problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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7. Given integers $p$ and $q$ satisfy $p+q=2010$, and the quadratic equation $67 x^{2}+p x+q=0$ has two positive integer roots. Then $p=$ $\qquad$ . | 7. -2278 .
Let the two positive integer roots of the equation be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$.
Then $x_{1}+x_{2}=-\frac{p}{67}, x_{1} x_{2}=\frac{q}{67}$.
Thus, $x_{1} x_{2}-x_{1}-x_{2}=\frac{p+q}{67}=\frac{2010}{67}=30$
$$
\begin{array}{l}
\Rightarrow\left(x_{1}-1\right)\left(x_{2}-1\right)=31 \\
... | -2278 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Place the nine digits $1,2, \cdots, 9$ into the nine small squares in Figure 4, so that the seven three-digit numbers $\overline{a b c} 、 \overline{d e f} 、 \overline{g h i} 、 \overline{a d g} 、 \overline{b e h} 、 \overline{c f i}$ and $\overline{a e i}$ are all divisible by 11. Find the maximum value of the three-... | 12. According to the problem, for modulo 11 we have
$$
\begin{array}{l}
a+c \equiv b, d+f \equiv e, g+i \equiv h, a+g \equiv d, \\
b+h \equiv e, c+i \equiv f, a+i \equiv e . \\
\text { Then }(a+c)+(d+f)+(g+i)+(b+h)+e \\
\equiv b+h+3 e \equiv 4 e(\bmod 11) .
\end{array}
$$
The left side of the above equation is
$$
1+2+... | 734 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. A natural number is called a "good number" if it is exactly 2007 more than the sum of its digits. Then the sum of all good numbers is $\qquad$ . | 5.20145.
Let $f(n)=n-S(n)$ ($S(n)$ is the sum of the digits of the natural number $n$). Then the function $f(n)$ is a non-strictly increasing function, and
$$
\begin{array}{l}
f(2009)<f(2010) \\
=f(2011)=\cdots=f(2019)=2007 \\
<f(2020) .
\end{array}
$$
Therefore, there are only 10 natural numbers that satisfy the con... | 20145 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the sum of 12 distinct positive integers is 2010. Then the maximum value of the greatest common divisor of these positive integers is . $\qquad$ | $-1.15$.
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d$, where $\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then $2010=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$S \geqsla... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n}+a_{n+1}=-n^{2} \text {. }
$$
then $a_{15}=$ $\qquad$ | 7. -104 .
Rewrite $a_{n}+a_{n+1}=-n^{2}$ as
$$
\left(a_{n}+\frac{n^{2}}{2}-\frac{n}{2}\right)+\left[a_{n+1}+\frac{(n+1)^{2}}{2}-\frac{n+1}{2}\right]=0 \text {. }
$$
Let $b_{n}=a_{n}+\frac{n^{2}}{2}-\frac{n}{2}$. Then
$$
b_{1}=1 \text {, and } b_{n+1}=-b_{n} \text {. }
$$
Therefore, $b_{2 k-1}=1, b_{2 k}=-1(k=1,2, \c... | -104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. If a four-digit number $n$ contains at most two different digits among its four digits, then $n$ is called a "simple four-digit number" (such as 5555 and 3313). Then, the number of simple four-digit numbers is | 8. 576.
If the four digits of a four-digit number are all the same, then there are nine such four-digit numbers.
If the four digits of a four-digit number have two different values, the first digit \( a \in \{1,2, \cdots, 9\} \) has 9 possible choices. After choosing \( a \), select \( b \in \{0,1, \cdots, 9\} \) and... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. The capacity of a set refers to the sum of its elements. Then the total capacity of all non-empty sets $A$ that satisfy the condition “ $A \subseteq\{1,2, \cdots, 7\}$, and if $a \in A$ then $8-a \in A$ ” is
(Answer with a specific number).
| 12.224.
First, find the single-element and two-element sets that satisfy the conditions:
$$
A_{1}=\{4\}, A_{2}=\{1,7\}, A_{3}=\{2,6\}, A_{4}=\{3,5\} .
$$
Then, any combination of elements from these four sets also meets the requirements.
Therefore, the total sum of elements in all sets $A$ that satisfy the condition... | 224 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3: There is an $8 \times 8$ chessboard, and at the start, each of the 64 small squares contains a "castle" chess piece. If a castle chess piece can attack an odd number of other castle chess pieces still on the board, it is removed. Question: What is the maximum number of castle chess pieces that can be removed (a cast... | First, prove that the pieces on the four corners of the chessboard will not be taken away. Since a castle on a corner can attack at most two other castles, if this castle is taken away, it means that the castle can only attack one. Without loss of generality, we can assume that the castle in the upper left corner is th... | 59 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Take the subset $A_{i}=\left\{a_{i}, a_{i+1}, \cdots, a_{i+59}\right\}(i=1,2, \cdots$, 70 ) of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}$, where $a_{70+i}=a_{i}$. If there exist $k$ sets among $A_{1}, A_{2}, \cdots, A_{70}$ such that the intersection of any seven of them is non-empty, fin... | Given:
$$
A_{1} \cap A_{2} \cap \cdots \cap A_{60}=\left\{a_{60}\right\} \text {, }
$$
Furthermore, the intersection of any seven sets from $A_{1}, A_{2}, \cdots, A_{\infty}$ is non-empty, hence $k \geqslant 60$.
We will now prove that if $k>60$, it cannot be guaranteed that the intersection of any seven sets from th... | 60 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
410 translators are invited to an international mathematics conference. Each translator is proficient in exactly two of the five languages: Greek, Slovenian, Vietnamese, Spanish, and German, and no two translators are proficient in the same pair of languages. The translators are to be assigned to five rooms, with two t... | Construct a graph $G$, where points $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ represent five languages, and any two points (such as $x_{1}, x_{2}$) are connected by an edge $(x_{1} x_{2})$ representing a translator proficient in these two languages.
By the problem statement, graph $G$ is a simple complete graph.
Next, orient... | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. A. Let $a=\sqrt{7}-1$. Then the value of the algebraic expression $3 a^{3}+12 a^{2}-$ $6 a-12$ is ( ).
(A) 24
(B) 25
(C) $4 \sqrt{7}+10$
(D) $4 \sqrt{7}+12$ | ,- 1. A. A.
Notice
$$
a=\sqrt{7}-1 \Rightarrow a^{2}+2 a-6=0 \text {. }
$$
Then $3 a^{3}+12 a^{2}-6 a-12$
$$
=\left(a^{2}+2 a-6\right)(3 a+6)+24=24 .
$$ | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. B. Given that the lengths of the two legs are integers $a$ and $b$ $(b<2011)$. Then the number of right triangles with the hypotenuse length $b+1$ is | 6. B. 31.
By the Pythagorean theorem, we have
$$
a^{2}=(b+1)^{2}-b^{2}=2 b+1 \text{. }
$$
Given $b<2011$, we know that $a$ is an odd number in the interval $(1, \sqrt{4023})$, so $a$ must be $3, 5, \cdots, 63$.
Therefore, there are 31 right-angled triangles that satisfy the conditions. | 31 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. A. As shown in Figure 3, in the right triangle $\triangle ABC$, the hypotenuse $AB$ is 35 units long, and the square $CDEF$ is inscribed in $\triangle ABC$ with a side length of 12. Then the perimeter of $\triangle ABC$ is $\qquad$ | 10. A. 84.
Let $BC = a, AC = b$. Then,
$$
a^{2} + b^{2} = 35^{2} = 1225.
$$
Since Rt $\triangle AFE \sim \text{Rt} \triangle ACB$, we have,
$$
\frac{FE}{CB} = \frac{AF}{AC} \Rightarrow \frac{12}{a} = \frac{b-12}{b}.
$$
Thus, $12(a + b) = ab$.
