the function $f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$ be defined as follows: Let
$$\frac{(2 n)!}{n!(n+1000)!}=\frac{A(n)}{B(n)}$$
where $n \in \mathbf{N}^{*}, A(n), B(n)$ are coprime positive integers. If $B(n)=1$, then $f(n)=1$; if $B(n)>1$, then $f(n)$ is the largest prime factor of $B(n)$.
Prove: The function $f$ is a bounded function, and find the maximum value of $f$. | Proof:
The idea of the proof is to find a constant such that for any $n \in \mathbf{N}^{*}$, the product of $\frac{A(n)}{B(n)}$ and this constant is a positive integer, thereby deducing that $f$ is a bounded function.
First, we prove a lemma: For any non-negative real numbers $x, y$,
$$[2 x]+[2 y] \geqslant[x]+[x+y]$$
In fact, let $x=m+\alpha, y=n+\beta$, where $m, n \in \mathbf{N}$, and $\alpha, \beta \in[0,1)$, then (2) is equivalent to proving
that is,
$$\begin{array}{c}
2 m+[2 \alpha]+2 n+[2 \beta] \geqslant m+(m+n)+[\alpha+\beta] \\
n+[2 \alpha]+[2 \beta] \geqslant[\alpha+\beta] .
\end{array}$$
Since $[\alpha+\beta] \leqslant 1$, and when $[\alpha+\beta]=1$, at least one of $\alpha, \beta$ is not less than $\frac{1}{2}$, thus in this case $[2 \alpha]+[2 \beta] \geqslant 1$. Therefore, (3) holds. The lemma is proved.
Returning to the original problem, we first prove: For any $n \in \mathbf{N}^{*}$, the number $\frac{(2 n)!\cdot 2000!}{n!(n+1000)!}$ is a positive integer. To do this, we only need to prove that for any prime $p$, the power of $p$ in the prime factorization of $(2 n)!\cdot 2000!$ is not less than the power of $p$ in the prime factorization of $n!(n+1000)!$. Using property 5, we only need to prove:
$$\sum_{k=1}^{+\infty}\left(\left[\frac{2 n}{p^{k}}\right]+\left[\frac{2000}{p^{k}}\right]\right)-\left(\left[\frac{n}{p^{k}}\right]+\left[\frac{n+1000}{p^{k}}\right]\right) \geqslant 0$$
If we let $x=\frac{n}{p^{k}}, y=\frac{1000}{p^{k}}$, using (2) we know that inequality (4) holds, so $\frac{(2 n)!\cdot 2000!}{n!(n+1000)!} \in \mathbf{N}^{*}$, i.e.,
$$\frac{A(n)}{B(n)} \cdot 2000!\in \mathbf{N}^{*}$$
The above discussion shows that for any $n \in \mathbf{N}^{*}$, we have $f(n) \mid 2000$!. Noting that $f(n)$ is 1 or a prime number, and by direct verification, 1999 is a prime number, so for any $n \in \mathbf{N}^{*}$, we have $f(n) \leqslant 1999$. Therefore, $f(n)$ is a bounded function.
Furthermore, when $n=999$, we know that
$$B(n)=1999 \cdot(999!)$$
At this time, $f(n)=1999$. Hence, the maximum value of $f(n)$ is 1999. | 1999 | Number Theory | proof | Yes | Yes | number_theory | false |
2. In decimal notation, how many $m \in\{1,2, \cdots, 2009\}$ are there such that there exists $n \in \mathbf{N}^{*}$, satisfying: $S\left(n^{2}\right)=m$? Here $S(x)$ denotes the sum of the digits of the positive integer $x$. | 2. Notice that, perfect squares $\equiv 0,1,4,7(\bmod 9)$, and for $k \in \mathbf{N}^{*}$, let $n=$ $\underbrace{11 \cdots 1} \underbrace{22 \cdots 25}$, then $S(n)=3 k+4$, and
$k-1$ 个 $\underbrace{2+0}_{k \text { 个 }}$
$$\begin{aligned}
n & =\frac{10^{k-1}-1}{9} \times 10^{k+1}+\frac{2 \times\left(10^{k}-1\right)}{9} \times 10+5 \\
& =\frac{1}{9}\left(10^{2 k}-10^{k+1}+2 \times 10^{k+1}-20+45\right) \\
& =\frac{1}{9}\left(10^{2 k}+10 \times 10^{k}+25\right) \\
& =\frac{1}{9}\left(10^{k}+5\right)^{2}=(\underbrace{33 \cdots 35}_{k-1 \uparrow})^{2}
\end{aligned}$$
That is, $n$ is a perfect square.
By setting $k=3 t+2,3 t, 3 t+1, t \in \mathbf{N}$, and combining $S\left(1^{2}\right)=1$, we can see that for any $m \in\{0,1,2, \cdots, 2009\}$, if $m \equiv 1,4$ or $7(\bmod 9)$, then there exists $n \in \mathbf{N}^{*}$, such that $S\left(n^{2}\right)=m$.
Furthermore, for $k \in \mathbf{N}^{*}, n=\underbrace{99 \cdots 9}_{k \text { 个 }}$, we have
$$\begin{aligned}
n^{2} & =\left(10^{k}-1\right)^{2}=10^{2 k}-2 \times 10^{k}+1 \\
& =\underbrace{99 \cdots 9}_{k-1 \uparrow} \underbrace{00 \cdots 01}_{k-1 \uparrow},
\end{aligned}$$
Thus, $S\left(n^{2}\right)=9 k$. Therefore, when $m \equiv 0(\bmod 9)$, there also exists $n \in \mathbf{N}^{*}$, such that $S\left(n^{2}\right)=$ $m$.
The above discussion shows that the number of numbers that meet the condition is $\frac{4}{9} \times 2007+1=893$. | 893 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6. Let \( x=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{100000}} \). Find the value of \([x]\). | 6. Notice that, when $k \geqslant 2$, we have
$$\begin{array}{l}
\frac{1}{\sqrt{k}}=\frac{2}{2 \sqrt{k}}\frac{2}{\sqrt{k}+\sqrt{k+1}}=2(\sqrt{k+1}-\sqrt{k}) .
\end{array}$$
Therefore,
$$\begin{aligned}
x & =1+\sum_{k=2}^{10^{6}} \frac{1}{\sqrt{k}}2 \sum_{k=1}^{10^{6}}(\sqrt{k+1}-\sqrt{k}) \\
& =2\left(\sqrt{10^{6}+1}-1\right)>1998 .
