problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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1. The number of positive integers $n$ that make $1^{2 \times 15}+2^{2005}+\cdots+n^{205}$ divisible by $n+2$ is | $=1.0$.
Let $S_{n}=1^{2008}+2^{2000}+\cdots+n^{2005}$, then
$$
S_{n}=n^{2008}+(n-1)^{2005}+\cdots+1^{2008} \text {. }
$$
By misalignment addition, we get
$$
\begin{aligned}
2 S_{n}= & 2+\left(2^{2005}+n^{2005}\right)+\left[3^{2005}+(n-1)^{2000}\right]+ \\
& \cdots+\left(n^{2005}+2^{2005}\right) .
\end{aligned}
$$
Fro... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Let $S=\{1,2, \cdots, n\}$. Find the smallest natural number $n$, such that when $S$ is arbitrarily divided into two subsets, there is always one subset that contains two different numbers $a$ and $b$, satisfying $(a+b) \mid a b$. | Solution: Two distinct numbers $a$ and $b$ that satisfy $(a+b) \mid a b$ are called a "good pair." Clearly, the answer to this problem is only related to these good pairs and has nothing to do with numbers that do not appear in any good pair. Therefore, we can list all the good pairs within a certain range, for example... | 40 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $B C$ respectively. Find the distance from point $D$ to the plane $B_{1} E F$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result dire... | Solution: As shown in Figure 10, in the rectangular coordinate system, we have
$$
\begin{array}{l}
D(0,0,0), \\
B_{1}(1,1,1), \\
E\left(1, \frac{1}{2}, 0\right), \\
E\left(\frac{1}{2}, 1,0\right) .
\end{array}
$$
Thus, $B_{1} E=\left(0,-\frac{1}{2},-1\right)$,
$$
\boldsymbol{B}_{1} \boldsymbol{F}=\left(-\frac{1}{2}, 0... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given that $x$ and $y$ are positive integers, and satisfy the conditions $x y + x + y = 71, x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$. | Solution: Let $x+y=u, xy=v$. From the given equation, we get $u+v=71, uv=880$.
By Vieta's formulas, $u$ and $v$ are the roots of the quadratic equation
$$
t^{2}-71t+880=0
$$
Solving this equation, we get $t=16$ or $t=55$.
Therefore, $\left\{\begin{array}{l}u=16, \\ v=55\end{array}\right.$ or $\left\{\begin{array}{l}u=... | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In a $9 \times 9$ grid, there are 81 small squares. In each small square, write a number. If in every row and every column, there are at most three different numbers, it can be guaranteed that there is a number in the grid that appears at least $n$ times in some row and at least $n$ times in some column. What is the... | 3. If a $9 \times 9$ grid is divided into 9 $3 \times 3$ grids, and each small cell in the same $3 \times 3$ grid is filled with the same number, and the numbers in any two different $3 \times 3$ grids are different, then each row and each column will have exactly three different numbers. Therefore, the maximum value o... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $\frac{(2 x+z)^{2}}{(x+y)(-2 y+z)}=8$. Then $2 x+$ $4 y-z+6=$ $\qquad$ | 1. Hint: $(2 x+4 y-z)^{2}=0, 2 x+4 y-z+6=6$. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If $2 x^{2}+7 x y-15 y^{2}+a x+b y+3$ can be factored into the product of two linear polynomials with integer coefficients, where $a$ and $b$ are real numbers, then the minimum value of $a+b$ is $\qquad$ | $$
\begin{array}{c}
\text { If the original expression is }=(x+5 y)(2 x-3 y)+a x+b y+3, \\
\text { then }(a, b)=(-5,-12),(5,12),(-7,4),(7,-4). \\
\text { Therefore, }(a+b)_{\text {min }}=-17.
\end{array}
$$ | -17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A computer user plans to purchase single-piece software and boxed disks, priced at 60 yuan and 70 yuan each, respectively, with a budget of no more than 500 yuan. According to the needs, the user must buy at least 3 pieces of software and at least 2 boxes of disks. How many different purchasing options are there? | 4. First buy 3 pieces of software and 2 boxes of disks. With the remaining 180 yuan, if you do not buy software, you can buy 0, 1, or 2 more boxes of disks; if you buy 1 more piece of software, you can buy 0 or 1 more box of disks; if you buy 2 more pieces of software, you cannot buy any more disks; if you buy 3 more p... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the equation $6 x^{2}+2(m-13) x+12-m$ $=0$ has exactly one positive integer solution. Then the value of the integer $m$ is | 5. Since $\Delta=4(m-13)^{2}-24(12-m)$ is a perfect square, there exists a non-negative integer $y$ such that
$$
(m-13)^{2}-6(12-m)=y^{2} \text {, }
$$
i.e., $(m-10-y)(m-10+y)=3$.
Since $m-10-y \leqslant m-10+y$, we have
$$
\left\{\begin{array} { l }
{ m - 1 0 - y = 1 , } \\
{ m - 1 0 + y = 3 }
\end{array} \text { or... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. The cost of a house consists of the cost of the above-ground part and the cost of the foundation part. For a house with an area of $N \mathrm{~m}^{2}$, the cost of the above-ground part is proportional to $N \sqrt{N}$, and the cost of the foundation part is proportional to $\sqrt{N}$. It is known that for a house w... | 10. Let the area of each house in square meters be $y$, and a total of $x$ identical houses are built, with the total cost being $S$. Then
$$
\left\{\begin{array}{l}
x y=80000, \\
S=(\alpha y \sqrt{y}+\beta \sqrt{y}) \cdot x, \\
\frac{\alpha \cdot 3600 \sqrt{3600}}{\beta \sqrt{3600}}=\frac{72}{100},
\end{array}\right.
... | 5000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. It is known that $\alpha^{2005}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$. Find the sum of the coefficients of this polynomial.
（Zhu Huawei provided the problem） | 1. Solution 1: In the expansion of $\alpha^{k}+\beta^{k}$, let $\alpha+\beta=1$, $\alpha \beta=1$, the sum of the coefficients we are looking for is $S_{k}=\alpha^{k}+\beta^{k}$. From
$$
\begin{array}{l}
(\alpha+\beta)\left(\alpha^{k-1}+\beta^{k-1}\right) \\
=\left(\alpha^{k}+\beta^{k}\right)+\alpha \beta\left(\alpha^{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $S=\{1,2, \cdots, 2005\}$. If any set of $n$ pairwise coprime numbers in $S$ contains at least one prime number, find the minimum value of $n$.
(Tang Lihua) | 3. First, we have $n \geqslant 16$.
In fact, take the set $A_{0}=\left\{1,2^{2}, 3^{2}, 5^{2}, \cdots, 41^{2}, 43^{2}\right\}$, then $A_{0} \subseteq S, \left|A_{0}\right|=15, A_{0}$ contains any two numbers that are coprime, but there are no primes in it, which shows that $n \geqslant 16$.
Next, we prove: For any $A... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $n$ students be such that among any 3 of them, 2 know each other, and among any 4 of them, 2 do not know each other. Find the maximum value of $n$.
(Tang Lihua | 8. The maximum value of $n$ is 8.
When $n=8$, the example shown in Figure 7 satisfies the requirements, where $A_{1}, A_{2}, \cdots, A_{8}$ represent 8 students, and the line between $A_{i}$ and $A_{j}$ indicates that $A_{i}$ and $A_{j}$ know each other.
Suppose $n$ students meet the requirements of the problem, we w... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. Let $r, s, t$ be integers, and the set $\left\{a \mid a=2^{r}+\right.$ $2^{s}+2^{t}, 0 \leqslant t<s<r \mid$ consists of numbers arranged in ascending order to form the sequence $\left\{a_{n}\right\}: 7,11,13,14, \cdots$. Then $a_{34}=$ $\qquad$ | 12.131.
Since $r, s, 1$ are integers $\mathrm{HL}, 0 \leqslant 1<s<r$, then $r$ takes the smallest value 2. At this time, the numbers that meet the condition are $C_{2}^{2}=1$.
When $r=3$, $s, 1$ can be chosen from $0,1,2$, the numbers that meet the condition are $C_{i}^{3}=3$.
