problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
9. 241 Let the complex number $z_{k}=x_{k}+i y_{k}, k=1,2, \cdots, n, x_{k}$ and $y_{k}$ be real numbers, $i=\sqrt{-1}$. Let $r$ denote the absolute value of the real part of $\sqrt{z_{1}^{2}+z_{2}^{2}+\cdots+z_{n}^{2}}$, prove that:
$$r \leqslant\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|$$ | [Proof] Let $a+i b=\sqrt{z_{1}^{2}+z_{2}^{2}+\cdots+z_{n}^{2}}$, where $a, b$ are real numbers, then
$$\begin{array}{l}
a^{2}-b^{2}=\sum_{k=1}^{n} x_{k}^{2}-\sum_{k=1}^{n} y_{k}^{2} \\
a b=\sum_{k=1}^{n} x_{k} y_{k}
\end{array}$$
If $r=|a|>\sum_{k=1}^{n}\left|x_{k}\right|$, since $\sum_{k=1}^{n}\left|x_{k}\right| \geq... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,178 |
9.242 Let $p$ and $n$ be positive integers, $C_{h, k} \in [0,1]$, $h=1,2, \cdots, n$, $k=1,2, \cdots, p h$. Prove that:
$$\sum_{h=1}^{n} \sum_{k=1}^{p h}\left(\frac{C_{h, k}}{h}\right)^{2} \leqslant 2 p \sum_{h=1}^{n} \sum_{k=1}^{p h} C_{h, k} .$$ | [Proof] Let $a_{h}=\frac{1}{h} \sum_{k=1}^{p h} C_{h, k}$, then $0 \leqslant a_{h} \leqslant p$. Since
$$\sum_{h=1}^{n} \sum_{k=1}^{\rho_{h}}\left(\frac{C_{h, k}}{h}\right)^{2} \leqslant\left(\sum_{h=1}^{n} \sum_{k=1}^{\rho_{h}} \frac{C_{h, k}}{h}\right)^{2}$$
it suffices to prove
$$\left(\sum_{h=1}^{n} a_{h}\right)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,179 |
9. 243 Prove: For any real numbers $a_{1}, a_{2}, \cdots, a_{n}$, there exists a natural number $k, 1 \leqslant k \leqslant$ $n$, such that for any $1 \geqslant b_{1} \geqslant b_{2} \cdots \geqslant b_{n} \geqslant 0$, we have
$$\left|\sum_{i=1}^{n} b_{i} a_{i}\right| \leqslant 1 \sum_{i=1}^{k} a_{i} \mid$$ | [Proof] Let $s_{0}=0, s_{i}=a_{1}+a_{2}+\cdots+a_{i}, i=1,2, \cdots, n$, then $a_{i}=s_{i}-s_{i-1}, i=1,2, \cdots, n$.
Thus, we have
$$\begin{array}{l}
\left|\sum_{i=1}^{n} b_{i} a_{i}\right|=\left|\sum_{i=1}^{n} b_{i}\left(s_{i}-s_{i-1}\right)\right|=\left|\sum_{i=1}^{n} b_{i} s_{i}-\sum_{i=1}^{n-1} b_{i+1} s_{i}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,180 |
9・244 Let real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ and $A$ satisfy
$$A+\sum_{i=1}^{n} a_{i}^{2}<\frac{1}{n-1}\left(\sum_{i=1}^{n} a_{i}\right)^{2}$$
Prove: For $1 \leqslant i<j \leqslant n$ we have
$$A<2 a_{i} a_{j} .$$ | [Proof] By contradiction. Suppose there exist $1 \leqslant i<j \leqslant n$, such that
$$A \geqslant 2 a_{i} a_{j}$$
Without loss of generality, let $i=1, j=2$. Then we have
$$A+\sum_{i=1}^{n} a_{i}^{2} \geqslant 2 a_{1} a_{2}+\sum_{i=1}^{n} a_{i}^{2}=\left(a_{1}+a_{2}\right)^{2}+a_{3}^{2}+\cdots+a_{n}^{2}$$
By the C... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,181 |
$$\begin{array}{l}
9 \cdot 245 \quad \text { Let } 0<x_{i} \leqslant 1,0<y_{i} \leqslant 1, i=1,2, \cdots, n \text {, and } \\
x_{i}+y_{j}=1, i=1,2, \cdots, n .
\end{array}$$
For any natural number $m$, prove:
$$\left(1-x_{1} \cdots x_{n}\right)^{m}+\left(1-y_{1}^{m}\right) \cdots\left(1-y_{n}^{m}\right) \geqslant 1$$ | [Proof] First, for any $a, b \in [0,1]$, use induction to prove that for any natural number $m$,
$$(a+b-ab)^{m} \geqslant a^{m}+b^{m}-a^{m} b^{m}$$
When $m=1$, it is obvious that both sides of (1) are equal. Assume that (1) holds for $m=k$. Then we have
$$\begin{aligned}
(a+b-ab)^{k+1} & =(a+b-ab)^{k}(a+b-ab) \\
& \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,182 |
9. 246 If a positive number $M$ and the array
$$\begin{array}{l}
a_{11}, a_{12}, \cdots, a_{1 n} \\
a_{21}, a_{22}, \cdots, a_{2 n} \\
a_{n 1}, a_{n 2}, \cdots, a_{n n}
\end{array}$$
satisfy that for any $x_{1}, x_{2}, \cdots, x_{n} \in\{-1,1\}$ we have
$$\sum_{k=1}^{n}\left|a_{k 1} x_{1}+a_{k 2} x_{2}+\cdots+a_{k n} ... | [Proof] Let the set
$$X=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) ; x_{k} \in\{-1,1\}, k=1,2, \cdots, n\right\}$$
Clearly, the number of elements in $X$ is $2^{n}$, and for any $\left(x_{1}, \cdots, x_{n}\right) \in X$ we have
$$\sum_{k=1}^{n}\left|a_{k 1} x_{1}+\cdots+a_{k n} x_{n}\right| \leqslant M$$
Thus,
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,183 |
$9 \cdot 247$ Prove: For any $a_{1}, a_{2}, \cdots, a_{n} \in[0,2], n \geqslant 2$, we have
$$\sum_{i, j=1}^{n}\left|a_{i}-a_{j}\right| \leqslant n^{2} .$$
Determine for what $a_{1}, a_{2}, \cdots, a_{n}$ the equality holds in the above inequality. | [Proof] Without loss of generality, let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$, then
$$\begin{aligned}
S & =\sum_{i, j=1}^{n}\left|a_{i}-a_{j}\right|=2 \sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right) \\
& =2 \sum_{i=1}^{n}(2 i-n-1) a_{i} \\
& =\left\{\begin{array}{ll}
2 \sum_{i=1}^{\frac{n-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,184 |
9. 248 For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, let $a_{n+1}=a_{1}$, is the inequality
$$\sum_{i=1}^{n}\left(\frac{a_{i}}{a_{i+1}}\right)^{n} \geqslant \sum_{i=1}^{n} \frac{a_{i+1}}{a_{i}}$$
true? | [Solution] By the AM-GM inequality, we have
$$\begin{aligned}
\frac{a_{2}}{a_{1}} & =\frac{a_{2}}{a_{3}} \cdot \frac{a_{3}}{a_{4}} \cdot \cdots \cdot \frac{a_{n}}{a_{n+1}} \cdot 1 \\
& \leqslant \frac{1}{n}\left[\left(\frac{a_{2}}{a_{3}}\right)^{n}+\left(\frac{a_{3}}{a_{4}}\right)^{n}+\cdots+\left(\frac{a_{n}}{a_{n+1}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,185 |
9. 249 For any positive numbers $a_{k}, b_{k}, k=1,2, \cdots, n$, prove that:
$$\sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}} \leqslant \frac{A B}{A+B} .$$
where
$$A=\sum_{k=1}^{n} a_{k}, B=\sum_{k=1}^{n} b_{k} .$$ | [Proof] First, prove that for any positive numbers $x_{1}, y_{1}, x_{2}, y_{2}$, we have
$$\frac{x_{1} y_{1}}{x_{1}+y_{1}}+\frac{x_{2} y_{2}}{x_{2}+y_{2}} \leqslant \frac{\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)}{x_{1}+x_{2}+y_{1}+y_{2}}$$
In fact (1)
$$\begin{array}{l}
\Leftrightarrow \quad \frac{x_{1} y_{1}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,186 |
9.250 Mark some intervals on a line segment of length 1 such that no two intervals have a common interior, and the distance between any two points, whether in the same interval or different intervals, is not equal to 0.1. Prove: The sum of the lengths of the marked intervals does not exceed 0.5. | [Proof] Without loss of generality, let the line segment of length 1 be $[0,1]$. Let $M$ be the union of points on the marked intervals, and $N=[0,1] \backslash M$, i.e., $N$ is the union of points on some intervals in $[0,1]$ that have no common interior. Suppose $k$ is a natural number and $1 \leqslant k \leqslant 10... | \sum_{k=1}^{10} x_{k} \leqslant \frac{1}{2} | Combinatorics | proof | Yes | Yes | inequalities | false | 735,187 |
1.149 Calculate the value of the following expression
$$\frac{\left(10^{4}+324\right)\left(22^{4}+324\right)\left(34^{4}+324\right)\left(46^{4}+324\right)\left(58^{4}+324\right)}{\left(4^{4}+324\right)\left(16^{4}+324\right)\left(28^{4}+324\right)\left(40^{4}+324\right)\left(52^{4}+324\right)}$$ | [Solution] Since $324=4 \times 81=4 \times 3^{4}$,
the factors in the numerator and denominator of this problem are all in the form of $x^{4}+4 y^{4}$. Since
$$\begin{aligned}
& x^{4}+4 y^{4} \\
= & x^{4}+4 x^{2} y^{2}+4 y^{4}-4 x^{2} y^{2} \\
= & \left(x^{2}+2 y^{2}\right)^{2}-(2 x y)^{2} \\
= & \left(x^{2}+2 x y+2 ... | 373 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,188 |
9. 251 For a set of vectors $v_{1}, v_{2}, \cdots, v_{1989}$ in the plane, each of length not exceeding 1, prove that there exist $\varepsilon_{k} \in\{-1,1\}, k=1,2, \cdots, 1989$, such that
$$\left|\sum_{k=1}^{1989} \varepsilon_{k} v_{k}\right| \leqslant \sqrt{3} \text {. }$$ | [Proof] Prove a stronger proposition by induction: Given $n$ vectors $v_{1}, v_{2}, \cdots, v_{n}$ in the plane, each of length not exceeding 1, there exist $\varepsilon_{k} \in\{-1,1\}, k=1,2, \cdots, n$, such that
$$\left|\sum_{k=1}^{n} \varepsilon_{k} v_{k}\right| \leqslant \sqrt{2} .$$
When $n=1$, (1) clearly hold... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,189 |
9. 252 In the plane, $m$ vectors $\vec{u}_{1}, \vec{u}_{2}, \cdots, \overrightarrow{u_{m}}$ satisfy
$$\left|\vec{u}_{i}\right| \leqslant 1, i=1,2, \cdots, m, \vec{u}_{1}+\vec{u}_{2}+\cdots+\overrightarrow{u_{m}}=\overrightarrow{0} .$$
Prove: These $m$ vectors can be rearranged as $\vec{v}_{1}, \vec{v}_{2}, \cdots, \ve... | [Proof] First, we prove a lemma: Let real numbers $x_{1}, x_{2}, \cdots, x_{m}$ satisfy:
$$\left|x_{i}\right| \leqslant 1, i=1,2, \cdots, m, x_{1}+x_{2}+\cdots+x_{m}=0$$
Then they can be rearranged as $y_{1}, y_{2}, \cdots, y_{m}$, such that
$$\begin{array}{l}
\left|y_{1}\right| \leqslant 1,\left|y_{1}+y_{2}\right| \l... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,190 |
$9 \cdot 253$ Given $n$ real numbers on a circle $(n \geqslant 2)$, assume their sum is 0 and one of the numbers is 1.
