problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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23 Let $k$ and $m$ be positive integers. Find the minimum possible value of $\left|36^{k}-5^{m}\right|$. | 23. Notice that
$$36^{k}-5^{m} \equiv \pm 1(\bmod 6)$$
and
$$36^{k}-5^{m} \equiv 1(\bmod 5)$$
Therefore, the positive integers that $36^{k}-5^{m}$ can take, in ascending order, are $1, 11, \cdots$.
If $36^{k}-5^{m}=1$, then $k>1$, so taking both sides modulo 8, we require
$$5^{m} \equiv-1(\bmod 8)$$
However, $5^{m} ... | 11 | Number Theory | math-word-problem | 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 $\squ... | 77 | Geometry | math-word-problem | Yes | Yes | number_theory | false |
40 Find the smallest positive integer $n$ such that the indeterminate equation
$$n=x^{3}-x^{2} y+y^{2}+x-y$$
has no positive integer solutions. | 40. Let $F(x, y)=x^{3}-x^{2} y+y^{2}+x-y$, then $F(1,1)=1, F(1,2)=2$.
Therefore, when $n=1,2$, the equation has positive integer solutions.
Next, we prove that $F(x, y)=3$ has no positive integer solutions.
Consider the equation $F(x, y)=3$ as a quadratic equation in $y$
$$y^{2}-\left(x^{2}+1\right) y+x^{3}+x-3=0$$
I... | 3 | Number Theory | 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 mu... | 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 prop... | 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; ... | 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 $m$ dig... | 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 $1$, $3$, $7$, $9$. Therefor... | 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
$$\begin{aligned}
a^{2}-b^{2}+c^{2} & \geqslan... | 1999 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
19 The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integers $m, n$, if $m \mid n, m<n$, then $a_{m} \mid a_{n}$, and $a_{m}<a_{n}$. Find the minimum possible value of $a_{2000}$. | 19. From the conditions, when $m \mid n$ and $m < n$, we have $a_{n} \geqslant 2 a_{m}$. Therefore, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{4} \geqslant 2 a_{2} \geqslant 2^{2}$, similarly, $a_{8} \geqslant 2^{3}, a_{16} \geqslant 2^{4}, a_{30} \geqslant 2^{5}, a_{400} \geqslant 2^{6}, a_{2000} \geqslant 2^{7}$, whic... | 128 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
22 Prove: For any positive integer $n$ and positive odd integer $m$, we have $\left(2^{m}-1,2^{n}+1\right)=1$.
| 22. Let $d=\left(2^{m}-1,2^{n}+1\right)$, then
hence
that is
$$\begin{array}{c}
d \mid 2^{m}-1 \\
d \mid\left(2^{m}\right)^{n}-1^{n} \\
d \mid 2^{n}-1
\end{array}$$
Additionally, $d \mid 2^{n}+1$, and since $m$ is odd, we have
$$2^{n}+1 \mid\left(2^{n}\right)^{m}+1^{m},$$
thus
$$d \mid 2^{m}+1$$
Comparing the two de... | 1 | Number Theory | proof | Yes | Yes | number_theory | false |
32 Given the pair of positive integers $(a, b)$ satisfies: the number $a^{a} \cdot b^{b}$ in decimal notation ends with exactly 98 zeros. Find the minimum value of $a b$.
| 32. Let the prime factorization of $a$ and $b$ have the powers of $2$ and $5$ as $\alpha_{1}$, $\beta_{1}$ and $\alpha_{2}$, $\beta_{2}$, respectively, then
$$\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.$$
and one ... | 7350 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
39 The numbers $1,2, \cdots, 33$ are written on the blackboard. Each time, it is allowed to perform the following operation: take any two numbers $x, y$ from the blackboard that satisfy $x \mid y$, remove them from the blackboard, and write the number $\frac{y}{x}$. Continue until there are no such two numbers on the b... | 39. Consider the objective function $S=$ the product of all numbers on the blackboard.