\end{aligned}$$
From this, we know that $[x]=1998$. | 1998 | Calculus | math-word-problem | Yes | Yes | number_theory | false |
25. Find the largest positive integer $m$, such that for $k \in \mathbf{N}^{*}$, if $1<k<m$ and $(k, m)=1$, then $k$ is a power of some prime.
---
The translation maintains the original format and line breaks as requested. | 25. The required maximum positive integer $m=60$.
On the one hand, it can be directly verified that $m=60$ meets the requirement (it is only necessary to note that the product of the smallest two primes coprime with 60, 7 and 11, is greater than 60).
On the other hand, let $p_{1}, p_{2}, \cdots$ represent the sequence of primes in ascending order. Suppose $m$ is a positive integer that meets the requirement. We first prove: if $p_{n} p_{n+1} \leqslant n$, then
$$p_{1} p_{2} \cdots p_{n} \leqslant m$$
In fact, if there are two different numbers $p_{u}, p_{v}$ in $p_{1}, p_{2}, \cdots, p_{n+1}$ that are not divisors of $m$, then $\left(p_{u} p_{v}, m\right)=1, p_{u} p_{v} \leqslant p_{n} p_{n+1} \leqslant m$, which contradicts the property of $m$. Therefore, at most one of $p_{1}, p_{2}, \cdots, p_{n+1}$ is not a divisor of $m$, so (5) holds.
Now, if there exists $m(>60)$ that meets the requirement, then when $m \geqslant 77=p_{4} p_{5}$, let $n$ be the largest positive integer such that $p_{1} p_{2} \cdots p_{n} \leqslant m$. Then, by (5), we know $n \geqslant 4$. Combining the conclusion from the previous problem, we have $p_{n+1} p_{n+2}5 \times 7$, so $2,3,5,7$ have at most one that is not a divisor of $m$ (see the proof of (5)), thus $m$ is a multiple of one of the numbers 105, 70, 42, 30, hence $m=70$. But in this case, $(33, m)=1$, which contradicts the property of $m$. | 60 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 By Fermat's Little Theorem, for any odd prime $p$, we have $2^{p-1} \equiv 1(\bmod p)$. Question: Does there exist a composite number $n$ such that $2^{n-1} \equiv 1(\bmod n)$ holds? | $$\begin{array}{l}
\text { Hence } \\
\text { Therefore }
\end{array}$$
$$\begin{array}{l}
2^{10}-1=1023=341 \times 3 \\
2^{10} \equiv 1(\bmod 341) \\
2^{340} \equiv 1^{34} \equiv 1(\bmod 341)
\end{array}$$
Hence 341 meets the requirement.
Furthermore, let $a$ be an odd composite number that meets the requirement, then $2^{a}-1$ is also an odd composite number (this can be known through factorization). Let $2^{a-1}-1=a \times q, q$ be an odd positive integer, then
$$\begin{aligned}
2^{2^{a}-1-1}-1 & =2^{2\left(2^{a-1}-1\right)}-1 \\
& =2^{2 a q}-1 \\
& =\left(2^{a}\right)^{2 q}-1 \\
& \equiv 1^{2 q}-1 \\
& \equiv 0\left(\bmod 2^{a}-1\right)
\end{aligned}$$
Therefore $2^{a}-1$ is also a number that meets the requirement. By this recursion (combined with 341 meeting the requirement), it is known that there are infinitely many composite numbers that meet the condition. | 341 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 6 A cube with a side length of 3 is divided into 27 unit cubes. The numbers $1, 2, \cdots$, 27 are randomly placed into the unit cubes, one number in each. Calculate the sum of the 3 numbers in each row (horizontal, vertical, and column), resulting in 27 sum numbers. Question: What is the maximum number of odd numbers among these 27 sum numbers? | Solve: To calculate the sum $S$ of these 27 sums, since each number appears in exactly 3 rows, we have
$$S=3 \times(1+2+\cdots+27)=3 \times 27 \times 14$$
Thus, $S$ is an even number, so the number of odd numbers among these 27 sums must be even.
If 26 of these 27 sums are odd, let's assume that the even number is the sum of the 3 numbers in the first row of Figure 1, i.e., $a_{1}+a_{2}+a_{3}$ is even, while the sums of the other 5 rows in Figure 1 are all odd. In this case, summing the numbers in Figure 1 by rows and columns, we get
$$\begin{aligned}
& \left(a_{1}+a_{2}+a_{3}\right)+\left(a_{4}+a_{5}+a_{6}\right)+\left(a_{7}+a_{8}+a_{9}\right) \\
= & \left(a_{1}+a_{4}+a_{7}\right)+\left(a_{2}+a_{5}+a_{8}\right)+\left(a_{3}+a_{6}+a_{9}\right)
\end{aligned}$$
However, the left side of this equation is the sum of two odd numbers and one even number, while the right side is the sum of 3 odd numbers, leading to a contradiction where the left side is even and the right side is odd.
Therefore, among these 27 sums, there can be at most 24 odd numbers.
The example below (as shown in Figure 2) demonstrates that there exists a way to fill the numbers such that 24 of the 27 sums can be odd. In the tables of Figure 2, 0 represents an even number, and 1 represents an odd number, from left to right, representing the top, middle, and bottom layers of the unit cubes.
Thus, among these 27 sums, the maximum number of odd numbers is 24. | 24 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Find the smallest positive integer $n$, such that there exist integers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$$ | From property 1, for any integer $a$, we know that
$$a^{2} \equiv 0(\bmod 4) \text { or } a^{2} \equiv 1(\bmod 8),$$
From this, we can deduce that
$$a^{4} \equiv 0 \text { or } 1(\bmod 16).$$
Using this conclusion, if $n<15$, let
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4} \equiv m(\bmod 16),$$
then
and
$$\begin{array}{c}
m \leqslant n<15 \\
1599 \equiv 15(\bmod 16)
\end{array}$$
This is a contradiction, so
$$n \geqslant 15.$$
Furthermore, when $n=15$, it is required that
$$x_{1}^{4} \equiv x_{2}^{4} \equiv \cdots \equiv x_{n}^{4} \equiv 1(\bmod 16)$$
This means that $x_{1}, x_{2}, \cdots, x_{n}$ must all be odd numbers, which points us in the right direction to find suitable numbers. In fact, among $x_{1}, x_{2}, \cdots, x_{15}$, if one number is 5, twelve are 3, and the other two are 1, then
$$\begin{aligned}
& x_{1}^{4}+x_{2}^{4}+\cdots+x_{15}^{4} \\
= & 5^{4}+12 \times 3^{4}+2 \\
= & 625+972+2 \\
= & 1599
\end{aligned}$$
Therefore, the minimum value of $n$ is 15. | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1 Let $p$ and $q$ both be prime numbers, and $7p + q$, $pq + 11$ are also prime numbers. Find the value of $\left(p^{2} + q^{p}\right)\left(q^{2} + p^{q}\right)$. | 1. If $p$ and $q$ are both odd, then $7p + q$ is even, and it is not a prime number. Therefore, one of $p$ and $q$ must be even.