(1) Prove that there exist two adjacent numbers whose difference is not less than $\frac{4}{n}$;
(2) Prove that there exists a number such that the difference between its two adjacent numbers' arithmet... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Algebra | proof | Yes | Yes | inequalities | false | 735,191 |
9・254 Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$ be $n$ real numbers satisfying the following condition: for any integer $k>0$, we have
$$a_{1}^{k}+a_{2}^{k}+\cdots+a_{n}^{k} \geqslant 0$$
Let
$$p=\max \left\{\left|a_{1}\right|,\left|a_{2}\right|, \cdots,\left|a_{n}\right|\right\} .$$
Prove: $p=a_{1... | [Proof] First, use proof by contradiction to prove $p=a_{1}$.
In fact, if $p \neq a_{1}$, then $p=\left|a_{n}\right|$, and $p>a_{1}, a_{n}a_{n}$,
where $1 \leqslant k \leqslant n-1$. Then
$$\begin{array}{c}
a_{1}^{2 l+1}+a_{2}^{2 l+1}+\cdots+a_{n}^{2 l+1} \\
=a_{n}^{2 l+1}\left[\left(\frac{a_{1}}{a_{n}}\right)^{2 l+1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,192 |
$\begin{array}{c}9 \cdot 255 \text { If } a_{1}+a_{2}+\cdots+a_{n}=1 \text {, prove: } \frac{a_{1}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}} \\ +\frac{a_{2}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}}+\cdots+\frac{a_{n}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{n} a_{1}^{2}+a_{1}^{3}} \geqslant \f... | [Proof] Let
$$\begin{aligned}
A= & \frac{a_{1}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}}+\frac{a_{2}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}}+\cdots+ \\
& \frac{a_{n}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{n} a_{1}^{2}+a_{1}^{3}}, \\
B= & \frac{a_{2}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,193 |
9-256 Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ non-negative real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=$ $n$. Prove:
$$\frac{a_{1}^{2}}{1+a_{1}^{4}}+\frac{a_{2}^{2}}{1+a_{2}^{4}}+\cdots+\frac{a_{n}^{2}}{1+a_{n}^{4}} \leqslant \frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\cdots+\frac{1}{1+a_{n}}$$ | [Proof] Since for any real number $a$, we have
$$2 a^{2} \leqslant 1+a^{4}$$
Therefore, the left side of the original inequality $\leqslant \frac{n}{2}$.
By the arithmetic-harmonic mean inequality, we get
$$\begin{array}{c}
\frac{\left(1+a_{1}\right)+\left(1+a_{2}\right)+\cdots+\left(1+a_{n}\right)}{n} \\
\geqslant \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,194 |
9.257 Given $a>2,\left\{a_{n}\right\}$ is recursively defined as follows: $a_{0}=1, a_{1}=a$, and $a_{n+1}=\left(\frac{a_{n}^{2}}{a_{n-1}^{2}}-2\right) a_{n}$.
Prove: For any $K \in N$, we have
$$\frac{1}{a_{0}}+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{k}}<\frac{1}{2}\left(2+a-\sqrt{a^{2}-4}\right) .$$ | [Proof] Let $f(x)=x^{2}-2$. Then
$$\frac{a_{n+1}}{a_{n}}=f\left(\frac{a_{n}}{a_{n-1}}\right)=f^{(2)}\left(\frac{a_{n-1}}{a_{n-2}}\right)=\cdots=f^{(n)}\left(\frac{a_{1}}{a_{0}}\right)=f^{(n)}(a) .$$
Thus,
$$\begin{aligned}
a_{n} & =\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdot \cdots \cdot \frac{a_{1}}{a_{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,195 |
$9 \cdot 258$ Let $2 n$ real numbers $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}(n \geqslant 3)$ satisfy the conditions: $\square$
(1) $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}$;
(2) $0<a_{1}=a_{2}, a_{i}+a_{i+1}=a_{i+2}(i=1,2, \cdots, n-2)$;
(3) $0<b_{1} \leqslant b_{2}, b_{i}+b_{i+1} \leqslant... | [Proof] If $a_{1} \leqslant b_{1}$, then by the known recurrence relation, we have $a_{i} \leqslant b_{i}(i=2,3$, $\cdots, n)$, and thus the conclusion holds.
If there exists $i_{0}$ such that $2 \leqslant i_{0} \leqslant n-1, a_{i_{0}} \leqslant b_{i_{0}}$ and $a_{i_{0}+1} \leqslant b_{i_{0}+1}$, then when $i_{0}=n-1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,196 |
9・260 Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers satisfying the following conditions.
$$\left|x_{1}+x_{2}+\cdots+x_{n}\right|=1$$
and $\quad\left|x_{i}\right| \leqslant \frac{n+1}{2}, i=1,2, \cdots, n$.
Prove: There exists a permutation $y_{1}, y_{2}, \cdots, y_{n}$ of $x_{1}, x_{2}, \cdots, x_{n}$, such that $... | [Proof] For any permutation $\pi=\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ of $x_{1}, x_{2}, \cdots, x_{n}$. Let
$$S(\pi)=y_{1}+2 y_{2}+\cdots+n y_{n}$$
and
$$r=\frac{n+1}{2} \text {. }$$
We need to prove that there exists a permutation $\pi$ such that $|S(\pi)| \leqslant r$.
$$\text { Let } \pi_{0}=\left(x_{1}, x_{2... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,198 |
1. 150 Prove that the arithmetic square root of the product of four consecutive natural numbers plus 1 is still a natural number. | [Proof] Let four consecutive natural numbers be $n-1, n, n+1, n+2(n \geqslant 2)$, then we have
$$\begin{aligned}
& \sqrt{(n-1) n(n+1)(n+2)+1}=\sqrt{n^{4}+2 n^{3}-n^{2}-2 n+1} \\
= & \sqrt{\left(n^{2}+n-1\right)^{2}} \\
= & n^{2}+n-1
\end{aligned}$$
which is a natural number. | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,199 |
9-262 Let $n$ be a positive integer, and $n \geqslant 3$. Also, let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers, where $2 \leqslant a_{i} \leqslant 3, i=1,2, \cdots, n$. If we take $S=a_{1}+a_{2}+\cdots+a_{n}$. Prove:
$$\frac{a_{1}^{2}+a_{2}^{2}-a_{3}^{2}}{a_{1}+a_{2}-a_{3}}+\frac{a_{2}^{2}+a_{3}^{2}-a_{4}^{2}}{a_{2}... | [Proof $] \quad \frac{a_{i}^{2}+a_{i+1}^{2}-a_{i+2}^{2}}{a_{i}+a_{i+1}-a_{i+2}}$
$$=a_{i}+a_{i+1}+a_{i+2}-\frac{2 a_{i} a_{i+1}}{a_{i}+a_{i+1}-a_{i+2}}$$
Notice that
$$1=2+2-3 \leqslant a_{i}+a_{i+1}-a_{i+2} \leqslant 3+3-2=4$$
and from $\left(a_{i}-2\right)\left(a_{i+1}-2\right) \geqslant 0$ we get
$$-2 a_{i} a_{i+1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,201 |
$9 \cdot 263$ Let $n \in N, x_{0}=0, x_{i}>0, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} x_{i}=1$.