Similarly, when $r=4$, the numbers that... | 131 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x+y=3, x^{2}+y^{2}-x y=4$. Then $x^{4}+y^{4}+$ $x^{3} y+x y^{3}$ is $\qquad$ | (Hint: Let $x+y=u, x y=v$. Answer: 36 .) | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a>b>0$. Then the minimum value of $a^{3}+\frac{12}{b(a-b)}$ is $\qquad$ .
| 6.20
$$
\begin{array}{l}
a^{3}+\frac{12}{b(a-b)} \geqslant a^{3}+\frac{12}{\frac{(b+a-b)^{2}}{4}} \\
=a^{3}+\frac{48}{a^{2}}=\frac{a^{3}}{2}+\frac{a^{3}}{2}+\frac{16}{a^{2}}+\frac{16}{a^{2}}+\frac{16}{a^{2}} \geqslant 20 .
\end{array}
$$ | 20 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) (1) Given $a+\log _{2}(2 a+6)=11$ and $b+2^{b-1}=14$. Find the value of $a+b$.
(2) Given $f(x)=\frac{2}{2^{x-2}+1}$. Find
$$
f(-1)+f(0)+f(1)+f(2)+f(3)+f(4)+f(5)
$$
the value. | (1) Let $t=\log _{2}(2 a+6)$, then we have $t+2^{t-1}=14$. Since the function $f(x)=x+2^{x-1}$ is an increasing function on $\mathbf{R}$, we have $b$ $=t$. Therefore, $a+t=11$, which means $a+b=11$.
(2) Since $f(x)+f(4-x)=\frac{2}{2^{x-2}+1}+\frac{2}{2^{2-x}+1}$
$$
=\frac{2\left(2^{x-2}+1\right)}{2^{x-2}+1}=2 \text {, ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six. (20 points) As shown in Figure 3, let
$$
P_{1}(1, \sqrt{3}), P_{2}(4,2 \sqrt{3}), \cdots, P_{n}\left(x_{n}, y_{n}\right)
$$
$\left(0<y_{1}<y_{2}<\cdots<y_{n}\right)$ be $n$ points on the curve $C: y^{2}=3 x(y \geqslant 0)$. Points $A_{i}(i=1,2, \cdots, n)$ are on the positive $x$-axis, satisfying that $\triangle A... | (1) Let point $A_{n}\left(a_{n}, 0\right)$. Using the graph, we can find that the x-coordinate of the point is the average of the x-coordinates of points $A_{n-1}$ and $A_{n}$, i.e., $x_{n}=\frac{a_{n-1}+a_{n}}{2}$. Therefore, the y-coordinate of point $P_{n}$ is $y_{n}=\sqrt{3\left(\frac{a_{n-1}+a_{n}}{2}\right)}$.
At... | 88 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given a three-digit number $\overline{a b c}$ that satisfies $a+$ $b+c=a b c$. Find the sum of all such three-digit numbers.
| Three, from the problem, we get that $a$, $b$, and $c$ are all non-zero, and without loss of generality, assume $a \geqslant b \geqslant c$.
When $b c>3$, $a+b+c=a b c>3 a$, which means $b+c>2 a$. This contradicts $b+c \leqslant 2 a$.
When $b c=3$, then $b=3, c=1$. In this case, $a=2$, which contradicts $a \geqslant b... | 1332 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given
$$
A=\{n \in \mathbf{N} \mid 1 \leqslant n \leqslant 2006 \text { and }(n+4,30) \neq 1\} \text {. }
$$
then $\operatorname{card}(A)=(\quad)$.
(A) 1605
(B) 1537
(C) 1471
(D) 1404 | -1.C.
Since $30=2 \times 3 \times 5, (n+4,30) \neq 1$, therefore, $n+4=2k$ or $3k$ or $5k\left(k \in \mathbf{N}_{+}\right)$.
Also, $1 \leqslant n \leqslant 2006$, then $5 \leqslant n+4 \leqslant 2010$. Thus, there are $\left[\frac{2010}{2}\right]-2=1003$ numbers of the form $2k$, $\left[\frac{2010}{3}\right]-1=669$ nu... | 1471 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
5. For any positive integer $k$, let $f_{1}(k)$ be the sum of the squares of the digits of $k$, and for $n \geqslant 2, f_{n}(k)=$ $f_{1}\left(f_{n-1}(k)\right)$. Then $f_{2006}(2006)=$ $\qquad$ | 5. 145 .
Notice that
$$
\begin{array}{l}
f_{1}(2006)=2^{2}+6^{2}=40, f_{2}(2006)=f_{1}(40)=16, \\
f_{3}(2006)=f_{1}(16)=37, f_{4}(2006)=f_{1}(37)=58, \\
f_{5}(2006)=f_{1}(58)=89, f_{6}(2006)=f_{1}(89)=145, \\
f_{7}(2006)=f_{1}(145)=42, f_{8}(2006)=f_{1}(42)=20, \\
f_{9}(2006)=f_{1}(20)=4, f_{10}(2006)=f_{1}(4)=16,
\en... | 145 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Arrange the digits $1,2,3,4,5,6,7,8,9$ in some order to form a nine-digit number abodefghi, and let
$$
A=\overline{a b c}+\overline{b c d}+\overline{c d e}+\overline{d e f}+\overline{e f g}+\overline{f g h}+\overline{g h i} .
$$
Then the maximum possible value of $A$ is | 2.4648 .
$A=111(c+d+e+f+g)+110 b+100 a+11 h+i$,
then when $c+d+e+f+g=9+8+7+6+5=35, b=4$, $a=3, h=2, i=1$, the maximum value of $A$ is 4648. | 4648 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If a two-digit number $\bar{x}$ and a three-digit number $\overline{3 y z}$ have a product of 29,400, then, $x+y+z=$ $\qquad$ | 3.18.
According to the problem, we have
$$
73<\frac{29400}{400}<\overline{x 5} \leqslant \frac{29400}{300}=98 .
$$
Thus, $\overline{x 5}=75$ or 85 or 95.
Since $85 \times 29400, 95 \times 29400$, it can only be $\overline{x 5}=75$.
When $\overline{x 5}=75$, $\overline{3 y z}=\frac{29400}{75}=392$.
Therefore, $x=7, y=... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure 2, in trapezoid
$A B C D$, $A B / / D C$,
$D C=2 A B=2 A D$. If
$B D=6, B C=4$, then
$S_{A B C D}=\ldots \quad\left(S_{A B C D}\right.$ represents
the area of quadrilateral $A B C D$, the same below. | 7. 18
As shown in Figure 6, take the midpoint $M$ of $CD$, and connect $BM$, $AM$.
From the given conditions, it is easy to see that quadrilateral $ADMB$ is a rhombus, and quadrilateral $AMCB$ is a parallelogram. Therefore, $AM = BC = 4$.
Noting that $\triangle ADM \cong \triangle MAB \cong \triangle BMC$, we have $S... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $p$ be a prime number, and let the equation $x^{2}-p x-580 p=0$ have two integer roots. Then the value of $p$ is $\qquad$ . | 10.29.
Since $x^{2}-p x-580 p=0$ has two integer roots, we have
$$
\Delta=(-p)^{2}+4 \times 580 p=p(p+2320)
$$
is a perfect square.
From this, we know that $p 12320$.
Since $2320=2^{4} \times 5 \times 29$, we have $p=2$ or 5 or 29.
When $p=2$, $\Delta=2^{2}(1+1160)=2^{2} \times 1161=2^{2} \times 3^{2} \times 129$ is ... | 29 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) Given that $a$, $b$, and $c$ are all prime numbers greater than 3, and $2a + 5b = c$.
(1) Prove that there exists a positive integer $n > 1$ such that the sum $a + b + c$ of all three prime numbers $a$, $b$, and $c$ that satisfy the given condition is divisible by $n$;
(2) Find the maximum value of $... | (1) Since $c=2a+5b$, we have
$$
a+b+c=3a+6b=3(a+2b).
$$
Also, since $a, b, c$ are all prime numbers greater than 3, it follows that $3 \mid (a+b+c)$, i.e., there exists a positive integer $n>1$ (for example, $n=3$) such that $n! \mid (a+b+c)$.
