problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
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3. (25 points) Write the 90 positive integers $10, 11, \cdots, 99$ on the blackboard, and erase $n$ of them so that the product of all the remaining numbers on the blackboard has a units digit of 1. Find the minimum value of $n$.
| 3. If the unit digit of the product of all remaining numbers on the blackboard is 1, then all even numbers between $10 \sim 99$ must be erased, and numbers with a unit digit of 5 must also be erased.
Thus, the unit digit of the remaining numbers must be one of $1, 3, 7, 9$.
Notice that the unit digit of $11 \times 13... | 55 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. The sum of all positive integers $n$ that satisfy $\frac{1}{4}<\sin \frac{\pi}{n}<\frac{1}{3}$ is . $\qquad$ | 7. 33 .
By the convexity of the sine function, we know that when $x \in\left(0, \frac{\pi}{6}\right)$, $\frac{3}{\pi} x < \sin x < x$. For example, $\frac{3}{\pi} \times \frac{\pi}{12}=\frac{1}{4}$, $\sin \frac{\pi}{10} < \frac{3}{\pi} \times \frac{\pi}{9}=\frac{1}{3}$.
Therefore, the positive integer values of $n$ t... | 33 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function $f(x)=\sqrt{x^{2}+2}(x>0)$. Then the integer part of $N=f(1002)+f(1003)+\cdots+f(2005)$ is ( ).
(A) 1506500
(B) 1509514
(C) 4010
(D) 3013 | 6. B.
Notice,
$$
\begin{array}{l}
f(x)-x=\sqrt{x^{2}+2}-x>0, \\
f(x)-x=\sqrt{x^{2}+2}-x=\frac{2}{\sqrt{x^{2}+2}+x} \\
1002+1003+\cdots+2005 \text {, }
$$
and $\square$
$$
\begin{aligned}
N- & (1002+1003+\cdots+2005) \\
= & \left(\sqrt{1002^{2}+2}-1002\right)+ \\
& \left(\sqrt{1003^{2}+2}-1003\right)+\cdots+ \\
& \lef... | 1509514 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given 10 pairwise distinct positive integers $a_{1}$, $a_{2}, \cdots, a_{10}$ that satisfy the conditions
$$
\begin{array}{l}
a_{2}=a_{1}+a_{5}, a_{3}=a_{2}+a_{6}, \\
a_{4}=a_{3}+a_{7}, a_{6}=a_{5}+a_{8}, \\
a_{7}=a_{6}+a_{9}, a_{9}=a_{8}+a_{10} .
\end{array}
$$
then the minimum possible value of $a_{4}$ is | Ni, 1.20.
It is easy to get
$$
\begin{array}{l}
a_{4}=a_{3}+a_{7} \\
=a_{1}+a_{5}+a_{6}+a_{6}+a_{8}+a_{10} \\
=\left(a_{1}+a_{10}\right)+3\left(a_{5}+a_{8}\right) .
\end{array}
$$
To make $a_{4}$ the smallest, then $a_{5}$ and $a_{8}$ should be as small as possible. And they are all different, so let's take $a_{5}=1, ... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 There are 2011 points in space and no three points are collinear. Now, connect each pair of points with a line of one color, such that for any point, any two lines originating from that point are of different colors. How many different colors of lines are needed at least? Prove your conclusion. If the 2011 po... | To generalize, replace 2011 with $n(n \geqslant 2)$, and denote the minimum number of colors for the line segments as $f(n)$. It is easy to see that
$$
\begin{array}{l}
f(2)=1, f(3)=3, f(4)=3, \\
f(5)=5, \cdots \cdots
\end{array}
$$
Below, we prove that in the general case,
$$
f(2 n+1)=2 n+1, f(2 n)=2 n-1 \text {. }
$... | 2011 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Three. (25 points) Let the pairs of positive integers $(m, n)$, where both are no more than 1000, satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} \text {. }
$$
Find the number of all such pairs $(m, n)$. | Three, 1706.
$$
\begin{array}{l}
\text { Given } \frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} \\
\Rightarrow \sqrt{2} n-1<m<\sqrt{2}(n+1) .
\end{array}
$$
For each $n$, the number of integers in the above range is
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1,
\end{array}
$$
where $[... | 1706 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For $n \in \mathbf{N}_{+}$, define
$$
S(n)=\left[\frac{n}{10^{[18 n]}}\right]+10\left(n-10^{[i \mid n]}\left[\frac{n}{10^{\left[1 / B^{n}\right]}}\right]\right),
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then, among $1,2, \cdots, 2012$, the number of positive integers $n$ that sati... | 6. 108.
Let $t=10^{[\lg n]}$. Then
$$
S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right) \text {. }
$$
Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$, and $\left[\frac{n}{t}\right]$ is the first digit of $n$.
We will discuss the cases below.
(1) If $n$ is a one-dig... | 108 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For example, $5 n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ have a sum of 2009. If these $n$ numbers can be divided into 41 groups with equal sums and also into 49 groups with equal sums, find the minimum value of $n$.
| Let the 41 groups be $A_{1}, A_{2}, \cdots, A_{41}$, where the sum of the numbers in each group is 49, and we call such groups "A-type groups"; and the 49 groups be $B_{1}, B_{2}, \cdots, B_{49}$, where the sum of the numbers in each group is 41, and we call such groups "B-type groups".
Clearly, each term $x_{k}$ belo... | 89 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 In the Mathematical Olympiad training team, there are 30 members, each of whom has the same number of friends in the team. It is known that in a test, everyone's scores are different. If a member scores higher than the majority of their friends, they are called a "pro". Question: What is the maximum number of... | Let each team member have $k$ friends, and this exam has produced $m$ experts, the best-performing member of the team, is the best in their $k$ "friend pairs," and is naturally an expert. Each of the other experts is at least the best in $\left[\frac{k}{2}\right]+1 \geqslant \frac{k+1}{2}$ (where [x] denotes the greate... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the set $S=\{1,2, \cdots, 50\}$. Find the smallest positive integer $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$, satisfying $(a+b) \mid a b$. | First, by enumeration, we obtain 23 pairs $(a, b)$, each of which satisfies $(a+b) \mid a b$.
Construct a 50-order graph $G$ (with the set $S$ of numbers $1,2, \cdots, 50$ as vertices, and if two numbers $a, b$ belong to the above pairs, then let $a, b$ be adjacent). Thus, the graph $G$ has exactly 23 edges (isolated ... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given
$$
a^{2}(b+c)=b^{2}(a+c)=2010 \text {, and } a \neq b \text {. }
$$
Then $c^{2}(a+b)=$ $\qquad$ [2]
$(2010$, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | 【Analysis】The given condition equation has the same structure as the algebraic expression to be evaluated. According to the known condition equation, it is impossible to determine the values of $a$, $b$, and $c$. We can only conjecture that there is an intrinsic relationship between $a$, $b$, and $c$. By constructing a... | 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let real numbers $x, y, z$ simultaneously satisfy
$$
\left\{\begin{array}{l}
x^{3}+y=3 x+4, \\
2 y^{3}+z=6 y+6, \\
3 z^{3}+x=9 z+8 .
\end{array}\right.
$$
Try to find the value of $2008(x-1)^{2}+2009(y-1)^{2}+$ $2010(z-2)^{2}$. ${ }^{[3]}$
(1st Youth Mathematical Week (Zonghu Cup) Mathematical Competition) | Solve: From the given, we have
$$
\left\{\begin{array}{l}
y-2=-x^{3}+3 x+2=-(x-2)(x+1)^{2}, \\
z-2=-2 y^{3}+6 y+4=-2(y-2)(y+1)^{2}, \\
x-2=-3 z^{3}+9 z+6=-3(z-2)(z+1)^{2} .
\end{array}\right.
$$
Multiplying the above three equations, we get
$$
(x-2)(y-2)(z-2)
$$
$$
\begin{aligned}
= & -6(x-2)(y-2)(z-2)(x+1)^{2}(y+1)^{... | 4017 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Use the digits $1,2, \cdots, 7$ to form a seven-digit number such that it is a multiple of 11. The number of seven-digit numbers that can be formed is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5. 576 .