Case one: Suppose $p$ is even, then $p = 2$. In this case, since $7p + q$ is a prime number, $q$ must be an odd prime. If $q \neq 3$, then $q \equiv 1$ or $2 \pmod{3}$.
If $q \equiv 1 \pmod{3}$, then
$$7p + q = 14 + q \equiv 0 \pmod{3}$$
This is a contradiction;
If $q \equiv 2 \pmod{3}$, then
$$pq + 11 = 2q + 11 \equiv 4 + 11 \equiv 0 \pmod{3}$$
This is also a contradiction. Therefore, $q = 3$. In this case,
$$7p + q = 17, \quad pq + 11 = 17$$
Both are prime numbers, so
$$\left(p^2 + q^p\right)\left(q^2 + p^q\right) = \left(2^2 + 3^2\right)\left(3^2 + 2^3\right) = 221$$
Case two: Suppose $q$ is even, then $q = 2$. Similarly, we find that $p = 3$. In this case,
$$\left(p^2 + q^p\right)\left(q^2 + p^q\right) = \left(3^2 + 2^3\right)\left(2^2 + 3^2\right) = 221$$
In conclusion, the desired value is 221. | 221 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 Let $p_{1}<p_{2}<p_{3}<p_{4}<p_{5}$ be 5 prime numbers, and $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}$ form an arithmetic sequence. Find the minimum value of $p_{5}$. | 2. Let $d$ be the common difference, then $p_{1}, p_{1}+d, p_{1}+2 d, p_{1}+3 d, p_{1}+4 d$ are all primes. If $2 \nmid d$, i.e., $d$ is odd, then one of $p_{1}+d$ and $p_{1}+2 d$ is even, and it is not a prime. If $3 \nmid d$, then one of $p_{1}+d, p_{1}+2 d, p_{1}+3 d$ is a multiple of 3 (they form a complete residue system modulo 3), which is a contradiction.
If $5 \nmid d$, then one of $p_{1}, p_{1}+d, \cdots, p_{1}+4 d$ is a multiple of 5, which can only be $p_{1}=$ 5, in this case the common difference $d$ is a multiple of 6.
And $5,11,17,23,29$ is a sequence of 5 primes in arithmetic progression, so, $p_{5}$ is at least 29. | 29 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8 Fibonacci sequence $\left\{F_{n}\right\}$ is defined as follows: $F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n}, n=1$, $2, \cdots$.
(1) Prove: The sum of any 10 consecutive terms of this sequence is a multiple of 11;
(2) Find the smallest positive integer $k$, such that the sum of any $k$ consecutive terms of this sequence is a multiple of 12. | 8. Consider the sequence $\left\{F_{n}\right\}$ where each term is taken modulo 11 (or 12).
(1) $\left\{F_{n}(\bmod 11)\right\}: 1,1,2,3,5,-3,2,-1,1,0,1,1, \cdots$, so $\left\{F_{n}(\bmod 11)\right\}$ is a purely periodic sequence with a period of 10. Therefore,
the sum of any 10 consecutive terms in $\left\{F_{n}\right\}$
$$\begin{array}{l}
\equiv 1+1+2+3+5+(-3)+2+(-1)+1+0 \\
=11 \equiv 0(\bmod 11)
\end{array}$$
The proposition is proved. $\square$
(2) $\left\{F_{n}(\bmod 12)\right\}: 1,1,2,3,5,-4,1,-3,-2,-5,5,0,1, 1, \cdots$ is a purely periodic sequence with a period of 12. Direct verification shows that the smallest positive integer $k$ satisfying the condition is $k=36$.
Note: If $k$ is the smallest positive integer satisfying (2), and $n$ is any positive integer satisfying (2), then $k \mid n$ (this conclusion is left for the reader to prove). Therefore, after finding that $n=36$ (where the sum of the numbers in each period of $\left\{F_{n}(\bmod 12)\right\}$ is $\equiv 4(\bmod 12)$), it is only necessary to verify that the positive divisors of 36 do not satisfy the condition to conclude that 36 is the smallest positive integer satisfying the condition. | 36 | Number Theory | proof | Yes | Yes | number_theory | false |
17 Find the number of all positive integers $a$ that satisfy the following condition: there exist non-negative integers $x_{0}, x_{1}, x_{2}, \cdots$, $x_{2001}$, such that $a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}}$. | 17. If $a$ is a number that satisfies the condition, then $a^{x_{0}}>1$, so $a>1$. At this point, taking
both sides modulo $a-1$, we know
so $\square$
$$\begin{array}{c}
a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}} \\
1 \equiv \underbrace{1+\cdots+1}_{2001 \uparrow}(\bmod a-1), \\
a-1 \mid 2000 .
\end{array}$$
On the other hand, if $a>1$ and $a-1 \mid 2000$, then we can take $x_{1}, x_{2}, \cdots, x_{2001}$ to be $a$ numbers as $0$, $a-1$ numbers as $1$, $a-1$ numbers as $2$, ..., $a-1$ numbers as $k-1$, where $k=\frac{2000}{a-1}$, and take $x_{0}=k$, then we have $a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}}$.
Therefore, $a$ is a number that satisfies the condition if and only if $a>1$ and $a-1 \mid 2000$, and there are 20 such $a$. | 20 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
20 Find the smallest positive integer $a$, such that for any integer $x$, we have $65 \mid\left(5 x^{13}+13 x^{5}+9 a x\right)$. | 20. From the condition, we know $65 \mid (18 + 9a)$ (take $x=1$), and since $(9,65)=1$, it follows that $65 \mid a+2$, hence $a \geq 63$.
