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33 Find all positive integers $m$ such that $m=d(m)^{4}$.
Find all positive integers $m$ such that $m=d(m)^{4}$. | 33. From the conditions, we know that \( m \) is a fourth power, so we can set
$$m=2^{4 a_{2}} \cdot 3^{4 a_{3}} \cdot 5^{4 a_{5}} \cdot 7^{4 a_{7}} \cdots \cdots$$
where \(\alpha_{2}, \alpha_{3}, \alpha_{5}, \alpha_{7}, \cdots\) are non-negative integers. And
$$d(m)=\left(4 \alpha_{2}+1\right)\left(4 \alpha_{3}+1\right) \cdots$$
is an odd number, hence \(\alpha_{2}=0\), and
$$1=\frac{4 \alpha_{3}+1}{3^{a_{3}}} \cdot \frac{4 \alpha_{5}+1}{5^{a_{5}}} \cdot \frac{4 \alpha_{7}+1}{7^{\alpha_{7}}} \cdots=x_{3} x_{5} x_{7} \cdots$$
where
$$x_{3}=\frac{4 \alpha_{3}+1}{3^{a_{3}}}, x_{5}=\frac{4 \alpha_{5}+1}{5^{a_{5}}}, \cdots$$
When \(\alpha_{3}=1\), \(x_{3}=\frac{5}{3}\); when \(\alpha_{3}=0\) or 2, \(x_{3}=1\); and when \(\alpha_{3} \geqslant 3\), \(3^{\alpha_{3}} > 4 \alpha_{3}+1\), so in this case \(x_{3}<1\). For \(p > 5\), where \(p\) is a prime, when \(\alpha_{p}=0\), \(x_{p}=1\), and when \(\alpha_{p}=1\), \(p^{\alpha_{p}}>5=4 \alpha_{p}+1\), so \(x_{p}<1\); and when \(\alpha_{p} \geqslant 2\), \(x_{p}<\frac{9}{25}\).
The above discussion shows: if \(\alpha_{3} \neq 1\), then
$$x_{3}=x_{5}=x_{7}=\cdots=1$$
Thus,
$$\alpha_{3}=0 \text{ or } 2, \alpha_{5}=0 \text{ or } 1,$$
and
$$\alpha_{7}=\alpha_{11}=\cdots=0$$
That is,
$$m=1,3^{8}, 5^{4} \text{ or } 45^{4} \text{. }$$
If \(\alpha_{3}=1\), then \(3 \mid m\), at this time, from \(m=d(m)^{4}\), we know
$$m=5^{4} \times\left(4 \alpha_{5}+1\right)^{4} \times\left(4 \alpha_{7}+1\right)^{4} \cdots$$
Thus, there exists a prime \(p \geqslant 5\) such that \(3 \mid 4 \alpha_{p}+1\), which requires \(\alpha_{p} \geqslant 2\), hence \(x_{p}<\frac{9}{25}\). This leads to
$$x_{3} x_{5} x_{7} \cdots \leqslant \frac{5}{3} \times \frac{9}{25}=\frac{3}{5}<1$$
a contradiction.
Therefore,
$$m=1,5^{4}, 3^{8}, 3^{8} \cdot 5^{4}$$
(Direct verification shows that they indeed satisfy the conditions). | m=1,5^{4}, 3^{8}, 3^{8} \cdot 5^{4} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
35 Find all positive integers $a$, $b$, and $c$, such that $a^{2}+1$ and $b^{2}+1$ are both prime, and satisfy
$$\left(a^{2}+1\right)\left(b^{2}+1\right)=c^{2}+1$$ | 35. Let's assume $a \leqslant b$, from the condition we have
$$a^{2}\left(b^{2}+1\right)=c^{2}+1-b^{2}-1=(c-b)(c+b)$$
Thus, $b^{2}+1 \mid c-b$ or $b^{2}+1 \mid c+b$ (here we use the fact that $b^{2}+1$ is a prime).
If
$$b^{2}+1 \mid c-b$$
then $\square$
$$c-b \geqslant b^{2}+1 \text { (note that } c>b \text { is obvious), }$$
i.e., $\square$
$$c \geqslant b^{2}+b+1$$
In this case,
$$c^{2}+1 \geqslant\left(b^{2}+b+1\right)^{2}+1>\left(b^{2}+1\right)^{2} \geqslant\left(a^{2}+1\right)\left(b^{2}+1\right)$$
Contradiction.
If
$$b^{2}+1 \mid c+b$$
then $\square$
$$c+b \geqslant b^{2}+1$$
i.e., $\square$
$$c \geqslant b^{2}-b+1$$
Thus,
$$\begin{aligned}
c^{2}+1 & \geqslant\left(b^{2}-b+1\right)^{2}+1 \\
& =\left(b^{2}+1\right)^{2}-2 b\left(b^{2}+1\right)+b^{2}+1 \\
& =\left(b^{2}+1\right)\left((b-1)^{2}+1\right)
\end{aligned}$$
Notice that, if $a=b$, then $c^{2}+1=\left(a^{2}+1\right)^{2}$, which cannot hold for positive integers $a$ and $c$ (since the difference of squares of two positive integers is at least 3), so $a<b$, i.e., $a \leqslant b-1$, thus
$$c^{2}+1 \geqslant\left(b^{2}+1\right)\left((b-1)^{2}+1\right) \geqslant\left(b^{2}+1\right)\left(a^{2}+1\right)$$
Combining with the condition, we have
$$a=b-1, c=b^{2}-b+1$$
At this point, since $a^{2}+1$ and $b^{2}+1$ are both primes, $b^{2}+1$ is odd, so $b$ is even, and thus $a=b-1$ is odd, and $a^{2}+1$ is even, so $a=1$, hence $b=2, c=3$. When $(a, b, c)=$ $(1,2,3)$ or $(2,1,3)$, the condition is satisfied, and these are the required solutions. | (1,2,3) \text{ or } (2,1,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
37 Let $a$, $b$, $c$ all be positive integers greater than 1. Find the minimum possible value of the algebraic expression $\frac{a+b+c}{2}-\frac{[a, b]+[b, c]+[c, a]}{a+b+c}$. | 37. By symmetry, we may assume $a \geqslant b \geqslant c$. Note that when $(a, b, c)=(2,2,2),(3,2,2),(3,3,2),(4,2,2)$, the value of the given algebraic expression $A$ is $2, \frac{3}{2}, \frac{17}{8}, \frac{11}{4}$, respectively. This indicates that when $a+b+c \leqslant 8$, $A \geqslant \frac{3}{2}$.
Next, we prove that when $a+b+c \geqslant 9$, we have $A \geqslant \frac{3}{2}$.
In fact,
$$\begin{aligned}
A \geqslant \frac{3}{2} & \Leftrightarrow (a+b+c)^{2}-2([a, b]+[b, c]+[c, a]) \geqslant 3(a+b+c) \\
& \Leftrightarrow a^{2}+b^{2}+c^{2}+2 \sum(a b-[a, b]) \geqslant 3(a+b+c)
\end{aligned}$$
Since for positive integers $x, y$, we have $x y \geqslant [x, y]$, it suffices to prove:
$$a^{2}+b^{2}+c^{2} \geqslant 3(a+b+c)$$
Given $a+b+c \geqslant 9$, to prove (1) holds, it suffices to prove:
$$\begin{aligned}
& a^{2}+b^{2}+c^{2} \geqslant \frac{1}{3}(a+b+c)^{2} \\
\Leftrightarrow & 3\left(a^{2}+b^{2}+c^{2}\right) \geqslant (a+b+c)^{2} \\
\Leftrightarrow & 2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a) \geqslant 0 \\
\Leftrightarrow & (a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0
\end{aligned}$$
The last inequality is obviously true.
Therefore, the minimum value of the given algebraic expression is $\frac{3}{2}$. | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | number_theory | false |
38 For any given prime $p$, how many integer triples $(a, b, c)$ are there such that
(1) $1 \leqslant a, b, c \leqslant 2 p^{2}$;
(2) $\frac{[a, c]+[b, c]}{a+b}=\frac{p^{2}+1}{p^{2}+2} \cdot c$. | 38. Let $u=(a, c), v=(b, c)$, then condition (2) becomes
$$\frac{\frac{a c}{u}+\frac{b c}{v}}{a+b}=\frac{p^{2}+1}{p^{2}+2} \cdot c$$
i.e.,
$$\frac{a}{u}+\frac{b}{v}=\frac{p^{2}+1}{p^{2}+2}(a+b)$$
When $\frac{1}{2}2$, the number of arrays satisfying the condition is $p^{2}-p$. | p^{2}-p | 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 blackboard. How many numbers will be left on the blackboard at least? | 39. Consider the objective function $S=$ the product of all numbers on the blackboard.
Initially, $S=33!=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$. Each operation targets $x, y (x \mid y)$, where $y=kx$. Removing $x, y$ and replacing them with $k$ results in $S$ becoming $\frac{S}{xy} \cdot k = \frac{S}{x^2}$. This indicates that after each operation, the parity of the exponent of each prime factor in $S$ remains unchanged. Specifically, $2, 3, 5,$ and $11$ always divide the $S$ obtained after each operation. Since $2 \times 3 \times 5 \times 11 > 33$, at least two numbers must remain on the blackboard such that their product is a multiple of $2 \times 3 \times 5 \times 11$.
Furthermore, note that the primes $17, 19, 23, 29, 31$ have no multiples greater than themselves that are less than or equal to 33. Therefore, no operation can remove any of these numbers.
The above discussion shows that at least 7 numbers must remain on the blackboard.
The following example demonstrates that exactly 7 numbers can remain:
$$\begin{array}{l}
(32,16) \rightarrow 2, (30,15) \rightarrow 2, (28,14) \rightarrow 2, (26,13) \rightarrow 2, (24,12) \rightarrow 2 \\
(22,11) \rightarrow 2; (27,9) \rightarrow 3, (21,7) \rightarrow 3, (18,6) \rightarrow 3; (25,5) \rightarrow 5 \\
(20,4) \rightarrow 5; (8,2) \rightarrow 4 \\
(5,5) \rightarrow 1; (4,2) \rightarrow 2; (3,3) \rightarrow 1, (3,3) \rightarrow 1, (2,2) \rightarrow 1, (2,2) \rightarrow 1 \\
(2,2) \rightarrow 1
\end{array}$$
Thus, the blackboard is left with $10, 17, 19, 23, 29, 31, 33$ and 7 ones. The 7 ones can be removed by pairing them with 17 in 7 operations.
