problem stringlengths 1 7.47k | solution stringlengths 0 13.5k | answer stringlengths 1 272 | problem_type stringclasses 8
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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
$$d(m)=\left(4 \alpha_{2}+1\right)\left(4 \alpha_{3}+1\rig... | m=1,5^{4}, 3^{8}, 3^{8} \cdot 5^{4} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
35 Find all positive integers $a$, $b$, and $c$, such that $a^{2}+1$ and $b^{2}+1$ are both prime, and satisfy
$$\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 obv... | (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 ... | \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 b... | 39. Consider the objective function $S=$ the product of all numbers on the blackboard.
Initially, $S=33!=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$. Each operation targets $x, y (x \mid y)$, where $y=kx$. Removing $x, y$ and replacing them with $... | 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 t... | 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}} \e... | 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 ... | 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(... | 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
$$\begin{aligned}
2^{p}+3^{p} & =2^{p}+(5-2)^{p}=5^{p}-\mathrm{C}_{p}^{1} 5^{p-1} \times 2+\cdots+5 \mathrm{C}_{p}^{p-1} 2^{p-1} \\
& =5^{2} u+5 p \times 2^{p-1}, u \text { is an integer. }
\end{alig... | p=2,5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6 Let $p$ be a given odd prime, find all positive integer solutions $x, y$ of $p^{x}-y^{p}=1$.
The text has been translated while preserving the original line breaks and format. | 6. From $p^{x}=y^{p}+1=(y+1)\left(y^{p-1}-y^{p-2}+\cdots-y+1\right)$, we know that $y+1=p^{n}$, where $n$ is an integer. Since $y>0$, it follows that $n>0$. Therefore,
$$\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}... | 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. The... | 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 ... | 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 standardizi... | 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)... | ( \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 $x=-5,-1,1,2,4,5,7,11$.... | (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 value... | (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
$$\left\{\... | (1,1),(16,11),(11,16) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all positive integer solutions to the indeterminate equation
$$(n-1)!=n^{k}-1$$ | When $n=2$, from (1) we get the solution $(n, k)=(2,1)$. When $n>2$, the left side of (1) is even, so its right side must also be even, hence $n$ is odd. When $n=3,5$, from (1) we solve $(n, k)=(3,1), (5,2)$.
Below, let $n>5$ and $n$ be odd. In this case, $\frac{n-1}{2}$ is an integer and $\frac{n-1}{2}(n-1)!$
This i... | (n, k)=(2,1),(3,1),(5,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Let $p$ be a given positive integer, try to determine the minimum positive value of $(2 p)^{2 m}-(2 p-1)^{n}$, where $m, n$ are any positive integers. | The smallest positive value sought is $(2 p)^{2}-(2 p-1)^{2}=4 p-1$. To prove this, we first note that by
and
$$\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)$$
Furthermore, we pr... | 4p-1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 Find all positive integers $n$, such that the decimal integer composed of $n-1$ digits 1 and one digit 7 is a prime number.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 2. A positive integer $N$ composed of $n-1$ digits 1 and one digit 7 can be expressed in the form $N = A_{n} + 6 \times 10^{k}$, where $0 \leqslant k \leqslant n-1$, and $A_{n}$ is an integer composed of $n$ ones.
When $3 \mid n$, the sum of the digits of $A_{n}$ is divisible by 3, so $3 \mid A_{n}$, and thus $3 \mid ... | n = 1,2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Let $m>n \geqslant 1$, find the minimum value of $m+n$ such that $: 1000 \mid 1978^{m}-1978^{n}$. | 3. Solution: When $n \geqslant 3$, from $1000 \mid 1978^{m}-1978^{n}$, we get
$125 \mid 1978^{m-n}-1$, and since $1978=15 \times 125+103$,
we have $125 \mid 103^{m-n}-1$, so $25 \mid 103^{m-n}-1$.
Thus, $25 \mid 3^{m-n}-1$.
Let $m-n=l$, then: $25 \mid 3^{l}-1$ (obviously $l$ is even, otherwise it is easy to see that $3... | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
9. Determine all integers $n > 1$ such that: for all integers $a, b \in \mathbf{Z}$ that are coprime with $n$, $a \equiv b(\bmod n) \Leftrightarrow a b \equiv 1(\bmod n)$ | 9. Solution: We say that $n$ has property $T$ if for all integers $a, b$ coprime with $n$, $a \equiv b(\bmod n)$ if and only if $a b \equiv 1(\bmod n)$. It is not hard to see that $n$ has property $T$ if and only if for any $a \in \mathbf{Z}$, if $(a, n)=1$, then $a^{2} \equiv 1$ $(\bmod n)$. We will prove: if $n$ has ... | 2,3,4,6,8,12,24 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Solve the system of congruences
$$\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.$$ | Let $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, then $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=$ $3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$
From $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$,
so $1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$, we can take $M_{1}^{-1}=1$.
