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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 standardizing the right side, we have
$$(z-y)(z+y)=2^{2} \times 17$$
Since $z-y$ and $z+y$ are both positive integers, and $z-y < z+y$, it follows from (1) and the unique factorization theorem (Unit 3 (5)) that we must have
$$\left\{\begin{array} { l }
{ z - y = 1 , } \\
{ z + y = 2 ^ { 2 } \times 1 7 ; }
\end{array} \left\{\begin{array}{l}
z-y=2, \\
z+y=2 \times 17 ;
\end{array} ;\left\{\begin{array}{l}
z-y=2^{2}, \\
z+y=17
\end{array}\right.\right.\right.$$
Solving these systems of linear equations one by one, we find $y=16, z=18$, hence $x=156$. | 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 $3^{l}-1 \equiv \pm 3-1 \equiv 4$ or $2(\bmod 5)$, which is a contradiction), let $l=2a$,
then $25 \mid 9^{a}-1$, and $9^{a}-1=(10-1)^{a}-1$
$$\begin{array}{l}
\equiv C_{a}^{2} 10^{2}(-1)^{a-2}+C_{a}^{1} 10(-1)^{a-1}+(-1)^{a}-1(\bmod 25) \\
\equiv 10 a(-1)^{a-1}+(-1)^{a}-1(\bmod 25)
\end{array}$$
Obviously, $a$ is even.
So $9^{a}-1 \equiv 10 a(-1)(\bmod 25)$.
Thus, $5 \mid a$, and since $a$ is even, $10 \mid a$.
Let $a=10b$, then $l=20b$.
So $103^{t}-1=(100+3)^{20b}-1$
$$\begin{array}{l}
\equiv C_{20b}^{1} 100 \cdot 3^{20b-1}+3^{20b}-1(\bmod 125) \\
\equiv 9^{10b}-1(\bmod 125) \\
\equiv(10-1)^{10b}-1(\bmod 125) \\
\equiv C_{10b}^{2} \cdot 10^{2}-C_{10b}^{1} 10(\bmod 125) \\
\equiv-100b(\bmod 125)
\end{array}$$
So $5 \mid b$, thus $l=20b \geqslant 100$. Therefore, $m+n=l+2n \geqslant 106$.
When $m=103, n=3$, the above equality holds.
If $n \leqslant 2$, it is easy to see that $8 \times 1978^{m}-1978^{n} \Rightarrow 1000 \times 1978^{m}-1978^{n}$, so the minimum value of $m+n$ is 106. | 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 $\frac{1}{p}=0 . a_{1} a_{2} \cdots a_{3 r} a_{1} a_{2} \cdots a_{3 r} \cdots$
For each $p \in S$ and any integer $k \geqslant 1$, define $f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2 r(p)}$.
(1) Prove that the set $S$ contains infinitely many primes;
(2) For $k \geqslant 1$ and $p \in S$, find the maximum value of $f(k, p)$. | 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 $p_{q}$ cannot be divisible by 3, because $N_{q}$ is a factor of $10^{3 q}-1$. The repeating cycle length of $\frac{1}{p_{q}}$ is $3 q$, so the length of the smallest repeating cycle of $\frac{1}{p_{q}}$ is a divisor of $3 q$. If the smallest repeating cycle length is $q$, then from $10^{q} \equiv 1\left(\bmod p_{q}\right)$ we get $N_{q}=10^{2 q}+10^{q}+1 \left(\bmod p_{q}\right)$, which is a contradiction! If the smallest repeating cycle length is 3, there is only one case, i.e., $p_{q}$ is a factor of $10^{3}-1=3^{3} \times 37$, i.e., $p_{q}=37$. In this case, $N_{q}=3 \times 37 \equiv 3(\bmod 4)$, while $N_{q}=10^{2 q}+10^{q}+1 \equiv 1(\bmod 4)$, which is a contradiction! Thus, for each prime $q$, we can find a prime $p_{q}$ such that the length of the smallest repeating cycle of the decimal part of $\frac{1}{p_{q}}$ is $3 q$.
(2) Let the prime $p \in S$, and $3 r(p)$ be the length of the smallest repeating cycle of $\frac{1}{p}$. Then $p$ is a factor of $10^{3 r(p)}-1$ but not a factor of $10^{r(p)}-1$, so $p$ is a factor of $N_{r}=10^{2 r(p)}+10^{r(p)}+1$.
