problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
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1. Among the natural numbers from $1 \sim 10000$, the integers that are neither perfect squares nor perfect cubes are $\qquad$ in number. | In the natural numbers from $1 \sim 10000$, there are 100 perfect squares.
$$
\begin{array}{l}
\text { Because } 22^{3}=10648>10000, \\
21^{3}=9261<10000,
\end{array}
$$
Therefore, there are 21 perfect cubes.
Next, consider the number of natural numbers from $1 \sim 10000$ that are both perfect squares and perfect cub... | 9883 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the sequence of natural numbers from $1 \sim 8$ be $a_{1}, a_{2}$, $\cdots, a_{8}$. Then
$$
\begin{array}{l}
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{4}\right|+\left|a_{4}-a_{5}\right|^{\prime}+ \\
\left|a_{5}-a_{6}\right|+\left|a_{6}-a_{7}\right|+\left|a_{7}-a_{8}\right|+\left|a_{8}-a_{1... | 5. 32 .
From the problem, we have
$$
\begin{aligned}
S= & \left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{4}\right|+ \\
& \left|a_{4}-a_{5}\right|+\left|a_{5}-a_{6}\right|+\left|a_{6}-a_{7}\right|+ \\
& \left|a_{7}-a_{8}\right|+\left|a_{8}-a_{1}\right| .
\end{aligned}
$$
Removing the absolute value s... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Five, (15 points) A school assigns numbers to the contestants participating in a math competition, with the smallest number being 0001 and the largest number being 2014. No matter which contestant steps forward to calculate the average of the numbers of all other contestants in the school, the average is always an inte... | Let the school have a total of $n$ participants, whose admission numbers are
$$
1=x_{1}<x_{2}<\cdots<x_{n-1}<x_{n}=2014 .
$$
According to the problem, we have
$$
S_{k}=\frac{x_{1}+x_{2}+\cdots+x_{n}-x_{k}}{n-1}(k=1,2, \cdots, n) \in \mathbf{Z}_{+} .
$$
For any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
S_{i}-S_{... | 34 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=a_{n}+a_{n-1}(n \geqslant 2) \text {. }
$$
If $a_{7}=8$, then $a_{1}+a_{2}+\cdots+a_{10}=$ $\qquad$ | 3.88.
From the problem, we know that $a_{7}=8 a_{2}+5 a_{1}$.
Therefore, $a_{1}+a_{2}+\cdots+a_{10}$
$$
=88 a_{2}+55 a_{1}=11 a_{7}=88 .
$$ | 88 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Arrange the numbers in the set $\left\{2^{x}+2^{y}+2^{z} \mid x 、 y 、 z \in \mathbf{N}, x<y<z\right\}$ in ascending order. The 100th number is $\qquad$ (answer with a number).
| 5.577.
Notice that the number of combinations $(x, y, z)$ such that $0 \leqslant x<y<z \leqslant n$ is $\mathrm{C}_{n+1}^{3}$.
Since $\mathrm{C}_{9}^{3}=84<100<120=\mathrm{C}_{10}^{3}$, the 100th number must satisfy $z=9$.
Also notice that the number of combinations $(x, y)$ such that $0 \leqslant x<y \leqslant m$ i... | 577 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function $f(x)$ satisfies
$$
f(x)=\left\{\begin{array}{ll}
x-3, & x \geqslant 1000 ; \\
f(f(x+5)), & x<1000 .
\end{array}\right.
$$
Then $f(84)=$ . $\qquad$ | 6.997.
Let $f^{(n)}(x)=\underbrace{f(f(\cdots f(x)))}_{n \uparrow}$. Then
$$
\begin{aligned}
& f(84)=f(f(89))=\cdots=f^{(184)}(999) \\
= & f^{(185)}(1004)=f^{(184)}(1001)=f^{(183)}(998) \\
= & f^{(184)}(1003)=f^{(183)}(1000)=f^{(182)}(997) \\
= & f^{(183)}(1002)=f^{(182)}(999)=f^{(183)}(1004) \\
= & f^{(182)}(1001)=f^... | 997 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $N>1$ be a positive integer, and $m$ denote the largest divisor of $N$ that is less than $N$. If $N+m$ is a power of 10, find $N$.
| 5. $N=75$.
Let $N=m p$. Then $p$ is the smallest prime factor of $N$.
By the problem, we know $m(p+1)=10^{k}$.
Since $10^{k}$ is not a multiple of 3, therefore, $p>2$.
Hence, $N$ and $m$ are both odd.
Thus, $m=5^{*}$.
If $s=0, N=p=10^{k}-1$ is a multiple of 9, which is a contradiction.
Then $s \geqslant 1,5 \mid N$.
T... | 75 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. In the figure shown in Figure 3, on both sides of square $P$, there are $a$ and $b$ squares to the left and right, and $c$ and $d$ squares above and below, where $a$, $b$, $c$, and $d$ are positive integers, satisfying
$$
(a-b)(c-d)=0 \text {. }
$$
The shape formed by these squares is called a "cross star".
There i... | 8. For a cross, the cell $P$ referred to in the problem is called the "center block" of the cross.
When $a=b$, the cross is called "standing"; when $c=d$, it is called "lying" (some crosses are both standing and lying).
If the union of a row and a column of a rectangle $R$ is exactly a cross $S$, then $R$ is called t... | 13483236 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $1 \leqslant x, y, z \leqslant 6$.
The number of cases where the product of the positive integers $x, y, z$ is divisible by 10 is
$\qquad$ kinds. | 8. 72 .
(1) The number of ways to choose $x, y, z$ is $6^{3}$;
(2) The number of ways to choose $x, y, z$ without taking $2, 4, 6$ is $3^{3}$; (3) The number of ways to choose $x, y, z$ without taking 5 is $5^{3}$;
(4) The number of ways to choose $x, y, z$ without taking $2, 4, 5, 6$ is $2^{3}$. Therefore, the number ... | 72 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}_{n \geqslant 0}$ satisfies $a_{0}=0$, $a_{1}=1$, and for all positive integers $n$,
$$
a_{n+1}=2 a_{n}+2013 a_{n-1} \text {. }
$$
Find the smallest positive integer $n$ such that $2014 \mid a_{n}$. | 10. Below are $a_{n}$ modulo 2014.
Then $a_{n+1} \equiv 2 a_{n}-a_{n-1} \Rightarrow a_{n+1}-a_{n} \equiv a_{n}-a_{n-1}$.
Therefore, the sequence $\left\{a_{n}\right\}$ has the characteristics of an arithmetic sequence modulo 2014.
Since $a_{0}=0, a_{1}=1$, we have $a_{n} \equiv n$.
Thus, the smallest positive integer ... | 2014 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have
$$
f(x+3) \leqslant f(x)+3, f(x+2) \geqslant f(x)+2 .
$$
Let $g(x)=f(x)-x$. If $f(4)=2014$, then
$$
f(2014)=
$$
$\qquad$ | 6.4024.
Let $g(x)=f(x)-x$, then we have
$$
\begin{array}{l}
g(x+2)=f(x+2)-x-2, \\
g(x+3)=f(x+3)-x-3 .
\end{array}
$$
Also, from $f(x+3) \leqslant f(x)+3$,
$$
f(x+2) \geqslant f(x)+2 \text {, }
$$
we get
$$
\begin{array}{l}
g(x+2) \geqslant f(x)+2-x-2=f(x)-x, \\
g(x+3) \leqslant f(x)+3-x-3=f(x)-x .
\end{array}
$$
Fr... | 4024 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. If non-negative integers $m, n$ add up with exactly one carry (in decimal), then the ordered pair $(m, n)$ is called "good". The number of all good ordered pairs whose sum is 2014 is $\qquad$ . | 7. 195 .
If the carry is in the units place, then the combination of units and tens is $5+9$, $6+8$, $7+7$, $8+6$, $9+5$, a total of 5 kinds; the hundreds place can only be $0+0$, a total of 1 kind; the thousands place is $0+2$, $1+1$, $2+0$, a total of 3 kinds. In this case, there are $5 \times 1 \times 3=15$ pairs.
... | 195 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Prove: There exists a set $S$ consisting of 2014 positive integers, with the following property: if a subset $A$ of $S$ satisfies that for any $a, a' \in A, a \neq a'$, we have $a + a' \notin S$, then $|A| \leq 152$.