From equations (1) and (2), we get
$$
\begin{array}{l}
(a + b)^{2} = a^{2}... | 84 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. B. If five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are randomly selected from $1,2, \cdots, n$, and one of these integers is always a prime number, find the maximum value of $n$. | 13. B. When $n \geqslant 49$, take the integers $1, 2^{2}, 3^{2}, 5^{2}, 7^{2}$. These five integers are five pairwise coprime distinct integers, but none of them are prime.
When $n=48$, in the integers $1, 2, \cdots, 48$, take any five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. If $a_{1},... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
12. (16 points) Given a positive integer $n$ that satisfies the following condition: for each positive integer $m$ in the open interval $(0,2009)$, there always exists a positive integer $k$, such that
$$
\frac{m}{2009}<\frac{k}{n}<\frac{m+1}{2010} \text {. }
$$
Find the minimum value of such $n$. | 12. Notice
$$
\begin{array}{l}
\frac{m}{2009}2010 k
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
m n+1 \leqslant 2009 k, \\
m n+n-1 \geqslant 2010 k
\end{array}\right. \\
\Rightarrow 2009(m n+n-1) \geqslant 2009 \times 2010 k \\
\geqslant 2010(m n+1) \\
\Rightarrow 2009 m n+2009 n-2009 \\
\geqslant 2010 m ... | 4019 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $k_{1}<k_{2}<\cdots<k_{n}$ be non-negative integers, satisfying $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=227$.
Then $k_{1}+k_{2}+\cdots+k_{n}=$ $\qquad$ | - 1. 19.
Notice that
$$
\begin{array}{l}
227=1+2+32+64+128 \\
=2^{0}+2^{1}+2^{5}+2^{6}+2^{7} .
\end{array}
$$
Therefore, $k_{1}+k_{2}+\cdots+k_{n}=0+1+5+6+7=19$. | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the sequence of rational numbers $\left\{a_{n}\right\}$ be defined as follows:
$a_{k}=\frac{x_{k}}{y_{k}}$, where $x_{1}=y_{1}=1$, and
if $y_{k}=1$, then $x_{k+1}=1, y_{k+1}=x_{k}+1$;
if $y_{k} \neq 1$, then $x_{k+1}=x_{k}+1, y_{k+1}=y_{k}-1$.
How many terms in the first 2011 terms of this sequence ... | 11. The sequence is
$$
\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{1}{3}, \frac{2}{2}, \frac{3}{1}, \cdots, \frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}, \cdots \text {. }
$$
Group it as follows:
$$
\begin{array}{l}
\left(\frac{1}{1}\right),\left(\frac{1}{2}, \frac{2}{1}\right),\left(\frac{1}{3}, \frac{2}{2}, \fra... | 213 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) A dance troupe has $n(n \geqslant 5)$ actors, and they have arranged some performances, each of which is performed by four actors on stage. In one performance, they found that: it is possible to appropriately arrange several performances so that every two actors in the troupe perform on stage togethe... | Three, use $n$ points to represent $n$ actors.
If two actors have performed on the same stage once, then connect the corresponding points with an edge. Thus, the condition of this problem is equivalent to:
Being able to partition the complete graph $K_{n}$ of order $n$ into several complete graphs $K_{4}$ of order 4, ... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. From the numbers $1,2, \cdots, 2014$, what is the maximum number of numbers that can be selected such that none of the selected numbers is 19 times another? | According to the problem, if $k$ and $19 k$ cannot both appear in $1,2, \cdots, 2014$, and since $2014=19 \times 106$, and $106=5 \times 19+11$, then choose
$$
1,2,3,4,5,106,107, \cdots, 2013 \text {, }
$$
These 1913 numbers satisfy the requirement.
The numbers not chosen are $6,7, \cdots, 105,2014$, a total of 101 nu... | 1913 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $T \subseteq\{1,2, \cdots, 25\}$. If for any two distinct elements $a, b (a \neq b)$ in $T$, their product $ab$ is not a perfect square, find the maximum number of elements in $T$, and the number of subsets $T$ that satisfy this condition. | Prompt: Divide the set $\{1,2, \cdots, 25\}$ into several subsets such that the product of any two elements in the same subset is a perfect square, and the product of any two elements in different subsets is not a perfect square.
To maximize the number of elements in $T$, let
$$
\begin{array}{l}
A_{1}=\{1,4,9,16,25\}, ... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the smallest positive integer $n$ such that for any $n$ integers, there exist at least two numbers whose sum or difference is divisible by 1991.
(1991, Australian Mathematical Olympiad) | Let $M=\left\{a_{i} \mid a_{i}=0,1, \cdots, 995\right\}$.
Since $a_{i}+a_{j} \leqslant 995+994=1989<1991$, $0<\left|a_{i}-a_{j}\right| \leqslant 995$,
then the sum and difference of any two numbers in $M$ are not multiples of 1991.
Therefore, $n \geqslant 997$.
Let $a_{1}, a_{2}, \cdots, a_{997}$ be any 997 integers.
I... | 997 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the side lengths of a trapezoid are $3,4,5,6$. Then the area of this trapezoid is $\qquad$ . | 3. 18.
First, determine the lengths of the two bases. As shown in Figure 7, let trapezoid $ABCD$ have $AD$ and $BC$ as the upper and lower bases, respectively. Draw $AE \parallel CD$. Then $BE$ is the difference between the upper and lower bases.
In $\triangle ABE$, $AB - AE = AB - CD < BE = BC - AD$. Therefore, $AB ... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. (25 points) For a set $M=\left\{p_{1}, p_{2}, \cdots, p_{2_{n}}\right\}$ consisting of $2 n$ prime numbers, its elements can be paired to form $n$ products, resulting in an $n$-element set. If
$$
\begin{aligned}
A & =\left\{a_{1} a_{2}, a_{3} a_{4}, \cdots, a_{2 n-1} a_{2 n}\right\} \\
\text { and } \quad B & =\lef... | 11. Six elements can form fifteen different "slips of paper," listed as follows:
$$
\begin{array}{l}
\{a b, c d, e f\},\{a b, c e, d f\},\{a b, c f, d e\}, \\
\{a c, b d, e f\},\{a c, b e, d f\},\{a c, b f, d e\}, \\
\{a d, b c, e f\},\{a d, b e, c f\},\{a d, b f, c e\}, \\
\{a e, b c, d f\},\{a e, b d, c f\},\{a e, b ... | 60 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let two fixed points in the plane be $A(-3,0)$ and $B(0,-4)$, and let $P$ be any point on the curve $y=\frac{12}{x}(x>0)$. Draw $PC \perp x$-axis and $PD \perp y$-axis, with the feet of the perpendiculars being $C$ and $D$, respectively. Then the minimum value of $S_{\text{quadrilateral } ACD}$ is | 4. 24.
Notice that
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \\
=2\left(x+\frac{9}{x}\right)+12 \geqslant 24 .
\end{array}
$$
The equality holds if and only if $x=3$. Therefore, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 24. | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 There are three types of goods, A, B, and C. If you buy 3 pieces of A, 7 pieces of B, and 1 piece of C, it costs a total of 315 yuan; if you buy 4 pieces of A, 10 pieces of B, and 1 piece of C, it costs a total of 420 yuan. Question: How much would it cost to buy one piece each of A, B, and C? | Let the unit prices of A, B, and C be $x$, $y$, and $z$ yuan, respectively. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
3 x+7 y+z=315, \\
4 x+10 y+z=420 .
\end{array}\right.
$$
The problem actually only requires finding the value of $x+y+z$, without necessarily solving for $x$, $y$, and $z$ indi... | 105 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For $\triangle A B C$, squares are constructed outward on its three sides $a, b, c$, with their areas denoted as $S_{a}, S_{b}, S_{c}$ respectively. If $a+b+c=18$, then the minimum value of $S_{a}+S_{b}+S_{c}$ is $\qquad$ | 2. 108.
$$
\begin{array}{l}
\text { Given }(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0 \\
\Rightarrow 2\left(a^{2}+b^{2}+c^{2}\right) \geqslant 2(a b+b c+c a) \\
\Rightarrow 3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2} \\
\Rightarrow a^{2}+b^{2}+c^{2} \geqslant \frac{18^{2}}{3}=108 . \\
\text { Therefore, }\left(... | 108 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. If the real numbers $x, y, z, w$ satisfy
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}=1, \\
\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}
$$
then $x^{2... | 3. 36 .
