problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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Example 2. If $x, y$ are real numbers, and
$$
y=\frac{\left(1-x^{2}\right)^{\frac{1}{2}}+\left(x^{2}-1\right)^{\frac{1}{6}}}{4 x-5} \text {, }
$$
find the value of $\log _{\frac{1}{7}}(x+y)$. | To make the expression for $y$ meaningful, $x$ must simultaneously satisfy the following three inequalities.
$$
\begin{array}{c}
1-x^{2} \geqslant 0, \\
x^{2}-1 \geqslant 0, \\
4 x-5 \neq 0 . \\
\therefore x=1, \quad \text { thus } y=0 . \\
\therefore \log _{\frac{1}{7}}(x+y)=\log _{\frac{1}{7}} 1=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, find the smallest positive integer $n$ (where $n>1$) such that the square mean of the first $n$ natural numbers is an integer.
(Note: The square mean of $n$ numbers $a_{1}, a_{2}, \cdots, a_{i}$ is defined as $\left.\left[\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}\right]^{\frac{1}{2}}\right)$ | $$
\begin{array}{c}
\text { Three, let } \frac{1^{2}+2^{2}+\cdots+n^{2}}{n} \\
=\frac{n(n+1)(2 n+1)}{6 n}=\frac{(n+1)(2 n+1)}{6} \text { be a }
\end{array}
$$
perfect square $m^{2}$, then $6 m^{2}=(n+1)(2 n+1)$. Since $6 m^{2}$ is even, it must be odd. We can set $n=3 p+q$, where $q=-1, 1$ or 3.
If $q=2$, then $6 m^{... | 337 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$\therefore 、 m$ distinct positive even numbers and $n$ distinct positive odd numbers have a total sum of 1987, for all such $m$ and $n$, what is the maximum value of $3 m+4 n$? Please prove your conclusion.
The sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 1987, for all such $m$ a... | The maximum value of $3 m+4 n$ is 221. The proof is as follows: Let $a_{1}+\cdots+a_{n}+b_{1}+\cdots+b_{n}=1987$. Here, $a_{i} (1 \leqslant i \leqslant m)$ are distinct positive even numbers, and $b_{j} (1 \leqslant j \leqslant n)$ are distinct positive odd numbers. Clearly, $u$ is always an odd number, and
$$
\begin{a... | 221 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3 . Find the integers $x, y$ that satisfy the equation $2 x-3 y=17$, and for which $x^{2}+y^{2}$ is minimized, as well as the minimum value of $x^{2}+y^{2}$. | 3. Answer: $x=4, y=-3, x^{2}+y^{2}$ has a minimum value of 25.
Hint: The general solution of $2 x-3 y=17$ is $x=3 t+7$, $y=2 t-1$. Substituting into $x^{2}+y^{2}$, we get $x^{2}+y^{2}$ $=13\left(t+\frac{19}{13}\right)^{2}+\frac{289}{13}$. The integer $t$ closest to $t=-\frac{19}{13}$ is -1. | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Indicate one method for each of the following to divide a square with a side length of 1986 mm:
(1) Divide it into 1986 smaller squares with integer side lengths;
(2) Divide it into 1986 smaller squares with not all side lengths being integers. | (1) $\because 1986=2 \times 993$, divide the two sides $A B$ and $A D$ of the square into 993 equal parts (each part being $2 \text{ mm}$). As shown in Figure 5, using the discussion from Proof 1, we get: $2 \times 993-1=1985$ small squares with a side length of $2 \text{ mm}$, and one square $F C E E_{1}$ with a side ... | 1986 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. Given a three-digit integer that is a multiple of 5, the sum of its digits is 20, and the sum of one digit and the hundreds digit is a multiple of 3, find this integer. | Solution: Let the required three-digit integer be
$$
N=100 x+10 y+z \text {. }
$$
Since $N$ is a multiple of 5, it must be that $z=0$ or $z=5$. It is also given that the sum of the digits of $N$ is 20, so
$$
x+y+z=20 \text { . }
$$
If $z=0$, then $x+y=20$.
Given $1 \leqslant x \leqslant 9,0 \leqslant y \leqslant 9$, ... | 785 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example. The infinite sequence
$$
a_{1}, a_{2}, a_{3}, \ldots
$$
has the relation
$$
a_{n}=4+\frac{1}{3} u_{n-1}(n=2,3,1, \cdots)
$$
Find $\lim _{n \rightarrow \infty} a_{n}$. | $$
\begin{array}{l}
a_{2}=4+\frac{1}{3} a_{1}, \\
a_{3}=4+\frac{1}{3} a_{2}=4+\frac{1}{3}\left(4+\frac{1}{3} a_{1}\right) \\
=4+4 \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{4}=4+\frac{1}{3} a_{3} \\
=4+4 \cdot \frac{1}{3}+4\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{n}=4+4 \cdo... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Lake 1. Given the arithmetic sequence with the first term $a_{1}=3, a_{n+1}=a_{n}+5$, find up to the 5th term. | Solution: From the recursive formula $a_{2}=a_{1}+5=3+5=8$,
$$
\begin{array}{l}
a_{3}=a_{2}+5=8+5=13, \\
a_{4}=a_{3}+5=13+5=18, \\
a_{5}=a_{4}+5=18+5=23 .
\end{array}
$$
From the recursive formula to find the general term, there are commonly four forms:
I $\cdot a_{n+1}=a_{n}+d$,
Arithmetic sequence type. | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Find the smallest positive integer $n$ that makes $\frac{n-13}{5 n+6}$ a non-zero reducible fraction. | Solution: Since $(1,5)=1,|1 \times 6-(-13) \times 5|$ $=71 \neq(1,5),(1,71)=1$, the fraction can be simplified by 71. From $\left.n-13=71 m_{1}, n=71 m_{1}+13\left(m_{1} \in Z\right)\right)$, and $n-13>0$, so we take $m_{1}=1$, obtaining the smallest $n=84$. | 84 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
{\left[\frac{1^{2}}{1980}\right],\left[\frac{2^{2}}{1980}\right],\left[\frac{3^{2}}{1980}\right], \cdots,} \\
{\left[\frac{1900^{2}}{1980}\right] \text { How many different numbers are there in the sequence? }}
\end{array}
$$ | Solution: First, note that when $\alpha-\beta>1$, the value of $[a] 1_{j} [\beta]$ is definitely different. Therefore, we solve the inequality
$$
\frac{(k+1)^{2}}{1980}-\frac{k^{2}}{1980}>1,
$$
which simplifies to
$$
\begin{array}{l}
2 k+1>1980, \\
k>989 .
\end{array}
$$
Thus, starting from the 990th term, these 1980... | 1486 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Given $\sin \alpha+\cos \alpha=a$.
(1) Find the value of $\sin ^{5} \alpha+\cos ^{5} \alpha$;
(2) If $a=1$, find the value of $\sin ^{n} \alpha+\cos ^{n} \alpha$. | Let $f(n)=\sin ^{n} \alpha+\cos ^{n} \alpha$, then (1) $=a$.
Given $f(2)=\sin ^{2} \alpha+\cos ^{2} \alpha=1$,
$f(3)=a f(2)-\frac{a^{2}-1}{2} \cdot a$
$=\frac{-a^{3}+3 a}{2}$,
$f(4)=a f(3)-\frac{a^{2}-1}{2} f(2)$
$=-a^{4}+2 a^{2}+1$
2
$\therefore f(5)=\sin ^{5} \alpha+\cos ^{5} \alpha$
$=a f(4)-\frac{a^{2}-1}{2} f(3)$
... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Learn, there are 2 teachers who can teach mathematics, and there are 4 teachers who can teach both English and Japanese. Now, 3 mathematics teachers and 3 Japanese teachers are being dispatched to teach outside the school during the holiday. How many ways are there to select them?
保留源文本的换行和格式，直接输出翻译结果。 | Let the 3 teachers competent in English teaching be set $A$, 2 teachers competent in Japanese teaching be set $B$, and 4 teachers competent in both English and Japanese teaching be set $C$.
