problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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5. Let $m, n \in \mathbf{N}_{+}$, and for any $k \in \mathbf{N}_{+}$, we have $(17 k-1, m)=(17 k-1, n)$. Prove: There exists $l \in \mathbf{Z}$, such that $m=17^{l} \cdot n$. | 5. Proof: Let $m=17^{i} p, n=17^{j} q$, where $i, j \in \mathbf{N}, p, q \in \mathbf{N}_{+},(p, 17)=(q, 17)=1$.
Below we prove: $p=q$.
Assume $p>q$ (the case $p<q$ can be discussed similarly), since $(p, 17)=1$, by the Chinese Remainder Theorem, there exists $a \in \mathbf{N}_{+}$, such that $a \equiv 0(\bmod p), a \e... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,190 |
1. (Euler's Theorem) If $(a, m)=1$, then: $a^{\phi(m)} \equiv 1(\bmod m)$. | Proof: Let $r_{1}, r_{2}, r_{3}, \cdots, r_{\alpha(m)}$ be a reduced residue system modulo $m$. Since $(a, m)=1$, it follows that $a r_{1}, a r_{2}, \cdots, a r_{\phi(m)}$ is also a reduced residue system modulo $m$.
Thus, $a r_{1} \cdot a r_{2} \cdots a r_{q(m)} \equiv r_{1} r_{2} \cdots r_{\rho^{(m)}}(\bmod m)$,
whic... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,191 |
From the definition of the order of $a$ modulo $m$ (denoted as $k$), it is easy to see that $k$ has the following properties:
(1) Suppose $(a, m)=1, k$ is the order of $a$ modulo $m$, and $u, v$ are any integers, then $a^{u} \equiv a^{v}(\bmod m)$ if and only if $u \equiv v(\bmod k)$. In particular, $a^{u} \equiv 1(\bm... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 742,192 |
3. (Fermat's Theorem) If $p$ is a prime, then $a^{p} \equiv a(\bmod p)$. | Proof: If $(a, p) \neq 1$, then $p \mid a$, in which case the conclusion is obviously true. If $(a, p)=1$, then by Euler's theorem $a^{\varphi(p)} \equiv 1(\bmod p)$, and since $\varphi(p)=p-1$, we have $a^{p-1} \equiv 1(\bmod p)$, so $a^{p} \equiv a(\bmod p)$.
Theorem proved. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,193 |
4. (Wilson's Theorem) If $p$ is a prime, then $(p-1)!\equiv-1(\bmod p)$. | Proof: For any integer $a, 1 \leqslant a \leqslant p-1$, by Bézout's theorem, there exists an integer $a^{\prime}$ such that $a a^{\prime} \equiv 1$ $(\bmod p)$. Without loss of generality, let $1 \leqslant a^{\prime} \leqslant p-1$, then $a^{\prime}$ is called the modular inverse of $a$.
If there is an integer $b$ su... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,194 |
Proof of a lemma: For a prime $p>3$ and $p \equiv 2(\bmod 3)$, if $x^{3} \equiv y^{3}(\bmod p)$, then $x \equiv y(\bmod p)$. | Lemma proof: (1) If one of $x, y$ is a multiple of $p$, then it is easy to see that the lemma holds in this case. (2) If $p \nmid x, p \nmid y$, then since $p$ is a prime, we have $(x, p)=(y, p)=1$.
By Fermat's Little Theorem:
$$x^{p-1} \equiv y^{p-1} \equiv 1(\bmod p)$$
Also, $x^{3} \equiv y^{3}(\bmod p), p \equiv 2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,197 |
Example 9 (2006 National Training Team Problem) Prove: For any positive integers $m, n$, there always exists a positive integer $k$, such that $2^{k} - m$ has at least $n$ distinct prime factors.
| Prove that for a fixed $m$, assuming $m$ is odd. We will prove that for any positive integer $n$, there exists $k_{n}$ such that $2^{k_{n}}-m$ has at least $n$ distinct prime factors.
(1)
We will prove (1) using mathematical induction:
(1) When $n=1$, $2^{3m}-m$ clearly has at least one prime factor;
(2) Assume $2^{k_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,198 |
3. Let $m>n \geqslant 1$, find the minimum value of $m+n$ such that $: 1000 \mid 1978^{m}-1978^{n}$. | 3. Solution: When $n \geqslant 3$, from $1000 \mid 1978^{m}-1978^{n}$, we get
$125 \mid 1978^{m-n}-1$, and since $1978=15 \times 125+103$,
we have $125 \mid 103^{m-n}-1$, so $25 \mid 103^{m-n}-1$.
Thus, $25 \mid 3^{m-n}-1$.
Let $m-n=l$, then: $25 \mid 3^{l}-1$ (obviously $l$ is even, otherwise it is easy to see that $3... | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,199 |
7. Let $p, q$ be two odd primes, $p>q$. Prove: For any positive integer $m, p q \nmid m^{p-q}+1$. | 7. Proof: (by contradiction) Suppose there exists $m \in \mathbf{N}_{+}$, such that $p q \mid m^{p-q}+1$, then $(m, p)=(m, q)=1$. From the condition, we have: $p\left|m^{p-q}+1 \Rightarrow p\right| m^{p-1}+m^{q-1}$.
By Fermat's Little Theorem, $m^{p-1} \equiv 1(\bmod p)$, so $p \mid m^{q-1}+1$.