Initially, $S=33!=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$. Each operation targets $x, y (x \mid y)$, where $y=kx$. Removing $x, y$ and replacing them with $... | 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 (1) Let $m, n$ be coprime positive integers, $m, n>1$. Let $a$ be an integer coprime to $m n$. Suppose the orders of $a$ modulo $m$ and modulo $n$ are $d_{1}, d_{2}$, respectively, then the order of $a$ modulo $m n$ is $\left[d_{1}, d_{2}\right]$;
(2) Find the order of 3 modulo $10^{4}$. | 2. (1) Let the order of $a$ modulo $mn$ be $r$. From $a^{r} \equiv 1(\bmod m n)$, we can deduce $a^{r} \equiv 1(\bmod m)$ and $a^{r} \equiv 1(\bmod n)$. Therefore, $d_{1} \mid r$ and $d_{2} \mid r$, which implies $\left[d_{1}, d_{2}\right] \mid r$. On the other hand, from $a^{d_{1}} \equiv 1(\bmod m)$ and $a^{d_{2}} \e... | 500 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 A positive integer, when added to 100, becomes a perfect square. If 168 is added to it, it becomes another perfect square. Find this number.
| Let the number to be found be $x$. By the problem, there exist positive integers $y, z$, such that
$$\left\{\begin{array}{l}
x+100=y^{2} \\
x+168=z^{2}
\end{array}\right.$$
Eliminating $x$ from the above two equations, we get
$$z^{2}-y^{2}=68$$
Factoring the left side of this binary quadratic equation and standardizi... | 156 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Let $m>n \geqslant 1$, find the minimum value of $m+n$ such that $: 1000 \mid 1978^{m}-1978^{n}$. | 3. Solution: When $n \geqslant 3$, from $1000 \mid 1978^{m}-1978^{n}$, we get
$125 \mid 1978^{m-n}-1$, and since $1978=15 \times 125+103$,
we have $125 \mid 103^{m-n}-1$, so $25 \mid 103^{m-n}-1$.
Thus, $25 \mid 3^{m-n}-1$.
Let $m-n=l$, then: $25 \mid 3^{l}-1$ (obviously $l$ is even, otherwise it is easy to see that $3... | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
14. Let $S$ be the set of all prime numbers $p$ that satisfy the following condition: the number of digits in the smallest repeating block of the decimal part of $\frac{1}{p}$ is a multiple of 3, i.e., for each $p \in S$, there exists the smallest positive integer $r=r(p)$, such that $\frac{1}{p}=0 . a_{1} a_{2} \cdots... | 14. Prove: (1) The length of the smallest repeating cycle of $\frac{1}{p}$ is the smallest integer $d (d \geqslant 1)$ such that $10^{d}-1$ is divisible by $p$.
Let $q$ be a prime number, and $N_{q}=10^{2 q}+10^{q}+1$. Then $N_{q} \equiv 3(\bmod q)$. Let $p_{q}$ be a prime factor of $\frac{N_{q}}{3}$. Then $p_{q}$ can... | 19 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 5 (2005 National High School Mathematics Competition Question) Define the function
$$f(k)=\left\{\begin{array}{l}
0, \text { if } k \text { is a perfect square } \\
{\left[\frac{1}{\{\sqrt{k}\}}\right], \text { if } k \text { is not a perfect square }}
\end{array} \text {, find } \sum_{k=1}^{240} f(k)\right. \t... | When $k$ is not a perfect square, $k$ must lie between two consecutive perfect squares. Divide $[1,240]$ into several intervals with perfect squares as boundaries, and sum $f(k)$ for each subinterval.
Solving $15^{2}<240<16^{2}$.
Since $f(k)=0$ when $k$ is a perfect square, we have
$$\begin{aligned}
\sum_{k=1}^{240} f... | 768 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
For non-negative integers $x$, the function $f(x)$ is defined as follows:
$$f(0)=0, f(x)=f\left(\left[\frac{x}{10}\right]\right)+\left[\lg \frac{10}{x-10\left[\frac{x-1}{10}\right]}\right]$$
What is the value of $x$ when $f(x)$ reaches its maximum in the range $0 \leqslant x \leqslant 2006$? | 1. Solution: Let $x=10 p+q$, where $p$ and $q$ are integers and $0 \leqslant q \leqslant 9$.
Then $\left[\frac{x}{10}\right]=\left[p+\frac{q}{10}\right]=p$.
Thus, $\left[\frac{x-1}{10}\right]=\left[p+\frac{q-1}{10}\right]$ has a value of $p-1$ (when $q=0$) or $p$ (when $q \neq 0$), and $x-10\left[\frac{x-1}{10}\right]... | 1111 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4. Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$, prove that: $k=5$. | 4. Prove: If $a=b$, then $k=\frac{2 a^{2}}{a^{2}-1}=2+\frac{2}{a^{2}-1}$.
Then $\left(a^{2}-1\right) \mid 2$ and $a^{2}-1>0$.
So $a^{2}-1=1$ or 2, but this contradicts $a \in \mathbf{N}_{+}$.
Therefore, $a \neq b$, without loss of generality, let $a>b$.