It is known that $2^{2}$ and $4^{2}$ are the two roots of the equation with respect to $t$:
$$
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}=1
$$
which means they are the two roots of the equation:
$$
t^{2}-\left(1^{2}+3^{2}+x^{2}+y^{2}\right) t+1^{2} \times 3^{2}+3^{2} x^{2}+1^{2} \times y^{2}=0
$$
Therefore,... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) In the set of numbers $1,2, \cdots, 2009$, what is the maximum number of numbers that can be selected such that the sum of any two selected numbers is divisible by 100? | Three, let the $n$ numbers that meet the conditions be,
$$
a_{1}, a_{2}, \cdots, a_{n} \text {, }
$$
Take any three of these numbers (let them be $a_{k} \backslash a_{m} \backslash a_{f}$). Then
$$
\begin{array}{l}
a_{k}+a_{m}=100 k_{1}, \\
a_{k}+a_{4}=100 k_{2}, \\
a_{m}+a_{4}=100 k_{3},
\end{array}
$$
where $k_{1} ... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) A company printed a batch of T-shirts, each T-shirt can have three different colors: red, yellow, and blue, and four different patterns. Now, this batch of T-shirts is to be distributed to $n$ new employees, with each employee receiving exactly 4 T-shirts with different patterns. Try to find the minim... | Four, the minimum value of $n$ is 19.
When $n=18$, the answer scenario shown in Table 1 does not meet the requirements.
[Note] In Table 1, (1), (2), (3), (4) are patterns, $A_{1}, A_{2}, \cdots, A_{18}$ are members, and $A$, $B$, $C$ represent red, yellow, and blue colors, respectively.
The proof below shows that when ... | 19 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In $\triangle A B C$, it is known that $A B=A C=2$, and there are 100 different points $P_{1}, P_{2}, \cdots, P_{100}$ on side $B C$. Let $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C(i=1,2, \cdots, 100)$.
Find the value of $m_{1}+m_{2}+\cdots+m_{100}$. | Solve As shown in Figure 2, since $\triangle A B C$ is an isosceles triangle, applying the property we get
$$
\begin{aligned}
& A P_{i}^{2} \\
= & A B^{2}-B P_{i} \cdot P_{i} C .
\end{aligned}
$$
Therefore, $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C=A B^{2}=4$.
Thus, $m_{1}+m_{2}+\cdots+m_{100}=400$. | 400 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, \text { and } \\
a_{n+2}=\left|a_{n+1}\right|-a_{n}
\end{array}\left(n \in \mathbf{N}_{+}\right) .
$$
Let $\left\{a_{n}\right\}$'s sum of the first $n$ terms be $S_{n}$. Then $S_{100}=$ | - 1. 89.
From the given, $a_{k+9}=a_{k}$.
Then $S_{100}=a_{1}+11\left(a_{1}+a_{2}+\cdots+a_{9}\right)=89$. | 89 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $n<100$. Then the largest integer $n$ such that the expansion of $(a+b)^{n}$ has three consecutive terms with coefficients in arithmetic progression is $\qquad$ . . . | 3. 98.
Let the coefficients of three consecutive terms in the expansion of $(a+b)^{n}$ be $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1} (1 \leqslant k \leqslant n-1)$.
By the problem, we have $2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1}$.
Expanding and rearranging according to the... | 98 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Among the positive integers less than 20, each time three numbers are taken without repetition, so that their sum is divisible by 3. Then the number of different ways to do this is $\qquad$ . | 4.327.
Divide these 19 numbers into three categories based on the remainder when divided by 3:
$$
\begin{array}{l}
A_{1}: 3,6,9,12,15,18 ; \\
A_{2}: 2,5,8,11,14,17 ; \\
A_{3}: 1,4,7,10,13,16,19 .
\end{array}
$$
Thus, the number of ways to satisfy the conditions of the problem are only four scenarios.
(1) Choose any t... | 327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Place 10 numbers on a given circle such that their total sum is 200, and the sum of any three consecutive numbers is not less than 58. Then the maximum value of the largest number among all sets of 10 numbers that satisfy the above requirements is $\qquad$ | 8. 26 .
Let the maximum number in all placements be $A$. Then
$$
A+3 \times 58 \leqslant 200 \Rightarrow A \leqslant 26 \text {. }
$$
In fact, $26,6,26,26,6,26,26,6,26,26$ satisfies. | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{0,1, \cdots, 6\}(i=0,1,2,3)$, and $a_{3} \neq 0$.
If positive integers $m 、 n \in A$, and $m+n=2010(m>n)$, then the number of positive integers $m$ that satisfy the condition is $\q... | 6. 662 .
According to the problem, we know that $m$ and $n$ are four-digit numbers in base 7, and the largest four-digit number in base 7 is
$$
6 \times 7^{3}+6 \times 7^{2}+6 \times 7+6=2400,
$$
the smallest one is $1 \times 7^{3}=343$.
Since $m+n=2010(m>n)$, therefore,
$$
1006 \leqslant m \leqslant 1667 \text {. }
... | 662 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange the real solutions of the equation $x^{3}-3[x]=4$ in ascending order to get $x_{1}, x_{2}, \cdots, x_{k}$. Then the value of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}$ is $\qquad$ ( $[x]$ denotes the greatest integer less than or equal to the real number $x$). | 8. 15 .
Notice that $x-1<[x] \leqslant x$.
Therefore, when $x \geqslant 3$,
$$
\begin{array}{l}
x^{3}-3[x] \geqslant x^{3}-3 x=x\left(x^{2}-3\right) \\
\geqslant 3 \times 6=18 ;
\end{array}
$$
When $x \leqslant-3$,
$$
\begin{array}{l}
x^{3}-3[x]<x^{3}-3(x-1)=x\left(x^{2}-3\right)+3 \\
\leqslant-3 \times 6+3=-15 .
\end... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. There are three numbers arranged in sequence: $3, 9, 8$. For any two adjacent numbers, the difference between the right number and the left number is written between these two numbers, resulting in a new sequence $3, 6, 9, -1, 8$, which is called the first operation; after the second similar operation, a new sequenc... | For convenience, let the sequence of $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. According to the problem, the newly added numbers are $a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{n}-a_{n-1}$. Therefore, the sum of the newly added numbers is
$$
\begin{array}{l}
\left(a_{2}-a_{1}\right) + \left(a_{3}-a_{2}\right) + \cdots + ... | 520 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
19. There are 10 red, 10 black, and 10 white balls. Now, all of them are to be placed into two bags, A and B, with the requirement that each bag must contain balls of all three colors, and the product of the number of balls of each color in bags A and B must be equal. How many ways are there to do this? | 19. Let the number of red, black, and white balls in bag A be $x$, $y$, and $z$ respectively. Then $1 \leqslant x, y, z \leqslant 9$, and
$$
x y z=(10-x)(10-y)(10-z) \text {, }
$$
i.e., $x y z=500-50(x+y+z)+5(x y+y z+z x)$.
Thus, $5 \mid x y z$.
Therefore, one of $x, y, z$ must be 5.