Let $q-1=2^{k} \cdot \a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,200 |
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 ... | 2,3,4,6,8,12,24 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,201 |
2. Let $a, b$ be positive integers, $a^{2}+a b+1$ is divisible by $b^{2}+a b+1$, 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. | 2. Proof: According to the problem, \(a^{2}+a b+1\) is divisible by \(b^{2}+a b+1\), then \(\left(a^{2}+a b+1, b^{2}+a b+1\right)=b^{2}+a b+1\),
Also, since \(b\left(a^{2}+a b+1\right)-a\left(b^{2}+a b+1\right)=b-a\), if \(a \neq b\), by Bézout's theorem, \(\left(b^{2}+a b+1\right) \mid (b-a)\),
i.e., \(b^{2}+a b+1 \l... | null | Number Theory | proof | Yes | Yes | number_theory | false | 742,202 |
12. Given that $p$ is an odd prime greater than 3, and $a>b>1$. Prove: $C_{a}^{b p} \equiv C_{a}^{b}\left(\bmod p^{3}\right)$. | 12. Proof: Let $f(x)=(x p-1)(x p-2) \cdots(x p-p+1)\left(x \in \mathbf{N}_{+}\right)$,
Then $C_{a p}^{b p}=\frac{a p \cdot(a p-p) \cdots(a p-b p+p)}{b p(b p-p) \cdots 2 p \cdot p} \cdot \frac{f(a) f(a-1) \cdots f(a-b+1)}{f(b) f(b-1) \cdots f(1)}$
$$=C_{a}^{b} \frac{f(a) f(a-1) \cdots f(a-b+1)}{f(b) f(b-1) \cdots f(2) ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,203 |
14. Let $S$ be the set of all prime numbers $p$ that satisfy the following condition: the number of digits in the smallest repeating block of the decimal part of $\frac{1}{p}$ is a multiple of 3, i.e., for each $p \in S$, there exists the smallest positive integer $r=r(p)$, such that $\frac{1}{p}=0 . a_{1} a_{2} \cdots... | 14. Prove: (1) The length of the smallest repeating cycle of $\frac{1}{p}$ is the smallest integer $d (d \geqslant 1)$ such that $10^{d}-1$ is divisible by $p$.
Let $q$ be a prime number, and $N_{q}=10^{2 q}+10^{q}+1$. Then $N_{q} \equiv 3(\bmod q)$. Let $p_{q}$ be a prime factor of $\frac{N_{q}}{3}$. Then $p_{q}$ can... | 19 | Number Theory | proof | Yes | Yes | number_theory | false | 742,204 |
Theorem 1 Let $m=m_{1} m_{2}, x_{i}^{(1)}$ be a complete residue system modulo $m_{1}$, and $x_{j}^{(2)}$ be a complete residue system modulo $m_{2}$. Then $x_{i j}$ $=x_{i}^{(1)}+m_{1} x_{j}^{(2)}$ is a complete residue system modulo $m$. That is, as $x^{(1)}, x^{(2)}$ run through the complete residue systems modulo $... | Prove that at this time, $x_{ij}$ has a total of $m=m_{1} m_{2}$ numbers, so it is only necessary to prove that they are pairwise distinct modulo $m$.
If $x_{i_{1}}^{(1)}+m_{1} x_{j_{1}}^{(2)} \equiv x_{i_{1} j_{1}} \equiv x_{i_{2} j_{2}} \equiv x_{i_{2}}^{(1)}+m_{1} x_{j_{2}}^{(2)}\left(\bmod m_{1} m_{2}\right)$,
then... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,205 |
Theorem 2 Let $m=m_{1} m_{2},\left(m_{1}, m_{2}\right)=1, x^{(1)}, x^{(2)}$ traverse the complete residue systems modulo $m_{1}, m_{2}$, respectively, then $x_{i j}=m_{2} x_{i}^{(1)}+m_{1} x_{j}^{(2)}$ traverses the complete residue system modulo $m_{1} m_{2}$. | Prove Theorem 2 for $k=2$.
Assume the theorem holds for $k=n (n \geqslant 2)$.
When $k=n+1$,
let $\bar{x}^{(n)}=\frac{m}{m_{1} m_{n+1}} x^{(1)}+\cdots+\frac{m}{m_{n} m_{n+1}} x^{(n)}$,
we have $x=m_{n+1} \bar{x}^{(n)}+\frac{m}{m_{n+1}} x^{(n+1)}$.
From the above two equations and the fact that $k=2$ holds, the conclusi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,206 |
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} \eq... | x \equiv 394(\bmod 1155) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,209 |
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... | x \equiv 3, 31 \pmod{56} \text{ for (I), no solution for (II)} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,210 |
4. For any positive integer $n$, there exist $n$ consecutive positive integers such that in the standard factorization of each positive integer, the exponent of each prime is less than or equal to 1. | 4. We only need to prove that there exist $n$ consecutive positive integers, each of which has at least one prime factor that appears exactly once in its standard factorization. We select $n$ distinct primes $p_{1}, p_{2}, \cdots, p_{n}$, and consider the system of congruences $x \equiv -i + p_{i} \left(\bmod p_{i}^{2}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,211 |
5. Let $n$ be any positive integer, prove: there must exist $n$ consecutive integers, none of which are prime numbers.
. | 5. Take $2 n$ distinct prime numbers $p_{1}, p_{2}, \cdots, p_{2 n}$. By the Chinese Remainder Theorem, the system of congruences $x \equiv -k$ $\left(\bmod p_{2 k-1} p_{2 k}\right), k=1,2, \cdots, n$ has a solution. Since for any $k, x+k$ has at least two different prime factors, it cannot be a prime number. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,212 |
6. Given a positive integer $n$, let $f(n)$ be the smallest positive integer such that $\sum_{k=1}^{(n)} k$ is divisible by $n$. Prove: $f(n)=2n-1$ if and only if $n$ is a power of 2. | 6. (1) When $n=2^{m}$, $\sum_{k=1}^{2 n-1} k=\frac{(2 n-1) \cdot 2 n}{2}=\left(2^{m+1}-1\right) \cdot 2^{m}$. On one hand, $2^{m}$ divides the above expression. On the other hand, if $r \leqslant 2 n-2$, then $\sum_{k=1}^{r} k=\frac{r(r+1)}{2}$ cannot be divided by $2^{m}$, because one of $r$ and $r+1$ is odd, and the ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,213 |
4. (32nd United States of America Mathematical Olympiad) Prove: For each positive integer $n$, there exists an $n$-digit positive integer that is divisible by $5^{n}$ and has only odd digits. | 4. Prove by induction: When $n=1$, $5 \mid 5$.
Assume when $n=m$, $5^{m} \mid \overline{a_{1} a_{2} \cdots a_{m}}$, where $a_{i}(i=1,2, \cdots, m)$ are single-digit odd numbers.