Prove:
$$\begin{aligned}
1 & \leqslant \sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{0}+x_{1}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+x_{i+1}+\cdots+x_{n}}} \\
& <\frac{\pi}{2}
\end{aligned}$$ | [Proof] Since
$$\sum_{i=1}^{n} x_{i}=1$$
Therefore, by the AM-GM inequality, we have
$$\begin{aligned}
& \sqrt{\left(1+x_{0}+x_{1}+\cdots+x_{i-1}\right)\left(x_{i}+x_{i+1}+\cdots+x_{n}\right)} \\
\leqslant & \frac{1+x_{0}+x_{1}+\cdots+x_{n}}{2}=1
\end{aligned}$$
Thus,
$$\begin{aligned}
& \sum_{i=1}^{n} \frac{x_{i}}{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,202 |
9・264 Let $r_{1}, r_{2}, \cdots, r_{n}$ be real numbers greater than or equal to 1. Prove:
$$\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}+\cdots+\frac{1}{r_{n}+1} \geqslant \frac{n}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1}$$ | [Proof] When $n=1$, the inequality obviously holds.
Below, we prove by mathematical induction that the inequality holds for $n=2^{k}$ (where $k$ is a non-negative integer). Assume that the inequality holds for $n=2^{m}$ (where $m$ is some non-negative integer). If $r_{1}, r_{2}, \cdots, r_{2 n} \geqslant 1$, then we ha... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,203 |
9・265 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and satisfy
$$a_{1}+a_{2}+\cdots+a_{n}<1$$
Prove:
$$\frac{a_{1} a_{2} \cdots a_{n}\left[1-\left(a_{1}+a_{2}+\cdots+a_{n}\right)\right]}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)} \leqslant \... | [Proof] Let
$$a_{n+1}=1-\left(a_{1}+a_{2}+\cdots+a_{n}\right) \text {. }$$
Clearly, $a_{n+1}>0$. Thus, we obtain $n+1$ positive numbers whose sum is 1. The inequality to be proved can be rewritten as
$$n^{n+1} a_{1} a_{2} \cdots a_{n} a_{n+1} \leqslant\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,204 |
9. 266 For each integer $n \geqslant 2$, and $x_{1}, x_{2}, \cdots, x_{n} \in[0,1]$, prove:
$$\sum_{k=1}^{n} x_{k}-\sum_{1 \leqslant k<j \leqslant n} x_{k} x_{j} \leqslant 1$$ | [Proof] Using mathematical induction.
When $n=2$, since $x_{1}, x_{2} \in [0,1]$, we have
$$x_{1}+x_{2}-x_{1} x_{2}=1-\left(1-x_{1}\right)\left(1-x_{2}\right)<1.$$
Assume that when $n=m$ (where $m \geqslant 2$ is a positive integer), for $x_{1}, x_{2}, \cdots, x_{m} \in [0,1]$, we have
$$\sum_{k=1}^{m} x_{k}-\sum_{1 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,205 |
9・267 $a, b, c, d$ are all positive integers, $r=1-\frac{a}{b}-\frac{c}{d}$, if $a+c \leqslant$ 1993, and $r>0$, prove: $r>\frac{1}{1993^{3}}$. | [Proof] From the given,
$$r=1-\frac{a}{b}-\frac{c}{d}=\frac{b d-a d-b c}{b d}$$
From $r>0$, we get
$$b d>a d+b c>a d$$
Thus, $b>a$.
Similarly, $b d>a d+b c>b c$,
so $d>c$.
Since $b d-(a d+b c) \geqslant 1$,
we have $r \geqslant \frac{1}{b d}$.
Without loss of generality, assume $b \leqslant d$ (the case for $b>d$ can... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,206 |
$9 \cdot 268$ Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in[1,2]$ and
$$\sum_{i=1}^{n} a_{i}^{2}=\sum_{i=1}^{n} b_{i}^{2},$$
Prove:
$$\sum_{i=1}^{n} \frac{a_{i}^{3}}{b_{i}} \leqslant \frac{17}{10} \sum_{i=1}^{n} a_{i}^{2} .$$
And find the necessary and sufficient condition for equality. | [Proof] Since $a_{i}, b_{i} \in [1,2], i=1,2, \cdots, n$, therefore
$$\frac{1}{2} \leqslant \frac{\sqrt{\frac{a_{i}^{3}}{b_{i}}}}{\sqrt{a_{i} b_{i}}}=\frac{a_{i}}{b_{i}} \leqslant 2$$
Thus, we have
$$\left(\frac{1}{2} \sqrt{a_{i} b_{i}}-\sqrt{\frac{a_{i}^{3}}{b_{i}}}\right)\left(2 \sqrt{a_{i} b_{i}}-\sqrt{\frac{a_{i}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,207 |
1-151 1 Prove that $2^{1992}-1$ can be expressed as the product of six integers, each greater than $2^{248}$.
| [Solution] $1992=8 \times 249 \quad$ and $249 \times 4=332 \times 3$
$$\begin{aligned}
2^{1992}-1= & \left(2^{249}\right)^{8}-1 \\
= & \left\{\left(2^{249}\right)^{4}-1\right\}\left\{\left(2^{249}\right)^{4}+1\right\} \\
= & \left(2^{249}-1\right)\left(2^{249}+1\right)\left\{\left(2^{249}\right)^{2}+1\right\}\left\{\le... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,208 |
1-155 $a, b$ and $n$ are fixed natural numbers, and for any natural number $k(k \neq b)$, $a-k^{n}$ is divisible by $b-k$. Prove $a=b^{n}$.
保留了源文本的换行和格式，这是翻译结果。 | [Solution] Since $k^{n}-b^{n}=(k-b)\left(k^{n-1}+k^{n-2} b+\cdots+b^{n-1}\right)$, $k^{n} - b^{n}$ is divisible by $k-b$.
Therefore, for any $k \neq b, \left(k^{n}-b^{n}\right)-\left(k^{n}-a\right)=a-b^{n}$ is divisible by $k-b$. By the arbitrariness of $k$, we get $a=b^{n}$. | a=b^{n} | Number Theory | proof | Yes | Yes | inequalities | false | 735,209 |
1. 156 Let $a$ and $b$ be two odd integers, prove: $a^{3}-b^{3}$ is divisible by $2^{n}$ if and only if $a-b$ is divisible by $2^{n}$. | [Proof] Since $a^{3}-b^{3}$ equals the product of the number $A=a^{2}+a b+b^{2}$ and the number $B=a-b$, if $B$ is divisible by $2^{n}$, then the product $A B$ is also divisible by $2^{n}$.
Since $a$ and $b$ are odd, $A=a^{2}+a b+b^{2}$ is also odd, so $A$ and $2^{n}$ are coprime. In this case, if the product $A B$ is... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,210 |
1. 158 (1) Prove: 10201 is a composite number in any base of numeration greater than 2. | ［Solution］（1）Let $x$ be the base. If $x>2$, then we have
$$\begin{aligned}
10201 & =x^{4}+2 x^{2}+1 \\
& =\left(x^{2}+1\right)^{2} .
\end{aligned}$$
It is clearly a composite number.
$$\text { (2) } \begin{aligned}
\text { Since } 10101 & =x^{4}+x^{2}+1 \\
& =\left(x^{2}+x+1\right)\left(x^{2}-x+1\right) .
\end{aligned... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,211 |
$1 \cdot 159$ Find the positive integer pairs $a, b$, such that:
(1) $a b(a+b)$ is not divisible by 7;
(2) $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$.
Verify your answer. | ［Solution］We have
$$(a+b)^{7}-a^{7}-b^{7}=7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2}$$
From (1) and (2), we know that $7^{7} \mid 7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2}$, which means $7^{3} \mid a^{2}+a b+b^{2}$.
Next, we need to find the positive integer pairs $(a, b)$ that satisfy
$$(a+b)^{2}-a b=a^{2}+a b+b^{2... | a=18, b=1 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,212 |
$1 \cdot 160$ Let $p$ and $q$ be natural numbers, satisfying
$$\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{1318}+\frac{1}{1319}$$
Prove that $p$ is divisible by 1979. | [Proof]
$$\begin{aligned}
\frac{p}{q} & =1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{1318}+\frac{1}{1319} \\
& =\left(1+\frac{1}{2}+\cdots+\frac{1}{1319}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right) \\
& =\left(1+\frac{1}{2}+\cdots+\frac{1}{1319}\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{659}\right... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,213 |
1. 161 Given that $k, m, n$ are positive integers, $m+k+1$ is a prime number greater than $n+1$, and let $C_{s}=s(s+1)$. Prove:
$\left(C_{m+1}-C_{k}\right)\left(C_{m+2}-C_{k}\right) \cdots\left(C_{m+n}-C_{k}\right)$ is divisible by $C_{1} C_{2} \cdots C_{n}$ | [Proof] If the product
$$\left(C_{m+1}-C_{k}\right)\left(C_{m+2}-C_{k}\right) \cdots\left(C_{m+n}-C_{k}\right)$$
equals 0, then the conclusion is obviously true.