When $a=63$, using Fermat's Little Theorem, we know that for any integer $x$, we have
$$\begin{aligned}
& 5 x^{13} + 13 x^{5} + 9 a x \\
\equiv & 13 x + 9 a x \\
\equiv & (3 + (-1) \times 3) x \\
\equiv & 0 \pmod{5}
\end{aligned}$$
$$\begin{aligned}
& 5 x^{13} + 13 x^{5} + 9 a x \\
\equiv & 5 x + 9 a x \\
\equiv & (5 + 9 \times (-2)) x \\
\equiv & 0 \pmod{13}
\end{aligned}$$
Therefore,
$$65 \mid 5 x^{13} + 13 x^{5} + 9 a x$$
In summary, the smallest positive integer $a=63$. | 63 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
23 Find the number of integer pairs $(a, b)$ that satisfy the following conditions: $0 \leqslant a, b \leqslant 36$, and $a^{2}+b^{2}=$ $0(\bmod 37)$. | 23. Notice that, $a^{2}+b^{2} \equiv a^{2}-36 b^{2}(\bmod 37)$, so from the condition we have
$$37 \mid a^{2}-36 b^{2},$$
which means
$$37 \mid(a-6 b)(a+6 b),$$
thus
$37 \mid a-6 b$ or $37 \mid a+6 b$.
Therefore, for each $1 \leqslant b \leqslant 36$, there are exactly two $a(a \equiv \pm 6 b(\bmod 37))$ that satisfy the condition, and when $b=0$, from $a^{2}+b^{2} \equiv 0(\bmod 37)$ we know $a=0$. Hence, the number of pairs $(a, b)$ that satisfy the condition is $2 \times 36+1=73$ (pairs). | 73 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
26 Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$.
Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$. | 26. Let $n=1000 x+y$, where $x$ is a positive integer, $y$ is an integer, and $0 \leqslant y \leqslant 999$. According to the problem,
$$x^{3}=1000 x+y$$
From $0 \leqslant y \leqslant 999$, we know
$$1000 x \leqslant x^{3}<1000 x+1000=1000(x+1)$$
Thus,
$$x^{2} \geqslant 1000, x^{3}+1 \leqslant 1000(x+1)$$
Therefore,
$$x^{2} \geqslant 1000, x^{2}-x+1 \leqslant 1000$$
So,
$$32 \leqslant x<33$$
Hence,
$$x=32,$$
Then,
$$y=768$$
Therefore,
$$n=32768$$ | 32768 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
28 Find the smallest positive integer $n$, such that in decimal notation $n^{3}$ ends with the digits 888. | 28. From the condition, we know $n^{3} \equiv 888(\bmod 1000)$, hence
$$n^{3} \equiv 888(\bmod 8), n^{3} \equiv 888(\bmod 125)$$
From the former, we know $n$ is even, let $n=2 m$, then
$$m^{3} \equiv 111(\bmod 125)$$
Therefore
$$m^{3} \equiv 111 \equiv 1(\bmod 5)$$
Noting that when $m=0,1,2,3,4(\bmod 5)$, correspondingly
$$m^{3} \equiv 0,1,3,2,4(\bmod 5)$$
So, from $m^{3} \equiv 1(\bmod 5)$ we know $m \equiv 1(\bmod 5)$, we can set $m=5 k+1$, at this time
$$m^{3}=(5 k+1)^{3}=125 k^{3}+75 k^{2}+15 k+1 \equiv 111(\bmod 125)$$
Hence
Thus
That is, $\square$
Therefore
This requires
Hence
$$\begin{array}{c}
75 k^{2}+15 k \equiv 110(\bmod 125) \\
15 k^{2}+3 k \equiv 22(\bmod 25) \\
15 k^{2}+3 k+3 \equiv 0(\bmod 25) \\
5 k^{2}+k+1 \equiv 0(\bmod 25) \\
5 k^{2}+k+1 \equiv 0(\bmod 5) \\
5 \mid k+1
\end{array}$$
We can set $k+1=5 l$, then
$$\begin{aligned}
5 k^{2}+k+1 & =5 \times(5 l-1)^{2}+5 l \\
& =125 l^{2}-50 l+5(l+1) \\
& \equiv 0(\bmod 25) \\
& 5 \mid l+1
\end{aligned}$$
We can set $l+1=5 r$, therefore
$$\begin{aligned}
n & =2 m=10 k+2=10(5 l-1)+2 \\
& =50 l-8=50(5 r-1)-8 \\
& =250 r-58
\end{aligned}$$
Combining that $n$ is a positive integer, we know
$$n \geqslant 250-58=192$$
Also, $192^{3}=7077888$ meets the requirement, hence the smallest positive integer satisfying the condition is 192. | 192 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
33 Find the largest positive integer that cannot be expressed as the sum of a positive multiple of 42 and a composite number.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 33. For any positive integer $n$ that cannot be expressed as a positive multiple of 42 and a composite number, consider the remainder $r$ when $n$ is divided by 42. If $r=0$ or $r$ is a composite number, then $n \leqslant 42$.
Now consider the case where $r=1$ or $r$ is a prime number.
If $r \equiv 1(\bmod 5)$, then
$$84+r \equiv 0(\bmod 5)$$
In this case,
$$n<3 \times 42=126$$
If $r \equiv 2(\bmod 5)$, then
$$4 \times 42+r \equiv 0(\bmod 5)$$
In this case,
$$n<5 \times 42=210$$
If $r \equiv 3(\bmod 5)$, then
$$42+r \equiv 0(\bmod 5)$$
In this case,
$$n<2 \times 42=84$$
If $r \equiv 4(\bmod 5)$, then
$$3 \times 42+r \equiv 0(\bmod 5)$$
In this case,
$$n<4 \times 42=168$$
If $r \equiv 0(\bmod 5)$, then
$$r=5,$$
In this case, since 5, 47, 89, 131, and 173 are all prime numbers, $n$ is at most 215.