When $n=m+1$, consider $\overline{1 a_{1} a_{2} \cdots a_{m}}, \overline{3 a_{1} a_{2} \cdots a_{m}}, \cdots, \overline{9 a_{1} a_{2} \cdots ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,214 |
Theorem 2 (Euler's Criterion)
Let prime $p>2, p>d$, then $d$ is a quadratic residue modulo $p$ if and only if
$$d^{\frac{p-1}{2}} \equiv 1(\bmod p)$$
$d$ is a non-quadratic residue modulo $p$ if and only if
$$d^{\frac{p-1}{2}} \equiv-1(\bmod p)$$ | First, we prove that for any $d, p \times d$, either equation (4) or (5) holds, but not both. By Fermat's Little Theorem, we know $d^{p-1} \equiv 1 \pmod{p}$, thus $\left(d^{\frac{p-1}{2}}-1\right)\left(d^{\frac{p-1}{2}}+1\right) \equiv 0 \pmod{p}$. Since the prime $p > 2$ and $\left(d^{\frac{p-1}{2}}-1, d^{\frac{p-1}{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,215 |
2. Let $p$ be an odd prime. Prove: the sum $S$ of all quadratic residues modulo $p$ in $1,2, \cdots, p-1$ is $\frac{p\left(p^{2}-1\right)}{24}-$ $p \sum_{j=1}^{\frac{p-1}{2}}\left[\frac{j^{2}}{p}\right]$ | 2. Let $j^{2}=p q_{j}+r_{j}, 1 \leqslant r_{j}<p, 1 \leqslant j \leqslant \frac{p-1}{2}$, from this and $q_{j}=\left[\frac{j^{2}}{p}\right]$ we get $S=\sum r_{j}=\sum j^{2}-$ $p \Sigma\left[\frac{j^{2}}{p}\right]$
Using (3) from the previous problem, we get
$$S=\frac{p\left(p^{2}-1\right)}{24}-p \sum_{j=1}^{\frac{p-1}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,217 |
5. Let $k$ be a positive integer. Prove: there exist infinitely many perfect squares of the form $n \cdot 2^{k}-7$, where $n$ is a positive integer. | 5. First prove that for any given $k$, there exists a positive integer $a_{k}$, such that $a_{k}^{2} \equiv -7 \pmod{2^{k}}$, and prove it by mathematical induction on $k$.
Direct observation shows that when $k \leqslant 3$, taking $a_{k}=1$ satisfies the condition. Suppose for some $k>3$, we have $a_{k}^{2} \equiv -7... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,218 |
9. Prove: There are infinitely many primes of the form $4k+1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the text is already in English, here is the same text with the requested format preserved:
9. Prove... | 9. Suppose there are only finitely many such primes, and let them be $p_{1}, p_{2}, \cdots, p_{k}$.
We consider $\left(2 p_{1} \cdots p_{k}\right)^{2}+1=p$, by the assumption and $p \equiv 1(\bmod 4)$, so $p$ is not a prime, let $p_{0}$ be a prime factor of $p$, $p_{0}$ is of course odd, so -1 is a quadratic residue m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,219 |
Theorem 2 If $(a, b)=1$, and $x_{0}, y_{0}$ is a solution of (1), then all solutions of equation (1) are $x=x_{0}+b t, y=y_{0}$
$$-a t(t \in \mathbf{Z})$$ | Let $x_{0}, y_{0}$ be a solution of (1), then we have
$$a x_{0}+b y_{0}=c \text {. }$$
Let $x, y$ be any solution of (1), subtracting (3) from (1) yields
$$a\left(x-x_{0}\right)+b\left(y-y_{0}\right)=0 .$$
Since $(a, b)=1$, it follows that $b \mid\left(x-x_{0}\right)$, i.e., $x-x_{0}=b t$, thus $x=x_{0}+b t$. Substit... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,220 |
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 n... | A | Number Theory | MCQ | Yes | Yes | number_theory | false | 742,221 |
9. Prove: There are infinitely many positive integers $n$ such that $[\sqrt{2} n]$ is a perfect square, ( $[x]$ denotes the greatest integer not exceeding $x$).
untranslated text remains the same as requested. | 9. Proof: Consider the Pell's equation $x^{2}-2 y^{2}=-1$ which has infinitely many positive integer solutions. Take any solution $u, v$, then $2 v^{2}=u^{2}+1$.
Multiply both sides of the above equation by $u^{2}$ to get $2(u v)^{2}=\left(u^{2}+1\right) u^{2}$, hence $u^{2}<\sqrt{2} u v<u^{2}+1$. Therefore, $[\sqrt{2}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,222 |
Example 1 Proof: The indeterminate equation $x^{2}+y^{2}-8 z^{3}=6$ has no integer solutions.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that $x, y$ have the same parity by modulo 2; modulo 4 shows that $x, y$ are both odd, thus $x^{2} \equiv 1(\bmod 8), y^{2} \equiv 1$ $(\bmod 8)$.
Taking both sides of the equation modulo 8, we have $6 \equiv x^{2}+y^{2}-8 z^{3} \equiv 2(\bmod 8)$, a contradiction. Therefore, the original equation has no integer... | null | Number Theory | proof | Yes | Yes | number_theory | false | 742,223 |
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... | x=y=3, z=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,224 |
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 la... | a=2, b=0, c=0, d=3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,225 |
Example 6 Proof: The only positive integer solution to the indeterminate equation $8^{x}+15^{y}=17^{z}$
is $x=y=z=2$. | First, we use congruence to prove that $y$ and $z$ are both even.
Taking equation (1) modulo 4, we get $(-1)^{y} \equiv 1(\bmod 4)$,
thus $y$ is even. Taking equation (1) modulo 16, we get $8^{x}+(-1)^{y} \equiv 1(\bmod 16)$, which simplifies to $8^{x} \equiv 0(\bmod 16)$, hence $x \geqslant 2$.
Note that $17^{2} \equ... | x=y=z=2 | Number Theory | proof | Yes | Yes | number_theory | false | 742,226 |
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. T... | a=3, b=3, c=4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,227 |
1. (2004 China Mathematical Olympiad) Prove: there does not exist a pair of positive integers $x, y$ satisfying $3 y^{2}=x^{4}+x$.
| 1. Proof: By contradiction
If the equation $3 y^{2}=\left(x^{3}+1\right) x$ has a set of integer solutions $x, y$, since $\left(x, x^{3}+1\right)=1$, there exist $u, v \in \mathbf{N}_{+}$, such that $y=u v$, and $3 u^{2}=x^{3}+1, v^{2}=x$ or $u^{2}=x^{3}+1,3 v^{2}=x$.