Let $\frac{1}{p}=0 . \overline{a_{1} a_{2} \cdots}, x_{j}=\frac{10^{j-1}}{p}, y_{j}=\left\{x_{j}\right\}=0 . \overline{a_{j} a_{j+1} \cdots}$, where $\{x\}$ denotes the fractional part of $x$, then $a_{j}<10 y_{j}$. Thus, $f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2 r(p)}<10\left(y_{k}+y_{k+r(p)}+y_{k+2 r(p)}\right)$. Since $x_{k}+x_{k+r(p)}+x_{k+2 r(p)}=\frac{10^{k-1} \cdot N_{r(p)}}{p}$ is an integer, $y_{k}+y_{k+r(p)}+y_{k+2 r(p)}$ is also an integer and is a number less than 3, so $y_{k}+y_{k+r(p)}+y_{k+2 r(p)} \leqslant 2$, thus $f(k, p)<20$, so $f(k, p) \leqslant 19$. Since $f(2,7)=4+8+7=19$, the maximum value sought is 19. | 19 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 5 (2005 National High School Mathematics Competition Question) Define the function
$$f(k)=\left\{\begin{array}{l}
0, \text { if } k \text { is a perfect square } \\
{\left[\frac{1}{\{\sqrt{k}\}}\right], \text { if } k \text { is not a perfect square }}
\end{array} \text {, find } \sum_{k=1}^{240} f(k)\right. \text {. }$$ | 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
$$\begin{aligned}
\sum_{k=1}^{240} f(k) & =\sum_{k=1+1}^{2^{2}-1} f(k)+\sum_{k=2^{2}+1}^{3^{2}-1} f(k)+\cdots+\sum_{k=14^{2}+1}^{15^{2}-1} f(k)+\sum_{k=226}^{240} f(k) \\
& =\sum_{n=1}^{14} \sum_{k=n^{2}+1}^{(n+1)^{2}-1} f(k)+\sum_{k=226}^{240} f(k) .
\end{aligned}$$
When $n^{2}+1 \leqslant k \leqslant(n+1)^{2}-1$, $[\sqrt{k}]=n,\{k\}=\sqrt{k}-[k]=\sqrt{k}-n$.
Thus, $\left[\frac{1}{\{\sqrt{k}\}}\right]=\left[\frac{1}{\sqrt{k}-n}\right]=\left[\frac{\sqrt{k}+n}{k-n^{2}}\right]=\left[\frac{[\sqrt{k}+n]}{k-n^{2}}\right]=\left[\frac{2 n}{k-n^{2}}\right]$.
So $\sum_{k=n^{2}+1}^{(n+1)^{2}-1} f(k)=\sum_{k=n^{2}+1}^{(n+1)^{2}-1}\left[\frac{2 n}{k-n^{2}}\right]=\sum_{i=1}^{2 n}\left[\frac{2 n}{i}\right]$,
$$\begin{aligned}
\sum_{k=226}^{240} f(k) & =\sum_{k=226}^{240}\left[\frac{2 \cdot 15}{k-15^{2}}\right]=\sum_{i=1}^{15}\left[\frac{30}{i}\right] . \\
\text { Therefore, } \sum_{k=1}^{240} f(k) & =\sum_{n=1}^{14} \sum_{i=1}^{2 n}\left[\frac{2 n}{i}\right]+\sum_{i=1}^{15}\left[\frac{30}{i}\right] \\
& =\sum_{n=1}^{14}\left(\sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+\sum_{i=n+1}^{2 n}\left[\frac{2 n}{i}\right]\right)+\sum_{i=1}^{15}\left[\frac{30}{i}\right] \\
& =\sum_{n=1}^{14}\left(\sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+n\right)+\sum_{i=1}^{15}\left[\frac{30}{i}\right] \\
& =\sum_{n=1}^{14} \sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+\sum_{i=1}^{15}\left[\frac{30}{i}\right]+\sum_{n=1}^{14} n \\
& =\sum_{n=1}^{15} \sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+105 .