In conclusion, at least 7 numbers remain. | 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find all prime numbers $p, q, r (p \leqslant q \leqslant r)$ such that
$$p q+r, p q+r^{2}, q r+p, q r+p^{2}, r p+q, r p+q^{2}$$
are all prime numbers. | If $p>2$, then $p, q, r$ are all odd numbers, in which case $pq+r$ is an even number greater than 2, leading to a contradiction. Therefore, $p=2$. Now, the numbers
$$2q+r, 2q+r^2, qr+2, qr+4, 2r+q, 2r+q^2$$
are all prime numbers.
If $q, r$ include an even number, then $qr+2$ is an even number greater than 2, leading to a contradiction. Therefore, $q, r$ are both odd prime numbers. If $q>3$, then $3 \nmid qr$. In this case, if $qr \equiv 1 \pmod{3}$, then $qr+2 \equiv 0 \pmod{3}$, contradicting the fact that $qr+2$ is a prime number; if $qr \equiv 2 \pmod{3}$, then $qr+4 \equiv 0 \pmod{3}$, contradicting the fact that $qr+4$ is a prime number. Therefore, $q=3$. Thus, the numbers
$$6+r, 6+r^2, 3r+2, 3r+4, 2r+3, 2r+9$$
are all prime numbers.
If $r \neq 5$, then $r \neq 0 \pmod{5}$. However, when $r \equiv 1, 2, 3, 4 \pmod{5}$, the corresponding numbers $3r+2, 3r+4, 2r+9, 6+r$ are multiples of 5, leading to a contradiction. Therefore, $r=5$.
Direct verification shows that they satisfy the conditions. The required prime numbers are
$$p=2, q=3, r=5$$ | p=2, q=3, r=5 | 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 |
Example 5 Find all positive integers $x>1, y>1, z>1$, such that
$$1!+2!+\cdots+x!=y^{z}$$ | The key step is to prove that when $x \geqslant 8$, it must be that $z=2$. Since the left side of (1) is divisible by 3, hence $3 \mid y^{z}$, which implies $3 \mid y$, and thus the right side of (1) is divisible by $3^{z}$. On the other hand,
$$1!+2!+\cdots+8!=46233$$
is divisible by $3^{2}$, but not by $3^{3}$; and for $n \geqslant 9$, $3^{3} \mid n!$. Therefore, when $x \geqslant 8$, the left side of (1) is divisible by $3^{2}$ but not by $3^{3}$, and thus the right side of (1) must also be so, i.e., $z=2$ must hold.
Now, we further prove that when $x \geqslant 8$, equation (1) has no solutions. Modulo 5: when $x \geqslant 8$, the left side of (1) $\equiv$ $1!+2!+3!+4!\equiv 3(\bmod 5)$; and since it has been proven that $z=2$ in this case, the right side of (2) $z^{2} \equiv 0$, or $1(\bmod 5)$, thus the above assertion holds.
Finally, when $x<8$, it is not difficult to find through inspection that the solution to (1) is $x=y=3, z=2$. | x=y=3, z=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 Find all positive integers $m, n$, such that $\left|12^{m}-5^{n}\right|=7$.
| 2. Mod 4 reveals that the equation
$$12^{m}-5^{n}=-7$$
has no positive integer solutions. The equation
$$12^{m}-5^{n}=7$$
obviously has the solution $m=n=1$. Below we prove that when $m>1$ it has no positive integer solutions. Taking (1) mod 3 yields $-(-1)^{n} \equiv 1(\bmod 3)$, so $n$ is odd, thus $5^{n} \equiv 5(\bmod 8)$. Also, $m \geqslant 2$, so $8 \mid 12^{m}$. Taking (1) mod 8 yields $-5 \equiv 7(\bmod 8)$, which is impossible. Therefore, $m=1$, hence $n=1$. | m=n=1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3 Find all prime numbers $p$, such that $2^{p}+3^{p}$ is a $k$-th power of an integer (where $k \geqslant 2$).
Find all prime numbers $p$, such that $2^{p}+3^{p}$ is a $k$-th power of an integer (where $k \geqslant 2$). | 3. $p=2,5$ do not meet the requirements. Let prime $p>2$ and $p \neq 5$. By the binomial theorem, it is easy to see that
$$\begin{aligned}
2^{p}+3^{p} & =2^{p}+(5-2)^{p}=5^{p}-\mathrm{C}_{p}^{1} 5^{p-1} \times 2+\cdots+5 \mathrm{C}_{p}^{p-1} 2^{p-1} \\
& =5^{2} u+5 p \times 2^{p-1}, u \text { is an integer. }
\end{aligned}$$
Thus $5 \|\left(2^{p}+3^{p}\right)$, hence $2^{p}+3^{p}$ cannot be a $k$-th power of an integer $(k>1)$. | p=2,5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6 Let $p$ be a given odd prime, find all positive integer solutions $x, y$ of $p^{x}-y^{p}=1$.
The text has been translated while preserving the original line breaks and format. | 6. From $p^{x}=y^{p}+1=(y+1)\left(y^{p-1}-y^{p-2}+\cdots-y+1\right)$, we know that $y+1=p^{n}$, where $n$ is an integer. Since $y>0$, it follows that $n>0$. Therefore,
$$\begin{aligned}
p^{x} & =\left(p^{n}-1\right)^{p}+1 \\
& =p^{n p}-p \cdot p^{n(p-1)}+\mathrm{C}_{p}^{2} p^{n(p-2)}-\cdots-\mathrm{C}_{p}^{p-2} p^{2 n}+p \cdot p^{n}
\end{aligned}$$
It is easy to see that, except for the last term, all terms on the right-hand side of the above equation are divisible by $p^{n+2}$ (note that since $p$ is a prime, all $\mathrm{C}_{p}^{i}$ for $i=1, \cdots, p-2$ are divisible by $p$), so $p^{n+1}$ is the highest power of $p$ on the right-hand side of (1), hence it must be that $x=n+1$. At this point, (1) becomes
$$p^{n p}-p \cdot p^{n(p-1)}+\mathrm{C}_{p}^{2} p^{n(p-2)}-\cdots-\mathrm{C}_{p}^{p-2} p^{2 n}=0$$
When $p=3$, (2) becomes $3^{3 n}-3 \cdot 3^{2 n}=0$, yielding $n=1$, so $x=y=2$. If $p \geqslant 5$, noting that $\mathrm{C}_{p}^{p-2}$ is not divisible by $p^{2}$, it is easy to see that all terms on the left-hand side of (2) except the last one are divisible by $p^{2 n+2}$, but the last term is not divisible by $p^{2 n+2}$, indicating that (2) cannot hold. Therefore, the only solution to this problem is when $p=3$, with $x=y=2$. | x=y=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find all integers $n>1$, such that $\frac{2^{n}+1}{n^{2}}$ is an integer. | It is easy to conjecture that $n=3$ is the only solution that meets the requirements. Below, we prove that this is indeed the case.
The proof requires several steps. The first point is to consider the smallest prime divisor $p$ of $n$, and from $n \mid (2^n + 1)$, derive that $p=3$, as seen in Example 1 of Unit 8. Therefore, we can assume
$$n=3^m c, \quad m \geqslant 1, \quad 3 \nmid c.$$
The second step is to prove that $m=1$. From $n^2 \mid (2^n + 1)$, we get $2^{3^m c} \equiv -1 \pmod{3^{2m}}$, hence
$$2^{2 \times 3^m c} \equiv 1 \pmod{3^{2m}}.$$
If $m \geqslant 2$, then by Example 5 of Unit 8, the order of 2 modulo $3^{2m}$ is $2 \times 3^{2m-1}$, so from (2) we deduce that $2 \times 3^{2m-1} \mid 2 \times 3^m c$, i.e., $3^{m-1} \mid c$, thus $3 \mid c$ (since $m \geqslant 2$), which contradicts (1) where $3 \nmid c$. Therefore, we must have $m=1$.
The third step is to prove that $c=1$ in (1). This can be done similarly to the first step, i.e., Example 1 of Unit 8: $\square$
If $c>1$, let $q$ be the smallest prime divisor of $c$, then
$$2^{3c} \equiv -1 \pmod{q}.$$
Let $r$ be the order of 2 modulo $q$. From (3), we get $2^{6c} \equiv 1 \pmod{q}$, and also $2^{q-1} \equiv 1 \pmod{q}$, so $r \mid 6c$ and $r \mid (q-1)$, hence $r \mid (6c, q-1)$. By the selection of $q$, we have $(q-1, c)=1$, so $r \mid 6$. From $2^r \equiv 1 \pmod{q}$, we deduce that $q=3$ or 7. It is easy to see that $q=3$ is impossible; and from (3), $q=7$ is also impossible. Therefore, we must have $c=1$. Thus, $n=3$.