(2) Since $a, b, c$ are all prime numbers greater than 3, $a, b, c$ are not... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
11. When seven dice are rolled simultaneously, the probability that the sum of the numbers on the seven faces is 10 is equal to the probability that the sum of the numbers on the seven faces is $a(a \neq 10)$. Then, $a=$ $\qquad$ | 11.39.
$$
\text { Given }\left(x_{1}, x_{2}, \cdots, x_{7}\right) \rightarrow\left(7-x_{1}, 7-x_{2}, \cdots, 7-x_{7}\right)
$$
we know that the probability of the sum of the points being 10 is the same as the probability of the sum of the points being $49-10$ $=39$. | 39 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. The positive integer solutions $(x, y)$ of the equation $2 x^{2}-x y-3 x+y+2006=0$ are $\qquad$ pairs.
| 12.4.
From $2 x^{2}-x y-3 x+y+2006=0$, we get $y=\frac{2 x^{2}-3 x+2006}{x-1}=2 x-1+\frac{2005}{x-1}$.
Therefore, $x-1$ can take the values $1,5,401,2005$. Hence, the equation has 4 pairs of positive integer solutions. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. As shown in Figure 5, there is an iron tower $A B$ at the top of a slope. Under the sunlight, the tower's shadow $D E$ is cast on the slope. It is known that the base width of the tower $C D=14 \mathrm{~m}$, the length of the tower's shadow $D E=36 \mathrm{~m}$, and both Xiao Ming and Xiao Hua are $1.6 \mathrm{~m}$... | 15.20.
Draw a perpendicular from point $D$ to $CD$ intersecting $AE$ at point $F$, and draw a perpendicular from point $F$ to $AB$, with the foot of the perpendicular being $G$. It is easy to see that
$$
\begin{array}{l}
D F=\frac{1.6}{4} D E=14.4(\mathrm{~m}), \\
A G=\frac{1.6}{2} F G=\frac{1.6}{2} B D=5.6(\mathrm{~m... | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
20. Several 1s and 2s are arranged in a row:
$$
1,2,1,2,2,1,2,2,2,1,2, \cdots
$$
The rule is: The 1st number is 1, the 2nd number is 2, the 3rd number is 1. Generally, first write a row of 1s, then insert $k$ 2s between the $k$-th 1 and the $(k+1)$-th 1 ($k=1,2$, $\cdots$). Try to answer:
(1) Is the 2005th number 1 or... | 20. (1) Divide the sequence of numbers into $n$ groups:
These $n$ groups have a total of $1+2+\cdots+n=\frac{1}{2} n(n+1)$ numbers.
When $n=62$, there are 1953 numbers;
When $n=63$, there are 2016 numbers.
It is clear that the 2005th number is in the 63rd group, and it is not the last number in that group, so the 2005... | 7789435 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. The sequence $x_{1}, x_{2}, \cdots, x_{100}$ satisfies the following conditions: For $k=1,2, \cdots, 100, x_{k}$ is less than the sum of the other 99 numbers by $k$. Given that $x_{S 0}=\frac{m}{n}, m, n$ are coprime positive integers. Then $m+n$ equals $(\quad)$.
(A) 50
(B) 100
(C) 165
(D) 173 | 4.D.
Let $S=x_{1}+x_{2}+\cdots+x_{100}$, then $x_{k}=\left(S-x_{k}\right)-k$, which means $k+2 x_{k}=S$.
Summing over $k$ we get
$$
(1+2+\cdots+100)+2 S=100 S \text {. }
$$
Therefore, $S=\frac{2525}{49}$. Thus, $x_{50}=\frac{S-50}{2}=\frac{75}{98}$.
Hence, $m+n=173$. | 173 | Algebra | MCQ | Yes | Yes | cn_contest | false |
10. The function $f(x)$ satisfies: for any real numbers $x, y$, we have
$$
\frac{f(x) f(y)-f(x y)}{3}=x+y+2 .
$$
Then the value of $f(36)$ is | 10.39.
Let $x=y=0$, then $\frac{f^{2}(0)-f(0)}{3}=2$.
Therefore, $f(0)=-2$ or $f(0)=3$.
If $f(0)=-2$, let $y=0$, then $\frac{f(x) f(0)-f(0)}{3}=x+2$. So, $f(x)=-\frac{3}{2} x-2$. Substituting into the original equation shows it does not satisfy the condition.
If $f(0)=3$, solving yields $f(x)=x+3$, substituting for v... | 39 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Let $a$, $b$, $c$ be positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$. | 14. Let the two real roots of the equation be $x_{1}, x_{2}$. By Vieta's formulas, we know that $x_{1}, x_{2}$ are both negative.
From $\frac{c}{a}=x_{1} x_{2}9$. Therefore, $b^{2} \geqslant 4 a c=4 \times \frac{a}{c} \times c^{2}>4 \times 9 \times 1^{2}=36$.
Solving this, we get $b>6$. Hence, $b \geqslant 7$.
Also, $\... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Let sets $A$ and $B$ be sets composed of positive integers, $|A|=10,|B|=9$, and set $A$ satisfies the following condition: if $x, y, u, v \in A, x+y=u+v$, then $\{x, y\}=\{u, v\}$. Let $A+B=\{a+b \mid a \in A$, $b \in B\}$. Prove: $|A+B| \geqslant 50(|X|$ represents the number of elements in set $X$). | 15. Consider the general case.
Let $|A|=m,|B|=n, A+B=\left\{s_{1}, s_{2}, \cdots, s_{k}\right\}$.
For any $1 \leqslant i \leqslant k$, let $s_{i}$ have $f(i)$ ways to be expressed in the form $a+b$, where $a \in A, b \in B$, i.e.,
$$
s_{i}=a_{i 1}+b_{i 1}=a_{i 2}+b_{i 2}=\cdots=a_{i f(i)}+b_{i f(i)} .
$$
Then it is c... | 50 | Combinatorics | proof | Yes | Yes | cn_contest | false |
8. A mapping $f$ from set $A$ to set $B$ is called a surjection if for every element $y$ in set $B$, there is at least one $x \in A$ such that $f(x)=y$. Given a five-element set $A=\{1,2,3,4,5\}$ and a three-element set $B=$ $\{\alpha, \beta, \gamma\}$. Then the number of surjections from set $A$ to set $B$ is. | 8. 150 .
The total number of mappings from a five-element set to a three-element set is $3^{5}$. Among these, the mappings that are not surjective can be divided into two categories for calculation: one category is where only one element in $B$ is the image of all elements in $A$, and there are $\mathrm{C}_{3}^{1}=3$ ... | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Let the set $M=\{1,2, \cdots, 2005\}$. If $X \subseteq M$, $X \neq \varnothing$, and $a_{X}$ is the sum of the largest and smallest numbers in $X$ (if the set $X$ contains only one element, then this element is both the largest and the smallest number), then, for all non-empty subsets $X$ of $M$, the average value ... | 11.2006.
For any non-empty subset $X \subseteq M$, define $X^{\prime}=\{2006-x \mid x \in X \}$. It is clear that $X^{\prime} \subseteq M$. Thus, $f: X \rightarrow X^{\prime}$ provides a one-to-one correspondence on the family of all non-empty subsets of $M$. Note that
$$
\begin{array}{l}
\max (X)+\min \left(X^{\prime... | 2006 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. A and B take turns tossing a fair coin, and whoever tosses heads first wins, at which point the game ends, and the loser gets to toss first in the next game.
(1) Find the probability that the first person to toss wins in any given game;
(2) Suppose they play a total of 10 games, and A tosses first in the first game... | 15. (1) In any given match, the probability of the first person to throw winning is $\frac{1}{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{5}+\cdots=\frac{2}{3}$.
(2) From (1), the probability of the second person to throw winning is $1-\frac{2}{3}=\frac{1}{3}$.
