problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
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2. Assign five college students to three villages in a certain town. If each village must have at least one student, then the number of different assignment schemes is $\qquad$ . | 2. 150 .
According to the number of college students allocated to each village, the only two types that meet the requirements are $1,1,3$ and $1, 2, 2$. Therefore, the number of different allocation schemes is
$$
C_{3}^{1} C_{5}^{3} C_{2}^{1}+C_{3}^{1} C_{5}^{2} C_{3}^{2}=60+90=150
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a_{n}=2^{n}, b_{n}=5 n-1\left(n \in \mathbf{Z}_{+}\right)$,
$$
S=\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{a_{2015}}\right\} \text {. }
$$
Then the number of elements in the set $S$ is | 5.504.
Since the set $\left\{b_{1}, b_{2}, \cdots, b_{2015}\right\}$ contains $2^{2015}$ elements, forming an arithmetic sequence with a common difference of 5, and each term has a remainder of 4 when divided by 5, we only need to consider the number of terms in the set $\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\}$ ... | 504 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $P(1,4,5)$ is a fixed point in the rectangular coordinate system $O-x y z$, a plane is drawn through $P$ intersecting the positive half-axes of the three coordinate axes at points $A$, $B$, and $C$ respectively. Then the minimum value of the volume $V$ of all such tetrahedrons $O-A B C$ is $\qquad$ | 4. 90 .
Let the plane equation be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where the positive numbers $a$, $b$, and $c$ are the intercepts of the plane on the $x$-axis, $y$-axis, and $z$-axis, respectively.
Given that point $P$ lies within plane $ABC$, we have $\frac{1}{a}+\frac{4}{b}+\frac{5}{c}=1$.
From $1=\frac{1}{... | 90 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) From the 2015 positive integers 1, 2, $\cdots$, 2015, select $k$ numbers such that the sum of any two different numbers is not a multiple of 50. Find the maximum value of $k$.
untranslated part:
在 1,2 , $\cdots, 2015$ 这 2015 个正整数中选出 $k$ 个数,使得其中任意两个不同的数之和均不为 50 的倍数. 求 $k$ 的最大值.
translated part:
From ... | 10. Classify $1 \sim 2015$ by their remainders when divided by 50.
Let $A_{i}$ represent the set of numbers from $1 \sim 2015$ that have a remainder of $i$ when divided by 50, where $i=0,1, \cdots, 49$. Then $A_{1}, A_{2}, \cdots, A_{15}$ each contain 41 numbers, and $A_{0}, A_{16}, A_{17}, \cdots, A_{49}$ each contai... | 977 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Real numbers $x, y, a$ satisfy $x+y=a+1$ and $xy=a^{2}-7a+16$. Then the maximum value of $x^{2}+y^{2}$ is $\qquad$ | 5. 32 .
Notice that,
$$
\begin{array}{l}
x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=(a+1)^{2}-2\left(a^{2}-7 a+16\right) \\
=-a^{2}+16 a-31=-(a-8)^{2}+33 . \\
\text { Also, }(x-y)^{2}=(a+1)^{2}-4\left(a^{2}-7 a+16\right) \geqslant 0 \\
\Rightarrow-3 a^{2}+30 a-63 \geqslant 0 \Rightarrow 3 \leqslant a \leqslant 7 .
\end{array}
$$... | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a triangle with sides as three consecutive natural numbers, the largest angle is twice the smallest angle. Then the perimeter of the triangle is $\qquad$ | II, 7.15.
Let the three sides of $\triangle ABC$ be $AB=n+1, BC=n, AC=n-1$, and the angle opposite to side $AC$ be $\theta$, and the angle opposite to side $AB$ be $2\theta$.
According to the Law of Sines and the Law of Cosines, we have
$$
\begin{array}{l}
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the minimum value of the function
$$
f(x)=|x-1|+|x-2|+\cdots+|x-10|
$$
. ${ }^{[1]}$ | 【Analysis】By the geometric meaning of absolute value, $\sum_{i=1}^{n}\left|x-a_{i}\right|$ represents the sum of distances from the point corresponding to $x$ on the number line to the points corresponding to $a_{i}(i=1,2, \cdots, n)$. It is easy to know that when the point corresponding to $x$ is in the middle of the ... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given a moving large circle $\odot O$ that is externally tangent to a fixed small circle $\odot O_{1}$ with radius 3 at point $P, AB$ is the external common tangent of the two circles, with $A, B$ being the points of tangency. A line $l$ parallel to $AB$ is tangent to $\odot O_{1}$ at point $C$ and intersects $\odot... | 4.36.
As shown in Figure 5, connect $AP$, $PB$, $O_{1}C$, $BO_{1}$, $PC$, and draw the common tangent line of the two circles through point $P$, intersecting $AB$ at point $Q$.
It is easy to prove that $\angle APB=90^{\circ}$, points $B$, $O_{1}$, and $C$ are collinear, $\angle BPC=90^{\circ}$, and points $A$, $P$, a... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let real numbers $x_{1}, x_{2}, \cdots, x_{1999}$ satisfy the condition $\sum_{i=1}^{1990}\left|x_{i}-x_{i+1}\right|=1991$.
And $y_{k}=\frac{1}{k} \sum_{i=1}^{k} x_{i}(k=1,2, \cdots, 1991)$. Try to find the maximum value of $\sum_{i=1}^{1990}\left|y_{i}-y_{i+1}\right|$. ${ }^{[3]}$ | 【Analysis】For $k=1,2, \cdots, 1990$, we have
$$
\begin{array}{l}
\left|y_{k}-y_{k+1}\right|=\left|\frac{1}{k} \sum_{i=1}^{k} x_{i}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}\right| \\
=\left|\frac{1}{k(k+1)}\left(\sum_{i=1}^{k} x_{i}-k x_{k+1}\right)\right| \\
\leqslant \frac{1}{k(k+1)} \sum_{i=1}^{k} i\left|x_{i}-x_{i+1}\ri... | 1990 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let positive integers $a_{1}, a_{2}, \cdots, a_{31}, b_{1}, b_{2}, \cdots, b_{31}$ satisfy
$$
\begin{array}{l}
\text { (1) } a_{1}<a_{2}<\cdots<a_{31} \leqslant 2015, \\
b_{1}<b_{2}<\cdots<b_{31} \leqslant 2015 ; \\
\text { (2) } a_{1}+a_{2}+\cdots+a_{31}=b_{1}+b_{2}+\cdots+b_{31} \text {. } \\
\text { Find } S=\lef... | 1. Define the sets
$$
\begin{array}{l}
A=\left\{m \mid a_{m}>b_{m}, 1 \leqslant m \leqslant 31\right\}, \\
B=\left\{n \mid a_{n}<b_{n}, 1 \leqslant n \leqslant 31\right\} . \\
\text { Let } S_{1}=\sum_{m \in A}\left(a_{m}-b_{m}\right), S_{2}=\sum_{m \in B}\left(b_{n}-a_{n}\right) .
\end{array}
$$
Then $S=S_{1}+S_{2}$.... | 30720 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. In the coordinate plane, points with both coordinates as integers are called integer points, and triangles with all three vertices as integer points are called integer point triangles. Find the number of integer point right triangles $OAB$ with a right angle at the origin $O$ and with $I(2015,7 \times 2015)$ as the... | 14. Let point $A$ be in the first quadrant.
Let $\angle x O I=\alpha$.
Then $\tan \alpha=7$,
$k_{O A}=\tan \left(\alpha-\frac{\pi}{4}\right)=\frac{\tan \alpha-1}{1+\tan \alpha}=\frac{3}{4}$
$\Rightarrow k_{O B}=-\frac{4}{3}$.
Since $A$ and $B$ are integer points, set
$$
A\left(4 t_{1}, 3 t_{1}\right), B\left(-3 t_{2},... | 54 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. For any positive integer $m$, the set
$$
\{m, m+1, m+2, \cdots, m+99\}
$$
any $n(n \geqslant 3)$-element subset of it, always contains three elements that are pairwise coprime. Find the minimum value of $n$. | 15. Consider the set $\{1,2, \cdots, 100\}$ with $m=1$, and its 67-element subset, whose elements are even numbers and odd numbers divisible by 3, i.e.,
$$
P=\{2,4, \cdots, 100,3,9, \cdots, 99\} .
$$
Clearly, there do not exist three pairwise coprime elements in set $P$.