In summary, the largest positive integer sought is 215. | 215 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
34 Find a positive integer $n$, such that each number $n, n+1, \cdots, n+20$ is not coprime with 30030. | 34. Since $30030=2 \times 3 \times 5 \times 7 \times 11 \times 13$, if we take $N=210 k$, then $N$ and $N \pm r$ are not coprime with 30030, where $r$ is a number in $2,3, \cdots, 10$. Now consider the numbers $N \pm 1$. We choose $k$ such that
$$210 k \equiv 1(\bmod 11) \text { and } 210 k \equiv-1(\bmod 13),$$
The first condition requires $k \equiv 1(\bmod 11)$, set $k=11 m+1$,
The second condition requires
$$210(11 m+1) \equiv-1(\bmod 13)$$
Solving this, we get
$$m \equiv 4(\bmod 13)$$
Therefore, let $k=45$, then the 21 numbers $9440,9441, \cdots, 9460$ are not coprime with 30030, so taking $n=9440$ is sufficient. | 9440 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
37 points are arranged on a circle, and one of the points is labeled with the number 1. Moving clockwise, label the next point with the number 2 after counting two points, then label the next point with the number 3 after counting three points, and so on, labeling the points with the numbers $1,2, \cdots, 2000$. In this way, some points are not labeled, and some points are labeled with more than one number. Question: What is the smallest number among the numbers labeled on the point that is labeled with 2000? | 37. Equivalent to finding the smallest positive integer $n$, such that
$$1+2+\cdots+n \equiv 1+2+\cdots+2000(\bmod 2000)$$
i.e. $\square$
$$\frac{n(n+1)}{2} \equiv 1000(\bmod 2000),$$
which is equivalent to
$$n(n+1) \equiv 2000(\bmod 4000),$$
This requires
$$2000 \mid n(n+1)$$
Notice that
$$(n, n+1)=1$$
and
$$2000=2^{4} \times 5^{3},$$
so $2^{4}\left|n, 5^{3}\right| n+1$; or $5^{3}\left|n, 2^{4}\right| n+1$; or one of $n$ and $n+1$ is a multiple of 2000. Solving these, the smallest $n$ is found to be $624, 1375, 1999$, with the smallest number satisfying (1) being 624. Therefore, the smallest number marked on the point labeled 2000 is 624. | 624 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
38 There are 800 points on a circle, labeled $1,2, \cdots, 800$ in a clockwise direction, dividing the circle into 800 gaps. Now, choose one of these points and color it red, then perform the following operation: if the $k$-th point is colored red, then move $k$ gaps in a clockwise direction and color the point reached red. Question: According to this rule, what is the maximum number of points that can be colored red on the circle? Prove your conclusion. | 38. This is equivalent to finding the maximum number of distinct numbers in the sequence $a, 2a, 2^2a, 2^3a, \cdots$ under modulo 800, where $a$ takes values in $1, 2, \cdots, 800$.
Notice that when $2^n \not\equiv 2^m \pmod{800}$, it is not necessarily true that $2^n a \not\equiv 2^m a \pmod{800}$. Conversely, when $2^n a \not\equiv 2^m a \pmod{800}$ holds, $2^n \equiv 2^m \pmod{800}$ must hold. Therefore, the maximum number of distinct elements in the sequence $a, 2a, 2^2a, \cdots$ under modulo 800 can be achieved when $a=1$. Thus, we only need to find the number of distinct elements in the sequence $1, 2, 2^2, \cdots$ under modulo 800.
Since $800 = 2^5 \times 5^2$, and for $n \geq 5$ we have
$$2^n \equiv 0 \pmod{2^5}$$
Additionally, $\{2^n \pmod{25}\}$ is: $2, 4, 8, 16, 7, 14, 3, 6, 12, -1, -2, -4, -8, -16, -7, -14, -3, -6, -12, 1, \cdots$, so $\{2^n \pmod{25}\}$ contains exactly 20 distinct elements. Combining this with $\{2^n \pmod{2^5}\}$ being $2, 4, 8, 16, 0, 0, \cdots$, we get that $\{2^n \pmod{800}\}$ contains $20 + 4 = 24$ (distinct) numbers.
Therefore, there are at most 24 points on the circle that are colored red. | 24 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the number of positive integer solutions to the indeterminate equation
$$7 x+19 y=2012$$ | Solve: First, find a particular solution of (1).
$$x=\frac{1}{7}(2012-19 y)=287-3 y+\frac{1}{7}(3+2 y) .$$
Therefore, $\frac{1}{7}(3+2 y)$ must be an integer. Taking $y_{0}=2$, then $x_{0}=282$.
Using the conclusion of Theorem 2, the general solution of equation (1) is
$$\left\{\begin{array}{l}
x=282-19 t, \\
y=2+7 t .
\end{array} \text { ( } t \text { is an integer }\right)$$
Combining $x>0, y>0$ and $t$ being an integer, we can solve to get $0 \leqslant t \leqslant 14$.
Therefore, equation (1) has 15 sets of positive integer solutions. | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find the number of positive integer solutions to the indeterminate equation
$$x+2 y+3 z=2012$$ | Let $(x, y, z)$ be a positive integer solution to (1), then $3 z \leqslant 2009$, i.e., $1 \leqslant z \leqslant 669$, which respectively yield
$$x+2 y=2009,2006, \cdots, 5$$
Correspondingly, the range of values for $y$ are
$$\begin{array}{l}
1 \leqslant y \leqslant 1004,1 \leqslant y \leqslant 1002,1 \leqslant y \leqslant 1001 \\
1 \leqslant y \leqslant 999, \cdots, 1 \leqslant y \leqslant 2
\end{array}$$
Since when $y$ and $z$ are determined, the value of $x$ is uniquely determined, the number of positive integer solutions to (1) is
$$\begin{aligned}
& (1004+1002)+(1001+999)+\cdots+(5+3)+2 \\
= & 2006+2000+\cdots+8+2 \\
= & \frac{1}{2}(2006+2) \times 335=336340
\end{aligned}$$
In summary, there are 336340 sets of positive integer solutions. | 336340 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
1 Given that the ages of A, B, and C are all positive integers, A's age does not exceed twice B's age, B is 7 years younger than C, the sum of the three people's ages is a prime number less than 70, and the sum of the digits of this prime number is 13. Question: What is the maximum age of A? | 1. Let Jia's age be $x$ years, and Yi's age be $y$ years, then Bing is $y+7$ years old, and $x \leqslant 2 y$.
Since the positive integers less than 70 and with a digit sum of 13 are only $49$, $58$, and $67$, the sum of the three people's ages (which is a prime number) can only be 67 years, i.e.,
$$x+y+(y+7)=67$$
This gives
$$x+2 y=60 .$$
Combining with
$$x \leqslant 2 y$$
We get
$$4 y \geqslant 60 \text{, i.e., } y \geqslant 15,$$
Thus,
$$x=60-2 y \leqslant 60-30=30,$$
Therefore, Jia is at most 30 years old (Note: Jia, Yi, and Bing being 30 years, 15 years, and 22 years old, respectively, meet the requirements). | 30 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4. Positive integers $a, b, c, d$ satisfy: $1<a<b<c<d<1000$, and $a+d=b+c$, $bc-ad=2004$. Find the number of all such positive integer tuples $(a, b, c, d)$. | 4. Let $b=a+x, c=a+y$, then $x<y$, and $d=a+x+y$ (this is obtained from $a+d=b+c$), thus
$$b c-a d=(a+x)(a+y)-a(a+x+y)=x y$$
That is
$$x y=2004$$
Combining $a+x+y<1000$ and $2004=2^{2} \times 3 \times 167$, we know that $(x, y)=(3,668),(4,501),(6,334),(12,167)$.