---
The translation maintains the original text's formatting and line breaks. | For $1<k<2014$, let
$2014=k q+r(0 \leqslant r<k)$.
For $i=1,2, \cdots, k$, let
$S_{i}=\left\{2^{i-1} m \mid q \leqslant m \leqslant 2 q-1\right\}$.
Then $\left|S_{i}\right|=q$, and for any
$1 \leqslant i<j \leqslant k, q \leqslant m_{1}, m_{2} \leqslant 2 q-1$,
we have $2^{i-1} m_{1}=2^{j-1} m_{2} \Leftrightarrow 2^{j... | 152 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 2 Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying $f(0) \neq 0$, and for any $x, y \in \mathbf{R}$ we have
$$
f\left((x-y)^{2}\right)=f^{2}(x)-2 x f(y)+y^{2} .
$$
Then $f(2012)=$ $\qquad$ | Let $x=y=0$.
Then $f(0)=f^{2}(0) \Rightarrow f(0)=1$ or 0 (discard 0).
Let $y=x$.
$$
\begin{array}{l}
\text { Then } f(0)=f^{2}(x)-2 x f(x)+x^{2}=(f(x)-x)^{2} \\
\Rightarrow f(x)=x \pm 1 .
\end{array}
$$
If there exists $x_{0}$ such that $f\left(x_{0}\right)=x_{0}-1$, let
$$
\begin{array}{l}
x=x_{0}, y=0 \text {. } \\... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
f(x)=x\left(\sqrt{36-x^{2}}+\sqrt{64-x^{2}}\right)
$$
Find the maximum value of the function. | Algebraic solution Using the Cauchy-Schwarz inequality, we get
$$
\begin{aligned}
f(x) & =x \sqrt{36-x^{2}}+x \sqrt{64-x^{2}} \\
& \leqslant \sqrt{\left(x^{2}+36-x^{2}\right)\left(64-x^{2}+x^{2}\right)}=48 .
\end{aligned}
$$
Geometric solution Construct $\triangle A B C, A D \perp B C$, and let $A B=6$, $A C=8, A D=x$... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 The function $f$ is defined on the set of ordered pairs of positive integers, and satisfies
$$
\begin{array}{c}
f(x, x)=x, f(x, y)=f(y, x), \\
(x+y) f(x, y)=y f(x, x+y) .
\end{array}
$$
Calculate $f(14,52)$. | Since $f(x, x+y)=\frac{x+y}{y} f(x, y)$, we have,
$$
\begin{aligned}
& f(14,52)=f(14,14+38)=\frac{52}{38} f(14,38) \\
= & \frac{26}{19} f(14,14+24)=\frac{13}{6} f(14,24) \\
= & \frac{13}{6} f(14,14+10)=\frac{26}{5} f(14,10) \\
= & \frac{26}{5} f(10,14)=\frac{26}{5} f(10,10+4) \\
= & \frac{91}{5} f(10,4)=\frac{91}{5} f(... | 364 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If positive numbers $a, b$ satisfy
$$
2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b) \text {, }
$$
then $\frac{1}{a}+\frac{1}{b}=$ $\qquad$ . | $-, 1.108$
Let $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$. Then
$$
\begin{array}{l}
a=2^{k-2}, b=3^{k-3}, a+b=6^{k} . \\
\text { Therefore } \frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}=\frac{6^{k}}{2^{k-2} \times 3^{k-3}} \\
=2^{2} \times 3^{3}=108 .
\end{array}
$$ | 108 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example: Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x, y \in \mathbf{R}$, we have
$$
f(x+3) \leqslant f(x)+3, f(x+2) \geqslant f(x)+2 .
$$
Let $g(x)=f(x)-x$.
(1) Prove: $g(x)$ is a periodic function;
(2) If $f(998)=1002$, find the value of $f(2000)$. | (1) Proof: From $g(x)=f(x)-x$, we get
$$
\begin{array}{l}
g(x+2)=f(x+2)-x-2, \\
g(x+3)=f(x+3)-x-3 .
\end{array}
$$
Substituting into the inequality in the problem, we get
$$
\begin{array}{l}
g(x+2) \geqslant f(x)+2-x-2=f(x)-x, \\
g(x+3) \leqslant f(x)+3-x-3=f(x)-x .
\end{array}
$$
From equation (1), we get
$$
\begin{... | 2004 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{\pi}{6}, a_{n+1}=\arctan \left(\sec a_{n}\right)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the positive integer $m$ such that
$$
\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}=\frac{1}{100} .
$$ | 10. From the problem, we know that for any positive integer $n$,
$$
a_{n+1} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),
$$
and $\tan a_{n+1}=\sec a_{n}$.
Since $\sec a_{n}>0$, then $a_{n+1} \in\left(0, \frac{\pi}{2}\right)$.
From equation (1), we get $\tan ^{2} a_{n+1}=\sec ^{2} a_{n}=1+\tan ^{2} a_{n}$.
Thus, $\ta... | 3333 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let $f(n)$ be a function defined on $\mathbf{N}$ taking non-negative integer values, and for all $m, n \in \mathbf{N}$ we have
$$
f(m+n)-f(m)-f(n)=0 \text{ or } 1 \text{, }
$$
and $f(2)=0, f(3)>0, f(6000)=2000$.
Find $f(5961)$. | Solve: From $0=f(2) \geqslant 2 f(1) \Rightarrow f(1)=0$;
From $f(3)-f(2)-f(1)=0$ or 1
$$
\Rightarrow 0 \leqslant f(3) \leqslant 1 \text {. }
$$
But $f(3)>0$, hence $f(3)=1$.
By the problem statement, we know
$$
\begin{array}{l}
f(3 n+3)=f(3 n)+3+0 \text { or } 1 \\
\Rightarrow f(3(n+1)) \geqslant f(3 n)+1 .
\end{arra... | 1987 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The function $f(x)(x \neq 1)$ defined on $\mathbf{R}$ satisfies $f(x)+2 f\left(\frac{x+2002}{x-1}\right)=4015-x$. Then $f(2004)=(\quad)$. | Let $x=2, x=2004$, we get
$$
f(2004)=2005 .
$$ | 2005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. People numbered $1,2, \cdots, 2015$ are arranged in a line, and a position-swapping game is played among them, with the rule that each swap can only occur between adjacent individuals. Now, the person numbered 100 and the person numbered 1000 are to swap positions, with the minimum number of swaps required being $\q... | $$
-, 1.1799 .
$$
Using the formula, the minimum number of swaps required is
$$
(1000-100) \times 2-1=1799
$$
times.
| 1799 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c=-1, a+b+c=4, \\
\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}=1 .
\end{array}
$$
Find the value of $a^{2}+b^{2}+c^{2}$. | From the conditions given in the problem, we have
$$
\frac{1}{a}=-b c, \quad a=4-b-c \text {. }
$$
Notice that,
$$
\begin{array}{l}
\frac{a}{a^{2}-3 a-1}=\frac{1}{a-3-\frac{1}{a}}=\frac{1}{b c-b-c+1} \\
=\frac{1}{(b-1)(c-1)} .
\end{array}
$$
Similarly, $\frac{b}{b^{2}-3 b-1}=\frac{1}{(c-1)(a-1)}$,
$\frac{c}{c^{2}-3 c... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the maximum value of the positive integer $r$ that satisfies the following condition: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. ${ }^{\text {[2] }}$
(2013, Romanian National Team Selection Exam) | 【Analysis】Similarly, map the five subsets of 500 elements each to five vectors in a 1000-dimensional linear space. Since the requirement is the number of elements rather than their parity, we can consider the Euclidean space.
Let $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ be the five vectors after transformation.
Notice,
$$
\... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Function
$$
y=\tan 2013 x-\tan 2014 x+\tan 2015 x
$$
The number of zeros of the function in the interval $[0, \pi]$ is $\qquad$. | 2. 2014 .
$$
\begin{aligned}
y & =\tan 2013 x-\tan 2014 x+\tan 2015 x \\
& =\frac{\sin (2013 x+2015 x)}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{\sin 4028 x}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{2 \sin 2014 x \cdot \cos 2014 x}{\cos 2013 x \cdot \c... | 2014 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $P_{1}$ and $P_{2}$ be two points on a plane, $P_{2 k+1}\left(k \in \mathbf{Z}_{+}\right)$ be the symmetric point of $P_{2 k}$ with respect to $P_{1}$, and $P_{2 k+2}$ be the symmetric point of $P_{2 k+1}$ with respect to $P_{2}$. If $\left|P_{1} P_{2}\right|=1$, then $\left|P_{2013} P_{2014}\right|=$ $\qquad$ . | 4.4024.