If the product (1) is not equal to 0, all factors must have the same sign. Without loss of generality, assume all factors are positive.
Since \( C_{S}=S(S+1... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,214 |
$1 \cdot 162$ (1) Determine all positive integers $n$ such that $2^{n}-1$ is divisible by 7.
(2) Prove that for all positive integers $n, 2^{n}+1$ is not divisible by 7. | [Solution] (1) If $n$ is a multiple of 3, then we can set $n=3k$ (where $k$ is a positive integer),
$$\begin{aligned}
2^{n}-1 & =2^{3k}-1 \\
& =8^{k}-1 \\
& =(8-1)\left(8^{k-1}+8^{k-2}+\cdots+1\right)
\end{aligned}$$
Therefore, $2^{n}-1$ is divisible by 7.
If $n$ is not a multiple of 3, then we can set $n=3k+1, n=3k+2... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,215 |
$1 \cdot 163$ integers $A$ and $B$ have the same last $k$ digits, prove: the numbers $A^{n}$ and $B^{n} (n$ being a natural number $)$ also have the same last $k$ digits. | ［Proof］ Since $A^{n}-B^{n}$
$$=(A-B)\left(A^{n-1}+A^{n-2} B+\cdots+A B^{n-2}+B^{n-1}\right),$$
but the last $k$ digits of $A-B$ are the same, i.e., $A-B$ is divisible by $10^{k}$. Therefore, $A^{n}-B^{n}$ is also divisible by $10^{k}$. This means that the last $k$ digits of $A^{n}$ and $B^{n}$ (where $n$ is a natural ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,216 |
1-17 Let $A$ be the sum of the digits of the decimal number $4444^{4444}$, and let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$ (all numbers here are in decimal). | [Solution] First, we estimate the number of digits in $10000^{4444}$, which has $4 \times 4444 + 1 = 17777$ digits. Since each digit can be at most 9, and $4444^{4444} < 10000^{4444}$, we have
$$A < 17777 \times 9 = 159993.$$
Now, let's represent $A$ as
$$A = a_{5} \cdot 10^{5} + a_{4} \cdot 10^{4} + a_{3} \cdot 10^{3... | 7 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,217 |
1. 164 A pure decimal repeating fraction refers to a decimal $0 . a_{1} \cdots a_{k}$, which starts repeating $k$ digits after the decimal point. For example: $0.243243243 \cdots=\frac{9}{37}$. A mixed repeating decimal $0 . b_{1} \cdots b_{m} \dot{a}_{1} \cdots \dot{a}_{k}$ refers to a decimal that eventually repeats ... | [Proof] Without loss of generality, assume we have shifted the cycle as far to the left as possible, so in $0 . b_{1} \cdots b_{m}$ $\dot{a}_{1} \cdots \dot{a}_{k}$, $m \geqslant 1$, and $b_{m} \neq a_{k}$.
By the given, this repeating decimal equals the fraction $\frac{p}{q}$, thus
$$\frac{p}{q}=\frac{\left(10^{k}-1\... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,218 |
1. 167 Find all such three-digit numbers $abc$, whose square ends with the digits $abc$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | [Solution] Let the three-digit number that satisfies the requirement be $A$, then $A^{2}-A=A(A-1)$ is divisible by 1000 $=8 \cdot 125$. Since $A$ and $A-1$ are coprime, it can only be that 125 divides $A$ and 8 divides $A-1$, or 125 divides $A-1$ and 8 divides $A$. It is easy to verify that the only three-digit numbers... | null | Geometry | math-word-problem | Yes | Yes | inequalities | false | 735,220 |
1・ 170 Proof: For each natural number $n, 13 \cdot(-50)^{n}+17 \cdot 40^{n}-30$ is always divisible by 1989. | [Proof] Let $f(n)=13 \cdot(-50)^{n}+17 \cdot 40^{n}-30$.
$$\begin{aligned}
g(n) & =f(n+1)-f(n) \text {. Then } \\
g(n) & =(-1)^{n+1} \cdot 13 \cdot 50^{n} \cdot 51+17 \cdot 40^{n} \cdot 39 \\
& =3 \times 13 \times 17 \times 10^{n}\left[(-1)^{n+1} \cdot 5^{n}+4^{n}\right]
\end{aligned}$$
Since $f(1)=0$, and for $n \geq... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,221 |
$1 \cdot 174$ On the blackboard, the numbers 1 and 2 are written. Now, it is allowed to write a new number according to the following rule: if the numbers $a$ and $b$ are already written on the blackboard, then the number $a b + a + b$ can be written. Can the following numbers be obtained by such a rule:
(1) 13121;
(2)... | [Solution] Let $C$ represent the new number $ab + a + b$. Then $c + 1 = ab + a + b + 1 = (a + 1)(b + 1)$. This means that if we do not consider the numbers already written on the board, but instead consider the numbers that are one greater than the written numbers, then each new number is equal to the product of two ex... | 13121 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,223 |
1. 18 Find all natural numbers $n$ : in decimal notation, every $n$-digit number consisting of $n-1$ ones and one 7 is a prime number. | [Solution] Let $m$ be an $n$-digit number consisting of $n-1$ ones and one 7, then it can be written in the following form:
$$m=A_{n}+6 \times 10^{k}$$
Here $A_{n}$ is an $n$-digit number consisting of $n$ ones, and $0 \leqslant k \leqslant n-1$.
If $n$ is divisible by 3, then the sum of the digits of $m$ is divisible... | n=1 \text{ or } 2 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,224 |
1. 175 In an $n \times n$ grid, each cell is filled with a real number, and the sum of the numbers in each row and each column is equal to 0. Now, we are allowed to perform the following operation: choose any row and add its numbers to the numbers in a certain column (i.e., add the first number of the row to the first ... | [Proof] Number the rows of the table from top to bottom and the columns from left to right. Let \( O_{j}^{i}, k \) denote the operation of adding the \( i \)-th row to the \( j \)-th column and subtracting the \( i \)-th row from the \( k \)-th column.
When we apply the operations:
\[ O_{n, 1}^{1}, O_{n, 2}^{2}, \cdot... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,225 |
1-176 Three automata print pairs of natural numbers on cards. The automata work as follows: the first automaton reads a card $(a ; b)$ and outputs a new card $(a+1 ; b+1)$, the second automaton reads a card $(a, b)$ and outputs a new card $\left(\frac{a}{2} ; \frac{b}{2}\right)$ (it only works when $a, b$ are even), an... | [Solution] (1) We can obtain $(1 ; 50)$. The method is as follows:
$$\begin{array}{l}
(5 ; 19) \rightarrow(6 ; 20) \rightarrow(3 ; 10) \rightarrow \cdots \rightarrow(10 ; 17) \rightarrow(3 ; 17) \rightarrow \\
(4 ; 18) \rightarrow(2 ; 9) \rightarrow \cdots \rightarrow(9 ; 16) \rightarrow(2 ; 16) \rightarrow(1 ; 8) \rig... | not found | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,226 |
1. 177 There are several numbers $0, 1, 2$ written on the blackboard. You can erase two of these numbers that are not equal and replace them with a number that is different from the erased numbers (for example, replace 0 and 1 with 2, replace 1 and 2 with 0, replace 0 and 2 with 1). Prove: If after several such operati... | [Proof] Suppose on the blackboard there are $P$ zeros, $q$ ones, and $r$ twos. After each step, all three numbers $p$, $q$, and $r$ increase or decrease by 1, thus the parity of $p$, $q$, and $r$ changes accordingly. When only one number remains on the blackboard, one of $p$, $q$, and $r$ becomes 1, and the other two b... | proof | Logic and Puzzles | proof | Yes | Yes | inequalities | false | 735,227 |
1. 178 Find all three-digit numbers $A$ such that the arithmetic mean of all numbers obtained by rearranging the digits of $A$ is still equal to $A$. | [Solution] Let $\overline{abc}$ be the number we are looking for. Then, according to the problem, we have
$$abc + \overline{acb} + \overline{bac} + \overline{bca} + \overline{cab} + \overline{cba} = 6 \overline{abc}$$,
which simplifies to $7a = 3b + 4c$.
$$7(a-b) = 4(c-b)$$
From this, either $a = b = c$; or $a - b = 4... | 111, 222, \cdots, 999, 407, 518, 629, 370, 481, 592 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,228 |
179 Several numbers are written on a circle. If for some 4 consecutive numbers $a, b, c, d$ there is $(a-d)(b-c)<0$, then $b$ and $c$ can be swapped. Prove: such operations can only be performed a finite number of times. | [Proof] If $(a-d)(b-c)<0$, then
$$a b+c d<a c+b d$$
Therefore, $a b+b c+c d<a c+c b+b d$.
This means that after swapping $b$ and $c$, the sum $s$ of the products of adjacent numbers increases. However, the sum $s$ can only take a finite number of different values, so such operations can only be performed a finite numb... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,229 |
1-180 Each card has a five-digit number from 11111 to 99999 written on it, and then these cards are arranged in any order to form a row. Prove: the resulting 444445-digit number cannot be a dwarf of 2's.