When $b=1$, $k=\frac{a^{2}+1}{a-1}=a+1+\frac{2}{a-1}$, then $(a-1... | 5 | Number Theory | proof | Yes | Yes | number_theory | false |
6. In the Cartesian coordinate system, the number of integer points $(x, y)$ that satisfy $(|x|-1)^{2}+(|y|-1)^{2}<2$ is $\qquad$ . | 6. 16 Since $(|x|-1)^{2} \leqslant(|x|-1)^{2}+(|y|-1)^{2}<2$.
Therefore, $(|x|-1)^{2}<2, -1 \leqslant|x|-1 \leqslant 1$, which means $0 \leqslant|x| \leqslant 2$, similarly $0 \leqslant|y| \leqslant 2$.
Upon inspection, $(x, y)=(-1, \pm 1),(-1,0),(1,0),(0, \pm 1),(-1, \pm 2),(1, \pm 2),(-2, \pm 1),(2, \pm 1)$ are the ... | 16 | Geometry | math-word-problem | Yes | Yes | number_theory | false |
Example 3 In the Cartesian coordinate system, the number of integer points that satisfy $(1) y \geqslant 3 x$; (2) $y \geqslant \frac{1}{3} x ;(3) x+y \leqslant 100$ is how many? | As shown in Figure 7-5, the region enclosed by the lines $y=3x$, $y=\frac{1}{3}x$, and $x+y=100$ forms a triangular region. The three vertices of this triangle are $O(0,0)$, $A(75,25)$, and $B(25,75)$. $\square$
Next, we calculate the number of integer points $N$ on the boundary and inside $\triangle OAC$. For a grid ... | 2551 | Inequalities | math-word-problem | Yes | Yes | number_theory | false |
5. In $1 \sim 1000$, the number of pairs $(x, y)$ that make $\frac{x^{2}+y^{2}}{7}$ an integer is $\qquad$ pairs. | 5. 10011 From $\frac{x^{2}+y^{2}}{7} \in \mathbf{Z}$ we know $7 \mid x^{2}+y^{2}$.
When $7 \mid x$ and $7 \mid y$, it is obvious that $7 \mid x^{2}+y^{2}$.
When $7 \nmid x$ or $7 \nmid y$, from $7 \mid x^{2}+y^{2}$ we know that $7 \nmid x$ and $7 \nmid y$. By Fermat's Little Theorem, $x^{6} \equiv 1(\bmod 7)$, which m... | 10011 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6. Let the planar region $T=\{(x, y) \mid x>0, y>0, x y \leqslant 48\}$, then the number of lattice points within $T$ is $\qquad$ . | $$\begin{array}{l}
\text { 6. } 202 \text { Let } T_{1}=\left\{(x, y) \mid 0<x \leqslant \sqrt{48}, 0<y \leqslant \frac{n}{x}\right\}, \\
T_{2}=\left\{(x, y) \mid 0<y \leqslant \sqrt{48}, 0<x \leqslant \frac{n}{y}\right\} .
\end{array}$$
Then $T=T_{1} \cup T_{2}, T_{1} \cap T_{2}=\{(x, y) \mid 0<x \leqslant \sqrt{48},... | 202 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 7 How many positive integer factors does 20! have? | Analyze writing 20! in its standard factorization form $n=\beta_{1}^{a_{1}} \beta_{2}^{a_{2}} \cdots \beta_{k}^{q_{k}}$, and then using $r(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots$ $\left(\alpha_{k}+1\right)$ to calculate the result.
Solution Since the prime numbers less than 20 are $2,3,5,7,11,13,1... | 41040 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 (2001 Irish Mathematical Olympiad) Find the smallest positive integer $a$ such that there exists a positive odd integer $n$ satisfying $2001 \mid$
$$55^{n}+a \cdot 32^{n}$$ | Analysis Using the properties of congruence, noting that $2001=23 \times 87$, we can find that $a \equiv 1(\bmod 87)$ and $a \equiv-1$ $(\bmod 23)$. Thus, we can obtain the smallest value of $a$ that meets the requirements.
Solution Since $2001=87 \times 23$. By the problem, there exists a positive odd number $n$ such... | 436 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
$$\text { 9. } 3333^{8888}+8888^{3333} \equiv$$
$$\qquad (\bmod 7)$$. | 9.0 Hint: $3333 \equiv 1(\bmod 7)$, so $3333^{8888} \equiv 1(\bmod 7)$. $8888 \equiv 5(\bmod 7)$, so $8888^{3} \equiv 5^{3}(\bmod 7) \equiv-1(\bmod 7)$. Therefore, $8888^{3333} \equiv-1(\bmod 7)$. | 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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