From the problem, we know
$$
\begin{array}{l}
\left\{\begin{array}{l}
P_{2 k+1}=2 P_{1}-P_{2 k}, \\
P_{2 k+2}=2 P_{2}-P_{2 k+1}
\end{array}\right. \\
\Rightarrow P_{2 k+2}=2\left(P_{2}-P_{1}\right)+P_{2 k} \\
\Rightarrow\left\{\begin{array}{l}
P_{2 k+2}=2 k\left(P_{2}-P_{1}\right)+P_{2}, \\
P_{2 k+1}=2 k\left(... | 4024 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=0, a_{2}=1$, and for all $n \geqslant 3, a_{n}$ is the smallest positive integer greater than $a_{n-1}$ such that there is no subsequence of $a_{1}, a_{2}, \cdots, a_{n}$ that forms an arithmetic sequence. Find $a_{2014}$. | 4. First, prove a lemma using mathematical induction.
Lemma A non-negative integer appears in the sequence if and only if its ternary expansion contains only 0 and 1.
Proof It is obvious that the proposition holds for 0.
Assume the proposition holds for all non-negative integers less than \( N \), and consider \( N \)... | 88327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Prove: In the prime factorization of the product of any 10 consecutive three-digit numbers, there are at most 23 distinct prime factors. | First, in the prime factorization of each three-digit number, at most two prime factors greater than 10 can appear; otherwise, their product would exceed 1000, which is impossible.
Second, in any sequence of 10 consecutive three-digit numbers, there is one that is a multiple of 10, and in its prime factorization, at m... | 23 | Number Theory | proof | Yes | Yes | cn_contest | false |
3. In the 100th year of Besmiki's tenure as the President of the Currency Authority, he decided to issue new gold coins. In this year, he put into circulation an unlimited number of gold coins with a face value of $2^{100}-1$ yuan. In the following year, he put into circulation an unlimited number of gold coins with a ... | 3. It happens in the 200th year of Besmiki's presidency.
Assume that the described scenario occurs in the $k$-th year of Besmiki's presidency. Then,
$$
2^{k}-1=a_{1}+a_{2}+\cdots+a_{n}=N-n,
$$
where $N$ is the sum of some powers of 2, all of which are divisible by $2^{100}$.
Since $2^{k}$ is also divisible by $2^{10... | 200 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let the decimal representation of the positive integer $N$ consist only of the digits 1 and 2. By deleting digits from $N$, one can obtain all 10000 different positive integers formed by 9999 digits 1 and 1 digit 2. Find the minimum possible number of digits in $N$. | 7. The minimum possible number of digits in $N$ is 10198.
For example, $N=\underbrace{1 \cdots 1}_{99 \uparrow} \underbrace{1 \cdots}_{100 \uparrow} 12 \underbrace{1 \cdots}_{100 \uparrow} \underbrace{2}_{98 \uparrow} \underbrace{1 \cdots 1}_{99 \uparrow}$.
For a number formed by 9999 digits 1 and 1 digit 2, if there ... | 10198 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. The integer $n$ that satisfies $\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{2014}\right)^{2014}$ is $=$. | 6. -2015 .
Notice that for any $x \in(-1,+\infty)$ we have
$$
\frac{x}{1+x} \leqslant \ln (1+x) \leqslant x \text {. }
$$
Then for $f(x)=\left(1+\frac{1}{x}\right)^{x+1}(x>0)$ and
$$
g(x)=\left(1+\frac{1}{x}\right)^{x}(x>0)
$$
the derivatives are respectively
$$
\begin{array}{l}
f^{\prime}(x)=\left(1+\frac{1}{x}\rig... | -2015 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. If $x, y, z > 0$ satisfy
$$
\left\{\begin{array}{l}
\frac{2}{5} \leqslant z \leqslant \min \{x, y\}, \\
x z \geqslant \frac{4}{15}, \\
y z \geqslant \frac{1}{5},
\end{array}\right.
$$ | 7.13.
From the problem, we have
$$
\frac{1}{\sqrt{x}} \leqslant \frac{\sqrt{15 z}}{2}, \frac{1}{z} \leqslant \frac{5}{2}, \frac{1}{\sqrt{y}} \leqslant \sqrt{5 z} \text {. }
$$
Then
$$
f=\frac{2}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}+\frac{1}{z}\left(1-\frac{z}{x}\right)+
$$
$$
\begin{aligned}
& 2\left[\frac{2}{\sqrt{y}... | 13 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. Rearrange the six-element array $(1,2,3,4,5,6)$ to
$$
\begin{array}{l}
A=\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right), \\
\text { and } \quad B=\left(b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\right) .
\end{array}
$$
Then the minimum value of $P=\sum_{i=1}^{6} i a_{i} b_{i}$ is | 8. 162 .
From the geometric mean
$$
G=\sqrt[6]{\prod_{i=1}^{6} i a_{i} b_{i}}=\sqrt[6]{(6!)^{3}}=12 \sqrt{5} \in(26,27),
$$
we know that there exists at least one term not less than 27, and at least one term not greater than 25.
Let $i_{1} a_{i_{1}} b_{i_{1}} \leqslant 25, i_{2} a_{i_{2}} b_{i_{2}} \geqslant 27$.
$$
... | 162 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. The largest positive integer $n$ for which the inequality $\frac{9}{17}<\frac{n}{n+k}<\frac{8}{15}$ holds for a unique integer $k$ is $\qquad$ | 2. 144 .
From the problem, we know that $\frac{7}{8}<\frac{k}{n}<\frac{8}{9}$.
By the uniqueness of $k$, we have
$$
\begin{array}{l}
\frac{k-1}{n} \leqslant \frac{7}{8}, \text { and } \frac{k+1}{n} \geqslant \frac{8}{9} . \\
\text { Therefore, } \frac{2}{n}=\frac{k+1}{n}-\frac{k-1}{n} \geqslant \frac{8}{9}-\frac{7}{8}... | 144 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given positive integers $a, b, c$ satisfy
$$
1<a<b<c, a+b+c=111, b^{2}=a c \text {. }
$$
then $b=$ $\qquad$ | 4. 36.
Let $(a, c)=d, a=a_{1} d, c=c_{1} d, a_{1}, c_{1}$ be positive integers, and $\left(a_{1}, c_{1}\right)=1, a_{1}<c_{1}$.
Then $b^{2}=a c=d^{2} a_{1} c_{1} \Rightarrow d^{2}\left|b^{2} \Rightarrow d\right| b$.
Let $b=b_{1} d\left(b_{1} \in \mathbf{Z}_{+}\right)$. Then $b_{1}^{2}=a_{1} c_{1}$.
Since $\left(a_{1},... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $n$ be an integer. If there exist integers $x, y, z$ satisfying
$$
n=x^{3}+y^{3}+z^{3}-3 x y z \text {, }
$$
then $n$ is said to have property $P$.
(1) Determine whether $1, 2, 3$ have property $P$;
(2) Among the 2014 consecutive integers $1, 2, \cdots, 2014$, how many do not have property $P$? | (1) Let $x=1, y=z=0$, we get
$$
1=1^{3}+0^{3}+0^{3}-3 \times 1 \times 0 \times 0 \text {. }
$$
Thus, 1 has property $P$.
Let $x=y=1, z=0$, we get
$$
2=1^{3}+1^{3}+0^{3}-3 \times 1 \times 1 \times 0 \text {. }
$$
Thus, 2 has property $P$.
If 3 has property $P$, then there exist integers $x, y, z$ such that
$$
\begin{a... | 448 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $n$ is a positive integer, then
$$
\sum_{n=1}^{2014}\left(\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]\right)=
$$
$\qquad$ | 7. 2027091.
Let $f(n)=\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]$. Then
$$
\begin{array}{l}
f(0)=0, f(1)=0, f(2)=1, \\
f(3)=2, f(4)=3, f(5)=3 .