Let $n$ be a seven-digit number satisfying the condition, and let $a$ and $b$ be the sums of the digits in the odd and even positions, respectively. Then $a+b=28$, and $a-b$ is a multiple of 11.
Since $a+b$ and $a-b$ have the same parity, they must both be even. Clearly, $|a-b| \neq 22$, so only $a-b=0$.
Thus... | 576 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (20 points) Question: In how many different ways can the elements of the set $M=\{1,2,3,4,5\}$ be assigned to three (ordered) sets $A$, $B$, and $C$, such that each element is contained in at least one of the sets, the intersection of these three sets is empty, and the intersection of any two of these sets is not... | As shown in Figure 2, consider the seven parts divided by the Venn diagram, represented by $x, u, v, w, a, b, c$ respectively.
Now, fill the elements of $M$ into these parts. According to the problem, $x$ cannot be filled with any number, while $u, v, w$ must be filled with numbers, and the numbers filled in these par... | 1230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let positive numbers $x, y, z$ satisfy
$$
\frac{1}{x^{3}}=\frac{8}{y^{3}}=\frac{27}{z^{3}}=\frac{k}{(x+y+z)^{3}} \text {. }
$$
Then $k=$ $\qquad$ | $$
\begin{array}{l}
\sqrt[3]{k}=\frac{x+y+z}{x}=\frac{2(x+y+z)}{y} \\
=\frac{3(x+y+z)}{z}=\frac{6(x+y+z)}{x+y+z}=6 .
\end{array}
$$
Therefore, $k=216$. | 216 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $x_{n}$ denote the unit digit of the number $n^{4}$. Then
$$
x_{1}+x_{2}+\cdots+x_{2012}=
$$
$\qquad$ | 4.6640 .
Notice that, the unit digit of $(10+n)^{4}$ is the same as that of $n^{4}$, and the unit digits of $1^{4}, 2^{4}, \cdots, 10^{4}$ are $1,6,1,6,5,6,1,6,1,0$ respectively.
Thus, $x_{1}+x_{2}+\cdots+x_{10}=33$.
Therefore, $x_{1}+x_{2}+\cdots+x_{2012}$
$$
=201 \times 33+(1+6)=6640 \text {. }
$$ | 6640 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Suppose there are 10 red, 10 yellow, and 10 blue small balls. Now, all of them are to be placed into two bags, A and B, such that each bag contains balls of two colors, and the sum of the squares of the number of balls of two colors in bags A and B are equal. There are $\qquad$ ways to do this. | 5.61.
Let the number of red, yellow, and blue balls in bag A be $x, y, z (1 \leqslant x, y, z \leqslant 9)$. Then the number of balls of corresponding colors in bag B are $10-x, 10-y, 10-z$.
First, assume $x \leqslant y \leqslant z$.
From the problem, we know
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=(10-x)^{2}+(10-y)^{2... | 61 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given two lines with a slope of 1, $l_{1}$ and $l_{2}$, passing through the two foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $l_{1}$ intersects the ellipse at points $A$ and $B$, $l_{2}$ intersects the ellipse at points $C$ and $D$. If quadrilateral $\square A B C D$ satisfies $A C \pe... | 7.2012.
It is known that $\square A B C D$ is symmetric about the origin $O$. As shown in Figure 1, let $\angle A \dot{F}_{1} F_{2}=\alpha$. Then
. $\tan \alpha=1$
. $\Rightarrow \alpha=45^{\circ}$.
Since $A C \perp A B$, we know
$A C \perp A F_{1}$.
Thus, $\triangle A F_{1} O$ is an
isosceles right triangle. Therefo... | 2012 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Rolling Dice (a uniform cube, with six faces marked with $1,2,3,4,5,6$) Game rules are as follows: First roll 9 dice, take out the dice showing 1 and set them aside; on the second roll, take out the dice showing 1 from the remaining dice; $\cdots \cdots \cdots$, until no dice show 1 or all dice are take... | 10. According to the game rules, if the game ends exactly after 9 rounds, then in the first eight rounds, each time exactly 1 die shows a 1, and the ninth round ends the game regardless of whether it shows a 1 or not. Among these, the probability that exactly 1 die shows a 1 in the $k(k=1,2, \cdots, 8)$-th round, where... | 2012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Let the number of all positive integers satisfying the following conditions be $N$:
(1) less than or equal to 2,012;
(2) the number of 1s in their binary representation is at least 2 more than the number of 0s.
Find the sum of the digits of $N$.
| Three, from $2012=(11111011100)_{2}$, we know that the numbers satisfying the conditions have at most 11 digits in binary representation.
The first digit must be 1, so the number of $d+1$-digit numbers with exactly $k+1$ digits being 1 is $\mathrm{C}_{d}^{k}$, and condition (2) is equivalent to
$$
\begin{array}{l}
k+1... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Let $n \in \mathbf{N}_{+}, f(n)$ be the number of all integer sequences $\left\{a_{k} \mid k=0,1, \cdots, n\right\}$ that satisfy the following conditions:
$$
\begin{array}{l}
\text { (1) } a_{0}=0, a_{n}=2 n, \text { and } \\
1 \leqslant a_{k+1}-a_{k} \leqslant 3(k=0,1, \cdots, n-1) ;
\end{array}
$$
... | Divide a circle of length $2 \cdot n$ into $2n$ equal parts, and label the points sequentially as $0,1, \cdots, 2n$. Then color the points labeled $a_{i} (i=0,1, \cdots, n-1)$ black, and the other $n$ points white. The sequence given in the problem corresponds one-to-one with the following coloring method:
(1) The poin... | 2012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let real numbers $a, b$ satisfy
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {. }
$$
Find the minimum value of $u=9 a^{2}+72 b+2$. | Notice,
$$
\begin{array}{l}
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \\
\Rightarrow(a-2 b)(3 a-4 b+5)=0 \\
\Rightarrow a-2 b=0 \text { or } 3 a-4 b+5=0 .
\end{array}
$$
(1) $a-2 b=0$.
Then $u=9 a^{2}+72 b+2=36 b^{2}+72 b+2$ $=36(b+1)^{2}-34$.
Thus, when $b=-1$, the minimum value of $u$ is -34.
$$
\text { (2) } 3 a-4 b+5=0 \t... | -34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac... | Solve: Consider the given system of equations as a fractional equation in terms of $t$
$$
\begin{array}{c}
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1 \\
\Rightarrow\left(t-1^{2}\right)\left(t-3^{2}\right)\left(t-5^{2}\right)\left(t-7^{2}\right)- \\
x^{2}\left(t-3^{2}\right... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (16 points) As shown in Figure 1, in a certain engineering project to measure the radius $R$ of an arc, two identical small balls are placed on the arc so that each contact point is tangent to the arc. The height difference between the balls is $h$, and the radius of the small balls is $r$. Try to express $R$ in te... | 14. Let the angle between the line connecting the centers of the two smaller circles and the line connecting the center of the larger circle be $\theta$. Then
$$
\cos \theta=\frac{(R-r)^{2}+(R-r)^{2}-(2 r)^{2}}{2(R-r)^{2}} .
$$
Also, $h=(R-r)-(R-r) \cos \theta=\frac{2 r^{2}}{R-r}$
$$
\Rightarrow R=r+\frac{2 r^{2}}{h} ... | 600 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given that $x_{1}, x_{2}, \cdots, x_{40}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{40}=58$. If the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ is $A$, and the minimum value is $B$, then $A+B=$ $\qquad$ | Solution: Since there are only a finite number of ways to write 58 as the sum of 40 positive integers, the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ exist.
Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{40}$.
If $x_{1}>1$, then
$$
x_{1}+x_{2}=\lef... | 494 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let there be a non-empty set $A \subseteq\{1,2, \cdots, 7\}$, and when $a \in A$, it must also be that $8-a \in A$. Then the number of such sets $A$ is $\qquad$ . | 3. 15 .