Method 1: Classify set $A$
(1) Select 3 English teachers (choose 3 Japanese teachers from $B$ and $C$), total $C_{3}^{3} \cdot C_... | 216 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 11. Find the number of consecutive zeros at the end of 1987!.
untranslated text:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
translated text:
Example 11. Find the number of consecutive zeros at the end of 1987!.
Note: The note at the end is not part of the original text and is provided for context. | To solve this problem, it is equivalent to finding the exponent of $1 \mathrm{C}$ in 1987!, which is the same as finding the exponent of 5 in 1987!, since
$$
\begin{array}{l}
{\left[\frac{1987}{5}\right]+\left[\frac{1987}{5^{2}}\right]+\left[\begin{array}{c}
1987 \\
5^{3}
\end{array}\right]} \\
+\left[\frac{1987}{5^{4}... | 494 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Let $x, y$ satisfy $3 x^{2}+2 y^{2}=6 x$, find the maximum value of $x^{2}+y^{2}$:
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | If not analyzed, from the known conditions:
$$
\begin{array}{l}
x^{2}+y^{2}=-\frac{1}{2} x^{2}+3 x \\
=-\frac{1}{2}(x-3)^{2}+\frac{9}{2},
\end{array}
$$
it would be incorrect to say that when $x=3$, $x^{2}+y^{2}$ achieves its maximum value $\frac{9}{2}$. This is because from $y^{2}=-\frac{3}{2} x^{2}+3 x \geqslant 0$,... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Find the maximum distance between two points, one on the surface of a sphere centered at $(-2$, $-10,5)$ with a radius of 19, and the other on the surface of a sphere centered at $(12,8,-16)$ with a radius of 87. | 2. Let $O$ and $O_{1}$ be the centers of two spheres, and $P, P_{1}$ be the intersection points of the extended line segment $O_{1}$ with the two spherical surfaces, such that $O$ is inside $P O_{1}$ and $O_{1}$ is inside $O P_{1}$. Clearly, the maximum distance between these two points is $P P_{1}=P O+O O_{1}+O_{1} P_... | 137 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. A natural number greater than 1, if it is exactly equal to the product of its distinct proper divisors (factors excluding 1 and itself), then it is called "good". Find the sum of the first ten "good" natural numbers. | 3. Let $k$ be a positive integer, and let $1, d_{1}, d_{2}, \cdots$, $d_{n_{-1}}, d_{n}, k$ be all its divisors, arranged in increasing order $\Rightarrow 1 \cdot k=d_{1} \cdot d_{n}=d_{2} \cdot d_{n_{-1}}=\cdots$. If $k$ is "good," then by definition, these products are also equal to $d_{1} \cdot d_{2} \cdots \cdots d... | 182 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the area of the region enclosed by the graph of the equation $|x-60|+|y|=\left|\frac{x}{4}\right|$.
untranslated text remains the same as requested. | 4. First, the graph of this equation is symmetric about the $x$-axis, so we only need to find the area enclosed by
$$
\left\{\begin{array}{l}
\left.y=\left|\frac{x}{4}\right|-1 x-60 \right\rvert\,, \\
y \geqslant 0
\end{array}\right.
$$
The region enclosed by the difference of $0 \cdots \times 1$ and the $x$-axis in t... | 480 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
$5 . x 、 y$ are integers that satisfy the equation $y^{2}+3 x^{2} y^{2}=30 x^{2} +517$, find the value of $3 x^{2} y^{2}$ . | 5. The original equation transforms to $\left(y^{2}-10\right)\left(3 x^{2}+1\right)$ $=3 \times 13^{2}>y^{2}-10=1 ， 3 ， 13 ， 39 ， 169$ or $507=>y^{2}=11,13,23,42,179$ or 517. Since $y$ is an integer $>y^{2}=49 \rightarrow y^{2}-10=39 \Rightarrow 3 x^{2}+1=13$ $\therefore 3 x^{2}=12>3 x^{2} y^{2}=12 \times 49=588$. | 588 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in the figure below, rectangle $A B C D$ is divided into 4 equal-area parts by 5 line segments. Given $X Y=Y B+B C$ $+C Z=Z W=W D+D A+A X, P Q / / A B$. If $B C=19 \mathrm{~cm}, P Q=87 \mathrm{~cm}$, find the length of $A B$ (in cm). | 6.18 The trapezoids $X Y Q P$ and $Z W P Q$ have equal areas, $H A Y-\| Z$. Both are equal to $\frac{B C}{2}$. Also, $X Y$ is $\frac{1}{4}$ of the perimeter of rectangle $A B C D$, so
$$
\begin{array}{l}
X Y=\frac{A B+B C}{2} \text { by } \frac{(P Q+X Y)}{2} \times \frac{B C}{2} \\
=\frac{A B \cdot B C}{4} \Rightarrow ... | 193 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Find the largest positive integer $n$, such that the inequality $\frac{8}{15}<$ $-\frac{n}{n+k}<\frac{7}{13}$ holds for exactly one integer $k$. | 8 . Transform the original inequality to $\frac{15}{8}>\frac{n+k}{n}>\frac{13}{7}$, which is equivalent to $\frac{7}{8}>\frac{k}{n}>\frac{6}{7} \Leftrightarrow 49 n>56 k>48 n$. Therefore, the problem is converted to finding the largest open interval ( $48 n$, $49 n$ ) that contains only one multiple of 56. Since the in... | 112 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
10. Pay for an escalator moving upwards. A walks down from its top to its bottom, totaling 150 steps, B walks up from its bottom to its top, totaling 75 steps. Assuming A's speed (number of steps walked per unit time) is 3 times B's speed, how many steps of the escalator are visible at any given moment? (Assume this nu... | 10. Let $v_{1}, v_{2}, v$ represent the speeds (number of steps walked per unit time) of $A, B$, and the automatic escalator, respectively, and let $t_{1}, t_{2}, t$ represent the time taken by $A, B$, and the automatic escalator, respectively. From the problem, we have: $v_{1}=3 v_{2}, v_{1} t_{1}=150, v_{2} t_{2}=75$... | 120 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Find the maximum value of $k$ such that $3^{11}$ can be expressed as the sum of $k$ consecutive positive integers. | i. Find the maximum positive integer $k$ such that $3^{11}=(n+1)+(n+2)+\cdots+(n+k)$ holds (where $n$ is a non-negative integer). From the right side, we have $=\frac{k(k+2 n+1)}{2}$ $->K \cdot(k+2 n+1)=2 \cdot 3^{11}$. To make the smaller factor $k$ as large as possible, $n$ must be non-negative $\Rightarrow k=2 \cdot... | 486 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
12. The square root of a number without $m$ is $\cdots$ numbers of the form $n+r$, where $n$ is a positive integer, and $r$ is a real number less than $\frac{1}{1000}$. If $m$ is the smallest positive integer satisfying the above condition, find the value of $n$ when $m$ is the smallest positive integer. | 12. From the problem: $\sqrt[3]{m}=n+r, 0\frac{1000}{3}
\end{array}
$
$\rightarrow n^{2} \approx \frac{1000}{3}$. Since $18^{2}<\frac{1000}{3}<19^{2}$, we can guess $n=18$ or $n=19$. Upon verification, when $n=18$, the inequality does not hold, but when $n=19$, the inequality is satisfied. Therefore, $n=19$ is the smal... | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. For a known sequence of distinct real numbers $r_{1}, r_{2}, r_{3}, \cdots, r_{\text {n}}$, a single operation involves comparing the second term with the first term, and swapping them if and only if the second term is smaller; then comparing the third term with the new second term, and swapping them if and only if... | 13. Notice that the operation defined in the problem, when applied to any sequence $r_{1}, r_{2}, \cdots, r_{k}$ once, will always result in the last number being the largest in the sequence. Therefore, for the initial sequence $r_{1}, r_{2}, \cdots, r_{20}, \cdots, r_{30}, r_{31}, \cdots, r_{40}$, $r_{20}$ can be move... | 931 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Find the minimum value of $|x-1|+|x-3|+|x-5|$. | To solve this type of problem, we can use the method of "fixing points, dividing segments, and discussing."