\end{array}
$$
For a positive integer $k$, we have
$$
\begin{array}{l}
f(6 k)=\left[\frac{6 k}{2}\right]+\left[\frac{6 k}{3}\right]+\left[\fra... | 2027091 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure 1, under the rules of Chinese chess, the "pawn" at point $A$ can reach point $B$ through a certain path (before crossing the river, the pawn can only move to the adjacent intersection directly in front of it each step; after crossing the river, it can move to the adjacent intersections in front, t... | 7. 6561.
Assume the chessboard has 10 horizontal lines from bottom to top, sequentially labeled as the 1st, 2nd, ..., 10th rows, and 9 vertical lines from left to right, sequentially labeled as the 1st, 2nd, ..., 9th columns. For example, point $A$ is located at the 4th row and 5th column.
Note that, during the movem... | 6561 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $n+2$ real numbers
$$
a_{1}, a_{2}, \cdots, a_{n}, 16, a_{n+2} \text {, }
$$
where the average of the first $n$ numbers is 8, the average of the first $n+1$ numbers is 9, and the average of these $n+2$ numbers is 10. Then the value of $a_{n+2}$ is $\qquad$ | 2. 18 .
From the condition, $\frac{8 n+16}{n+1}=9 \Rightarrow n=7$.
Also, $\frac{8 \times 7+16+a_{n+2}}{7+1+1}=10 \Rightarrow a_{n+2}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. If $a \in A$, and $a-1 \notin A, a+1 \notin A$, then $a$ is called an isolated element of set $A$. Therefore, the number of four-element subsets of set $M=\{1,2, \cdots, 9\}$ without isolated elements is $\qquad$ . | 9. 21 .
Consider the smallest element $i$ and the largest element $j$ in a set that satisfies the condition.
Let this set be $A$. Then $i+1 \in A, j-1 \in A$ (otherwise, $i$ or $j$ would be an isolated element).
Thus, $A=\{i, i+1, j-1, j\}$.
And $2 \leqslant i+1 < j-1 \leqslant 8$, so the number of ways to choose $i+... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Divide the sequence of positive integers $1,2, \cdots$ from left to right into segments such that the first segment has $1 \times 2$ numbers, the second segment has $2 \times 3$ numbers, $\cdots$, the $n$-th segment has $n \times(n+1)$ numbers, $\cdots$. Then 2014 is in the $\qquad$ segment. | $-, 1.18$.
$$
\begin{array}{l}
\text { Let } S_{n}=1 \times 2+2 \times 3+\cdots+n(n+1) \\
=\left(1^{2}+2^{2}+\cdots+n^{2}\right)+(1+2+\cdots+n) \\
=\frac{n(n+1)(n+2)}{3} .
\end{array}
$$
If 2014 is in the $(n+1)$-th segment, since there are $S_{n}$ numbers before this segment, then $S_{n}<2014 \leqslant S_{n+1}$.
Sinc... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. If the fraction $\frac{p}{q}\left(p, q \in \mathbf{Z}_{+}\right)$ is converted to a decimal as
$$
\frac{p}{q}=0.198 \cdots,
$$
then when $q$ takes the minimum value, $p+q=$ . $\qquad$ | 6. 121 .
Given $\frac{p}{q}=0.198 \cdots5 p$.
Let $q=5 p+m\left(m \in \mathbf{Z}_{+}\right)$. Then
$$
\begin{array}{l}
\frac{p}{5 p+m}=0.198 \cdots \\
\Rightarrow 0.198(5 p+m)<p<0.199(5 p+m) \\
\Rightarrow 19.8 m<p<39.8 m .
\end{array}
$$
When $m=1$, $20 \leqslant p \leqslant 39$, taking $p=20, m=1$, $q$ is minimized... | 121 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. $A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{0114}}{9}\right]$ when divided by 63 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 9. 56 .
Notice that, for any positive integer $k$, $\frac{8^{2 k-1}}{9}$ and $\frac{8^{2 k}}{9}$ are not integers, and
$$
\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}=8^{2 k-1} \text {. }
$$
Therefore, for any positive integer $k$, we have
$$
\begin{array}{l}
{\left[\frac{8^{2 k-1}}{9}\right]+\left[\frac{8^{2 k}}{9}\right]=... | 56 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For any positive integer $n$, the function $f(n)$ is the sum of the digits (i.e., the digital sum) of $n^{2}+3 n+1$ in decimal notation. Question: Does there exist an integer $n$ such that
$$
f(n)=2013 \text { or } 2014 \text { or } 2015 \text { ? }
$$ | When $3 \mid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 1(\bmod 3) ;
$$
When $3 \nmid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 2(\bmod 3) \text {. }
$$
Thus, $3 \nmid f(n)$.
Since 312013, there does not exist an integer $n$ such that
$$
f(n)=2013 \text {. }
$$
If there exists an integer $n$ such that $f(n)=2014$, then by $2... | 2015 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A. The minimum value of the algebraic expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$ is | $=、 6$. A. 13.
This problem can be transformed into finding the minimum value of the sum of distances from a point $(x, 0)$ on the $x$-axis to the points $(0,2)$ and $(12,3)$ in a Cartesian coordinate system.
Since the symmetric point of $(0,2)$ with respect to the $x$-axis is $(0,-2)$, the length of the line segment ... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let real numbers $x_{1}, x_{2}, \cdots, x_{2014}$ satisfy
$$
\left|x_{1}\right|=99,\left|x_{n}\right|=\left|x_{n-1}+1\right| \text {, }
$$
where, $n=2,3, \cdots, 2014$. Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$. | 11. From the given, we have
$$
x_{n}^{2}=x_{n-1}^{2}+2 x_{n}+1(n=2,3, \cdots, 2014) \text {. }
$$
Adding the above 2013 equations, we get
$$
\begin{array}{l}
x_{2014}^{2}=x_{1}^{2}+2\left(x_{1}+x_{2}+\cdots+x_{2013}\right)+2013 \\
\Rightarrow 2\left(x_{1}+x_{2}+\cdots+x_{2014}\right) \\
\quad=x_{2014}^{2}+2 x_{2014}-2... | -5907 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Consider each permutation of $1,2, \cdots, 8$ as an eight-digit number. Then the number of eight-digit numbers that are multiples of 11 is $\qquad$ | 8. 4608 .
For each such eight-digit number $\overline{a_{1} a_{2} \cdots a_{8}}$, let
$$
A=\left\{a_{1}, a_{3}, a_{5}, a_{7}\right\}, B=\left\{a_{2}, a_{4}, a_{6}, a_{8}\right\} .
$$
Let $S(A)$ and $S(B)$ denote their digit sums, and assume $S(A) \geqslant S(B)$.
Then $S(A)+S(B)=36$.
Thus, $S(A)$ and $S(B)$ have the ... | 4608 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given natural numbers $a, b, c$ whose sum is $S$, satisfying $a+b=1014, c-b=497, a>b$. Then the maximum value of $S$ is ( ).
(A) 2014
(B) 2015
(C) 2016
(D) 2017 | -1. D.
From the given, we have $a \geqslant b+1$.
Then $1014=a+b \geqslant 2 b+1$
$$
\begin{array}{l}
\Rightarrow b \leqslant 506.5 \Rightarrow b \leqslant 506 . \\
\text { Also } S=(a+b)+(c-b)+b \\
=1014+497+b=1511+b \\
\leqslant 1511+506=2017,
\end{array}
$$
Therefore, the maximum value of $S$ is 2017. | 2017 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given a set of data consisting of seven positive integers, the only mode is 6, and the median is 4. Then the minimum value of the sum of these seven positive integers is $\qquad$ | 2, 1.26.
Arrange these seven positive integers in ascending order, it is clear that the fourth number is 4. If 6 appears twice, then the other four numbers are $1, 2, 3, 5$, their sum is smaller, being 27; if 6 appears three times, the other three numbers are $1, 1, 2$, their sum is 26.