Given $P_{1}=\frac{2}{3}$, for $2 \leqslant... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
17. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1} a_{n}-1=a_{n}^{2} \text {. }
$$
(1) Prove: $\sqrt{2 n-1} \leqslant a_{n} \leqslant \sqrt{3 n-2}$;
(2) Find the integer $m$, such that $\left|a_{2 \cos }-m\right|$ is minimized. | 17. (1) From the given information, it is easy to see that $\{a_n\}$ is an increasing sequence, and all its terms are positive. Since $a_{n+1} a_{n} - 1 = a_{n}^2$, we have, for $k \geq 2$,
$$
a_{k} = a_{k-1} + \frac{1}{a_{k-1}}.
$$
From this, we get
$$
a_{k}^2 = \left(a_{k-1} + \frac{1}{a_{k-1}}\right)^2 = a_{k-1}^2 ... | 63 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the three-digit number $\bar{b} c$ be divisible by 3, and let $a$, $b$, and $c$ be the lengths of the sides of an isosceles triangle (including equilateral triangles). Then the number of such three-digit numbers is ( ) .
(A) 21
(B) 36
(C) 45
(D) 63 | 4.C.
From the problem, we know that $a+b+c$ is a multiple of 3, and $a, b, c$ are positive integers from 1 to 9.
When $a=b=c$, there are 9 three-digit numbers that satisfy the condition;
When two of $a, b, c$ are equal, let's assume $a=b$, then there are $441, 447, 552, 558, 663, 669, 771, 774, 882, 885, 993$, 996, a ... | 45 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
3. Given the polynomial
$$
\begin{aligned}
a_{0}+ & \left(a_{1}+4\right) x+ \\
& \left(a_{2}-10\right) x^{2}+\left(a_{3}+6\right) x^{3}+\left(a_{4}-1\right) x^{4}+ \\
& \left(a_{5}-1\right) x^{5}+a_{6} x^{6}+\cdots+a_{2 \alpha 5} x^{2 \omega 5}
\end{aligned}
$$
can be divided by $x^{2}+3 x-2$, and $\alpha^{2}+3 \alpha... | 3.0 .
Since $\alpha^{2}+3 \alpha-2=0$, $\alpha$ is a root of the equation $x^{2}+3 x-2=0$. Therefore,
$$
\begin{array}{l}
a_{0}+\left(a_{1}+4\right) \alpha+\left(a_{2}-10\right) \alpha^{2}+\left(a_{3}+6\right) \alpha^{3}+ \\
\left(a_{4}-1\right) \alpha^{4}+\left(a_{5}-1\right) \alpha^{5}+a_{6} \alpha^{6}+\cdots+a_{2 \... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the quadratic function $y=a x^{2}+13 x$ passes through two lattice points (points with integer coordinates) in the first quadrant, and their y-coordinates are both prime numbers. Then $a=$ $\qquad$ - | 4. -6 .
Let the lattice point satisfying the given conditions be $(m, n)$.
Then $n=a m^{2}+13 m=m(a m+13)$.
Since $n$ is a prime number, we have $m_{1}=1$ and $a m_{1}+13$ is a prime number, or $a m_{2}+13=1$ and $m_{2}$ is a prime number.
Therefore, $m_{1}=1$ and $a+13$ is a prime number, and $a m_{2}=-12=-2 \times ... | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. On the positive direction of the $x$-axis, take a sequence of points $\left\{A_{j}\right\}, j=1,2, \cdots$, and in the first quadrant on the parabola $y^{2}=\frac{3}{2} x$, take a sequence of points $\left\{B_{k}\right\}, k=1,2, \cdots$, such that $\triangle A_{k-1} B_{k} A_{k}(k=1,2, \cdots)$ are all equilateral t... | 11.2005.
Let the side length of the $n$-th equilateral triangle be $a_{n}$. Then the coordinates of the vertex $B_{n}$ of the $n$-th equilateral triangle on the parabola are
$$
\left(a_{1}+a_{2}+\cdots+a_{n-1}+\frac{a_{n}}{2}, \sqrt{\frac{3}{2}\left(a_{1}+a_{2}+\cdots+a_{n-1}+\frac{a_{n}}{2}\right)}\right) .
$$
From ... | 2005 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. In a live military exercise, the Red side has set up 20 positions along a straight line. To test 5 different new weapons, it is planned to equip 5 positions with these new weapons, with the requirement that the first and last positions are not equipped with new weapons, and that among every 5 consecutive positions,... | 15. Let 20 positions be sequentially numbered as $1,2, \cdots, 20$, and let the sequence number of the $k$-th new weapon be $a_{k}(k=1,2,3,4,5)$.
$$
\begin{array}{l}
\text { Let } x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}=a_{3}-a_{2}, \\
x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}, x_{6}=20-a_{5} .
\end{array}
$$
Then we have $x... | 69600 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. The smallest natural number $n$ that satisfies $n \sin 1 > 1 + 5 \cos 1$ is $\qquad$ . | 6.5 .
Since $\frac{\pi}{4}n \sin 1>1+5 \cos 1>1+5 \cos \frac{\pi}{3}=\frac{7}{2} \text {. }$
Thus, $n>\frac{7}{\sqrt{3}}>4$.
When $n=5$, it is easy to prove that $5 \sin 1>1+5 \cos 1$, which means
$$
5(\sin 1-\cos 1)>1 \text {, }
$$
Squaring it gives $\sin 2<\frac{24}{25}$, which is obviously true.
In fact, because $... | 5 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. A certain cargo yard has 2005 vehicles waiting in line to be loaded, with the requirement that the first vehicle must load 9 boxes of cargo, and every 4 adjacent vehicles must load a total of 34 boxes. To meet the above requirements, the minimum number of boxes of cargo should be ( ).
(A) 17043
(B) 17044
(C) 17045
(... | 8. A.
Except for the first vehicle which carries 9 boxes, among the remaining 2004 vehicles, every 4 vehicles carry 34 boxes, so there are $\frac{2004}{4} \times 34+9=17043$ boxes. | 17043 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
14. The elements in set $X$ are positive integers, and have the property: if $x \in X$, then $12-x \in X$. The number of such sets $X$ is $\qquad$. | 14.63.
Let $Y=\{(1,11),(2,10),(3,9),(4,8),(5,7), 6\}$. Therefore, the number of sets that satisfy the condition is $2^{6}-1=63$. | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. (12 points) The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}(n \in \mathbf{N}), \\
b_{n}=\frac{1}{1+a_{n}}, S_{n}=b_{1}+b_{2}+\cdots+b_{n}, \\
P_{n}=b_{1} b_{2} \cdots b_{n} .
\end{array}
$$
Try to find the value of $2 P_{n}+S_{n}$. | Three, 15. Since $a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}, n \in \mathbf{N}$, therefore, $a_{n+1}=a_{n}\left(a_{n}+1\right)$. Then
$$
\begin{array}{l}
b_{n}=\frac{1}{1+a_{n}}=\frac{a_{n}^{2}}{a_{n} a_{n+1}}=\frac{a_{n+1}-a_{n}}{a_{n} a_{n+1}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}, \\
P_{n}=b_{1} b_{2} \cdots b_{n}=\frac... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. (12 points) Draw a line through point $P(3+2 \sqrt{2}, 4)$ that intersects the $x$-axis and $y$-axis at points $M$ and $N$, respectively. Find the maximum value of $OM + ON - MN$ (where $O$ is the origin). | 17. A circle is drawn through the point $P(3+2 \sqrt{2}, 4)$, tangent to the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and such that point $P$ lies on the major arc $\overparen{A B}$. The equation of the circle is $(x-3)^{2}+(y-3)^{2}=9$. Thus, the tangent line through point $P$ intersects the $x$-axis ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four, (16 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$. Find the number of elements in the set
$$
\left\{n \left\lvert\, n=\left[\frac{k^{2}}{2005}\right]\right., 1 \leqslant k \leqslant 2004, k \in \mathbf{N}\right\}
$$ | From $\frac{(k+1)^{2}}{2005}-\frac{k^{2}}{2005}=\frac{2 k+1}{2005} \leqslant 1$, we solve to get
$k \leqslant 1002$.
When $k=1,2, \cdots, 1002$, we have
$$
\left[\frac{(k+1)^{2}}{2005}\right]=\left[\frac{k^{2}}{2005}\right]
$$
or $\left[\frac{(k+1)^{2}}{2005}\right]=\left[\frac{k^{2}}{2005}\right]+1$.