The first case is obviously impossible, now assum... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,228 |
5. Prove that the indeterminate equation $5^{x}-3^{y}=2$ has only the positive integer solution $x=y=1$.
Translate the text above 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 neede... | 5. Proof: The equation clearly has a solution $x=y=1$. Taking the equation modulo 4, it is easy to see that $y$ is odd. If $y>1$, taking the equation modulo 9 yields
$$5^{x} \equiv 2(\bmod 9)$$
It is not difficult to find that for $x=1,2, \cdots, 5^{x}$ modulo 9, the cycle is $5,7,8,4,2,1$. Therefore, by (1), $x$ must... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,229 |
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
$$3^{x}-1=3^{2^{m}(2 n+1)}-1=\left(3^{2 n... | (x, y)=(1,1),(2,2),(4,5) | Geometry | math-word-problem | Yes | Yes | number_theory | false | 742,230 |
11. (2005 USA Mathematical Olympiad Problem) Prove: The system of equations $\left\{\begin{array}{l}x^{6}+x^{3}+x^{3} y+y=147^{157} \\ x^{3}+x^{3} y+y^{2}+y+z^{9}=157^{147}\end{array}\right.$ has no integer solutions $x, y, z$. | 11. Proof: Adding both sides of the equation and then adding 1, we get
$$\left(x^{3}+y+1\right)^{2}+z^{9}=147^{157}+157^{147}+1$$
We will now prove that both sides of equation (1) are not congruent modulo 19.
The choice of modulo 19 is because the least common multiple of 2 and 9 is 18, and by Fermat's Little Theorem,... | proof | Algebra | proof | Yes | Yes | number_theory | false | 742,231 |
10. Write the first $2 n$ natural numbers in any order in a row, and number them sequentially as $1,2, \cdots, 2 n$. Then, find the sum of each number and its position number, resulting in $2 n$ sum numbers. Next, divide each sum number by $2 n$, obtaining $2 n$ quotients and remainders. Prove that at least two of the ... | 10. Proof: Let the first $2n$ natural numbers be written in any order as $a_{1}, a_{2}, a_{3}, \cdots, a_{2n-1}, a_{2n}$, where $1 \leqslant a_{i} \leqslant 2n$, $a_{i} \neq a_{j}(i \neq j),(i, j=1,2,3, \cdots, 2n)$, then the corresponding sum numbers are $1+a_{1}, 2+a_{2}, \cdots, 2n-1+a_{2n-1}$, $2n+a_{2n}$.
These s... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,232 |
Example 2 (2004 Taiwan Mathematical Olympiad Problem) Find all pairs of positive integers $(a, b)$, satisfying
$$\sqrt{\frac{a b}{2 b^{2}-a}}=$$
$$\frac{a+2 b}{4 b} .$$ | Solve: Squaring both sides of $\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b}$ and simplifying, we get
$$a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right),$$
which is $a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2}$.
Since the left side of the above equation is even, let $a=2 t^{2}, t \in \mathbf{N}_{+}$, sub... | null | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,234 |
Example 5 (2004 Belarusian Mathematical Olympiad Problem) Positive integers $a, b, c$ satisfy the equation $c(a c+1)^{2}=(5 c+2 b)(2 c+b)$,
(1) Prove that if $c$ is odd, then $c$ is a perfect square;
(2) For some $a, b$, does there exist an even $c$ that satisfies (1);
(3) Prove: Equation (1) has infinitely many positi... | Solve (2) first.
(2) Assume $c$ is even, let $c=2 c_{1}$. Then the known equation can be written as $c_{1}\left(2 a c_{1}+1\right)^{2}=\left(5 c_{1}+b\right)\left(4 c_{1}+b\right)$. Let $d=\left(c_{1}, b\right)$, then $c_{1}=d c_{0}, b=d b_{0}$, where $\left(c_{0}, b_{0}\right)=1$.
Thus, we have $c_{0}\left(2 a d c_{0}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,235 |
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(... | (2^{2 n+1}+2^{n}+1,1, n),(23,2,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,236 |
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 ... | (0,2,1), (2,0,1), (1,2,1), (2,1,1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,237 |
Example 5 (2005 National High School Mathematics Competition Question) Define the function
$$f(k)=\left\{\begin{array}{l}
0, \text { if } k \text { is a perfect square } \\
{\left[\frac{1}{\{\sqrt{k}\}}\right], \text { if } k \text { is not a perfect square }}
\end{array} \text {, find } \sum_{k=1}^{240} f(k)\right. \t... | When $k$ is not a perfect square, $k$ must lie between two consecutive perfect squares. Divide $[1,240]$ into several intervals with perfect squares as boundaries, and sum $f(k)$ for each subinterval.
Solving $15^{2}<240<16^{2}$.
Since $f(k)=0$ when $k$ is a perfect square, we have
$$\begin{aligned}
\sum_{k=1}^{240} f... | 768 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,239 |
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 | 742,240 |
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 $... | D | Algebra | MCQ | Yes | Yes | number_theory | false | 742,241 |
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... | 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... | C | Algebra | MCQ | Yes | Yes | number_theory | false | 742,242 |
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, $\frac{1}{[\sqrt{k}]}=\frac{1}{n}$.
$$\begin{aligned}
t^{2} \sum_{k=1}^{2006} \frac{1}{[\sqrt{k}]} & =\sum_{1... | \frac{1927}{22} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,243 |
Example 2 (2003 Croatian National Mathematical Olympiad) For any integer $n(n>2)$, prove: $\left[\frac{n(n+1)}{4 n-2}\right]=$
$$\left[\frac{n+1}{4}\right]$$ | Analysis When a fraction with a small positive integer as the denominator appears inside a Gaussian function, in many cases, it is necessary to discuss the remainder of the unknown divided by this denominator.
Proof Since $\frac{n(n+1)}{4 n-2}=\frac{n(n+1)-\frac{n+1}{4} \times(4 n-2)}{4 n-2}+\frac{n+1}{4}$
$$=\frac{\f... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,244 |
Example 3 Prove the inequality $[\sqrt{\alpha}]+[\sqrt{\alpha+\beta}]+[\sqrt{\beta}] \geqslant[\sqrt{2 \alpha}]+[\sqrt{2 \beta}]$ for any real numbers $\alpha$ and $\beta$ not less than 1.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result... | Analyzing the original inequality, which contains 5 floor functions, is quite complex and it is not easy to segment the values of $\alpha$ and $\beta$ directly. Therefore, we should first appropriately relax the inequality to remove some of the floor functions.