Therefore, the minimum sum of th... | 26 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given the sequence $\left\{a_{n}\right\}$ satisfies
$a_{1}=1, a_{2 n}=\left\{\begin{array}{ll}a_{n}, & n \text { is even; } \\ 2 a_{n}, & n \text { is odd, }\end{array}\right.$
$a_{2 n+1}=\left\{\begin{array}{ll}2 a_{n}+1, & n \text { is even; } \\ a_{n}, & n \text { is odd. }\end{array}\right.$
Find the number of posi... | Solve: From the given, we have
$$
\begin{array}{l}
a_{2014}=2 a_{1007}=2 a_{503}=2 a_{251}=2 a_{125} \\
=2\left(2 a_{62}+1\right)=2\left(4 a_{31}+1\right) \\
=2\left(4 a_{15}+1\right)=2\left(4 a_{7}+1\right) \\
=2\left(4 a_{3}+1\right)=2\left(4 a_{1}+1\right)=10 .
\end{array}
$$
In fact, $a_{n}$ can be obtained throug... | 320 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that Figure 1 is the graph of an even function $f(x)$, and Figure 2 is the graph of an odd function $g(x)$.
Let the number of real roots of the equations $f(f(x))=0, f(g(x))=0$,
$$
g(g(x))=0, g(f(x))=0
$$
be $a, b, c, d$ respectively. Then
$$
a+b+c+d=
$$
$\qquad$ | 3. 30 .
From the graph, we know that the range of the function $y=f(x)$ is $[-1,1]$, and the range of $y=g(x)$ is $[-2,2]$.
Notice that, the roots of the equation $f(x)=0$ are $0, x_{1}, x_{2}$, with $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$;
The roots of the equation $g(x)=0$ are $0, x_{3}, x_{4}$, with $\le... | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set $P=\{1,2, \cdots, 2014\}, A \cong P$. If any two numbers in set $A$ have a difference that is not a multiple of 99, and the sum of any two numbers is also not a multiple of 99, then the set $A$ can contain at most $\qquad$ elements. | 5.50.
Let the set
$$
B_{i}=\{99 \times 1+i, 99 \times 2+i, \cdots, 99 \times 20+i\} \text {, }
$$
where, $i=0,1, \cdots, 34$;
$$
B_{j}=\{99 \times 1+j, 99 \times 2+j, \cdots, 99 \times 19+j\},
$$
where, $j=35,36, \cdots, 98$.
Take any $a, b \in A$.
Since the difference between any two numbers in set $A$ is not a mul... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $0<x<\frac{\pi}{2}, \sin x-\cos x=\frac{\pi}{4}$. If $\tan x+\frac{1}{\tan x}$ can be expressed in the form $\frac{a}{b-\pi^{c}}$ ($a$, $b$, $c$ are positive integers), then $a+b+c=$ $\qquad$ . | 8. 50 .
Squaring both sides of $\sin x-\cos x=\frac{\pi}{4}$ and rearranging yields $\sin x \cdot \cos x=\frac{16-\pi^{2}}{32}$.
Therefore, $\tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$
$$
=\frac{1}{\sin x \cdot \cos x}=\frac{32}{16-\pi^{2}} \text {. }
$$
Thus, $a=32, b=16, c=2$.
Hence, $a+b+... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the positive integer $n$ satisfy $31 \mid\left(5^{n}+n\right)$. Then the minimum value of $n$ is $\qquad$ . | 9. 30 .
Given $5^{3} \equiv 1(\bmod 31)$, when $n=3 k\left(k \in \mathbf{Z}_{+}\right)$, $5^{n}+n \equiv 1+n \equiv 0(\bmod 31)$, at this time, the minimum value of $n$ is $n_{\text {min }}=30$;
When $n=3 k+1\left(k \in \mathbf{Z}_{+}\right)$,
$$
5^{n}+n \equiv 5+n \equiv 0(\bmod 31),
$$
at this time, the minimum val... | 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Given
$$
S_{n}=|n-1|+2|n-2|+\cdots+10|n-10| \text {, }
$$
where, $n \in \mathbf{Z}_{+}$. Then the minimum value of $S_{n}$ is $\qquad$ | 10. 112 .
From the problem, we know
$$
\begin{array}{l}
S_{n+1}-S_{n} \\
=|n|+2|n-1|+\cdots+10|n-9|- \\
{[|n-1|+2|n-2|+\cdots+10|n-10|] } \\
=|n|+|n-1|+\cdots+|n-9|-10|n-10| .
\end{array}
$$
When $n \geqslant 10$, $S_{n+1}-S_{n}>0$, thus, $S_{n}$ is monotonically increasing; $\square$
When $n=0$, $S_{1}-S_{0}>0$;
Wh... | 112 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $15 \mid \overline{\text { xaxax }}$, then the sum of all five-digit numbers $\overline{x a x a x}$ that satisfy the requirement is ( ).
(A) 50505
(B) 59595
(C) 110100
(D) 220200 | - 1. D.
Notice that, $15| \overline{\text { xaxax }} \Leftrightarrow\left\{\begin{array}{l}5 \mid \overline{\text { xaxax }}, \\ 3 \mid \overline{\text { xaxax }} .\end{array}\right.$
From $51 \overline{\text { xaxax }} \Rightarrow x=5$;
From $31 \overline{x a x a x} \Rightarrow 31(x+a+x+a+x)$
$$
\Rightarrow 3|2 a \Ri... | 220200 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. Let $a_{1}, a_{2}, \cdots, a_{2015}$ be a sequence of numbers taking values from $-1, 0, 1$, satisfying
$$
\sum_{i=1}^{2015} a_{i}=5 \text {, and } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040,
$$
where $\sum_{i=1}^{n} a_{i}$ denotes the sum of $a_{1}, a_{2}, \cdots, a_{n}$.
Then the number of 1's in this sequenc... | $=, 1.510$.
Let the number of -1's be $x$, and the number of 0's be $y$.
Then, from $\sum_{i=1}^{2015} a_{i}=5$, we know the number of 1's is $x+5$. Combining this with
$$
\begin{array}{c}
\text { the equation } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040, \text { we get } \\
\left\{\begin{array}{l}
x+(x+5)+y=2015, ... | 510 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If the integer $n$
satisfies
$$
(n+1)^{2} \mid\left(n^{2015}+1\right) \text {, }
$$
then the minimum value of $n$ is $\qquad$ | 3. -2016 .
From $n^{2015}+1=(n+1) \sum_{i=0}^{2014}(-1)^{i} n^{2014-i}$, we know
$$
\begin{array}{l}
(n+1)^{2} \mid\left(n^{2015}+1\right) \\
\left.\Leftrightarrow(n+1)\right|_{i=0} ^{2014}(-1)^{i} n^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2014}(-1)^{i}(-1)^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2... | -2016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. To color the eight vertices of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with four different colors, such that the two endpoints of the same edge have different colors, there are a total of coloring methods. | 8. 2652 .
First, color the four points $A, B, C, D$ above, with 84 coloring methods. Then consider the four points below, using the principle of inclusion-exclusion, the total number of methods is
$$
\begin{array}{l}
84\left\{84-C_{4}^{1}[3 \times(3+2 \times 2)+\right. \\
(3+2 \times 2)]+2\left(\frac{36}{84} \times 9+... | 2652 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[5]}$
$(2012$, Girls' Mathematical Olympiad) | 【Analysis】Perform prime factorization $2012=2^{2} \times 503$.
First consider 503 I $\mathrm{C}_{2012}^{k}$.
By Corollary 3, we know that when and only when $k$ and $2012-k$ do not produce a carry when added in base 503, $\left(503, \mathrm{C}_{2012}^{k}\right)=1$.
Writing 2012 in base 503 gives $2012=(40)_{503}$.
Let ... | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a_{1}, a_{2}, \cdots, a_{2014}$ be a permutation of the positive integers $1,2, \cdots$, 2014. Denote
$$
S_{k}=a_{1}+a_{2}+\cdots+a_{k}(k=1,2, \cdots, 2014) \text {. }
$$
Then the maximum number of odd numbers in $S_{1}, S_{2}, \cdots, S_{2014}$ is $\qquad$ | 6.1511.
If $a_{i}(2 \leqslant i \leqslant 2014)$ is odd, then $S_{i}$ and $S_{i-1}$ have different parities. From $a_{2}$ to $a_{2014}$, there are at least $1007-1=1006$ odd numbers. Therefore, $S_{1}, S_{2}, \cdots, S_{2014}$ must change parity at least 1006 times.