Note that if we first prove that $c=1$ in (1), it will not be effective. The order of the arguments here is quite important. Additionally, in the second step, $m=1$ can also be proven by comparing powers of primes:
By the binomial theorem,
$$2^n + 1 = (3-1)^n + 1 = 3n + \sum_{k=2}^n (-1)^k \binom{n}{k} 3^k.$$
Let $3^a \parallel k!$, then
$$\alpha = \sum_{l=1}^{\infty} \left[\frac{k}{3^l}\right] k + m - \frac{k}{2} \geqslant m + 1.$$
Thus, $\beta \geqslant m + 2$, so the sum on the right-hand side of (4) is divisible by $3^{m+2}$. If $m>1$, then $2m \geqslant m+2$, so from $3^{2m} \mid (2^n + 1)$ and (4), we get $3^{m+2} \mid 3n$, i.e., $3^{m+1} \mid n$, which contradicts (1). Therefore, we must have $m=1$. | n=3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3 Let positive integers $a, b$ satisfy $a^{2} b \mid\left(a^{3}+b^{3}\right)$, prove $a=b$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. Let $a^{3}+b^{3}=m a^{2} b$, then $\left(\frac{a}{b}\right)^{3}-m\left(\frac{a}{b}\right)^{2}+1=0$, which means the rational number $\frac{a}{b}$ is a root of the integral coefficient equation with leading coefficient 1:
$$x^{3}-m x^{2}+1=0$$
Therefore, $\frac{a}{b}$ must be an integer (Exercise 2, Question 4). On the other hand, any integer root of equation (1) must divide the constant term 1, and thus can only be $\pm 1$; since $a, b$ are positive, we have $\frac{a}{b}=1$, i.e., $a=b$. | a=b | Combinatorics | MCQ | 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 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 |
Example 6 Find all integer solutions of $\left(x^{2}-y^{2}\right)^{2}=1+16 y$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: Since the left side of the equation $\geqslant 0$, the right side $\geqslant 0$, thus $y \geqslant 0$. It is also clear that $x^{2}-y^{2} \neq 0$, and since $x_{1} y$ is an integer, we have $|x| \geqslant y+1$, or $|x| \leqslant y-1$.
When $|x| \geqslant y+1$, the left side of the equation $\geqslant\left((y+1)^{2}-y^{2}\right)^{2}=(2 y+1)^{2}$.
When $|x| \leqslant y-1$, at this time $y-1 \geqslant 0$, and $y^{2}-x^{2} \geqslant y^{2}-(y-1)^{2}=2 y-$ $1>0$, so the left side of the equation $\geqslant(2 y-1)^{2}$.
Therefore, from the original equation, we get
$$(2 y-1)^{2} \leqslant 1+16 y$$
Thus, $0 \leqslant y \leqslant 5$. Checking each value, we find all integer solutions to be $\left(x_{1} y\right)=( \pm 1,0),( \pm 4,3)$, $( \pm 4,5)$. | ( \pm 1,0),( \pm 4,3), ( \pm 4,5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Find all integer solutions of the indefinite system of equations
$$\left\{\begin{array}{l}
x+y+z=3 \\
x^{3}+y^{3}+z^{3}=3
\end{array}\right.$$ | 3. Eliminate $z$ from the system of equations, to get
$$8-9 x-9 z+3 x^{2}+6 x y+3 y^{2}-x^{2} y-x y^{2}=0$$
Transform it into
$$8-3 x(3-x)-3 y(3-x)+x y(3-x)+y^{2}(3-x)=0,$$
which is $(3-x)\left(3 x+3 y-x y-y^{2}\right)=8$. Hence $(3-x) \mid 8$, so $3-x= \pm 1, \pm 2, \pm 4, \pm 8$, which means $x=-5,-1,1,2,4,5,7,11$. Substitute each value into the original system of equations to verify, and the complete set of integer solutions is $(x, y, z)=(1,1,1),(-5,4,4),(4,-5,4),(4,4,-5)$. | (x, y, z)=(1,1,1),(-5,4,4),(4,-5,4),(4,4,-5) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
4 Find all integer solutions to $x^{3}=y^{3}+2 y^{2}+1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. First note that if $y^{2}+3 y>0$, then from the original equation we deduce that $(y+1)^{3}>x^{3}>y^{3}$, which means $x^{3}$ lies between two consecutive perfect cubes, which is impossible. Therefore, it must be that $y^{2}+3 y \leqslant 0$, yielding the integer solutions $y=$ $-3,-2,-1,0$. Substituting these values into the original equation, we find all integer solutions to be $(x, y)=(1,0)$, $(1,-2),(-2,-3)$ | (x, y)=(1,0), (1,-2), (-2,-3) | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false |
Find all positive integers $x, y$ such that $x^{2}+3 y, y^{2}+3 x$ are both perfect squares.
The text above has been translated into English, preserving the original text's line breaks and format. | 5. Let $\left\{\begin{array}{l}x^{2}+3 y=u^{2} \\ y^{2}+3 x=v^{2}\end{array}\right.$ Since $x, y$ are positive integers, we have $u>x, v>y$. We set $u=x+a, v=y+b$, where $a, b$ are positive integers. From
$$\left\{\begin{array}{l}
x^{2}+3 y=(x+a)^{2} \\
y^{2}+3 x=(y+b)^{2}
\end{array}\right.$$
we can derive
$$\left\{\begin{array}{l}
3 y=2 a x+a^{2} \\
3 x=2 b y+b^{2}
\end{array}\right.$$
Solving this system of linear equations in $x, y$ yields $\left\{\begin{array}{l}x=\frac{2 a^{2} b+3 b^{2}}{9-4 a b}, \\ y=\frac{2 b^{2} a+3 a^{2}}{9-4 a b} .\end{array}\right.$ Since $x, y$ are positive integers, we have $9-4 a b>0$. Since $a, b$ are positive integers, we have $a b=1$ or 2, i.e., $(a, b)=(1,1),(1,2)$, $(2,1)$. Correspondingly, we obtain $(x, y)=(1,1),(16,11),(11,16)$. | (1,1),(16,11),(11,16) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all positive integer solutions to the indeterminate equation
$$(n-1)!=n^{k}-1$$ | When $n=2$, from (1) we get the solution $(n, k)=(2,1)$. When $n>2$, the left side of (1) is even, so its right side must also be even, hence $n$ is odd. When $n=3,5$, from (1) we solve $(n, k)=(3,1), (5,2)$.
Below, let $n>5$ and $n$ be odd. In this case, $\frac{n-1}{2}$ is an integer and $\frac{n-1}{2}(n-1)!$
This indicates that when $n>5$, equation (1) has no positive integer solutions, i.e., the complete set of positive integer solutions to (1) is $(n, k)=(2,1),(3,1),(5,2)$. | (n, k)=(2,1),(3,1),(5,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Let $p$ be a given positive integer, try to determine the minimum positive value of $(2 p)^{2 m}-(2 p-1)^{n}$, where $m, n$ are any positive integers. | The smallest positive value sought is $(2 p)^{2}-(2 p-1)^{2}=4 p-1$. To prove this, we first note that by
and
$$\begin{array}{c}
(2 p)^{2}=(4 p-2) p+2 p \\
(2 p-1)^{2}=(4 p-2)(p-1)+(2 p-1)
\end{array}$$
it is easy to deduce that
$$(2 p)^{2 m}-(2 p-1)^{n} \equiv(2 p)-(2 p-1) \equiv 1(\bmod 4 p-2)$$
Furthermore, we prove that there are no positive integers $m, n$ such that $(2 p)^{2 m}-(2 p-1)^{n}=1$. Suppose the contrary, then we have
$$\left((2 p)^{m}-1\right)\left((2 p)^{m}+1\right)=(2 p-1)^{n}$$
The two factors on the left side are clearly coprime, while the right side is the $n$-th power of a positive integer, hence
$$(2 p)^{m}+1=a^{n}$$
where $a$ is a positive integer, and $a \mid(2 p-1)$. Taking (2) modulo $a$, the left side becomes
$$(2 p-1+1)^{m}+1 \equiv 1+1 \equiv 2(\bmod a)$$
leading to $2 \equiv 0(\bmod a)$, but since $a \mid 2 p-1$, $a$ is an odd number greater than 1, which leads to a contradiction.
Combining (1), if $(2 p)^{2 m}-(2 p-1)^{n}>0$, then $(2 p)^{2 m}-(2 p-1)^{n} \geqslant 4 p-1$, and equality is achieved when $m=1, n=2$, thus proving our conclusion. | 4p-1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 Find all positive integers $n$, such that the decimal integer composed of $n-1$ digits 1 and one digit 7 is a prime number.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 2. A positive integer $N$ composed of $n-1$ digits 1 and one digit 7 can be expressed in the form $N = A_{n} + 6 \times 10^{k}$, where $0 \leqslant k \leqslant n-1$, and $A_{n}$ is an integer composed of $n$ ones.
When $3 \mid n$, the sum of the digits of $A_{n}$ is divisible by 3, so $3 \mid A_{n}$, and thus $3 \mid N$. Since $N > 3$, $N$ is not a prime number in this case.
Now, let's assume $3 \nmid n$. Note that $10^{6} \equiv 1(\bmod 7)$. We will classify $n$ modulo 6 to discuss the value of $A_{n}$ modulo 7 (the cases $n \equiv 0,3(\bmod 6)$ do not need to be considered). It is easy to see that for $l \geqslant 0$,
$$\begin{aligned}
A_{6l+1} & =\frac{1}{9} \times\left(10^{6l+1}-1\right)=\frac{1}{9} \times\left(10^{6l}-1\right) \times 10+\frac{1}{9} \times(10-1) \\
& \equiv 1(\bmod 7) \\
A_{6l+2} & \equiv 4, \quad A_{6l+4} \equiv 5, \quad A_{6l+5} \equiv 2(\bmod 7)
\end{aligned}$$
Additionally, $10^{0}, 10^{2}, 10^{4}, 10^{5}$ are congruent to $1,2,4,5$ modulo 7, respectively. Therefore, when $n > 6$, for $n \equiv 1,2,4,5(\bmod 6)$, we can take $k=0,4,5,2$ respectively, and we find that
$$N = A_{n} + 6 \times 10^{k} \equiv A_{n} - 10^{k} \equiv 0(\bmod 7),$$
hence $N$ is not a prime number, and thus $n > 5$ does not meet the requirement. When $n \leqslant 5$, it is not difficult to verify that only $n = 1,2$ meet the requirement. | n = 1,2 | 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 |
9. Determine all integers $n > 1$ such that: for all integers $a, b \in \mathbf{Z}$ that are coprime with $n$, $a \equiv b(\bmod n) \Leftrightarrow a b \equiv 1(\bmod n)$ | 9. Solution: We say that $n$ has property $T$ if for all integers $a, b$ coprime with $n$, $a \equiv b(\bmod n)$ if and only if $a b \equiv 1(\bmod n)$. It is not hard to see that $n$ has property $T$ if and only if for any $a \in \mathbf{Z}$, if $(a, n)=1$, then $a^{2} \equiv 1$ $(\bmod n)$. We will prove: if $n$ has property $T$, then any prime factor of $n$ also has property $T$.
