problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
2. As shown in Figure 1, given that $G$ is the centroid of $\triangle A B O$. If $P Q$ passes through point $G$, and
$$
\begin{array}{l}
O A=a, O B=b, \\
O P=m a, O Q=n b,
\end{array}
$$
then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$ | 2.3.
From $O M=\frac{1}{2}(a+b)$, we know
$$
O G=\frac{2}{3} O M=\frac{1}{3}(a+b) .
$$
From the collinearity of points $P, G, Q$, we have $\boldsymbol{P G}=\lambda \boldsymbol{G} \boldsymbol{Q}$.
And $P G=O G-O P=\frac{1}{3}(a+b)-\dot{m} a$
$$
\begin{array}{l}
=\left(\frac{1}{3}-m\right) a+\frac{1}{3} b, \\
G Q=O Q-O... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. If $p$, $q$, $\frac{2 p-1}{q}$, $\frac{2 q-1}{p}$ are all integers, and $p>1$, $q>1$. Then $p+q=$ $\qquad$ . | 3.8 .
If $p=q$, then
$$
\frac{2 p-1}{q}=\frac{2 p-1}{p}=2-\frac{1}{p} \text {. }
$$
Given $p>1$, then $\frac{2 p-1}{q}$ is not an integer, which contradicts the problem statement. Therefore, $p \neq q$.
By symmetry, without loss of generality, assume $p>q$. Let
$$
\begin{array}{l}
\frac{2 p-1}{q}=m, \\
\frac{2 q-1}{p... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (20 points) Find the smallest real number $A$, such that for each quadratic trinomial $f(x)$ satisfying the condition $|f(x)| \leqslant 1(0 \leqslant x \leqslant 1)$, the inequality $f^{\prime}(0) \leqslant A$ holds. | Three, let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {. }
$$
From the problem, we know
$$
|f(0)| \leqslant 1,\left|f\left(\frac{1}{2}\right)\right| \leqslant 1,|f(1)| \leqslant 1 .
$$
Notice that $f(0)=c, f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$,
$$
\begin{array}{l}
f(1)=a+b+c, \\
f... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 As shown in Figure 1, in square $A B C D$, $E$ is a fixed point on $B C$, and $B E=10, E C=14, P$ is a moving point on $B D$. Then the minimum value of $P E+P C$ is $\qquad$
(2006, Zhejiang Province Junior High School Mathematics Competition) | Analysis: Considering the minimum value of the sum of two line segments, it is relatively easy to think of using the property "the shortest distance between two points is a straight line." This requires transforming one of the line segments to the other side. Therefore, by the axial symmetry of the square (as shown in ... | 26 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. In a $7 \times 7$ unit square grid, there are 64 grid points, and there are many squares with these grid points as vertices. How many different values can the areas of these squares have?
(21st Jiangsu Province Junior High School Mathematics Competition) | (提示: Consider the number of different chordal graphs in a square grid with side length less than or equal to 7 (side length is an integer) (including extreme cases, excluding those with the same area). As shown in Figure 18, the area of the shaded square is $a^{2}+$ $b^{2}$, where $0 \leqslant a+b \leqslant 7$. Without... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $\alpha^{2008}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$. Find the sum of the coefficients of this polynomial.
(2005, China Western Mathematical Olympiad) | Explanation: Let $S_{n}=\alpha^{n}+\beta^{n}\left(n \in \mathbf{N}_{+}\right)$,
$$
\left(\sigma_{1}, \sigma_{2}\right)=(\alpha+\beta, \alpha \beta) \text {. }
$$
Then $S_{1}=\sigma_{1}$,
$$
S_{2}=\sigma_{1} S_{1}+2(-1)^{3} \sigma_{2}=\sigma_{1}^{2}-2 \sigma_{2}, \cdots \cdots
$$
By Newton's formula, we have
$$
\alpha... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given a positive integer $n(n>3)$, let real numbers $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n} \geqslant n, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2} .
\end{array}
$$
Find the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$.
(28th Uni... | The above equation holds for $a \geqslant 2$.
Try taking $a=2$, at this point, $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ $=2$. From this, we can conjecture that the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ is 2.
Solution: First, we prove that for any real numbers $a_{1}, a_{2}, \cdots... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 A $98 \times 98$ chessboard is displayed on a computer screen, with its squares colored like a chessboard. You are allowed to select any rectangle with the mouse (the sides of the rectangle must lie on the grid lines), and then click the mouse button, which will change the color of each square in the rectangl... | Analysis: If we consider "all cells of the same color," it would be very difficult (because too many objects can obscure the essential relationships), leading to the idea of adopting a local thinking strategy.
First local thinking: Transform "all cells of the same color" into "any two adjacent cells of the same color.... | 98 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given real numbers $x, y$ satisfy $\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$. Then the value of $3 x^{2}-2 y^{2}+3 x-3 y-2007$ is ( ).
(A) -2008
(B) 2008
(C) -1
(D) 1 | 6.D.