Assume $x=5$, substituting into eq... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. It is known that fresh shiitake mushrooms contain $90 \% \sim 99 \%$ water, while dried shiitake mushrooms contain $30 \% \sim 45 \%$ water. Then, under the influence of drying, by what maximum factor can the weight of fresh shiitake mushrooms be reduced? | 2. Let the weights of fresh mushrooms, baked mushrooms be $m_{1} \mathrm{~g}, m_{2} \mathrm{~g}$, and the weight of dried mushrooms be $x \mathrm{~g}$ (unknown). Then the range of the proportion of dried mushrooms in fresh mushrooms and baked mushrooms is
$$
\begin{array}{l}
0.01=1-0.99 \\
\leqslant \frac{x}{m_{1}} \le... | 70 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given two circles $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent at point $A$, circle $\Gamma$ is externally tangent to $\Gamma_{1}$ and $\Gamma_{2}$ at points $B$ and $C$ respectively. Extend the chord $B A$ of circle $\Gamma_{1}$ to intersect circle $\Gamma_{2}$ at point $D$, extend the chord $C A$ of circle... | 5. First, prove that quadrilateral $B C G F$ is a rectangle:
In fact, let $K L, B M, C N$ be the common tangents of circles $\Gamma_{1}$ and $\Gamma_{2}$, $\Gamma_{1}$ and $\Gamma$, $\Gamma_{2}$ and $\Gamma$ (Figure 2). According to the inscribed angle and the angle between a tangent and a chord, we have
$$
\angle A B... | 13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Let $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ be a function, and for any positive integers $m, n$, we have
$$
f(f(m)+f(n))=m+n .
$$
Find the value of $f(2011)$. | Since $f(x)$ is a function from the set of positive integers to the set of positive integers, let
$$
f(1)=p\left(p \in \mathbf{N}_{+}\right) \text {. }
$$
If $p>1$, then $p \geqslant 2$.
$$
\text { Let } p=1+b\left(b \in \mathbf{N}_{+}\right), f(b)=c\left(c \in \mathbf{N}_{+}\right) \text {. }
$$
On one hand,
$$
\beg... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 2, given that $A B$ is the diameter of $\odot O$, chord $C D$ intersects $A B$ at point $E$, a tangent line through $A$ intersects the extension of $C D$ at point $F$, and $D$ is the midpoint of $E F$. If $D E=\frac{3}{4} C E, A C=$ $8 \sqrt{5}$, then $A B=$ . $\qquad$ | 4.24.
Let $C E=4 x, A E=y$. Then
$$
D F=D E=3 x, E F=6 x \text {. }
$$
Connect $A D, B C$.
Since $A B$ is the diameter of $\odot O$ and $A F$ is the tangent of $\odot O$, we have
$$
\angle E A F=90^{\circ}, \angle A C D=\angle D A F \text {. }
$$
Since $D$ is the midpoint of the hypotenuse $E F$ of the right triangl... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. If the constant term in the expansion of $\left(a \sqrt{x}-\frac{1}{\sqrt{x}}\right)^{6}$ is -160, then $\int\left(3 x^{2}-1\right) \mathrm{d} x=$. $\qquad$ | $$
\begin{array}{l}
T_{r+1}=\mathrm{C}_{6}^{r}(a \sqrt{x})^{6-r}\left(-\frac{1}{\sqrt{x}}\right)^{r} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{\frac{6-r}{2}-\frac{r}{2}} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{3-r} .
\end{array}
$$
Let $3-r=0$. Then $r=3$.
The constant term is
$$
\begin{array}{l}
-\mathrm{C}_{6}^{3}... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $x_{1}$ and $x_{2}$ are the two real roots of the equation
$$
x^{2}-2009 x+2011=0
$$
real numbers $m$ and $n$ satisfy
$$
\begin{array}{l}
2009 m x_{1}+2009 n x_{2}=2009, \\
2010 m x_{1}+2010 n x_{2}=2010 .
\end{array}
$$
Then $2011 m x_{1}+2011 n x_{2}=$ $\qquad$ | 2. -2009 .
From the given, we know
$$
\begin{array}{l}
=2010 \times 2009-2011 \times 2009=-2009 . \\
\end{array}
$$ | -2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the set $M$ is a subset of $\{1,2, \cdots, 2011\}$, and the sum of any four elements in $M$ cannot be divisible by 3. Then $|M|_{\text {max }}=$ $\qquad$ | - 1. 672.
Consider the set $A=\{3,6,9, \cdots, 2010\}$,
$$
\begin{array}{l}
B=\{1,4,7, \cdots, 2011\}, \\
C=\{2,5,8, \cdots, 2009\} .
\end{array}
$$
If $M \cap A \neq \varnothing$, then $|M \cap B|<3,|M \cap C|<3,|M \cap A|<4$. Therefore, $|M|<10$.
Now assume $M \cap A=\varnothing$.
If $|M \cap B| \geqslant 2$, then ... | 672 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. For any positive integer $n$, let $a_{n}$ be the smallest positive integer such that $n \mid a_{n}$!. If $\frac{a_{n}}{n}=\frac{2}{5}$, then $n=$ $\qquad$ . | 4.25.
From $\frac{a_{n}}{n}=\frac{2}{5} \Rightarrow a_{n}=\frac{2 n}{5} \Rightarrow 51 n$.
Let $n=5 k\left(k \in \mathbf{N}_{+}\right)$. If $k>5$, then $5 k \mid k!$.
Thus, $a_{\mathrm{n}} \leqslant k<\frac{2 n}{5}$, a contradiction.
Clearly, when $k=2,3,4$, $a_{n} \neq \frac{2 n}{5}$.
Also, $a_{25}=10=\frac{2}{5} \ti... | 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that on the side $C A$ of $\angle A C B$ there are 2011 points $A_{1}, A_{2}, \cdots, A_{2011}$, and on the side $C B$ there are 2011 points $B_{1}, B_{2}, \cdots, B_{2011}$, satisfying
$$
\begin{array}{l}
A_{1} A_{2}=A_{2} A_{3}=\cdots=A_{2010} A_{2011}, \\
B_{1} B_{2}=B_{2} B_{3}=\cdots=B_{2010} B_{2011} .
\... | 7.2010 .
$$
\begin{array}{l}
\text { Let } \angle A C B=\alpha, C A_{1}=a, C B_{1}=b, \\
A_{i} A_{i+1}=s, B_{i} B_{i+1}=t(i=1,2, \cdots, 2010), \\
S_{\text {trapezoid } A_{i} A_{i+1} B_{i+2} B_{i}}=S_{i}(i=1,2, \cdots, 2009) . \\
\text { Then } S_{i}=\frac{1}{2}\{[a+(i+1) s][b+(i+1) t]- \\
\quad[a+(i-1) s][b+(i-1) t]\}... | 2010 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given positive integers $a, b, c$ satisfy
$$
(a!)(b!)=a!+b!+c! \text {. }
$$
then $c\left(a^{5}+b^{5}+c^{2}\right)+3=$ $\qquad$ | 8.2011 .
$$
\begin{array}{l}
\text { Given }(a!)(b!)=a!+b!+c! \\
\Rightarrow(a!-1)(b!-1)=c!+1 \text {. }
\end{array}
$$
Assume without loss of generality that $a \geqslant b$. Clearly, $c > a$. Then $(a!-1) \mid(c!+1)$.
Also, $c!+1=\frac{c!}{a!}(a!-1)+\frac{c!}{a!}+1$, so $(a!-1) \left\lvert\,\left(\frac{c!}{a!}+1\rig... | 2011 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. At a concert, there are 20 singers who will perform. For each singer, there is a set of other singers (possibly an empty set) such that he wishes to perform later than all the singers in this set. Question: Is there a way to have exactly 2,010 ways to order the singers so that all their wishes are satisfied? | 1. Such examples exist.
A ranking of singers that satisfies everyone's wishes is called "good".
If for a set of wishes of $k$ singers there exist $N$ good rankings, then $N$ is called "achievable by $k$ singers" (or simply "$k$-achievable").
Next, we prove that 2010 is "20-achievable".
First, we prove a lemma.
Lemma ... | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. (50 points) Given that $p$ is a prime number, the fractional part of $\sqrt{p}$ is $x$, and the fractional part of $\frac{1}{x}$ is $\frac{\sqrt{p}-31}{75}$. Find all prime numbers $p$ that satisfy the condition. | 3. Let $p=k^{2}+r$, where $k, r$ are integers, and satisfy $0 \leqslant r \leqslant 2 k$.