In fact, for any $a \in \mathbf{Z}$, if $(a, p)=1$, then there exists $a^{\prime} \in \mathbf{Z},\left(a, a^{\prime}\right)=1$ and $a \equiv a^{\prime}(\bmod p)$, thus
$a^{2} \equiv\left(a^{\prime}\right)^{2}(\bmod p),\left(a^{\prime}\right)^{2} \equiv 1(\bmod n)$,
so $\left(a^{\prime}\right)^{2} \equiv 1(\bmod p) \Rightarrow a^{2} \equiv 1(\bmod p)$.
By the choice of $a$, we know that $p$ has property $T$.
Next, consider a prime $p$ with property $T$. If $p \geqslant 5$, then $(2, p)=1$, but $2^{2} \neq 1(\bmod p)$, so $p$ does not have property $T$. Therefore, $p$ can only be 2 or 3.
Thus, any positive integer $n$ with property $T$ must be of the form $2^{\alpha} \cdot 3^{\beta}(\alpha, \beta \geqslant 0)$, so we can take $(5, n)=$ 1. Therefore, $n \mid 5^{2}-1=24$.
So the positive integers greater than 1 that have property $T$ can only be: $2,3,4,6,8,12,24$.
Upon verification, these 7 natural numbers indeed have property $T$, and they are the ones we seek. | 2,3,4,6,8,12,24 | 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 $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, then $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=$ $3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$
From $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$,
so $1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$, we can take $M_{1}^{-1}=1$.
From $M_{2} \equiv(-2) \cdot(2) \cdot(1) \equiv 1(\bmod 5)$,
so $1 \equiv M_{2} M_{2}^{-1} \equiv M_{2}^{-1}(\bmod 5)$, we can take $M_{2}^{-1}=1$.
From $M_{3} \equiv 3 \times 5 \times 4 \equiv 4(\bmod 7)$,
so $1 \equiv M_{3} M_{3}^{-1} \equiv 4 M_{3}^{-1}(\bmod 7)$, we can take $M_{3}^{-1}=2$.
From $M_{4} \equiv 3 \times 5 \times 7 \equiv 4 \times 7 \equiv 6(\bmod 11)$,
so $1 \equiv M_{4} M_{4}^{-1} \equiv 6 M_{4}^{-1}(\bmod 11)$, we can take $M_{4}^{-1}=2$.
By the Chinese Remainder Theorem, the solution to the system of congruences is
$$x=(5 \cdot 7 \cdot 11) \cdot 1 \cdot 1+(3 \cdot 7 \cdot 11) \cdot 1 \cdot(-1)+(3 \cdot 5 \cdot 11) \cdot 2 \cdot 2+(3 \cdot 5 \cdot 7)$$
- $2 \cdot(-2)(\bmod 3 \cdot 5 \cdot 7 \cdot 11)$
which is $x \equiv 394(\bmod 1155)$. | x \equiv 394(\bmod 1155) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1. Solve the system of congruences: (1) $\left\{\begin{array}{l}x \equiv 3(\bmod 7) \\ 6 x \equiv 10(\bmod 8)\end{array}\right.$
(1) (2) $\left\{\begin{array}{l}3 x \equiv 1(\bmod 10) \\ 4 x \equiv 7(\bmod 15)\end{array}\right.$ | For the system of equations (I), first solve (2), we get $x \equiv -1, 3 \pmod{8}$, so the solutions to (1) are the solutions to the following two systems of congruences: $x \equiv 3 \pmod{7}, x \equiv -1 \pmod{8}$, or $x \equiv 3 \pmod{7}, x \equiv 3 \pmod{8}$.
Using the Chinese Remainder Theorem, the solutions to (I) are $x \equiv 3, 31 \pmod{56}$.
For the system of equations (II), it is equivalent to the following system of congruences:
$$\left\{\begin{array}{l}
3 x \equiv 1 \pmod{2}, \\
3 x \equiv 1 \pmod{5}, \\
4 x \equiv 7 \pmod{3}, \\
4 x \equiv 7 \pmod{5}.
\end{array}\right.$$
But from $3 x \equiv 1 \pmod{5} \Rightarrow x \equiv 2 \pmod{5}$,
and from $4 x \equiv 7 \pmod{5} \Rightarrow x \equiv 3 \pmod{5}$, which is a contradiction!
Therefore, the system of equations (II) has no solution. | x \equiv 3, 31 \pmod{56} \text{ for (I), no solution for (II)} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. The indeterminate equation $96 x_{1}+97 x_{2}=1000$ has $(\quad)$ sets of non-negative integer solutions.
A. 0
B. 1
C. 2
D. 3 | 2. A The complete set of integer solutions for the equation can be obtained as
$$\left\{\begin{array}{l}
x=-30+97 t, \\
y=40-96 t
\end{array} \quad(t \in \mathbf{Z})\right.$$
Let $x, y \geqslant 0$, we get $\frac{30}{97} \leqslant t \leqslant \frac{40}{96}$, and $t$ has no integer values. Therefore, the equation has no non-negative integer solutions. | A | Number Theory | MCQ | Yes | Yes | number_theory | false |
Example 2 Find all positive integers $x>1, y>1, z>1$, such that $1!+2!+\cdots+x!=y^{z}$. | The key step is to prove that when $x \geqslant 8$, it must be that $z=2$. Since the left side of (1) is divisible by 3, hence $3 \mid y^{z}$, which implies $3 \mid y$, so the right side of (1) is divisible by $3^{z}$. On the other hand,
$$1!+2!+\cdots+8!=46233$$
is divisible by $3^{2}$, but not by $3^{3}$; and for $n \geqslant 9$, $3^{3} \mid n!$, so when $x \geqslant 8$, the left side of (1) is divisible by $3^{2}$, but not by $3^{3}$, thus the right side of (1) must also be so, i.e., $z=2$ must hold.
Now, we further prove that when $x \geqslant 8$, equation (1) has no solutions. Taking both sides of (1) modulo 5, when $x \geqslant 8$, the left side of (1) $\equiv 1! + 2! + 3! + 4! \equiv 3 (\bmod 5)$; and since $z=2$ is known, the right side of (1) $\equiv 0, \pm 1 (\bmod 5)$, thus the above assertion holds. Finally, when $x<8$, by verification, the solution to (1) is $x=y=3, z=2$. | x=y=3, z=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 (2003 British Mathematical Olympiad)
Given that $34!=295232799 c d 96041408476186096435 a b 000000$, find the values of the digits $a, b, c, d$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given that $34!=k \times 11^{3} \times 7^{4} \times 5^{7} \times 3^{15} \times 2^{32}=k \times 11^{3} \times 7^{4} \times 3^{15} \times 2^{25} \times 10^{7}$,
so $b=0, a \neq 0$.
Considering the scenario when the last 7 zeros are removed, we know that the number can be divided by $2^{25}$, and also by 8, meaning the last three digits $\overline{55 a}$ can be divided by 8 (since 1000 can be divided by 8), thus $a=2$.
Since 34! can be divided by 9, the sum of its digits can also be divided by 9, so, $141+c+d \equiv 0(\bmod 9)$, which means $c+d \equiv 3(\bmod 9)$.
And 34! can also be divided by 11, so the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions can be divided by 11, thus, $80+d \equiv 61+c(\bmod 11)$, which means $8+d \equiv c(\bmod 11)$,
When $c+d=3,8+d=c$, there is no solution;
When $c+d=12,8+d=c$, there is no solution;
When $c+d=12, d=c+3$, there is no solution;
When $c+d=3, d=c+3$, we get $c=0, d=3$.
In conclusion, $a=2, b=0, c=0, d=3$. | a=2, b=0, c=0, d=3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 (2003 British Mathematical Olympiad) Find all positive integers $a, b, c$ satisfying $(a!) \cdot(b!)=a! + b! + c!$
untranslated text remains in its original format and line breaks are preserved. | Without loss of generality, assume $a \geqslant b$, then the original equation becomes $a!=\frac{a!}{b!}+1+\frac{c!}{b!}$.
Since the three terms $a!, \frac{a!}{b!}, 1$ in the above equation are integers, it follows that $c \geqslant b$.
Since each term on the right side is a positive integer, their sum is at least 3. Therefore, $a \geqslant 3$, and $a!$ is even. Thus, $\frac{a!}{b!}$ and $\frac{c!}{b!}$ have exactly one term that is odd.
(1) Assume $\frac{a!}{b!}$ is odd, then either $a=b$, or $\frac{a!}{b!}=b+1$ and $b+1$ is odd, $a=b+1$.
(i) If $a=b$, then $a!=2+\frac{c!}{a!}$,
When $a=3$, we have $b=3, c=4$,
When $a>3$, since $a!-2 \equiv-2(\bmod 3)$, so
$c=a+1$ or $a+2$ (otherwise $\frac{c!}{a!} \equiv 0(\bmod 3)$).
$\frac{c!}{a!}=a+1$ or $(a+1)(a+2)$,
$a!=(a+3)$ or $(a+1)(a+2)+2$,
When $a=4$ or 5, it does not satisfy (1).
When $a \geqslant 6$, it is clear that the left side of (1) is greater than the right side, so the original equation has no solution.
(ii) If $a=b+1$, where $b$ is even,
then the original equation becomes $(b+1)!=b+2+\frac{c!}{b!}$,
Taking both sides modulo $b+1$ gives $\frac{c!}{b!} \equiv-1(\bmod (b+1))$.
Since $\frac{a!}{b!}$ is odd, $\frac{c!}{b!}$ is even, hence $c>b$.
Thus, $\frac{c!}{b!} \equiv 0(\bmod (b+1))$, which is a contradiction.
(2) Assume $\frac{a!}{b!}$ is even, $\frac{c!}{b!}$ is odd, then $c=b$ or $c=b+1$ ($b$ is even).
(i) If $c=b$, then the equation becomes $(a!) \cdot(b!)=(a!)+2 \times(b!)$,
So $\frac{a!}{b!}(b!-1)=2$. Hence $\frac{a!}{b!}=2, b!-1=1$.
Thus $b=2, a!=4$, which is a contradiction.
(ii) If $c=b+1$, then the equation becomes $(a!)(b!)=(a!)+(b+2)(b!)$.