Note: The term "dwarf of 2's" is not standard mathematical terminology and may need clarification. If it refers t... | [Proof] We prove that any 444445-digit number obtained can be divided by 11111. Therefore, it cannot be a power of 2.
$$10^{5}=9 \times 11111+1,$$
Thus, $10^{5}$ has a remainder of 1 when divided by 11111.
Since any number $S$ obtained can be written as
$$S=11111 \times\left(10^{5}\right)^{k_{1}}+11112 \cdot\left(10^{... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,230 |
1. 181 Take a seventeen-digit number, write down all its digits in reverse order to get a new number, and then add this number to the original number. Prove: the sum obtained contains at least one even digit. | [Proof] Suppose that each number in the sum is odd. We write one of the two numbers under the other (aligning the digits), and add the digits in the corresponding positions. Since the last digit of the sum is odd, the sum of the first digits also yields an odd number. Therefore, when adding, 1 cannot be carried from th... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,231 |
1・183 Given any set of $n$ integers $a_{1}, a_{2}, \cdots, a_{n}$, from it we can obtain a new set: $\square$
$$\frac{a_{1}+a_{2}}{2}, \frac{a_{2}+a_{3}}{2}, \cdots, \frac{a_{n-1}+a_{n}}{2}, \frac{a_{1}+a_{n}}{2};$$
From this set, according to the same rule, we can obtain a new set $\cdots \cdots$. Prove that if all t... | [Proof] If the $n$ initial numbers are not all equal, then after $n$ steps, the maximum number in this set must decrease, and the minimum number must increase.
Since the maximum and minimum numbers obtained at each step are integers, after a finite number of steps, a set of equal numbers $(a, a, \cdots, a)$ can certai... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,232 |
1・184 (1) Given a set of positive numbers $(a, b, c, d)$, a new array $(a b, b c, c d, d a)$ can be obtained according to the following rule: each number is multiplied by the next one, and the 4th number is multiplied by the first one. According to this rule, a third array, a fourth array, etc., can also be obtained. P... | [Proof] (1) Suppose four positive numbers $(a, b, c, d)$ reappear.
First, we prove: In this case, $a b c d=1$.
Let $a b c d=p$, then the product of the second set of four positive numbers equals $p^{2}$, the third set is $p^{4}$, and the fourth set is $p^{8}, \cdots$ Clearly, when $p \neq 1$, there will be no two ident... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,233 |
1. 19 Below is a division of an eight-digit number by a three-digit number. Try to find the quotient and explain your reasoning.
The above text is translated into English, preserving the original text's line breaks and format. | [Solution] The original expression is labeled as follows:
Where all characters represent decimal digits, and characters with subscript 1 are not equal to 0.
(1) Since after the first trial division by 7, two digits $\overline{a_{5} a_{6}}$ are simultaneously moved down, it indicates that $\overline{c_{1} c_{2} a_{5}} ... | 70709 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,234 |
1. 186 In the exercise book, several numbers are written. If the arithmetic mean of two or more of these numbers does not equal any of the numbers, then this mean can be written next to the already written numbers. Prove: starting from the number 0.1, using this method of writing, you can obtain the numbers:
(1) $\frac... | [Solution] First, we can obtain all rational numbers between 0 and 1 with denominators that are positive integer powers of 2.
In fact, from the number line, using the method of finding the arithmetic mean, we can obtain the midpoint of any line segment with two known points as endpoints. That is, from 0 and 1, we can ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,235 |
1- 187 In a game, the "magician" asks a person to think of a three-digit number $(abc)$ (where $a, b, c$ are the decimal digits of the number in order), and asks this person to select 5 numbers $(acb), (bac), (bca), (cab)$, and $(cba)$, and to find the sum $N$ of these 5 numbers, and tell the sum $N$ to the "magician".... | [Solution] $(a b c)+N$
$$\begin{array}{l}
=(a b c)+(a c b)+(b a c)+(b c a)+(c a b)+(c b a) \\
=222(a+b+c) .
\end{array}$$
Therefore, we should find a multiple of 222, $222k$, such that it is greater than $N$ (since $(a b c) \neq 0$), and less than $N+1000$ (since $(a b c)$ is a three-digit number), and the sum of the ... | 358 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 735,236 |
'1.188 Write the numbers $1,2,3, \cdots, 1986,1987$ on the blackboard. At each step, determine some of the numbers written and replace them with the remainder of their sum divided by 7. After several steps, two numbers remain on the blackboard, one of which is 987. What is the second remaining number? | [Solution] We notice that after each "erasing" and writing of the "remainder", the sum of all numbers still written on the blackboard remains unchanged modulo 7. Let the last remaining number be $x$, then $x+987$ and the following sum
$$1+2+\cdots+1986+1987=1987 \cdot 7 \cdot 142$$
have the same remainder when divided... | 0 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,237 |
1. 190 Let $x=a_{1} a_{2} \cdots a_{n}$ be an $n$-digit number in decimal notation, where $a_{1}, a_{2}, \cdots, a_{n}\left(a_{1} \neq 0\right)$ are non-negative integers not exceeding 9. The following $n$ numbers are obtained by cyclically permuting $x$:
$$\begin{array}{l}
x_{1}=x=a_{1} a_{2} \cdots a_{n}, \\
x_{2}=a_... | [Solution]
$$\text { Let } \begin{aligned}
x_{1}=x & =a_{1} a_{2} \cdots a_{n} \\
& =a_{1} 10^{n-1}+a_{2} 10^{n-2}+\cdots+a_{n-1} 10+a_{n},
\end{aligned}$$
then $x_{2}=a_{n} a_{1} a_{2} \cdots a_{n-1}$,
thus we have $10 x_{2}-x_{1}=\left(10^{n}-1\right) a_{n}$.
Given $1989\left|x_{2}, 1989\right| x_{1}, 1989=9 \times ... | 48 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,239 |
1-191 Write down all natural numbers from 1 to 1988 on a blackboard. Perform operations $A$ and $B$ repeatedly on these numbers: first $A$, then $B$, followed by $A$ again, and then $B$, and so on. Operation $A$ involves subtracting the same natural number from each number written on the blackboard (the subtrahend can ... | [Solution] Each time operations $A$ and $B$ are performed once, one number is erased from the blackboard. Therefore, after performing operations $A$ and $B$ 1987 times each, only one number remains on the blackboard.
Let the natural number $d_{k}$ be the subtrahend when the $k$-th operation $A$ is performed, where $k=... | 1 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,240 |
1・192 Konya and Betya divide $2 n+1$ nuts, $n \geqslant 2$, and each wants to get as many as possible. Assume there are three ways to divide (each way has three steps).
First step: Betya divides all the nuts into two parts, each part having no less than two nuts.
Second step: Konya then divides each part into two part... | [Solution] The most favorable method for Konya is the first one, and the least favorable method is the third one.
In fact, after Betia's first move, regardless of the size of the walnut piles (each with $a$ and $b$ walnuts, $a<b$), Konya can always divide the larger pile into two parts, each containing 1 and $b-1$ wal... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,241 |
1-194 $a$, $b$, $p$ are any integers, prove: there must exist two coprime numbers $k$, $l$, such that $a k + b l$ is divisible by $p$. | [Proof] Let the greatest common divisor of the numbers $b$ and $p-a$ be $d$, and $b=k d, p-a=l d$, then the numbers $k$ and $l$ are coprime. At this point,
$$a k+b l=\frac{a b}{d}+\frac{(p-a) b}{d}=p k$$
Thus, $a k+b l$ is divisible by $p$. | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,242 |
$1 \cdot 20$ An integer is called divisible by its digit sum if: (1) none of its digits are 0; (2) it can be divided by its digit sum (for example, 322 is divisible by its digit sum). Prove: there are infinitely many integers that are divisible by their digit sum. | [Proof] We will prove that for any natural number $n$, the number $11 \cdots 1$ can be divided by $3^{n}$.
Indeed, when $n=1$, 111 can be divided by the sum of its digits $1+1+1=3$, so the conclusion holds. Assume that when $n=k$, $11 \cdots 1$ can be divided by the sum of its digits $3^{k}$. Then, since
$$\underbrace{... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,243 |
1-195 1 to 1982 natural numbers are arranged in a certain order, one after another. The computer reads two adjacent numbers from left to right (the 1st and the 2nd, the 2nd and the 3rd, etc.), until the last two numbers, and if the larger number is on the left, the computer swaps their positions. Then the computer read... | [Solution] Let the largest number among the first 99 numbers be $a$, the smallest number among the last 1882 numbers be $b$, and the 100th number be $c$. If $a>c$, then when the computer reads from left to right, it will move the position of $a$ to the left of $c$, causing the number $c$ to change its position, which c... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,244 |
1-196 There is a table with 4 rows, in the first row, any natural numbers are written, and these numbers may be the same. In the second row, fill in the numbers as follows: from left to right, when seeing $a$ in the first row, if $a$ has already appeared $k$ times in the first row, then fill in $k$ below $a$. Similarly... | [Proof] If the first 3 numbers from the top in a column of the table are $a$, $m$, and $n$, then each number in the sequence $1, 2, \cdots, m-1, m$ appears at least $n$ times to the left of the number $m$ in the second row, while any number greater than $m$ appears fewer than $n$ times. Therefore, the number in the fou... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,245 |
1. 197 Write 1000 numbers in a row, and then write the second row below it according to the following rule: write a natural number below each number $a$ in the first row, indicating the number of times $a$ appears in the first row. The same method can be used to get the third row from the second row: write a natural nu... | [Proof] (1) From the problem, we know:
(1) In the $m$-th row $(m \geqslant 2)$, each number $a$ is followed by a number not less than $a$;
(2) Each such number does not exceed 1000.
