problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
3. The real number $x$ satisfies
$$
\sqrt{x^{2}+3 \sqrt{2} x+2}-\sqrt{x^{2}-\sqrt{2} x+2}=2 \sqrt{2} x \text {. }
$$
Then the value of the algebraic expression $x^{4}+x^{-4}$ is $\qquad$ | 3.14.
Obviously $x \neq 0$.
Let $\sqrt{x^{2}+3 \sqrt{2} x+2}=a$,
$$
\sqrt{x^{2}-\sqrt{2} x+2}=b \text {. }
$$
Then $a^{2}-b^{2}=(a+b)(a-b)=4 \sqrt{2} x$.
Also, $a-b=2 \sqrt{2} x$, so, $a+b=2$.
Thus, $a=\sqrt{2} x+1$.
Then $(\sqrt{2} x+1)^{2}=x^{2}+3 \sqrt{2} x+2$.
Rearranging gives $x^{2}-\sqrt{2} x-1=0$, which is $x... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) A positive integer $M$ has 8 positive divisors. Xiao Feng, while calculating the sum of the 8 positive divisors of $M$, forgot to add one of the divisors, resulting in a sum of 2776. Find all possible values of the positive integer $M$.
| Three, if $M$ has at least four different prime factors, then the number of positive divisors of $M$ is at least $(1+1)^{4}=16$, which contradicts the problem statement.
Therefore, $M$ has at most three different prime factors.
If all prime factors of $M$ are odd, then all positive divisors of $M$ are odd. In this case... | 2008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $\triangle A B C$ with side lengths $a=17, b=18, c$ $=19$, a point $O$ inside $\triangle A B C$ is drawn perpendiculars to the three sides of $\triangle A B C$, intersecting at points $D$, $E$, and $F$, such that $B D+C E+A F=27$. Then $B D+B F$ $=$ $\qquad$ | 2.18.
As shown in Figure 5, let \( BD = x \), \( CE = y \), \( AF = z \). Then
\[
\begin{array}{l}
CD = 17 - x, \\
AE = 18 - y, \\
BF = 19 - z.
\end{array}
\]
Connect \( OA \), \( OB \), \( OC \).
In right triangles \( \triangle OBD \), \( \triangle OCE \), \( \triangle OAF \),
\[
\left\{\begin{array}{l}
x^{2} + OD^{... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. A positive integer $M$, when added to 36, 300, and 596 respectively, results in the squares of three different positive integers. The sum of the smallest and the largest of these three different positive integers is twice the middle one. Then this positive integer $M=$ | 3.925 .
According to the problem, let $M+36=(k-m)^{2}$,
$M+300=k^{2}$,
$M+596=(k+m)^{2}(m>0)$.
Then $(M+36)+(M+596)=2\left(k^{2}+m^{2}\right)$.
Therefore, $M+316=M+300+16=k^{2}+m^{2}$.
Solving this, we get $m=4$.
$$
\begin{array}{l}
\text { Also, }(M+596)-(M+36) \\
=(k+m)^{2}-(k-m)^{2}=4 k m,
\end{array}
$$
Thus, $k=... | 925 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If $a-2$ is a positive integer and a divisor of $3 a^{2}-2 a+10$, then the sum of all possible values of $a$ is $\qquad$ . | 2.51.
$$
\begin{array}{l}
\text { Given } \frac{3 a^{2}-2 a+10}{a-2} \\
=\frac{\left(3 a^{2}-6 a\right)+(4 a-8)+18}{a-2} \\
=3 a+4+\frac{18}{a-2}(a \neq 2),
\end{array}
$$
we know that $a-2$ must be a divisor of 18.
Thus, $a=3,4,5,8,11,20$.
Therefore, the sum of all possible values of $a$ is
$$
3+4+5+8+11+20=51 \text ... | 51 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $x$ is a real number. Then the maximum value of $\sqrt{2000-x}+$ $\sqrt{x-2000}$ is $\qquad$ . | 4.4.
$$
\begin{array}{l}
\text { Let } t_{1}=\sqrt{2008-x}+\sqrt{x-2000} \text {. Then } \\
t_{1}^{2}=8+2 \sqrt{(2008-x)(x-2000)} \\
\leqslant 8+8=16 .
\end{array}
$$
Therefore, $t \leqslant 4$, i.e., $t_{\max }=4$. At this point, $x=2004$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. A ten-digit number, whose digits can only be 2 or 3, and there are no two adjacent 3s. How many such ten-digit numbers are there?
保留源文本的换行和格式，直接输出翻译结果。 | 5.144.
Solution 1: Consider all positive integers with the given property (using a recursive method).
The number of ten-digit numbers starting with 2 is the same as the number of nine-digit numbers;
Ten-digit numbers starting with 3 must start with 32, and their number is the same as the number of eight-digit number... | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. There are $n(n>3)$ integers on a circle with a sum of 94, each of which is equal to the absolute value of the difference between the two numbers that follow it (in a clockwise direction). Then all possible values of $n$ are | 6.141 .
Obviously, these $n$ integers are all non-negative.
Let $a$ be the largest one, and the next four numbers be $b, c, d, e$ respectively. Then $a=|b-c|$.
Thus, $b=a$ or $c=a$.
When $b=a$, $c=0$.
From $b=|c-d|$, we know $d=a$.
Also, from $c=|d-e|$, we get $d=e$.
Therefore, $n$ is a multiple of 3.
Assume $n=3m$. T... | 141 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that $M$ is a four-digit perfect square. If the thousand's digit of $M$ is reduced by 3 and the unit's digit is increased by 3, another four-digit perfect square can be obtained. Then the value of $M$ is $\qquad$ | 7.4761 .
Let $M=\overline{a b c d}=A^{2}$ and $M$'s thousands digit decreases by 3 and the units digit increases by 3 to equal $B^{2}$. Then
$$
\left\{\begin{array}{l}
A^{2}=1000 a+100 b+10 c+d, \\
B^{2}=1000(a-3)+100 b+10 c+(d+3) .
\end{array}\right.
$$
Therefore, $A^{2}-B^{2}=2997$.
Hence $(A-B)(A+B)=3^{4} \times 3... | 4761 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 2, each segment of the broken line $A-B-C-D$ is parallel to the sides of the rectangle, and it divides the rectangle into two equal areas. Point $E$ is on the side of the rectangle such that segment $A E$ also bisects the area of the rectangle.
Given that segment $A B=30, B C=$
$24, C D=10$. Then ... | 8.12.
Solution 1: Let line segment $A E$ intersect $B C$ at point $M$, and draw $E P \perp B C$, with the foot of the perpendicular being $P$.
Let $E D=x, B M=y$. Then $M P=24-x-y$.
By the problem, $S_{\triangle B B Y}=S_{\text {quadrilateral } U C D E}$, then
$15 y=120-5 y+5 x$.
Thus, $4 y=24+x$.
Also, $\triangle A M... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Given the function $f(x)=a x^{2}-c(a, c$ are real numbers). If $-4 \leqslant f(1) \leqslant-1,-1 \leqslant f(2) \leqslant 2$, then the maximum value of $f(8)$ is $\qquad$ . | 9.122.
Since $f(1)=a-c, f(2)=4a-c$, therefore, $a=\frac{1}{3}(f(2)-f(1)), c=\frac{1}{3}(f(2)-4f(1))$.
Thus, $f(8)=64a-c=21f(2)-20f(1)$.
Hence $-1=21 \times(-1)-20 \times(-1) \leqslant f(8)$ $\leqslant 21 \times 2-20 \times(-4)=122$.
When $f(x)=2x^{2}-6$, it satisfies the given conditions, and $f(8)=122$. Therefore, t... | 122 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. The smallest positive integer $n$ for which $n^{2}-n+11$ has four prime factors (not necessarily distinct) is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 11.132.