Points. Let $x-1=0, x-3=0, x-5=0$, to determine $x_{1}=1, x_{2}=3, x_{3}=5$ three points.
Dividing segments. The above three points divide the number line into four segments: $(-\infty, 1]$, $(1,3]$, $(3,5]$, $... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $y=\sqrt[3]{1+3 x}, x=\frac{1}{3}\left(y^{3}-1\right)$, when $x \rightarrow 0$, $y \rightarrow 1$.
Original expression $=\frac{2}{3} \lim _{y \rightarrow 1} \frac{y^{3}-2 y^{3}+1}{y^{3}-3 y+2}$ is still of the form $\frac{0}{0}$. Let $z=y-1$, then,
Original expression $=\frac{2}{3} \lim _{z \rightarrow 0} \frac{\l... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Some people stand in a row, among them, A does not stand at the head, and B does not stand at the end. How many ways are there to arrange them? | (1) 䒴
$P_{4}^{4}$
etc. transmission
(2) one of the middle three positions $P_{3}^{\frac{1}{3}} P \frac{1}{3} P_{3}^{3}$, so the number of ways for not standing at the head and not standing at the tail is $P_{4}^{4}+P{ }_{3}^{1} P_{3}^{1} P_{3}^{3}=78$ (benefit).
Solution two (using the properties 2 and 3 mentioned ea... | 78 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Five people stand in a row, requiring that A does not stand at the head, B does not stand at the end, and C and D do not stand together. How many ways are there to arrange them? | Let $E=\{$ permutations of five people standing in a row $\}$,
$A_{1}=\{$ permutations where A stands at the head $\}$,
$A_{2}=\{$ permutations where B stands at the tail $\}$,
$A_{3}=\{$ permutations where C and D stand together $\}$.
Then the number of permutations required in the problem is
$$
\begin{align... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. Five people stand in a row. When they reline up, none of them stands in their original position. How many different formations are there? | Let the original formation of the five people be
$$
a_{1} a_{2} a_{3} a_{4} a_{5}
$$
$E=\{$ all permutations of five people standing in a line $\}$,
$A_{i}=\{a$ standing in his original position in the permutations $\}$
$$
\text { ( } i=1,2,3,4,5) \text {, }
$$
Then the number of desired permutations is
$$
\begin{arra... | 44 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
甽6. If $x+3y=10$, find the extremum of $x^{2}+3y^{2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6. If $x+3y=10$, find the extremum of $x^{2}+3y^{2}$. | Solve $x+3 y$
$=10$ as the line in Figure 6. Let $x^{2}+3 y^{2}=c$, then
$$
\frac{x^{2}}{(\sqrt{c})^{2}}+\frac{y^{2}}{\left(\sqrt{\frac{c}{3}}\right)^{2}}=1 \text {. }
$$
From Figure 6, we know that when the line $x+3 y=10$ is tangent to the ellipse $x^{2}+3 y^{2}=c$, the semi-axes of the ellipse are the shortest. We ... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. To complete a certain project, the number of days required for $A$ to work alone is $m$ times that of $B$ and $C$ working together, the number of days required for $B$ to work alone is $n$ times that of $A$ and $C$ working together, and the number of days required for $C$ to work alone is $p$ times that of $A$ and ... | Let $x$, $y$, $z$ be the number of days required for $A$, $B$, $C$ to complete the work individually, respectively, then we have
$$
\frac{m}{x}=\frac{1}{y}+\frac{1}{z}
$$
Therefore, $\frac{m+1}{x}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
Similarly, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{n+1}{y}=\frac{p+1}{z}$.
Thus, ... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
14. A number is the product of three prime factors. The squares of these three prime factors; the sum is 2331, the number 7560 is less than this number and is coprime with it, and the sum of its divisors (excluding 1 itself) is 10560. Find this number. | Let $c, b, c$ be the prime factors of a number, then
$$
a^{2}+b^{2}+c^{2}=2331.
$$
In addition, we know that the number of integers less than this number and coprime to it is
$$
\begin{array}{l}
a b c\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right). \\
\text { That is, }(a-1)(b-1)(c-1)=75... | 8987 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. Given $f\left(1-x^{2}\right)=\frac{1-x^{2}}{x^{2}}$. Find $f\left(\frac{1}{2}\right)$. | $\begin{array}{l}\text { Solve } f\left(1-x^{2}\right)=\frac{1-x^{2}}{1-\left(1-x^{2}\right)}, \\ \text { hence } f(x)=\frac{x}{1-x}, f\left(\frac{1}{2}\right)=1.\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Given an integer $n$, if $n$ plus 38 is a perfect square, prove that $n$ plus 38 is a perfect square, find $n$.
The text seems to have some typographical errors or unclear phrasing. Here is a corrected version for clarity:
Example 2. Given an integer $n$, if $n$ plus 38 is a perfect square, prove that $n$ ... | Let $n-51=x^{2},(x, y \in Z)$
$$
n+38=y^{2} .
$$
Subtracting the equations yields
$$
\begin{array}{l}
(y-x)(y+x)=89=1 \cdot 89 \text { ? } \\
=(-1)(-89) .
\end{array}
$$
Thus, $\left\{\begin{array}{l}y+x=1,89,-1,-89, \\ y-x=89,1,-89,-1 .\end{array}\right.$
Therefore, $\left\{\begin{array}{l}x=-44,41,41, \quad 41, \\ ... | 1987 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The page numbers of a book range from 1 to $n$. When all these page numbers are added together, one of the page numbers was mistakenly added twice. The incorrect sum obtained is 1987. What is the page number that was added twice? | Answer: 34 | 34 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9. Given the dihedral angle $C-A B=D=120^{\circ}$, $\angle C B A=60^{\circ}, \angle D A B=30^{\circ}, A B=\sqrt{37} \mathrm{~cm}$ (Figure 17). Find the distance $d$ between the skew lines $B C, A D$. | Let the angle between $BC$ and $AD$ be $\theta$. By the triple product formula,
$$
\begin{aligned}
\cos \theta= & \cos 60^{\circ} \cos 150^{\circ} \\
& +\sin 60^{\circ} \sin 150^{\circ} \cos 120^{\circ} \\
= & -\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{8}=-\frac{3 \sqrt{3}}{8} .
\end{aligned}
$$
Thus, $\sin \theta=\frac{\sqr... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Find a three-digit number such that the ratio of the number to the sum of its digits is minimized.
untranslated text remains unchanged:
例2. 求一个三位数, 便它与它的各位数字之和的比为最小.
However, for a proper translation, it should be:
Example 2. Find a three-digit number such that the ratio of the number to the sum of its d... | Solve: Arrange the ratios of all three-digit numbers to the sum of their respective digits in a $9 \times 100$ grid (Table 1).
We examine the numbers in each row and select the smaller ones. By the inequality $\frac{a}{b}>\frac{a+1}{b+1}(a>b)$, from left to right, every 10 numbers form a group, and the last number in ... | 159 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. For the equation $(1984 x)^{2}-1983 \times 1985 x-1$ $=0$, the larger root is $r$, and for the equation $x^{2}+1983 x-1984=0$, the smaller root is $s$. What is $r-s$? (1984 Beijing Junior High School Mathematics Competition Question) | Solve $\because 1984^{2}+(-1983) \times 1985$ $+(-1)=0$, according to property 1, we can get the roots of the first equation $x_{1}=1, \quad x_{2}=-\frac{1}{1984^{2}}$.