Since $\frac{\sqrt{p}-31}{75}$ is the fractional part of $\frac{1}{x}$, we have $0 \leqslant \frac{\sqrt{p}-31}{75} < 1$. Let
$\frac{1}{x}=\frac{1}{\sqrt{p}-k}=N+\frac{\sqrt{p}-31}{75}(N \geqslant 1)$.
Then $\frac{\sqrt{p}+k}{r}=N... | 2011 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. (50 points) A scientist stored the design blueprint of a time machine on a computer, setting the file opening password as a permutation of $\{1,2, \cdots, 64\}$. They also designed a program that, when eight positive integers between 1 and 64 are input each time, the computer will indicate the order (from left to ri... | 8. Prepare $n^{2}(n=8)$ cards, the front side of which are numbered $1,2, \cdots, n^{2}$, and the back side corresponds to the position of the number in the password (counting from the left). Of course, the operator does not know the numbers on the back side in advance.
First, divide the $n^{2}$ cards into $n$ groups ... | 45 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. In the array of numbers shown in Figure 1, the three numbers in each row form an arithmetic sequence, and the three numbers in each column also form an arithmetic sequence. If $a_{22}=2$, then the sum of all nine numbers is equal to
保留源文本的换行和格式，直接输出翻译结果如下：
5. In the array of numbers shown in Figure 1, the three nu... | 5.18.
From the problem, we have
$$
\begin{array}{l}
a_{11}+a_{13}=2 a_{12}, a_{21}+a_{23}=2 a_{22}, \\
a_{31}+a_{33}=2 a_{32}, a_{12}+a_{32}=2 a_{22} .
\end{array}
$$
Thus, the sum of all nine numbers is $9 a_{n}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $p$ and $q$ are both prime numbers, and $7p+q$, $2q+11$ are also prime numbers. Then $p^{q}+q^{p}=$ $\qquad$ . | 8. 17 .
Since $7 p+q$ is a prime number, and $7 p+q>2$, $7 p+q$ must be an odd number.
Therefore, one of $p$ or $q$ must be even (which can only be 2).
Clearly, $q \neq 2$ (otherwise, $2 q+11=15$, which is not a prime number). Thus, $p=2$.
At this point, $14+q$ and $2 q+11$ are both prime numbers.
If $q=3 k+1\left(k \... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 As shown in Figure 7, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3$, the distance from point $P$ to line $A N$ is $P D=12, A D=$
30, and a line is drawn through $P$ intersecting
$A N$ and $A M$ at points
$B$ and $C$ respectively. Find the minimum
value of the area of... | Solve As shown in Figure 7, it can be proven: when the line moves to $B_{0} P=P C_{0}$, the area of $\triangle A B C$ is minimized.
Draw $C_{0} Q / / A N$, intersecting $C B$ at point $Q$.
Thus, $\triangle P C_{0} Q \cong \triangle P B_{0} B$.
Also, $S_{\triangle P C_{0} C} \geqslant S_{\triangle P C_{0} Q}=S_{\triangl... | 624 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. From the set $\{1,2, \cdots, 10\}$, any two non-adjacent numbers are taken and multiplied. Then the sum of all such products is equal to
| 5. 990 .
Take any two numbers, multiply them, and then find their sum:
$$
\begin{array}{l}
S_{1}=\frac{1}{2} \sum_{k=1}^{10} k\left(\sum_{i=1}^{10} i-k\right) \\
=\frac{1}{2} \sum_{k=1}^{10} k(55-k)=\frac{1}{2} \sum_{k=1}^{10}\left(55 k-k^{2}\right) \\
=\frac{1}{2}(55 \times 55-385)=1320,
\end{array}
$$
Among them, t... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given positive integers $a_{1}, a_{2}, \cdots, a_{18}$ satisfying
$$
\begin{array}{l}
a_{1}<a_{2}<\cdots<a_{18}, \\
a_{1}+a_{2}+\cdots+a_{18}=2011 .
\end{array}
$$
Then the maximum value of $a_{9}$ is | 7.193.
To maximize $a_{9}$, $a_{1}, a_{2}, \cdots, a_{8}$ should be as small as possible, and $a_{10}, a_{11}, \cdots, a_{18}$ should be as close to $a_{9}$ as possible. Therefore, we take $a_{1}, a_{2}, \cdots, a_{8}$ to be $1, 2, \cdots, 8$, respectively, with their sum being 36.
Let $a_{9}=n$. Then
$$
\begin{array}... | 193 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Define the sequence $\left\{a_{n}\right\}: a_{n}=n^{3}+4\left(n \in \mathbf{N}_{+}\right)$, let $d_{n}=\left(a_{n}, a_{n+1}\right)$. Then the maximum value of $d_{n}$ is $\qquad$ | 8.433.
Given $d_{n} \mid\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(n^{3}+4,3 n^{2}+3 n+1\right)$.
Then $d_{n} \mid\left[-3\left(n^{3}+4\right)+n\left(3 n^{2}+3 n+1\right)\right]$,
and $\square$
$$
\begin{array}{l}
d_{n} \mid\left(3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}... | 433 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9.3. There is a convex 2011-gon on the blackboard. Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw? | 9.3.4016.
Use induction to prove: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. We can sequentially draw $2n-6$ diagonals as follows: $A_{2} A_{4}, A_{3} A_{5}$, $A_{4} A_{6}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_... | 4016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let four distinct real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\left(a^{2011}-c^{2011}\right)\left(a^{2011}-d^{2011}\right)=2011, \\
\left(b^{2011}-c^{2011}\right)\left(b^{2011}-d^{2011}\right)=2011 . \\
\text { Then }(a b)^{2011}-(c d)^{2011}=(\quad) .
\end{array}
$$
Then $(a b)^{2011}-(c d)^{2011}=(\quad... | 4. B.
From $a \neq b$, we know $a^{2011} \neq b^{2011}$.
Thus, $a^{2011}$ and $b^{2011}$ are the two distinct real roots of the quadratic equation in $x$:
$$
\left(x-c^{2011}\right)\left(x-d^{2011}\right)=2011,
$$
which is
$$
x^{2}-\left(c^{2011} \div d^{2011}\right) x+(c d)^{2011}-2011=0.
$$
By Vieta's formulas, we g... | -2011 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Figure 2 is a part of the HZ district map. A river runs through the district, with the two banks being the broken lines $A-B-C$ and $D-O-E$, and there are two locations $M$ and $N$. Two bridges perpendicular to the riverbanks and roads are to be built to connect $M$ and $N$ to both banks of the river, making the tot... | 6. B.
Translate point $M(-16,12)$ 3 units to the left to get point $M^{\prime}(-13,12)$, and translate point $N(11,-1)$ 3 units upwards to get point $N^{\prime}(11,2)$. Connect $M^{\prime} N^{\prime}$, which intersects $A B$ and $B C$ at points $P$ and $Q$ respectively.
Then the minimum length of the road is
$$
M^{\pr... | 32 | Geometry | MCQ | Yes | Yes | cn_contest | false |
4. Given a prime number $p$ such that $p^{3}-6 p^{2}+9 p$ has exactly 30 positive divisors. Then the smallest value of $p$ is $\qquad$ . | 4. 23 .
Obviously, when $p=2$ or 3, it does not meet the requirements of the problem.
Therefore, $p>3$.
Also, $p^{3}-6 p^{2}+9 p=p(p-3)^{2}$, at this point,
$(p, p-3)=(p, 3)=1$.
Since $p$ has two factors, $(p-3)^{2}$ has 15 factors.
And $15=5 \times 3$, to make $p$ the smallest, $p-3$ is also even, so it can only be
... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. From the set $\{1,2, \cdots, 2011\}$, any two different numbers $a, b$ are selected such that $a+b=n$ (where $n$ is some positive integer) with a probability of $\frac{1}{2011}$. Then the minimum value of $ab$ is | 6.2010.