Since $\left[\fr... | 1503 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. Given the inequality $|a x-3| \leqslant b$ has the solution set $\left[-\frac{1}{2}, \frac{7}{2}\right]$. Then $a+b=$ $\qquad$ . | 13.6.
From $|a x-3| \leqslant b$, we get $3-b \leqslant a x \leqslant 3+b$. It is easy to see that $a \neq 0$, so $\frac{3-b}{a}+\frac{3+b}{a}=-\frac{1}{2}+\frac{7}{2}$. Solving this, we get $a=2$.
Therefore, $\frac{3-b}{a}=-\frac{1}{2}$. Solving this, we get $b=4$. Hence, $a+b=6$. | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
18. The inequality $x+2 \sqrt{2 x y} \leqslant a(x+y)$ holds for all positive numbers $x, y$. Then the minimum value of the real number $a$ is $\qquad$ | 18.2.
Since $x+2 \sqrt{2 x y} \leqslant x+(x+2 y)=2(x+y)$, then the minimum value of $a$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
19. As shown in Figure 1, a
small boat is traveling northward at a speed of $10 \mathrm{~m} / \mathrm{s}$ over the lake, and on a bridge $20 \mathrm{~m}$ above the lake, a car is moving eastward at a speed of $20 \mathrm{~m} / \mathrm{s}$. Currently, the boat is $40 \mathrm{~m}$ south of point $P$ on the water, and t... | 19.30.
Assuming after $t \mathrm{~s}$, the distance between the boat and the car is $y \mathrm{~m}$, then
$$
\begin{array}{l}
y=\sqrt{(40-10 t)^{2}+(-30+20 t)^{2}+400} \\
=\sqrt{500(t-2)^{2}+900} \text {. } \\
\end{array}
$$
When $t=2 \mathrm{~s}$, $y$ has a minimum value of $30 \mathrm{~m}$. | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
25. The coordinates of point $P(x, y)$ satisfy the following relations:
$$
\left\{\begin{array}{l}
2 x+y \geqslant 15, \\
x+3 y \geqslant 27, \\
x \geqslant 2, \\
y \geqslant 3,
\end{array}\right.
$$
and $x, y$ are both integers. Then the minimum value of $x+y$ is
$\qquad$, and at this time, the coordinates of point $... | $25.12,(3,9)$ or $(4,8)$.
Adding equations (1), (2), and (3), we get $x+y \geqslant 11$.
If $x+y=11$, then the equalities in equations (1), (2), and (3) hold, and the system of equations has no integer solutions.
If $x+y=12$, from $2 x+y \geqslant 15$, we get $x \geqslant 3$;
from $x+3 y \geqslant 27$, we get $2 y \geq... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. Use $1,2,3,4,5$ to form a five-digit number, such that the difference between any two adjacent digits is at least 2. Then the number of such five-digit numbers is $\qquad$ . | 2.1.14.
Consider $\square$ classified by the middle number $a$:
(1) If $a=1$, then 2 and 3 are on opposite sides of $a$, and 4 and 5 are on opposite sides of $a$, resulting in 4 permutations:
$$
24135,24153,35142,53142 \text {; }
$$
(2) If $a=2$, then 4 and 5 are on opposite sides of $a$, and 1 and 3 are not adjacent t... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The graphs of the functions $y=x^{2}-x$ and $y=\cos 10 \pi x(x \geqslant 0)$ intersect at $\qquad$ points. | 4.17.
When $x \geqslant 0$, from $\left|x^{2}-x\right| \leqslant 1$, we get
$$
0 \leqslant x \leqslant \frac{1+\sqrt{5}}{2}=1.618 \cdots \text {. }
$$
The period of $y=\cos 10 \pi x$ is $\frac{1}{5}$. Within each period, if $\left|x^{2}-x\right|<1$, then the graphs of $y=x^{2}-x$ and $y=\cos 10 \pi x$ have 2 intersect... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If $n \in\{1,2, \cdots, 100\}$, and $n$ is a multiple of the sum of its digits, then there are $\qquad$ such $n$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The blank space represented by $\qquad$ is kept as is in the translation. | 6.33.
Let $n=\overline{a b}=10 a+b,(a+b) \mid (10 a+b)$.
When $a, b$ is 0, it obviously meets the requirement. Therefore, all one-digit numbers and two-digit numbers ending in 0, as well as the three-digit number 100, meet the requirement, making a total of 19 such $n$.
When $a, b$ are both not 0, since $\frac{10 a+b... | 33 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) The sequence $\left\{a_{n}\right\}$ is
$$
1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots \text {. }
$$
Its construction method is:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, thus, we get $a_{2}=1, a_{3}=2$;
Then copy all the previous items $1,1,2$, and add the successor numb... | Five, according to the construction method of $\left\{a_{n}\right\}$, it is easy to know
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \text {. }
$$
Generally, there is $a_{2^{n}-1}=n$, that is, the number $n$ first appears at the $2^{n}-1$ term, and if
$m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, then ... | 3961 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. For an integer $m$, its unit digit is denoted by $f(m)$, and let $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$. Then $a_{2006}$ $=$ . $\qquad$ | 3.7.
It is known that $f\left(2^{m}\right)=f\left(2^{m+4}\right)$. Therefore,
$$
a_{2 \omega 0}=f\left(2^{2 \omega 1}-1\right)=f\left(2^{2 \omega 07}\right)-1=f(8)-1=7 .
$$ | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $S$ is a set composed of $n(n \geqslant 3)$ positive numbers. If there exist three distinct elements in $S$ that can form the three sides of a triangle, then $S$ is called a "triangle number set". Consider the set of consecutive positive integers $\{4,5, \cdots, m\}$, where all its 10-element subsets are ... | 6. C.
Let there be three positive numbers $a_{1}, a_{2}, a_{3} \left(a_{1}<a_{2}<a_{3}\right)$.
If $a_{1}+a_{2} \leqslant a_{3}$, then these three numbers cannot form the sides of a triangle. To make $a_{3}$ the smallest, take $a_{1}+a_{2}=a_{3}$.
$$
\begin{array}{l}
\text { Let } a_{1}=4, a_{2}=5, a_{3}=a_{1}+a_{2}=9... | 253 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. Let $A$ be a finite set, for any $x, y \in A$, if $x \neq y$, then $x+y \in A$. Then, the maximum number of elements in $A$ is $\qquad$ . | Let the number of elements in $A$ be $n$.
If $n>3$, then there must be two elements with the same sign. Without loss of generality, assume there are two positive numbers. Let the largest two positive numbers be $a, b$ (with $a < b$). Then, $a+b \notin A$, which is a contradiction.
Therefore, $n \leqslant 3$.
Also, $A=\... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Positive integers $a, b, c$ satisfy
$$
\log _{6} a+\log _{6} b+\log _{6} c=6 \text {, }
$$
$a, b, c$ form an increasing geometric sequence, and $b-a$ is a perfect square. Then the value of $a+b+c$ is $\qquad$ | $=.1 .111$.
From the given, we have $abc=6$, and $b^2=ac$. Therefore, $b=36$.
Also, $b-a$ is a perfect square, so $a$ can take the values $11, 20, 27, 32$.
Upon verification, only $a=27$ meets the conditions. In this case, $c=48$.
Thus, $a+b+c=27+36+48=111$. | 111 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $1 \leqslant a \leqslant b \leqslant 100$, and $57 \mid a b$. The number of positive integer pairs $(a, b)$ is ( ).
(A) 225
(B) 228
(C) $\overline{231}$
(D) 234 | 6.B.
Obviously, $57=3 \times 19$.
We proceed with the classification and counting.
(1) If $57 \mid a$, then $a=57, b=57,58, \cdots, 100$, i.e., there are $100-56=44$ pairs $(a, b)$.
(2) If $57 \mid b$, and $57 \nmid a$, then $b=57, a=1,2, \cdots, 56$, i.e., there are 56 pairs $(a, b)$.
(3) If $3 \mid a, 19 \mid b$, an... | 228 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
4. Let the solution set of the equation $x^{2}-x-1=\left(x^{2}-1\right) \pi^{x}-$ $x \pi^{x^{2}-1}$ be $M$. Then the sum of the cubes of all elements in $M$ is ( ).