Thus, $n \leqslant 67$ does not meet the requir... | 68 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. After removing all perfect squares from the sequence of positive integers $\{1,2, \cdots\}$, the remaining numbers form a sequence $\left\{a_{n}\right\}$ in their original order. Then $a_{2015}=$ $\qquad$ . | 6.2060.
Let $k^{2}2014
\end{array}\right. \\
\Rightarrow k=45, a_{2015}=2060 .
\end{array}
$ | 2060 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$, and
$$
|b-c| \cos \frac{A}{2}=8,(b+c) \sin \frac{A}{2}=15 \text {. }
$$
Then $a=$ | 2. 17 .
From the cosine theorem, we get
$$
\begin{aligned}
a^{2}= & b^{2}+c^{2}-2 b c \cos A \\
= & \left(b^{2}+c^{2}\right)\left(\sin ^{2} \frac{A}{2}+\cos ^{2} \frac{A}{2}\right)- \\
& 2 b c\left(\cos ^{2} \frac{A}{2}-\sin ^{2} \frac{A}{2}\right) \\
= & (b+c)^{2} \sin ^{2} \frac{A}{2}+(b-c)^{2} \cos ^{2} \frac{A}{2}... | 17 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $\left\{a_{n}\right\}$ be a monotonically increasing sequence of positive integers, satisfying
$$
a_{n+2}=3 a_{n+1}-a_{n}, a_{6}=280 \text {. }
$$
Then $a_{7}=$ | 7.733.
From the problem, we have
$$
\begin{array}{l}
a_{3}=3 a_{2}-a_{1}, \\
a_{4}=3 a_{3}-a_{2}=8 a_{2}-3 a_{1}, \\
a_{5}=3 a_{4}-a_{3}=21 a_{2}-8 a_{1}, \\
a_{6}=3 a_{5}-a_{4}=55 a_{2}-21 a_{1}=280=5 \times 7 \times 8, \\
a_{7}=3 a_{6}-a_{5}=144 a_{2}-55 a_{1} .
\end{array}
$$
Since $(21,55)=1$, and $a_{2}>a_{1}$ a... | 733 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given positive integers $a$, $b$, $c$, $d$ satisfy $a^{2}=c(d+29)$, $b^{2}=c(d-29)$. Then the value of $d$ is $\qquad$. | $=, 1.421$
Let $(a, b)=d^{\prime}, a=d^{\prime} a_{1}, b=d^{\prime} b_{1},\left(a_{1}, b_{1}\right)=1$.
From the given conditions, dividing the two equations yields
$$
\begin{array}{l}
\frac{a^{2}}{b^{2}}=\frac{d+29}{d-29} \Rightarrow \frac{a_{1}^{2}}{b_{1}^{2}}=\frac{d+29}{d-29} \\
\Rightarrow\left\{\begin{array}{l}
d... | 421 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $a$ and $b$ are positive integers, the fraction $\frac{a}{b}$, when converted to a decimal, contains the consecutive digits $\overline{2015}$. Then the minimum value of $b$ is . $\qquad$ | 5. 129.
Given that multiplying a fraction by an appropriate power of 10 can result in the first four digits after the decimal point being $\overline{2015}$, we only need to find the minimum value of $b$ for the fraction $\frac{a}{b}=0.2015 \cdots$.
On one hand,
$\frac{a}{b}-\frac{1}{5}=0.2015 \cdots-0.2=0.0015$
$$
125... | 129 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the sequence $\left\{a_{n}\right\}$ :
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, a_{3}=5, \\
a_{n}=a_{n-1}-a_{n-2}+a_{n-3}(n=4,5, \cdots) .
\end{array}
$$
Then the sum of the first 2015 terms of this sequence $S_{2015}=$ | -、1.6045.
From the given information, we have $a_{4}=3$, and the sequence $\left\{a_{n}\right\}$ has a period of 4. Therefore, $S_{2015}$
$$
\begin{array}{l}
=503\left(a_{1}+a_{2}+a_{3}+a_{4}\right)+a_{1}+a_{2}+a_{3} \\
=503(1+3+5+3)+1+3+5=6045 .
\end{array}
$$ | 6045 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $f(x)=a x^{5}+b x^{3}+c x+10$, and $f(3)$ $=3$. Then $f(-3)=$ $\qquad$ | 5. 17 .
Notice that, for any $x$, we have $f(x)+f(-x)=20$.
Therefore, $f(-3)=17$. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Try to find the largest integer $k$, such that for each positive integer $n$, we have
$$
1980^{k} \left\lvert\, \frac{(1980 n)!}{(n!)^{1980}} .\right.
$$ | Notice,
$$
\begin{array}{l}
1980=2^{2} \times 3^{2} \times 5 \times 11, \\
v_{11}(1980!)=\sum_{i \geqslant 1}\left[\frac{1980}{11^{i}}\right]=197 . \\
\text { Since }\left[\frac{m}{5^{i}}\right] \geqslant\left[2 \times \frac{m}{11^{i}}\right] \geqslant 2\left[\frac{m}{11^{i}}\right], \text { hence } \\
v_{5}(m!) \geqsl... | 197 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Polynomial
$$
p(x)=x^{3}-224 x^{2}+2016 x-d
$$
has three roots that form a geometric progression. Then the value of $d$ is $\qquad$ | 2. 729 .
Let the three roots of the polynomial be $a$, $b$, and $c$, and $b^{2}=a c$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a+b+c=224, \\
a b+b c+c a=2016, \\
a b c=d .
\end{array}\right.
$$
Then $b=\frac{b(a+b+c)}{a+b+c}=\frac{a b+b c+a c}{a+b+c}=9$.
Thus, $d=a b c=b^{3}=729$. | 729 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $a, b, c, d$ are prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. [1] | Let
\[
\begin{array}{l}
a b c d=x+(x+1)+\cdots+(x+34) \\
=5 \times 7(x+17) .
\end{array}
\]
Assume \( a=5, b=7, c \leqslant d, c d=x+17 \).
Thus, \( d_{\text {min }}=5 \).
If \( d=5 \), then \( c_{\text {min }}=5 \), at this point,
\[
x=8, a+b+c+d=22 \text {; }
\]
If \( d=7 \), then \( c_{\text {min }}=3 \), at this ... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given positive integer $n=a b c<10000, a, b, c$ are all prime numbers, and $2 a+3 b=c, 4 a+c+1=4 b$. Find the value of $n$.
| From the given equations, we have
$$
b=6 a+1, c=20 a+3 \text{. }
$$
Then $a(6 a+1)(20 a+3)<10000$
$\Rightarrow 12 a^{3}<10000 \Rightarrow$ prime $a<5$.
Combining $b=6 a+1, c=20 a+3$ being primes, we can determine
$$
a=2, b=13, c=43, n=1118 \text{. }
$$ | 1118 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $S=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{980100}}$. Find the greatest positive integer $[S]$ that does not exceed the real number $S$. | Notice that, $\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}$.
Therefore, $\frac{1}{\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$,
$2(\sqrt{n+1}-\sqrt{n})=\frac{2}{\sqrt{n+1+\sqrt{n}}}<\frac{1}{\sqrt{n}}$.
Thus, $1977<2(\sqrt{980101}-\sqrt{2})$
$$
<S<2(\sqrt{980100}-1)=1978 \text {. }
$$
Hence, $[S]=... | 1977 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 For integer pairs $(a, b)(0<a<b<1000)$, a set $S \subseteq\{1,2, \cdots, 2003\}$ is called a "jump set" of the pair $(a, b)$: if for any element pair $\left(s_{1}, s_{2}\right)$, $s_{1} 、 s_{2} \in$ $S,\left|s_{1}-s_{2}\right| \notin\{a, b\}$.
Let $f(a, b)$ be the maximum number of elements in a jump set of ... | 【Analysis】This method is quite typical, the minimum value is obtained using the greedy idea, while the maximum value is derived using the pigeonhole principle. Unfortunately, during the exam, most candidates only answered one part correctly.