Proof: Let $a=\sqrt{\alpha}, b=\sqrt{\beta}$, then the or... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 742,245 |
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^... | 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 | 742,246 |
Example 7 Let $[x]$ denote the greatest integer not exceeding $x$, and let $a$ be the positive root of the equation $x^{2}-m x-1=0$, where $m$ is a given positive integer. Prove that there exist infinitely many positive integers $n$ satisfying the equation
$$[a n+m a[a n]]=m n+\left(m^{2}+1\right)[a n] .$$ | Given that $a$ satisfies $a^{2}-m a-1=0$, i.e., $a=m+\frac{1}{a}$, use this equation to simplify the expression inside the floor function, and finally obtain infinitely many positive integers $n$ that are related to $a$.
Proof: Since $a$ is the positive root of the equation $x^{2}-m x-1=0$,
we have $a^{2}-m a-1=0$, i.... | proof | Algebra | proof | Yes | Yes | number_theory | false | 742,247 |
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 p... | 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, $4(a n-1)n+a(a n-1)-1$... | a=2+\sqrt{3} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,248 |
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 | Algebra | MCQ | Yes | Yes | number_theory | false | 742,250 |
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 \leq... | B | Algebra | MCQ | Yes | Yes | number_theory | false | 742,251 |
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... | B | Number Theory | MCQ | Yes | Yes | number_theory | false | 742,252 |
7. The real solution of the equation $3 x^{3}-[x]=3$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7. $x=\sqrt[3]{\frac{4}{3}} \quad(1)$ If $x \leqslant -1$, then $3 x^{3}-[x] \leqslant 3 x-x+13 x-x \geqslant 4$; (4) If $0 \leqslant x<1$, then $3 x^{3}$ $-[x]=3 x^{3}<3$; (5) If $1 \leqslant x<2$, then $3 x^{3}-[x]=3 x^{3}-1=3, x=\sqrt[3]{\frac{4}{3}}$. From this, we know that the equation has only one solution $x=\s... | null | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,253 |
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 \ge... | x=1, \sqrt{33}, \sqrt{41}, 7 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,254 |
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 $\{a\} x<1$. ... | \left[1+\frac{1}{n}, 1+\frac{1}{n-1}\right) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,255 |
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}{... | x=k\pi+\frac{\pi}{4}, k \in \mathbf{Z} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,257 |
4. $n \in \mathbf{N}_{+}, n \geqslant 2$, then $\sum_{k=2}^{n}[\sqrt[k]{n}]=\sum_{k=2}^{n}\left[\log _{k} n\right]$. | Prove the construction of the planar region $D=\left\{(x, y) \mid y^{x} \leqslant n, x \geqslant 2, y \geqslant 2\right\}$, and consider the number of integer points in $D$.
If counted column by column, when $x=2$, there are $[\sqrt{n}]$ points, when $x=3$, there are $[\sqrt[3]{n}]$ points, $\cdots$, when $x=n$, there... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,258 |
Example 5 Let $p_{1}, p_{2}, p_{3}, \cdots$ be the prime numbers in increasing order, and let $x_{0}$ be a real number between 0 and 1. For a positive integer $k$, define
$$x_{k}=\left\{\begin{array}{ll}
0, & \text { if } x_{k-1}=0 \\
\left\{\frac{p_{k}}{x_{k-1}}\right\} . & \text { if } x_{k-1} \neq 0
\end{array}\righ... | Analysis When $x$ is an irrational number, $\frac{p}{x}$ is also an irrational number $(p \in \mathbf{Q}$ and $p \neq 0)$, thus $\left\{\frac{p}{x}\right\}$ is also an irrational number. Therefore, we can conjecture that $x_{0}$ can take all rational numbers.
Solution We prove that the sequence eventually appears 0 if... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,260 |
Example 8 For real numbers $x, y$, let $s(x, y)=\{s \mid s=[n x+y], n \in \mathbf{N}\}$. Prove that if $r>1$ is a rational number, then there exist real numbers $u, v$, such that
$$s(r, 0) \cap s(u, v)=\varnothing, s(r, 0) \cup s(u, v)=\mathbf{N}$$ | Proof: Let $r=\frac{p}{q}, p, q \in \mathbf{N}, p>q$, then
$u=\frac{p}{p-q}$ and satisfies $-\frac{1}{p-q} \leqslant vq$, then when $v(p-q)m$, $[n r]>[m r],[n u+v]>[m u+v]$.
Therefore, there are no identical elements in $s(u, v)$.
Thus, the number of elements in $s(u, v) \bigcap\{1,2, \cdots, k-1\}$ equals $[m u+v]-\f... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,262 |
For non-negative integers $x$, the function $f(x)$ is defined as follows:
$$f(0)=0, f(x)=f\left(\left[\frac{x}{10}\right]\right)+\left[\lg \frac{10}{x-10\left[\frac{x-1}{10}\right]}\right]$$
What is the value of $x$ when $f(x)$ reaches its maximum in the range $0 \leqslant x \leqslant 2006$? | 1. Solution: Let $x=10 p+q$, where $p$ and $q$ are integers and $0 \leqslant q \leqslant 9$.
Then $\left[\frac{x}{10}\right]=\left[p+\frac{q}{10}\right]=p$.
Thus, $\left[\frac{x-1}{10}\right]=\left[p+\frac{q-1}{10}\right]$ has a value of $p-1$ (when $q=0$) or $p$ (when $q \neq 0$), and $x-10\left[\frac{x-1}{10}\right]... | 1111 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,263 |
4. Let $m$ be a positive integer, and the number of factor 2s in $m!$ is denoted as $n(m)$. Prove that there exists a natural number $m > 2006^{2006}$, such that $m = 3^{2006} + n(m)$. | 4. Proof: Let $m=2^{a}-1$, then the number of factor 2 in $m!$ is $\left[\frac{2^{a}-1}{2}\right]+\left[\frac{2^{a}-1}{2^{2}}\right]+\cdots+\left[\frac{2^{a}-1}{2^{a}}\right]$
$$\begin{array}{l}
=\left(2^{a-1}-1\right)+\left(2^{a-2}-1\right)+\cdots+(1-1) \\
=\left(2^{a-1}+2^{a-2}+\cdots+1\right)-a=2^{a}-1-a
\end{array}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,264 |
Example 1 (1) Let $n$ be a positive integer, prove: $(21 n+4,14 n+3)=1$; (2) Prove: $(n!+1,(n+1)!+$ $1)=1$; (3) Let $F_{k}=2^{2^{k}}+1, k \geqslant 0$, prove: for $m \neq n$, $(F_{m}, F_{n})=1$. | (1) and (2) can be solved using the Euclidean algorithm, while (3) requires the use of Bézout's identity.