Find the single element or binary element set that is congruent to $8-a$ in $A$:
$$
\begin{array}{l}
A_{1}=\{4\}, A_{2}=\{1,7\}, \\
A_{3}=\{2,6\}, A_{4}=\{3,5\} .
\end{array}
$$
The problem is equivalent to finding the number of non-empty subsets of $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$.
Thus, there are... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the first seven digits of an 11-digit mobile phone number are 1390931. If the remaining four digits can only be 1, 3, 5 and each must appear at least once, then there are such mobile phone numbers.
untranslated: 个.
Note: The word "个" at the end of the sentence is not translated as it is a placeholder for t... | 5. 36 . | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
\left\{\begin{array}{l}
x-999 \geqslant 1000, \\
x+1 \leqslant a
\end{array}\right.
$$
has a finite number of real solutions. Then the value of $a$ is $\qquad$ . | Ni, 1.2000.
The solution set of the inequality is $1999 \leqslant x \leqslant a-1$.
From the problem, we know $a-1=1999 \Rightarrow a=2000$. | 2000 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. On the way from Xiaoming's house to the swimming pool, there are 200 trees. When going to the swimming pool and returning, Xiaoming ties red ribbons on some trees: 1: to make marks. When going to the swimming pool, he marks the 1st tree, the 6th tree, the 11th tree, …, each time skipping 4 trees without marking. On ... | 2. 140 .
According to the problem, the trees marked are the $5 x+1(x=0,1, \cdots, 39)$ and $8 y(y=1,2, \cdots, 25)$ ones, among which exactly 5 trees are marked twice.
Therefore, the number of trees that are not marked is
$$
200-(40+25-5)=140 \text {. }
$$ | 140 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given that the positive integer $M$ when divided by the positive integer $N$ leaves a remainder of 2, and the sum of the reciprocals of all values of $N$ is $\frac{627}{670}$, with the number of all values of $N$ being less than 16. Find all possible values of $M$.
| Obviously, $N>2, N \mid (M-2)$.
Since $670=2 \times 5 \times 67$, $M-2$ has prime factors 2, 5, and 67. Therefore, the sum of the reciprocals of all positive divisors of $M-2$ is
$$
\frac{1}{1}+\frac{1}{2}+\frac{627}{670}=\frac{816}{335} .
$$
The number of positive divisors of $M-2$ is less than 18.
Since $(1+1)^{5}=32... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a_{1}, a_{2}, \cdots, a_{6}$ be any permutation of $1,2, \cdots, 6$, and $f$ be a one-to-one mapping from $\{1,2, \cdots, 6\}$ to $\{1,2, \cdots, 6\}$, satisfying
$$
f(i) \neq i, f(f(i))=i(i=1,2, \cdots, 6) .
$$
Consider the number table
$$
A=\left[\begin{array}{cccccc}
a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & ... | 5. 10800 .
For a permutation $a_{1}, a_{2}, \cdots, a_{6}$, consider the one-to-one mapping satisfying
$$
f(i) \neq i, f(f(i))=i(i=1,2, \cdots, 6)
$$
For each such mapping $f$, the elements of set $A$ can be paired as $\{i, j\}$, such that
$$
f(i)=j, f(j)=i \text {. }
$$
Thus, for each permutation, the number of map... | 10800 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $a_{1}, a_{2}, \cdots$ is a geometric sequence with the first term $a_{1}=a\left(a \in \mathbf{Z}_{+}\right)$ and common ratio $r\left(r \in \mathbf{Z}_{+}\right)$. Suppose
$$
\log _{4} a_{2}+\log _{4} a_{3}+\cdots+\log _{4} a_{12}=2013 .
$$
Then the number of ordered pairs $(a, r)$ that satisfy the condition... | 6. 62.
From the given, we have $a_{n}=a r r^{n-1}$, substituting into the given equation yields
$$
a^{11} r^{66}=2^{4026} \Rightarrow a r^{6}=2^{366} \text{. }
$$
Let $a=2^{x}, r=2^{y}(x, y \in \mathrm{N})$. Then $x+6y=366$.
Thus, the number of pairs satisfying the condition is 62.
_ 171
2. $\frac{1}{3}$.
By the uni... | 62 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than 1. Then
$\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=$ $\qquad$ [2]
(2012, Zhejiang Province High School Mathematics Competition) | When the common ratio $q=1$,
$$
\begin{array}{l}
a_{n}=a_{1}, \\
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011 .
\end{array}
$$
When the common ratio $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1}... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given positive integers $a_{1}, a_{2}, \cdots, a_{10}$ satisfy
$$
\frac{a_{j}}{a_{i}}>\frac{2}{3}(1 \leqslant i \leqslant j \leqslant 10) \text {. }
$$
Then the minimum possible value of $a_{10}$ is $\qquad$ . | 2. 92.
From $a_{1} \geqslant 1, a_{2}>\frac{3}{2} a_{1} \geqslant \frac{3}{2}\left(a_{2} \in \mathbf{N}_{+}\right)$, we get $a_{2} \geqslant 2$.
Similarly, $a_{3}>\frac{3}{2} a_{2} \geqslant 3, a_{3} \geqslant 4$;
$$
\begin{array}{l}
a_{4}>\frac{3}{2} a_{3} \geqslant 6, a_{4} \geqslant 7 ; \\
a_{5}>\frac{3}{2} a_{4} \... | 92 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, \\
a_{n+2}=\frac{2(n+1)}{n+2} a_{n+1}-\frac{n}{n+2} a_{n}(n=1,2, \cdots) .
\end{array}
$$
If $a_{m}>2+\frac{2011}{2012}$, then the smallest positive integer $m$ is . $\qquad$ | 8.4025.
$$
\begin{array}{l}
\text { Given } a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
\begin{array}{l}
\Rightarrow a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right) \\
=\frac{n-2}{n} \cdot \frac{n-3}{n-1}\left(a_{n-2}-a_{n-3}\right)=\cdots \\
= \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdots \fr... | 4025 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. (16 points) Given an integer $n(n \geqslant 3)$, let $f(n)$ be the minimum number of elements in a subset $A$ of the set $\left\{1,2, \cdots, 2^{n}-1\right\}$ that satisfies the following two conditions:
(i) $1 \in A, 2^{n}-1 \in A$;
(ii) Each element in subset $A$ (except 1) is the sum of two (possibly the same) e... | 12. (1) Let set $A \subseteq\left\{1,2, \cdots, 2^{3}-1\right\}$, and $A$ satisfies (i) and (ii). Then $1 \in A, 7 \in A$.
Since $\{1, m, 7\}(m=2,3, \cdots, 6)$ does not satisfy (ii), hence $|A|>3$.
Also, $\{1,2,3,7\},\{1,2,4,7\},\{1,2,5,7\}$,
$\{1,2,6,7\},\{1,3,4,7\},\{1,3,5,7\}$,
$\{1,3,6,7\},\{1,4,5,7\},\{1,4,6,7\}... | 108 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Arrange the positive integers whose sum of digits is 5 in ascending order to form a sequence. Then 2012 is the $\qquad$th term of this sequence. | 7.38.
To represent 5 as the sum of no more than four positive integers, there are six methods, that is
$$
\begin{array}{l}
5=1+4=2+3=1+1+3 \\
=1+2+2=1+1+1+2 .
\end{array}
$$
When filling them into a $1 \times 4$ grid, positions that are not filled are supplemented with 0. Then $\{5\}$ has 3 ways of filling; $\{1,4\}$... | 38 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $18^{2}=324, 24^{2}=576$, they are formed by the permutation of two consecutive digits $2,3,4$ and $5,6,7$ respectively; and $66^{2}=4356$ is formed by the permutation of four consecutive digits $3, 4, 5, 6$. Then the next such square number is $\qquad$
| 8.5476.