For the minimum value, take $a=1, b=2$, then
$$
\begin{array}{l}
\{1,2,3\},\{4... | 668 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Let positive integers $a_{1}, a_{2}, \cdots, a_{31}, b_{1}, b_{2}$, $\cdots, b_{31}$ satisfy
$$
\begin{array}{l}
\text { (1) } a_{1}<a_{2}<\cdots<a_{31} \leqslant 2015, \\
b_{1}<b_{2}<\cdots<b_{31} \leqslant 2015 ;
\end{array}
$$
$$
\text { (2) } a_{1}+a_{2}+\cdots+a_{31}=b_{1}+b_{2}+\cdots+b_{31} \text {. }
$$
Find t... | 【Detailed Estimation】It is known that, $a_{31} \leqslant 2015, a_{30} \leqslant 2014, \cdots \cdots$
Thus, for all $1 \leqslant i \leqslant 31$, we have $a_{i} \leqslant 1984+i$.
Similarly, $b_{i} \leqslant 1984+i$.
Also, $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots \cdots$
Thus, for all $1 \leqslant i \leqslant 31$, ... | 30720 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the set $M=\{1,2, \cdots, 12\}$, and the three-element set $A=$ $\{a, b, c\}$ satisfies $A \subset M$, and $a+b+c$ is a perfect square. Then the number of sets $A$ is $\qquad$. | ,- 1.26 .
From $6 \leqslant a+b+c \leqslant 33$, we know that the square numbers within this range are $9, 16, 25$. Let's assume $a<b<c$.
If $a+b+c=9$, then the possible values for $c$ are 6, 5, 4, in which case,
$$
A=\{1,2,6\},\{1,3,5\},\{2,3,4\} \text {; }
$$
If $a+b+c=16$, then $7 \leqslant c \leqslant 12$, we can... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The volume of a rectangular prism is 8 cubic centimeters, and the total surface area is 32 square centimeters. If the length, width, and height form a geometric sequence, then the sum of all the edges of this rectangular prism is $\qquad$ | 4.32 cm.
Let the common ratio be $q$, and the length, width, and height be $q a, a, \frac{a}{q}$, respectively. Then $a^{3}=8 \Rightarrow a=2$.
$$
\begin{array}{l}
\text { Also } 2\left(q a \cdot a+a \cdot \frac{a}{q}+q a \cdot \frac{a}{q}\right)=32 \\
\Rightarrow q+\frac{1}{q}+1=4 .
\end{array}
$$
Therefore, the sum... | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) If the subset $A$ of the set $M=\{1,2, \cdots, 200\}$ consists of elements each of which can be expressed as the sum of the squares of two natural numbers (allowing the same number), find the maximum number of elements in the set $A$.
| 11. Notice that, the squares not exceeding 200 are $0^{2}, 1^{2}, 2^{2}, \cdots, 14^{2}$.
First, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form $k^{2}+0^{2}$, and there are 14 such numbers; while the sum of each pair of numbers in $1^{2}$, $2^{2}, \cdots, 10^{2}$ (allowing the same n... | 79 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. In the plane, $m$ points have no three points collinear, and their convex hull is an $n$-sided polygon. By appropriately connecting lines, a grid region composed of triangles can be obtained. Let the number of non-overlapping triangles be $f(m, n)$. Then $f(2016,30)=$ $\qquad$ | One, 1.4000.
Since no three points are collinear, we have
$$
f(m, n)=(n-2)+2(m-n)=2 m-n-2 \text {. }
$$
Therefore, $f(2016,30)=2 \times 2016-30-2=4000$. | 4000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the set $T=\{1,2, \cdots, 2010\}$, for each non-empty subset of $T$, calculate the reciprocal of the product of all its elements. Then the sum of all such reciprocals is $\qquad$ | 6. 2010 .
Notice that, in the expansion of the product
$$
(1+1)\left(1+\frac{1}{2}\right) \cdots\left(1+\frac{1}{2010}\right)
$$
the $2^{2010}$ terms are all the reciprocals of positive integers, and each is precisely the reciprocal of the product of the numbers in one of the $2^{2010}$ subsets of set $T$. Removing t... | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For integers $n \geqslant 3$, let
$$
f(n)=\log _{2} 3 \times \log _{3} 4 \times \cdots \times \log _{n-1} n \text {. }
$$
Then $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)=$ $\qquad$ | 2. 54
Notice,
$$
\begin{array}{l}
f(n)=\log _{2} 3 \times \frac{\log _{2} 4}{\log _{2} 3} \times \cdots \times \frac{\log _{2} n}{\log _{2}(n-1)} \\
=\log _{2} n .
\end{array}
$$
Thus, $f\left(2^{k}\right)=k$.
Therefore, $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)$
$$
=2+3+\cdots+10=54 \text ... | 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of t... | 3. 126.
First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls.
Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by ... | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. (14 points) A flower shop purchases a certain number of roses from a farm at a price of 5 yuan per stem every day, and then sells them at a price of 10 yuan per stem. If they are not sold on the same day, the remaining roses are treated as garbage.
(1) If the flower shop purchases 16 stems of roses one day, find th... | 12. (1) When the daily demand $n \geqslant 16$, the profit $y=80$. When the daily demand $n<16$, the profit $y=10 n-80$. Therefore, the function expression of $y$ with respect to the positive integer $n$ is
$$
y=\left\{\begin{array}{ll}
10 n-80, & n<16 ; \\
80, & n \geqslant 16 .
\end{array}\right.
$$
(2) (i) The possi... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Problem 5: In an $n \times n$ grid, 101 cells are colored blue. It is known that there is a unique way to cut the grid along the grid lines into some rectangles, such that each rectangle contains exactly one blue cell. Find the minimum possible value of $n$.
(2015, Bulgarian Mathematical Olympiad) | The minimum possible value of $n$ is 101.
First, prove the more general conclusion below.
Lemma Given an $n \times n$ grid $P$ with $m$ cells colored blue. A "good" partition of the grid $P$ is defined as: if the grid is divided along the grid lines into $m$ rectangles, and each rectangle contains exactly one blue cell... | 101 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given
$$
5 x+16 y+33 z \geqslant 136\left(x 、 y 、 z \in \mathbf{R}_{+}\right) \text {. }
$$
then the minimum value of $x^{3}+y^{3}+z^{3}+x^{2}+y^{2}+z^{2}$ is | 3. 50 .
Notice that,
$$
\begin{array}{l}
x^{3}+x^{2}-5 x+3=(x+3)(x-1)^{2} \geqslant 0 \\
\Rightarrow x^{3}+x^{2} \geqslant 5 x-3, \\
y^{3}+y^{2}-16 y+20=(y+5)(y-2)^{2} \geqslant 0 \\
\Rightarrow y^{3}+y^{2} \geqslant 16 y-20, \\
z^{3}+z^{2}-33 z+63=(z+7)(z-3)^{2} \geqslant 0 \\
\Rightarrow z^{3}+z^{2} \geqslant 33 z-6... | 50 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
For any real number $x$, $[x]$ represents the greatest integer not exceeding $x$, and $\{x\}$ represents the fractional part of $x$. Then
$$
\begin{array}{l}
\left\{\frac{2014}{2015}\right\}+\left\{\frac{2014^{2}}{2015}\right\}+\cdots+\left\{\frac{2014^{2014}}{2015}\right\} \\
=
\end{array}
$$ | $-, 1.1007$.
Notice that,
$$
\frac{2014^{n}}{2015}=\frac{(2015-1)^{n}}{2015}=M+\frac{(-1)^{n}}{2015} \text {, }
$$
where $M$ is an integer.
Then $\left\{\frac{2014^{n}}{2015}\right\}=\left\{\begin{array}{ll}\frac{1}{2015}, & n \text { is even; } \\ \frac{2014}{2015}, & n \text { is odd. }\end{array}\right.$
Therefore,... | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a_{1}, a_{2}, \cdots, a_{9}$ as any permutation of $1,2, \cdots, 9$. Then the minimum value of $a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9}$ is $\qquad$ | 2. 214.