From $M_{2} \eq... | x \equiv 394(\bmod 1155) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1. Solve the system of congruences: (1) $\left\{\begin{array}{l}x \equiv 3(\bmod 7) \\ 6 x \equiv 10(\bmod 8)\end{array}\right.$
(1) (2) $\left\{\begin{array}{l}3 x \equiv 1(\bmod 10) \\ 4 x \equiv 7(\bmod 15)\end{array}\right.$ | For the system of equations (I), first solve (2), we get $x \equiv -1, 3 \pmod{8}$, so the solutions to (1) are the solutions to the following two systems of congruences: $x \equiv 3 \pmod{7}, x \equiv -1 \pmod{8}$, or $x \equiv 3 \pmod{7}, x \equiv 3 \pmod{8}$.
Using the Chinese Remainder Theorem, the solutions to (I... | x \equiv 3, 31 \pmod{56} \text{ for (I), no solution for (II)} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. The indeterminate equation $96 x_{1}+97 x_{2}=1000$ has $(\quad)$ sets of non-negative integer solutions.
A. 0
B. 1
C. 2
D. 3 | 2. A The complete set of integer solutions for the equation can be obtained as
$$\left\{\begin{array}{l}
x=-30+97 t, \\
y=40-96 t
\end{array} \quad(t \in \mathbf{Z})\right.$$
Let $x, y \geqslant 0$, we get $\frac{30}{97} \leqslant t \leqslant \frac{40}{96}$, and $t$ has no integer values. Therefore, the equation has n... | A | Number Theory | MCQ | Yes | Yes | number_theory | false |
Example 2 Find all positive integers $x>1, y>1, z>1$, such that $1!+2!+\cdots+x!=y^{z}$. | The key step is to prove that when $x \geqslant 8$, it must be that $z=2$. Since the left side of (1) is divisible by 3, hence $3 \mid y^{z}$, which implies $3 \mid y$, so the right side of (1) is divisible by $3^{z}$. On the other hand,
$$1!+2!+\cdots+8!=46233$$
is divisible by $3^{2}$, but not by $3^{3}$; and for $n... | x=y=3, z=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 (2003 British Mathematical Olympiad)
Given that $34!=295232799 c d 96041408476186096435 a b 000000$, find the values of the digits $a, b, c, d$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given that $34!=k \times 11^{3} \times 7^{4} \times 5^{7} \times 3^{15} \times 2^{32}=k \times 11^{3} \times 7^{4} \times 3^{15} \times 2^{25} \times 10^{7}$,
so $b=0, a \neq 0$.
Considering the scenario when the last 7 zeros are removed, we know that the number can be divided by $2^{25}$, and also by 8, meaning the la... | a=2, b=0, c=0, d=3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 (2003 British Mathematical Olympiad) Find all positive integers $a, b, c$ satisfying $(a!) \cdot(b!)=a! + b! + c!$
untranslated text remains in its original format and line breaks are preserved. | Without loss of generality, assume $a \geqslant b$, then the original equation becomes $a!=\frac{a!}{b!}+1+\frac{c!}{b!}$.
Since the three terms $a!, \frac{a!}{b!}, 1$ in the above equation are integers, it follows that $c \geqslant b$.
Since each term on the right side is a positive integer, their sum is at least 3. T... | a=3, b=3, c=4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10. Find the positive integer solutions of the equation $3^{x}=2^{x} y+1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 10. Solution: Rewrite the equation as $3^{x}-1=2^{x} y$
(1)
This indicates that if $(x, y)$ is a solution, then $x$ cannot exceed the exponent of the factor 2 in the standard factorization of $3^{x}-1$. Let $x=2^{m}(2 n+1)$, where $m, n$ are non-negative integers. Thus, we have
$$3^{x}-1=3^{2^{m}(2 n+1)}-1=\left(3^{2 n... | (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(... | (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 ... | (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
$$f(k)=\left\{\begin{array}{l}
0, \text { if } k \text { is a perfect square } \\
{\left[\frac{1}{\{\sqrt{k}\}}\right], \text { if } k \text { is not a perfect square }}
\end{array} \text {, find } \sum_{k=1}^{240} f(k)\right. \t... | When $k$ is not a perfect square, $k$ must lie between two consecutive perfect squares. Divide $[1,240]$ into several intervals with perfect squares as boundaries, and sum $f(k)$ for each subinterval.
Solving $15^{2}<240<16^{2}$.
Since $f(k)=0$ when $k$ is a perfect square, we have
$$\begin{aligned}
\sum_{k=1}^{240} f... | 768 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 $... | 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... | 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 |
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, $\frac{1}{[\sqrt{k}]}=\frac{1}{n}$.
$$\begin{aligned}
t^{2} \sum_{k=1}^{2006} \frac{1}{[\sqrt{k}]} & =\sum_{1... | \frac{1927}{22} | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 5 (2003 2004 Swedish Mathematical Competition) Find all real numbers $x$ that satisfy the equation $\left[x^{2}-2 x\right]+2[x]=$ $[x]^{2}$, where $[a]$ denotes the greatest integer less than or equal to $a$.
| Analyzing $\left[x^{2}-2 x\right]$ is relatively complex, so we consider using substitution to simplify it.