Thus, $S_{1}, S_{2}, \cdots, S_{2014}$ must have at... | 1511 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Print 90000 five-digit numbers
$$
10000,10001, \cdots, 99999
$$
on cards, with one five-digit number on each card. Some cards (such as 19806, which reads 90861 when flipped) have numbers that can be read in two different ways, causing confusion. The number of cards that will not cause confusion is $\qquad$ cards. | 8. 88060 .
Among the ten digits $0 \sim 9$, the digits that can still represent numbers when inverted are $0, 1, 6, 8, 9$.
Since the first digit cannot be 0 and the last digit cannot be 0, the number of such five-digit numbers that can be read when inverted is $4 \times 5 \times 5 \times 5 \times 4=2000$.
Among these... | 88060 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $A$ be a set composed of any 100 distinct positive integers, and let
$$
B=\left\{\left.\frac{a}{b} \right\rvert\, a 、 b \in A \text { and } a \neq b\right\},
$$
$f(A)$ denotes the number of elements in set $B$. Then the sum of the maximum and minimum values of $f(A)$ is $\qquad$ . | 11. 10098.
From the problem, when the elements in set $B$ are pairwise coprime, $f(A)$ reaches its maximum value $\mathrm{A}_{100}^{2}=9900$; when the elements in set $B$ form a geometric sequence with a common ratio not equal to 1, $f(A)$ reaches its minimum value of $99 \times 2=198$.
Therefore, the sum of the maxim... | 10098 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let
$$
\begin{array}{l}
A=\{1,2, \cdots, 2014\}, \\
B_{i}=\left\{x_{i}, y_{i}\right\}(i=1,2, \cdots, t)
\end{array}
$$
be $t$ pairwise disjoint binary subsets of $A$, and satisfy the conditions
$$
\begin{array}{l}
x_{i}+y_{i} \leqslant 2014(i=1,2, \cdots, t), \\
x_{i}+y_{i} \neq x_{j}+y_{j}(1 \leqslant i<j \leqslan... | 8. 805 .
On the one hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\left\{x_{i}, y_{i}\right\} \cap\left\{x_{j}, y_{j}\right\}=\varnothing \text {. }
$$
Thus, $x_{1}, x_{2}, \cdots, x_{t}, y_{1}, y_{2}, \cdots, y_{t}$ are $2 t$ distinct integers. Therefore,
$$
\sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \geqslant... | 805 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the set $A=\{1,2,3\}, f$ and $g$ are functions from set $A$ to $A$. Then the number of function pairs $(f, g)$ whose image sets intersect at the empty set is $\qquad$ . | 4. 42 .
When the image set of function $f$ is 1 element, if the image set of function $f$ is $\{1\}$, at this time the image set of function $g$ is a subset of $\{2,3\}$, there are $2^{3}=8$ kinds, so there are $3 \times 8=24$ pairs of functions $(f, g)$ that meet the requirements.
When the image set of function $f$ ... | 42 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $f(x)=\left(x^{2}+3 x+2\right)^{\cos \pi x}$. Then the sum of all $n$ that satisfy the equation
$$
\left|\sum_{k=1}^{n} \log _{10} f(k)\right|=1
$$
is | 5. 21 .
It is known that for integer $x$, we have
$$
f(x)=[(x+1)(x+2)]^{(-1)^{x}} \text {. }
$$
Thus, when $n$ is odd,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2-\log _{10}(n+2) \text {; }
$$
When $n$ is even,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2+\log _{10}(n+2) \text {. }
$$
Therefore, the $n$ that... | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given real numbers $a_{0}, a_{1}, \cdots, a_{2015}$, $b_{0}, b_{1}, \cdots, b_{2011}$ satisfy
$$
\begin{array}{l}
a_{n}=\frac{1}{65} \sqrt{2 n+2}+a_{n-1}, \\
b_{n}=\frac{1}{1009} \sqrt{2 n+2}-b_{n-1},
\end{array}
$$
where $n=1,2, \cdots, 2015$.
If $a_{0}=b_{2015}$, and $b_{0}=a_{2015}$, find the value ... | 10. Notice that, for any $k=1,2 \cdots, 2015$, we have
$$
\begin{array}{l}
a_{k}-a_{k-1}=\frac{1}{65} \sqrt{2 k+2}, \\
b_{k}+b_{k-1}=\frac{1}{1009} \sqrt{2 k+2} .
\end{array}
$$
Multiplying the two equations, we get
$$
\begin{array}{l}
a_{k} b_{k}-a_{k-1} b_{k-1}+a_{k} b_{k-1}-a_{k-1} b_{k} \\
=\frac{2 k+2}{65 \times ... | 62 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A positive integer that can be expressed as the difference of squares of two positive integers is called a "wise number". For example, $9=5^{2}-4^{2}, 9$ is a wise number.
(1) Try to determine which numbers among the positive integers are wise numbers, and explain the reason;
(2) In the sequence of wise numbers arr... | 14. (1) It is easy to know that the positive integer 1 cannot be expressed as the difference of squares of two positive integers, i.e., 1 is not a wise number.
For odd numbers greater than 1, we have
$$
2 k+1=(k+1)^{2}-k^{2}(k=1,2, \cdots),
$$
which means that all odd numbers greater than 1 are wise numbers.
When $k=2... | 2689 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
16. The basketball league has eight teams. Each season, each team plays two games (home and away) against each of the other teams in the league, and each team also plays four games against opponents outside the league. Therefore, in one season, the eight teams in the league play a total of ( ) games.
(A) 60
(B) 88
(C) ... | 16. B.
There are $\mathrm{C}_{8}^{2}=28$ pairs that can be formed from the eight teams in the league.
Since each pair of teams plays both a home and an away match, the eight teams in the league play a total of $28 \times 2=56$ matches.
Since each team plays four matches against opponents outside the league, the eight... | 88 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
14. Let the angle between vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ be $\frac{\pi}{3}$, the angle between vectors $\boldsymbol{c}-\boldsymbol{a}$ and $\boldsymbol{c}-\boldsymbol{b}$ be $\frac{2 \pi}{3}$, $|\boldsymbol{a}-\boldsymbol{b}|=5$, and $|\boldsymbol{c}-\boldsymbol{a}|=2 \sqrt{3}$. Then the maximum value of... | 14. 24 .
Let $\overrightarrow{O A}=a, \overrightarrow{O B}=b, \overrightarrow{O C}=c$. Then
$$
|\overrightarrow{A C}|=|c-a|=2 \sqrt{3},|\overrightarrow{A B}|=|a-b|=5 \text {. }
$$
Also, $\angle A O B=\frac{\pi}{3}, \angle A C B=\frac{2 \pi}{3}$, at this time, $O, A, C, B$ are concyclic.
By the Law of Sines, we get
$$... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Allocate 24 volunteer slots to 3 schools. Then the number of allocation methods where each school gets at least one slot and the number of slots for each school is different is $\qquad$ kinds. ${ }^{[2]}$ | Let the quotas allocated to schools A, B, and C be $x$, $y$, and $z$ respectively.
First, without considering that $x, y, z$ are pairwise distinct.
From $x+y+z=24$, we get a total number of combinations $\mathrm{C}_{23}^{2}$.
Next, we separate out the number of positive integer solutions $(x, y, z)$ that are not pairwi... | 222 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. The sum of the ages of three people, A, B, and C, represented by $x, y, z$ is 120, and $x, y, z \in (20,60)$. Then the number of ordered triples $(x, y, z)$ is $\qquad$ | 8. 1141 .
Notice a basic conclusion:
The number of positive integer solutions $(a, b, c)$ to the indeterminate equation $a+b+c=n$ is $\mathrm{C}_{n-1}^{2}$.
Thus, the number of solutions to the indeterminate equation
$$
(x-20)+(y-20)+(z-20)=60
$$
satisfying $x, y, z>20$ is $\mathrm{C}_{59}^{2}$.
Among these $\mathrm{... | 1141 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The sequence $\left\{a_{n}\right\}$ has 9 terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. Find the number of such sequences. | Prompt: Categorized Count. Answer: 491. | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If numbers $1,2, \cdots, 14$ are taken in ascending order as $a_{1}, a_{2}, a_{3}$, such that $a_{2}-a_{1} \geqslant 3$, and $a_{3}-a_{2} \geqslant 3$, find the number of different ways to choose them. | From the given information, we have
$$
\begin{aligned}
a_{1} & \leqslant a_{2}-3 \leqslant a_{3}-6 \\
& \Rightarrow 1 \leqslant a_{1}<a_{2}-2<a_{3}-4 \leqslant 10 .