For any square number, its last digit can only be $0, 1, 4, 5, 6, 9$, and
$$
\begin{array}{l}
(10 a)^{2}=100 a^{2}, \\
(10 a+5)^{2}=100 a^{2}+100 a+25, \\
(10 a \pm 4)^{2}=100 a^{2} \pm 80 a+16 ; \\
(10 a \pm 1)^{2}=100 a^{2} \pm 20 a+1, \\
(10 a \pm 3)^{2}=100 a^{2} \pm 60 a+9 ; \\
(10 a \pm 2)^{2}=100 a^{2} ... | 5476 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $4^{a}-3 a^{b}=16, \log _{2} a=\frac{a+1}{b}$. Then $a^{b}=$ $\qquad$ . | - 1. 16.
From $\log _{2} a=\frac{a+1}{b} \Rightarrow a^{b}=2^{a+1}$.
Substituting into $4^{a}-3 a^{b}=16$, we get
$2^{2 a}-6 \times 2^{a}-16=0$.
Solving, we get $2^{a}=8$ or -2 (discard).
Therefore, $a^{b}=2^{a+1}=16$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (16 points) Let the side lengths opposite to two interior angles of $\triangle A B C$ be $a, b, c$ respectively, and $a+b+c=16$. Find
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
the value. | $$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 4 R^{2}\left(\sin ^{2} B \cdot \cos ^{2} \frac{C}{2}+\sin ^{2} C \cdot \cos ^{2} \frac{B}{2}+\right. \\
\left.2 \sin B \cdot \sin C \cdot \cos \frac{B}{2} \cdot \cos \frac... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
16. Let $P$ be any point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ other than the endpoints of the major axis, $F_{1}$ and $F_{2}$ be the left and right foci respectively, and $O$ be the center. Then
$\left|P F_{1}\right|\left|P F_{2}\right|+|O P|^{2}=$ $\qquad$ . | 16. 25.
According to the definition of an ellipse and the cosine rule, the solution can be found. | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 (An Ancient Chinese Mathematical Problem) Emperor Taizong of Tang ordered the counting of soldiers: if 1,001 soldiers make up one battalion, then one person remains; if 1,002 soldiers make up one battalion, then four people remain. This time, the counting of soldiers has at least $\qquad$ people. | Let the first troop count be 1001 people per battalion, totaling $x$ battalions, then the total number of soldiers is $1001 x + 1$ people; let the second troop count be 1002 people per battalion, totaling $y$ battalions, then the total number of soldiers is $1002 y + 4$ people.
From the equality of the total number of ... | 1000000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Mother's Day is coming, and Xiao Hong, Xiao Li, and Xiao Meng went to the flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the three types... | Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $\sqrt{z}$ yuan, respectively. Then,
$$
\left\{\begin{array}{l}
3 x+7 y+z=14, \\
4 x+10 y+z=16 .
\end{array}\right.
$$
Eliminating $z$ gives
$$
x=2-3 y \text{. }
$$
Substituting equation (2) into equation (1) gives
$$
z=8+2 y \text{. }
$$... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 In the tetrahedron $A-B C D$, it is known that
$$
\begin{array}{l}
\angle A C B=\angle C B D \\
\angle A C D=\angle A D C=\angle B C D=\angle B D C=\theta,
\end{array}
$$
and $\cos \theta=\frac{\sqrt{10}}{10}$.
If the length of edge $A B$ is $6 \sqrt{2}$, then the volume of this pyramid is
$\qquad$ | Solve As shown in Figure 2, from the problem, we know $\triangle A C D \cong \triangle B C D$,
and $\square$
$$
\begin{array}{l}
A C=A D \\
=B C=B D .
\end{array}
$$
Then $\triangle A C D$
$\cong \triangle B C D$
$\cong \triangle C A B$
$\cong \triangle D A B$.
Therefore, $\angle A C B$
$$
\begin{array}{l}
\because \a... | 144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Two boxes of candy have a total of 176 pieces. 16 pieces are taken from the second box and placed into the first box, at which point, the number of pieces of candy in the first box is 31 more than $m($ an integer $m>1)$ times the number of pieces of candy in the second box. Then, the first box originally had ... | Let the first box originally contain $x$ candies, and the second box originally contain $y$ candies.
According to the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
x+y=176, \\
x+16=m(y-16)+31 .
\end{array}\right.
$$
Rearranging, we get
$$
x+16=m(176-16-x)+31 \text {, }
$$
which simplifies to ... | 131 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given three points $A, B, C$ in a plane satisfying
$$
|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=5,|\overrightarrow{C A}|=6 \text {. }
$$
Then the value of $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$ is ( ). | 5. C.
From the cosine theorem, we know
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}=-\overrightarrow{B A} \cdot \overrightarrow{B C} \\
=-\frac{|\overrightarrow{B A}|^{2}+|\overrightarrow{B C}|^{2}-|\overrightarrow{A C}|^{2}}{2} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\overrightarrow{B ... | -35 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Every day at 5 PM when school is over, Xiao Ming's father always drives from home to pick him up on time and take him back. One day, the school dismissed an hour early, and Xiao Ming walked home by himself. On the way, he met his father who was coming to pick him up, and as a result, they arrived home 20 minutes ear... | 二、1.50 minutes.
As shown in Figure 6, Xiao Ming
starts walking home from point $A$
and meets the car coming to pick him up
at point $C$. As a result, the car returns from $C$ to $B$ 20 minutes earlier than usual. This indicates that the car takes 20 minutes to travel from $C$ to $A$ and back to $C$.
Therefore, the car ... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. Let real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Denote the maximum value of $x+y+z+w$ as $a$, and the minimum value as $b$. Then $a+b=$ $\qquad$ | 2. 45 .
From $x \geqslant y \geqslant z \geqslant w \geqslant 0$, we know
$$
\begin{array}{l}
100=5 x+4 y+3 z+6 w \geqslant 4(x+y+z+w) \\
\Rightarrow x+y+z+w \leqslant 25 .
\end{array}
$$
When $x=y=z=\frac{25}{3}, w=0$, the equality holds.
$$
\begin{array}{l}
\text { Also } 100=5 x+4 y+3 z+6 w \leqslant 5(x+y+z+w) \\... | 45 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. After rotating any positive integer by $180^{\circ}$, some interesting phenomena can be observed, such as 808 still being 808 after a $180^{\circ}$ rotation, 169 becoming 691 after a $180^{\circ}$ rotation, and 37 not being a number after a $180^{\circ}$ rotation. Then, among all five-digit numbers, the number of fi... | 2. 60.
Among the ten digits from $0$ to $9$, $(0,0)$, $(1,1)$, $(8,8)$, and $(6,9)$ can be placed in the symmetric positions at the beginning and end of a five-digit number. When rotated $180^{\circ}$, the resulting number is the same as the original number, while other digits cannot appear in the five-digit number. S... | 60 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the sum of the first $n(n>1)$ terms of an arithmetic sequence is 2013, the common difference is 2, and the first term is an integer. Then the sum of all possible values of $n$ is . $\qquad$ | 3. 2975 .
Let the first term of the sequence be $a_{1}$, and the common difference $d=2$. Then
$$
\begin{array}{l}
S_{n}=n a_{1}+n(n-1) \\
=n\left(a_{1}+n-1\right)=2013 .
\end{array}
$$
Also, $2013=3 \times 11 \times 61$, and $n$ is a divisor of 2013, so the sum of all possible values of $n$ is
$$
(1+3) \times(1+11) ... | 2975 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=p, a_{2}=p+1, \\
a_{n+2}-2 a_{n+1}+a_{n}=n-20,
\end{array}
$$
where $p$ is a given real number, and $n$ is a positive integer. Try to find the value of $n$ that minimizes $a_{n}$. | 10. Let $b_{n}=a_{n+1}-a_{n}$.
From the problem, we have $b_{n+1}-b_{n}=n-20$, and $b_{1}=1$. Then, $b_{n}-b_{1}=\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)=\sum_{i=1}^{n-1}(i-20)$.
Thus, $b_{n}=\frac{(n-1)(n-40)}{2}+1$.