So $(a!)(b!-1)=(b+2)(b!)$.
Since $(b!, b!-1)=1$, it follows that $(b!-1) \mid b+2$.
Since $b$ is even, so $b=2, a!=8$, which is impossible.
In conclusion, the original equation has a unique solution $a=3, b=3, c=4$. | a=3, b=3, c=4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10. Find the positive integer solutions of the equation $3^{x}=2^{x} y+1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 10. Solution: Rewrite the equation as $3^{x}-1=2^{x} y$
(1)
This indicates that if $(x, y)$ is a solution, then $x$ cannot exceed the exponent of the factor 2 in the standard factorization of $3^{x}-1$. Let $x=2^{m}(2 n+1)$, where $m, n$ are non-negative integers. Thus, we have
$$3^{x}-1=3^{2^{m}(2 n+1)}-1=\left(3^{2 n+1}\right)^{2^{m}}-1=\left(3^{2 n+1}-1\right) \prod_{k=0}^{m-1}\left(\left(3^{2 n+1}\right)^{2^{k}}+1\right)$$
Since $3^{2 n+1}=(1+2)^{2 n+1}$
$\equiv 1+2(2 n+1)+4 n(2 n+1)(\bmod 8)($ Binomial Theorem)
$\equiv 3(\bmod 8)$,
for the required exponent, it is 1 when $m=0$, and $m+2$ when $m=1,2, \cdots$. Therefore, we can conclude that $x$ cannot exceed $m+2$.
Thus, $2^{m} \leqslant 2^{m}(2 n+1)=x \leqslant m+2$. Hence, $m \in\{0,1,2\}$ and $n=0$.
From this, we can conclude that the positive integer solutions to the given equation are $(x, y)=(1,1),(2,2),(4,5)$. | (x, y)=(1,1),(2,2),(4,5) | Geometry | math-word-problem | Yes | Yes | number_theory | false |
4. (2003 Turkey Mathematical Olympiad) Find the triple positive integer solutions $(x, m, n)$ of the equation $x^{m}=2^{2 n+1}+2^{n}+1$. | 4. Solution: Clearly, $x$ is an odd number, and let the power of 2 in $t$ be $V_{2}(t)$.
If $m$ is odd, let $y=x-1$, then $x^{m}-1=(y+1)^{m}-1=y^{m}+C_{m}^{1} y^{m-1}+C_{m}^{2} y^{m-2}+\cdots+C_{m}^{m-1} y$.
In the term $C_{m}^{m-1} y$, the power of 2 is the power of 2 in $y$, and the other terms satisfy $V_{2}\left(C_{m}^{i} y^{i}\right)=V_{2}(y)+(i-1) V_{2}(y)+V_{2}\left(C_{m}^{i}\right)>V_{2}(y)$, so $V_{2}\left(x^{m}-1\right)=V_{2}(y)=V_{2}(x-1)$.
Also, $V_{2}\left(x^{m}-1\right)=V_{2}\left(2^{2 n+1}+2^{n}\right)=n$, thus $V_{2}(x-1)=n$.
Then $2^{n} \mid x-1$. Therefore, $x-1 \geqslant 2^{n}, x \geqslant 2^{n}+1$.
And $x^{3} \geqslant\left(2^{n}+1\right)^{3}=2^{3 n}+3 \times 2^{2 n}+3 \times 2^{n}+1>2^{2 n+1}+2^{n}+1=x^{m}$.
So, $m<3$. Hence $m=1$, at this time $x=2^{2 n+1}+2^{n}+1$.
If $m$ is even, let $m=2 m_{0}$. Then $\left(x^{m_{0}}\right)^{2}=7 \times 2^{2 n-2}+\left(2^{n-1}+1\right)^{2}$,
which means $\left(x^{m_{0}}-2^{n-1}-1\right)\left(x^{m_{0}}+2^{n-1}+1\right)=7 \times 2^{2 n-2}$.
If $n=1$, then $x^{m}=2^{3}+2^{1}+1=8+2+1=11$ is not a square number, which is impossible.
Therefore, $n \geqslant 2$.
So, $x^{m_{0}}-2^{n-1}-1 \not\equiv x^{m_{0}}+2^{n-1}+1(\bmod 4)$.
Since they are both even, one of them has a power of 2 of 1, and there are only the following four cases:
$$\begin{array}{l}
\left\{\begin{array}{l}
x^{m_{0}}-2^{n-1}-1=14 \\
x^{m_{0}}+2^{n-1}+1=2^{2 n-3}
\end{array}\right. \\
\left\{\begin{array}{l}
x^{m_{0}}-2^{n-1}-1=2^{2 n-3}, \\
x^{m_{0}}+2^{n-1}+1=14 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x^{m_{0}}-2^{n-1}-1=7 \times 2^{2 n-3} \\
x^{m_{0}}+2^{n-1}+1=2 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x^{m_{0}}-2^{n-1}-1=2, \\
x^{m_{0}}+2^{n-1}+1=7 \times 2^{2 n-3}
\end{array}\right.
\end{array}$$
From (1), we have $2^{n-1}+1=2^{2 n-4}-7$, i.e.,
$2^{n-4}+1=2^{2 n-7}$, solving gives $n=4, x^{m_{0}}=23$. Hence $x=23, m_{0}=1$.
From (2), we have $2^{n-1}+1=7-2^{2 n-4}$, i.e., $2^{n-1}+2^{2 n-4}=6$.
Since $6=4+2$, there is no solution.
From (3), we have $2^{n-1}+1=1-7 \times 2^{2 n-4}$, which is also impossible.
From (4), we have $2^{n-1}+1=7 \times 2^{2 n-4}-1$, i.e., $2^{n-2}+1=7 \times 2^{2 n-5}$.
Considering the number of 1s in the binary representation, it is also impossible.
Therefore, $(x, m, n)=\left(2^{2 n+1}+2^{n}+1,1, n\right),(23,2,4)$. | (2^{2 n+1}+2^{n}+1,1, n),(23,2,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10. (2005 Vietnam Mathematical Olympiad) Find all natural number triples $(x, y, n)$ that satisfy the relation $\frac{x!+y!}{n!}=3^{n}$ (with the convention that $0!=1$). | 10. Solution: Suppose $(x, y, n)$ is a natural number triplet that satisfies
\[
\frac{x!+y!}{n!}=3^{n}
\]
It is easy to see that $n \geqslant 1$.
Assume $x \leqslant y$, and consider the following two cases:
(1) $x \leqslant n$, it is easy to see that $\frac{x!+y!}{n!}=3^{n} \Leftrightarrow 1+\frac{y!}{x!}=3^{n} \cdot \frac{n!}{x!}$
From equation (2), we can deduce that $1+\frac{y!}{x!}=0(\bmod 3)$.
Since the product of three consecutive integers is divisible by 3, and because $n \geqslant 1$, we have $x \leqslant n$, and it is easy to see that $\frac{x!+y!}{n!}=3^{n} \Leftrightarrow \frac{x!}{n!}\left(1+\frac{y!}{x!}\right)=3^{n}$.
Thus, equation (7) becomes $n+1+\frac{y!}{n!}=3^{n}$.
Since $y \geqslant x$, then $y \geqslant n+1$.
Let $A=\frac{y!}{(n+1)!}$, substituting into equation (8) gives $(n+1)(A+1)=3^{n}$.
If $y \geqslant n+4$, then $A \equiv 0(\bmod 3)$, so $A+1$ cannot be a power of 3.
Therefore, from equation (9), we get $y \leqslant n+3$, so $n+1 \leqslant y \leqslant n+3$.
(i) If $y=n+3$, then $A=(n+2)(n+3)$,
From equation (9), we get $(n+1)[1+(n+2)(n+3)]=3^{n}$, i.e., $(n+2)^{3}-1=3^{n}$
This implies $n>2, n+2 \equiv 1(\bmod 3)$.
Let $n+2=3 k+1, k \geqslant 2$, from equation (11) we have
$9 k\left(3 k^{2}+3 k+1\right)=3^{3 k-1}$, which contradicts $\left(3 k^{2}+3 k+1,3\right)=1$, so $y \neq n+3$.
(ii) If $y=n+2$, then $A=n+2$.
From equation (9), we get $(n+1)(n+3)=3^{n}$.
For $n \geqslant 1$, $(n+1)$ and $(n+3)$ cannot both be powers of 3, so $y \neq n+2$.
(iii) If $y=n+1$, then $A=1$. From equation (9), we get $2(n+1)=3^{n}$.
It is clear that there is no $n$ that satisfies the above equation, so $y \neq n+1$.
In summary, the triplets that satisfy equation (1) are $(x, y, n)=(0,2,1)$ or $(2,0,1)$ or $(1,2,1)$ or $(2,1,1)$.
Upon verification, these four solutions satisfy the condition. | (0,2,1), (2,0,1), (1,2,1), (2,1,1) | Number Theory | math-word-problem | 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 |
1. The condition $[x]=[y]$ is the condition $|x-y|<1$ ( $\quad$ )
A. Sufficient but not necessary condition
B. Necessary but not sufficient condition
C. Sufficient and necessary condition
D. Neither sufficient nor necessary condition | 1. Let $a=[x]=[y]$, then $a \leqslant x, y < a+1$, naturally we have $|x-y|<1$. Conversely, take $x=-\frac{1}{3}, y=$ $\frac{1}{3}$, then $|x-y|<1$, but $[x] \neq [y]$. | A | Algebra | MCQ | Yes | Yes | number_theory | false |
5. Find $\sum_{i=1}^{2006}\left(\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}-\frac{1}{[\sqrt[4]{i(i+5)}]}\right)$, the result is ( )
A. -1
B. 1
C. 0
D. $-\frac{1}{2}$ | 5. D When $i=1$, $\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}-\frac{1}{[\sqrt[4]{i(i+5)}]}=\frac{1}{\sqrt[4]{12}}-\frac{1}{\sqrt[4]{6}}=\frac{1}{1}-\frac{1}{1}=0$;
When $i=2$, $\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}-\frac{1}{[\sqrt[4]{i(i+5)}]}=\frac{1}{\sqrt[4]{20}}-\frac{1}{\sqrt[4]{10}}=\frac{1}{2}-\frac{1}{1}=-\frac{1}{2}$;
When $i=3$ or $i=4$, $\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}-\frac{1}{[\sqrt[4]{i(i+5)}]}=\frac{1}{2}-\frac{1}{2}=0$;
When $i \geqslant 5$, $(i+2)^{2}<i(i+5)<(i+2)(i+3)<(i+3)^{2}$, at this time $\sqrt{i+2}<\sqrt[4]{i(i+5)}<$ $\sqrt[4]{(i+2)(i+3)}<\sqrt{i+3}$. Assume at this time $[\sqrt[4]{i(i+5)}] \neq[\sqrt[4]{(i+2)(i+3)}]$, then there exists $n \in \mathbf{N}_{+}$ such that $\sqrt[4]{i(i+5)}<n \leqslant \sqrt[4]{(i+2)(i+3)}$. Thus $\sqrt{i+2}<n<\sqrt{i+3}$, i.e., $i+2<n^{2}<i+3$, which also means $n^{2}-3$ $<i<n^{2}-2$. But the interval $\left(n^{2}-3, n^{2}-2\right)$ contains no integers, contradicting $i \in \mathbf{N}_{+}$. This proves that when $i \geqslant 5$ we must have $\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}=\frac{1}{[\sqrt[4]{i(i+5)}]}$.