Also $\because 1+1983-1984=0$, we can get the roots of the second equation $x_{1}^{\prime}=1, x_{2}^{\prime}=-1984$.
$$
\text { Therefore } r-s=1985 \... | 1985 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Given that 1 is a root of the equation $a x^{2}+b x+c=0$, find the value of $\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}+\frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$. | Given that $1$ is a root of the equation $a x^{2}+b x+c=0$,
$$
\begin{array}{c}
\therefore a+b+c=0 . \\
\text { Also, } a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c) \\
.\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
From (1) and (2), we can deduce
$$
\begin{array}{c}
a^{3}+b^{3}+c^{3}=3 a b c . \\
\text { Therefore,... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the sets $M=\{x, x y, \lg (x y)\}$ and $N=\{0,|x|, y\}$, and $M=N$. Then, $\left(x+\frac{1}{y}\right)+\left(x^{2}+\frac{1}{y^{2}}\right)+\left(x^{3}+\right.$ $\left.\frac{1}{y^{3}}\right)+\cdots+\left(x^{2001}+\frac{1}{y^{2001}}\right)$ is equal to | Given $\because M=N$,
$\therefore 0 \in M$, and $x, y$ cannot both be 0, then
$\lg (x y)=0, x y=1$.
Thus, $1 \in N$. If $y=1$, then $x=1$, which contradicts the distinctness of the elements in the set. Therefore, $|x|=1$, and $x=-1$, so $y=-1$.
Therefore, $x+\frac{1}{y}=-2 \cdot x^{2}+\frac{1}{y^{2}}=2$, $x^{3}+\frac{... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, given $x, y \in N$, find the largest $y$ value such that there exists a unique $x$ value satisfying the following inequality:
$$
\frac{9}{17}<\frac{x}{x+y}<\frac{8}{15} \text {. }
$$ | Solve the original system of inequalities $\Longleftrightarrow\left\{\begin{array}{l}8 x-9 y>0, \\ 7 x-8 y>0\end{array}\right.$ $\longleftrightarrow \frac{9 y}{8}<x < \frac{8 y}{7}$.
For the interval $\left(\frac{9 y}{\circ}, \frac{8 y}{7}\right)$ to contain a unique integer, the necessary and sufficient condition is e... | 112 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $\mathrm{f}$ be a function satisfying the following conditions:
(i) If $x>y$, and $f(y)-y \geqslant v \geqslant f(x)-x$, then for some number $z$ between $x$ and $y$, $f(z)=v+z$;
(ii) The equation $\mathrm{f}(\mathrm{x})=0$ has at least one solution, and among the solutions, there is one that is not less than al... | 1. Solution
(1) From (ii), we know that $f(x)=0$ has at least one solution. Let's assume $u$ is a solution of $f(x)$.
(2) Assume $u>0$.
(3) From (i) and (iii), we know that there exists a number $z$ between 0 and $u$ such that $f(z)=z$.
(4) From (v), we get:
$$
\begin{array}{l}
0=f(z) \cdot f(u)=f(z \cdot f(u)+u \cdot... | 1988 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
List 4. (IMO-4-1) Find the smallest natural number $n$ whose decimal representation ends with 6, such that when the last digit 6 is deleted and written at the beginning of the remaining digits, it becomes four times $n$. | Solve for $n$ whose decimal representation ends with 6, meaning that it leaves a remainder of 6 when divided by 10. By the division algorithm, we have
$$
n=10 m+6 \text {. }
$$
where $m$ is a positive integer, and we assume it is an $l$-digit number. According to the problem,
$$
4(10 m+6)=6 \times 10^{l}+m .
$$
Thus,... | 153815 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
12. (IMO-20-1) The last three digits of the numbers $1978^{n}$ and $1978^{m}$ are equal. Find the positive integers $n$ and $m$ such that $n+m$ is minimized, where $n>m \geqslant 1$. | From the given, we have
$$
\begin{array}{l}
1978^{m}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right) \\
=2^{m} \times 989^{m} \times\left(1978^{m}-1\right) \\
\equiv 0\left(\bmod 10^{3}\right) .
\end{array}
$$
Since
$$
10^{3}=2^{3} \times 5^{3},
$$
and $989^{m}$ and $1978^{\mathrm{m}}-1$ are both odd, it follows that $m \... | 106 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. In the corner of the room, there are several identical cubes piled up. How many of the cubes are not visible? | In the second layer, there is 1 that is invisible,
in the third layer, there are $(1+2)$ that are invisible:
in the fourth layer, there are $(1+2+3)$ that are invisible, in the fifth layer, there are $(1+2+3+4)$ that are invisible, so the total number of invisible ones is $1+3+6+10=20$. | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. For a regular $n$-sided polygon, construct $n$ squares outside the polygon, each sharing one side with the polygon. It is known that the $2n$ outer vertices of these $n$ squares form a regular $2n$-sided polygon. For what value of $n$ is this possible? | Solve As shown in Figure 1, $P, Q, R$ are three vertices of a regular $n$-sided polygon; $M, N, K, L$ are four vertices of a $2n$-sided polygon. According to the given conditions, we have $MN = NQ, QK = KL, NK = KL = MN$, which means $\triangle NQK$ is an equilateral triangle. Therefore, $\angle NQK = 60^{\circ}$. From... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. On the blackboard are the numbers $1,2, \cdots, 1987$. Perform the following transformation: erase some of the numbers on the blackboard and add the remainder when the sum of the erased numbers is divided by 7. After several such transformations, only two numbers remain on the blackboard, one of which is 987. Find t... | Note the fact that with each transformation: the sum of all numbers on the blackboard, when divided by 7, leaves the same remainder. $1+2+\cdots+1987=1987 \cdot 7 \cdot 142$, which indicates that the original sum on the blackboard is divisible by 7, so the sum of the two final numbers can also be divided by 7. Among th... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Find the smallest natural number such that when the last digit of this number is moved to the first position, the resulting number is 5 times the original number. | Let the required number be $a_{1} a_{2} \cdots a_{n-1} a_{n}$ - According to the given conditions, we have
$$
\overline{a_{n} a_{1} a_{2} \cdots a_{n-1}}=5 \cdot \overline{a_{1} a_{2} \cdots a_{n-1} a_{n}} \text {. }
$$
Then
$$
\begin{array}{l}
a_{n} \cdot 10^{n-1}+\overline{a_{1} a_{2} \cdots a_{n-1}} \\
=5\left(\ove... | 142857 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. At least how many circles with a radius of 1 are needed to cover a circle with a radius of 2. | Let $O$ be the center of a circle with radius 2, and let $A B C D E F$ be a regular hexagon inscribed in the circle (Figure 8). The side length of the hexagon is 2. Construct six circles with the sides of the hexagon as diameters. The intersection points of these six circles, other than $A, B, C, D, E, F$, are denoted ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 9. Primary School One and Primary School Two have the same number of students participating in the Golden Cup Competition. The schools use cars to transport the students to the examination site. Primary School One uses cars that can seat 15 people each; Primary School Two uses cars that can seat 13 people each.... | Let's assume that at the beginning, both schools had $x$ participants. According to the problem, the first school sent $\frac{x}{15}=a \cdots 0$ ($a$ cars),
the second school sent $\frac{x}{13}=a \cdots 12(a+1$ cars $)$; the second time, the first school sent $\frac{x+1}{15}=a \cdots 1(a+1$ cars $)$,
the second schoo... | 184 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In a hand clapping competition, the winner of each match gets 2 points, the loser gets 0 points; if it's a draw, each player gets 1 point. Now the scores are: $1979$, $1980$, $198$, $1985$. After verification, how many players participated in the competition? | Since $n(n-1)$ is the product of two consecutive natural numbers, it must be even, so 1979 and 1985 are impossible. And the last digit of $n(n-1)$, being the product of two consecutive natural numbers, can only be 0, 2, or 6, so 1984 is also impossible. The only possibility left is 1980. For $n(n-1)=1980$, solving give... | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. If $a>1, b$ is a positive rational number, $a^{b}+a^{-0}$ $=2 \sqrt{2}$, find the value of $a^{b}-a^{-b}$. | Consider the following solution:
Let $a^{b}=x$, then $x+\frac{1}{x}=2 \sqrt{2}$.
Transform it into $x^{2}-2 \sqrt{2} x+1=0$.