Let the number of ways such that $a+b=n$ be $k$. Then
$$
\frac{k}{\mathrm{C}_{2011}^{2}}=\frac{1}{2011} \Rightarrow k=1005 \text {. }
$$
Consider $a b$ to be as small as possible, and the number of ways such that $a+b=n$ is 1005.
Take $n=2011$. Then
$$
1+2010=2+2009=\cdots=1005+1006 \text {. }
$$
At this poi... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.
---
The above text translated into English, preserving the original text's line breaks and format, is as follows:
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, ... | Solve: It is easy to know that $x=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
Let $d=(a, b), a=d a_{0}, b=d b_{0},\left(a_{0}, b_{0}\right)=1$.
Then $x=\frac{d^{2} a_{0}^{2} b_{0}^{2}}{a_{0}^{2}+b_{0}^{2}} \in \mathbf{N}_{+}$.
By $\left(a_{0}^{2}+b_{0}^{2}, a_{0}^{2} b_{0}^{2}\right)=1$, we know $\left(a_{0}^{2}+b_{0}^{2}\right)... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Find the smallest positive integer $n$, such that there exist $n$ distinct positive integers $s_{1}, s_{2}, \cdots, s_{n}$, satisfying
$$
\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \cdots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010} .
$$ | 1. Suppose the positive integer $n$ satisfies the condition, and let $s_{1}=39$.
Thus, $n \geqslant 39$.
Below, we provide an example to show that $n=39$ satisfies the condition.
Take 39 different positive integers:
$$
2,3, \cdots, 33,35,36, \cdots, 40,67,
$$
which satisfy the given equation.
In conclusion, the minimu... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a_{1}, a_{2}, \cdots$ be integers, for any positive integer $n$ we have
$$
a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right] \text {. }
$$
If $2001 a_{199}$, find the smallest positive integer $n(n>1)$, such that $200 \mathrm{l} a_{n}$. | 3. From the given, we have
$$
a_{\mathrm{T99}}=198\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right] \text {. }
$$
From $2001 a_{199}$, we get
$$
100 \left\lvert\, 99\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right]\right. \text {. }
$$
Thus, $a_{2}$ is an even number.
Let $a_{2}=2 m$. Then
$$
1001[(m-1)... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $0<a<b<c<d<500$, and $a+d=b+c$. Also, $bc-ad=93$.
Then the number of ordered quadruples of integers ( $\dot{a}, b, c$,
$d)$ that satisfy the conditions is . $\qquad$ | 7.870.
Since $a+d=b+c$, we set
$$
(a, b, c, d)=(a, a+x, a+y, a+x+y) \text {, }
$$
where $x$ and $y$ are integers, and $0<x<y$.
Then $93=b c-a d$
$$
=(a+x)(a+y)-a(a+x+y)=x y \text {. }
$$
Therefore, $(x, y)=(1,93)$ or $(3,31)$.
First case
$$
(a, b, c, d)=(a, a+1, a+93, a+94) \text {, }
$$
where $a=1,2, \cdots, 405$;... | 870 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Each cell of a $2011 \times 2011$ grid is labeled with an integer from $1,2, \cdots, 2011^{2}$, such that each number is used exactly once. Now, the left and right boundaries, as well as the top and bottom boundaries of the grid, are considered the same, forming a torus (which can be viewed as the surface of a "doug... | 6. Let $N=2011$.
Consider a general $N \times N$ table.
When $N=2$, the conclusion is obvious, and the required $M=2$. An example is shown in Table 1.
Table 1
\begin{tabular}{|l|l|}
\hline 1 & 2 \\
\hline 3 & 4 \\
\hline
\end{tabular}
When $N \geqslant 3$, first prove:
$M \geqslant 2 N-1$.
Starting from a state where... | 4021 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.1. A table of numbers consisting of $n$ rows and 10 columns, with each element being an integer from $0 \sim 9$, satisfies the following condition: for any row $A$ and any two columns $B$ and $C$, there exists another row $D$ such that $D$ differs from $A$ only in the numbers in columns $B$ and $C$. Prove: $n \geqsl... | 10.1. Let $R_{0}$ be the first row. Arbitrarily select $2 m$ columns $C_{1}$, $C_{2}, \cdots, C_{2 m}$. By the given condition, there exists a row $R_{1}$, which differs from $R_{0}$ only in $C_{1}$ and $C_{2}$; further, there exists a row $R_{2}$, which differs from $R_{1}$ only in $C_{3}$ and $C_{4}$; $\cdots \cdots$... | 512 | Combinatorics | proof | Yes | Yes | cn_contest | false |
3. If the length, width, and height of a rectangular prism are all prime numbers, and the sum of the areas of two adjacent sides is 341, then the volume of this rectangular prism $V=$ $\qquad$ . | 3. 638.
Let the length, width, and height of the rectangular prism be $x, y, z$. From the problem, we have
$$
\begin{array}{l}
x(y+z)=341=11 \times 31 \\
\Rightarrow(x, y+z)=(11,31),(31,11) .
\end{array}
$$
Since $y+z$ is odd, one of $y, z$ must be 2 (let's assume $z=2$).
Also, $11-2=9$ is not a prime number, so
$$
(... | 638 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let
$$
\begin{array}{l}
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \\
\text { then } f(1)+f(2)+\cdots+f(50)=
\end{array}
$$ | $$
\begin{array}{l}
x^{2}-53 x+196=(x-4)(x-49) . \\
\text { Therefore, when } 4 \leqslant x \leqslant 49, f(x)=0 . \\
\text { Then } f(1)+f(2)+\cdots+f(50) \\
=f(1)+f(2)+f(3)+f(50)=660 \text {. }
\end{array}
$$ | 660 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Two boxes are filled with black and white balls. The total number of balls in both boxes is 25. Each time, a ball is randomly drawn from each box. The probability that both balls are black is $\frac{27}{50}$, and the probability that both balls are white is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$.... | 5.26.
Let the first box contain $x$ balls, of which $p$ are black, and the second box contain $25-x$ balls, of which $q$ are black. Then
$$
\begin{array}{l}
\frac{p}{x} \cdot \frac{q}{25-x}=\frac{27}{50}, \\
50 p q=27 x(25-x) .
\end{array}
$$
Thus, $x$ is a multiple of 5.
Substituting $x=5$ and $x=10$ into the two e... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. The equation $x+y+z=2011$ satisfies $x<y<z$ to prove: the number of solutions $(x, y, z)$ is $\qquad$ groups. | 5. 336005 .
The equation $x+y+z=2011$ has $\mathrm{C}_{2010}^{2}$ sets of positive integer solutions. In each solution, $x, y, z$ cannot all be equal, and there are $3 \times 1005$ sets of solutions where two of $x, y, z$ are equal. Therefore, the number of solutions that meet the requirements is
$$
\frac{C_{2010}^{2}... | 336005 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given positive integers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a b+b c+c a+2(a+b+c)=8045, \\
a b c-a-b-c=-2 .
\end{array}\right.
$$
then $a+b+c=$ $\qquad$ | $$
-1.2012 .
$$
Note that
$$
\begin{array}{l}
(a+1)(b+1)(c+1) \\
=a b c+a b+b c+c a+a+b+c+1 \\
=8045+(-2)+1=8044 .
\end{array}
$$
Since $a, b, c$ are positive integers, we have
$$
a+1 \geqslant 2, b+1 \geqslant 2, c+1 \geqslant 2 \text{. }
$$
Therefore, 8044 can only be factored as $2 \times 2 \times 2011$. Hence $a... | 2012 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In a class, there are two types of students: one type always lies, and the other type never lies. Each student knows what type the other students are. During a gathering today, each student has to state what type the other students are, and all students together said "liar" 240 times. At a similar gathering yesterda... | 2. 22 .
Consider four possible scenarios:
(1) If student $A$ is a liar and student $B$ is not a liar, then $A$ will say $B$ is a liar;
(2) If student $A$ is a liar and student $B$ is also a liar, then $A$ will say $B$ is not a liar;
(3) If student $A$ is not a liar and student $B$ is also not a liar, then $A$ will say... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
5. The positive integer $n$ has exactly 4 positive divisors (including 1 and $n$). It is known that $n+1$ is four times the sum of the other two divisors. Then $n=$ | 5.95.