In summary, the original expression $=0+\left(-\frac{1}{2}\right)+0 \times 2+0 \times 2002=-\frac{1}{2}$. | D | Algebra | MCQ | Yes | Yes | number_theory | false |
6. Define the sequence $\left\{a_{n}\right\}: a_{1}=1, a_{2}=2, a_{n+2}=a_{n}+a_{n+1}, n \in \mathbf{N}_{+}$, then $\left[\frac{a_{2}}{a_{1}}\right] \cdot\left\{\frac{a_{3}}{a_{2}}\right\} \cdot\left\{\frac{a_{4}}{a_{3}}\right\} \cdot \cdots$
- $\left\{\frac{a_{99}}{a_{98}}\right\} \cdot\left[\frac{a_{98}}{a_{2}}\right]$ is ( )
A. 2
B. 0
C. 1
D. $\frac{1}{2}$ | 6. C From $a_{1}=1, a_{2}=2, a_{n+2}=a_{n}+a_{n+1}, n \in \mathbf{N}_{+}$, it is easy to know that for any $n \in \mathbf{N}_{+}$, $a_{n} \in \mathbf{N}_{+}$, and $a_{1} < a_{2} < \cdots < a_{n} < \cdots$. Therefore, $a_{n+2}=a_{n}+a_{n+1}<a_{n+1}+a_{n+1}=2 a_{n+1}$, which means $\frac{a_{n+2}}{a_{n+1}}<2$, so $1<\frac{a_{n+2}}{a_{n+1}}<2$. Thus, $\left\{\frac{a_{n+2}}{a_{n+1}}\right\}=\frac{a_{n+2}}{a_{n+1}}-1=\frac{a_{n+1}+a_{n}}{a_{n+1}}-1=\frac{a_{n}}{a_{n+1}}, n \in \mathbf{N}_{+}$. Therefore, the original expression $=\left[\frac{a_{2}}{a_{1}}\right] \cdot\left[\frac{a_{98}}{a_{2}}\right] \cdot \frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdots \cdots \cdot \frac{a_{97}}{a_{98}}=\left[\frac{a_{2}}{a_{1}}\right] \cdot\left[\frac{a_{98}}{a_{2}}\right] \cdot \frac{a_{1}}{a_{98}}=2 \cdot\left[\frac{a_{98}}{2}\right] \cdot \frac{1}{a_{98}}$. From the recurrence relation $a_{n+2}=a_{n}+a_{n+1}$ and $a_{1}=1$, $a_{2}=2$, it is easy to prove by mathematical induction that $a_{3 k+2}$ is even, and $a_{3 k+1}$ and $a_{3 k+3}$ are odd $(k \in \mathbf{N})$, so $a_{98}$ is even, and the original expression $=2 \cdot \frac{a_{98}}{2} \cdot \frac{1}{a_{98}}=1$. | C | Algebra | MCQ | Yes | Yes | number_theory | false |
8. $\sum_{k=1}^{2006} \frac{1}{[\sqrt{k}]}-\sum_{k=1}^{44} \frac{1}{k}$ The value is $\qquad$ | 8. $\frac{1927}{22}$ From $44^{2}=1936<2006<2025=45^{2}$, we know that for any $k \in\{1,2, \cdots, 2006\}$, there exists some $n$ $\in\{1,2, \cdots, 44\}$ such that $n^{2} \leqslant k<(n+1)^{2}$. At this time, $\frac{1}{[\sqrt{k}]}=\frac{1}{n}$.
$$\begin{aligned}
t^{2} \sum_{k=1}^{2006} \frac{1}{[\sqrt{k}]} & =\sum_{1 \leqslant k<2^{2}} \frac{1}{[\sqrt{k}]}+\sum_{2^{2} \leqslant k<3^{2}} \frac{1}{[\sqrt{k}]}+\cdots+\sum_{43^{2} \leqslant k<44^{2}} \frac{1}{[\sqrt{k}]} \\
& +\sum_{4^{2} \leqslant k<2006} \frac{1}{[\sqrt{k}]}=\sum_{1<k<2^{2}} \frac{1}{1}+\sum_{2^{2} \leqslant k<3^{2}} \frac{1}{2}+\cdots+\sum_{43^{2} \leqslant k<44^{2}} \frac{1}{43}+\sum_{44^{2} \leqslant k \leqslant 2006} \frac{1}{44} \\
& =\frac{2^{2}-1}{1}+\frac{3^{2}-2^{2}}{2}+\cdots+\frac{44^{2}-43^{2}}{43}+\frac{2006-44^{2}+1}{44}=\frac{3}{1}+\frac{5}{2}+\cdots+\frac{87}{43}+\frac{71}{44} \\
& =2 \times 43+\frac{71}{44}+\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{43} .
\end{aligned}$$
Therefore, the original expression $=86+\frac{71}{44}+\sum_{k=1}^{43} \frac{1}{k}-\sum_{k=1}^{44} \frac{1}{k}=86+\frac{70}{44}=\frac{1927}{22}$. | \frac{1927}{22} | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 5 (2003 2004 Swedish Mathematical Competition) Find all real numbers $x$ that satisfy the equation $\left[x^{2}-2 x\right]+2[x]=$ $[x]^{2}$, where $[a]$ denotes the greatest integer less than or equal to $a$.
| Analyzing $\left[x^{2}-2 x\right]$ is relatively complex, so we consider using substitution to simplify it.
Let $x=y+1$, then the original equation becomes
$$\left[(y+1)^{2}-2(y+1)\right]+2[y+1]=[y+1]^{2},$$
which is $\left[y^{2}-1\right]+2[y+1]=[y+1]^{2}$,
or $\left[y^{2}\right]-1+2[y]+2=[y]^{2}+2[y]+1$.
or $\left[y^{2}\right]=[y]^{2}$.
Let $n=[y]$, then $\left[y^{2}\right]=n^{2}$, so $n^{2} \leqslant y^{2}<n^{2}+1$.
Thus $|n| \leqslant y<\sqrt{n^{2}+1}$ or $-\sqrt{n^{2}+1}<y \leqslant-|n|$.
When $n \in \mathbf{N}$, $n \leqslant y<\sqrt{n^{2}+1}$, satisfying $n=[y]$;
When $n \notin \mathbf{N}$, i.e., $n \leqslant-1$, $-\sqrt{n^{2}+1}<y \leqslant n$, combined with $n=[y]$ we know that in this case it can only be $y=n$. Therefore, $y \in \mathbf{Z} \cup\left[n, \sqrt{n^{2}+1}\right)(n \in \mathbf{N})$.
Thus, $x \in \mathbf{Z} \cup\left[n+1, \sqrt{n^{2}+1}+1\right)(n \in \mathbf{N})$. | x \in \mathbf{Z} \cup\left[n+1, \sqrt{n^{2}+1}+1\right)(n \in \mathbf{N}) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 8 (2003 Bulgaria National Mathematical Olympiad) Find all real numbers $a$ that satisfy the condition $4[a n]=n+[a[a n]]$, where $n$ is any positive integer.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The provided text is already in English, so no translation is needed. If you intended to translate the problem statement into English, it has been done as requested.)
Example 8 (2003 Bulgaria National Mathematical Olympiad) Find all real numbers $a$ that satisfy the condition $4[a n]=n+[a[a n]]$, where $n$ is any positive integer. | Analysis Since $n$ is any positive integer, we should assign appropriate values to $n$ to obtain the required equation.
Solution If $a \leqslant 0$, then $[a n] \leqslant 0, a[a n] \geqslant 0$.
Thus, $n+[a[a n]] \geqslant n+0>0 \geqslant 4[a n]$, which is a contradiction!
Therefore, $a>0$.
Thus, $4(a n-1)n+a(a n-1)-1$,
So $4 a>1+a^{2}-\frac{a+1}{n}$.
And $4 a<1+a^{2}+\frac{4}{n}$, i.e., $1+a^{2}-\frac{a+1}{n}<4 a<1+a^{2}+\frac{4}{n}$.
Let $n \rightarrow+\infty$, we get $1+a^{2}=4 a$,
So $a=2-\sqrt{3}$ or $a=2+\sqrt{3}$.
Let $n=1$, then $4[a]=1+[a[a]]$, and $a=2-\sqrt{3}$ does not satisfy $4[a]=1+[a[a]]$. Hence $a \neq 2-\sqrt{3}$.
When $a=2+\sqrt{3}$, for any $n \in \mathbf{N}_{+}$, let $\left[\frac{n}{a}\right]=b,\left\{\frac{n}{a}\right\}=c$.
Then, from $1+a^{2}=4 a$, we know $a=4-\frac{1}{a}$, so
$$\begin{array}{l}
n+[a[a n]]=\left[n+a\left[4 n-\frac{n}{a}\right]\right] \\
=[n+a(4 n-b-1)] \\
=[a(4 n-1+c)]=\left[\left(4-\frac{1}{a}\right)(4 n-1+c)\right] \\
=\left[4(4 n-1)-4\left(\frac{n}{a}-c\right)+\frac{1-c}{a}\right] \\
=4(4 n-1-b)=4\left[4 n-\frac{n}{a}\right]=4[a n] .