Solving yields $x_{1}=\sqrt{2}+1, x_{2}=\sqrt{2}-1$.
$$
\text { When } \begin{aligned}
x & =\sqrt{2}+1, \\
& a^{b}-a^{-b} \\
& =\sqrt{2}+1-\frac{1}{\sqrt{2}+1} \\
& =\sqrt{2}+1-(\sqrt{2}-1) \\
& ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. If $\lg ^{2} x \lg 10 x<0$, find the value of $\frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}}$. | From the known condition $\lg ^{2} x \lg 10 x<0$, it is not difficult to conclude that $\lg x \neq 0$ and $\lg 10 x<0$, which also implicitly indicates the existence of $\lg x$.
$$
\text { Hence } \begin{aligned}
& \frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}} \\
= & \frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+2 \lg x+\lg ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. (1MO-27 Preliminary Question) Let $A, B, C$ be three points on the edge of a circular pool, with $B$ due west of $C$, and $A B C$ forming an equilateral triangle with side lengths of 86 meters. A swimmer starts from $A$ and swims directly to $B$. After swimming $x$ meters, he reaches point $\boldsymbol{E}$, ... | Given the figure, since $\triangle AEF$ is an equilateral triangle, we have
$$
AF = AE = x.
$$
By symmetry, $FG = DE = y$. Also, since
$$
AE \cdot EB = DE \cdot EG,
$$
we have
$$
\begin{aligned}
& x(86 - x) \\
= & y(x + y).
\end{aligned}
$$
Notice that if $x$ is odd, then $x(86 - x)$ is also odd, while $y(x + y)$ is... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 14. Find the number of integer solutions $(x, y, z)$ that satisfy $0<x<y<z$, and $\sqrt{1984}=\sqrt{ } x+\sqrt{ } y+\sqrt{ } z$. | $$
\text { Stele } \begin{aligned}
& \sqrt{1984}=8 \sqrt{31} \\
& =\sqrt{31}+2 \sqrt{31}+5 \sqrt{31} \\
& =\sqrt{31}+\sqrt{2} \times 31+\sqrt{5} \times 31 \\
& =\sqrt{31}+3 \sqrt{31}+4 \sqrt{31} \\
& =\sqrt{31}+\sqrt{3} \times 31+\sqrt{2} \times 1 .
\end{aligned}
$$
There are two kinds of correct statements in total.
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
oi1. Find the number of positive integer solutions for the equation
$$
x+y+z=20
$$ | Solving for $z$: $=1$, substituting into (1), we get $x+y=19$. For this boundary, when $x$ takes the values $1, 2, \cdots, 18$ $=2$, there are 17 sets of positive integer solutions. Continuing this way, when $z=18$, there is only 1 set of positive integer solutions. Therefore, the total number of positive integer solut... | 171 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4. Find the indefinite solution of the equation
$$
x+y+z=20
$$ | Find the number of positive integer solutions for (1). For this, we have 20 a's and 19 spaces (denoted by $s$) arranged in one line as follows:
asasasasasasasasasasasasasasasasas asisa.
Now, choose any 2 out of the 19 s's, and remove the rest of the $s$'s, for example:
caaaasaaaaaasaaauaaaaa. This results in three seg... | 231 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
List 7. Let $x+y+z+w=20$. (8) Try to find (i) the number of positive integer solutions to equation (8); the number of groups;
(iii) the number of positive integer solutions to (8) satisfying $x \geq 7, y=8$;
(iv) the number of positive integer solutions to (8) satisfying $x \leqslant 6$;
(v) the number of positive inte... | (i) From the last statement of Example 4, the number of positive integer solutions of equation (8) is $C_{2}^{4-1}=C_{1}^{3}=969$.
(ii) Let $x^{\prime}=x-6$, then equation (8) becomes $x^{\prime}+y+z+w=14$.
Now, we find the number of positive integer solutions of (9) under the condition that $x \geqslant 7$ for equati... | 483 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. The boss hands the letters to be printed to the secretary, one at a time, and places each on top of the pile. The secretary, whenever free, takes the top letter to type. One day, there are 9 letters to be typed, and the boss hands them over in the order of the first, second, $\cdots$, to the ninth. During lunch, th... | 15. ( 704$)$
We consider two cases.
The first case: The ninth letter is finally typed by the secretary before noon.
This may be the order of the subset of the set $T=\{1,2$, ..., 6, 7, 9\} of possible subsets. In fact, the secretary can type a letter as soon as it arrives if its number is not in the subset, and not t... | 704 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Go to the room, take a card at random, and ask which number is most likely to be the unit digit of the number drawn by Jack and Jill.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 12
I
A
12. The list below, the numbers in the table represent the units digit of the sum of two numbers.
\begin{tabular}{ccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline 2 & 3 & 4 & 3 & 6 & 7 & 8 & 0 & 3 \\
3 & 4 & 0 & 3 & 7 & 3 & 9 & 1 & 1 \\
4 & 5 & 6 & 7 & 3 & 9 & 0 & 1 & 2 \\
5 & 6 & 1 & 3 & 3 & 0 & 1 & 2 & 3... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. A commercial lock with 10 buttons, which can be opened by pressing the correct five numbers, regardless of the order. The figure below is an example using $\{1,2,3,6,9\}$ as its combination. If these locks are reprogrammed to allow combinations of one to nine digits, how many additional combinations (i.e., not using... | 1. ( 770$)$
There are $2^{10}$ ways to choose from ten buttons. From this, we need to subtract the case where all 10 digits are selected, the case where no digits are selected, and the case where exactly five digits are selected. Therefore, the total number is
$$
2^{10}-1-1-C_{10}^{5}=1024-2-252=770 .
$$ | 770 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3 . If $\mathrm{iog}_{2}\left(\log _{8} x\right)=\log _{8}\left(\log _{2} x\right)$, find $\left(\log _{2} x\right)^{2}$. | $$
\begin{array}{l}
3 .(27) \\
\because \log _{8} x=\frac{1}{\log _{x} 3}=\frac{1}{3 \log _{x} 2}=\frac{1}{3} \log _{2} x, \\
\log _{8}\left(\log _{2} x\right)=\frac{1}{3} \log _{2}\left(\log _{2} x\right),
\end{array}
$$
Let $y=\log _{2} x$, then the equation in the problem is equivalent to
$$
\log _{2} \frac{y}{3}=\... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in the figure, in a 5 $\times$ 5 square, there are 21 empty cells. It is known that positive integers can be filled in the empty cells so that each row and each column forms an arithmetic sequence. What number must be filled in the cell marked with a star?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
6. As shown in... | $6 .(142)$
Let $a$ and $b \%$ be the numbers that fill in the blanks. From $b-2a$, and the common difference of the first row being $2b-a-74$, we get
$$
2a+4(b-2a)=186
$$
and $a+2(2b-a-74)=103$.
Solving these equations, we can find the numbers that fill in the blanks in the figure. Therefore, the number that should be... | 142 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. In $\triangle \mathrm{ABC}$, $\operatorname{tg} \angle \mathrm{CAB}=\frac{22}{7}$, a line is drawn from $\mathrm{A}$ to $\mathrm{BC}$ dividing $BC$ into segments of lengths 3 and 17. What is the area of $\triangle \mathrm{ABC}$? | 7. (110)
Let $x$ be the length of the altitude, intersecting $B C$ at $D$. Then, from $\angle B A C=\angle B A D+\angle D A C$, we get $\operatorname{arctg} \frac{22}{7}=\operatorname{arctg} \frac{3}{x}+\operatorname{arctg} \frac{17}{x}$.