Also, $a_{3}=a_{2}+b_{2}=p-17<a_{1}<a_{2}$, so when the value of $a_{n}$ is the smallest, we have $... | 40 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given $x, y \in \mathbf{N}_{+}$. Find the minimum value of $\sqrt{512^{x}-7^{2}-1}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $z=512^{x}-7^{2 y-1}$.
Obviously, $z \geqslant 0$, and
$z \equiv 1(\bmod 7), z \equiv 1(\bmod 8)$.
(1) $x=2 x_{1}-1$ is an odd number.
Then $z \equiv(-1)^{x}-1 \equiv-2 \equiv 1(\bmod 3)$.
Thus, $z \equiv 1(\bmod 3 \times 7 \times 8)$.
Let $z=1$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=1$. Then $7^{2 y-1}=512^{2 x_{1}-1}... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given
$$
A=\left\{z \mid z^{18}=1\right\} \text { and } B=\left\{\omega \mid \omega^{48}=1\right\}
$$
are sets of complex roots of unity,
$$
C=\{z w \mid z \in A, w \in B\}
$$
is also a set of complex roots of unity. How many distinct elements are there in the set $C$? ${ }^{[3]}$ | Notice that, $z=\cos \frac{2 k \pi}{18}+\mathrm{i} \sin \frac{2 k \pi}{18}(k \in \mathbf{Z})$ (18 distinct elements),
$\omega=\cos \frac{2 t \pi}{48}+\mathrm{i} \sin \frac{2 t \pi}{48}(t \in \mathbf{Z})$ (48 distinct elements),
$$
\begin{array}{l}
z \omega=\cos \frac{2 \pi(8 k+3 t)}{144}+\mathrm{i} \sin \frac{2 \pi(8 k... | 144 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that the upper base, height, and lower base of a trapezoid are three consecutive positive integers, and these three numbers make the value of the polynomial $x^{3}-30 x^{2}+a x$ (where $a$ is a constant) also three consecutive positive integers in the same order. Then the area of this trapezoid is $\qquad$ | 7. 100 .
Let the upper base, height, and lower base be $n-1$, $n$, and $n+1$ respectively, and the other three consecutive integers be $m-1$, $m$, and $m+1$. Then,
$$
\begin{array}{l}
(n-1)^{3}-30(n-1)^{2}+a(n-1)=m-1, \\
n^{3}-30 n^{2}+a n=m, \\
(n+1)^{3}-30(n+1)^{2}+a(n+1)=m+1 . \\
\text { (1) }+(3)-2 \times(2) \text... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange all positive integers that leave a remainder of 2 and 3 when divided by 4 in ascending order. Let $S_{n}$ denote the sum of the first $n$ terms of this sequence. Then $\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right]$ $=$ $\qquad$ ([ $x]$ denotes the greatest integer no... | 8. 2025078.
Given that this sequence is
$$
2,3,6,7, \cdots, 4 n-2,4 n-1, \cdots \text {. }
$$
From $4 n-2+4 n-1=8 n-3$, we know
$$
\begin{array}{l}
S_{2 n}=5+13+\cdots+(8 n-3) \\
=4 n(n+1)-3 n=4 n^{2}+n . \\
\text { Also, } 4 n^{2}<S_{2 n}<(2 n+1)^{2} \text {, then } \\
2 n<\sqrt{S_{2 n}}<2 n+1 .
\end{array}
$$
Ther... | 2025078 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 A scientist stored the design blueprint of his time machine in a computer, setting the password to open the file as a permutation of $\{1,2, \cdots, 64\}$. He also designed a program that, when eight positive integers between $1 \sim 64$ are input each time, the computer will indicate the order (from left to ... | Let the password be denoted as $a_{1} a_{2} \cdots a_{n^{2}} (n=8)$.
First, write the numbers $1, 2, \cdots, n^{2}$ arbitrarily into an $n \times n$ grid (one number per cell). After the first $n$ operations, input each row of numbers once, and rearrange the numbers in each row from left to right according to the compu... | 45 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the set
$$
A=\{x \mid 5 x-a \leqslant 0, a \in \mathbf{N}\} \text {. }
$$
If $5 \in A \cap \mathbf{Z}$, then the minimum value of $a$ is | $$
-, 1.25
$$
From $A \left\lvert\,=\left(-\infty, \frac{a}{5}\right]\right.$, and $5 \in A \cap \mathbf{Z}$, we know
$$
\frac{a}{5} \geqslant 5 \Rightarrow a \geqslant 25 \text {. }
$$
Therefore, the minimum value of $a$ is 25. | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
the graph always passes through a fixed point $A$. If point $A$ lies on the line
$$
\frac{x}{m}+\frac{y}{n}+1=0(m, n>0)
$$
then the minimum value of $3 m+n$ is | 5. 16 .
Note that the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
always passes through the fixed point $(-3,-1)$.
So point $A(-3,-1)$.
Then $-\frac{3}{m}-\frac{1}{n}+1=0 \Rightarrow 1=\frac{3}{m}+\frac{1}{n}$.
Thus, $3 m+n=(3 m+n)\left(\frac{3}{m}+\frac{1}{n}\right)$
$$
=10+\frac{3 n}{m}+\frac{3 m}{n} \geqslant 16 \te... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. (15 points) As shown in Figure 2, in the Cartesian coordinate system, the equation of circle $\odot M$ is
$$
x^{2}+y^{2}+D x+E y+F=0,
$$
and the quadrilateral $A B C D$ inscribed in $\odot M$ has diagonals $A C$ and $B D$ that are perpendicular to each other, with $A C$ and $B D$ lying on the $x$-axis and $y$-axis... | 13. (1) Let $A(a, 0), C(c, 0)$.
From the problem, points $A$ and $C$ are on the negative and positive halves of the $x$-axis, respectively. Thus, $a c<0$.
When $y=0$, the equation becomes
$$
x^{2}+D x+F=0 \text{, }
$$
where the two roots of the equation are the $x$-coordinates of points $A$ and $C$.
Therefore, $x_{1}... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
I. Fill in the Blanks (8 points each, total 64 points)
1. Among the positive integers less than 20, choose three different numbers such that their sum is divisible by 3. The number of different ways to choose these numbers is $\qquad$. | $-, 1.327$
$$
C_{6}^{3}+C_{6}^{3}+C_{7}^{3}+6 \times 6 \times 7=327
$$ | 327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given a geometric sequence $\left\{a_{n}\right\}$ with all terms being positive. If $2 a_{4}+a_{3}-2 a_{2}-a_{1}=8$, then the minimum value of $2 a_{8}+a_{7}$ is $-2 x-2$ $\qquad$ | 6. 54.
Let $\left\{a_{n}\right\}$ be a geometric sequence with common ratio $q(q>0)$. Then, according to the problem,
$$
\begin{array}{l}
2 a_{2} q^{2}+a_{1} q^{2}-\left(2 a_{2}+a_{1}\right)=8 \\
\Rightarrow\left(2 a_{2}+a_{1}\right)\left(q^{2}-1\right)=8 \\
\Rightarrow 2 a_{2}+a_{1}=\frac{8}{q^{2}-1},
\end{array}
$$
... | 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
\begin{array}{l}
{\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 2012\right]} \\
=
\end{array}
$$ | 8. 18084.
When $2^{k} \leqslant x<2^{k+1}$, then $\left[\log _{2} x\right]=k$.
Given $1024=2^{10}<2012<2^{11}=2048$, we know
$$
\begin{array}{l}
{\left[\log _{2} 1024\right]+\left[\log _{2} 1025\right]+\cdots+\left[\log _{2} 2012\right]} \\
=10 \times(2012-1023)=9890 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence ... | 18084 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $[x]$ denote the integer part of the real number $x$. Then $[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]=(\quad)$.