\end{aligned}
$$
Substitute $\left(a_{1}, a_{2}-2, a_{3}-4\right)=(x, y, z)$.
Then the number of tuples $\left(a_{1}, a_{2}, a_{3}\right)$ equals the numb... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 For any non-empty subset $X$ of the set $M=\{1,2, \cdots, 1000\}$, let $\alpha_{X}$ denote the sum of the maximum and minimum numbers in $X$. Find the arithmetic mean of all such $\alpha_{X}$. | 【Analysis】According to the problem, the required average is
$$
f=\frac{\sum_{\varnothing \neq X \subseteq M} \alpha_{X}}{2^{1000}-1} \text {. }
$$
The key to solving this is to calculate the sum in the numerator
$$
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \text {. }
$$
To compute such an "unordered sum", on... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 and 8. Find the number of different ways to select such a group of eight points. | 【Analysis】First analyze the essential structure of the data: Label 24 points in a clockwise direction as $1,2, \cdots, 24$. Then arrange them into the following $3 \times 8$ number table:
$$
\left(\begin{array}{cccccccc}
1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
17 & 20 & 23 & 2 & 5 & ... | 258 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Fill two $a$s and two $b$s into the 16 cells shown in Figure 3, with at most one letter per cell. If the same letters must not be in the same row or column, find the number of different ways to fill the cells. | Prompt: Categorized Count. Answer: 3960. | 3960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its positive divisors $1, 2, 17, 34, 59, 118, 1003, 2006$ is 3240. Find the smallest good number... | Let $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $\alpha_{i} \in \mathbf{Z}_{+}, p_{i} (i=1,2, \cdots, k)$ are prime numbers, and $p_{1}<p_{2}<\cdots<p_{k}$.
From $\tau(n)=\prod_{i=1}^{k}\left(1+\alpha_{i}\right)=8=2^{3}$, we know $k=1,2,3$.
(1) If $k=1$, then
$\alpha_{1}=7, n=p^{7}$ ($p$ is a prime number), $\sum_{i=... | 1614 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Let positive real numbers $a, b, c$ satisfy $ab + bc + ca = 48$. Try to find the minimum value of $f = \left(a^{2} + 5\right)\left(b^{2} + 5\right)\left(c^{2} + 5\right)$. | Given $A=48, k=5$, hence
$$
f \geqslant 5(48-5)^{2}=9245 \text {. }
$$
When $a=5,\{b, c\}=\left\{\frac{43+\sqrt{97}}{12}, \frac{43-\sqrt{97}}{12}\right\}$,
the equality holds.
Therefore, $f_{\min }=9245$. | 9245 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 1, in the right $\triangle A B C$, it is known that $\angle A C B$ $=90^{\circ}, A C=21, B C=$ 28, and a square $A B D E$ is constructed outward with $A B$ as one side. The angle bisector of $\angle A C B$ intersects $D E$ at point $F$. Then the length of line segment $D F$ is $\qquad$ | 8. 15 .
As shown in Figure 6, let $CF$ intersect $AB$ and $AD$ at points $G$ and $O$ respectively.
Then $\angle BCO = 45^{\circ}$
$= \angle OAB$.
Thus, $O, A, C, B$
are concyclic.
Hence $\angle ABO$
$$
= \angle OCA = 45^{\circ}.
$$
Therefore, $OA = OB$.
Thus, $O$ is the center of the square $ABDE$.
By symmetry, $DF ... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Given that $a$ and $b$ are real numbers, the system of inequalities about $x$
$$
\left\{\begin{array}{l}
20 x+a>0, \\
15 x-b \leqslant 0
\end{array}\right.
$$
has only the integer solutions $2, 3, 4$. Then the maximum value of $ab$ is . $\qquad$ | 10. -1200 .
From the conditions, we know that $-\frac{a}{20}<x \leqslant \frac{b}{15}$.
The integer solutions of the inequality system are only $2, 3, 4$, so $1 \leqslant-\frac{a}{20}<2,4 \leqslant \frac{b}{15}<5$, which means $-40<a \leqslant-20,60 \leqslant b<75$.
Therefore, $-3000<a b \leqslant-1200$.
Thus, when $a... | -1200 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
16. (25 points) Given $A \subseteq\{1,2, \cdots, 2014\}$, let real numbers $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} 、 x_{1} 、 x_{2} 、 x_{3}$ satisfy
(i) $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} \in\{-1,0,1\}$ and not all are 0;
(ii) $x_{1}, x_{2} 、 x_{3} \in A$;
(iii) If $x_{i}=x_{j}$, then $\lambda_{i} \lambda_{j} \ne... | 16. (1) Construct a good set $A$ with 503 elements.
Let $A=\{1,3,5, \cdots, 1005\}$.
If $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ are all non-zero, then
$$
\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3} \equiv x_{1}+x_{2}+x_{3} \equiv 1(\bmod 2) \text {. }
$$
Thus, $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_... | 503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. If positive integers $m, n$ satisfy $\frac{(m+n)!}{n!}=5040$, then the value of $m!n$ is $\qquad$ . | 9. 144 .
$$
\begin{array}{l}
\text { Given } \frac{(m+n)!}{n!} \\
=(m+n)(m+n-1) \cdots(n+1), \\
5040=10 \times 9 \times 8 \times 7,
\end{array}
$$
we know $\left\{\begin{array}{l}m+n=10, \\ n+1=7\end{array} \Rightarrow\left\{\begin{array}{l}m=4, \\ n=6 .\end{array}\right.\right.$
Therefore, $m!n=144$. | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. Let the monotonic increasing sequence $\left\{a_{n}\right\}$ consist of positive integers, and $a_{7}=120, a_{n+2}=a_{n}+a_{n+1}\left(n \in \mathbf{Z}_{+}\right)$. Then $a_{8}=$ . $\qquad$ | 10. 194 .
From $a_{n+2}=a_{n}+a_{n+1}$, we get $a_{7}=5 a_{1}+8 a_{2}=120, a_{8}=8 a_{1}+13 a_{2}$. Since $(5,8)=1$, and $a_{1}, a_{2}$ are both positive integers, it follows that $8\left|a_{1}, 5\right| a_{2}$.
Let $a_{1}=8 k, a_{2}=5 m\left(k, m \in \mathbf{Z}_{+}\right)$.
Then $k+m=3$.
Also, $a_{1}<a_{2}$, so, $k=1... | 194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $n$ be the smallest positive integer satisfying the following conditions:
(1) $n$ is a multiple of 75;
(2) $n$ has exactly 75 positive divisors (including 1 and itself).
Find $\frac{n}{75}$.
(Eighth American Mathematical Invitational) | ```
Given: $n=75 k=3 \times 5^{2} k$.
\[
\begin{array}{l}
\text { By } 75=3 \times 5 \times 5 \\
=(2+1)(4+1)(4+1),
\end{array}
\]
we know the number of prime factors is at most three.
By discussing the number of prime factors, we can solve to get
\[
\left(\frac{n}{75}\right)_{\min }=432 \text {. }
\]
``` | 432 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Choose three different angles from $1^{\circ}, 2^{\circ}, \cdots, 179^{\circ}$ to form the three interior angles of a triangle. There are $\qquad$ different ways to do this. | 3.2611.
Notice that, the equation $x+y+z=180$ has $\mathrm{C}_{179}^{2}=$ 15931 sets of positive integer solutions, among which, the solution where $x=y=z$ is 1 set, and the solutions where exactly two of $x, y, z$ are equal are $88 \times 3=$ 264 sets.
Therefore, the number of selection methods is $\frac{15931-1-264}... | 2611 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $f_{0}(x)=|x|-2015$,
$$
f_{n}(x)=\left|f_{n-1}(x)\right|-1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Then the number of zeros of the function $y=f_{2015}(x)$ is
$\qquad$ | 8.4031 .