Let $x=y+1$, then the original equation becomes
$$\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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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)^... | 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... | 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 |
6. In the Cartesian coordinate system, the number of integer points $(x, y)$ that satisfy $(|x|-1)^{2}+(|y|-1)^{2}<2$ is $\qquad$ . | 6. 16 Since $(|x|-1)^{2} \leqslant(|x|-1)^{2}+(|y|-1)^{2}<2$.
Therefore, $(|x|-1)^{2}<2, -1 \leqslant|x|-1 \leqslant 1$, which means $0 \leqslant|x| \leqslant 2$, similarly $0 \leqslant|y| \leqslant 2$.
Upon inspection, $(x, y)=(-1, \pm 1),(-1,0),(1,0),(0, \pm 1),(-1, \pm 2),(1, \pm 2),(-2, \pm 1),(2, \pm 1)$ are the ... | 16 | Geometry | math-word-problem | Yes | Yes | number_theory | false |
Example 3 In the Cartesian coordinate system, the number of integer points that satisfy $(1) y \geqslant 3 x$; (2) $y \geqslant \frac{1}{3} x ;(3) x+y \leqslant 100$ is how many? | As shown in Figure 7-5, the region enclosed by the lines $y=3x$, $y=\frac{1}{3}x$, and $x+y=100$ forms a triangular region. The three vertices of this triangle are $O(0,0)$, $A(75,25)$, and $B(25,75)$. $\square$
Next, we calculate the number of integer points $N$ on the boundary and inside $\triangle OAC$. For a grid ... | 2551 | Inequalities | math-word-problem | Yes | Yes | number_theory | false |
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 |
5. In $1 \sim 1000$, the number of pairs $(x, y)$ that make $\frac{x^{2}+y^{2}}{7}$ an integer is $\qquad$ pairs. | 5. 10011 From $\frac{x^{2}+y^{2}}{7} \in \mathbf{Z}$ we know $7 \mid x^{2}+y^{2}$.
When $7 \mid x$ and $7 \mid y$, it is obvious that $7 \mid x^{2}+y^{2}$.
When $7 \nmid x$ or $7 \nmid y$, from $7 \mid x^{2}+y^{2}$ we know that $7 \nmid x$ and $7 \nmid y$. By Fermat's Little Theorem, $x^{6} \equiv 1(\bmod 7)$, which m... | 10011 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6. Let the planar region $T=\{(x, y) \mid x>0, y>0, x y \leqslant 48\}$, then the number of lattice points within $T$ is $\qquad$ . | $$\begin{array}{l}
\text { 6. } 202 \text { Let } T_{1}=\left\{(x, y) \mid 0<x \leqslant \sqrt{48}, 0<y \leqslant \frac{n}{x}\right\}, \\
T_{2}=\left\{(x, y) \mid 0<y \leqslant \sqrt{48}, 0<x \leqslant \frac{n}{y}\right\} .
\end{array}$$
Then $T=T_{1} \cup T_{2}, T_{1} \cap T_{2}=\{(x, y) \mid 0<x \leqslant \sqrt{48},... | 202 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Given that $m, n$ are integers greater than 7, consider a rectangular array of $m \times n$ points. Color $k$ of these points red, such that the three vertices of any right-angled triangle with its two legs parallel to the sides of the rectangle are not all red points. Find the maximum value of $k$. | Assume there are $m$ rows and $n (n \leqslant m)$ columns, then the maximum value is at least $m+n-2$, because by selecting any row and column, and coloring all points in these row and column except their common point in red, it satisfies the condition.
Below, we use mathematical induction to prove that in an $m \time... | m+n-2 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
8. If $n$ is a natural number less than 50, find all values of $n$ such that the values of the algebraic expressions $4 n+5$ and $7 n+6$ have a common divisor greater than 1.
Let the above text be translated into English, please retain the original text's line breaks and format, and output the translation result direc... | 8. Solution: Let $(4 n+5,7 n+6)=d>1$, then $d|(4 n+5), d|(7 n+6)$
Thus $d \mid(7 n+6-(4 n+5))=3 n+1$,
$$\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 |
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 |
Example 7 How many positive integer factors does 20! have? | Analyze writing 20! in its standard factorization form $n=\beta_{1}^{a_{1}} \beta_{2}^{a_{2}} \cdots \beta_{k}^{q_{k}}$, and then using $r(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots$ $\left(\alpha_{k}+1\right)$ to calculate the result.
Solution Since the prime numbers less than 20 are $2,3,5,7,11,13,1... | 41040 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 ge... | 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}$$ | Analysis Using the properties of congruence, noting that $2001=23 \times 87$, we can find that $a \equiv 1(\bmod 87)$ and $a \equiv-1$ $(\bmod 23)$. Thus, we can obtain the smallest value of $a$ that meets the requirements.
Solution Since $2001=87 \times 23$. By the problem, there exists a positive odd number $n$ such... | 436 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
$$\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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