Since $a_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, 9)$, by the AM-GM inequality, we have
$$
\begin{array}{l}
a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9} \\
\geqslant 3 \sqrt[3]{a_{1} a_{2} \cdots a_{9}}=3 \sqrt[3]{9!} \\
=3 \sqrt[3]{(2 \times 5 \times 7) \times(1 \times 8 \times 9) \times(3 \times 4 \tim... | 214 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$.
Then the number of such $n$ is $\qquad$. | From $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right] \leqslant \frac{n}{3}+\frac{n}{6}=\frac{n}{2}$, knowing that equality holds, we find that $n$ is a common multiple of $3$ and $6$, i.e., a multiple of $6$. Therefore, the number of such $n$ is $\left[\frac{2006}{6}\right]=334$. | 334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
23. For any real numbers $a$, $b$, $c$, define an operation ※ with the following properties:
(1) $a ※(b ※ c)=(a ※ b) \cdot c$,
(2) $a ※ a=1$,
where “$\cdot$” denotes multiplication.
If the solution to the equation $2016 \times(6 ※ x)=100$ is $x=\frac{p}{q}(p, q$ are positive integers, $(p, q)=1)$, then the value of $p... | 23. A.
$$
\begin{array}{l}
\text { Given } 2016 ※(6 ※ x)=(2016 ※ 6) \cdot x \\
\Rightarrow(2016 ※ 6) \cdot 6=\frac{600}{x} \\
\Rightarrow 2016 ※(6 ※ 6)=\frac{600}{x} .
\end{array}
$$
Since $a ※ a=1$, we have,
$$
\begin{array}{l}
6 ※ 6=2016 ※ 2016=1 . \\
\text { Therefore } 2016 ※(2016 ※ 2016)=\frac{600}{x} \\
\Rightar... | 109 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given positive integers $a, b, c (a < b < c)$ form a geometric sequence, and
$$
\log _{2016} a+\log _{2016} b+\log _{2016} c=3 .
$$
Then the maximum value of $a+b+c$ is $\qquad$ | One, 1.4066 273.
From the given, we know
$$
\begin{array}{l}
b^{2}=a c, a b c=2016^{3} \\
\Rightarrow b=2016, a c=2016^{2} .
\end{array}
$$
Since $a$, $b$, and $c$ are positive integers, when $a=1$, $c=2016^{2}$, $a+b+c$ reaches its maximum value
$$
2016^{2}+2017=4066273 .
$$ | 4066273 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
492 Select five subsets $A_{1}, A_{2}, \cdots, A_{5}$ from the set $\{1,2, \cdots, 1000\}$ such that $\left|A_{i}\right|=500(i=1,2$, $\cdots, 5)$. Find the maximum value of the number of common elements in any three of these subsets. | Let $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ be five 1000-dimensional vectors, and let $\alpha_{i}(j)$ denote the $j$-th element of $\alpha_{i}$. Then $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ satisfy
$$
\alpha_{i}(j)=\left\{\begin{array}{l}
1, j \in A_{i} ; \\
0, j \notin A_{i} .
\end{array}\right.
$$
Let the ... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. For a finite set $A$ consisting of positive integers, if $A$ is divided into two non-empty disjoint subsets $A_{1}$ and $A_{2}$, and the least common multiple (LCM) of the elements in $A_{1}$ equals the greatest common divisor (GCD) of the elements in $A_{2}$, then such a partition is called "good". Find the minimum... | 3. 3024.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\left(a_{1}<a_{2}<\cdots<a_{n}\right)$.
For any non-empty finite set of positive integers $B$, let $\operatorname{lcm} B$ and $\operatorname{gcd} B$ denote the least common multiple and greatest common divisor of the elements in $B$, respectively.
Consider any... | 3024 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $m$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=m, a_{1}=\varphi(m), \\
a_{2}=\varphi^{(2)}(m)=\varphi(\varphi(m)), \cdots, \\
a_{n}=\varphi^{(n)}(m)=\varphi\left(\varphi^{(n-1)}(m)\right),
\end{array}
$$
where $\varphi(m)$ is the Euler'... | 3. It is known that if $a \mid b$, then $\varphi(a) \mid \varphi(b)$.
Moreover, when $m>2$, $\varphi(m)$ is even.
Therefore, it is only necessary to find integers $m$ such that $a_{1} \mid a_{0}$, which implies $a_{k+1} \mid a_{k}$.
If $m$ is an odd number greater than 2, it is impossible for $a_{1} \mid m$.
Let $m=2^... | 1944 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the set
$$
I=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{i} \in\{1,2, \cdots, 11\}\right\} \text {, }
$$
$A$ is a subset of $I$ and satisfies: for any
$$
\left(x_{1}, x_{2}, x_{3}, x_{4}\right) 、\left(y_{1}, y_{2}, y_{3}, y_{4}\right) \in A \text {, }
$$
there exist $i 、 j(1 \leqslant i<j \leqslant... | 8. First, consider the set satisfying $x_{1}+x_{2}+x_{3}+x_{4}=24$
$$
\begin{aligned}
A= & \left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{i} \in\{1,2, \cdots, 11\},\right. \\
& i=1,2,3,4\},
\end{aligned}
$$
The number of its elements is
$$
\begin{array}{l}
\mathrm{C}_{23}^{3}-4\left(\mathrm{C}_{2}^{2}+\mathrm{... | 891 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ and $b$ are integers, $\frac{127}{a}-\frac{16}{b}=1$. Then the maximum value of $b$ is $\qquad$ . | $$
\text { Two, 1.2 } 016 .
$$
The original equation is transformed into $b=\frac{16 a}{127-a}=\frac{127 \times 16}{127-a}-16$. When and only when $127-a=1$, $b$ reaches its maximum value, at this point, $b=127 \times 16-16=2016$. | 2016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 If a natural number $K$ can be expressed as
$$
\begin{aligned}
K & =V(a, b, c) \\
& =a^{3}+b^{3}+c^{3}-3 a b c \quad (a, b, c \in \mathbf{N})
\end{aligned}
$$
then $K$ is called a "Water Cube Number".
Now, arrange all different Water Cube Numbers in ascending order to form a sequence $\left\{K_{n}\right\}$, ... | 【Analysis】Let $M=\{3 x \mid x \in \mathbf{N},(3, x)=1\}$.
Lemma A natural number $k$ is a water cube number if and only if $k \in \mathbf{N} \backslash M$.
Proof Note that,
$V(a, b, c)=a^{3}+b^{3}+c^{3}-3 a b c$
$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$=(a+b+c)^{3}-3(a+b+c)(a b+b c+c a)$.
Then when $3 \mid... | 2614032 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$, is called a "fractal sequence", and its construction method is as follows:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;
Then copy all the previous items $1, 1, 2$, and add... | Solving the construction method of the sequence $\left\{a_{n}\right\}$, we easily know
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$
Generally, $a_{2^{n-1}}=n$, which means the number $n$ first appears at the $2^{n}-1$ term, and if
$$
m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right),
$$
then $a_{m}=... | 3950 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the function be
$$
f(x)=\sin ^{4} \frac{k x}{10}+\cos ^{4} \frac{k x}{10}\left(k \in \mathbf{Z}_{+}\right) .
$$
If for any real number $a$, we have
$$
\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\} \text {, }
$$
then the minimum value of $k$ is $\qquad$ | 6. 16 .
From the given conditions, we have
$$
\begin{array}{l}
f(x)=\left(\sin ^{2} \frac{k x}{10}+\cos ^{2} \frac{k x}{10}\right)^{2}-2 \sin ^{2} \frac{k x}{10} \cdot \cos ^{2} \frac{k x}{10} \\
=1-\frac{1}{2} \sin ^{2} \frac{k x}{5}=\frac{1}{4} \cos \frac{2 k x}{5}+\frac{3}{4},
\end{array}
$$
The function $f(x)$ re... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four distinct numbers from $1, 2, \cdots, 100$, satisfying
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \\
=\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2} .
\end{array}
$$
Then the number of such ordered quadruples... | 8. 40 .
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \\
\geqslant\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2},
\end{array}
$$
Equality holds if and only if
$$
\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\frac{a_{... | 40 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. | Three, using these ten points as vertices and the connected line segments as edges, we get a simple graph $G$ of order 10.