\end{aligned}$$
Let $T_{n}=\sum_{i=1}^{n}\left[\frac{2 n}{i}\right], 1 \leqslant n \leqslant 15$.
$$\begin{array}{l}
\text { Then } T_{1}=\left[\frac{2}{1}\right]=2, T_{2}=\left[\frac{4}{1}\right]+\left[\frac{4}{2}\right]=4+2=6, \\
T_{3}=\left[\frac{6}{1}\right]+\left[\frac{6}{2}\right]+\left[\frac{6}{3}\right]=6+3+2=11, \\
T_{4}=\left[\frac{8}{1}\right]+\left[\frac{8}{2}\right]+\left[\frac{8}{3}\right]+\left[\frac{8}{4}\right]=8+4+2 \times 2=16, \\
T_{5}=\left[\frac{10}{1}\right]+\left[\frac{10}{2}\right]+\cdots+\left[\frac{10}{5}\right]=10+5+3+2 \times 2=22, \\
T_{6}=\left[\frac{12}{1}\right]+\left[\frac{12}{2}\right]+\cdots+\left[\frac{12}{6}\right]=12+6+4+3+2 \times 2=29, \\
T_{7}=\left[\frac{14}{1}\right]+\left[\frac{14}{2}\right]+\cdots+\left[\frac{14}{7}\right]=14+7+4+3+2 \times 3=34, \\
T_{8}=\left[\frac{16}{1}\right]+\left[\frac{16}{2}\right]+\cdots+\left[\frac{16}{8}\right]=16+8+5+4+3+2 \times 3=42, \\
T_{9}=\left[\frac{18}{1}\right]+\left[\frac{18}{2}\right]+\cdots+\left[\frac{18}{9}\right]=18+9+6+4+3 \times 2+2 \times 3=49, \\
T_{10}=\left[\frac{20}{1}\right]+\left[\frac{20}{2}\right]+\cdots+\left[\frac{20}{10}\right]=20+10+6+5+4+3+2 \times 4=56, \\
T_{11}=\left[\frac{22}{1}\right]+\left[\frac{22}{2}\right]+\cdots+\left[\frac{22}{11}\right]=22+11+7+5+4+3 \times 2+2 \times 4=63, \\
T_{12}=\left[\frac{24}{1}\right]+\left[\frac{24}{2}\right]+\cdots+\left[\frac{24}{12}\right]=24+12+8+6+4 \times 2+3 \times 2+2 \times 4=72, \\
T_{13}=\left[\frac{26}{1}\right]+\left[\frac{26}{2}\right]+\cdots+\left[\frac{26}{13}\right]=26+13+8+6+5+4+3 \times 2+2 \times 5=78, \\
T_{14}=\left[\frac{28}{1}\right]+\left[\frac{28}{2}\right]+\cdots+\left[\frac{28}{14}\right]=28+14+9+7+5+4 \times 2+3 \times 2+2 \times 5=87, \\
T_{15}=\left[\frac{30}{1}\right]+\left[\frac{30}{2}\right]+\cdots+\left[\frac{30}{15}\right]=30+15+10+7+6+5+4+3 \times 3+2 \times 5=96, \\
t_{2} \sum_{k=1}^{240} f(k)=\sum_{n=1}^{15} T_{n}+105 \\
=2+6+11+16+22+29+34+42+49+56+63+72+78+87+96+105 \\
=768
\end{array}$$ | 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 $x-10\left[\frac{x-1}{10}\right]$ has a value of 10 (when $q=0$) or $q$ (when $q \neq 0$),
$\left[\lg \frac{10}{x-10\left[\frac{x-1}{10}\right]}\right]$ has a value of 1 (when $q=1$) or $0$ (when $q \neq 1$).
Therefore, $f(x)=f(p)+1$ (when $q=1$) or $f(p)$ (when $q \neq 1$).
Hence, the value of $f(x)$ is the number of digit 1s in the decimal representation of $x$, so when $0 \leqslant x \leqslant 2006$, the maximum value of $f(x)$ is 4, at which point $x=1111$. | 1111 | Number Theory | math-word-problem | 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 integer solutions that satisfy the original inequality. | 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 point $M(m, n)$ on the boundary or inside $\triangle OAC$, when $0 \leqslant m < 75$, $0 \leqslant n \leqslant \frac{1}{3} m$. When $75 \leqslant m \leqslant 100$, $0 \leqslant n \leqslant 100-m$. Therefore,
$$\begin{aligned}
N= & \sum_{m=0}^{74}\left(\left[\frac{1}{3} m\right]+1\right)+\sum_{m=75}^{100}(100-m+1) \\
& =\sum_{k=0}^{24}\left(\left[\frac{1}{3} \cdot 3 k\right]+1+\left[\frac{1}{3} \cdot(3 k+1)\right]+1+\left[\frac{1}{3} \cdot(3 k+2)\right]+1\right)+\sum_{m=75}^{100}(100-m+1) \\
& =\sum_{k=0}^{24}(3 k+3)+\sum_{m=75}^{100}(100-m+1) \\
& =975+351=1326
\end{aligned}$$
Similarly, the number of integer points on the boundary and inside $\triangle OBD$ is also $N=1326$.