From Table 1, it can be verified that for any positive integer $n, n^{2}-$ $n+11$ is not a multiple of $2,3,5,7$.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $\bmod 2$ & $\bmod 3$ & \multicolumn{3}{|c|}{$\bmod 5$} & \multicolumn{3}{|c|}{$\bmod 7$} \\
\hline$n$ & 0 & 1 & 0 & 1 & 2 & ... | 132 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. (40 points) If the simplest fraction $\frac{p}{q}$ is written in decimal form as $0 . a b a b a b a b \cdots$ (non-negative integers $a$ and $b$ can be equal, but at least one is non-zero), then, among the fractions that meet the condition, how many different numerators are there? | 1. Decimal numbers in the form of 0.abababab $\cdots$ can all be written as $\frac{k}{99}$ $\left(k \in \mathbf{N}_{+}, k=1,2, \cdots, 98\right)$. Among the numbers $1,2, \cdots$, 99, those that are multiples of 3 or 11 are
$$
\left[\frac{99}{3}\right]+\left[\frac{99}{11}\right]-\left[\frac{99}{3 \times 11}\right]=39 \... | 63 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. (40 points) A rectangular box with dimensions $a_{1} \times b_{1} \times c_{1}$ can fit into another rectangular box with dimensions $a_{2} \times b_{2} \times c_{2}$ if and only if $a_{1} \leqslant a_{2} 、 b_{1} \leqslant b_{2} 、 c_{1} \leqslant c_{2}$. Therefore, among the rectangular boxes with dimensions $a \tim... | 4. Among the boxes that meet the conditions, there are 5 types of boxes that are cubes, and 20 types of boxes that are square-based but not cubic prisms, because they have five different heights and four different choices for the square side length.
In addition, there are 10 types of boxes where the length, width, and... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. (40 points) On the blackboard, the numbers 1 and 4 are originally written. For the numbers on the blackboard, an operation can be performed: each time, any two numbers can be chosen from the blackboard, and a new number \( c = ab + a + b \) can be added to the blackboard. After several operations, what is the smalle... | 6. From $c_{1}=a b+a+b=(a+1)(b+1)-1$, we have
$$
c_{1}+1=(a+1)(b+1) \text {. }
$$
Taking numbers $a$ and $c_{1}$, we get
$$
c_{2}+1=(a+1)\left(c_{1}+1\right)=(a+1)^{2}(b+1) \text {; }
$$
Taking numbers $b$ and $c_{1}$, we get
$$
c_{2}+1=(b+1)\left(c_{1}+1\right)=(a+1)(b+1)^{2} \text {; }
$$
$\qquad$
Suppose after sev... | 2047 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7.(40 points) As shown in Figure 7, quadrilateral $ABCD$ is inscribed in a circle, $AB=AD$, and its diagonals intersect at point $E$. Point $F$ lies on segment $AC$ such that $\angle BFC = \angle BAD$. If $\angle BAD = 2 \angle DFC$, find the value of $\frac{BE}{DE}$.
保留源文本的换行和格式，直接输出翻译结果。 | 7. From $AB=AD$, we know $\angle ABD=\angle ADB=\theta$. By the property of equal arcs subtending equal angles at the circumference, we have
$$
\angle ACD=\angle ACB=\theta.
$$
Let $\angle DFC=\varphi$. Then
$$
\angle BAD=\angle BFC=2\varphi.
$$
Therefore, $\angle ABD + \angle ADB + \angle BAD$
$$
=\theta + \theta + ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. A sequence of numbers, the first three numbers are $1, 9, 9$, and each subsequent number is the remainder of the sum of the three preceding numbers divided by 3. What is the 1999th number in this sequence? | (Tip: Apart from the first three numbers $1,9,9$, this sequence repeats every 13 terms (i.e., $1,1,2,1,1,1,0,2,0,2,1,0,0$). Since $1999-3=13 \times 153+7$, the 1999th number is the 7th number in the 154th cycle, which is exactly 0.) | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Observe the array:
(1),(3,5),(7,9,11),(13,15,17,19), .....
In which group is 2003? | (Tip: Use the trial method. The first 45 groups have a total of $1+2+\cdots$ $+45=1035$ numbers. The last number of the 45th group is 2069, and the first number is 1981, so, 2003 is in the 45th group.) | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Several 1s and 2s are arranged in a row
$$
1,2,1,2,2,1,2,2,2,1,2, \cdots
$$
The rule is: the 1st number is 1, the 2nd number is 2, the 3rd number is 1, ... Generally, first write a row of 1s, then insert $k$ 2s between the $k$th 1 and the $(k+1)$th 1 ($k=1$, $2, \cdots$). Try to answer:
(1) Is the 2005th num... | Explanation: Clearly, the position of 1 is somewhat special. For the convenience of calculation, we might as well divide this sequence of numbers into $n$ groups: the 1st group has 1 number, the 2nd group has 2 numbers, the 3rd group has 3 numbers, ..., the $n$th group has $n$ numbers, and the last number of each group... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. If $a, b$ are integers, and the equation $a x^{2}+b x-$ $2008=0$ has two distinct prime roots, then the value of $3 a+b$ is ( ).
(A) 100
(B) 400
(C) 700
(D) 1000 | 6.D.
Let the roots of the equation be $x_{1}, x_{2}\left(x_{1}<x_{2}\right)$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, \\
x_{1} x_{2}=-\frac{2008}{a},
\end{array}\right.
$$
which means
$$
\left\{\begin{array}{l}
b=-a\left(x_{1}+x_{2}\right), \\
a x_{1} x_{2}=-2008=-2 \times 2 \times 2 \times 251.
\en... | 1000 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given real numbers $a>0, b>0$, satisfying $a+\sqrt{a}=2008, b^{2}+b=2008$. Then the value of $a+b$ is $\qquad$ | $=, 1.2008$
For the convenience of calculation, let $2008=m$.
From $a+\sqrt{a}-m=0$, we get
$$
\sqrt{a}=\frac{-1+\sqrt{1+4 m}}{2} \text{, }
$$
which means $a=\frac{1+2 m-\sqrt{1+4 m}}{2}$.
From $b^{2}+b-m=0$, we get
$$
b=\frac{-1+\sqrt{1+4 m}}{2} \text{. }
$$
Therefore, $a+b=m=2008$. | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Arrange positive integers starting from 1 according to the pattern shown in Figure 2, where 2 is at the first turn, 3 is at the second turn, 5 is at the third turn, 7 is at the fourth turn, … then the number at the 2007th turn is $\qquad$ | To explore the relationship between the number of turns and the numbers at the turns, let $a_{n}$ represent the number corresponding to the $n$-th turn. If we observe in the order of natural numbers, then
$$
\begin{array}{l}
a_{1}=2, a_{2}=3, a_{3}=5, a_{4}=7, a_{5}=10, \\
a_{6}=13, a_{7}=17, a_{8}=21, \cdots \cdots
\e... | 1008017 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11.2. Given that the weight ratio of any two out of $N$ weights belongs to $\left[\frac{4}{5}, \frac{5}{4}\right]$, and these $N$ weights can be divided into 10 groups of equal weight, as well as 11 groups of equal weight. Find the minimum possible value of $N$. | 11.2. $N_{\min }=50$.
An example for $N=50$.
In fact, 20 weights of $50 \mathrm{~g}$ and 30 weights of $40 \mathrm{~g}$ satisfy the requirement. These 50 weights can be divided into 10 equal groups, each consisting of 2 weights of $50 \mathrm{~g}$ and 3 weights of $40 \mathrm{~g}$. They can also be divided into 11 equa... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If $\sqrt{m^{2}+1203}$ is an integer, then the sum of all positive integers $m$ that satisfy the condition is ( ).
(A)401
(B) 800
(C) 601
(D) 1203 | 4.B.
Let $\sqrt{m^{2}+1203}=n\left(n \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
n^{2}-m^{2}=1203 \\
\Rightarrow(n+m)(n-m)=401 \times 3=1203 \times 1 .
\end{array}
$$
Therefore, $\left\{\begin{array}{l}n+m=401, \\ n-m=3\end{array}\right.$ or $\left\{\begin{array}{l}n+m=1203, \\ n-m=1 .\end{array}\right.$
Sol... | 800 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
4. As shown in Figure $2, \odot O$ is internally tangent to $\odot O^{\prime}$ at point $P, \odot O$'s chord $A B$ is tangent to $\odot O^{\prime}$ at point $C$, and $A B / / O O^{\prime}$. If the area of the shaded part is $4 \pi$, then the length of $A B$ is $\qquad$ | 4.4.
As shown in Figure 10, connect $O^{\prime} C$ and $O A$. Draw $O D \perp A B$ at $D$. Then quadrilateral $O O^{\prime} C D$ is a rectangle.
Therefore, $O^{\prime} C=O D$.
Thus, $S_{\text {fill }}$
$=\pi O A^{2}-\pi O^{\prime} C^{2}$ $=\pi\left(O A^{2}-O D^{2}\right)=\pi A D^{2}=4 \pi$.