The following proves: The number of edges in graph $G$ does not exceed 15.
Let the vertices of graph $G$ be $v_{1}, v_{2}, \cdots, v_{10}$, with a total of $k$ edges, and use $\operatorname{deg}\le... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Several pairwise non-overlapping isosceles right triangles with leg lengths of 1 are placed on a $100 \times 100$ grid paper. It is known that the hypotenuse of any right triangle is the diagonal of some unit square; each side of a unit square is the leg of a unique right triangle. A unit square whose diagonals are ... | 3. For a general $2 n \times 2 n$ grid paper, the maximum number of empty cells is $n(n-1)$.
In fact, the $2 n \times 2 n$ grid paper is enclosed by $2 n+1$ horizontal lines and $2 n+1$ vertical lines:
$$
\begin{array}{l}
\{(x, y) \mid x=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\}, \\
\{(x, y) \mid y=k, 0 \leqs... | 2450 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set
$$
S=\{1,2, \cdots, 12\}, A=\left\{a_{1}, a_{2}, a_{3}\right\}
$$
satisfy $a_{1}<a_{2}<a_{3}, a_{3}-a_{2} \leqslant 5, A \subseteq S$. Then the number of sets $A$ that satisfy the conditions is | 5. 185 .
Notice that, the number of all three-element subsets of set $S$ is
$$
\mathrm{C}_{12}^{3}=220 \text {. }
$$
The number of subsets satisfying $1 \leqslant a_{1}<a_{2}<a_{3}-5 \leqslant 7$ is $\mathrm{C}_{7}^{3}=35$, so the number of sets $A$ that meet the conditions of the problem is $220-35=185$. | 185 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If a positive integer can be expressed as the difference of cubes of two consecutive odd numbers, then the positive integer is called a "harmonious number" (for example, $2=1^{3}-(-1)^{3}, 26=3^{3}-1^{3}, 2, 26$ are both harmonious numbers). Among the positive integers not exceeding 2016, the sum of all harmonious n... | 3. B.
Notice that,
$$
(2 k+1)^{3}-(2 k-1)^{3}=2\left(12 k^{2}+1\right) \text {. }
$$
From $2\left(12 k^{2}+1\right) \leqslant 2016 \Rightarrow|k|<10$.
Taking $k=0,1, \cdots, 9$, we get all the harmonious numbers not exceeding 2016, and their sum is
$$
\begin{array}{l}
{\left[1^{3}-(-1)^{3}\right]+\left(3^{3}-1^{3}\ri... | 6860 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
a+b+c=5, a^{2}+b^{2}+c^{2}=15, \\
a^{3}+b^{3}+c^{3}=47 . \\
\text { Find }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
\end{array}
$$ | Given $a+b+c=5, a^{2}+b^{2}+c^{2}=15$, we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
Then $47-3... | 625 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=0,5 a_{n+1}=4 a_{n}+3 \sqrt{1-a_{n}^{2}}(n \in \mathbf{N}) \text {. }
$$
Let $S_{n}=\sum_{i=0}^{n} a_{i}$. Then $S_{51}=$ $\qquad$ | 2. 48 .
Let $a_{n}=\sin \alpha_{n}$.
Suppose $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$.
Then $30^{\circ}0 ; \\
\sin \left(\alpha_{n}-\theta\right), \cos \alpha_{n}<0 .
\end{array}\right.
$
Thus, $a_{n}=\left\{\begin{array}{l}\sin n \theta, n=0,1 ; \\ \sin 2 \theta, n=2 k\left(k \in \mathbf{Z}_{+}\right) ; \\... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. A drawer contains red and blue socks, with a total number not exceeding 2016. If two socks are randomly drawn, the probability that they are the same color is $\frac{1}{2}$. Then the maximum number of red socks in the drawer is . $\qquad$ | 5.990.
Let $x$ and $y$ be the number of red and blue socks in the drawer, respectively. Then
$$
\frac{x y}{\mathrm{C}_{x+y}^{2}}=\frac{1}{2} \Rightarrow (x-y)^{2}=x+y.
$$
Thus, the total number of socks is a perfect square.
Let $n=x-y$, i.e., $n^{2}=x+y$. Therefore, $x=\frac{n^{2}+n}{2}$.
Since $x+y \leqslant 2016$, ... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the set $I=\{1,2, \cdots, n\}(n \geqslant 3)$. If two non-empty proper subsets $A$ and $B$ of $I$ satisfy $A \cap B=\varnothing, A \cup$ $B=I$, then $A$ and $B$ are called a partition of $I$. If for any partition $A, B$ of the set $I$, there exist two numbers in $A$ or $B$ such that their sum is a perfect square... | 6. 15 .
When $n=14$, take
$A=\{1,2,4,6,9,11,13\}$,
$B=\{3,5,7,8,10,12,14\}$.
It is easy to see that the sum of any two numbers in $A$ and $B$ is not a perfect square.
Thus, $n=14$ does not meet the requirement.
Therefore, $n<14$ also does not meet the requirement.
Now consider $n=15$.
Use proof by contradiction. Assum... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the set
$$
\{[x]+[2 x]+[3 x] \mid x \in \mathbf{R}\} \cap\{1,2, \cdots, 100\}
$$
has elements. | 4. 67 .
Let $f(x)=[x]+[2 x]+[3 x]$.
Then $f(x+1)=f(x)+6$.
When $0 \leqslant x<1$, all possible values of $f(x)$ are 0, 1, 2, 3.
Therefore, the range of $f(x)$ is
$$
S=\{6 k, 6 k+1,6 k+2,6 k+3 । k \in \mathbf{Z}\} \text {. }
$$
Thus, $S \cap\{1,2, \cdots, 100\}$ has $4 \times 17-1=67$ elements. | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the sequence $\left\{\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the number of rational terms in the first 2016 terms of the sequence $\left\{S_{n}\right\}$ is | - 1.43.
$$
\begin{array}{l}
\text { Given } \frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1} \\
\Rightarrow S_{n}=1-\frac{\sqrt{n+1}}{n+1} .
\end{array}
$$
Since $44<\sqrt{2016}<45$, the number of values for which $S$ is rational is 43. | 43 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 One evening, 21 people made $n$ phone calls to each other. It is known that among them, there are $m$ ($m$ is an odd number) people $a_{1}, a_{2}$, $\cdots, a_{m}$ such that $a_{i}$ and $a_{i+1}\left(i=1,2, \cdots, m ; a_{m+1}=a_{1}\right)$ made phone calls. If no three people among these 21 people made phone... | Solve: Represent 21 people with 21 points.
If two people communicate by phone, connect the corresponding two points with an edge; otherwise, do not connect an edge, resulting in a graph $G$. It is known that graph $G$ contains an odd cycle, and there are no triangles in graph $G$. The task is to find the maximum number... | 101 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the sum of the first $n$ terms denoted as $S_{n}$. If $S_{6}=26, a_{7}=2$, then the maximum value of $n S_{n}$ is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result direct... | 3. 338 .
$$
\begin{array}{l}
\text { Given } S_{6}=26, a_{7}=2 \\
\Rightarrow a_{1}=6, d=-\frac{2}{3} \\
\Rightarrow S_{n}=-\frac{1}{3} n^{2}+\frac{19}{3} n \\
\Rightarrow n S_{n}=-\frac{1}{3} n^{3}+\frac{19}{3} n^{2} . \\
\text { Let } f(x)=-\frac{1}{3} x^{3}+\frac{19}{3} x^{2}(x \in \mathbf{R}) .
\end{array}
$$
Then... | 338 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
9. (16 points) When $x \in [1,2017]$, find
$$
f(x)=\sum_{i=1}^{2017} i|x-i|
$$
the minimum value. | When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2016)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2017} i(i-x) \\
=\left(k^{2}+k-2017 \times 1009\right) x+ \\
\frac{2017 \times 2018 \times 4035}{6}-\frac{k(k+1)(2 k+1)}{3}
\end{array}
$$
is a linear function, and its minimum value is attain... | 801730806 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (40 points) Given a $2016 \times 2016$ grid. Find the smallest positive integer $M$, such that it is possible to draw $M$ rectangles (with their sides on the grid lines) in the grid, and each small square in the grid has its sides included in the sides of one of the $M$ rectangles. | Second, $M=2017$.