The number of integer points on the line segment $OA$ is $25+1=26$.
Similarly, there are 26 integer points on the line segment $OB$.
Let $L$ be the number of integer points inside $\triangle OAB$. Then $L+2N-26 \times 2$ represents the number of integer points on the boundary and inside $\triangle OCD$, which numerically equals $\sum_{x=1}^{100}([100-x]+1)=\sum_{x=0}^{100}(101-x)=5151$.
Therefore, $L+2 \times 1326-2 \times 26=5151$.
Thus, the number of grid points $L=2551$. | 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 means $\left(x^{2}\right)^{3} \equiv 1(\bmod 7)$. Also, $x^{2} \equiv -y^{2}(\bmod 7)$, so $\left(-y^{2}\right)^{3} \equiv 1(\bmod 7)$, which means $-y^{6} \equiv 1(\bmod 7)$. By Fermat's Little Theorem, $y^{6} \equiv 1(\bmod 7)$, which contradicts $-y^{6} \equiv 1(\bmod 7)$.
In summary, $\frac{x^{2}+y^{2}}{7} \in \mathbf{Z}(x, y \in \mathbf{Z}) \Leftrightarrow 7|x, 7| y$.
Therefore, the number of integer pairs $(x, y)$ is $C_{142}^{2}=10011$. | 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 $T=T_{1} \cup T_{2}, T_{1} \cap T_{2}=\{(x, y) \mid 0<x \leqslant \sqrt{48}, 0<y<\sqrt{48}\}$, using $\|x\|$ to denote the number of lattice points in set $x$, thus according to the principle of inclusion-exclusion, we have
$$\begin{aligned}
\|T\| & =\left\|T_{1} \cup T_{2}\right\| \\
& =\left\|T_{1}\right\|+\left\|T_{2}\right\|-\left\|T_{1} \cap T_{2}\right\| \\
& =\sum_{0<x \leqslant \sqrt{48}}\left[\frac{48}{x}\right]+\sum_{0<y \leqslant \sqrt{48}}\left[\frac{48}{y}\right]-[\sqrt{48}]^{2} \\
& =2 \sum_{0<k \leqslant 6}\left[\frac{48}{k}\right]-36=119 \times 2-36=202
\end{aligned}$$ | 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 $2,3,5,7,11,13,17,19$, in the standard factorization of 20!, the highest power of 2 $=\left[\frac{20}{2}\right]+\left[\frac{20}{4}\right]+\left[\frac{20}{8}\right]+\left[\frac{20}{16}\right]=18$,
the highest power of 3 $=\left[\frac{20}{3}\right]+\left[\frac{20}{9}\right]=8$,
the highest power of 5 $=\left[\frac{20}{5}\right]=4$,
the highest powers of $7,11,13,17,19$ are $2,1,1,1,1$ respectively.
Thus, the standard factorization of 20! is
$$20!=2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19$$
Therefore, the number of positive divisors $r(20!)$ of 20! is:
$$\begin{aligned}
r(20!) & =(18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1) \\
& =19 \cdot 9 \cdot 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \\
& =41040
\end{aligned}$$
Thus, the number of positive divisors of $20!$ is 41040. | 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 that
$87 \mid 55^{n}+a \cdot 32^{n}$,
and $23 \mid 55^{n}+a \cdot 32^{n}$.
From (1), we have: $0 \equiv 55^{n}+a \cdot 32^{n} \quad(\bmod 87)$
$$\begin{array}{l}
\equiv(-32)^{n}+a \cdot 32^{n} \quad(\bmod 87) \\
\equiv 32^{n}(a-1) \quad(\bmod 87)
\end{array}$$
Since $\left(32^{n}, 87\right)=1$, it follows that $a-1 \equiv 0(\bmod 87)$
From (2), we have: $0 \equiv 55^{n}+a \cdot 32^{n}(\bmod 23)$
$$\equiv 32^{n}+a \cdot 32^{n} \quad(\bmod 23)$$
Since $\left(32^{n}, 23\right)=1$, it follows that $a+1 \equiv 0(\bmod 23)$,
hence $a \equiv-1(\bmod 23)$
From (3), let $a=87 k+1(k \in \mathbf{N})$ and substitute into (4) to get
$$23|87 k+2 \Rightarrow 23| 18 k+2 \Rightarrow 23|-5 k+25 \Rightarrow 23| k-5$$
Since $k \geqslant 0$, it follows that $k \geqslant 5$.
Thus, the smallest value of $a$ is $87 \times 5+1=436$. | 436 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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