So, $A D=2$.
By the Perpend... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The digit at the 2007th position after the decimal point of the irrational number $0.2342343423434342343434342 \cdots$ is $\qquad$ . | (Observation: Note the position of the digit 2 after the decimal point. The $n$th 2 is at the $n^{2}$th position after the decimal point. Since $44^{2}<2007<45^{2}$, the digit at the 2007th position after the decimal point is between the 44th and 45th 2, with 44 instances of 34 in between. The digit 3 is in the odd pos... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given a height of 241, find the minimum value of the function
$$
S_{n}=|n-1|+2|n-2|+\cdots+100|n-100|
$$
$\left(n \in \mathbf{N}_{+}\right)$. | Solution: When $n \geqslant 100$,
$$
\begin{array}{l}
S_{n}=(n-1)+2(n-2)+\cdots+100(n-100) \\
=n(1+2+\cdots+100)-\left(1^{2}+2^{2}+\cdots+10^{2}\right) \\
\geqslant S_{100} .
\end{array}
$$
Therefore, the minimum value of $S_{n}$ can only be obtained when $1 \leqslant n \leqslant 100$.
When $2 \leqslant n \leqslant 10... | 99080 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find four distinct natural numbers such that the sum of any two of them can be divided by their difference. If the sum of the largest and smallest of these four numbers is to be minimized, what is the sum of the middle two numbers?
(3rd Hua Luogeng Cup) | Analysis: Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four numbers that meet the conditions, and $a_{1}8$, which contradicts the condition that $a_{1}+a_{4}$ is the smallest.
In summary, $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(2,3,4,6)$, the sum of the middle two numbers is 7. . | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If a natural number $N$ is appended to the right of any natural number, the resulting number can be divided by $N$ (for example, 2 appended to 35 results in 352, which is divisible by 2), then $N$ is called a "magic number". Among the natural numbers less than 130, how many magic numbers are there? | Analysis: To calculate how many magic numbers there are, we first need to clarify what constitutes a magic number. Although the problem provides the definition of a magic number, it is not convenient to directly use this definition to determine whether a number is a magic number. Therefore, we need to explore a simplif... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Initially 243 Hexagon $A B C D E F$ is inscribed in $\odot O$, $A F // D C, F E // C B, E D // B A, A B+B C=$ $2 C D$. Constructing squares on each of the six sides, the sum of the areas of these six squares is 2008. Find the perimeter of hexagon $A B C D E F$. | Solution: As shown in Figure 3, from $A F$
$/ / D C, F E / / C B, E D / /$
$B A$, we know that $A D, B E, C F$ are
all diameters of $\odot O$, thus
$$
\begin{array}{l}
D E=A B, \\
E F=B C, \\
F A=C D .
\end{array}
$$
Let $C D=x, A B=x-d$. Then
$$
\begin{array}{l}
B C=x+d, D E=x-d, \\
A F=x, E F=x+d .
\end{array}
$$
A... | 108 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
244 Given real numbers $a, b, c, d$ satisfy
$$
a+b+c+d=ab+ac+ad+bc+bd+cd=3 \text{. }
$$
Find the maximum real number $k$, such that the inequality
$$
a+b+c+2ab+2bc+2ca \geqslant k d
$$
always holds. | Given:
$$
\begin{aligned}
& (a+b+c+d)^{2}=3^{2} \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}+2 a b+2 a c+2 a d+ \\
& 2 b c+2 b d+2 c d=9 \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}=3 .
\end{aligned}
$$
Let \( y=(x-a)^{2}+(x-b)^{2}+(x-c)^{2} \).
Then \( y=3 x^{2}-2(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right) \).
Since \( y \geqsl... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given the function $f(x)=x+\frac{t}{x}(t>0)$ and the point $P(1,0)$, draw two tangent lines $P M$ and $P N$ from $P$ to the curve $y=f(x)$, with the points of tangency being $M$ and $N$.
(1) Let $|M N|=g(t)$, find the expression for the function $g(t)$.
(2) Does there exist a $t$ such that $M$, $N$, and $A(0,... | Solution: (1) The equation of the chord of tangents $MN$ passing through the external point $P(1,0)$ of the curve $y=x+\frac{t}{x}(t>0)$ is
$$
y=2 x+2 t \text {. }
$$
Substituting into $y=x+\frac{t}{x}(t>0)$, we get $x^{2}+2 t x-t=0$.
Thus, $x_{1}+x_{2}=-2 t, x_{1} x_{2}=-t \ldots$
Therefore, $g(t)=\sqrt{\left(x_{1}-x... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
110 players participate in a table tennis tournament, with each pair of players playing one match. If player $i$ beats player $j$, player $j$ beats player $k$, and player $k$ beats player $i$, this is called a "triangle." Let $W_{i}$ and $L_{i}$ represent the number of wins and losses, respectively, for the $i$-th play... | Solution: Let the points $-v_{1}, v_{2}, \cdots, v_{10}$ represent the 10 players. If $v_{i}$ beats $v_{j}$, then draw a directed edge between the corresponding points: $v_{i} \rightarrow v_{j}$. Thus, we obtain a 10-vertex directed graph $G$, then
$$
\text { and } \begin{aligned}
d^{+}\left(v_{i}\right) & =W_{i}, d^{-... | 40 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 5 Choose $n$ numbers from $1,2, \cdots, 9$, among which there must be several numbers (at least one, or all), the sum of which can be divisible by 10. Find the minimum value of $n$.
(2008, National Junior High School Mathematics Competition) | Solution: When $n \leqslant 4$, since no combination of $n$ numbers from $1,3,5,8$ can sum to a multiple of 10, we have $n \geqslant 5$.
Below, we prove that $n=5$ meets the requirement, i.e., we prove that from $1,2, \cdots, 9$, any 5 numbers chosen will always include some numbers whose sum is a multiple of 10.
We u... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. As shown in Figure 5, in quadrilateral $A B C D$, $\angle A=$ $\angle B C D=90^{\circ}, B C=$ $C D, E$ is a point on the extension of $A D$. If $D E=A B$ $=3, C E=4 \sqrt{2}$, then the length of $A D$ is | 11.5.
In Figure 5, connect $A C$. It is easy to prove that
$\triangle C D E \cong \triangle C B A, \angle A C E=90^{\circ}$.
Since $C A=C E=4 \sqrt{2}$, therefore, $A E=8$.
Thus, $A D=5$. | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. Xiao Wang walks along the street at a constant speed, and he notices that a No. 18 bus passes him from behind every $6 \mathrm{~min}$, and a No. 18 bus comes towards him every $3 \mathrm{~min}$. Assuming that each No. 18 bus travels at the same speed, and the No. 18 bus terminal dispatches a bus at fixed intervals,... | 13.4 .
Let the speed of bus No. 18 be $x \mathrm{~m} / \mathrm{min}$, the walking speed of Xiao Wang be $y \mathrm{~m} / \mathrm{min}$, and the distance between two consecutive buses traveling in the same direction be $s \mathrm{~m}$. From the problem, we have
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s .
\end{... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. There are 6 seats arranged in a row, and three people are to be seated, with exactly two adjacent empty seats. The number of different seating arrangements is ( ).
(A) 48
(B) 60
(C) 72
(D) 96 | 7.C.
Arrange the seats in order as $1,2, \cdots, 6$.
If the adjacent empty seats are 1 and 2 or 5 and 6, there are $\mathrm{C}_{3}^{1} \cdot \mathrm{A}_{3}^{3}$ arrangements each; if the adjacent empty seats are 2 and 3, 3 and 4, or 4 and 5, there are $\mathrm{C}_{2}^{1} \cdot \mathrm{A}_{3}^{3}$ arrangements each.
T... | 72 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
2. Given 99 | $\overline{62 x y 427}$. Find $950 x+24 y+1$. (1997, Hope Cup Mathematics Invitational)
Find the value of $950 x+24 y+1$ if 99 divides $\overline{62 x y 427}$. (1997, Hope Cup Mathematics Invitational) | (提示: From $91 \overline{62 x y 427}$, we get $91(x+y+21)$. Therefore, $9 \mid(x+y+3)$. Noting that $3 \leqslant x+y+3 \leqslant$ 21, hence $x+y=6$ or 15. From $11 \overline{62 x y 427}$, we get $x-$ $y=-2$ or 9. Also note that $x+y$ and $x-y$ have the same parity, solving gives $(x, y)=(2,4)$. Therefore, $950 x+24 y+1=... | 1997 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. For a real number $x$, $[x]$ denotes the greatest integer not exceeding the real number $x$. It is known that the sequence of positive numbers $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right),
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\... | 10. B.