\end{array}$$
This means that $a=2+\sqrt{3}$ satisfies the condition.
Therefore, $a=2+\sqrt{3}$ is the only solution. | a=2+\sqrt{3} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Let $m$ be a positive integer, the equation $x^{2}-\left[x^{2}\right]=\{x\}^{2}$ has ( ) solutions on $[1, m]$.
A. 0
B. $m(m-1)+1$
C. $m^{2}$
D. infinitely many | 2. B Let $[x]=a,\{x\}=b$, then from $x^{2}-\left[x^{2}\right]=\{x\}^{2}$ we get $(a+b)^{2}-\left[(a+b)^{2}\right]=b^{2}$, which means $(a+b)^{2}-a^{2}-\left[b^{2}+2 a b\right]=b^{2}$. Therefore, $2 a b=\left[b^{2}+2 a b\right]$.
From this, we know $2 a b \in \mathbf{Z}$, and when $2 a b \in \mathbf{Z}$, by $0 \leqslant b<1$ we know $0 \leqslant b^{2}<1$, so at this time $2 a b=\left[b^{2}+2 a b\right]$, which means $x$ is a solution of the equation $\Leftrightarrow 2 a b \in \mathbf{Z}$. For any $a \in\{1,2, \cdots, m-1\}$, there are $a$ values of $b$ that satisfy the requirement, plus $x=m$, there are $m(m-1)+1$ solutions. | B | Algebra | MCQ | Yes | Yes | number_theory | false |
4. The sum of all roots of the equation $[3 x+1]=2 x-\frac{1}{2}$ is ( )
A. -1
B. -2
C. $-\frac{5}{4}$
D. 0 | 4. B Let $[3 x+1]=t$, then $t$ is an integer and $0 \leqslant(3 x+1)-t<1$.
Thus, the original equation becomes $t=2 x-\frac{1}{2}$, which means $x=\frac{1}{2} t+\frac{1}{4}$.
Substituting the above equation into (1) yields $0 \leqslant \frac{3}{2} t+\frac{3}{4}+1-t<1$.
This simplifies to $-\frac{7}{2} \leqslant t \leqslant-\frac{3}{2}$
From (2) and $t \in \mathbf{Z}$, we know $t=-2$ or $t=-3$.
When $t=-2$, $x_{1}=-\frac{3}{4}$; when $t=-3$, $x_{2}=-\frac{5}{4}$.
Therefore, $x_{1}+x_{2}=\left(-\frac{3}{4}\right)+\left(-\frac{5}{4}\right)=-2$. | B | Algebra | MCQ | Yes | Yes | number_theory | false |
5. The smallest natural number $n$ for which the equation $\left[\frac{10^{n}}{x}\right]=2006$ has an integer solution $x$ lies in the interval $(\quad)$
A. $[1,6]$
B. $[7,12]$
C. $[13,18]$
D. $[19,24]$ | 5. B Since $\left[\frac{10^{n}}{x}\right]=2006$, then $2006 \leqslant \frac{10^{n}}{x}<2007$, so $\frac{10^{n}}{2006}<x \leqslant \frac{10^{n}}{2007}$, which means 0.00049850 $\cdots<x<0.00049825 \cdots$ Therefore, when $n \geqslant 7$, $\left(\frac{10^{n}}{2006}, \frac{10^{n}}{2007}\right]$ contains an integer, so the minimum value of $n$ is 7. | B | Number Theory | MCQ | Yes | Yes | number_theory | false |
9. The solutions to the equation $x^{2}-8[x]+7=0$ are
The translation maintains the original text's line breaks and format. | 9. $x=1, \sqrt{33}, \sqrt{41}, 7$
Let $[x]=n$, then the known equation can be written as $x^{2}+7=8 n$, so $n>0$.
By $n \leqslant x0$.
Solving, we get $1 \leqslant n<2$ or $4<n \leqslant 7$, hence $n=1,5,6,7$. Then we have $x^{2}+7=8,40,48,56$ solving for $x= \pm 1$, $\pm \sqrt{33}, \pm \sqrt{41}, \pm 7$
Since $n \geqslant 0$, so $x \geqslant 0$, hence only $x=1, \sqrt{33}, \sqrt{41}, 7$. | x=1, \sqrt{33}, \sqrt{41}, 7 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
10. Let $a>1$ be a positive real number, and $n \geqslant 2$ be a natural number, and the equation $[a x]=x$ has exactly $n$ distinct solutions, then the range of values for $a$ is $\qquad$ . | 10. $\left[1+\frac{1}{n}, 1+\frac{1}{n-1}\right)$
By the problem statement, $x$ must be an integer. When $x<0$, $[a x]<a x<x$, so $x \geqslant 0$. Let $\{a\}=a-[a]$, then the original equation becomes $x=[a x]=[a] x+[\{a\} x]$. Since $[a] \geqslant 1$, the above equation holds $\Leftrightarrow[a]=1$, and $\{a\} x<1$. Also, $x=0$ clearly satisfies the condition.
Therefore, $x<\frac{1}{\{a\}}$ holds exactly for $n-1$ positive integers $1,2, \cdots, n-1$, thus $\frac{1}{n} \leqslant\{a\}<\frac{1}{n-1}$. That is, $1+\frac{1}{n} \leqslant a<1+\frac{1}{n-1}$. | \left[1+\frac{1}{n}, 1+\frac{1}{n-1}\right) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
11. The solution set of the equation $[\tan x]=2 \cos ^{2} x$ is $\qquad$ | 11. $\left\{x \left\lvert\, x=k \pi+\frac{\pi}{4}\right., k \in \mathbf{Z}\right\}$
Let $\tan x=y$, then from $2 \cos ^{2} x=\cos 2 x+1=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}$, we know $[y]=\frac{1-y^{2}}{1+y^{2}}+1=\frac{2}{1+y^{2}}>0$, hence $[y] \geqslant 1$, so $y \geqslant 1$. From $y \geqslant 1$ we know $\frac{2}{1+y^{2}} \leqslant \frac{2}{1+1^{2}}=1$, thus $1 \leqslant[y] \leqslant 1$ which means $[y]=1$, hence $y=1$, i.e., $\tan x=1$. | x=k\pi+\frac{\pi}{4}, k \in \mathbf{Z} | Algebra | 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 |
Example 6 (2006 National Training Team Test) Find all positive integer pairs $(a, n)$ such that $\frac{(a+1)^{n}-a^{n}}{n}$ is an integer.
Find all positive integer pairs $(a, n)$ such that $\frac{(a+1)^{n}-a^{n}}{n}$ is an integer. | First, we point out that $(a, 1)$ is clearly a solution to the original problem (here $a$ is any positive integer). Below, we prove that the original problem has no other solutions.
Assume $(a, n) (n \geqslant 2)$ is a solution to the original problem, then there must exist a positive integer $k$ such that $(a+1)^{n}-a^{n}=k n$. Since $a+1$ and $a$ are coprime, it follows from the above equation that $n$ is coprime with both $a$ and $a+1$. By Euler's theorem, we have: $(a+1)^{\varphi(n)} \equiv a^{\varphi(n)} \equiv 1(\bmod n)$.
Let $d=(n, \varphi(n))$, then by Bézout's theorem, there exist integers $\alpha$ and $\beta$ such that
$$d=\alpha n+\beta \varphi(n),$$
From $(a+1)^{n} \equiv a^{n}(\bmod n)$ and $(a+1)^{\varphi(n)} \equiv a^{\phi(n)} \equiv 1(\bmod n)$ (since $(a+1)^{n}-a^{n}=k n$, we have $(a+1)^{n} \equiv a^{n}(\bmod n)$), we can deduce
$$(a+1)^{d} \equiv (a+1)^{\alpha n+\beta \varphi(n)} \equiv a^{\alpha n+\beta \varphi(n)} \equiv a^{d}(\bmod n)$$
Clearly, $d>1$ (otherwise $a+1 \equiv a(\bmod n)$, then $n=1$, which contradicts the assumption), and note that $\varphi(n) > 1$ for $n > 1$. Therefore, the only positive integer pairs $(a, n)$ that make $\frac{(a+1)^{n}-a^{n}}{n}$ an integer are $(a, 1)$ (where $a$ can be any positive integer). | (a, 1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1. Regardless of the shape, the smallest positive integer $n$ that makes the total number of lattice points inside and on the boundary of a convex $n$-sided polygon $\geqslant n+1$ is ( )
A. 4
B. 5
C. 6
D. 7 | 1. B Consider the square $A B C D$ with vertices $A(0,0), B(0,1), C(1,1), D(1,0)$. There are no lattice points inside this square, so the required $n \geqslant 5$.
On the other hand, by classifying lattice points based on their parity, we can divide them into four categories: (odd, odd), (odd, even), (even, odd), (even, even). Among the five vertices of the convex pentagon, there must be two that belong to the same category, and the midpoint of these two vertices is also a lattice point, and this lattice point lies inside the convex pentagon, so the required $n \leqslant 5$. | B | Geometry | MCQ | Yes | Yes | number_theory | false |
2. Among all lines passing through the point $\left(10, \frac{1}{2}\right)$, the number of lines that pass through at least two lattice points is ( )
A. one
B. none
C. a finite number but no less than 2
D. infinitely many | 2. D For any $m \in \mathbf{Z}$, the line $m(2 y-1)=x-10$ passes through the point $\left(10, \frac{1}{2}\right)$, and also through the lattice point $(10+$ $m(2 t-1), t)(t \in \mathbf{Z})$. Therefore, by the arbitrariness of $m$, we know that there are infinitely many such lines. | D | Number Theory | MCQ | Yes | Yes | number_theory | false |
4. $C$ is a circle with radius $r$, centered at the point $(\sqrt{2}, \sqrt{3})$, where $r$ is a positive real number. Then the maximum number of integer points on $C$ is ( ) .