Proof (1) By the Euclidean algorithm, we get
$$\begin{aligned}
(21 n+4,14 n+3) & =(21 n+4-14 n-3,14 n+3) \\
& =(7 n+1,14 n+3)=(7 n+1,14 n+3-2(7 n+1)) \\
& =(7 n+1,1)=1
\end{aligned}$$
(2) Since $(n+1,(n+1)!+1)=1$,
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,265 |
Example 3 Let $a$ be a given positive integer, and $A$ and $B$ be two real numbers. Determine the necessary and sufficient condition for the system of equations
$$\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=(13 a)^{2} \\
x^{2}\left(A x^{2}+B y^{2}\right)+y^{2}\left(A y^{2}+B z^{2}\right)+z^{2}\left(A z^{2}+B x^{2}\right)... | The necessary and sufficient condition for the original system of equations to have positive integer solutions is $B=2A$. Below is the proof of this conclusion.
First, we prove the sufficiency, i.e., given $B=2A$, we prove that the original system of equations has positive integer solutions.
In this case, we can take ... | B = 2A | Algebra | proof | Yes | Yes | number_theory | false | 742,268 |
2. Find the integer solutions of the equation $x^{2}+y^{2}+z^{2}=x^{2} y^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 2. Solution: $x_{0}=y_{0}=z_{0}$ is a solution to the equation. Suppose the equation has another solution $(x, y, z)$. If $x, y, z$ include three or two odd numbers, then $x^{2}+y^{2}+z^{2} \equiv 2$ or $3(\bmod 4)$, while $x^{2} y^{2} \equiv 0$ or $1(\bmod 4)$, which is obviously impossible; if $x, y, z$ include exact... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,270 |
4. Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$, prove that: $k=5$. | 4. Prove: If $a=b$, then $k=\frac{2 a^{2}}{a^{2}-1}=2+\frac{2}{a^{2}-1}$.
Then $\left(a^{2}-1\right) \mid 2$ and $a^{2}-1>0$.
So $a^{2}-1=1$ or 2, but this contradicts $a \in \mathbf{N}_{+}$.
Therefore, $a \neq b$, without loss of generality, let $a>b$.
When $b=1$, $k=\frac{a^{2}+1}{a-1}=a+1+\frac{2}{a-1}$, then $(a-1... | 5 | Number Theory | proof | Yes | Yes | number_theory | false | 742,271 |
Example 3 On the coordinate plane, if the coordinates $x_{0}, y_{0}$ of a point $\left(x_{0}, y_{0}\right)$ are both integers, then the point is called an integer point. Try to prove: on the coordinate plane, there does not exist a regular $n$-gon $(n \geqslant 7)$, such that all its vertices are integer points. | Proof Assume there exists a regular $n$-sided polygon $A_{1} A_{2} A_{3} \cdots A_{n}(n \geqslant 7)$, whose vertices $A_{1}, A_{2}, \cdots, A_{n}$ are all integer points.
Take any integer point $M$ in the coordinate plane, and draw vectors $\overrightarrow{M B_{1}}, \overrightarrow{M B_{2}}, \overrightarrow{M B_{3}},... | proof | Geometry | proof | Yes | Yes | number_theory | false | 742,272 |
Example 4 Assume that $2 n+1$ positive integers $a_{1}, a_{2}, \cdots, a_{2 n+1}$ have the property: any $2 n$ of these numbers can be divided into two disjoint $n$-element subsets, and the sums of the $n$ numbers in the two subsets are equal, then these $2 n+1$ numbers must all be equal. | To prove that if there are $2n+1$ positive integers satisfying the property described in the problem, we call these $2n+1$ positive integers a "good array". Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2 n+1}$, and $a_{1}=1$ (because reducing the smallest number to 1 and subtracting the same number... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 742,273 |
Example 5 Let $f(x)=x^{2}+x+p, p \in \mathbf{N}$, prove: If $f(0), f(1), f(2), \cdots, f\left(\left[\sqrt{\frac{p}{3}}\right]\right)$ are prime numbers, then the numbers $f(0), f(1), \cdots, f(p-2)$ are all prime numbers. | Analysis If there exists some $m \in\{0,1,2, \cdots, p-2\}$, such that $m^{2}+m+p$ is a composite number, and we can find a way to decrease $m$, i.e., find a smaller number $m^{\prime}\left(m^{\prime} \in\{0,1,2, \cdots, p-2\}\right)$, such that $f\left(m^{\prime}\right)$ is a composite number, then by the method of in... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,274 |
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, 1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,275 |
Example 8 (29th Russian Mathematical Olympiad) The sequence of positive integers $\left\{a_{n}\right\}$ is constructed as follows: $a_{0}$ is some positive integer, if $a_{n}$ is divisible by 5, then $a_{n+1}=\frac{a_{n}}{5}$; if $a_{n}$ is not divisible by 5, then $a_{n+1}=\left[\sqrt{5} a_{n}\right]$ (where $[x]$ den... | Prove that for any positive integer $x$, we have $[\sqrt{5} x] \geqslant x+1$.
This is because when $x \in \mathbf{N}_{+}$, $\sqrt{5} x > 2 x \geqslant x+1$.
Therefore, $[\sqrt{5} x] \geqslant [x+1] = x+1$.
From this, we can see that the conclusion to be proven is equivalent to: there exists some positive integer $m$, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,276 |
2. Given $2 n$ points in the plane, no three of which are collinear, $n$ points represent farms $F=\left\{F_{1}, F_{2}, \cdots, F_{n}\right\}$, and the other $n$ points represent reservoirs $W=\left\{W_{1}, W_{2}, \cdots, W_{n}\right\}$. Each farm is connected to a reservoir by a straight road. Prove that there exists ... | 2. Proof: For any allocation method, if there exist two roads $F_{i} W_{m}$ and $F_{k} W_{j}$ that intersect (as shown in Figure 2 of Question 2), where $1 \leqslant i, k \leqslant n, 1 \leqslant j, m \leqslant n$ and $i \neq k, j \neq m$.