Accordingly, $1<a<329,1<a<495,1<a<660,1<a<821$. Thus, the number of qualifying arrays is $327+493+658+819=2297$ (groups). | 2297 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
7 rectangles can be divided into $n$ identical squares, and it can also be divided into $n+76$ identical squares, find the value of the positive integer $n$.
---
Note: The translation keeps the original format and line breaks as requested. However, the phrase "7-个长方形" is translated as "7 rectangles" for clarity in English, assuming it refers to 7 rectangles. If it refers to a single rectangle, the correct translation would be "A rectangle can be divided into $n$ identical squares...". Please clarify if needed. | 7. Let the side length of the squares when divided into $n$ squares be $x$, and the side length when divided into $n+76$ squares be $y$, then
$$n x^{2}=(n+76) y^{2}$$
Since both divisions are performed on the same rectangle, $\frac{x}{y}$ is a rational number (this can be seen by considering one side of the rectangle), thus $\frac{n+76}{n}=\left(\frac{x}{y}\right)^{2}$ is the square of a rational number. Therefore, we can set
$$n=k a^{2}, n+76=k b^{2}$$
where $k$, $a$, and $b$ are all positive integers. Consequently,
that is
$$\begin{array}{c}
k\left(b^{2}-a^{2}\right)=76 \\
k(b-a)(b+a)=76=2^{2} \times 19
\end{array}$$
Noting that $b-a$ and $b+a$ have the same parity, we have
$$(k, b-a, b+a)=(1,2,38),(4,1,19)$$
which gives
$$(k, a, b)=(1,18,20),(4,9,10)$$
The resulting $n$ is 324 in both cases. Therefore, $n=324$. | 324 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8 Positive integers $x, y, z$ satisfy $\left\{\begin{array}{l}7 x^{2}-3 y^{2}+4 z^{2}=8, \\ 16 x^{2}-7 y^{2}+9 z^{2}=-3 .\end{array}\right.$ Find the value of $x^{2}+y^{2}+z^{2}$. | 8. Let $x, y, z$ satisfy
$$\left\{\begin{array}{l}
7 x^{2}-3 y^{2}+4 z^{2}=8 \\
16 x^{2}-7 y^{2}+9 z^{2}=-3
\end{array}\right.$$
Multiplying (1) by 7 and (2) by 3, we get
Substituting back into (1) yields
$$\begin{array}{c}
x^{2}+z^{2}=65 \\
3 x^{2}-3 y^{2}=8-260
\end{array}$$
Thus,
$$\begin{aligned}
y^{2}-x^{2} & =84 \\
(y-x)(y+x) & =2^{2} \times 3 \times 7
\end{aligned}$$
Since $y-x$ and $y+x$ have the same parity, we have
$$(y-x, y+x)=(2,42),(6,14)$$
Solving these, we get
$$(x, y)=(20,22),(4,10)$$
However,
$$x^{2}+z^{2}=65$$
Thus, it can only be
$$(x, y)=(4,10)$$
In this case,
$$z=7$$
Therefore,
$$x^{2}+y^{2}+z^{2}=165$$ | 165 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
15 Positive integers $a, b, c$ satisfy: $[a, b]=1000,[b, c]=2000,[c, a]=2000$. Find the number of such ordered positive integer triples $(a, b, c)$. | 15. From the conditions, we can set
$$a=2^{\alpha_{1}} \cdot 5^{\beta_{1}}, b=2^{\alpha_{2}} \cdot 5^{\beta_{2}}, c=2^{\alpha_{3}} \cdot 5^{\beta_{3}}$$
Then
$$\begin{array}{l}
\max \left\{\alpha_{1}, \alpha_{2}\right\}=3, \max \left\{\alpha_{2}, \alpha_{3}\right\}=4, \max \left\{\alpha_{3}, \alpha_{1}\right\}=4 \\
\max \left\{\beta_{1}, \beta_{2}\right\}=\max \left\{\beta_{2}, \beta_{3}\right\}=\max \left\{\beta_{3}, \beta_{1}\right\}=3
\end{array}$$
Note that, the array $(a, b, c)$ is determined only when the non-negative integer arrays $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ and $\left(\beta_{1}, \beta_{2}, \beta_{3}\right)$ are both determined. From the conditions above, we know that $\alpha_{3}=4$, and at least one of $\alpha_{1}$ or $\alpha_{2}$ is 3, indicating that $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ has 7 possible combinations; at least two of $\beta_{1}$, $\beta_{2}$, and $\beta_{3}$ are equal to 3, so $\left(\beta_{1}, \beta_{2}, \beta_{3}\right)$ has 10 possible combinations.
In summary, the number of ordered arrays $(a, b, c)$ that satisfy the conditions is $7 \times 10 = 70$ (groups). | 70 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
16. Let $x, y$ be positive integers, and $y>3, x^{2}+y^{4}=2\left((x-6)^{2}+(y+1)^{2}\right)$. Prove: $x^{2}+y^{4}=1994$. | 16. Transpose and expand, we get
that is
$$\begin{array}{c}
y^{4}-2 y^{2}-4 y-2=x^{2}-24 x+72 \\
(x-12)^{2}=y^{4}-2 y^{2}-4 y+70
\end{array}$$
Notice that
$$\begin{aligned}
\left(y^{2}-2\right)^{2} & =y^{4}-4 y^{2}+4 \\
& 3 \text{ holds, and } y^{4}-2 y^{2}-4 y+70 \text{ is a perfect square, so it can only be or }
$$\begin{array}{c}
y^{4}-2 y^{2}-4 y+70=\left(y^{2}-1\right)^{2} \\
y^{4}-2 y^{2}-4 y+70=\left(y^{2}\right)^{2}
\end{array}$$
The former has no positive integer solutions, while the latter has a unique positive integer solution \( y=5 \). In this case, we require \((x-12)^{2}=5^{4}\), and the positive integer \( x=37 \). Therefore, \( x^{2}+y^{4}=37^{2}+5^{4}=1994 \). | 1994 | Algebra | proof | Yes | Yes | number_theory | false |
30 Given that the three sides of $\triangle A B C$ are all integers, $\angle A=2 \angle B, \angle C>90^{\circ}$. Find the minimum perimeter of $\triangle A B C$. | 30. Let the corresponding side lengths of $\triangle ABC$ be $a, b, c$. Draw the angle bisector of $\angle A$ intersecting $BC$ at point $D$, then
$$C D=\frac{a b}{b+c},$$
Using $\triangle A C D \backsim \triangle B C A$, we know
$$\begin{array}{c}
\frac{C D}{b}=\frac{b}{a} \\
a^{2}=b(b+c)
\end{array}$$
That is $\square$
And
Therefore
By
Let
$$\begin{array}{c}
\angle C>90^{\circ} \\
c^{2}>a^{2}+b^{2} \\
a^{2}=b(b+c) \\
(b, b+c)=d
\end{array}$$
Then $(b, c)=d$, and $d^{2} \mid a^{2}$, hence $d \mid a$.