(A) 146
(B) 161
(C) 210
(D) 365 | 4. C.
$$
\begin{array}{l}
{[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]} \\
=1 \times 3+2 \times 5+3 \times 7+4 \times 9+5 \times 11+6 \times 13+7 \\
=210
\end{array}
$$ | 210 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example 1 From the 205 positive integers $1,2, \cdots, 205$, what is the maximum number of integers that can be selected such that for any three selected numbers $a, b, c (a<b<c)$, we have
$$
a b \neq c ?^{[1]}
$$
(2005, (Casio Cup) National Junior High School Mathematics Competition) | Estimate first.
Since $14 \times 15=210>205$, then $14,15, \cdots$, 205 satisfy that for any three numbers $a 、 b 、 c(a<b<c)$, we have $a b \neq c$.
Because 1 multiplied by any number equals the number itself, $1,14,15, \cdots, 205$ satisfy the condition.
Therefore, there are $205-14+1+1=193$ numbers in total.
If we se... | 193 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2: 9 judges score 12 athletes participating in a bodybuilding competition. Each judge gives 1 point to the athlete they consider to be in 1st place, 2 points to the athlete in 2nd place, ..., and 12 points to the athlete in 12th place. The final scoring shows: the difference between the highest and lowest score... | Explanation: It is impossible for 9 judges to give 1 point to five or more athletes, because among five or more athletes, at least one athlete must be rated no less than 5 by a judge. However, according to the problem, each of these five athletes is rated no more than 4 by each judge, which is a contradiction.
Therefo... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
II. (40 points) Given the sequence of real numbers $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right],
$$
where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
$$
\text { Find } \sum_{i=1}^{2012} a_{i} \text {. }
$$ | $$
\text { II. } a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}, a_{4}=\frac{5}{3} .
$$
By mathematical induction, it is easy to prove
$$
\left\{\begin{array}{l}
a_{2 k+1}=\frac{3 k+1}{3}, \\
a_{2 k+2}=\frac{3 k+2}{3} .
\end{array}\right.
$$
Then $a_{2 k+1}+a_{2 k+2}=2 k+1$.
Therefore, $\sum_{i=1}^{2012} a_{... | 1012036 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Choose $k$ numbers from $1,2, \cdots, 2004$, such that among the chosen $k$ numbers, there are definitely three numbers that can form the side lengths of a triangle (the three numbers must be distinct). What is the minimum value of $k$ that satisfies this condition? | When selecting three numbers from 1 to 2004 to form the sides of a triangle, there are too many possibilities. Instead, let's approach it from the opposite direction and list all sets of three numbers that cannot form the sides of a triangle.
First, 1, 2, 3 cannot form the sides of a triangle. Adding 5, the set \(1, 2... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. A. Given positive integers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a+b^{2}-2 c-2=0, \\
3 a^{2}-8 b+c=0 .
\end{array}
$$
Then the maximum value of $a b c$ is $\qquad$ | 8. A. 2013.
The two equations are simplified and rearranged to get
$$
(b-8)^{2}+6 a^{2}+a=66 \text {. }
$$
Given that $a$ is a positive integer and $6 a^{2}+a \leqslant 66$, we have $1 \leqslant a \leqslant 3$. If $a=1$, then $(b-8)^{2}=59$, which has no positive integer solutions; if $a=2$, then $(b-8)^{2}=40$, whic... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
B. If $k$ numbers are chosen from 2, $, 8, \cdots, 101$ these 34 numbers, where the sum of at least two of them is 43, then the minimum value of $k$ is: $\qquad$ | B. 28.
Divide the 14 numbers less than 43 into the following seven groups: $(2,41),(5,38),(8,35),(11,32)$, $(14,29),(17,26),(20,23)$.
The sum of the two numbers in each group is 43. After selecting one number from each group and then taking all numbers greater than 43, a total of 27 numbers are selected. The sum of a... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. A. Xiaoming volunteered to sell pens at a stationery store one day. Pencils were sold at 4 yuan each, and ballpoint pens at 7 yuan each. At the beginning, it was known that he had a total of 350 pencils and ballpoint pens. Although he did not sell them all that day, his sales revenue was 2013 yuan. Then he sold at ... | 10. A. 207.
Let $x$ and $y$ represent the number of pencils and ballpoint pens sold, respectively. Then
\[
\begin{array}{l}
\left\{\begin{array}{l}
4 x+7 y=2013 ; \\
x+y=204
\end{array}\right.
\end{array}
\]
Thus, $y_{\text {min }}=207$, at which point, $x=141$. | 207 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. A. As shown in Figure 5, given that $A B$ is the diameter of $\odot O$, $C$ is a point on the circumference, and $D$ is a point on the line segment $O B$ (not at the endpoints), satisfying $C D \perp A B$ and $D E \perp C O$ at point $E$. If $C E = 10$, and the lengths of $A D$ and $D B$ are both positive integers,... | 12. A. Connect $A C$ and $B C$, then $\angle A C B=90^{\circ}$.
From $\mathrm{Rt} \triangle C D E \backsim \mathrm{Rt} \triangle C O D$, we know $C E \cdot C O=C D^{2}$.
From $\mathrm{Rt} \triangle A C D \backsim \mathrm{Rt} \triangle C B D$, we know $C D^{2}=A D \cdot B D$.
Therefore, $C E \cdot C O=A D \cdot B D$.
L... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$,
$f(x)=2^{x}+2 x+b$ ( $b$ is a constant).
Then $f(-10)=$ $\qquad$ . | 2. -1043 .
From the given condition, we easily know that
$$
f(0)=2^{0}+2 \times 0+b=0 \text {. }
$$
Solving for $b$ yields $b=-1$.
By the property of odd functions $f(-x)=-f(x)$, we have
$$
\begin{array}{l}
f(-10)=-f(10)=-2^{10}-2 \times 10+1 \\
=-1043 .
\end{array}
$$ | -1043 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 3, square $ABCD$ is divided into 8 triangles of equal area. If $AG=\sqrt{50}$, then the area $S$ of square $ABCD$ is $ \qquad $. | 4. 128 .
As shown in Figure 5, draw $K L / / D C$ through point $F$, take the midpoint $N$ of $A B$, and connect $G N$ with $A H$ intersecting at point $P$.
Let the side length of the square $A B C D$ be $a$.
Given $S_{\triangle D C I}=S_{\triangle M B H}=\frac{1}{8} S$, we know
$C I=B H=\frac{1}{4} B C=\frac{a}{4}$.
... | 128 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Given three distinct integers $x, y, z$ whose sum lies between 40 and 44. If $x, y, z$ form an arithmetic sequence with a common difference of $d$, and $x+y, y+z, z+x$ form a geometric sequence with a common ratio of $q$, then $d q=$ $\qquad$ | 5. 42 .
$$
\begin{array}{l}
\text { Given } x=y-d, z=y+d \\
\Rightarrow x+y=2 y-d, y+z=2 y+d \\
\Rightarrow z+x=2 y . \\
\text { Also, }(x+y)(z+x)=(y+z)^{2} \\
\Rightarrow 2 y(2 y-d)=(2 y+d)^{2} \\
\Rightarrow d(d+6 y)=0 .
\end{array}
$$
Since $d \neq 0$, we have $d=-6 y$.
$$
\begin{array}{l}
\text { Also, } 40<x+y+z=... | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. There are three sets of cards in red, yellow, and blue, each set containing five cards, marked with the letters $A, B, C, D, E$. If five cards are drawn from these 15 cards, with the requirement that the letters are all different and all three colors are included, then the number of different ways to draw the cards ... | 4. 150.
Divide into two categories: $3, 1,1$ and $2,2,1$.
Calculate respectively:
$$
\frac{C_{3}^{1} C_{5}^{3} C_{2}^{1} C_{2}^{1} C_{1}^{1}}{A_{2}^{2}}+\frac{C_{3}^{1} C_{5}^{2} C_{2}^{1} C_{3}^{2} C_{1}^{1}}{A_{2}^{2}}=150
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
14. (15 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$.
(1) Find the minimum value of $f(n)$;
(2) When $n=2 \times 10^{k}-1\left(k \in \mathbf{N}_{+}\right)$, find $f(n)$;
(3) Does there exist a positive integer $n$ such that
$$
f(n)=201... | 14. (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, we know that $f(n) \neq 1$.