First, we prove: All grid lines of a $2016 \times 2016$ square grid can be covered by the edges of 2017 rectangles.
In fact, all horizontal grid lines can be covered by the edges of 1008 $1 \times 2016$ rectangles and one $2016 \times 2016$ rectangle; all vertical grid lines can be covered by the edg... | 2017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. A certain unit distributes a year-end bonus of 1 million yuan, with first prize at 15,000 yuan per person, second prize at 10,000 yuan per person, and third prize at 5,000 yuan per person. If the difference in the number of people between third prize and first prize is no less than 93 but less than 96, then the tota... | 3. 147.
Let the number of first prize winners be $x$, the number of second prize winners be $y$, and the number of third prize winners be $z$.
Then $1.5 x + y + 0.5 z = 100$
$$
\begin{array}{l}
\Rightarrow (x + y + z) + 0.5 x - 0.5 z = 100 \\
\Rightarrow x + y + z = 100 + 0.5(z - x). \\
\text{Given } 93 \leqslant z - ... | 147 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the curve
$$
(x-20)^{2}+(y-16)^{2}=r^{2} \text { and } y=\sin x
$$
have exactly one common point $P\left(x_{0}, y_{0}\right)$. Then
$$
\frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=
$$
$\qquad$ | 2. 20 .
From the given information, there is a common tangent line at point $P\left(x_{0}, y_{0}\right)$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\sin x_{0}-16}{x_{0}-20} \cos x_{0}=-1 \\
\Rightarrow \frac{1}{2} \sin 2 x_{0}-16 \cos x_{0}+x_{0}=20 .
\end{array}
$$ | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (20 points) Given sets $A$ and $B$ are both sets of positive integers, and $|A|=20,|B|=16$. Set $A$ satisfies the following condition: if $a, b, m, n \in A$, and $a+b=$ $m+n$, then $\{a, b\}=\{m, n\}$. Define
$$
A+B=\{a+b \mid a \in A, b \in B\} \text {. }
$$
Determine the minimum value of $|A+B|$. | 14. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{20}\right\}$,
$$
\begin{array}{l}
B=\left\{b_{1}, b_{2}, \cdots, b_{16}\right\}, \\
C_{j}=\left\{a_{i}+b_{j} \mid i=1,2, \cdots, 20\right\},
\end{array}
$$
where $j=1,2, \cdots, 16$.
Thus, $A+B=\bigcup_{j=1}^{16} C_{j}$.
We now prove: $\left|C_{m} \cap C_{n}\right| \leqslant ... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. If the sum of 12 distinct positive integers is 2016, then the maximum value of the greatest common divisor of these positive integers is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. 24 .
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d\left(\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1\right)$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then, $2016=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$$
S \... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $m$ and $n$ be positive integers, and satisfy $24 m=n^{4}$. Then the minimum value of $m$ is $\qquad$ | 10. 54 .
$$
\begin{array}{l}
\text { Given } n^{4}=24 m=2^{3} \times 3 m \text {, we know that } \\
m_{\min }=2 \times 3^{3}=54 \text {. }
\end{array}
$$ | 54 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let four complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ satisfy
$$
\begin{array}{l}
\left|z_{1}-z_{2}\right|=1,\left|z_{3}-z_{4}\right|=2, \\
\left|z_{1}-z_{4}\right|=3,\left|z_{2}-z_{3}\right|=4, \\
z=\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right) .
\end{array}
$$
Then the maximum value of $|z|$ is | 5.14 .
Notice,
$$
\begin{array}{l}
|z|=\left|\left(z_{1}-z_{3}\right)\left(z_{2}-z_{4}\right)\right| \\
=\left|z_{1} z_{2}-z_{1} z_{4}-z_{3} z_{2}+z_{3} z_{4}\right| \\
=\left|\left(z_{1}-z_{2}\right)\left(z_{3}-z_{4}\right)+\left(z_{1}-z_{4}\right)\left(z_{2}-z_{3}\right)\right| \\
\leqslant\left|\left(z_{1}-z_{2}\ri... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Find the number of all positive integer solutions $(x, y, z)$ to the equation $\arctan \frac{1}{x}+\arctan \frac{1}{y}+\arctan \frac{1}{z}=\frac{\pi}{4}$. | 10. By symmetry, let $x \leqslant y \leqslant z$.
Taking the tangent of both sides of the given equation, we get
$\frac{\frac{1}{y}+\frac{1}{z}}{1-\frac{1}{y z}}=\frac{1-\frac{1}{x}}{1+\frac{1}{x}}$
$\Rightarrow \frac{y+z}{y z-1}=\frac{x-1}{x+1}=1-\frac{2}{x+1}$.
If $x \geqslant 5$, then
$1-\frac{2}{x+1} \geqslant 1-\... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $f(x)$ be a function defined on $\mathbf{R}$, if $f(0)$ $=1008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant 2(x+1), \\
f(x+12)-f(x) \geqslant 6(x+5) .
\end{array}
$$
Then $\frac{f(2016)}{2016}=$ $\qquad$ . | 9. 504 .
From the conditions, we have
$$
\begin{array}{l}
f(x+12)-f(x) \\
=(f(x+12)-f(x+8))+ \\
\quad(f(x+8)-f(x+4))+(f(x+4)-f(x)) \\
\leqslant 2((x+8)+1)+2((x+4)+1)+2(x+1) \\
=6 x+30=6(x+5) . \\
\text { Also, } f(x+12)-f(x) \geqslant 6(x+5), \text { thus, } \\
f(x+12)-f(x)=6(x+5) .
\end{array}
$$
Then, $f(2016)$
$$
... | 504 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. It is known that Team A and Team B each have several people. If 90 people are transferred from Team A to Team B, then the total number of people in Team B will be twice that of Team A; if some people are transferred from Team B to Team A, then the total number of people in Team A will be 6 times that of Team B. Then... | 8. 153.
Let the original number of people in team A and team B be $a$ and $b$ respectively.
Then $2(a-90)=b+90$.
Suppose $c$ people are transferred from team B to team A. Then $a+c=6(b-c)$.
From equations (1) and (2), eliminating $b$ and simplifying, we get
$$
\begin{array}{l}
11 a-7 c=1620 \\
\Rightarrow c=\frac{11 a... | 153 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
518 On a circle, initially write 1 and 2 at opposite positions. Each operation involves writing the sum of two adjacent numbers between them, for example, the first operation writes two 3s, the second operation writes two 4s and two 5s. After each operation, the sum of all numbers becomes three times the previous sum. ... | Observe the pattern, and conjecture that after a sufficient number of operations, the $n$ numbers written equal $\varphi(n)$. After each operation, the property that adjacent numbers are coprime remains unchanged, and the new number written each time is the sum of its two neighbors. Therefore, the number we are looking... | 2016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Define the sequence $\left\{a_{n}\right\}: a_{n}$ is the last digit of $1+2+\cdots+n$, and $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then $S_{2016}=$ $\qquad$ . | 6.7 066 .
From the problem, we know
$$
\begin{array}{l}
\frac{(n+20)(n+20+1)}{2}=\frac{n^{2}+41 n+420}{2} \\
=\frac{n(n+1)}{2}+20 n+210 .
\end{array}
$$
Then $\frac{(n+20)(n+21)}{2}$ and $\frac{n(n+1)}{2}$ have the same last digit, i.e., $a_{n+20}=a_{n}$.
$$
\begin{array}{l}
\text { Therefore, } S_{2016}=S_{16}+100 S... | 7066 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. From five positive integers $a, b, c, d, e$, any four are taken to find their sum, resulting in the set of sums $\{44,45,46,47\}$, then $a+b+c+d+e=$ $\qquad$ . | 2. 57 .
From five positive integers, if we take any four to find their sum, there are five possible ways, which should result in five sum values. Since the set $\{44,45, 46,47\}$ contains only four elements, there must be two sum values that are equal. Therefore,
$$
\begin{array}{l}
44+44+45+46+47 \\
\leqslant 4(a+b+c... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the numbers $2, 3, 4, 6, 8, 9, 12, 15$ in a row so that the greatest common divisor of any two adjacent numbers is greater than 1. The total number of possible arrangements is ( ) .