To find the minimum value of $a+b+c$, we can set $d=1$, in which case both $b$ and $b+c$ are perfect squares. Let
$$b=m^{2}, b+c=n^{2}, m, n \in \mathbf{N}^{*}$$
Then $a=m n$. Using $a+b>c$ and $c^{2}>a^{2}+b^{2}$, we know
And $\square$
$$\begin{array}{c}
m n+m^{2}>n^{2}-m^{2} \\
\left(n^{2}-m^{2}\right)^{2}>(m n)^{2}+m^{4}
\end{array}$$
Thus
$$m>n-m$$
That is $\square$
$$n3 m^{2} n^{2},$$
That is $\square$
$$n^{2}>3 m^{2},$$
So $\square$
$$3 m^{2}<n^{2}<4 m^{2}$$
Thus, there is a perfect square between $3 m^{2}$ and $4 m^{2}$, which requires $m \geqslant 4$, at this time $n \geqslant 7$, hence
$$a+b+c \geqslant 4 \times 7+7^{2}=77$$
Clearly, $(a, b, c)=(28,16,33)$ satisfies the conditions, so the minimum perimeter of $\triangle A B C$ is 77. | 77 | Geometry | math-word-problem | Yes | Yes | number_theory | false |
Example 1 There are 2012 lamps, numbered $1, 2, \cdots, 2012$, arranged in a row in a corridor, and initially, each lamp is on. A mischievous student performed the following 2012 operations: for $1 \leqslant k \leqslant 2012$, during the $k$-th operation, the student toggled the switch of all lamps whose numbers are multiples of $k$. Question: How many lamps are still on at the end? | Let $1 \leqslant n \leqslant 2012$, we examine the state of the $n$-th lamp. According to the problem, the switch of this lamp is pulled $d(n)$ times. An even number of pulls does not change the initial state of the lamp, while an odd number of pulls changes the state of the lamp from its initial state.
Using the properties of $d(n)$ and the previous discussion, since there are exactly 44 perfect squares among the numbers $1,2, \cdots, 2012$, it follows that finally, $2012-44=1968$ lamps are still on. | 1968 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
11 Given a positive integer $n$, among its positive divisors, there is at least one positive integer ending in each of the digits $0,1,2, \cdots, 9$. Find the smallest $n$ that satisfies this condition. | 11. The smallest $n$ that satisfies the condition is $270$.
In fact, from the condition, we know that $10 \mid n$, and we start the discussion from the factor of the last digit of $n$ being 9. If $9 \mid n$, then $90 \mid n$, and it can be directly verified that 90 and 180 are not multiples of any number ending in 7; if $19 \mid n$, then $190 \mid n$, and $n=190$ is not a multiple of any number ending in 7. However, 270 is a multiple of $10, 1, 2, 3, 54, 5, 6, 27, 18, 9$, which meets the conditions. Therefore, the smallest $n$ is 270. | 270 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12 Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the left $m$ digits of $M$ are multiples of $m$.
Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the number formed by the left $m$ digits of $M$ is a multiple of $m$. | 12. Let $M=\overline{a_{1} a_{2} \cdots a_{9}}$ be a number that satisfies the given conditions. From the conditions, we know that $a_{5}=5$, and $a_{2}$, $a_{4}$, $a_{6}$, $a_{8}$ are a permutation of $2$, $4$, $6$, $8$. Consequently, $a_{1}$, $a_{3}$, $a_{7}$, $a_{9}$ are a permutation of $1$, $3$, $7$, $9$. Therefore,
$$a_{4}=2 \text { or } 6\left(\text { because } 4 \mid \overline{a_{3} a_{4}}\right),$$
Furthermore,
$$8 \mid \overline{a_{7} a_{8}}$$
Thus,
$$a_{8}=2,6$$
Hence,
$$\left(a_{4}, a_{8}\right)=(2,6),(6,2)$$
By analyzing these two cases further, we can find a number $M=381654729$ that satisfies the conditions. | 381654729 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
17 Let $a, b, c, d$ all be prime numbers, and $a>3b>6c>12d, a^{2}-b^{2}+c^{2}-d^{2}=1749$. Find all possible values of $a^{2}+b^{2}+c^{2}+d^{2}$. | 17. From $a^{2}-b^{2}+c^{2}-d^{2}=1749$ being odd, we know that one of $a, b, c, d$ must be even, indicating that $d=2$. Then,
from
$$\begin{array}{l}
a^{2}-b^{2}+c^{2}=1753 \\
a>3 b>6 c>12 d
\end{array}$$
we know $c \geqslant 5, b \geqslant 2 c+1, a \geqslant 3 b+1$, so
$$\begin{aligned}
a^{2}-b^{2}+c^{2} & \geqslant(3 b+1)^{2}-b^{2}+c^{2} \\
& =8 b^{2}+6 b+c^{2}+1 \\
& \geqslant 8(2 c+1)^{2}+6(2 c+1)+c^{2}+1 \\
& =33 c^{2}+44 c+15
\end{aligned}$$
Thus,
$$33 c^{2}+44 c+15 \leqslant 1753$$
Therefore, $c<7$, and combining $c \geqslant 5$ and $c$ being a prime number, we have $c=5$, leading to
$$a^{2}-b^{2}=1728=2^{6} \times 3^{3}$$
Using
$$b \geqslant 2 c+1=11, a \geqslant 3 b+1$$
we know
$$a-b \geqslant 2 b+1 \geqslant 23, a+b \geqslant 4 b+1 \geqslant 45$$
From $(a-b)(a+b)=2^{6} \times 3^{3}$ and $a, b$ both being odd primes, we have
$$(a-b, a+b)=(32,54)$$
Thus,
$$\begin{array}{c}
(a, b)=(43,11) \\
a^{2}+b^{2}+c^{2}+d^{2}=1749+2 \times\left(11^{2}+2^{2}\right)=1999
\end{array}$$ | 1999 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
19 The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integers $m, n$, if $m \mid n, m<n$, then $a_{m} \mid a_{n}$, and $a_{m}<a_{n}$. Find the minimum possible value of $a_{2000}$. | 19. From the conditions, when $m \mid n$ and $m < n$, we have $a_{n} \geqslant 2 a_{m}$. Therefore, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{4} \geqslant 2 a_{2} \geqslant 2^{2}$, similarly, $a_{8} \geqslant 2^{3}, a_{16} \geqslant 2^{4}, a_{30} \geqslant 2^{5}, a_{400} \geqslant 2^{6}, a_{2000} \geqslant 2^{7}$, which means $a_{2000} \geqslant 128$.