Assume $f(n)=2$. Then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e.,
$$
3 n^{2}+n+1=10^{k}+1 \text {. }
$$
Clearly, when $k=1$, $n$ does not exist.
Therefore, $k$ is ... | 2012 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $A=\{x \mid x \geqslant 10, x \in \mathbf{N}\}, B \subseteq A$, and the elements in $B$ satisfy:
(i) The digits of any element are all different;
(ii) The sum of any two digits of any element is not equal to 9.
(1) Find the number of two-digit and three-digit numbers in $B$;
(2) Does there exist a five-di... | (1) For two-digit numbers, the digit in the tens place can be $1,2, \cdots, 9$; the digit in the units place, since it cannot be the same as the digit in the tens place and the sum of the two digits cannot be 9, has 8 possible choices for each digit in the tens place.
Therefore, the total number of two-digit numbers th... | 4012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 For a four-digit number, at most two of its digits are different. How many such four-digit numbers are there?
保留源文本的换行和格式，所以翻译结果如下：
Example 2 For a four-digit number, at most two of its digits are different.
Ask: How many such four-digit numbers are there? | Solution: Clearly, there are exactly 9 four-digit numbers where all four digits are the same.
Below, we consider four-digit numbers with exactly two different digits in three steps.
(1) First, consider the thousands place, which has 9 possible choices: $1,2, \cdots, 9$.
(2) Next, consider the hundreds, tens, and units... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $A \cup B \cup C=\{1,2, \cdots, 6\}$, and $A \cap B=\{1,2\},\{1,2,3,4\} \subseteq B \cup C$.
Then the number of $(A, B, C)$ that satisfy the conditions is $\qquad$ groups (different orders of $A, B, C$ are considered different groups). | As shown in Figure 1, for $1$ and $2$, they can belong to regions I and II, which gives $2^{2}$ possibilities; for $3$ and $4$, they can belong to $B \cup C$ except for regions I and II, i.e., regions III, IV, VI, and VII, which gives $4^{2}$ possibilities; for $5$ and $6$, they can belong to $A \cup B$ except for regi... | 1600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If 6 college graduates apply to three employers, and each employer hires at least one of them, then the number of different hiring scenarios is $\qquad$ kinds. | The number of ways to hire 3 people is $\mathrm{A}_{6}^{3}=120$; the number of ways to hire 4 people is $\frac{\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{1} \mathrm{C}_{3}^{1}}{2!} \times \mathrm{A}_{3}^{3}=540$; the number of ways to hire 5 people is
$$
\frac{C_{6}^{2} C_{4}^{2} C_{2}^{1}}{2!} \times A_{3}^{3}+\frac{C_{6}^{3}... | 2100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 In a $6 \times 6$ grid, three identical red cars and three identical black cars are parked, with one car in each row and each column, and each car occupies one cell. The number of ways to park the cars is ( ).
(A) 720
(B) 20
(C) 518400
(D) 14400 | Assume first that the three red cars are distinct and the three black cars are also distinct. The first car can obviously be placed in any of the 36 squares, giving 36 ways. The second car, which cannot be in the same row or column as the first car, has 25 ways to be placed.
Similarly, the third, fourth, fifth, and si... | 14400 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Example 7: There are 4 red cards, 3 blue cards, 2 yellow cards, and 1 white card. Cards of the same color are indistinguishable. Questions:
(1) How many ways are there to arrange these 10 cards in a row from left to right?
(2) How many ways are there to arrange the cards so that the first 3 cards from the left are of t... | (1) $\frac{10!}{4!\times 3!\times 2!\times 1!}=12600$ ways.
(2) For the left 3 cards being red, there are
$$
\frac{7!}{1!\times 3!\times 2!\times 1!}=420 \text { (ways); }
$$
For the left 3 cards being blue, there are
$$
\frac{7!}{4!\times 2!\times 1!}=105 \text { (ways). }
$$
Thus, there are $420+105=525$ ways in to... | 525 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If numbers $a_{1}, a_{2}, a_{3}$ are taken in increasing order from the set $1, 2, \cdots, 14$, such that the following conditions are satisfied:
$$
a_{2}-a_{1} \geqslant 3 \text { and } a_{3}-a_{2} \geqslant 3 \text {. }
$$
Then the number of different ways to choose such numbers is $\qquad$ kinds. | Let $a_{1}=x_{1}, a_{2}-a_{1}=x_{2}$,
$$
a_{3}-a_{2}=x_{3}, 14-a_{3}=x_{4} \text {. }
$$
Then $x_{1}+x_{2}+x_{3}+x_{4}=14$.
Thus, the problem is transformed into finding the number of integer solutions to the equation under the conditions
$$
x_{1} \geqslant 1, x_{2} \geqslant 3, x_{3} \geqslant 3, x_{4} \geqslant 0
$$... | 120 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $A$ and $B$ be two sets, and $(A, B)$ is called a "pair". When $A \neq B$, $(A, B)$ and $(B, A)$ are considered different pairs. Then the number of different pairs satisfying the condition $A \cup B=\{1,2,3,4\}$ is $\qquad$ | Prompt: Following Example 3, we know there are $3^{4}=81$ pairs.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Prompt: Following Example 3, we know there are $3^{4}=81$ pairs. | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5.2011 is a four-digit number whose sum of digits is 4. Then the total number of four-digit numbers whose sum of digits is 4 is $\qquad$. | In fact, the number of four-digit numbers $\overline{x_{1} x_{2} x_{3} x_{4}}$ is the number of integer solutions to the indeterminate equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=4
$$
satisfying the conditions $x_{1} \geqslant 1, x_{2}, x_{3}, x_{4} \geqslant 0$. It is easy to see that there are $\mathrm{C}_{6}^{3}=20$ such s... | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. There are 8 English letters $K, Z, A, I, G, A, K$, and $\mathrm{U}$, each written on 8 cards. Ask:
(1) How many ways are there to arrange these cards in a row?
(2) How many ways are there to arrange 7 of these cards in a row? | (1) $\frac{8!}{2!\times 2!}=10080$ ways.
(2) If the letter taken away is $K$ or $A$, then there are
$$
2 \times \frac{7!}{2!}=5040 \text { (ways); }
$$
If the letter taken away is $Z$, $I$, $G$, or $U$, then there are
$$
4 \times \frac{7!}{2!\times 2!}=5040 \text { (ways). }
$$
Therefore, there are $5040+5040=10080$ ... | 10080 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $n$ be a positive integer, and call $n$ a "good number" if the number of prime numbers not exceeding $n$ equals the number of composite numbers not exceeding $n$. Then the sum of all good numbers is ( ).
(A) 33
(B) 34
(C) 2013
(D) 2014 | 6. B.
Since 1 is neither a prime number nor a composite number, a good number must be an odd number.
Let the number of prime numbers not exceeding $n$ be $a_{n}$, and the number of composite numbers be $b_{n}$. When $n \leqslant 15$, only consider the case where $n$ is odd (as shown in Table 1).
Table 1
\begin{tabula... | 34 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Three. (25 points) Given that $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$
If positive integers $a$, $b$, and $m$ satisfy the equation
$$
(a t+m)(b t+m)=31 m
$$
find the value of $a b$. | Since $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$
$t$ is an irrational number, and $t^{2}=1-t$.
From the problem, we have
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=31 m \\
\Rightarrow a b(1-t)+m(a+b) t+m^{2}=31 m \\
\Rightarrow[m(a+b)-a b] t+\left(a b+m^{2}-31 m\right)=0 .
\end{array}
$$
Since $a, b,... | 150 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Given $t=\sqrt{2}-1$. If positive integers $a$, $b$, and $m$ satisfy
$$
(a t+m)(b t+m)=17 m
$$
find the value of $a b$. | Given that $t=\sqrt{2}-1$, we have
$$
t^{2}=3-2 \sqrt{2} \text {. }
$$
From the problem, we know
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=17 m \\
\Rightarrow a b(3-2 \sqrt{2})+m(a+b)(\sqrt{2}-1)+m^{2}=17 m \\
\Rightarrow \sqrt{2}[m(a+b)-2 a b]+ \\
\quad\left[3 a b-m(a+b)+m^{2}-17 m\right]=0 .