Notice that a positive integer with exactly four positive divisors must be of the form $p^{3}$ or $p q$ (where $p$ and $q$ are primes, $p \neq q$).
In the first case, all positive divisors are $1, p, p^{2}, p^{3}$, then $1+p^{3}=4\left(p+p^{2}\right)$, but $p \nmid \left(1+p^{3}\right)$, which is a contradictio... | 95 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Given a positive integer $n$ that satisfies the following conditions:
(1) It is an eight-digit number, and all its digits are 0 or 1;
(2) Its first digit is 1;
(3) The sum of the digits in the even positions equals the sum of the digits in the odd positions.
How many such $n$ are there? | 10.35.
From the fact that the sum of the digits in the even positions equals the sum of the digits in the odd positions, we know that the number of 1s in the even positions equals the number of 1s in the odd positions.
Since the first digit is fixed, there are only three positions in the odd positions that can be fre... | 35 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. Given that 15 rays share a common endpoint. Question: What is the maximum number of obtuse angles (considering the angle between any two rays to be the one not greater than $180^{\circ}$) that these 15 rays can form? | 15. First, it is explained that constructing 75 obtuse angles is achievable.
The position of the rays is represented by their inclination angles, with 15 rays placed near the positions of $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$. Within each group, the five rays are sufficiently close to each other (as shown in F... | 75 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Now arrange for 7 students to participate in 5 sports events, requiring that students A and B cannot participate in the same event, each event must have participants, and each person can only participate in one event. The number of different arrangements that meet the above requirements is $\qquad$ (answer in number... | 5. 15000 .
According to the problem, there are two scenarios that meet the conditions:
(1) One project has 3 participants, with a total of
$$
C_{7}^{3} \times 5! - C_{5}^{1} \times 5! = 3600
$$
schemes;
(2) Two projects each have 2 participants, with a total of
$$
\frac{1}{2}\left(\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2... | 15000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given
$$
a_{n}=\mathrm{C}_{200}^{n}(\sqrt[3]{6})^{200-n}\left(\frac{1}{\sqrt{2}}\right)^{n}(n=1,2, \cdots, 95) \text {. }
$$
The number of integer terms in the sequence $\left\{a_{n}\right\}$ is $\qquad$ | 8. 15 .
Notice that $a_{n}=\mathrm{C}_{200}^{n} \times 3^{\frac{200-n}{3}} \times 2^{\frac{200-5 n}{6}}$.
To make $a_{n}(1 \leqslant n \leqslant 95)$ an integer, it must be that $\frac{200-n}{3}$ and $\frac{400-5 n}{6}$ are both integers, i.e., $61(n+4)$.
When $n=6 k+2(k=0,1, \cdots, 13)$, $\frac{200-n}{3}, \frac{400... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Let $A$ be a $3 \times 9$ grid, with each small cell filled with a positive integer. If the sum of all numbers in an $m \times n (1 \leqslant m \leqslant 3, 1 \leqslant n \leqslant 9)$ subgrid of $A$ is a multiple of 10, then it is called a "good rectangle"; if a $1 \times 1$ cell in $A$ is not contai... | First, we prove by contradiction that there are no more than 25 bad cells in $A$.
Assume the conclusion is not true. Then, in the grid $A$, there is at most 1 cell that is not a bad cell. By the symmetry of the grid, we can assume that all cells in the first row are bad cells.
Let the numbers filled in the $i$-th col... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given a sequence of numbers
$$
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \cdots, \frac{k}{1}, \frac{k-1}{2}, \cdots, \frac{1}{k} \text {. }
$$
In this sequence, the index of the 40th term that equals 1 is ( ).
(A) 3120
(B) 3121
(C) 3200
(D) 3201 | - 1. B.
For the terms where the sum of the numerator and denominator is $k+1$, we denote them as the $k$-th group. According to the arrangement rule, the 40th term with a value of 1 should be the 40th number in the $2 \times 40-1=79$ group, with the sequence number being
$$
\begin{array}{l}
(1+2+\cdots+78)+40 \\
=\fra... | 3121 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{199}-\frac{1}{200}}{\frac{1}{201^{2}-1^{2}}+\frac{1}{202^{2}-2^{2}}+\cdots+\frac{1}{300^{2}-100^{2}}} \\
= \\
\end{array}
$$ | 3. 400 .
Original expression
$$
\begin{array}{l}
=\frac{\left(1+\frac{1}{2}+\cdots+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{200}\right)}{\frac{1}{202 \times 200}+\frac{1}{204 \times 200}+\cdots+\frac{1}{400 \times 200}} \\
=\frac{\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}}{\frac{1}{400... | 400 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. A real-coefficient polynomial $P(x)$ of degree not exceeding 2011 takes integer values for any integer $x$, and the remainders when $P(x)$ is divided by $x-1, x-2, \cdots, x-2011$ are $1, 2, \cdots, 2011$ respectively. Then $\max _{x \in \{-1, -2, \cdots, -2011\}}|P(x)|$ has the minimum value of $\qquad$ | $-1.2011$
First, $P(x)-x$. can be divided by $x-1, x-2, \cdots$, $x-2011$ respectively, so we can assume
$$
P(x)-x=q(x-1)(x-2) \cdots(x-2011) \text {, }
$$
where, $q \in \mathbf{Q}$.
If $q \neq 0$, then
$$
\begin{array}{l}
|P(-2011)|=\left|-2011-q \cdot \frac{4022!}{2011!}\right| \\
\geqslant\left|q \cdot \frac{4022!}... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $x$ is a positive integer, and $2011-x$ is a perfect cube. Then the minimum value of $x$ is $\qquad$ . | 1. 283.
From $12^{3}=1728<2011<13^{3}=2197$, we know the minimum value of $x$ is
$$
2011-1728=283 \text {. }
$$ | 283 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Divide a cube with an edge length of a positive integer into 99 smaller cubes, among which, 98 smaller cubes are unit cubes. Find the surface area of the original cube. | Let the side length of the original cube be $x$, and the side lengths of the 99 smaller cubes be 1 and $y(x>y>1)$. Then
$$
\begin{array}{l}
x^{3}-y^{3}=98 \\
\Rightarrow(x-y)\left[(x-y)^{2}+3 x y\right]=98 \\
\Rightarrow(x-y) \mid 98=7^{2} \times 2 \\
\Rightarrow x-y=1,2,7,14,49,98 . \\
\text { By }(x-y)^{3}<x^{3}-y^{3... | 150 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the number of ordered integer pairs $(a, b)$ such that
$$
x^{2}+a x+b=167 y
$$
has integer solutions $(x, y)$, where $1 \leqslant a, b \leqslant 2004 .^{[4]}$ (2004, Singapore Mathematical Olympiad) | If $x^{2}+a x+b \equiv 0(\bmod 167)$, then completing the square gives
$$
a^{2}-4 b \equiv(2 x+a)^{2}(\bmod 167) .
$$
Therefore, for a fixed $a$, $a^{2}-4 b$ is a quadratic residue modulo 167.
Hence, by the lemma, $b$ can take $\frac{167-1}{2}+1=84$ values modulo 167.
And $\frac{2004}{167}=12$, so each $a$ correspon... | 2020032 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Figure 2 is a rectangle composed of 6 squares. If the area of the smallest square is 1, then the area of this rectangle is $\qquad$ .
| 3. 143 .
Let the side lengths of the six squares, from smallest to largest, be
$$
1, x, x, x+1, x+2, x+3 \text{.}
$$
Then, by the equality of the top and bottom sides of the rectangle, we have
$$
\begin{array}{l}
x+x+(x+1)=(x+2)+(x+3) \\
\Rightarrow x=4 .