(A) 0
(B) 2
(C) 4
(D) 5 | 4.C.
Obviously, $x=0, \pm 1$ are all solutions to the original equation.
When $x \neq 0, \pm 1$, from
$$
\left(x^{2}-1\right)\left(\pi^{x}-1\right)=x\left(\pi^{x^{2}-1}-1\right)
$$
and $\pi$ being a transcendental number, we know $\pi^{x}-1=\pi^{x^{2}-1}-1$, which means $x=x^{2}-1$. At this point, there are two real ... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Question 5 In a $12 \times 9$ rectangular grid, $k$ cells' centers are colored red, such that no 4 red points form the 4 vertices of a rectangle with sides parallel to the grid lines. Find the maximum value of $k$.
In a $12 \times 9$ rectangular grid, $k$ cells' centers are colored red, such that no 4 red points form ... | Solution: Let the $i$-th row of the grid be marked with $x_{i}$ red dots, then we have $x_{1}+x_{2}+\cdots+x_{12}=k$.
Similarly to the first problem, we have
$$
\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{12}}^{2} \leqslant \mathrm{C}_{9}^{2}=36 \text {. }
$$
Simplifying and using the Cauchy-Sc... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Find the ordered integer pairs $(x, y)$ that satisfy $(|x|-2)^{2}+(|y|-2)^{2}<5$.
untranslated text remains unchanged. | 2. Clearly, $(|x|-2)^{2}$ and $(|y|-2)^{2}$ are both perfect squares.
Since the sum of these two perfect square terms is less than 5, each term must be no greater than $2^{2}=4$.
If $(|x|-2)^{2}=0$, then $|x|=2$. Thus, $x= \pm 2$. Therefore, $x$ has two possible values.
If $(|x|-2)^{2}=1$, then $|x|=1$ or $|x|=3$. T... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the area of $\triangle A B C$ is 2000, points $P$, $Q$, and $R$ are the midpoints of $B C$, $C A$, and $A B$ respectively, points $U$, $V$, and $W$ are the midpoints of segments $Q R$, $R P$, and $P Q$ respectively, and the lengths of segments $A U$, $B V$, and $C W$ are $x$, $y$, and $z$ respectively. Pr... | 4. As shown in Figure 1, since $Q$ and $R$ are the midpoints of $CA$ and $AB$ respectively, $\triangle AQR \sim \triangle ACB$, with point $A$ as the similarity center and a similarity ratio of $\frac{1}{2}$.
Since $P$ is the midpoint of $BC$, and $U$ is the midpoint of $RQ$, it is easy to see that $U$ is also the mid... | 375 | Geometry | proof | Yes | Yes | cn_contest | false |
4. A square piece of paper is cut into two parts along a straight line that does not pass through any vertex; take one of the parts and cut it into two parts along a straight line that does not pass through any vertex; then take one of the resulting three parts and cut it into two parts along a straight line that does ... | 4. B.
According to the problem, when cutting along a line that does not pass through the vertices with the front knife, each cut increases the sum of the interior angles of each part by \(360^{\circ}\). Thus, after \(k\) cuts, \(k+1\) polygons can be obtained, and the sum of the interior angles of these polygons is \(... | 2005 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
6. Given that $a$, $b$, and $c$ are integers, and $a+b=2006$, $c-a=2005$. If $a<b$, then the maximum value of $a+b+c$ is $\qquad$ | II, 6.5 013.
From $a+b=2006, c-a=2005$, we get $a+b+c=a+4011$.
Since $a+b=2006(a<b, a$ is an integer $)$, therefore, the maximum value of $a$ is 1002.
Thus, the maximum value of $a+b+c$ is 5013. | 5013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that the perimeter of the regular pentagon square $A B C D E$ is $2000 \mathrm{~m}$, two people, A and B, start from points $A$ and $C$ respectively at the same time, walking around the square in the direction of $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E \rightarrow A \rightarrow \cdots$. Pers... | 8. 104 .
Suppose that when person A has walked $x$ edges, A and B start walking on the same edge for the first time. At this point, A has walked $400 x \mathrm{~m}$, and B has walked $46 \times$ $\frac{400 x}{50}=368 x \mathrm{~m}$. Therefore,
$$
368(x-1)+800-400(x-1)>400 \text {, }
$$
and $(368 x+800)-400 x \leqslan... | 104 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. [ $x$ ] represents the greatest integer not exceeding $x$ (for example, $[3.2]=3$). Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$. Then the number of such $n$ is $\qquad$.
| 9.334.
Let $\left[\frac{n}{6}\right]=m$. Then $\frac{n}{6}=m+\alpha(0 \leqslant \alpha<1)$. When $0 \leqslant \alpha<\frac{1}{2}$, we have
$$
\frac{n}{3}=2 m+2 \alpha,\left[\frac{n}{3}\right]=2 m \text {. }
$$
According to $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$, we get
$$
2 m+m=3 m+3 \alpha \... | 334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Xiaoming's home phone number was originally a six-digit number. The first upgrade was to add the digit 8 between the first and second digits, making it a seven-digit phone number; the second upgrade was to add the digit 2 before the first digit, making it an eight-digit phone number. Xiaoming found that the eight-d... | 10.282500.
Let the original six-digit telephone number be $\overline{a b c d e f}$. Then, after two upgrades, the eight-digit telephone number is $\overline{2 a 8 b c c l e f}$.
According to the problem, we have $81 \times \overline{a b c d e f}=\overline{2 a 8 b c d e f}$. Let $x=b \times 10^{1}+c \times 10^{3}+d \ti... | 282500 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14.2006 positive integers $a_{1}, a_{2}, \cdots, a_{2006}$, none of which are equal to 119, are arranged in a row, where the sum of any consecutive several terms is not equal to 119. Find the minimum value of $a_{1}+a_{2}+\cdots+$ $a_{2006}$. | 14. First prove the proposition:
For any 119 positive integers $b_{1}, b_{2}, \cdots, b_{119}$, there must exist some (at least one, or possibly all) whose sum is a multiple of 119.
In fact, consider the following 119 positive integers:
$$
b_{1}, b_{1}+b_{2}, \cdots, b_{1}+b_{2}+\cdots+b_{119} \text {. }
$$
If one of... | 3910 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7: Xiao Li and his brother attended a gathering, along with two other pairs of brothers. After meeting, some people greeted each other with handshakes, but no one shook hands with their own brother, and no one shook hands with the same person twice. At this point, Xiao Li noticed that, apart from himself, every... | Explanation: From the problem, it is easy to see that the maximum number of handshakes for each person is 4, and the minimum is 0. Also, except for Xiao Li, each person has a different number of handshakes. Therefore, the 5 people must have handshakes 0 times, 1 time, 2 times, 3 times, and 4 times respectively.
Using ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
13. Let $f(x)$ be an odd function with a period of 2, and $f\left(-\frac{2}{5}\right)=3$. If $\sin \alpha=\frac{\sqrt{5}}{5}$, then the value of $f(4 \cos 2 \alpha)$ is | 13. -3.
Since $\sin \alpha=\frac{\sqrt{5}}{5}$, therefore,
$$
\cos 2 \alpha=1-2 \sin ^{2} \alpha=\frac{3}{5} \text{. }
$$
Also, $f(x)$ is an odd function with a period of 2, and $f\left(-\frac{2}{5}\right)=$
3, so,
$$
f(4 \cos 2 \alpha)=f\left(\frac{12}{5}\right)=f\left(\frac{2}{5}\right)=-f\left(-\frac{2}{5}\right)=... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Let $x>1, y>1, S=\min \left\{\log _{x} 2, \log _{2} y\right.$ , $\left.\log _{y}\left(8 x^{2}\right)\right\}$. Then the maximum value of $S$ is $\qquad$ . | 15.2.