Taking the tangent of both sides of (1) and using the $\operatorname{tg}(\alpha+... | 110 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. The function $f$ defined on ordered pairs of positive integers satisfies the following three properties:
$f(x, x)=x, f(x, y)=f(y, x)$ and $(x+$ $y) f(x, y)=y f(x, x+y)$. Try to compute $f(14$, 52 ). | 8. (364)
By property three, we get $f(x, x+y)=\frac{x+y}{y} f(x, y)$. We will repeatedly use this property, applying property two when the first number of the ordered pair is greater than the second, and finally applying property one, to get
$$
\begin{aligned}
& f(14,52)=f(14,14+38)=\frac{52}{38} f(14,38) \\
= & \frac... | 364 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Find the smallest positive integer $n$, such that the last three digits of its cube are 888. | 9. (192)
If a positive integer ends with 8, then this number itself must end with 2. That is, it can be written in the form $10k + 2$. Thus,
$$
n^{3} = (10k + 2)^{3} = 1000k^{3} + 600k^{2} + 120k + 8.
$$
The term $120k$ determines the tens digit of $n^{3}$, which is 8. Therefore, $12k$ must end with 8, meaning $k$ is... | 192 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. The surface of a convex polyhedron is composed of 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex, one square, one octagon, and one hexagon meet. How many line segments connecting the vertices of the polyhedron lie inside the polyhedron, and not on the faces or edges of the polyhedron? | 10. $(840)$
Let $a, b, c, d$ be the number of vertices, edges, face diagonals, and body diagonals, respectively.
Since at each vertex there is one of each type of regular polygon, and the four vertices of each square must be four different vertices of the polyhedron, we have
$c=4 \times$ the number of squares $=4 \ti... | 840 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $w_{1}, w_{4}, \cdots, m_{n}$ be complex numbers. If a straight line $l$ passes through points (complex numbers) $z_{1}: z_{2}, \cdots, z_{n}$, such that $\sum_{k=1}^{n}\left(z_{k}-w_{k}\right)=0$, then $l$ is called the “average line” of $w_{1}, w_{i}$, $\cdots, ~ w w_{\mathrm{n}}$.
For $w_{1}=32+170 i, w_{2}... | 11. (163)
Let $y=m x+b$ be the "average line" of the complex numbers $w_{\mathrm{K}}=u_{\mathrm{K}}+i v_{\mathrm{K}}$, where $u_{\mathrm{K}}, v_{\mathrm{K}} \in R, k=1,2, \cdots, n$. Assume that on the line $y=m x+b$, we select complex numbers $z_{\mathrm{K}}=x_{\mathrm{K}}+i y_{\mathrm{K}}\left(x_{\mathrm{K}}, y_{\ma... | 163 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Let $P$ be an interior point of $\triangle A B C$, and extend $A P, B P, C P$ to intersect the opposite sides. In the figure, $a, b, c$, $d$ are the lengths of the respective line segments. Given that $a+b+c=43$, $d=3$, find $a b c=$ ? | 12. $(441)$
$$
\begin{array}{l}
\because \frac{S_{\triangle B P C}}{S_{\triangle B A C}}=\frac{d}{d+a}, \frac{S_{\triangle C P A}}{S_{\triangle C B A}}=\frac{d}{d+b}, \\
\frac{S_{\triangle A P B}}{S_{\triangle A C B}}=\frac{d}{d+c},
\end{array}
$$
and $S_{\triangle B P C}+S_{\triangle C P A}+S_{\triangle A P B}=S_{\t... | 441 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}+b x^{10}+1$, try to find the value of $a$.
| 13. ( 987 )
The roots of $x^{2}-x-1=0$ are $p=\frac{1}{2}(1+\sqrt{5})$, $q=\frac{1}{2}(1-\sqrt{5})$. They must be the roots of $\boldsymbol{a} x^{17}$
$+b x^{18}+1=0$.
$$
\therefore a p^{17}+b p^{16}=-1, a q^{17}+b q^{18}=-1 \text {. }
$$
Multiplying the first equation by $q^{16}$, and the second equation by $p^{16}$,... | 987 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. $C$ is the graph of $x y=1$, and the symmetric image of $C$ about the line $y=2 x$ is $C^{\prime}$. It is known that $C^{\prime}$ can be expressed as
$$
12 x^{2}+b x y+c y^{2}+d=0
$$
Use this form to find the value of $b c$ | 14. $(84)$ points), connect $P^{\prime} P^{\prime}$ (as shown in the figure).
$\because F P^{\prime \prime} \perp$ line $y=2 x$,
$\therefore$ the slope of $P P^{\prime}$ is $-\frac{1}{2}$,
i.e., $-\frac{y-v}{x-u}=-\frac{1}{2}$.
Also, $\because$ the midpoint of $P P^{\prime}$ is on $y=2 x$,
$\therefore \quad \frac{y+v}... | 84 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. $N$ is the set of positive integers. Define the function $f$ on $N$ as follows:
$f(1)=1, f(3)=3$, and for $n \in N$ we have $f(2 n)=f(n)$,
$f(4 n+1)=2 f(2 n+1)-f(n)$,
$f(4 n+3)=3 f(2 n+1)-2 f(n)$.
Question: How many $n \in N$, and $n \leqslant 1988$ such that $f(n) = n$? | 3. According to these formulas, we can find
The pattern shown in this table seems to be
$$
\begin{array}{l}
f\left(2^{k}\right)=1, f\left(2^{k}-1\right)=2^{k}-1, \\
f\left(2^{k}+1\right)=2^{k}+1 .
\end{array}
$$
This suggests that we should consider the "binary" expansion of natural numbers. Our conjecture is:
$f(n)$... | 92 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
54. (Mongolia 4) Find the smallest natural number $\boldsymbol{n}$ such that if the set $\{1,2, \cdots, n\}$ is arbitrarily divided into two non-intersecting subsets, then one of the subsets contains 3 different numbers, where the sum of two of them equals the third. | 20. (Mongolia 4)
Solution: The required minimum value is 96.
Let $A_{\mathrm{s}}=\{1,2, \cdots, n\}$ be divided into two subsets $B_{\mathrm{n}}$ and $C_{\mathrm{s}}$, and neither $B_{\mathrm{s}}$ nor $C_{\mathrm{s}}$ contains three distinct numbers where the product of two equals the third. If $n \geqslant 96$, then ... | 96 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. In a "Battleship" game on a grid with $7 \times 7$ cells, what is the minimum number of shots needed to guarantee hitting a four-deck ship? It is known that the ship (1) has shape $A$ (Figure 2); (2) consists of four adjacent cells, but is not a square. | Solution: (1) Since 12 ships of shape $A$ can be placed in a $7 \times 7$ grid such that no two ships share a common cell, the number of shots required is at least 12. From figures $3, a, b$, it is shown that 12 shots are sufficient.
(2) 20 steps.
As shown in Figure 4, 20 shots can complete the task. Suppose we shoot ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Place one more of these shapes inside the square.
Keep the original text's line breaks and format, and output the translation result directly.
Note: The second sentence is a directive for the translation task and should not be included in the translation output. Here is the translation:
Place one more of these shap... | Solution: For each $2 \times 2$ square, at least two cells should be covered. An $8 \times 8$ square can be divided into 16 non-overlapping $2 \times 2$ squares. Therefore, at least 11 figures should be covered. With 11 figures, it is feasible. | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Uncle Qiel Yuemuer every night selects 9 or 10 warriors from 33 warriors to be on duty according to his own wishes. At least how many days are needed to ensure that each warrior has the same number of duty shifts? | The minimum number of days is 7 days, during which each warrior is on duty 2 times.
Let the required minimum number of days be $N$, and each warrior is on duty $n$ times within these $N$ days. We number the warriors sequentially as $1,2, \cdots, 33$, and introduce the number $a_{1 \times}(i=1,2, \cdots, N ; k=1,2, \cd... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. A number plus 79 becomes a perfect square, and when this number plus 204 is added, it becomes another perfect square. Then this number is $\qquad$ . | $$
\begin{array}{c}
204=y^{2}, \text { where } x \text { and } y \text { are non-negative integers. Thus, } \\
y^{2}-x^{2}=125=5^{3}, \text { i.e., } (y+x)(y-x)=5^{3}. \\
\therefore \quad\left\{\begin{array} { l }
{ y + x = 125, } \\
{ y - x = 1; }
\end{array} \text { or } \left\{\begin{array} { l }
{ y + x = 25, } \... | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $E(n)$ denote the greatest integer $k$ that is a divisor of the product $1^{1}, 2^{2}, 3^{3} \cdots n^{n}$. Then $E(150)=$ | $9.1,2, \cdots, 150$ contains 30 multiples of 5, each having one factor of 5 removed, along with their respective exponents, the exponent of 5 is
$$
\begin{array}{l}
5+5 \cdot 2+5 \cdot 3+\cdots+5 \cdot 30=5(1+2+3 \\
+\cdots+30)=2325 .