If every row is different from its next row, then only a finite number of rows can be written. This contradicts the problem statement. Th... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,246 |
198 Construct a triangular table according to the following rules: in the top row, write the natural number $a$, and below each number, write $a^{2}$ in the left lower position and $a+1$ in the right lower position. For example, when $a=2$, the resulting table is shown in the figure on the next page. Prove: in each row... | [Proof] Assume that there are identical numbers in some rows of the table. Let $n$ be the number of the topmost row among these rows, and let $p$ and $q$ be the identical numbers in this row.
Since there are no identical numbers in the $(n-1)$-th row, $p$ and $q$ must be derived from different numbers $r$ and $s$ in t... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,247 |
1. For a natural number $k$ in decimal with $n$ digits, round this number to the nearest ten: if the last digit is greater than 4, replace the last digit with 0 and increase the digit in the tens place by 1, then apply the same rounding method to the nearest hundred, and so on. After the $(n-1)$-th rounding, the result... | [Proof] If $k \geqslant a \underset{n-2}{44 \cdots 45} (a \geqslant 1$ is the leading digit $)$, then $\bar{k}=(a+1)10^{n-1}$, thus we get
$$\frac{\bar{k}}{k} \leqslant \frac{(a+1) 10^{n-1}}{a \underbrace{44 \cdots 45}_{n-2}}<\frac{a+1}{a+\frac{4}{9}} \leqslant \frac{18}{13}$$
If $k \leqslant a \underset{n 4 \cdots 4}... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,248 |
$1 \cdot 200$ Write the number $1234567891011 \cdots \cdots \cdot 19941995$ on the blackboard, forming the integer $N_{1}$, erase the digits in the even positions of $N_{1}$, the remaining digits form the integer $N_{2}$, erase the digits in the odd positions of $N_{2}$, the remaining digits form the integer $N_{3}$, e... | [Solution] Since
$$9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873$$
Therefore, the integer $N_{1}$ consists of 6873 digits, which we number from left to right as 1, $2,3, \cdots, 6873$. We examine the set of these numbers $\{1,2,3, \cdots, 6873\}$.
Erasing the digits in even positions from $N_{1}$ is equivalen... | 9 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,249 |
$1 \cdot 201$ In a $2000 \times 2000$ grid, each small square is filled with either a 1 or -1. It is known that the sum of all the numbers in the grid is non-negative. Prove: It is possible to find 1000 rows and 1000 columns such that the sum of the numbers in the squares where they intersect is not less than 1000. | [Solution] Since the total sum of the numbers filled in the table is non-negative, there exists a row where the sum of 2000 numbers is non-negative, hence in this row there are at least 1000 ones.
By swapping the columns of the number table, make the first 1000 numbers in this row all 1. Denote the $2000 \times 1000$ ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,250 |
1. 202 There are three piles of stones. Each time, A moves one stone from one pile to another, and A can receive payment from B after each move, the amount of which equals the difference between the number of stones in the pile where the stone is placed and the number of stones in the pile where the stone is taken from... | [Solution] The amount of money A earns is 0.
In fact, we can imagine that initially, in each pile of stones, there is exactly one line segment connecting any two stones. When A moves a stone, he first unties the line segments connecting this stone to other stones, and then uses these line segments to connect this stone... | 0 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,251 |
$1 \cdot 203$ Let $n$ be an odd number greater than 1. Given
$$x_{0}=\left(x_{1}^{(0)}, x_{2}^{(0)}, \cdots, x_{n}^{(0)}\right)=(1,0, \cdots, 0,1) .$$
Set $\quad x_{i}^{(k)}=\left\{\begin{array}{l}0, x_{i}^{(k-1)}=x_{i+1}^{(k-1)} \\ 1, x_{i}^{(k-1)} \neq x_{i+1}^{(k-1)}\end{array}\right.$ when, $\quad(i=1,2, \cdots n)... | [Proof] Divide a circle into $n$ equal parts, and place $x_{1}$, $x_{2}, \cdots, x_{n}$ in these points in a clockwise direction (as shown in the figure). The diameter of the circle passing through the midpoint of $x_{1}$ and $x_{n}$ has the property that the numbers on either side of it are symmetric about it. A diame... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,252 |
1.204 $a_{1}, a_{2}, \cdots, a_{n}$ is a sequence, for each $k, 1 \leqslant k \leqslant n, a_{k} \in$ $\{0,1\}$. If $a_{k}, a_{k+1}$ are the same, let $b_{k}=0$; if $a_{k}, a_{k+1}$ are different, let $b_{k}$ $=1$, thus obtaining a new sequence $b_{1}, b_{2}, \cdots, b_{n-1}, b_{k} \in\{0,1\}$, where $k=1,2$, $\cdots, ... | [Solution] Let $x_{n}$ denote the maximum value sought.
Obviously, $x_{1}=1, x_{2}=2$.
When $n=3$, the first row of the number table $a_{1}, a_{2}, a_{3}$ has only 8 possibilities, so the sum of 1s in these number tables is $3,4,4,4,3,3,3,0$. Therefore, we get
Now, we seek the relationship between $x_{n}$ and $x_{n+3}$... | x_{n}=\left[\frac{n(n+1)+1}{3}\right], n \in N | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,253 |
1. 205 Starting from any three-digit number $n$, we get a new number $f(n)$, which is the sum of the three digits of $n$, the pairwise products of the three digits, and the product of the three digits.
(1) When $n=625$, find $\frac{n}{f(n)}$ (take the integer value).
(2) Find all three-digit numbers $n$ such that $\fra... | ［Solution］Let
$$n=\overline{a b c}=100 a+10 b+c$$
Here $a, b, c \in\{0,1,2, \cdots, 9\}$, and $a \geqslant 1$. By the problem statement,
$$f(n)=a+b+c+a b+b c+c a+a b c$$
(1) When $n=625$,
$$\begin{aligned}
f(625) & =6+2+5+6 \times 2+2 \times 5+5 \times 6+6 \times 2 \times 5 \\
& =125, \\
\frac{625}{f(625)} & =5 .
\end... | 199,299,399,499,599,699,799,899,999 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,255 |
$2 \cdot 1$ set $S$ has an operation +, for all elements $a, b, c$ in $S$ there is $(a+c)+(b+c)=a+b$; and there is an element $a$ in $S$, such that $a+e=a, a+a=e$, define $a * b = a+(e+b)$. Prove: for all elements $a, b, c$ in $S$,
$$(a * b) * c = a * (b * c).$$ | [Proof]
$$\begin{aligned}
(a * b) * c & =(a * b)+(e+c) \\
& =[a+(e+b)]+(e+c) \\
a *(b * c) & =a+[e+(b * c)] \\
& =a+[e+(b+(e+c))]
\end{aligned}$$
Since $e+(b+a)=(a+a)+(b+a)=a+b$, we have
$$\begin{aligned}
& a+[e+(b+(e+c))]=a+[(e+c)+b] \\
= & {[a+(e+b)]+\{[(e+c)+b]+(e+b)\} } \\
= & {[a+(e+b)]+[(e+c)+e] } \\
= & {[a+(e+... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,256 |
$2 \cdot 2 S$ is a set, defined with addition, subtraction, and multiplication operations, satisfying the following axioms:
(1) If $x, y$ belong to $S$, then $x+y, xy$ belong to $S$.
(2) For all $x, y \in S, x+y=y+x, xy=yx$.
(3) For all $x, y, z \in S$,
$$x+(y+z)=(x+y)+z, x(yz)=(xy)z.$$
(4) For all $x, y, z \in S$,
$$x... | ［Sol］(a) Obviously.
(b) Since $x+x+x=0$, then $-x=x+x$,
$$\begin{array}{l}
x(-y)=x(y+y)=x y+x y=-x y \\
x \cdot 0=x(x+(-x))=x^{2}+x(-x)=x^{2}-x^{2}=0
\end{array}$$
Therefore, $(-x-y)+(x+y)=-x+(x+y)-y=y-y=0$,
so $\square$
$$-(x+y)=-x-y .$$
If $x \leqslant y$, then
$$x^{2} y-x y^{2}-x y+x^{2}=0$$
Multiplying (4) by $x... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,257 |
2. 4 For $\{1,2,3, \cdots, n\}$ and each of its non-empty subsets, we define the "alternating sum" as follows: arrange the numbers in the subset in decreasing order, then alternately add and subtract the numbers starting from the largest (for example, the alternating sum of $\{1,2,4,6,9\}$ is $9-6+4-2+1=6$, and the alt... | [Solution] Obviously, the non-empty subsets of the set $\{1,2,3,4,5,6,7\}$ total $2^{7}-1=127$, and each element appears $2^{6}=64$ times in the subsets.