On the other hand, for any positive integer $n$, let the prime factorization of $n$ be
$$n=p_{1}^{q_{1}} p_{2}^{q_{2}} \cdots p_{k^{\prime}}^{q_{k}}$$
where $p_{1}<p_{2}<\cdots<p_{k}$ are prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are positive integers. Define
$$a_{n}=2^{a_{1}+a_{2}+\cdots+a_{k}}$$
Then the sequence $\left\{a_{n}\right\}$ satisfies the requirements of the problem, and
$$a_{2000}=2^{4+3} \leqslant 2^{7}$$
Therefore, the minimum value of $a_{2000}$ is 128. | 128 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
32 Given the pair of positive integers $(a, b)$ satisfies: the number $a^{a} \cdot b^{b}$ in decimal notation ends with exactly 98 zeros. Find the minimum value of $a b$.
| 32. Let the prime factorization of $a$ and $b$ have the powers of $2$ and $5$ as $\alpha_{1}$, $\beta_{1}$ and $\alpha_{2}$, $\beta_{2}$, respectively, then
$$\left\{\begin{array}{l}
a \cdot \alpha_{1} + b \cdot \alpha_{2} \geqslant 98 \\
a \cdot \beta_{1} + b \cdot \beta_{2} \geqslant 98
\end{array}\right.$$
and one of (1) and (2) must be an equality.
If (2) is an equality, i.e., $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$, then when $\beta_{1}$ and $\beta_{2}$ are both positive integers, the left side is a multiple of 5. When $\beta_{1}$ or $\beta_{2}$ is zero, the other must be greater than zero, in which case the left side is still a multiple of 5, leading to a contradiction. Therefore, (1) must be an equality.
From $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$, if $\alpha_{1}$ or $\alpha_{2}$ is zero, without loss of generality, let $\alpha_{2} = 0$, then $\alpha_{1} > 0$. In this case, $a \cdot \alpha_{1} = 98$. If $\alpha_{1} \geqslant 2$, then $4 \mid a$, which is a contradiction. Hence, $\alpha_{1} = 1$, and thus $a = 98$. Substituting $a = 98$ into (2), we know $\beta_{1} = 0$, so $b \cdot \beta_{2} > 98$. Combining $\alpha_{2} = 0$, we find that the minimum value of $b$ is 75.
If $\alpha_{1}$ and $\alpha_{2}$ are both positive integers, without loss of generality, let $\alpha_{1} \geqslant \alpha_{2}$. If $\alpha_{2} \geqslant 2$, then $4 \mid a$ and $4 \mid b$, leading to $4 \mid 98$, which is a contradiction. Hence, $\alpha_{2} = 1$. Further, if $\alpha_{1} = 1$, then $a + b = 98$, but $\frac{a}{2}$ and $\frac{b}{2}$ are both odd, so $\frac{a}{2} + \frac{b}{2}$ is even, which is a contradiction. Therefore, $\alpha_{1} > 1$. In this case, if $\beta_{1}$ and $\beta_{2}$ are both positive integers, then $5 \mid a$ and $5 \mid b$, which contradicts $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$. Hence, one of $\beta_{1}$ and $\beta_{2}$ must be zero. If $\beta_{1} = 0$, then from (2) we know $b \cdot \beta_{2} > 98$, in which case the number of trailing zeros in $b^b$ is greater than 98 (since, in this case, $10 \mid b$. When $\beta_{2} = 1$, $b \geqslant 100$, so $10^{100} \mid b^b$. When $\beta_{2} \geqslant 2$, $50 \mid b$, if $b > 50$, then $10^{100} \mid b^b$; if $b = 50$, then $a \cdot \alpha_{1} = 48$, in which case when $\alpha_{1} \geqslant 4$, $2^5 \mid a \cdot \alpha_{1}$, and when $\alpha_{1} \leqslant 3$, $2^4 \nmid a \cdot \alpha_{1}$, both leading to a contradiction, so the number of trailing zeros in $b^b$ is greater than 98).
Similarly, if $\beta_{2} = 0$, then $a \cdot \beta_{1} > 98$, and similarly, the number of trailing zeros in $a^a$ is greater than 98, which is a contradiction. In summary, the minimum value of $ab$ is 7350 (when $(a, b) = (98, 75)$ or $(75, 98)$). | 7350 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 (1) Let $m, n$ be coprime positive integers, $m, n>1$. Let $a$ be an integer coprime to $m n$. Suppose the orders of $a$ modulo $m$ and modulo $n$ are $d_{1}, d_{2}$, respectively, then the order of $a$ modulo $m n$ is $\left[d_{1}, d_{2}\right]$;
(2) Find the order of 3 modulo $10^{4}$. | 2. (1) Let the order of $a$ modulo $mn$ be $r$. From $a^{r} \equiv 1(\bmod m n)$, we can deduce $a^{r} \equiv 1(\bmod m)$ and $a^{r} \equiv 1(\bmod n)$. Therefore, $d_{1} \mid r$ and $d_{2} \mid r$, which implies $\left[d_{1}, d_{2}\right] \mid r$. On the other hand, from $a^{d_{1}} \equiv 1(\bmod m)$ and $a^{d_{2}} \equiv 1(\bmod n)$, we can infer $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod m)$ and $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod n)$. Since $(m, n)=1$, it follows that $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod m n)$, thus $r \mid \left[d_{1}, d_{2}\right]$. Combining both results, we conclude $r=\left[d_{1}, d_{2}\right]$.
(2) Direct calculation shows that the order of 3 modulo $2^{4}$ is 4. It is also easy to see that the order of 3 modulo 5 is 4, hence by part (1) of Example 5, the order of 3 modulo $5^{4}$ is $4 \times 5^{3}$. Therefore, by part (1) of this problem, the order of 3 modulo $10^{4}$ is $[4, 4 \times 5^{3}]=500$. | 500 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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