Notice
$$
\begin{array}{l}
S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)=\frac{1}{2}\left[\left(S_{n}-S_{n-1}\right)+\frac{1}{S_{n}-S_{n-1}}\right], \\
S_{n}+S_{n-1}=\frac{1}{S_{n}-S_{n-1}}, \\
S_{n}^{2}=S_{n-1}^{2}+1 .
\end{array}
$$
Since $S_{1}=a_{1}=1$, we have $S_{n}^{2}=n, S_{n}=\sqrt{n}$. Therefor... | 18 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. For a positive integer $n$, its decimal representation consists only of the digits 0 and 1, and it is divisible by 225. Find the minimum value of $n$.
| (Tip: From $25 \mid n$, we know that the last two digits of $n$ are multiples of 25. But since all the digits of $n$ are 0 and 1, the last two digits of $n$ must be 00. From $91n$, we know that the sum of the digits of $n$ is a multiple of 9, so $n$ must contain at least 9 ones. Therefore, the smallest value of $n$ is ... | 11111111100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$1 . \underbrace{66 \cdots}_{100 \uparrow} \underbrace{77 \cdots 7}_{100 \uparrow} \underbrace{88 \cdots 8}_{100 \uparrow} 8$ divided by 72 has a remainder of | $$
\begin{array}{l}
\underbrace{66 \cdots 6}_{100 \uparrow} \underbrace{77 \cdots}_{100 \uparrow} \underbrace{88 \cdots 8}_{100 \uparrow} \\
=\underbrace{66 \cdots 677 \cdots}_{10 \uparrow} \underbrace{77}_{100 \uparrow} \underbrace{88 \cdots 82000}_{96 \uparrow}+72 \times 95+48 . \\
\underbrace{66 \cdots}_{100 \uparro... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. In trapezoid $A B C D$, $A D / / B C, A B=A C$, $B C=B D=(\sqrt{2}+1) C D$. Then the degree measure of $\angle B A C+\angle B D C$ is $\qquad$ . | 2.180 .
As shown in Figure 5, draw $A E \perp B C$ at $E$ and $D F \perp B C$ at $F$. Then
$$
\begin{array}{l}
B E=E C=\frac{1}{2} B C, \\
\angle C A E=\frac{1}{2} \angle B A C,
\end{array}
$$
and quadrilateral $A E F D$ is a rectangle. | 180 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $n$ is an integer, and the quadratic equation in $x$
$$
(n-1)^{2} x^{2}-5 n(n-1) x+\left(6 n^{2}-n-1\right)=0
$$
has at least one integer root. Then the sum of all possible values of $n$ is $\qquad$ | $$
[(n-1) x-(2 n-1)][(n-1) x-(3 n+1)]=0 .
$$
Since $n \neq 1$, then
$$
\begin{array}{l}
x_{1}=\frac{2 n-1}{n-1}=2+\frac{1}{n-1}, \\
x_{2}=\frac{3 n+1}{n-1}=3+\frac{4}{n-1} .
\end{array}
$$
Because the original equation has at least one integer root, and $n$ is an integer, so, $n-1= \pm 1, \pm 2, \pm 4$.
Solving gives... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) In any permutation of $1,2, \cdots, 200$, there always exists a sum of 20 consecutive numbers that is not less than $a$. Find the maximum value of $a$.
| Three, take any permutation, and divide it into 10 groups, each containing 20 numbers. Let $A_{1}=\left(b_{1}, b_{2}, \cdots, b_{20}\right), A_{2}=\left(b_{21}, b_{2}\right.$, $\left.\cdots, b_{40}\right), \cdots, A_{10}=\left(b_{181}, b_{182}, \cdots, b_{200}\right)$.
Take the average of the 10 groups of numbers
$$
\b... | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9}$, $x \in[0,1]$. Provide the following conclusions:
(1) $f(x)>0$;
(2) $f(x)<0$;
(3) There exists $x_{0} \in[0,1)$, such that $f\left(x_{0}\right)=0$;
(4) There exists $x_{0} \in[0,1]$, such that $f\left(x_{0}\right)<0$.
Among them, the correct conclusion numbers are... | 2. (1)
$$
\begin{array}{l}
f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9} \\
=x^{3}(1-x)+3 x^{3}-3 x^{2}+(x-1)^{2}+\frac{4}{9} \\
=x^{3}(1-x)+(x-1)^{2}+\frac{3}{2} x^{3}+\frac{3}{2} x^{3}+\frac{4}{9}-3 x^{2} \\
\geqslant x^{3}(1-x)+(x-1)^{2}+3 \sqrt[3]{\frac{3}{2} x^{3} \cdot \frac{3}{2} x^{3} \cdot \frac{4}{9}}-3 x^{2} ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $x$ is a
four-digit number, the sum of its digits is $y$. If the value of $\frac{x}{y}$ is minimized, then $x=$ $\qquad$ | 4.1099.
Let $x=\overline{a_{1} a_{2} a_{3} a_{4}}$, where $a_{1} \in \mathbf{N}_{+}, a_{2}, a_{3}, a_{4} \in \mathbf{N}$. Therefore,
$$
\begin{array}{l}
\frac{x}{y}=\frac{1000 a_{1}+100 a_{2}+10 a_{3}+a_{4}}{a_{1}+a_{2}+a_{3}+a_{4}} . \\
=1+9 \times \frac{111 a_{1}+11 a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}+a_{4}} .
\end{array... | 1099 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in Figure 3, there are 16 points, and the distance between any two adjacent points, whether left and right or up and down, is equal to 1. If these points are used as the vertices of triangles, then, a total of right-angled triangles can be obtained. | 5.200.
As shown in Figure 6, the number of right-angled triangles with $A$ as the right-angle vertex is $C_{3}^{1} \cdot C_{3}^{1}=9$;
the number of right-angled triangles with $B$ as the right-angle vertex is
$$
C_{3}^{1} \cdot C_{3}^{1}+C_{2}^{1}+1=12 \text {; }
$$
the number of right-angled triangles with $C$ as t... | 200 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $f(x)=x^{2}+2 x+1$, there exists a real number $t$ such that when $x \in[1, m]$, $f(x+t) \leqslant x$ always holds, then the maximum value of $m$ is $\qquad$ . | 2.4.
Translate the graph of $f(x)$ to the right by $-t$ units, and from the graphical analysis, the maximum value of $m$ is the larger of the x-coordinates of the two intersection points of $\left\{\begin{array}{l}y=x, \\ y=f(x+t)\end{array}\right.$
From $f(1+t)=1$, we get $t=-1, t=-3$.
Then, from $f(x-3)=x$, we have
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $n$ be a positive integer. If
$$
A(n)=1986^{n}+1987^{n}+1988^{n}+1989^{n}
$$
is not divisible by 5, find $n$. | (Tip: Only consider the unit digit $G(A(n))$ of $A(n)$. When $n=4 k\left(k \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
G(A(n))=G\left(6^{n}+7^{n}+8^{n}+9^{n}\right) \\
=G\left(6^{4}+7^{4}+8^{4}+9^{4}\right) \\
=G(6+1+6+1)=4,
\end{array}
$$
at this time, 5 does not divide $A(n)$;
$$
\begin{array}{l}
\text { When } ... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) On an infinite grid paper, some cells are colored red, and the rest are colored blue. In every $2 \times 3$ rectangle of six cells, there are exactly two red cells. How many red cells are there in a $9 \times 11$ rectangle of 99 cells? | II. 33 Red Squares.