In this case, we make the following adjustment: remove roads $F_{i} W_{m}$ and ... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 742,277 |
5. Let there be $2 n$ integers $a_{1}, a_{2}, \cdots, a_{2 n}$, and if any one of these numbers is taken out, the remaining $2 n-1$ numbers can always be divided into two disjoint subsets, such that the sum of the numbers in each subset is equal, then $a_{1}, a_{2}, \cdots, a_{2 n}$ must all be zero. (Here, "two disjoi... | 5. Proof: According to the problem, we can know that $\sum_{i \neq k} a_{i} (k=1,2,3, \cdots, 2n)$ must be even. From this, we can conclude that for any $k \in \{1,2,3, \cdots, 2n\}, \sum_{i=1}^{2n} a_{i}$ and $a_{k}$ have the same parity.
Obviously, $a_{1}, a_{2}, \cdots, a_{2n}$ must all be even (otherwise, $a_{1}, a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,278 |
6. Prove: The area of a Pythagorean triangle cannot be a perfect square.
| 6. Proof: First, we provide a lemma.
Lemma: The system of equations $\left\{\begin{array}{l}x^{2}+y^{2}=z^{2}, \\ x^{2}-y^{2}=w^{2}\end{array}\right.$ has no positive integer solutions.
The proof of the lemma is omitted (refer to Chapter 6, Section 1, Exercise 5).
We now prove the original proposition:
By contradictio... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,279 |
Example 3 (2004 Croatian Mathematical Competition Problem) A frog starts at point $(1,1)$ on a Cartesian coordinate plane and jumps according to the following rules: (1) The frog can jump from any point $(a, b)$ to point $(2a, b)$ or $(a, 2b)$; (2) If $a > b$, the frog can jump from $(a, b)$ to $(a-b, b)$, and if $a < ... | Analyzing the common odd factor of the two coordinates of the grid points that the frog can reach remains unchanged, whether jumping by rule (1) or rule (2).
Solution: For (1) and (4), the answer is affirmative. We provide an example of the frog's path from point $(1,1)$ to the given point:
The path for (1) is
$$\begi... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,280 |
Example 4 Please design a method to color all integer points, each point being colored one of white, red, or black, such that: (1) Points of each color appear on infinitely many lines parallel to the x-axis.
(2) For any white point $A$, red point $B$, and black point $C$, there always exists a red point $D$ such that $... | Analysis: Color the grid points according to their classification to satisfy condition (1). Label the points $A$, $B$, and $C$ with coordinates, and use the property that the diagonals of a parallelogram bisect each other and the midpoint formula to find the coordinates of $D$. This should yield a parallelogram $ABCD$ ... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 742,281 |
Example 5 Take any 6 lattice points $p_{i}\left(x_{i}, y_{i}\right)(i=1,2,3,4,5,6)$, satisfying
(1) $\left|x_{i}\right| \leqslant 2,\left|y_{i}\right| \leqslant 2(i=1,2,3,4,5,6)$;
(2) No three points are collinear.
Try to prove: Among all triangles with $p_{i}(i=1,2,3,4,5,6)$ as vertices, there must be one triangle wh... | Proof Assume there exist 6 lattice points $p_{1}, p_{2}, \cdots, p_{6}$ within the region $S=\{(x, y)|| x|\leqslant 2| y \mid, \leqslant 2\}$, and any 3 of these points form a triangle with an area greater than 2. Let $p=\left\{p_{1}, p_{2}, \cdots, p_{6}\right\}$.
(1) If the number of points in $p$ on the $x$-axis is ... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 742,282 |
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), (eve... | B | Geometry | MCQ | Yes | Yes | number_theory | false | 742,283 |
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 | 742,284 |
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)^... | B | Geometry | MCQ | Yes | Yes | number_theory | false | 742,285 |
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... | 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}=7... | (44,19) | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 742,286 |
6. In the Cartesian coordinate system, the number of integer points $(x, y)$ that satisfy $(|x|-1)^{2}+(|y|-1)^{2}<2$ is $\qquad$ . | 6. 16 Since $(|x|-1)^{2} \leqslant(|x|-1)^{2}+(|y|-1)^{2}<2$.
Therefore, $(|x|-1)^{2}<2, -1 \leqslant|x|-1 \leqslant 1$, which means $0 \leqslant|x| \leqslant 2$, similarly $0 \leqslant|y| \leqslant 2$.
Upon inspection, $(x, y)=(-1, \pm 1),(-1,0),(1,0),(0, \pm 1),(-1, \pm 2),(1, \pm 2),(-2, \pm 1),(2, \pm 1)$ are the ... | 16 | Geometry | math-word-problem | Yes | Yes | number_theory | false | 742,287 |
Theorem (1) Suppose the function $f(x)$ is continuous and non-negative on the closed interval $[a, b]$ (as shown in Figure 7-2), then the sum $\sum_{a<r \leqslant b}[f(t)]$ represents the number of lattice points in the planar region $a<x \leqslant b, 0<y \leqslant f(x)$, where $t$ denotes the integers in $(a, b]$, and... | The proof is as follows: As shown in Figure 7-3, because $(p, q)=1$, there are no lattice points on the line segment $O F$ except for $O$. Therefore, $\sum_{0<x<\frac{p}{2}}\left[\frac{q}{p} x\right]$ represents the number of lattice points inside $\triangle O D F$, and $\sum_{0<y<\frac{q}{2}}\left[\frac{p}{q} y\right]... | \left[\frac{p-1}{2}\right] \cdot \left[\frac{q-1}{2}\right] | Number Theory | proof | Yes | Yes | number_theory | false | 742,288 |
Example 3 In the Cartesian coordinate system, the number of integer points that satisfy $(1) y \geqslant 3 x$; (2) $y \geqslant \frac{1}{3} x ;(3) x+y \leqslant 100$ is how many? | As shown in Figure 7-5, the region enclosed by the lines $y=3x$, $y=\frac{1}{3}x$, and $x+y=100$ forms a triangular region. The three vertices of this triangle are $O(0,0)$, $A(75,25)$, and $B(25,75)$. $\square$
Next, we calculate the number of integer points $N$ on the boundary and inside $\triangle OAC$. For a grid ... | 2551 | Inequalities | math-word-problem | Yes | Yes | number_theory | false | 742,289 |
Example 5 For a lattice point $X$ on the coordinate plane, if the line segment $O X$ does not contain any other lattice points, then the point $X$ is said to be visible from the origin $O$. Prove that for any positive integer $n$, there exists a square $A B C D$ with area $n^{2}$ such that there are no lattice points v... | Notice that a lattice point $(x, y)$ is visible from the origin if and only if $x, y$ are coprime. Thus, the problem is reduced to finding a square $A B C D$ such that every lattice point $(x, y)$ inside it satisfies that $x$ and $y$ are not coprime. Let $P_{1}, P_{2}, \cdots$ be a sequence of distinct primes. In an $n... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,290 |
Example 6 If $a, b$ are coprime positive integers, prove:
$$\left[\frac{a}{b}\right]+\left[\frac{2 a}{b}\right]+\cdots+\left[\frac{b-1}{b} a\right]=\frac{1}{2}(a-1)(b-1) .$$ | Prove that, as shown in Figure 7-6, the four vertices of rectangle $O A B C$ are $O(0,0), A(b, 0), B(b, a)$, and $C(0, a)$. All integer points $(x, y)$ inside this rectangle satisfy $1 \leqslant x \leqslant b-1, 1 \leqslant y \leqslant a-1$, and there are a total of $(a-1) \cdot (b-1)$ such integer points.