(A) 720
(B) 1014
(C) 576
(D) 1296 | 4. D.
First, divide the eight numbers into three groups:
I $(2,4,8)$, II $(3,9,15)$, III $(6,12)$.
Since the numbers in group I and group II have no common factors, the arrangement that satisfies the condition must be:
(1) After removing the numbers 6 and 12, the remaining numbers are divided into three parts and arra... | 1296 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
3. If three numbers are taken simultaneously from the 14 integers $1,2, \cdots, 14$, such that the absolute difference between any two numbers is not less than 3, then the number of different ways to choose is $\qquad$ | 3. 120 .
Let the three integers taken out be $x, y, z (x<y<z)$.
$$
\begin{array}{l}
\text { Let } a=x, b=y-x-2, \\
c=z-y-2, d=15-z .
\end{array}
$$
Thus, $a, b, c, d \geqslant 1$.
If $a, b, c, d$ are determined, then $x, y, z$ are uniquely determined.
Since $a+b+c+d=11$, it is equivalent to dividing 11 identical ball... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given ten points in space, where no four points lie on the same plane. Some points are connected by line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$
(2016, National High School Mathematics Joint Compe... | Proof Let $v$ be a vertex in graph $G$, and $N_{i}(v)$ denote the set of points at a distance $i$ from $v$. For example,
$$
N_{0}(v)=\{v\},
$$
$N_{1}(v)=\{u \mid u$ is adjacent to $v\}$,
$N_{2}(v)=\{w \mid w$ is adjacent to $u$, not adjacent to $v$, and $u$ is adjacent to $v\}$.
First, we prove a lemma.
Lemma Let $G=(... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Arrange the numbers in the set $\left\{2^{x}+2^{y} \mid x 、 y \in \mathbf{N}, x<y\right\}$ in ascending order. Then the 60th number is $\qquad$ (answer with a number).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3.2064.
It is known that the number of combinations $(x, y)$ satisfying $0 \leqslant x<y \leqslant n$ is $\mathrm{C}_{n+1}^{2}$.
Notice that, $\mathrm{C}_{11}^{2}=55<60<66=\mathrm{C}_{12}^{2}$.
Therefore, the 60th number satisfies $y=11, x=4$, which means the 60th number is $2^{11}+2^{4}=2064$. | 2064 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the maximum value of $n$ such that there exists an arithmetic sequence $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfying
$$
\sum_{i=1}^{n}\left|a_{i}\right|=\sum_{i=1}^{n}\left|a_{i}+1\right|=\sum_{i=1}^{n}\left|a_{i}-2\right|=507 .
$$
$(2005$, China Southeast Mathematical Olympiad) | 【Analysis】Let's set $a_{i}=a-i d(d>0, i=1,2$, $\cdots, n)$. Then the given system of equations becomes
$$
\left\{\begin{array}{l}
\sum_{i=1}^{n}|a-i d|=507, \\
\sum_{i=1}^{n}|a+1-i d|=507, \\
\sum_{i=1}^{n}|a-2-i d|=507 .
\end{array}\right.
$$
Thus, the absolute value sum function $f(x)=\sum_{i=1}^{n}|x-i d|$ has thre... | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 There are 68 pairs of non-zero integers on the blackboard. For a positive integer $k$, at most one of the pairs $(k, k)$ and $(-k, -k)$ appears on the blackboard. A student erases some of these 136 numbers so that the sum of any two erased numbers is not 0. It is stipulated that if at least one number from a ... | Given that $(j, j)$ and $(-j, -j)$ can appear at most as one pair, we can assume that if $(j, j)$ appears, then $j > 0$ (otherwise, replace $j$ with $-j$). For a positive integer $k$, all $k$ or $-k$ can be deleted from the blackboard, but not both. For each $k > 0$, delete $k$ with probability $p$ and $-k$ with probab... | 43 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) A function calculator has a display screen and two operation keys. If the first operation key is pressed once, the number on the display screen will change to $\left[\frac{x}{2}\right]$ (where $[x]$ represents the greatest integer not exceeding the real number $x$); if the second operation key is pre... | Three, (1) Impossible.
Convert the number to binary. Then pressing the first operation key means removing the last digit of the number on the display; pressing the second operation key means appending 01 to the number on the display.
When the initial number is 1, after performing the above two operations, the resultin... | 233 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) There are $12k$ people attending a meeting, each of whom has shaken hands with exactly $3k+6$ others, and for any two of them, the number of people who have shaken hands with both is the same. How many people attended the meeting? Prove your conclusion. | Three, abstracting a person as a point and two people shaking hands as a line connecting two points, then $A$ shaking hands with $B$ and $C$ corresponds to $\angle BAC$ in the graph. Since each person has shaken hands with $3k+6$ people, there are $12k \mathrm{C}_{3k+6}^{2}$ such angles in the graph. The number of pair... | 36 | Combinatorics | proof | Yes | Yes | cn_contest | false |
2. A four-digit number divided by 433 has a quotient of $a$ and a remainder of $r$ $(a 、 r \in \mathbf{N})$. Then the maximum value of $a+r$ is $\qquad$ . | 2.454.
Let the four-digit number be $433 a+r(0 \leqslant r \leqslant 432)$. Since $433 \times 24=10392>9999$, then $a \leqslant 23$.
When $a=23$, $433 \times 23+40=9999$. At this point, $a+r=23+40=63$. When $a=22$, $433 \times 22+432=9958$. At this point, $a+r=22+432=454$. In summary, the maximum value of $a+r$ is 454... | 454 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
\begin{array}{c}
S=\left[\frac{1}{1}\right]+\left[\frac{2}{1}\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\left[\frac{3}{2}\right]+ \\
{\left[\frac{4}{2}\right]+\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{3}{3}... | 6.1078.
$$
2+4+\cdots+2 \times 44=1980 \text {. }
$$
For any integer \( k \) satisfying \( 1 \leqslant k \leqslant 44 \), the sum includes \( 2k \) terms with the denominator \( k \): \(\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]\), whose sum is \( k+2 \).
Also, \( 2016-1980=3... | 1078 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. If real numbers $a, b, c$ make the quadratic function $f(x) = a x^{2} + b x + c$ such that when $0 \leqslant x \leqslant 1$, always $|f(x)| \leqslant 1$. Then the maximum value of $|a| + |b| + |c|$ is $\qquad$ | 7. 17.
Take $x=0, \frac{1}{2}, 1$.
From the problem, we have
$$
\begin{array}{l}
|c| \leqslant 1,|a+2 b+4 c| \leqslant 4, \\
|a+b+c| \leqslant 1 .
\end{array}
$$
Let $m=a+2 b+4 c, n=a+b+c$.
Then $a=-m+2 n+2 c, b=m-n-3 c$
$$
\begin{aligned}
\Rightarrow & |a| \leqslant|m|+2|n|+2|c| \leqslant 8, \\
& |b| \leqslant|m|+|n... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. | Let the ten points be $A_{1}, A_{2}, \cdots, A_{10}$. Using these ten points as vertices and the line segments connecting them as edges, we obtain a simple graph $G$ of order 10. Let the degree of point $A_{i} (i=1,2, \cdots, 10)$ be $d_{i}$. Then the total number of edges in graph $G$ is $\frac{1}{2} \sum_{i=1}^{10} d... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the sum of the digits of the natural number $x$ be $S(x)$. Then the solution set of the equation $x+S(x)+S(S(x))+S(S(S(x)))=2016$ is $\qquad$ | $-1 .\{1980\}$.
It is easy to see that $x<2016$.
Note that, the sum of the digits of natural numbers less than 2016 is at most 28, for example, $S(1999)=28$, which indicates,
$$
S(x) \leqslant 28 \text {. }
$$
Furthermore, $S(S(x)) \leqslant S(19)=10$.
Finally, $S(S(S(x))) \leqslant 9$.