As shown in Figure 3, take any red square $K_{0}$ as the center of a $3 \times 3$ square; it is not allowed to color $K$ red. If $K$ is colored red, then in the $2 \times 3$ rectangles $A F H D$, $A B S T$, and $M N C D$, there will be two red squares each. To ensure that the shape $B C G E$ contain... | 33 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. There are 10 positive integers arranged from smallest to largest: $1, 4, 8$, $10, 16, 19, 21, 25, 30, 43$. How many groups of consecutive numbers have a sum that is divisible by 11? | Let 10 positive integers be $a_{1}, a_{2}$, $\cdots, a_{10}$, and let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Then
$$
a_{i+1}+a_{i+2}+\cdots+a_{j}=S_{j}-S_{i} .
$$
The sequence $a_{n}(n=1,2, \cdots, 10)$ has remainders modulo 11 of $1,4,-3,-1,5,-3,-1,3,-3,-1$; the sum of the first $n$ terms $S_{n}$ has remainders modulo 11 ... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. If $p$ and $q$ are both prime numbers, and $7p + q$ and $pq + 11$ are also prime numbers, find the value of $p^q + q^p$.
(1997, Jingzhou City, Hubei Province, Junior High School Mathematics Competition) | (Tip: Since $7 p+q$ is a prime number, and $7 p+q>7$, therefore, $7 p+q$ is odd. Thus, $p, q$ do not share the same parity.
If $p$ is even, then $p=2$. According to the problem, $14+$ $q, 2 q+11$ are both prime numbers. Therefore,
$$
14+q \neq \equiv 0(\bmod 3), 2 q+11 \not \equiv 0(\bmod 3),
$$
which means $q \neq 1... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If
$$
\frac{4^{5}+4^{5}+4^{5}+4^{5}}{3^{5}+3^{5}+3^{5}} \times \frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5}}{2^{5}+2^{5}}=2^{n},
$$
then, $n=$ . $\qquad$ | $-1.12$
The left side of the original equation can be transformed into
$$
\frac{4 \times 4^{5}}{3 \times 3^{5}} \times \frac{6 \times 6^{5}}{2 \times 2^{5}}=\frac{4^{6}}{3^{6}} \times \frac{6^{6}}{2^{6}}=4^{6}=2^{12} .
$$
Thus, $2^{12}=2^{n}$. Therefore, $n=12$. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let the integer $a$ divided by 7 leave a remainder of 3, and the integer $b$ divided by 7 leave a remainder of 5. If $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7.
(1994, Tianjin City Junior High School Mathematics Competition) | Solution: Let $a=7 m+3, b=7 n+5$. Then
$$
\begin{array}{l}
a^{2}-4 b=(7 m+3)^{2}-4(7 n+5) \\
=49 m^{2}+42 m+9-28 n-20 \\
=7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{array}
$$
Therefore, the remainder when $a^{2}-4 b$ is divided by 7 is 3. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $p$ is a prime number, and the equation
$$
x^{2}+p x-444 p=0
$$
has two integer roots. Then $p=$ $\qquad$ | 3.37.
According to
$$
x^{2}=p(444-x)
$$
we know that $p(444-x)$ is a perfect square.
Since $p$ is a prime number, it follows that $p\left|x^{2} \Rightarrow p\right| x$.
Let $x=n p(n \in \mathbf{Z})$ and substitute into equation (1) to get
$$
(n p)^{2}=p(444-n p) .
$$
Since $p \neq 0$, we have $n^{2} p=444-n p$, whic... | 37 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $a, b, c$ simultaneously satisfy $a-7b+8c=4$ and $8a+4b-c=7$. Then, $a^{2}-b^{2}+c^{2}=$ $\qquad$ . | 4.1.
According to the conditions $a+8c=4+7b, 8a-c=7-4b$. Squaring both sides of the two equations and then adding them yields
$$
\begin{array}{l}
(a+8c)^{2}+(8a-c)^{2} \\
=(7+4b)^{2}+(7-4b)^{2} .
\end{array}
$$
Simplifying and organizing, we get $65\left(a^{2}+c^{2}\right)=65\left(1+b^{2}\right)$.
Therefore, $a^{2}-b... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In trapezoid $A B C D$, $A B / / C D$, the base angles $\angle D A B=36^{\circ}, \angle C B A=54^{\circ}, M$ and $N$ are the midpoints of sides $A B$ and $C D$, respectively. If the lower base $A B$ is exactly 2008 units longer than the upper base $C D$, then the line segment $M N=$ $\qquad$ | 5.1004.
As shown in Figure 4, draw $N S / / A D$ and $N T / / C B$ through point $N$.
Then, from $\square A S N D$ and $\square B T N C$, we get
$D N=A S, N C=T B$,
and
$$
\begin{array}{l}
\angle N S T=\angle D A B=36^{\circ}, \\
\angle N T S=\angle C B A=54^{\circ} . \\
\text { Hence, } \angle S N T=180^{\circ}-(\a... | 1004 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Calculate: $1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+$ $\cdots+\frac{1}{20}(1+2+\cdots+20)$.
| $\begin{array}{l}\text { 1. Original expression }=1+\frac{1}{2} \times \frac{2 \times 3}{2}+\frac{1}{3} \times \frac{3 \times 4}{2}+ \\ \cdots+\frac{1}{20} \times \frac{20 \times 21}{2} \\ =\frac{1}{2}(2+3+4+\cdots+21) \\ =\frac{1}{2} \times \frac{23 \times 20}{2}=115 .\end{array}$ | 115 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let real numbers $x, y, z$ simultaneously satisfy
$$
\begin{array}{l}
x^{3}+y=3 x+4, \\
2 y^{3}+z=6 y+6, \\
3 z^{3}+x=9 z+8 .
\end{array}
$$
Try to find the value of $2008(x-1)^{2}+2009(y-1)^{2}+$ $2010(z-1)^{2}$. | 4. Transform to get
$$
\begin{array}{l}
y-2=-x^{3}+3 x+2=-(x-2)(x+1)^{2}, \\
z-2=-2 y^{3}+6 y+4=-2(y-2)(y+1)^{2}, \\
x-2=-3 z^{3}+9 z+6=-3(z-2)(z+1)^{2} .
\end{array}
$$
Multiplying the above three equations yields
$$
\begin{array}{l}
(x-2)(y-2)(z-2) \\
=-6(x-2)(y-2)(z-2) . \\
(x+1)^{2}(y+1)^{2}(z+1)^{2} .
\end{array}... | 6027 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. In an equilateral triangle $\triangle A B C$ with side length $2 \sqrt{7}$, $M$ is the midpoint of side $B C$, and $P$ is a point on $A C$.
(1) For what value of $P C$ is $B P+P M$ minimized?
(2) Find the minimum value of $B P+P M$. | 7. As shown in Figure 8, connect $A M$, and flip $\triangle A B C$ along the axis $A C$ by $180^{\circ}$ to get $\triangle A D C$. Let $N$ be the symmetric point of $M$ (about $A C$). At this point, we always have $P M = P N$. Since $B$ and $N$ are fixed points, $B P + P N \geqslant B N$, with equality holding if and o... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 There is a sequence of numbers: $1,3,4,7,11,18, \cdots$, starting from the third number, each number is the sum of the two preceding numbers.
(1) What is the remainder when the 1991st number is divided by 6?
(2) Group the above sequence as follows:
$(1),(3,4),(7,11,18), \cdots$,
where the $n$-th group has ex... | Solution: Let the $n$-th number of the sequence be $a_{n}$. Then
$$
a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3) .
$$
Through experimentation, it is known that the remainders of $a_{1}, a_{2}, \cdots, a_{26}$ when divided by 6 are
$$
\begin{array}{l}
1,3,4,1,5,0,5,5,4,3,1,4,5, \\
3,2,5,1,0,1,1,2,3,5,2,1,3 .
\end{array}
$$
No... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (15 points) The function $f(x)$ defined on the interval $[0,1]$ satisfies $f(0)=f(1)=0$, and for any $x_{1}, x_{2} \in [0,1]$, we have
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant f\left(x_{1}\right)+f\left(x_{2}\right) \text {. }
$$
(1) Prove: For any $x \in [0,1]$, we have
$$
f(x) \geqslant 0 \text {; }
$$
... | (1) Take $x_{1}=x_{2}=x \in[0,1]$, then
$$
f\left(\frac{2 x}{2}\right) \leqslant f(x)+f(x),
$$
i.e., $f(x) \leqslant 2 f(x)$.
Thus, $f(x) \geqslant 0$.
Therefore, for any $x \in[0,1]$, $f(x) \geqslant 0$.
(2) From $f(0)=f(1)=0$, we get
$$
f\left(\frac{0+1}{2}\right) \leqslant f(0)+f(1)=0+0=0.
$$
Thus, $f\left(\frac{1... | 0 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 8 A positive integer, if it can be expressed as the difference of squares of two positive integers, is called a "wise number". Arrange all the wise numbers in ascending order. Find the wise number at the 2009th position. | Solution: Let $n$ be any positive integer.