According to the problem, the numbers $1,2,3,4,5,6$ each appear 32 times in odd positions and 32 times in even positions in the subsets, so the alternating sum of th... | 448 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,258 |
2. $6 S$ is the set formed by $m$ unordered pairs of positive integers $(a, b)(1 \leqslant a<b \leqslant n)$. Prove that there are at least $4 m \cdot \frac{m-\frac{n^{2}}{4}}{3 n}$ unordered triples $(a, b, c)$, such that $(a, b),(b, c), (c, a)$ all belong to $S$. | [Proof] Consider $n$ points $1,2, \cdots, n$. If $(i, j) \in S$, then draw a line between $i$ and $j$. We aim to find the number of triangles in this graph.
Let $(i, j) \in S$, and suppose there are $d_{i}$ lines emanating from $i$. Then the number of triangles with $(i, j)$ as an edge is at least $d_{i}+d_{j}-n$. Sin... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,259 |
2 - 9 Assume necklace $A$ has 14 beads; $B$ has 19 beads. Prove: for every odd number $n \geqslant 1$, it is possible to label the beads with the sequence
$$\{n, n+1, n+2, \cdots, n+32\}$$
such that each number is used exactly once and the numbers on adjacent beads are coprime. | [Proof] Take an integer $m$ as a parameter to be determined, $1 \leqslant m \leqslant 5$.
Imagine labeling the beads of necklace $A$ with the 14 consecutive natural numbers $n+m, n+m+1, \cdots, n+m+13$. Since any pair of consecutive integers are coprime, it suffices to ensure that $n+m$ and $n+m+13$ are coprime, i.e.,
... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,260 |
2・ 10 (1) For what natural numbers $n>2$, is there a set of $n$ consecutive natural numbers such that the largest number in the set is a divisor of the least common multiple of the other $n-1$ numbers?
(2) For what $n>2$, is there exactly one set with the above property? | [Solution] (1) First, we point out that $n \neq 3$. Otherwise, let $\{r, r+1, r+2\}$ have the property required by the problem: $(r+2) \mid r(r+1)$. Since $(r+1, r+2)=1$, it follows that $(r+2) \mid r$, which is a contradiction.
Below, we discuss the cases for $n \geqslant 4$ by considering whether $n$ is odd or even.... | 4 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,261 |
2・ 11 Let $M$ be a set consisting of 1985 distinct positive integers, each of whose prime factors does not exceed 26. Prove that from $M$ at least one subset of four distinct elements can be found such that the product of these four numbers is equal to the fourth power of some positive integer. | [Proof] Since there are only 9 prime numbers not exceeding 26: $2,3,5,7,11,13,17,19,23$, for any $\alpha \in M$, it can certainly be written as $\alpha=2^{\alpha_{1}} \cdot 3^{a_{2}} \cdots \cdots \cdot 23^{2 \alpha}$. Clearly, $\alpha \in M$ is uniquely determined by the ordered array $\left(a_{1}, a_{2}, \cdots, a_{9... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,262 |
2-12 Let $d$ be any positive integer different from $2,5,13$. Prove that in the set $\{2,5,13, d\}$, there can be found two distinct elements $a, b$ such that $a b-1$ is not a perfect square. | [Proof] Obviously, $2 \cdot 5-1, 2 \cdot 13-1, 5 \cdot 13-1$ are all perfect squares, so we can only search among the three numbers $2 d-1, 5 d-1, 13 d-1$.
Assume that the three numbers are all perfect squares, i.e., $2 d-1=x^{2}, 5 d-1=y^{2}, 13 d-1=z^{2}$, where $x, y, z$ are all positive integers.
From $2 d-1=x^{2... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,263 |
2・13 Let $S=\{1,2, \cdots, n\}, A$ be an arithmetic sequence with at least two terms, a positive common difference, all of whose terms are in $S$, and adding any other element of $S$ to $A$ does not form an arithmetic sequence with the same common difference as $A$. Find the number of such $A$ (here, a sequence with on... | ［Solution］ Let the common difference of $A$ be $d$, then $1 \leqslant d \leqslant n-1$.
We discuss in two cases:
(1) Suppose $n$ is even, then
when $1 \leqslant d \leqslant \frac{n}{2}$, there are $d$ $A$s with common difference $d$;
when $\frac{n}{2}+1 \leqslant d \leqslant n-1$, there are $n-d$ $A$s with common diffe... | \left[\frac{n^2}{4}\right] | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,264 |
2- 15 (1) Use a small calculator to verify that for each integer $k=0,1,2,3,4,5,6,7$, the sum of the $k$-th powers of the numbers in the set $\{1,5,10,24,28,42,47,51\}$ is equal to the sum of the $k$-th powers of the numbers in the set $\{2,3,12,21,31,40,49,50\}$.
(2) Let $n$ be a positive integer. Can two sets $\left\... | [Solution] (1) Omitted.
(2) If $a_{1}^{k}+a_{2}^{k}+\cdots+a_{n}^{k}=b_{1}^{k}+b_{2}^{k}+\cdots+b_{n}^{k}, 1 \leqslant k \leqslant n$, then by Newton's formulas for power sums, the elementary symmetric polynomials of $a_{1}, a_{2}, \cdots, a_{n}$
$$\begin{array}{l}
\sigma_{1}=a_{1}+a_{2}+\cdots+a_{n}, \\
\sigma_{2}=a_{... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,266 |
2・16 Let $n$ be a given natural number greater than 2, and let $V_{n}$ be the set of numbers of the form 1 $+k n$ (where $k=1,2, \cdots$). A number $m \in V_{n}$ is called irreducible in $V_{n}$ if there do not exist two numbers $p, q \in V_{n}$ such that $p q=m$. Prove that there exists a number $r \in V_{n}$ that can... | [Proof] $V_{n}=\{1+n, 1+2 n, \cdots, 1+k n, \cdots\}$, where $n>2, k=1$, $2, \cdots$.
Since $n>2$, it is obvious that $n-1 \notin V_{n}, 2 n-1 \notin V_{n}$; and $n-1$ and $2 n-1$ cannot be decomposed into the product of $n$ numbers in $V_{n}$. And
$$\begin{array}{l}
(n-1)(2 n-1)=1+(2 n-3) n \in V_{n} \\
(n-1)^{2}=1+(... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,267 |
2. 17 For any non-empty set of numbers $S$, let $\sigma(S)$ and $\pi(S)$ denote the sum and product of all elements in $S$, respectively. Prove that:
$$\sum_{S} \frac{\sigma(S)}{\pi(S)}=\left(n^{2}+2 n\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)(n+1) .$$
where " $\sum_{S} $" denotes the sum over a... | [Proof] Since $\sigma(S)=\sum_{K \in S} K$, we have
$$\sum_{S} \frac{\sigma(s)}{\pi(s)}=\sum_{S} \sum_{k \in S} \frac{k}{\pi(s)}=\sum_{k=1}^{n} \sum_{S \ni k} \frac{k}{\pi(s)}$$
where " $\sum_{S \ni k}$ " denotes the summation over all subsets $S$ containing $k$ for fixed $k=1,2,3, \cdots, n$, then
$$\begin{aligned}
\... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,268 |
2. 18 The sum of a set of numbers is the sum of all its elements. Let $S$ be a set of positive integers not exceeding 15, such that the sums of any two disjoint subsets of $S$ are not equal, and among all sets with the above property, the sum of $S$ is the largest. Find the sum of the set $S$. | [Solution] We first prove that $S$ has at most 5 elements.
In fact, if $S$ has at least 6 elements, then the number of non-empty subsets of $S$ with at most 4 elements is at least
$$C_{6}^{1}+C_{6}^{2}+C_{6}^{3}+C_{6}^{4}=56 \text { (subsets) }$$
The sums of these subsets do not exceed 54 (i.e., $15+14+13+12=54$). By ... | 61 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,269 |
2. 19 Given any $h=2^{r}$ (where $r$ is a non-negative integer). Find all natural numbers $K$ that satisfy the following conditions: for each such $K$, there exist an odd natural number $m>1$ and a natural number $n$ such that
$$K\left|m^{h}-1, m\right| n^{\frac{m^{h}-1}{k}}+1$$ | [Solution] For $h=2^{r}$, we agree to denote the set of all $K$ that satisfy the conditions of the problem as $K(h)$. We will prove:
$$K(h)=\left\{2^{r+s} t \mid S, t \in N, 2 \times t\right\}$$
We will use the following fact:
$$m \equiv 1(\bmod 4) \Rightarrow 2^{r} \| \frac{m^{2^{r}}-1}{m-1}$$
This fact is obvious b... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,270 |
$2 \cdot 20$ Let $R_{1}, R_{2}, \cdots$ be a family of finite sequences of positive integers defined by the following rules:
$$R(1)=(1) \text {; }$$
If $R_{n-1}=\left(x_{1}, \cdots, x_{s}\right)$, then
$$R_{n}=\left(1,2, \cdots, x_{1}, 1,2, \cdots, x_{2}, \cdots, 1,2, \cdots, x_{s}, n\right)$$
For example: $R_{2}=(1,... | [Proof] For the sequence $w=\left(x_{1}, \cdots, x_{m}\right)$, define its "expansion"
$\bar{w}=\left(1,2, \cdots, x_{1}, 1,2, \cdots, x_{2}, \cdots, 1,2, \cdots, x_{m}\right)$.