From the graph, it is easy to see that the function $y=f_{1}(x)$ has 4 zeros, the function $y=f_{2}(x)$ has 6 zeros, $\cdots \cdots$ and so on, the function $y=f_{2014}(x)$ has 4030 zeros. However, the intersection point of the function $y=f_{2014}(x)$ with the y-axis is $(0,1)$, therefore, the function $y=f_... | 4031 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. There are infinitely many cards, each with a real number written on it. For each real number $x$, there is exactly one card with the number $x$ written on it. Two players each select a set of 100 cards, denoted as $A$ and $B$, such that the sets are disjoint. Formulate a rule to determine which of the two players wi... | 6. There are 100 ways.
To prove the more general case, where each person selects $n$ cards, there are $n$ ways that satisfy the conditions. Let $A>B$ or $B>A$ if and only if $a_{k}>b_{k}$.
Such rules satisfy all three conditions, and different $k$ correspond to different rules, hence there are at least $n$ different r... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given odd prime numbers $x, y, z$ satisfying
$$
x \mid \left(y^{5}+1\right), y \mid \left(z^{5}+1\right), z \mid \left(x^{5}+1\right) \text {. }
$$
Find the minimum value of the product $x y z$. (Cheng Chuanping, problem contributor) | 7. Let's assume $x$ is the minimum of $x, y, z$.
(1) If $x=3$, then
$$
3^{5}+1=244=2^{2} \times 61 \Rightarrow z=61.
$$
Since $3 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{3}$.
Clearly, $5 \nmid (61^{5}+1)$.
After calculation, we find $11 \mid (61^{5}+1)$.
Thus, $y_{\text{min}}=11$.
Therefore, $(x y z)_{\min}=3 \time... | 2013 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A coin collector has 100 coins that look the same. The collector knows that 30 of them are genuine, and 70 are fake, and that all genuine coins weigh the same, while all fake coins weigh different and are heavier than the genuine ones. The collector has a balance scale that can be used to compare the weight of two g... | 8. First, it is clear that 70 weighings can certainly find at least one genuine coin.
In fact, each time, one coin is placed on each side of the balance. If they weigh the same, both are genuine; if they do not, the heavier one must be a counterfeit. Thus, each weighing either finds two genuine coins or one counterfei... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
1. The function $f(x)$ defined on $\mathbf{R}$ satisfies $f\left(\frac{2 a+b}{3}\right)=\frac{2 f(a)+f(b)}{3}(a, b \in \mathbf{R})$,
and $f(1)=1, f(4)=7$.
Then $f(2015)=$ | $$
-, 1.4029 .
$$
From the function $f(x)$ being concave or convex in reverse on $\mathbf{R}$, we know its graph is a straight line, $f(x)=a x+b$.
Given $f(1)=1$ and $f(4)=7$, we have
$$
\begin{array}{l}
a=2, b=-1 \Rightarrow f(x)=2 x-1 \\
\Rightarrow f(2015)=4029 .
\end{array}
$$ | 4029 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Fill the numbers $1,2, \cdots, 36$ in a $6 \times 6$ grid, with each cell containing one number, such that the numbers in each row are in increasing order from left to right. Then the minimum value of the sum of the six numbers in the third column is $\qquad$ | 4. 63.
Let the six numbers filled in the third column, arranged in ascending order, be $A, B, C, D, E, F$.
Since the row where $A$ is located needs to fill in two numbers smaller than $A$ before it, then $A \geqslant 3$; since the row where $B$ is located needs to fill in two numbers smaller than $B$, and $A$ and the... | 63 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
In the plane rectangular coordinate system $x O y$, the set of points
$$
\begin{aligned}
K= & \{(x, y) \mid(|x|+|3 y|-6) . \\
& (|3 x|+|y|-6) \leqslant 0\}
\end{aligned}
$$
corresponds to a plane region whose area is $\qquad$ | 6. 24 .
Let $K_{1}=\{(x, y)|| x|+| 3 y |-6 \leqslant 0\}$.
First, consider the part of the point set $K_{1}$ in the first quadrant, at this time, $x+3 y \leqslant 6$. Therefore, these points correspond to $\triangle O C D$ and its interior in Figure 2.
By symmetry, the region corresponding to the point set $K_{1}$ is... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. For a four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant$ $9)$, if $a>b, b<c, c>d$, then $\overline{a b c d}$ is called a $P$ class number; if $a<c, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbe... | 8. 285.
Let the sets of all P-type numbers and Q-type numbers be denoted as $A$ and $B$, respectively. Let the set of all P-type numbers whose unit digit is zero be denoted as $A_{0}$, and the set of all P-type numbers whose unit digit is not zero be denoted as $A_{1}$.
For any four-digit number $\overline{a b c d} \... | 285 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ be an arithmetic sequence, and
$$
\sum_{i=1}^{n}\left|a_{i}+j\right|=2028(j=0,1,2,3) \text {. }
$$
Then the maximum value of the number of terms $n$ is | 6. 52.
Since the equation $|x|=|x+1|=|x+2|$ has no solution, we have $n \geqslant 2$ and the common difference is not 0.
Assume the general term of the sequence is $a-k d(1 \leqslant k \leqslant n, d>0)$.
Construct the function $f(x)=\sum_{k=1}^{n}|x-k d|$.
The given condition is equivalent to $f(x)=2028$ having at le... | 52 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example: $10 f(x)$ is a continuous function defined on the interval $[0,2015]$, and $f(0)=f(2015)$. Find the minimum number of real number pairs $(x, y)$ that satisfy the following conditions:
(1) $f(x)=f(y)$;
(2) $x-y \in \mathbf{Z}_{+}$.
$(2015$, Peking University Mathematics Summer Camp) | 【Analysis】Consider a continuous function $f(x)$ on the interval $[0, n]$ that satisfies $f(0)=f(n)$. Let the minimum number of real number pairs $(x, y)$ that satisfy (1) and (2) be $F(n)$.
Obviously, $F(1)=1, F(2)=2$.
We will prove: $F(n)=n\left(n \in \mathbf{Z}_{+}\right)$.
Assume that for $n \leqslant k-1$, equation... | 2015 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 In $\triangle ABC$ with a fixed perimeter, it is known that $AB=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
(1) Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ intersecting the c... | (1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of segment $AB$ as the $y$-axis.
Let $|CA| + |CB| = 2a (a > 3)$ be a constant. Then the locus of point $C$ is an ellipse with foci at $A$ and $B$.
Thus, the focal distance $2c = |AB| = 6$.
Notice,
$$
\begin{ar... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. How many different right-angled triangles with integer side lengths have an area that is 999 times their perimeter (considering congruent triangles as the same)? (Provided by Lin Chang)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result d... | 5. Let the three sides of a right-angled triangle be \(a, b, c\) (with \(c\) being the hypotenuse).
From the Pythagorean triple formula, we have
\[
a = k \cdot 2uv, \quad b = k(u^2 - v^2), \quad c = k(u^2 + v^2),
\]
where the greatest common divisor of the three sides \(k\) is a positive integer, \(u\) and \(v\) are co... | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The function $f(x)$ defined on $\mathbf{R}$, for any real number $x$, satisfies
$$
\begin{array}{l}
f(x+3) \leqslant f(x)+3, \\
f(x+2) \geqslant f(x)+2,
\end{array}
$$
and $f(1)=2$. Let $a_{n}=f(n)\left(n \in \mathbf{Z}_{+}\right)$, then
$$
f(2015)=
$$
$\qquad$ | 5.2016.
Notice,
$$
\begin{array}{l}
f(x)+3 \geqslant f(x+3) \\
=f(x+1+2) \geqslant f(x+1)+2 \\
\Rightarrow f(x)+1 \geqslant f(x+1) . \\
\text { Also } f(x)+4 \leqslant f(x+2)+2 \leqslant f(x+4) \\
=f(x+1+3) \leqslant f(x+1)+3 \\
\Rightarrow f(x+1) \geqslant f(x)+1 .
\end{array}
$$
Therefore, $f(x+1)=f(x)+1$.
Thus, $f... | 2016 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \text {. }
$$
Then the sum of the first 12 terms $S_{12}=$ $\qquad$ | 7. 104832 .
Notice that,
$$
\begin{array}{l}
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \\
=n(n+1)(n+2)(n+3) . \\
\text { Let } f(n)=\frac{1}{5} n(n+1)(n+2)(n+3)(n+4) .
\end{array}
$$
Then $a_{n}=f(n)-f(n-1), S_{n}=f(n)$.
Therefore, $S_{12}=f(12)=104832$. | 104832 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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