\end{array}
$$
Thus, the length and width of the rectangle ar... | 143 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four. (25 points) There are several (more than enough) socks in red, yellow, blue, and white. If any two socks of the same color can make 1 pair, the question is: What is the minimum number of socks needed to ensure that 10 pairs of socks can be formed? | Solution 1 Since there are 4 colors, among 5 socks, there must be 1 pair.
After taking out 1 pair, 3 socks remain. By adding 2 more socks, another pair can be formed.
Following this logic, the number of pairs of socks $(x)$ and the number of socks needed $(y)$ have the following relationship:
$$
y=2 x+3 \text {. }
$$... | 23 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 As shown in Figure 5, in $\triangle A B C$, $\angle C=90^{\circ}$, $I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$
(2004, Beijing Middle School Mathematics Competit... | Solve As shown in Figure 5, construct the symmetric points $F, G$ of points $E, D$ with respect to $AD, BE$ respectively. Then $F, G$ lie on $AB$. Connect $IF, IG$. It is easy to know that
$$
\angle AIB=90^{\circ}+\frac{1}{2} \angle C=135^{\circ}.
$$
By the properties of axial symmetry, we have
$$
\begin{array}{l}
IF=... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the set $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can be found two elements $a, b$, such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$ (where $|M|$ denotes the number of elements in the set $M$). (Supplied by Feng Zhigang) | 2. When
$$
M=\left\{1,2,2^{2}, \cdots, 2^{10}, 3,3 \times 2,3 \times 2^{2}, \cdots, 3 \times 2^{9}\right\}
$$
it satisfies the condition, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011,
\end{array}
$$
contradiction.
In summary, the maximum value of $|M|$ is 21. | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Nine positive real numbers $a_{1}, a_{2}, \cdots, a_{9}$ form a geometric sequence, and
$$
a_{1}+a_{2}=\frac{3}{4}, a_{3}+a_{4}+a_{5}+a_{6}=15 .
$$
Then $a_{7}+a_{8}+a_{9}=$ . $\qquad$ | 1. 112.
Let the common ratio be $q$. Then, from the given conditions, we have
$$
\begin{array}{l}
a_{1}(1+q)=\frac{3}{4} \\
a_{1} q^{2}\left(1+q+q^{2}+q^{3}\right)=15
\end{array}
$$
Dividing the above two equations yields $q^{2}\left(1+q^{2}\right)=20$.
Thus, $q=2, a_{1}=\frac{1}{4}$.
Therefore, $a_{7}+a_{8}+a_{9}=a_{... | 112 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 10, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition) | As shown in Figure 10, construct the symmetric point $B^{\prime}$ of point $B$ with respect to line $A C$, and let $B B^{\prime}$ intersect $A C$ at point $E$. Draw $B^{\prime} N \perp A B$ at point $N$, and let $B^{\prime} N$ intersect $A C$ at point $M$. Then, $M$ and $N$ are the required points. The minimum value so... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the set of positive real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$, and the set $S=\{(a, b) \mid a \in A, b \in A, a-b \in A\}$. Then the set $S$ can have at most $\qquad$ elements. | 9.4950 .
The number of ordered pairs of real numbers $(a, b)$ formed by the elements of set $A$ is $100^{2}=10000$.
Since $a_{i}-a_{i}=0 \notin A$, we have $\left(a_{i}, a_{i}\right) \notin S(i=1,2, \cdots, 100)$.
When $\left(a_{i}, a_{j}\right) \in S$, then $\left(a_{j}, a_{i}\right) \notin S$.
Therefore, the maximum... | 4950 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. Each vertex of the convex pentagon $A B C D E$ is colored with one of five colors, such that the two endpoints of each diagonal have different colors. The number of such coloring methods is $\qquad$ (answer with a number), | 12. 1020 .
The number of coloring ways where all vertices have different colors is $A_{5}^{5}=$ 120. The number of coloring ways where two adjacent vertices have the same color and the rest of the vertices have different colors is
$$
A_{5}^{1} A_{5}^{4}=5(5 \times 4 \times 3 \times 2)=600 \text { ways. }
$$
The numbe... | 1020 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given positive integers $x, y, z$ satisfying $x y z=(14-x)(14-y)(14-z)$, and $x+y+z<28$. Then the maximum value of $x^{2}+y^{2}+z^{2}$ is . $\qquad$ | 7. 219 .
From the problem, we know that $x$, $y$, and $z$ are all positive integers less than 14. On the other hand, expanding the given equation, we get
$$
2 x y z=14^{3}-14^{2}(x+y+z)+14(x y+y z+z x) \text {. }
$$
Thus, $71 x y z$.
Since $x$, $y$, and $z$ are all less than 14, at least one of $x$, $y$, or $z$ must ... | 219 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. For a positive integer $n$, let $x_{n}$ be the real root of the equation
$$
n x^{3}+2 x-n=0
$$
with respect to $x$, and let
$$
a_{n}=\left[(n+1) x_{n}\right](n=2,3, \cdots),
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
\frac{1}{1005}\left(a_{2}+a_{3}+\cdots+a_{2011}\right)... | 9.2013.
Let $f(x)=n x^{3}+2 x-n$.
It is easy to see that when $n$ is a positive integer, $f(x)$ is an increasing function.
When $n \geqslant 2$,
$$
\begin{array}{l}
f\left(\frac{n}{n+1}\right)=n\left(\frac{n}{n+1}\right)^{3}+2 \times \frac{n}{n+1}-n \\
=\frac{n}{(n+1)^{3}}\left(-n^{2}+n+1\right)0$.
Therefore, when $n ... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. In the Cartesian coordinate system, given the point set $I=\{(x, y) \mid x, y$ are integers, and $0 \leqslant x, y \leqslant 5\}$. Then the number of different squares with vertices in the set $I$ is . $\qquad$ | 10. 105.
It is easy to know that there are only two types of squares that meet the conditions: squares whose sides lie on lines perpendicular to the coordinate axes, called "standard squares," and squares whose sides lie on lines not perpendicular to the coordinate axes, called "oblique squares."
(1) In standard squar... | 105 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. There are 4 colors of light bulbs (with enough of each color), and we need to install a light bulb at each vertex of the triangular prism $A B C-A_{1} B_{1} C_{1}$. The requirement is that the light bulbs at the two endpoints of the same edge must be of different colors, and each color of light bulb must be used at ... | 9.216 .
We can first install $A$, $B$, and $C$, which has $\mathrm{A}_{4}^{3}$ ways; then select one vertex from $A_{1}$, $B_{1}$, and $C_{1}$ to install the fourth color of the light bulb, which has $\mathrm{C}_{3}^{1}$ ways; finally, there are 3 ways to install the remaining two vertices.
Therefore, there are 216 di... | 216 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. The smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers is $\qquad$ . | 10. 495 .
$$
\begin{array}{l}
\text { Let } t=l+(l+1)+\cdots+(l+8) \\
=m+(m+1)+\cdots+(m+9) \\
=n+(n+1)+\cdots+(n+10)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $l=n+2+\frac{2 n+1}{9}$,
$$
m=\frac{n}{10}+n+1 .
$$
Therefore, $2 n+1 \equiv 0(\bmod 9)$,
$$
n \equiv 0(\bmod 10) \text {. }
$$
Thus, the smalle... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=4$, the sum of the first $n$ terms is $S_{n}$, and it satisfies
$$
S_{n+1}-5 S_{n}-4 n-4=0\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the last four digits of $a_{2 \text { on }}$ are $\qquad$ | 7.8124.
Given $S_{n+1}-5 S_{n}-4 n-4=0$, and
$$
S_{n}-5 S_{n+1}-4 n=0 \text {, }
$$
subtracting the two equations yields
$$
\begin{array}{l}
a_{n+1}-5 a_{n}-4=0 \\
\Rightarrow a_{n+1}+1=5\left(a_{n}+1\right) \\
\Rightarrow a_{n}=5^{n}-1 .
\end{array}
$$
Notice that, when $k \in \mathbf{N}_{+}$,
$$
\begin{aligned}
5^... | 8124 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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