From $\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$,
we get
$$
\begin{array}{l}
x-\sqrt{x^{2}-2008} \\
=\frac{2008}{y-\sqrt{y^{2}-2008}}=y+\sqrt{y^{2}-2008}, \\
y-\sqrt{y^{2}-2008}=\frac{2008}{x-\sqrt{x^{2}-2008}} \\
=x+\sqrt{x^{2}-2008} .
\end{array}
$$
From the above two equations, we ... | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ | $$
\begin{array}{l}
\text { Given } a^{2}=\left(\frac{\sqrt{5}-1}{2}\right)^{2}=\frac{3-\sqrt{5}}{2}=1-a \\
\Rightarrow a^{2}+a=1 . \\
\text { Therefore, } \frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a} \\
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a \cdot a^{2}-a} \\
=\frac{a^{3}-2 a^{3}-1+2}{a(1-... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the squares of positive integers $1,2, \cdots$ in a sequence: $149162536496481100121144 \cdots$, the digit at the 1st position is 1, the digit at the 5th position is 6, the digit at the 10th position is 4, the digit at the 2008th position is $\qquad$. | 4.1.
$1^{2}$ to $3^{2}$, each result occupies 1 digit, totaling $1 \times 3$ $=3$ digits;
$4^{2}$ to $9^{2}$, each result occupies 2 digits, totaling $2 \times 6$ $=12$ digits;
$10^{2}$ to $31^{2}$, each result occupies 3 digits, totaling $3 \times$ $22=66$ digits;
$32^{2}$ to $99^{2}$, each result occupies 4 digits, t... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $a$ be a prime number, $b$ and $c$ be positive integers, and satisfy
$$
\left\{\begin{array}{l}
9(2 a+2 b-c)^{2}=509(4 a+1022 b-511 c), \\
b-c=2 .
\end{array}\right.
$$
Find the value of $a(b+c)$. | Three, Equation (1) is
$$
\left(\frac{6 a+6 b-3 c}{509}\right)^{2}=\frac{4 a+1022 b-511 c}{509} \text {. }
$$
Let $m=\frac{6 a+6 b-3 c}{509}$,
$$
n=\frac{4 a+1022 b-511 c}{509} \text {. }
$$
Then $2 b-c=\frac{509 m-6 a}{3}=\frac{509 n-4 a}{511}$.
Thus, $3 n-511 m+6 a=0$.
Since $n=m^{2}$, we have
$$
3 m^{2}-511 m+6 a=... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. The number of tetrahedra that can be formed with the vertices of a regular hexagonal prism is $\qquad$.
untranslated part: $\qquad$ | 11.426.
Among 12 vertices, choosing any 4, there are $\mathrm{C}_{12}^{4}$ different ways, but when the 4 chosen points are coplanar, they cannot form a tetrahedron and should be excluded.
The 4 chosen points being coplanar can be divided into 4 cases:
(1) 4 points are on the same side, with 6 sides, and the side is a ... | 426 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. There is a 6-row $n$-column matrix composed of 0s and 1s, where each row contains exactly 5 ones, and the number of columns in which any two rows both have a 1 is at most 2. Find the minimum value of $n$. | 15. First, the total number of 1s in the matrix is $5 \times 6=$ 30.
Let the number of 1s in the $k$-th column be $l_{k}$. Then
$$
\sum_{k=1}^{n} l_{k}=30 \text {. }
$$
For any $1 \leqslant i<j \leqslant 6$ and $1 \leqslant k \leqslant n$, consider the triplet $\{i, j, k\}$: such that the $i$-th row and the $j$-th row... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The integer part of the real number $\left(\tan 60^{\circ}+2 \cos 45^{\circ}\right)^{6}$ is ( ).
(A) 969
(B) 970
(C) 971
(D) None of the above conclusions is correct | $-1 . \mathrm{A}$.
$$
\left(\tan 60^{\circ}+2 \cos 45^{\circ}\right)^{6}=(\sqrt{3}+\sqrt{2})^{6} \text {. }
$$
Let $\sqrt{3}+\sqrt{2}=a, \sqrt{3}-\sqrt{2}=b$. Then
$$
a+b=2 \sqrt{3}, a b=1 \text {. }
$$
Thus, $a^{2}+b^{2}=(a+b)^{2}-2 a b=10$,
$$
a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)=18 \sqrt{3} \text {. }
$$
... | 969 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Given a rhombus $A B C D$ with side $A B=10$, diagonal $B D=12$, and 100 different points $P_{1}, P_{2}$, $\cdots, P_{100}$ on $B D$. Draw $P_{i} E_{i} \perp A B$ at $E_{i}$, and $P_{i} F_{i} \perp A D$ at $F_{i}$. Then
$$
P_{1} E_{1}+P_{1} F_{1}+P_{2} E_{2}+P_{2} F_{2}+\cdots+P_{100} E_{100}+P_{100} F_{100}
$$
is $... | 2. C.