\end{array}
$$
1. $2, \cdots, 150$ contains 6 multiples of $5^{2}$, each having one ... | 2975 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8 . All three-digit numbers $a b c$ that satisfy $a b c=(a+b+c)^{3}$ are $\qquad$ | 8. $\because 100 \leqslant a b c=(a+b+c)^{3} \leqslant 999$,
$\therefore 5<a+b+c$ S 9 . Directly calculating $5^{3}=125$, $6^{3}=216,7^{3}=343,8^{3}=512,9^{3}=729$, we can see that only the three-digit number 512 meets the conditions of the problem. | 512 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. In a regular $\triangle A B C$, $D$ and $E$ are the midpoints of sides $A B$ and $A C$ respectively. Fold $\triangle A B C$ along $D E$ to form a dihedral angle $A-D E-C B=60^{\circ}$. If $B C=10 \sqrt{13}$, then the distance between the skew lines $A E$ and $B D$ is $\qquad$ | 10. Let $M$ be the midpoint of $BC$, and the median $AM$ of $\triangle ABC$ intersects $DE$ at $N$. After folding into a dihedral angle, $\angle ANM$ is the plane angle of the dihedral angle $A-DE-CB$, i.e., $\angle ANM = 60^{\circ}$.
Connect $EM, AM, DM. \because EM \parallel DB$, $\therefore$ the distance $d$ betwee... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. Divisible by 3, and the digits of each number are limited to $1, 2, 3$ (1, 2, 3 do not have to be all used) all natural numbers less than 200000 are $\qquad$
$\qquad$ | 12. The only one-digit number that can be divided by 3 and is composed of only the digits 1, 2, 3 is 3.
The two-digit numbers that can be divided by 3 and are composed of only the digits $1, 2, 3$ can be obtained as follows: arbitrarily write one of $1, 2, 3$ in the tens place, then configure the units digit based on ... | 202 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 11. Let $\mathrm{a}$ be a real number, find the minimum value of the quadratic function
$$
y=x^{2}-4 a x+5 a^{2}-3 a
$$
denoted as $\mathrm{m}$.
When $a$ varies in $0 \leqslant a^{2}-4 a-2 \leqslant 10$, find the maximum value of $m$. | $$
\text{Solve} \begin{aligned}
y & = x^{2} - 4ax + 5a^{2} - 3a \\
& = (x - 2a)^{2} + a^{2} - 3a
\end{aligned}
$$
The above expression achieves its minimum value when $x = 2a$, so
$$
m = a^{2} - 3a = \left(a - \frac{3}{2}\right)^{2} - \frac{9}{4}.
$$
Furthermore, $0 \leqslant a^{2} - 4a - 2 \leqslant 10$ implies
$$
0... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. (1MO-23-1)
The function $f(n)$ is defined for all positive integers $n$, taking non-negative integer values. For all positive integers $m, n$, $f(m+n)-f(m)-f(n)=0$ or 1;
and $f(2)=0, f(3)>0, f(9999)=3333$. Find $f(1982)$. | Solve $0=f(2) \geqslant 2 f(1)$, we get $f(1)=0$.
From $f(3)-f(2)-f(1)=0$ or 1, we get $0 \leqslant f(3) \leqslant 1$.
Given $f(3)>0$, we get $f(3)=1$.
First, we prove that for $k3333,
\end{aligned}
$$
Contradicting the given condition, so it must be that $f(3k)=k$.
It is clear that $660 \leqslant \hat{j}(1332) \leqs... | 660 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7. (IMO-20-3)
Let $f, g: Z^{+} \rightarrow Z^{+}$ be strictly increasing functions, and $f\left(Z^{+}\right) \cup g\left(Z^{+}\right)=Z^{+}, f\left(Z^{+}\right) \cap g\left(Z^{+}\right)=\phi$, $g(n)=f[f(n)]+1$. Find $f(240)$. Here $Z^{+}$ is the set of positive integers. | Let
$$
F=\{f(n)\}, G=\{g(n)\} .
$$
If $F \cup G=Z^{+}, F \cap G=\phi$, then $F$ and $G$ are complementary.
We prove $f(n)=(\sqrt{5}+1) \cdots$ by induction. For: $g(n)=f(f(n))+1>1$. So we can set $f(1)=1=\left(\frac{\sqrt{5}+1}{2}\right)$, and $g(1)$ $=2$.
If $v_{n}<m$, then $f(n)=\left\{\frac{\sqrt{5}+1}{2}\right\}... | 388 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For example, solving the equation $\frac{x^{2}-7}{6}=\sqrt{6 x+7}$. This is the result of transitioning from concrete to abstract thinking.
From a geometric perspective, this is equivalent to finding the x-coordinates of the intersection points of the line $y=\frac{x^{2}-7}{6}$ and the curve $y=\sqrt{6 x+7}$. These tw... | Solve $\sqrt{6 x+7}=x$ to get
$$
\begin{array}{l}
x^{2}-6 x-7=0, \\
\therefore x_{1}=-1 \text { (discard), } x_{2}=7 .
\end{array}
$$
Upon verification, the root of the original equation is $x=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. There is a four-digit number. It is known that its tens digit minus 1 equals the units digit, and its units digit plus 2 equals the hundreds digit. The sum of this four-digit number and the number formed by reversing the order of its four digits equals 9878. Try to find this four-digit number. | Solution: Let the required four-digit number be $a b c d$. According to the problem, we have $\overline{a b c \bar{d}}+\overline{d c b a}=9878$,
which means $\left(10^{3} \cdot a+10^{2} \cdot b+10 \cdot c+d\right)+\left(10^{3} \cdot d\right.$ $\left.+10^{2} \cdot c+10 \cdot b+a\right)=9878$, or $10^{3} \cdot(a+d)+10^{... | 1987 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Rearrange the digits of a three-digit number to form the largest possible three-digit number, and subtract the smallest digit from it, which is exactly equal to the original number. Find these three digits. (Hua Luo Geng Math Contest 1988 Junior High School Level) | Let the three-digit number be $a, b, c$. If $abc$ is the largest, and $cb_{2}$ is the smallest, and
$$
\begin{array}{l}
\overline{a b c}-cba \\
=\left(10^{2} a+10 b+c\right)-\left(10^{2} \cdot c+10 b+a\right) \\
=99(a-c),
\end{array}
$$
i.e., the required three-digit number is a multiple of 99. Among such three-digit ... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5, find a four-digit number that can be divided by 11, and the quotient obtained is 10 times the sum of the digits of this four-digit number. | Let $\overline{a b c d}=110(a+b+c+d)$, then we have
$$
890 a=10 b+100 c+109 d \text {. }
$$
Also, since 11 divides $\overline{a b c d}$, we have $a-b+c-d= \pm 11$ or $a-b+c-d=0$.
If $a-b+c-d=0$,
then from $\left\{\begin{array}{l}a+c=b+d, \\ 890 a=10 b+100 c+109 d\end{array}\right.$
(2) $-10 \times(1)$ we get
$880 a=1... | 1980 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6. In the six-digit number $31 x y 13$ !, $x 、 y$ are both greater than 5, and this number is divisible by 13, find the four-digit number $\overline{1 x y 9}$. | $$
\begin{array}{l}
\text { Solution: Let } n=31 x y 13 \\
=3 \cdot 10^{5}+10^{4}+x \cdot 10^{3}+y \cdot 10^{2} \\
+13 \\
=310013+10^{3} x+10^{2} y, \\
\end{array}
$$
where $14,6 \leqslant x, y \leqslant 9$.