For any two sequences $u=\left(x_{1}, \cdots, x_{m}\right)$ and $v=\left(y_{1}, \cdots, y_{k}\right)$, define their concatenation
$u v=\left(x... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,271 |
$1 \cdot 23$ If a $2n$-digit number is a perfect square, and the numbers formed by its first $n$ digits and its last $n$ digits are also perfect squares (here the second $n$-digit number can start with 0, but cannot be 0, and the first $n$-digit number formed by the first $n$ digits cannot start with 0), then this numb... | ［Solution］(1) It is easy to verify that there is only one two-digit strange number, 49.
Let $x^{2}$ be a 4-digit strange number, then
$$(10 x+t)^{2}=100 x^{2}+20 x t+t^{2}$$
where $20 x t+t^{2}$ is the square of a natural number less than 10 and $x^{2} \geqslant 10$, thus $x \geqslant 4, x t \leqslant 4$, from which w... | not found | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,272 |
2. 21 Consider the representation of a real number $x$ in base 3. $K$ is the set of all such numbers $x$ in $[0,1]$ whose digits are 0 or 2. If
$$S=\{x+y \mid x, y \in K\},$$
prove that:
$$S=\{z \mid 0 \leqslant z \leqslant 2\}=[0,2]$$ | [Proof] In $K$, each digit of $x$ and $y$ is 0 or 2, so each digit of $\frac{x}{2}$ and $\frac{y}{2}$ is 0 or 1, and thus each digit of $\frac{x}{2}+\frac{y}{2}$ in base 3 is 0, 1, or 2, and by
$$x \in[0,1], y \in[0,1]$$
we know
$$\frac{x}{2}+\frac{y}{2} \in[0,1] .$$
Conversely, for any number in $[0,1]$, each digit ... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,273 |
2・22 Let $S=\{1,2, \cdots, 1990\}$. If the sum of the elements of a 31-element subset of $S$ is divisible by 5, it is called a good subset of $S$. Find the number of good subsets of $S$. | [Solution] For $k=0,1,2,3,4$ define
$F_{k}=\{A: A \subseteq S,|A|=31, A$'s element sum $\equiv k(\bmod 5)\}$, where $|A|$ denotes the number of elements in set $A$. Since $31 \equiv 1(\bmod 5)$, if $\left\{x_{1}, x_{2}, \cdots, x_{31}\right\}$ is in $F_{0}$, then $\left\{x_{1}+k, x_{2}+k, \cdots, x_{31}+k\right\}$ is i... | \frac{1}{5} C_{1990}^{31} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,274 |
$2 \cdot 26$ Try to divide the set $\{1,2, \cdots, 1989\}$ into 117 mutually disjoint subsets $A_{i}$ $(i=1,2, \cdots, 117\}$, such that
(i) each $A_{i}$ contains 17 elements;
(ii) the sum of the elements in each $A_{i}$ is the same. | [Solution] Since $1989=117 \times 17$, the set $\{1,2, \cdots, 1989\}$ can be sequentially divided into 17 segments, each containing 117 numbers. Now we appropriately place each segment of 117 numbers into $A_{1}$, $A_{2}, \cdots, A_{117}$ to satisfy condition (ii).
Starting from the fourth segment, place the numbers ... | 528 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,276 |
1.24 Does there exist a natural number $n$ such that: (in decimal notation) the sum of the digits of the number $n$ is equal to 1000, and the sum of the digits of the number $n^{2}$ is equal to $1000^{2} ?$ | [Proof] We will prove that for any natural number $m$, there exists a natural number $n$ composed entirely of the digits 1 or 0, such that the sum of the digits of $n$ equals $m$, and the sum of the digits of $n^{2}$ equals $m^{2}$.
Indeed, when $m=1$, there exists $n=1$ that satisfies the condition, so the conclusion... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,278 |
2. 31 Let $\sigma(S)$ denote the sum of all elements in a non-empty set of integers $S$. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{11}\right\}$ be a set of positive integers, and $a_{1}<a_{2}<\cdots<a_{11}$. If for every positive integer $n \leqslant 1500$, there exists a subset $S$ of $A$ such that $\sigma(S)=n$. Find th... | Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k}(1 \leqslant k \leqslant 11)$. By the problem statement, there exists an index $m$ such that $S_{m-1}S_{k-1}+1$, and $a_{1}+\cdots+a_{k-1}=S_{k-1}$. This means there does not exist $S \subset A$ such that $\sigma(S)=S_{k-1}+1$. Therefore,
$$S_{k}=S_{k-1}+a_{k} \leqslant 2 S_{k-1}+1$$
... | 248 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,279 |
2. 32 Let $Q$ be the set of rational numbers, and let $S$ be a non-empty subset of $Q$ with the following properties:
(1) $0 \notin S$.
(2) If $s_{1} \in S, s_{2} \in S$, then $s_{1} / s_{2} \in S$;
(3) There exists a non-zero rational number $q, q \notin S$ and every non-zero rational number not in $S$ can be written ... | [Proof] Suppose $s \in S$, let $s_{1}=s_{2}=S$, then $s_{1} / s_{2}=1 \in S$.
Let $s_{1}=1, s_{2}=s$, then $1 / s \in S$.
If $t \in S$, let $s_{1}=t, s_{2}=1 / s$, then
$s_{1} s_{2}=t /(1 / s)=s t \in S$ (thus $s$ is a solution in the multiplicative sense).
Assume $u$ is a non-zero rational number, if $u \notin S$, the... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,280 |
$2 \cdot 34$ Let $n, m, k$ be natural numbers, and $m \geqslant n$. Prove: If
$$1+2+\cdots+n=m k,$$
then the numbers $1,2, \cdots, n$ can be divided into $k$ groups, such that the sum of the numbers in each group is equal to $m$. | [Proof] When $n=1$, the conclusion is obviously true.
Assume the conclusion holds for all parameters less than $n$, we will examine the case for the set $S_{n}=\{1, 2, \cdots, n\}$.
If $m=n$, then $\frac{1}{2}(n+1)=k$ is an integer, and we can group as follows: $\{n\},\{1, n-1\},\{2, n-2\}, \cdots,\left\{\frac{1}{2}(n... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,281 |
2・36 $S$ is a subset of $\{1,2, \cdots, 1989\}$, and the difference between any two numbers in $S$ cannot be 4 or 7. How many elements can $S$ have at most? | [Solution] First, we prove: In any 11 consecutive integers in the set $\{1,2, \cdots, 1989\}$, at most 5 can be elements of $S$.
In fact, let $T=\{1,2, \cdots, 11\}$. We examine the subsets $\{1,5\},\{2,9\}$, $\{3,7\},\{4,11\},\{6,10\},\{8\}$. Clearly, each of these subsets can have at most one element in $S$.
If $T$... | 905 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,282 |
$2 \cdot 37$ Given the set
$$S=\left\{z_{1}, z_{2}, \cdots, z_{1993}\right\}$$
where $z_{1}, z_{2}, \cdots, z_{1993}$ are all non-zero complex numbers (which can be viewed as non-zero vectors in the plane). Prove that the elements of $S$ can be divided into several groups such that
(i) each element of $S$ belongs to a... | [Solution] Consider all subsets of the set $S$ and calculate the modulus of the sum of all elements in each subset. The modulus values obtained in this way are only finitely many, so there must be a maximum value. Let one of the subsets that achieves this maximum modulus be denoted as $A$. If $S \backslash A \neq \phi$... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,283 |
2・38 Let the set $A=\{1,2,3, \cdots, 366\}$. If a binary subset $B$ $=\{a, b\}$ of $A$ satisfies $17 \mid(a+b)$, then $B$ is said to have property $P$.
(1) Find the number of all binary subsets of $A$ that have property $P$;
(2) Find the number of all pairwise disjoint binary subsets of $A$ that have property $P$. | [Solution] (1) $17 \mid(a+b)$ if and only if $a+b \equiv 0(\bmod 17)$,
which means $a \equiv k(\bmod 17)$ and $b \equiv 17-k(\bmod 17)$ for $k=0,1,2, \cdots, 16$.
Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: [0$],[1], \cdots,[16]$. Since $366=17 \times 21+9$, the class [0$] cont... | 3928 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,284 |
2・39 Let $n>6$ be a natural number. Given an $n$-element set $X$, take any $m$ distinct 5-element subsets $A_{1}, A_{2}, \cdots, A_{m}$ of $X$. Prove that as long as
$$m>\frac{n(n-1)(n-2)(n-3)(4 n-15)}{600}$$
there must exist $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{6}}\left(1 \leqslant i_{1}<i_{2}<\cdots<i_{6} \leqslant ... | [Proof] By contradiction.
For natural numbers $m$ satisfying the inequality in the problem, assume there exist $m$ distinct 5-element subsets of $X$ such that the union of any 6 of them is not a 6-element set. Let the class of these $m$ sets be denoted as $\mathbb{A}$. Define
$$\mathscr{A}=\{B \subset X, |B|=4 \text{ ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,285 |
$1 \cdot 25$ Find all natural numbers $x$ that satisfy the following conditions: the product of the digits of $x$ equals $44x-86868$, and the sum of the digits is a perfect cube. | [Solution] Since $44 x \geqslant 86868$, hence $x \geqslant\left[\frac{86868+43}{44}\right]=1975$.
Thus, $x$ must be at least a four-digit number. On the other hand, if the number of digits $k \geqslant 5$ in $x$, then
$$44 x-86868>4 \times 10^{k}-10^{5} \geqslant 3 \times 10^{k}>9^{k}$$
Thus, we get $44 x-86868>p(x)$... | 1989 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,286 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.