When $n=2 k+1\left(k \in \mathbf{N}_{+}\right)$, we have $n=2 k+1=(k+1)^{2}-k^{2}$.
Thus, every odd number greater than 1 is a wise number.
When $n=4 k\left(k \in \mathbf{N}_{+}, k \geqslant 2\right)$, we have
$$
n=4 k=(k+1)^{2}-(k-1)^{2} \text {. }
$$
Thus, every number grea... | 2681 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Calculate:
(1) $\sqrt{627953481}+\sqrt{672935481}=$ $\qquad$
(2) $\sqrt{\sqrt{254817369}-\sqrt{152843769}}=$
$\qquad$ . | 2. (1) $51000(2) 60$ | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Find the smallest positive integer $n$, such that the last three digits of $n^{3}$ are 888. | Solution: Since the unit digit of $n^{3}$ is 8, then the unit digit of $n$ is 2. Let $n=10 k+2$. Then
$$
\begin{array}{l}
n^{3}=(10 k+2)^{3}=1000 k^{3}+600 k^{2}+120 k+8 \\
=1000 k^{3}+100 k(6 k+1)+20 k+8 .
\end{array}
$$
According to the problem, we know that the last three digits of $100\left(6 k^{2}+k\right)+20 k+8... | 192 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Sure, here is the translated text:
```
II. (20 points) Find all positive integers $n$ such that
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{4}\right]+\left[\frac{n}{5}\right]=69 \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
``` | $$
\begin{array}{l}
\frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}-4 \\
<\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{4}\right]+\left[\frac{n}{5}\right] \\
\leqslant \frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5} . \\
\text { Therefore, } \frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}-4 \\
<69 \l... | 55 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) For a set $S\left(S \subseteq \mathbf{N}_{+}\right)$, if for any $x \in S, \dot{x}$ cannot divide the sum of elements of any non-empty subset of $S \backslash\{x\}$, then $S$ is called a "good set" ($S \backslash\{x\}$ represents the set $S$ after removing the element $x$).
(1) If $\{3,4, n\}$ is a g... | Three, (1) Obviously, $n>4$.
If $n=5$, then $4 \mid (3+5)$; if $n=6$, then $3 \mid 6$; if $n=7$, then $7 \mid (3+4)$; if $n=8$, then $4 \mid 8$; if $n=9$, then $3 \mid 9$.
When $n=10$, it is verified that $\{3,4,10\}$ is a good set.
Therefore, the minimum value of $n$ is $n_{0}=10$.
(2) If $\{3,4,10, m\}$ is a good set... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Reverse the order of the four digits of a four-digit number, and add the resulting number to the original number. If the sum obtained is divisible by 35, then this four-digit number is called a "good number". Then, among all four-digit numbers, the number of good numbers is ( ).
(A) 234
(B) 252
(C) 270
(D) 369 | 3. A.
Let the four-digit number be $\overline{a_{1} a_{2} a_{3} a_{4}}$. Then
$$
\begin{array}{l}
A=\overline{a_{1} a_{2} a_{3} a_{4}}+\overline{a_{4} a_{3} a_{2} a_{1}} \\
=1001\left(a_{1}+a_{4}\right)+110\left(a_{2}+a_{3}\right) .
\end{array}
$$
Since $1001=7 \times 143$, we have
$$
\begin{array}{l}
7 \mid A \Leftr... | 234 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example 10 Given Theorem: "If three prime numbers $a, b, c$ greater than 3 satisfy the equation $2a + 5b = c$, then $a + b + c$ is a multiple of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion.
(1997, National Junior High School Mathematics League) | Solution: Let $a=3 k_{1}+r_{1}, b=3 k_{2}+r_{2}$. Then
$$
\begin{array}{l}
a+b+c=3(a+2 b) \\
=3\left(3 k_{1}+r_{1}+6 k_{2}+2 r_{2}\right) \\
=9\left(k_{1}+2 k_{2}\right)+3\left(r_{1}+2 r_{2}\right) .
\end{array}
$$
Since $a, b$ are both primes greater than 3, we have
$$
r_{1} r_{2} \neq 0.
$$
If $r_{1} \neq r_{2}$, t... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
4. Polynomial
$$
\begin{array}{l}
|x+1|+|x-2|+|x+3|+\cdots+ \\
|x+2007|+|x-2008|+|x+2009|
\end{array}
$$
The minimum value of the above expression is ( ).
(A) 2019044
(B) 2017035
(C) 2009
(D) 0 | 4. A.
$$
\begin{array}{l}
| x+ 1|+| x-2|+| x+3 |+\cdots+ \\
|x+2007|+|x-2008|+|x+2009| \\
=|x+1|+(|x-2|+|x+3|)+\cdots+ \\
(|x-2008|+|x+2009|) \\
\geqslant|x+1|+|(x-2)-(x+3)|+\cdots+ \\
|(x-2008)-(x+2009)| \\
=|x+1|+2+3+\cdots+2009 \\
\geqslant 0+2+3+\cdots+2009 \\
= \frac{2009 \times 2010}{2}-1 \\
= 2019044 .
\end{arr... | 2019044 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given $n$ is a natural number, $n^{2}+4 n+2009$ can be expressed as the sum of the squares of four consecutive natural numbers. Then the sum of all $n$ that satisfy this condition is $\qquad$ . | 2.1.712.
Let $n^{2}+4 n+2009$
$$
=(m-1)^{2}+m^{2}+(m+1)^{2}+(m+2)^{2} \text {, }
$$
where $m$ is a positive integer. Then
$$
(n+2)^{2}+2005=(2 m+1)^{2}+5 \text {, }
$$
which implies $(2 m+1)^{2}-(n+2)^{2}=2000$.
Factoring, we get
$$
(2 m+n+3)(2 m-n-1)=2000 \text {. }
$$
Since $2 m+n+3$ and $2 m-n-1$ have the same pa... | 712 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Find the remainder when $47^{37^{2}}$ is divided by 7.
Try to find the remainder of $47^{37^{2}}$ when divided by 7. | (Note that $47^{6} \equiv(-2)^{6} \equiv 2^{6} \equiv 8^{2} \equiv 1^{2}$ $\equiv 1(\bmod 7)$. Also, $37^{23} \equiv 1^{23} \equiv 1(\bmod 6)$, let $37^{23}$ $=6k+1$. Then $47^{37^{23}}=47^{k+1}=\left(47^{6}\right)^{k} \times 47 \equiv$ $\left.1^{k} \times 47 \equiv 47 \equiv 5(\bmod 7).\right)$ | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2009}\right]$ when divided by 7 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.6.
Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$. Then
$$
-1<\beta<0,1<\alpha<2 \text {, }
$$
and $\alpha+\beta=1, \alpha \beta=-1$.
Therefore, $\alpha, \beta$ are the two distinct real roots of $x^{2}-x-1=0$.
Let $A_{n}=\alpha^{n}+\beta^{n}$. Then $A_{n+2}=A_{n+1}+A_{n}$.
By $A_{1}=\alpha+\beta=1$,... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that the three non-zero real roots of the equation $x^{3}+a x^{2}+b x+c$ $=0$ form a geometric progression. Then $a^{3} c-b^{3}$ $=$ . $\qquad$ | 8.0.
Let the three roots be $d$, $d q$, and $d q^{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
d+d q+d q^{2}=-a, \\
d^{2} q+d^{2} q^{2}+d^{2} q^{3}=b, \\
d^{3} q^{3}=-c .
\end{array}\right.
$$
Dividing (2) by (1) gives $d q=-\frac{b}{a}$.
Substituting into (3) yields $\left(-\frac{b}{a}\right)^{3}=-c$... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) Given two lines
$$
\begin{array}{l}
l_{1}: 3 x+4 y-25=0, \\
l_{2}: 117 x-44 y-175=0,
\end{array}
$$
Point $A$ has projections $B$ and $C$ on $l_{1}$ and $l_{2}$, respectively.