From the equation we get
$$
\be... | 1980 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set $A=\left[\frac{7}{8}\right]+\left[\frac{7^{2}}{8}\right]+\cdots+\left[\frac{7^{2016}}{8}\right]$. Then the remainder when $A$ is divided by 50 is $\qquad$ | 3. 42.
Since $\frac{7^{2 k-1}}{8}$ and $\frac{7^{2 k}}{8}$ are not integers, and
$$
\frac{7^{2 k-1}}{8}+\frac{7^{2 k}}{8}=7^{2 k-1},
$$
for any $k \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
{\left[\frac{7^{2 k-1}}{8}\right]+\left[\frac{7^{2 k}}{8}\right]=7^{2 k-1}-1} \\
\equiv 7(-1)^{k-1}-1(\bmod 50) .
\end{arr... | 42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Rectangle $R$ is divided into 2016 small rectangles, with each small rectangle's sides parallel to the sides of rectangle $R$. The vertices of the small rectangles are called "nodes". For a line segment on the side of a small rectangle, if both endpoints are nodes and its interior does not contain any other n... | Consider a graph $G$ with all nodes as vertices and basic segments as edges. Let the number of vertices in graph $G$ be $v$, and the number of edges be $e$. Treat the external region of rectangle $R$ as one face (region). Then, the total number of faces in graph $G$ is $f=2017$.
By Euler's formula, we have $v+f-e=2$.
T... | 4122 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the length of intervals $(m, n)$, $[m, n)$, $(m, n]$, and $[m, n]$ to be $n-m$ (where $n, m \in \mathbf{R}$, and $n > m$). Then the sum of the lengths of the intervals of real numbers $x$ that satisfy
$$
\frac{1}{x-20}+\frac{1}{x-17} \geqslant \frac{1}{512}
$$
is $\qquad$ . | $-1.1024$.
Let $a=20, b=17, c=\frac{1}{512}$.
Then $a>b>c>0$.
The original inequality is equivalent to $\frac{2 x-(a+b)}{(x-a)(x-b)} \geqslant c$.
When $x>a$ or $x0, f(a)=b-a<0$.
Let the two real roots of $f(x)=0$ be $x_{1}$ and $x_{2}$ $\left(x_{1}<x_{2}\right)$. Then the interval of $x$ that satisfies $f(x) \leqslant... | 1024 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the function
$$
\begin{aligned}
f(x)= & a \tan ^{2017} x+b x^{2017}+ \\
& c \ln \left(x+\sqrt{x^{2}+1}\right)+20,
\end{aligned}
$$
where $a$, $b$, and $c$ are real numbers. If $f\left(\ln \log _{5} 21\right)=17$, then $f\left(\ln \log _{21} 5\right)=$ $\qquad$ | 3. 23 .
Let $g(x)$
$$
=a \tan ^{2017} x+b x^{2017}+c \ln \left(x+\sqrt{x^{2}+1}\right) \text {. }
$$
Then $g(-x)=-g(x)$
$$
\begin{array}{l}
\Rightarrow f(-x)-20=-(f(x)-20) \\
\Rightarrow f(-x)=40-f(x) .
\end{array}
$$
Therefore, $f\left(\ln \log _{21} 5\right)=f\left(-\ln \log _{5} 21\right)$
$$
=40-f\left(\ln \log ... | 23 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. For any positive integer $n$, define
$$
S(n)=\left[\frac{n}{10^{[\lg n]}}\right]+10\left(n-10^{[\lg n]}\left[\frac{n}{10^{[\lg n]}}\right]\right) \text {. }
$$
Then among the positive integers $1,2, \cdots, 5000$, the number of positive integers $n$ that satisfy $S(S(n))=n$ is $\qquad$ . | 7. 135.
Let $t=10^{[18 n]}$, then
$$
S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right)
$$
Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$, and $\left[\frac{n}{t}\right]$ is the leading digit of $n$.
We will discuss the cases separately.
(1) If $n$ is a one-digit nu... | 135 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given the ellipse $\Gamma: \frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, a line passing through the left focus $F(-2,0)$ of the ellipse $\Gamma$ with a slope of $k_{1}\left(k_{1} \notin\{0\right.$, $\infty\})$ intersects the ellipse $\Gamma$ at points $A$ and $B$. Let point $R(1,0)$, and extend $A R$ and $B R$ to intersect th... | 9. 305 .
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, $C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$,
$l_{A R}: x=\frac{x_{1}-1}{y_{1}} y+1$.
Substitute into the equation of the ellipse $\Gamma$, eliminate $x$ to get
$$
\frac{5-x_{1}}{y_{1}^{2}} y^{2}+\frac{x_{1}-1}{y_{1}} y-4=0 \text {. }
$... | 305 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the maximum value of the positive integer $r$ such that: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. | Three, first explain $r \leqslant 200$.
Take $k \in\{1,2, \cdots, 10\}$. Let
$$
A_{k}=\{100 k-99,100 k-98, \cdots, 100 k\} \text {. }
$$
Consider the set
$$
\begin{array}{l}
A_{1} \cup A_{5} \cup A_{6} \cup A_{7} \cup A_{9}, A_{1} \cup A_{2} \cup A_{7} \cup A_{8} \cup A_{10}, \\
A_{2} \cup A_{3} \cup A_{6} \cup A_{8} ... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the number of integers in the set $\left\{\left.\frac{2015[a, b]}{a+b} \right\rvert\, a 、 b \in \mathbf{Z}_{+}\right\}$. | Let $d=(a, b)$, and $a=A d, b=B d$, where $A$ and $B$ are coprime positive integers.
Since $[a, b]=A B(a, b)$, then
$(a+b)|2015[a, b] \Leftrightarrow(A+B)| 2015 A B$.
Because $(A, B)=1$, we have
$(A+B, A)=1,(A+B, B)=1$,
$(A+B, A B)=1$.
Thus, $(A+B) \mid 2015$.
For a fixed divisor $k$ of 2015 greater than 1, and $k$ is ... | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Red, blue, green, and white four dice, each die's six faces have numbers $1, 2, 3, 4, 5, 6$. Simultaneously roll these four dice so that the product of the numbers facing up on the four dice equals 36, there are $\qquad$ possible ways. | 4. 48 .
$$
\begin{array}{l}
36=6 \times 6 \times 1 \times 1=6 \times 3 \times 2 \times 1 \\
=4 \times 3 \times 3 \times 1=3 \times 3 \times 2 \times 2 .
\end{array}
$$
For each of the above cases, there are respectively
$$
\begin{array}{l}
\frac{4!}{(2!)(2!)}=6,4!=24, \\
\frac{4!}{2!}=12, \frac{4!}{(2!)(2!)}=6
\end{ar... | 48 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given natural numbers $a, b, c$ whose sum is $S$, satisfying $a+b=1014, c-b=497, a>b$. Then the maximum value of $S$ is ( ).
(A) 1511
(B) 2015
(C) 22017
(D) 2018 | $\begin{array}{l}\text { I. 1.C. } \\ \text { Given } S=a+b+c=1014+b+497 \text {, and } a>b \\ \Rightarrow 1014=a+b \geqslant b+1+b \\ \Rightarrow b \leqslant 506.6 \Rightarrow b_{\max }=506 \\ \Rightarrow S_{\text {max }}=1014+506+497=2017 .\end{array}$ | 2017 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Four. (50 points) Let $A=\{0,1, \cdots, 2016\}$. If a surjective function $f: \mathbf{N} \rightarrow A$ satisfies: for any $i \in \mathbf{N}$,
$$
f(i+2017)=f(i),
$$
then $f$ is called a "harmonious function".
$$
\begin{array}{l}
\text { Let } f^{(1)}(x)=f(x), \\
f^{(k+1)}(x)=f\left(f^{(k)}(x)\right)\left(k \in \mathbf... | On the one hand, note that 2017 is a prime number.
Let $g$ be a primitive root modulo 2017, then the half-order of $g$ modulo 2017 is 1008.
$$
\text{Let } f(i) \equiv g(i-1)+1(\bmod 2017) \text{.}
$$
Since $(g, 2017)=1$, $g(i-1)+1$ runs through a complete residue system modulo 2017.
Thus, the mapping $f: \mathbf{N} \r... | 1008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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