A. 0
B. 1
C. 4
D. Infinity | 4. B The equation of circle $C$ is: $(x-\sqrt{2})^{2}+(y-\sqrt{3})^{2}=r^{2}$.
If circle $C$ has two lattice points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then $\left(x_{1}-\sqrt{2}\right)^{2}+\left(y_{1}-\sqrt{3}\right)^{2}=r^{2},\left(x_{2}\right.$ $-\sqrt{2})^{2}+\left(y_{2}-\sqrt{3}\right)^{2}=r^{2}$
Subtracting the above two equations yields $2 \sqrt{2}\left(x_{2}-x_{1}\right)+2 \sqrt{3}\left(y_{2}-y_{1}\right)+\left(x_{1}^{2}+y_{1}^{2}\right)-\left(x_{2}^{2}+y_{2}^{2}\right)=0$.
However, $x_{1}=x_{2}, y_{1}=y_{2}$ cannot hold simultaneously. Therefore, $2 \sqrt{2}\left(x_{2}-x_{1}\right)+2 \sqrt{3}\left(y_{2}-y_{1}\right) \notin \mathbf{Q}(\mathbf{Q}$ is the set of rational numbers), which contradicts the above equation. Hence, circle $C$ can pass through at most one lattice point.
Furthermore, when $r=\sqrt{5}$, circle $C$ passes through the origin, so circle $C$ can pass through at most one lattice point. | B | Geometry | MCQ | Yes | Yes | number_theory | false |
5. In the Cartesian coordinate system, grid points are numbered as follows: $(0,0)$ is No. 1, $(1,0)$ is No. 2, $(1,1)$ is No. 3, $(0,1)$ is No. 4, $(0,2)$ is No. 5, $(1,2)$ is No. 6, $(2,2)$ is No. 7, $(2,1)$ is No. 8, $(2,0)$ is No. 9, $\cdots$ (as shown in Figure 7-1). Following the order of the arrows in the figure, the coordinates of the 2006th point are . $\qquad$ | 5. $(44,19)$ Since the number of lattice points in the region $0 \leqslant x \leqslant k, 0 \leqslant y \leqslant k$ is $(k+1)^{2}$, and $44^{2}=19362006$, the x-coordinate or y-coordinate of the point with number 2006 is 44. Since 44 is even, the point should be counted from $(0,44)$ to the right. Since $2006-44^{2}=70>44$, the x-coordinate of the point with number 2006 is 44, and the y-coordinate is $44-(70-45)=19$. Therefore, the coordinates of the point with number 2006 are $(44,19)$. | (44,19) | Combinatorics | 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 |
1. A triangle has three sides of integer lengths, with the longest side being 11. The number of such triangles is ( ) .
A. 32
B. 34
C. 36
D. 40 | 1. C Let the three sides of a triangle be integers $x, y, z$, and $x \leqslant y \leqslant z$, then $z=11, x+y \geqslant 12, x \leqslant y \leqslant 11$.
Therefore, we only need to find the number of integer points within the triangle (including the boundary) formed by the lines $y=12-x, y=11, y=x$. It is easy to calculate that the number of integer points is 36. | C | Geometry | MCQ | 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 4 Given that $m, n$ are integers greater than 7, consider a rectangular array of $m \times n$ points. Color $k$ of these points red, such that the three vertices of any right-angled triangle with its two legs parallel to the sides of the rectangle are not all red points. Find the maximum value of $k$. | Assume there are $m$ rows and $n (n \leqslant m)$ columns, then the maximum value is at least $m+n-2$, because by selecting any row and column, and coloring all points in these row and column except their common point in red, it satisfies the condition.
Below, we use mathematical induction to prove that in an $m \times n$ grid, the number of red points $\leqslant m+n-2$.
When $n=2$, if the number of red points is $m+n-1=m+1$, then there must be a row with 2 red points, which means the other points in these two columns cannot be red, leading to a contradiction!
When $n=3$, if the number of red points is $m+n-1=m+2$, then there must be a row with 2 red points, which means the other points in these two columns cannot be red, and the other column can have at most $m-1$ red points, leading to a contradiction.
If for $n \leqslant k$, in an $m \times n$ grid, the number of red points $\leqslant m+n-2$, assume for $n=k+1$, in an $m \times n$ grid, there are $m+n-1=m+k$ red points. By the pigeonhole principle, there must be a row with 2 red points, and in the columns of these 2 red points, there are no other red points. Removing these two columns, we get an $m \times (k-1)$ grid with $m-k-2$ red points, which contradicts the induction hypothesis.
Therefore, in an $m \times n$ grid, the number of red points $\leqslant m+n-2$.
In conclusion, the maximum value of $k$ is $m+n-2$. | m+n-2 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
8. If $n$ is a natural number less than 50, find all values of $n$ such that the values of the algebraic expressions $4 n+5$ and $7 n+6$ have a common divisor greater than 1.
Let the above text be translated into English, please retain the original text's line breaks and format, and output the translation result directly. | 8. Solution: Let $(4 n+5,7 n+6)=d>1$, then $d|(4 n+5), d|(7 n+6)$
Thus $d \mid(7 n+6-(4 n+5))=3 n+1$,
$$\begin{array}{l}
d \mid((4 n+5)-(3 n+1))=n+4 \\
d \mid((3 n+1)-2(n+4))=n-7 \\
d \mid((n+4)-(n-7))=11
\end{array}$$
Since 11 is a prime number, then $d=11$.
Let $n-7=11 k$, then
$0<n=11 k+7<50$. Solving for $k$ gives $k=0,1,2,3$.
Thus $n=7,18,29,40$. | n=7,18,29,40 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10. (2005 Canadian Mathematical Olympiad) If an ordered triple of positive integers $\{a, b, c\}$ satisfies $a \leqslant b \leqslant c, (a, b, c)=1, a^{n}+b^{n}+c^{n}$ is divisible by $a+b+c$, then $\{a, b, c\}$ is called $n$-energetic. For example, $\{1,2,3\}$ is 5-energetic.
(1) Find all ordered triples of positive integers that satisfy: for any $n \geqslant 1$, the ordered triple of positive integers is $n$-energetic;
(2) Find all ordered triples of positive integers that are both 2004-energetic and 2005-energetic but not 2007-energetic. | 10. Solution: (1) Since $(a+b+c)\left|\left(a^{2}+b^{2}+c^{2}\right),(a+b+c)\right|\left(a^{3}+b^{3}+c^{3}\right)$, then $(a+b+c) \mid\left[(a+b+c)^{2}-a^{2}-b^{2}-c^{2}\right]$,
i.e., $(a+b+c) \mid(2 a b+2 b c+2 c a)$.
Also, $(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3}-3 a b c$.
Thus, $(a+b+c) \mid 3 a b c$.
Let the prime $p$ satisfy $p^{a} \|(a+b+c)(\alpha \geqslant 1)$ (notation \| see page 18 of this book). If there exists $p \geqslant 5$, then $p \mid a b c$, assume without loss of generality that $p \mid a$.
Since $p \mid 2(a b+b c+c a)$, then $p \mid b c$.
Assume without loss of generality that $p \mid b$, since $p \mid(a+b+c)$, then $p \mid c$.
Thus, $p \mid(a, b, c)$, which is a contradiction.
Therefore, $p=2$ or 3.
Hence, $a+b+c=2^{m} \times 3^{n}(m, n \geqslant 0)$.
If $n \geqslant 2$, then $3|(a+b+c), 3| a b c, 3 \mid(a b+b c+c a)$.
Similarly, it can be deduced that $3 \mid(a, b, c)$, which is a contradiction.
Thus, $n=0$ or 1.
Let $a+b+c=2^{m} \cdot k(k=1$ or 3$)$, then $2^{m} \mid a b c$,
Since $(a, b, c)=1$, assume without loss of generality that $a$ is odd.
Also, $2^{m} \mid(a+b+c)$, assume without loss of generality that $b$ is odd, and $c$ is even, so $2^{m} \mid c$.
From $2^{m} \mid(2 a b+2 b c+2 c a)$, then $2^{m-1} \mid a b$.
Thus, $m=0$ or 1.
Also, $a+b+c \geqslant 3$, so, $a+b+c=3$ or 6.
By verification, $(a, b, c)=(1,1,1)$ or $(1,1,4)$.
(2) It is easy to get $a^{n}+b^{n}+c^{n}=(a+b+c)\left(a^{n-1}+b^{n-1}+c^{n-1}\right)-(a b+b c+c a)\left(a^{n-2}+b^{n-2}+\right.$ $\left.c^{n-2}\right)+a b c\left(a^{n-3}+b^{n-3}+c^{n-3}\right)$
Also, $(a+b+c)\left|\left(a^{2004}+b^{2004}+c^{2004}\right),(a+b+c)\right|\left(a^{2005}+b^{2005}+c^{2005}\right)$.
Substituting $n=2007$ into equation (1) yields
$(a, b, c) \mid\left(a^{2007}+b^{2007}+c^{2007}\right)$, which contradicts the given condition.
Therefore, there does not exist a set of $(a, b, c)$ that satisfies the condition. | (1,1,1) \text{ or } (1,1,4) | Number Theory | 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 |
3. Given the equation $x^{4}-p x^{3}+q=0$ has an integer root, find the prime numbers $p$ and $q$.
untranslated text:
已知方程 $x^{4}-p x^{3}+q=0$ 有一整数根, 求素数 $p 、 q$.
translated text:
Given the equation $x^{4}-p x^{3}+q=0$ has an integer root, find the prime numbers $p$ and $q$. | 3. Solution: Let the integer $x$ satisfy $q=x^{3}(p-x)$.
(1)
Then $x \mid q$.
Since $q$ is a prime, it can only be that $x= \pm 1, x= \pm q$.
When $x= \pm q$, from (1) we have
$q^{2} \mid 1$, which is impossible (since $q$ is a prime).
When $x=-1$, we have $p+q+1=0$, which is also impossible.
When $x=1$, from (1) we get $p=q+1$,
Since $p, q$ are both primes, then $p=3, q=2$. | p=3, q=2 | Algebra | 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 |
$$\text { 9. } 3333^{8888}+8888^{3333} \equiv$$
$$\qquad (\bmod 7)$$. | 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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