Since the remainders of $310013,10^{3}, 10^{2}$ when divided by 13 are $2,-1,-4$ respectively, we can set $3100... | 1989 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1. The nine-digit number in the form of $1234 \times \times \times \times \times$ composed of $1,2,3,4,5,6,7,8,9$, where the ten-thousands place is not 5, the thousands place is not 6, the hundreds place is not 7, the tens place is not 8, and the units place is not 9. How many such nine-digit numbers are there? | Solution: Obviously, this is a problem of five letters being placed in the wrong envelopes. And
$$
\begin{aligned}
\overline{5} & =5!-C_{5}^{1} 4!+C_{5}^{2} 3!-C_{5}^{3} 2! \\
& +C_{5}^{4} 1!-C_{5}^{5} 0! \\
& =120-120+60-20+5-1 \\
& =44 .
\end{aligned}
$$
The number of derangements it is asking for is 44. | 44 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Prove: $\quad(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$ has a minimum value of 8 on $0<u<\sqrt{2}, v>0$. | $(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$ can be seen as the square of the distance between point $P_{1}(u, \sqrt{2-u^{2}})$ and $P_{2}\left(v, \frac{9}{v}\right)$.
$P_{1}$ satisfies the equation of the circle:
$$
x^{2}+y^{2}=2 \text{. }
$$
$P_{2}$ satisfies the equation of the hyperbola:
$$
x y=9, x>0 \t... | 8 | Algebra | proof | Yes | Yes | cn_contest | false |
2. Let $\alpha, \beta$ be acute angles. When
$$
-\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}
$$
takes the minimum value, the value of $\operatorname{tg}^{2} \alpha+\operatorname{tg}^{2} \beta$ is | To find the minimum value of the original expression, we need
$$
\sin ^{2} \beta \cos ^{2} \beta=\frac{1}{4} \sin ^{2} 2 \beta
$$
to take the maximum value. Since $\beta$ is an acute angle, we know $\beta=\frac{\pi}{4}$.
When $\beta=\frac{\pi}{4}$,
$$
\begin{aligned}
\text { the original expression } & \geqslant \frac... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let [x] denote the greatest integer not exceeding $x$, then the sum of the squares of all real roots of the equation $x^{2}-5[x]+4=0$ is
保留源文本的换行和格式，直接输出翻译结果。 | Solve: The intersection points of the graphs of the functions $y=x^{2}$ and $y=5[x]-4$ (see the figure below) are the solutions to the original equation.
From $5[x]=x^{2}+4>0$, we know that $[x]>0$.
Therefore, we only
need to consider the case where $x \geqslant 1$.
Since $x \geqslant[x]$,
substituting into t... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. A bag of peanuts has a total of 1988 peanuts. On the first day, a monkey takes one peanut. Starting from the second day, the number of peanuts taken each day is the sum of all the peanuts taken in previous days. If on a certain day the number of peanuts left in the bag is less than the total number of peanuts alread... | Let's assume the monkey used $k$ days in the first round, meaning that on the $k+1$-th day, it started again by taking one peanut. Therefore, the total number of peanuts taken in the first $k$ days is
$$
1+1+2+2^{2}+2^{3}+\cdots+2^{k-1}=?^{k-1}
$$
peanuts, and $1988-2^{k-1}<2-1$. Also, $1988-2^{k-1}-2^{k-1}<2^{t-1}, t... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. The general term of the sequence is $a_{\mathrm{n}}=b[\sqrt{n+c}]+d$, and the terms are calculated successively as
$$
1,3,3,3,5,5,5,5,5, \cdots \text {. }
$$
where each positive odd number $m$ appears exactly $m$ times consecutively. The above $b, c, d$ are undetermined integers. Then, the value of $b+c+d$ is $\qqu... | Solve: First determine $b$.
Since $a_{n}$ is odd, and $a_{n}+1 \geqslant a_{n}$, we know that $a_{n}+1-a_{n}$ $\in\{0,2\}$,
i.e., $b[\sqrt{n+1+c}]-b[\sqrt{n+c}]$ is 0 or 2.
For any natural number $n$, it always holds that
$[\sqrt{n+1+c}]-[\sqrt{n+c}] \in\{0,1\}$.
Clearly, $b \neq 0$.
When $-[\sqrt{n+c}])=2$, by (1), it... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. $f(n)$ is a function defined on the set of natural numbers, when $p$ is a prime number, $f(p)=1$, and for any natural numbers $r$, $s$, we have
$$
f(r s)=r f(s)+s f(r) .
$$
Then, the sum of all $n$ that satisfy the condition
$$
f(n)=n, 1 \leqslant n \leqslant 10^{4}
$$
is. | Given the problem, for any natural numbers $r, s$, we have
$$
\frac{f(r s)}{r s}=\frac{f(r)}{r}+\frac{f(s)}{s} \text {. }
$$
Thus,
$$
\begin{array}{l}
\frac{f\left(p^{m}\right)}{p^{m}}=\frac{f(p)}{p}+\frac{f(p)}{p}+\cdots+\frac{f(p)}{p} \\
\quad=m \cdot \frac{f(p)}{p} .
\end{array}
$$
When $p$ is a prime number, we g... | 3156 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Tank A contains 4 liters of liquid A, Tank B contains 2 kilograms of liquid B, and Tank C contains 2 kilograms of liquid C. These liquids can all be mixed. First, 1 liter of liquid from Tank A is poured into Tank B, and then 1 liter of the mixed liquid from Tank B is poured into Tank C. Finally, 1 liter of the mixe... | Let $a_{n}, b_{n}, c_{n}$ be the amounts of liquid A in containers 甲, 乙, and 丙, respectively, after $n$ mixings. Then,
$$
a_{n}+b_{n}+c_{n}=4 \text{. }
$$
By symmetry, we have
$$
b_{n}=c_{n} \text{. }
$$
Now, let's examine the amount of liquid A in container 甲 after the $(n+1)$-th mixing. According to the problem,
$$... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3. Write the numbers $1, 2, 3, \cdots$, 1986, 1987 on the blackboard. At each step, determine some numbers from those written and erase them, replacing 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 ... | Solution: Clearly, at each step, the sum of all the numbers written down modulo 7 is preserved.
Let the remaining number be $x$, then $x+987$ is congruent to $1+2+\cdots+1987$ modulo 7.
Since $1+2+\cdots+1987=1987 \times 7 \times 142$ is divisible by 7, the remainder is 0, so $x+987$ is also divisible by 7. Since 987... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 ・At
7:00 AM, depart from port $A$,
take a motorboat at a constant speed of $v_{1}$ kilometers/hour (4
$\leqslant v_{1} \leqslant 20$) to
head to port $B$ 50 kilometers
away from port $A$, then immediately switch to a steamship at a constant speed of $v_{2}$ kilometers/hour $\left(30 \leqslant v_{2} \leqslant ... | (1) From the given conditions, we have
\[
\left\{
\begin{array}{l}
x=\frac{300}{v_{2}}, \quad 30 \leqslant v_{2} \leqslant 100, \\
y=\frac{50}{v_{1}}, \quad 4 \leqslant v_{1} \leqslant 20,
\end{array}
\right.
\]
and
\[
9 \leqslant x+y \leqslant 14,
\]
which implies
\[
\left\{
\begin{array}{l}
3 \leqslant x \leqslant 10... | 930 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Calculate $\sqrt{(31)(30)(29)(28)+1}$. | \begin{aligned} \text { 1. } & \because(k+1) \cdot k \cdot(k-1) \cdot(k-2)+1 \\ & =[(k+1)(k-2)][k(k-1)]+1 \\ & =\left(k^{2}-k-2\right)\left(k^{2}-k\right)+1 \\ & =\left(k^{2}-k\right)^{2}-2\left(k^{2} \cdots k\right)+1 \\ & =\left[\left(k^{2}-k\right)-1\right]^{2} \\ \therefore \quad & v^{\prime}(31) \cdot(30) \cdot(29... | 869 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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