From the problem, we have $\log _{x} 2 \geqslant S, \log _{2} y \geqslant S, \log _{r}\left(8 x^{2}\right) \geqslant S$, then $S \leqslant \log ,\left(8 x^{2}\right)=\frac{3+2 \log _{2} x}{\log _{2} y}=\frac{3+\frac{2}{\log _{x} 2}}{\log _{2} y} \leqslant \frac{3+\frac{2}{S}}{S}$. Therefore, $S^{3}-3 S-2 \leqsla... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let $a, b, c > 0$. Prove:
$$
f=\sum \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant 8 \text {. }
$$
(32nd United States of America Mathematical Olympiad) | Explanation: By homogeneity, we may assume $a+b+c=3$. Then
$$
\begin{aligned}
f & =\sum \frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}}=\sum \frac{a^{2}+6 a+9}{3\left(a^{2}-2 a+3\right)} \\
& =\frac{1}{3} \sum\left(1+\frac{8 a+6}{(a-1)^{2}+2}\right) \\
& \leqslant \frac{1}{3} \sum(4 a+4)=8 .
\end{aligned}
$$ | 8 | Inequalities | proof | Yes | Yes | cn_contest | false |
2. If the positive integer $n \geqslant 2006$, and 122 divides $91 n-37$, then the minimum value of $n$ is $\qquad$ . | 2.2061 .
Since 122 is even and 91 and 37 are odd, $n$ must be odd.
Let $n=2k-1(k \in \mathbf{Z})$, then
122 | $(91n-37) \Leftrightarrow 122 | [91(2k-1)-37]$
$\Leftrightarrow \frac{182k-128}{122} \in \mathbf{Z} \Leftrightarrow \frac{91k-64}{61} \in \mathbf{Z}$
$\Leftrightarrow \frac{30k-3}{61} \in \mathbf{Z} \Leftright... | 2061 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 3, in $\triangle A B C$, $M$ is the midpoint of side $B C$, and $M D \perp A B, M E \perp A C$, with $D$ and $E$ being the feet of the perpendiculars. If $B D=2, C E=1$, and $D E \parallel B C$, then $D M^{2}$ equals $\qquad$ | 4.I.
Let $D M=x, M E=y, A D=z, A E=w$.
Since $D E / / B C$, we have $\frac{z}{2}=\frac{w}{1}$, i.e., $z=2 w$.
Because $B M=M C$, we have $S_{\triangle A K M}=S_{\triangle A M}$, i.e., $\frac{1}{2} x(z+2)=\frac{1}{2} y(w+1)$.
Thus, $2 x+x z=y+y w$.
By the Pythagorean theorem, we get
$$
\left\{\begin{array}{l}
2^{2}+x^{... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $m$ be a real number not less than -1, such that the equation in $x$
$$
x^{2}+2(m-2) x+m^{2}-3 m+3=0
$$
has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$, find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$.
(2000, N... | Explanation: Since the equation has two distinct real roots, we have
$$
\begin{array}{l}
\Delta=4(m-2)^{2}-4\left(m^{2}-3 m+3\right) \\
=-4 m+4>0 .
\end{array}
$$
Thus, $m<1$.
Also, $m \geqslant-1$, so, $-1 \leqslant m<1$.
By the relationship between roots and coefficients, we get
$$
x_{1}+x_{2}=-2(m-2), x_{1} x_{2}=m... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that $x$ and $y$ are real numbers, and satisfy $x y + x + y = 17, x^{2} y + x y^{2} = 66$.
Find the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$. (2000, Shandong Province Junior High School Mathematics Competition) | Explanation: According to the structural characteristics of the known equations, they can be transformed into
$$
x y+(x+y)=17, x y(x+y)=66 \text{. }
$$
Thus, a quadratic equation can be constructed with $x y$ and $x+y$ as its roots:
$$
t^{2}-17 t+66=0 \text{. }
$$
Solving this, we get $t_{1}=6, t_{2}=11$.
When $x y=6... | 12499 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $x, y$ are positive integers. And they satisfy the conditions $x y+x+y=71, x^{2} y+x y^{2}=880$.
Find the value of $x^{2}+y^{2}$.
(1999, Jiangsu Province Junior High School Mathematics Competition) | Answer: 146.) | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. If $\sin \alpha \cdot \cos \beta=1$, then $\cos \alpha \cdot \sin \beta=$ | 12.0.
From $\sin \alpha \cdot \cos \beta=1$ and the boundedness of sine and cosine functions, we know that $\sin \alpha=\cos \beta=1$ or $\sin \alpha=\cos \beta=-1$. Therefore, $\cos \alpha=\sin \beta=0$, which means $\cos \alpha \cdot \sin \beta=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Simplify $\log _{\sqrt{2}} \sin \frac{7 \pi}{8}+\log _{\sqrt{2}} \sin \frac{3 \pi}{8}$, the result is $\qquad$ . | 13. -3 .
$$
\begin{array}{l}
\text { Original expression }=\log _{\sqrt{2}}\left(\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}\right)=\log _{\sqrt{2}}\left(\frac{1}{2} \sin \frac{\pi}{4}\right) \\
=\log _{\sqrt{2}} \frac{\sqrt{2}}{4}=\log _{2} \frac{1}{2} 2^{-\frac{3}{2}}=-3 .
\end{array}
$$ | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
18. Calculate $1^{2}-2^{2}+3^{2}-4^{2}+\cdots+2005^{2}-$ $2006^{2}=$ $\qquad$ . | 18. -2013021 .
$$
\begin{array}{l}
1^{2}-2^{2}+3^{2}-4^{2}+\cdots+2005^{2}-2006^{2} \\
=\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right)+\cdots+\left(2005^{2}-2006^{2}\right) \\
=(1+2)(1-2)+(3+4)(3-4)+\cdots+ \\
(2005+2006)(2005-2006) \\
=-(1+2+3+\cdots+2006)=-2013021 .
\end{array}
$$ | -2013021 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $\sqrt{x^{2}+2008}+\sqrt{x^{2}-2006}=$ 2007. Then
$$
3013 \sqrt{x^{2}-2006}-3005 \sqrt{x^{2}+2008}
$$
is ( ).
(A) 2008
(B) 2009
(C) 2010
(D) 2011 | 2.C.
Let $a=\sqrt{x^{2}+2008}, b=\sqrt{x^{2}-2006}$, then $a^{2}-b^{2}=4014$,
which means $(a+b)(a-b)=4014$.
Given $a+b=2007$. Therefore, $a-b=2$.
Solving, we get $a=1004.5, b=1002.5$.
Thus, the original expression $=3013 b-3005 a=8 b-3005(a-b)$
$$
=8 \times 1002.5-3005 \times 2=2010 \text {. }
$$ | 2010 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. Given that $x, y$ are integers, and
$$
y=\frac{4012}{\sqrt{x+2005}-\sqrt{x-2007}} \text {. }
$$
Then the maximum value of $y$ is $\qquad$ . | 3.2006 .
Rationalizing the denominator of the given equation, we get
$$
y=\sqrt{x+2005}+\sqrt{x-2007} \text {. }
$$
Let $m=\sqrt{x+2005}, n=\sqrt{x-2007}$.
Then $y=m+n, m^{2}=x+2005, n^{2}=x-2007$ $(m>n \geqslant 0)$, obviously $m, n$ are integers.
Eliminating $x$ yields $m^{2}-n^{2}=4012$, that is,
$$
(m+n)(m-n)=2^{... | 2006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) According to the "Personal Income Tax Law of the People's Republic of China," citizens do not need to pay tax on their monthly wages and salaries up to 800 yuan, and the portion exceeding 800 yuan is the taxable income for the whole month. In October 2005, the Standing Committee of the National People'... | Let's assume that Mr. A's total monthly salary and wage income in January 2006 is $x$ yuan, the tax due is $y_{1}$ yuan, the tax due in the past is $y_{2}$ yuan, and the underpaid tax is $y$ yuan, then $y=y_{2}-y_{1}$.
When $0 \leqslant x \leqslant 1600$, $y_{1}=0$.
When $1600<x \leqslant 2100$, $y_{1}=5 \% \cdot(x-160... | 1590 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Find the smallest positive integer $k$, such that from the 4006 numbers $1,2, \cdots, 4006$, any $k$ different numbers chosen will always include 4 numbers whose sum is 8013. | Three, from $1,2, \cdots, 4006$ take out the last 2005 numbers: $2002,2003, \cdots, 4006$,
then the sum of any four different numbers is not less than
$$
2002+2003+2004+2005=8014>8013 \text {. }
$$
Thus, $k \geqslant 2006$.