The equatio... | \frac{1}{2}(a-1)(b-1) | Number Theory | proof | Yes | Yes | number_theory | false | 742,291 |
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 calc... | C | Geometry | MCQ | Yes | Yes | number_theory | false | 742,292 |
5. In $1 \sim 1000$, the number of pairs $(x, y)$ that make $\frac{x^{2}+y^{2}}{7}$ an integer is $\qquad$ pairs. | 5. 10011 From $\frac{x^{2}+y^{2}}{7} \in \mathbf{Z}$ we know $7 \mid x^{2}+y^{2}$.
When $7 \mid x$ and $7 \mid y$, it is obvious that $7 \mid x^{2}+y^{2}$.
When $7 \nmid x$ or $7 \nmid y$, from $7 \mid x^{2}+y^{2}$ we know that $7 \nmid x$ and $7 \nmid y$. By Fermat's Little Theorem, $x^{6} \equiv 1(\bmod 7)$, which m... | 10011 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,293 |
6. Let the planar region $T=\{(x, y) \mid x>0, y>0, x y \leqslant 48\}$, then the number of lattice points within $T$ is $\qquad$ . | $$\begin{array}{l}
\text { 6. } 202 \text { Let } T_{1}=\left\{(x, y) \mid 0<x \leqslant \sqrt{48}, 0<y \leqslant \frac{n}{x}\right\}, \\
T_{2}=\left\{(x, y) \mid 0<y \leqslant \sqrt{48}, 0<x \leqslant \frac{n}{y}\right\} .
\end{array}$$
Then $T=T_{1} \cup T_{2}, T_{1} \cap T_{2}=\{(x, y) \mid 0<x \leqslant \sqrt{48},... | 202 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 742,294 |
8. Prove: There exist 2008 lattice points in the plane, not all on a straight line, such that the distance between any two points is an integer.
| 8. Proof: Let $B(0, b)$, where $b=2 p_{1} p_{2} \cdots p_{2008}$. ($p_{i}$ are distinct factors, $i=1,2, \cdots, 2008$), and let $A_{j}\left(a_{j}, 0\right)$, where $a_{j}=\left|\left(p_{1} p_{2} \cdots p_{j}\right)^{2}-\left(p_{j+1} p_{j+2} \cdots p_{2008}\right)^{2}\right|, j=1,2, \cdots, 2007$.
By the Pythagorean t... | proof | Geometry | proof | Yes | Yes | number_theory | false | 742,295 |
Minkowski's Theorem: Let $K$ be a convex figure in the $x O y$ plane, symmetric with respect to the origin, and with an area greater than 4. Prove: The figure $K$ must cover a lattice point other than the origin. | Prove that using lines parallel to the coordinate axes to divide the plane into some $2 \times 2$ squares, and then translating all the squares that cover the points of $K$ to the same square, since the area of $K$ is greater than 4, there must be a point in this square that is covered by the squares containing points ... | proof | Geometry | proof | Yes | Yes | number_theory | false | 742,296 |
Example 3 Let $L$ be a subset of the Cartesian plane, defined as follows:
$$L=\{(41 x+2 y, 59 x+15 y) \mid x, y \in \mathbf{Z}\}.$$
Prove that every parallelogram centered at the origin with an area of 2008 contains at least two points from $L$. | Prove that taking $(x, y)=(0,0),(0,1),(1,0),(1,1)$, we get four points in $L$: $(0,0),(2,15)$, $(41,59),(43,73)$.
Let these four integer points form the fundamental region $F$, which is a parallelogram containing no other integer points.
It is easy to see that the area of $F$ is 497.
Translating $F$ to form a network ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,297 |
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 \time... | m+n-2 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 742,298 |
3. In a Cartesian coordinate system, construct a closed broken line such that each node of the broken line is a lattice point, and the lengths of each segment of the broken line are equal. Prove: the number of segments of such a broken line must be even. | 3. Proof: Take a node on a grid paper as the origin to establish a Cartesian coordinate system, so that all nodes on the grid paper are lattice points.
Thus, the coordinates of each vertex $\left(x_{i}, y_{i}\right)(i=1,2, \cdots, n)$ of the closed polyline are integers.
Since it is a closed polyline, the projections ... | proof | Geometry | proof | Yes | Yes | number_theory | false | 742,299 |
5. Let $n, k$ be positive integers, and $n>k$, prove: the greatest common divisor of $C_{n}^{k}, C_{n+1}^{k}, \cdots, C_{n+k}^{k}$ is 1. | 5. Prove: Let $d \in \mathbf{N}$ be a common divisor of $C_{n}^{k}, C_{n+1}^{k}, \cdots, C_{n+k}^{k}$, then $d$ is also a common divisor of $C_{n}^{k-1}=C_{n+1}^{k}-C_{n}^{k}, C_{n+1}^{k-1}=$ $C_{n+2}^{k}-C_{n+1}^{k}, \cdots, C_{n+k-1}^{k-1}=C_{n+k}^{k}-C_{n+k-1}^{k}$.
Similarly, $d$ is also a common divisor of $C_{n}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,300 |
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 direc... | 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... | n=7,18,29,40 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,301 |
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 i... | 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) ... | (1,1,1) \text{ or } (1,1,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,302 |
1. Definition of prime numbers and some basic properties An integer $n$ greater than 1 has at least two distinct positive divisors: 1 and $n$. If $n$ has no divisor greater than 1 and less than $n$, then $n$ is called a prime number. If $n$ has a divisor greater than 1 and less than $n$, i.e., $n$ can be expressed in t... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 742,303 |
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