\end{array}
$$
Since... | 72 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the function $f(x)=\frac{3+x}{1+x}$. Let
$$
\begin{array}{l}
f(1)+f(2)+f(4)+\cdots+f(1024)=m, \\
f\left(\frac{1}{2}\right)+f\left(\frac{1}{4}\right)+\cdots+f\left(\frac{1}{1024}\right)=n .
\end{array}
$$
Then $m+n=$ . $\qquad$ | Ni, 7.42.
From $f(x)=1+\frac{2}{1+x}$, we know $f\left(\frac{1}{x}\right)=1+\frac{2 x}{1+x}$.
Therefore, $f(x)+f\left(\frac{1}{x}\right)=4$.
Also, $f(1)=2$, so, $m+n=4 \times 10+2=42$. | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $n$ be a positive integer less than 100, and satisfies $\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n$ is an integer. Then the sum of all positive integers $n$ that meet the condition is $\qquad$ | 11. 635 .
Notice that,
$$
\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n=\frac{5 n^{2}+3 n-5}{15}
$$
is an integer, so, $15 \mid \left(5 n^{2}+3 n-5\right)$.
Thus, $5 \mid n$, and $3 \mid \left(n^{2}-1\right)$.
Therefore, $n=15 k+5$ or $15 k+10$.
Hence, the sum of all positive integers $n$ that satisfy the condition i... | 635 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Suppose $N$ consecutive positive integers satisfy the following conditions: the sum of the digits of the 1st number is divisible by 1, the sum of the digits of the 2nd number is divisible by 2, $\cdots$. The sum of the digits of the $N$th number is divisible by $N$. Find the maximum possible value of $N$.
| Let the $N$ numbers be $a_{1}, a_{2}, \cdots, a_{N}$.
If $N \geqslant 22$, then among $a_{2}, a_{3}, \cdots, a_{21}$, these 20 numbers, at least two numbers have a units digit of 9, and among these two numbers, at least one has a tens digit that is not 9, let this number be $a_{i}$.
Thus, $i \leqslant 21$.
If $i$ is ev... | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Let $f(x)=x+\frac{1}{x}(x>0)$. If for any positive number $a$, there exist $m+1$ real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in the interval $\left[1, a+\frac{2013}{a}\right]$, such that the inequality
$$
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{m}\right)<f\left(a_{m+1}\right)
$$
holds, ... | II. 9. Let $a=\sqrt{2013}$, then there exist $m+1$ real numbers that meet the requirements in the interval $[1,2 \sqrt{2013}]$.
Notice that $[1,2 \sqrt{2013}] \subseteq\left[1, a+\frac{2013}{a}\right]$.
Therefore, we only need to consider the existence of real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in $[1,2 \sqrt{2013... | 44 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) In an annual super football round-robin tournament, 2013 teams each play one match against every other team. Each match awards 3 points to the winner, 0 points to the loser, and 1 point to each team in the event of a draw. After the tournament, Jia told Yi the total points of his team, and Yi immediatel... | 10. Consider the case with $n$ teams.
Let the team where Jia is located win $x$ games, draw $y$ games, and lose $z$ games. Then the total score of the team is
$$
S=3 x+y \quad (x+y+z=n-1, x, y, z \geqslant 0).
$$
Consider the region $\Omega:\left\{\begin{array}{l}x+y \leqslant n-1, \\ x \geqslant 0, \\ y \geqslant 0 ... | 6034 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. A set of 4027 points in the plane is called a "Colombian point set", where no three points are collinear, and 2013 points are red, 2014 points are blue. Draw a set of lines in the plane, which can divide the plane into several regions. If a set of lines for a Colombian point set satisfies the following two condition... | 2. $k=2013$.
Solution 1 First, give an example to show that $k \geqslant 2013$.
Mark 2013 red points and 2013 blue points alternately on a circle, and color another point in the plane blue. This circle is divided into 4026 arcs, each with endpoints of different colors. If the requirement of the problem is to be met, ea... | 2013 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a_{1}, a_{2}, \cdots, a_{100}$ are 100 distinct positive integers. For any positive integer $i \in\{1,2, \cdots, 100\}, d_{i}$ represents the greatest common divisor of the 99 numbers $a_{j}(j \neq i)$, and $b_{i}=a_{i}+$ $d_{i}$. Question: How many different positive integers are there at least in $b_{1}, b_... | 3. Contains at least 99 different positive integers.
If we let $a_{100}=1, a_{i}=2 i(1 \leqslant i \leqslant 99)$; then $b_{1}=b_{100}=3$.
This indicates that there are at most 99 different positive integers in $b_{i}$.
Next, we prove: there are at least 99 different positive integers in $b_{i}$.
Without loss of gene... | 99 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $A$ be a set of ten real-coefficient quadratic polynomials. It is known that there exist $k$ consecutive positive integers $n+1$, $n+2, \cdots, n+k$, and $f_{i}(x) \in A(1 \leqslant i \leqslant k)$, such that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence. Find the maximum possible valu... | 6. Given that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence, we know there exist real numbers $a$ and $b$ such that
$$
f_{i}(n+i)=a i+b .
$$
Notice that for any quadratic polynomial $f$, the equation
$$
f(n+x)=a x+b
$$
has at most two real roots. Therefore, each polynomial in $A$ appears at... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. Alex has 75 red cards and 75 blue cards. It is known that Alex can exchange 2 red cards for 1 silver card and 1 blue card at one stall, and can exchange 3 blue cards for 1 silver card and 1 red card at another stall. If he continues to exchange according to the above methods until he can no longer exchange any card... | 17. E.
If Alex has $a$ red cards, $b$ blue cards, and $c$ silver cards, denoted as $(a, b, c)$, then
$$
\begin{array}{l}
(75,75,0) \rightarrow(1,112,37) \\
\rightarrow(38,1,74) \rightarrow(0,20,93) \\
\rightarrow(6,2,99) \rightarrow(0,5,102) \\
\rightarrow(1,2,103) .
\end{array}
$$
Therefore, Alex has 103 silver card... | 103 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
23. In $\triangle A B C$, it is known that $A B=13, B C=14$, $C A=15$, points $D, E, F$ are on sides $B C, C A, D E$ respectively, satisfying $A D \perp B C, D E \perp A C, A F \perp B F$, the length of segment $D F$ is a reduced fraction $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$. Then $m+n=(\quad)$.
(... | 23. B.
As shown in Figure 5.
Let $p$ be the semi-perimeter of $\triangle ABC$. Then
$$
p=\frac{a+b+c}{2}=\frac{13+14+15}{2}=21.
$$
Thus, $S_{\triangle ABC}=\sqrt{p(p-a)(p-b)(p-c)}$
$$
\begin{array}{l}
=\sqrt{21 \times 8 \times 7 \times 6}=84 \\
\Rightarrow AD=\frac{2 S_{\triangle ABC}}{BC}=\frac{2 \times 84}{14}=12.
... | 21 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $x, y, z \in \mathbf{R}_{+}$,
$$
\begin{array}{l}
S=\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}, \\
T=\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1} .
\end{array}
$$
Then the minimum value of $S^{2}-T^{2}$ is | 4. 36 .
$$
\begin{array}{l}
S^{2}-T^{2}=(S+T)(S-T) \\
=(\sqrt{x+2}+\sqrt{x+1}+\sqrt{y+5}+ \\
\quad \sqrt{y+1}+\sqrt{z+10}+\sqrt{z+1}) \cdot \\
\quad\left(\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{4}{\sqrt{y+5}+\sqrt{y+1}}+\frac{9}{\sqrt{z+10}+\sqrt{z+1}}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$
The equality hold... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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