(1) Find the locus curve $\Gamma$ of point $A$ such that $S_{\triangle A B C}=\frac{1728}{625}$;
(2) If $\odot T:\left(x-\frac{... | 10. (1) Let $A(x, y)$. It is easy to know
$$
\begin{array}{l}
A B=\frac{|3 x+4 y-25|}{\sqrt{3^{2}+4^{2}}}=\frac{|3(x-3)+4(y-4)|}{5}, \\
A C=\frac{|117 x-44 y-175|}{\sqrt{117^{2}+44^{2}}} \\
=\frac{|117(x-3)-44(y-4)|}{125} .
\end{array}
$$
Let the angles of inclination of $l_{1}$ and $l_{2}$ be $\dot{\alpha}_{1}$ and $... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange $1, 2, 3$ in a cyclic order from left to right, forming a 2009-digit number: $x=123123 \cdots 12312$. Find the remainder when $x$ is divided by 101. | (Tip: Since $100 \equiv -1 \pmod{101}$, we have
$$
\begin{array}{l}
123123 = 123 \times 100 + 123 \\
\equiv -123 + 123 \equiv 0 \pmod{101}.
\end{array}
$$
Therefore, $123123 \cdots 12312 = 123123 \cdots 12300 + 12$
$$
\equiv 0 + 12 \equiv 12 \pmod{101}.
$$
) | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $p$ and $8 p^{2}+1$ both be prime numbers. Prove that $8 p^{2}-p+2$ is also a prime number. | (Tip: If $p \equiv 0(\bmod 3)$, then $p=3$. In this case, $8 p^{2}-p+2=72-3+2=71$ is a prime number.
If $p \equiv \pm 1(\bmod 3), 8 p^{2}+1 \equiv 8+1 \equiv$ $0(\bmod 3)$, i.e., $3 \mid\left(8 p^{2}+1\right)$, which is a contradiction.) | 71 | Number Theory | proof | Yes | Yes | cn_contest | false |
7. Let $p$ be a prime number, and $q=4^{p}+p^{4}+4$ is also a prime number. Find the value of $p+q$.
The text above is translated into English, keeping the original text's line breaks and format. | (Tip: Mod 3 we get $q=4^{p}+p^{4}+4 \equiv 1^{p}+p^{4}+$ $1 \equiv p^{4}+2(\bmod 3)$. Also, $p^{2} \equiv 0,1(\bmod 3)$, so $p^{4}$ $\equiv 0,1(\bmod 3)$.
If $p^{4} \equiv 1(\bmod 3)$, then $q \equiv p^{4}+2 \equiv 1+2 \equiv$ $0(\bmod 3)$. But $q=4^{p}+p^{4}+4>3$, which contradicts the fact that $q$ is a prime number... | 152 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find the unit digit of the natural number $2^{100}+3^{101}+4^{100}$. | Solution: Notice that
$$
\begin{array}{l}
2^{100} \equiv 2^{1 \times 25} \equiv\left(2^{4}\right)^{25} \equiv 16^{25} \\
\equiv 6^{25} \equiv 6(\bmod 10), \\
3^{101} \equiv 3^{4 \times 25+1} \equiv\left(3^{4}\right)^{25} \times 3^{1} \\
\equiv 1^{25} \times 3^{1} \equiv 3(\bmod 10), \\
4^{102} \equiv 4^{2 \times 51} \e... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The inequality $x^{2}+|2 x-6| \geqslant a$ holds for all real numbers $x$. Then the maximum value of the real number $a$ is $\qquad$ | 2.5.
When $x \geqslant 3$,
left side $=x^{2}+2 x-6=(x+1)^{2}-7 \geqslant 9$;
When $x<3$,
left side $=x^{2}-2 x+6=(x-1)^{2}+5 \geqslant 5$.
Therefore, for the given inequality to hold for all real numbers $x$, the maximum value of $a$ should be 5. | 5 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a_{n}$ denote the last digit of the number $n^{4}$. Then $a_{1}+a_{2}+\cdots+a_{2008}=$ $\qquad$ | 3.6632 .
Obviously, the unit digit of $(10+n)^{4}$ is the same as that of $n^{4}$, that is,
$$
1^{4}, 2^{4}, 3^{4}, 4^{4}, 5^{4}, 6^{4}, 7^{4}, 8^{4}, 9^{4}, 10^{4}
$$
the unit digits are
$$
\begin{array}{l}
1,6,1,6,5,6,1,6,1,0 \text {. } \\
\text { Then } a_{1}+a_{2}+\cdots+a_{10} \\
=1+6+1+6+5+6+1+6+1+0=33 \text {.... | 6632 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For the positive integer $n$, it is defined that $n!=1 \times 2 \times \cdots \times n$. Then, among all the divisors of the product $1! \times 2! \times \cdots \times 9!$, the number of divisors that are perfect squares is $\qquad$.
| 7.672.
Given $1!\times 2!\times \cdots \times 9!=2^{30} \times 3^{13} \times 5^{5} \times 7^{3}$, then the divisors of its perfect square must be in the form of $2^{2 a} \times 3^{2 b} \times$ $5^{2 c} \times 7^{2 d}$, where $a, b, c, d$ are non-negative integers, and $0 \leqslant a \leqslant 15,0 \leqslant b \leqslan... | 672 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $k$ is a positive integer not exceeding 2008, such that the equation $x^{2}-x-k=0$ has two integer roots. Then the sum of all such positive integers $k$ is $\qquad$ . | 8.30360
Since the roots of $x^{2}-x-k=0$ are $x=\frac{1 \pm \sqrt{4 k+1}}{2}$, therefore,
$x^{2}-x-k=0$ has two integer roots $\Leftrightarrow 4 k+1$ is a square of an odd number.
Let $1+4 k=(2 a+1)^{2}\left(a \in \mathbf{N}_{+}\right)$. Then $k=a(a+1)$.
Also, $k \leqslant 2008$, so $a \leqslant 44$.
Therefore, the su... | 30360 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (15 points) Find the largest positive integer $n$ that satisfies the inequality
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{11}\right]+\left[\frac{n}{13}\right]<n
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. | Five, Proof: For any integers $x, k$, we have
$$
\left[\frac{x}{k}\right] \geqslant \frac{x-k+1}{k} \text {. }
$$
Fact .. By the division algorithm, there exist integers $q, r$ such that $x=k q+r(0 \leqslant r \leqslant k-1)$. Thus,
$$
\begin{array}{l}
{\left[\frac{x}{k}\right]=\left[\frac{k q+r}{k}\right]=q+\left[\fr... | 1715 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Find the largest positive number $\lambda$, such that for any real numbers $x, y, z$ satisfying $x^{2}+y^{2}+z^{2}=1$, the inequality
$|\lambda x y+y z| \leqslant \frac{\sqrt{5}}{2}$ holds.
(Zhang Zhengjie) | 5. Notice that
$$
\begin{array}{l}
1=x^{2}+y^{2}+z^{2}=x^{2}+\frac{\lambda^{2}}{1+\lambda^{2}} y^{2}+\frac{1}{1+\lambda^{2}} y^{2}+z^{2} \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(\lambda|x y|+|y z|) \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(|\lambda x y+y z|),
\end{array}
$$
and when $y=\frac{\sqrt{2}}{2}, x=\frac... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $n$ be a positive integer, and let $f(n)$ denote the number of $n$-digit numbers (called wave numbers) $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfy the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and $a_{i} \neq$ $a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, the signs of $a_{i}-a_{i... | And the $n$-digit waveform number $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfies $a_{1}>a_{2}$ is called a "B-type number". According to symmetry, when $n \geqslant 2$, the number of B-type numbers is also $g(n)$. Therefore, $f(n)=2 g(n)$.
Next, we find $g(n)$: Let $m_{k}(i)$ represent the number of $k$-digit A-... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Given that positive integers $p, q$ are both prime numbers, and $7p+q, pq+11$ are also prime numbers. Find $p+q$.
保留源文本的换行和格式，直接输出翻译结果。 | When $p \equiv q \equiv 1(\bmod 2)$,
$$
7 p+q \equiv 7+1 \equiv 0(\bmod 2) \text{. }
$$
But $7 p+q \geqslant 7>2$, which contradicts the fact that $7 p+q$ is a prime number. Therefore, one of $p, q$ must be even.
Since $p, q$ are both primes, then $p=2$ or $q=2$.
(1) When $p=2$, $14+q$ and $2 q+11$ are both primes.
If... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Consider the set $S=\{1,2, \cdots, 10\}$ and all its non-empty subsets. If the number of even numbers in a non-empty subset is no less than the number of odd numbers, then this subset is called a "good subset". Therefore, the number of good subsets is ( ) .