Let $x_{1}, x_{2}, \cdots, x_{2006}$ be any 2006 numbers from $1,2, \cdots, 4006$.
First, divi... | 2006 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the complex numbers $z_{1}, z_{2}$ satisfy
$$
\left|z_{1}-z_{2}\right|=\sqrt{3}, z_{1}^{2}+z_{1}+q=0, z_{2}^{2}+z_{2}+q=0 \text {. }
$$
then the real number $q=$ $\qquad$ | 2.1.
From $\left|z_{1}-z_{2}\right|=\sqrt{3}$, we know $z_{1} \neq z_{2}$.
By the definition of the roots of the equation, $z_{1}$ and $z_{2}$ are the two imaginary roots of the quadratic equation $x^{2}+x+q=0$, then we have
$$
\Delta=1-4 q<0 \text{. }
$$
Solving the equation, we get $z_{1,2}=\frac{-1 \pm \sqrt{4 q-1... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $A$ is a subset of $S=\{1,2,3,4,5,6\}$ with at least 2 elements, and $a, b$ are two distinct elements in $A$. When $A$ ranges over $S$ and $a, b$ range over $A$, the total sum of the product $ab$ is $M=$ $\qquad$ | 6.2800 .
For any $\{a, b\} \subset S$ (assuming $a<b$), the number of $A$ that satisfies $\{a, b\} \subseteq A \subseteq S$ is $2^{6-2}=2^{4}$. Therefore, the contribution of $\{a, b\}$ to the total sum $M$ is $a b \times 2^{4}$.
When $a$ takes all values of $1,2,3,4,5$ and $b$ takes all values of $2,3,4,5,6$, we have... | 2800 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
f(n)=\left\{\begin{array}{ll}
1, & n=1 \text { when; } \\
2, & 1<n \leqslant 3 \text { when; } \\
3, & 3<n \leqslant 6 \text { when; } \\
\cdots \cdots . & \\
m, & \frac{m(m-1)}{2}<n \leqslant \frac{m(m+1)}{2} \text { when; } \\
\cdots \cdots . &
\end{array}\right.
$$
If $S_{n}=\sum_{k=1}^{n} f(k)=2001$, find the v... | By $1^{2}+2^{2}+\cdots+17^{2}=1785$,
$$
1^{2}+2^{2}+\cdots+18^{2}=2109,
$$
and $1785<2001<2109$,
we know that the last term of the sum $S_{n}=f(1)+f(2)+\cdots+f(n)$ appears in the 18th row, thus, $f(n)=18$.
Also, $1+2+\cdots+17=153$, so $S_{n}$ can be divided into two parts: the first part is the sum of the numbers i... | 165 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Place the numbers $1,2, \cdots, 8$ in the 8 squares around the perimeter of a $3 \times 3$ chessboard (as shown in Figure 1), such that the sum of the absolute values of the differences between adjacent numbers (numbers in squares that share a common edge) is maximized. Find this maximum value. | To make the description convenient, we define the difference between two adjacent numbers as the larger number minus the smaller number, thus avoiding the concept of absolute value.
To maximize the sum of the 8 differences, we want each difference to be as large as possible. For example, the largest number 8 should be... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Fill the numbers $1,2, \cdots, 9$ into the squares of a $3 \times 3$ chessboard (as shown in Figure 1), so that the sum of the absolute values of the differences between numbers in adjacent (sharing a common edge) squares is maximized. Find this maximum value. | Solution: To avoid the concept of absolute value, we now define each difference as the larger number minus the smaller number.
To maximize the sum of the 12 differences, we hope each difference is as large as possible, which means the minuend should be as large as possible and the subtrahend as small as possible.
Not... | 58 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Fill the numbers $1 \sim 8$ into the 8 squares surrounding the four edges of a $3 \times 3$ chessboard (as shown in Figure 1), so that the sum of the differences (the larger number minus the smaller number) of the adjacent (sharing a common edge) two numbers in these 8 squares is minimized. Find this minimum ... | To make the sum of the 8 differences as small as possible, we hope that each difference is as small as possible. For example, the largest number 8 should have 7 and 6 on its sides, making the differences 1 and 2, which are the smallest (as shown in Figure 6). Therefore, 5 should be placed next to 6 to minimize the diff... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Fill the 9 cells in Figure 8 with the numbers $1 \sim 9$ respectively, so that the sum of the differences (the larger number minus the smaller number) of the adjacent (having a common edge) two numbers in these 9 cells is maximized. Find this maximum value. | To maximize the sum of the 9 differences, we hope that the 9 minuends are as large as possible and the 9 subtrahends are as small as possible. For example, the largest number 9 should be adjacent to the smallest number 1 and the second smallest number 2, the second largest number 8 should be adjacent to 1, and then to ... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
5. $a$ and $b$ are rational numbers, and satisfy the equation
$$
a+b \sqrt{3}=\sqrt{6} \times \sqrt{1+\sqrt{4+2 \sqrt{3}}} \text {. }
$$
Then the value of $a+b$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 5. B.
Since $\sqrt{6} \times \sqrt{1+\sqrt{4+2 \sqrt{3}}}=\sqrt{6} \times \sqrt{1+(1+\sqrt{3})}$ $=\sqrt{12+6 \sqrt{3}}=3+\sqrt{3}$,
thus, $a+b \sqrt{3}=3+\sqrt{3}$, which means $(a-3)+(b-1) \sqrt{3}=0$.
Since $a, b$ are rational numbers, then $a=3, b=1$, hence $a+b=4$. | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. The sum of the x-coordinates of the x-axis intercepts of the graph of the function $y=x^{2}-2006|x|+2008$ is $\qquad$ | $=.1 .0$.
The original problem is transformed into solving the equation
$$
x^{2}-2006|x|+2008=0
$$
We are to find the sum of all real roots.
If a number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1). Therefore, the sum of all real roots of the equation is 0, i.e., ... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
The coordinates of the top points are $0(0,0), A(100,0)$, $B(100,100), C(0,100)$. If a lattice point $P$ is inside the square $O A B C$,
then the lattice point $P$ is called a "good point". The number of good points inside the square $O A B C$ is . $\qquad$ | 4. 197.
As shown in Figure 6, through point $P$, draw
$P I, P E, P F, P G$ perpendicular to sides
$O A, A B, B C, O C$ at $D, E$,
$F, G$. It is easy to see that
$$
\begin{array}{l}
P F+P D=100, \\
P E+P G=100 \text {. } \\
\text { If| } S_{\text {эхи }} \cdot S_{\text {эми: }} \\
=S_{\triangle P U} \cdot S_{\triangle ... | 197 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given the equation in terms of $x$
$$
x^{2}+2(a+2 b+3) x+\left(a^{2}+4 b^{2}+99\right)=0
$$
has no distinct real roots. How many ordered pairs of positive integers $(a, b)$ satisfy this condition? | Given that $x^{2}+2(a+2 b+3) x+\left(a^{2}+4 b^{2}+99\right)$ $=0$ has no two distinct real roots, therefore,
$$
\Delta=[2(a+2 b+3)]^{2}-4\left(a^{2}+4 b^{2}+99\right) \leqslant 0 \text {. }
$$
Simplifying to $2 a b+3 a+6 b \leqslant 45$, then
$$
(a+3)(2 b+3) \leqslant 54 \text {. }
$$
Since $a, b$ are positive integ... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If $(x+1)(y+1)=x^{2} y+x y^{2}=6$, then, $x^{2}+y^{2}$ equals $(\quad)$.
(A) 6
(B) 5
(C) 4
(D) 3 | 5.B.
From $\left\{\begin{array}{l}x y+(x+y)=5, \\ x y(x+y)=6,\end{array}\right.$ we get $\left\{\begin{array}{l}x+y=3, \\ x y=2\end{array}\right.$ or $\left\{\begin{array}{l}x+y=2, \\ x y=3\end{array}\right.$ (discard). Therefore, $x^{2}+y^{2}=(x+y)^{2}-2 x y=9-4=5$. | 5 | Algebra | MCQ | Yes | Yes | cn_contest | false |
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