(A) 631
(B) 633
(C) 635
(D) 637 | 5.D.
Suppose a good subset contains $i(i=1,2,3,4,5)$ even numbers. Then the number of odd numbers can be $j(j=0,1, \cdots, i)$.
Therefore, the number of good subsets is
$$
\begin{array}{l}
\sum_{i=1}^{5}\left(C_{5}^{i} \sum_{j=0}^{i} C_{5}^{j}\right) \\
= C_{5}^{1}\left(C_{5}^{0}+C_{5}^{1}\right)+C_{5}^{2}\left(C_{5}... | 637 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
8. If the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is $\qquad$ . | 8. -1 .
Let $z$ be a root satisfying the condition. Then the original equation is equivalent to
$$
z^{2008}(z+1)=-1 \text {. }
$$
Taking the modulus on both sides, we get $|z+1|=1$.
Since $|z|=1$, all roots with modulus 1 can only be the complex numbers corresponding to the intersection points of the circle I: with c... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Consider the 25 grid points in a $4 \times 4$ square grid. The number of different lines passing through at least 3 grid points is $\qquad$ .
| 9.32.
There are 10 horizontal and vertical lines, 10 lines parallel to the two diagonals, and 12 other lines that meet the conditions. In total, there are 32 lines. | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the value of $\sum_{k=1}^{2008}\left[\frac{2008 k}{2009}\right]$ is $\qquad$ . | 10.2015028.
For $k=1,2, \cdots, 2008$, since $\frac{2008 k}{2009}$ is not an integer, so,
$$
\begin{array}{l}
{\left[\frac{2008 k}{2009}\right]+\left[\frac{2008(2009-k)}{2009}\right]} \\
=\left[\frac{2008 k}{2009}\right]+\left[2008-\frac{2008 k}{2009}\right]=2007 . \\
\text { Therefore, } \sum_{k=1}^{2008}\left[\frac{... | 2015028 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. Given the rectangular cuboid $A B C D-A_{1} B_{1} C_{1} D_{1}$ satisfies $A A_{1}=2, A D=3, A B=251$, the plane $A_{1} B D$ intersects $C C_{1} 、 C_{1} B_{1} 、 C_{1} D_{1}$ at points $L 、 M 、 N$. Then the volume of the tetrahedron $C_{1} L M N$ is $\qquad$ . | 11.2008.
Since $B D / / M N, A_{1} D / / L M, A_{1} B / / L N$, therefore, $A_{1} 、 B 、 D$ are the midpoints of $M N 、 L M 、 L N$ respectively. Thus,
$$
\begin{array}{l}
C_{1} L=2 C C_{1}=4, C_{1} M=2 C_{1} B_{1}=6, \\
C_{1} N=2 C_{1} D_{1}=502 .
\end{array}
$$
Therefore, the volume of the tetrahedron $C_{1} L M N$ i... | 2008 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. There are 10 players $A_{1}, A_{2}, \cdots, A_{10}$, whose points are $9,8,7,6,5,4,3,2,1,0$, and their rankings are 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th. Now a round-robin tournament (i.e., each pair of players plays exactly one match) is held, and each match must have a winner. If the player with a hi... | 15. The minimum cumulative score of the new champion is 12.
If the new champion's score does not exceed 11 points, then
$A_{1}$ can win at most 2 games; $A_{2}$ can win at most 3 games;
$A_{3}$ can win at most 4 games; $A_{4}$ can win at most 5 games.
$A_{5}$ can increase by at most 6 points, but there are only 5 play... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If a positive integer cannot be expressed as the difference of squares of two positive integers, then this positive integer is called a "non-wise number". If these non-wise numbers are arranged in ascending order, then the 2009th non-wise number is $\qquad$ | ニ、1.8026.
1 cannot be expressed as the difference of squares of two positive integers, so 1 is the 1st non-wise number.
Odd numbers greater than 1, $2 n+1\left(n \in N_{+}\right)$, can be expressed as $(n+1)^{2}-n^{2}$.
Even numbers divisible by 4, $4 m$, can be expressed as $(m+1)^{2}-(m-1)^{2}(m>1)$. However, 4 can... | 8026 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $x$ and $y$ are integers, $y=\sqrt{x+2003}-$ $\sqrt{x-2009}$. Then the minimum value of $y$ is $\qquad$ . | 3.2 .
Let $\sqrt{x+2003}=a, \sqrt{x-2009}=b$.
Since $x, y$ are integers,
$y=a-b=\frac{a^{2}-b^{2}}{a+b}=\frac{4012}{a+b} \in \mathbf{Z}_{+}$.
Thus, $a-b \in \mathbf{Q}, a+b \in \mathbf{Q}$
$\Rightarrow a \in \mathbf{Q}, b \in \mathbf{Q} \Rightarrow a, b \in \mathbf{Z}$.
Then $(a+b)(a-b)=4012=2006 \times 2$.
Since $a+b... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In Rt $\triangle A B C$, $\angle A C B=90^{\circ}$, on the hypotenuse $A B$ respectively intercept $A D=$
$$
A C, B E=B C, D E=6 \text{, }
$$
$O$ is the circumcenter of $\triangle C D E$, as shown in Figure 2. Then the sum of the distances from $O$ to the three sides of $\triangle A B C$ is . $\qquad$ | 4.9.
As shown in Figure 8, connect
$$
O A, O B, O C, O D \text {, and }
$$
$O E$.
Since $O C=O D$,
we know that point $O$ lies on the perpendicular bisector of segment $C D$.
Also, $A D=A C$, so $O A$ bisects $\angle C A D$.
Similarly, $O B$ bisects $\angle C B D$.
Therefore, $O$ is the incenter of $\triangle A B C$.... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sum of $2 n+1\left(n \in \mathbf{N}_{+}\right)$ consecutive positive integers is $a$, and the difference between the sum of the squares of the last $n$ numbers and the sum of the squares of the first $n$ numbers is $b$. If $\frac{a}{b}=\frac{11}{60}$, then the value of $n$ is | 2.5
Let the $(n+1)$-th number be $m$. Then
$$
\begin{array}{l}
a=(2 n+1) m, \\
b=\sum_{i=1}^{n}(m+i)^{2}-\sum_{i=1}^{n}(m-i)^{2} \\
=2 m \sum_{i=1}^{n} 2 i=2 m n(n+1) . \\
\text { Hence } \frac{a}{b}=\frac{2 n+1}{2 n(n+1)}=\frac{11}{60} .
\end{array}
$$
Solving gives $n=5$ or $-\frac{6}{11}$ (discard).
| 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. For a right-angled triangle with a hypotenuse of 2009, if the two legs are also integers, then its area is $\qquad$ . | $-.1 .432180$.
Let the two legs be $x, y$. Then $x^{2}+y^{2}=2009^{2}$. According to Pythagorean triples and the symmetry of $x, y$, there exist positive integers $m, n, k (m>n)$, such that
$$
\begin{array}{l}
x=2 m n k, y=\left(m^{2}-n^{2}\right) k, \\
2009=\left(m^{2}+n^{2}\right) k .
\end{array}
$$
Since $2009=41 \... | 432180 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. From the arithmetic sequence $2,5,8,11, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ | 2.8 .
First, take $2,5,8,11,20,41,110,1640$, it is easy to see that the sum of their reciprocals is 1, i.e., $k=8$ satisfies the requirement.
Second, suppose we take $x_{1}, x_{2}, \cdots, x_{k}$ from the sequence, such that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1$.
Let $y_{i}=\frac{x_{1} x_{2} \cdot... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$ are any two points (which can coincide) on the graph of the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{2 x}{1-2 x}, & x \neq \frac{1}{2} \\
-1, & x=\frac{1}{2}
\end{array}\right.
$$
Point $M$ lies on the line $x=\frac{1}{2}$, and $\overrightarrow{... | 4. -2 .
Given that point $M$ is on the line $x=\frac{1}{2}$, let $M\left(\frac{1}{2}, y_{M}\right)$.
Also, $\overrightarrow{A M}=\overrightarrow{M B}$, that is,
$$
\begin{array}{l}
\overrightarrow{A M}=\left(\frac{1}{2}-x_{1}, y_{M}-y_{1}\right), \\
\overrightarrow{M B}=\left(x_{2}-\frac{1}{2}, y_{2}-y_{M}\right) .
\e... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.