As shown in Figure 5, extend $E_{i} P_{i}$ to intersect $C D$ at point $G_{i}$. Then,
$$
P_{i} G_{i} \perp C D \text {. }
$$
Thus, $P_{i} F_{i}=P_{i} G_{i}$. Connect $A C$ to intersect $B D$ at point $O$. Therefore,
$$
\begin{array}{l}
A C \perp B D, \\
B O=\frac{1}{2} B D=6 .
\end{array}
$$
Then $A O=\sqrt{A ... | 960 | Geometry | MCQ | Yes | Yes | cn_contest | false |
Three. (25 points) Given $n(n>1)$ integers (which can be the same) $x_{1}, x_{2}, \cdots, x_{n}$ satisfy
$$
x_{1}+x_{2}+\cdots+x_{n}=x_{1} x_{2} \cdots x_{n}=9111 \text {. }
$$
Find the maximum value of $x_{1}, x_{2}, \cdots, x_{n}$ when $n$ takes the minimum value. | Three, from $x_{1} x_{2} \cdots x_{n}=9111$, we know that $x_{1}, x_{2}, \cdots, x_{n}$ are all odd numbers.
Since $x_{1}+x_{2}+\cdots+x_{n}=9111$ is odd, $n (n>1)$ must be odd.
If $n=3$, then
$$
x_{1}+x_{2}+x_{3}=x_{1} x_{2} x_{3}=9111 \text {. }
$$
By the symmetry of the condition equations, without loss of general... | 9111 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. As shown in Figure 15, in trapezoid $A B C D$, $A D / / B C (B C > A D), \angle D=90^{\circ}, B C=$ $C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$ .
(2004, National Junior High School Mathematics Competition) | (Hint: First, complete the trapezoid $ABCD$ into a square $CBFD$, then rotate $\triangle ABF$ $90^{\circ}$ around point $B$ to the position of $\triangle BCG$. It is easy to see that $\triangle ABE \cong \triangle GBE$. Therefore, $AE = EG = EC + AF = 10$. Let $EC = x$. Then $AF = 10 - x$, $DE = 12 - x$, $AD = 2 + x$. ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Notice that $6!=8 \times 9 \times 10$. Then, the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers is $\qquad$ . | 5.23.
According to the requirements of the problem, write $n!$ as
$$
n!=n \times(n-1) \times \cdots \times 5 \times 24 \text {. }
$$
Therefore, $n+1 \leqslant 24, n \leqslant 23$.
Hence, the maximum value of $n$ is 23. At this point,
$$
23!=24 \times 23 \times \cdots \times 5 \text {. }
$$
The right side of the abov... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the quadratic function $f(x)=a x^{2}+b x+c, a$ $\in \mathbf{N}_{+}, c \geqslant 1, a+b+c \geqslant 1$, the equation $a x^{2}+b x+c$ $=0$ has two distinct positive roots less than 1. Then the minimum value of $a$ is | 6.5.
Let the two roots of the equation $a x^{2}+b x+c=0$ be $x_{1}$ and $x_{2}$, and $0<x_{1}<x_{2}<1, a>4$.
Therefore, $a \geqslant 5$.
Take $f(x)=5\left(x-\frac{1}{2}\right)\left(x-\frac{3}{5}\right)$, which meets the conditions. Hence, the minimum value of $a$ is 5. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. A polygon is cut with one cut (the cut does not pass through the vertices of the polygon) to divide it into 2 polygons, then one of the polygons is cut with another cut (the cut does not pass through the vertices of the polygon) to divide it into another polygon, ... and so on. If starting from a square, to cut out ... | 5.2011014.
Let a total of $k$ cuts be made. Since each cut increases the number of polygons by one, there are a total of $k+1$ polygons. Apart from one triangle, one quadrilateral, one pentagon, ... one 2008-sided polygon, there are
$$
(k+1)-2006=k-2005
$$
polygons.
Considering the total sum $S$ of the number of side... | 2011014 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four. (20 points) As shown in Figure 3, let $F$ be one of the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, and $A$ be the vertex of the ellipse that is farthest from $F$. Take 2008 distinct points $P_{i}(i=1,2, \cdots, 2008)$ on the minor axis $B C$ of the ellipse. Let $F P_{i}$ intersect $A B$ or $A C$ a... | For any point $P_{i}(i=1,2, \cdots, 2008)$, by symmetry, we can assume $P_{i}$ is on the positive half of the $y$-axis, as shown in Figure 7.
In $\triangle A F C$, since $A N_{i}$, $C O$, and $F M_{i}$ are concurrent, by Ceva's Theorem, we have
$$
\frac{A M_{i}}{M_{i} C} \cdot \frac{C N_{i}}{N_{i} F} \cdot \frac{F O}{... | 2009 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Figure 3 is a Pythagorean tree grown by squares and isosceles right triangles according to a certain pattern. Observe the figure and answer the questions.
(1) At the far right end of each layer of the tree, there is a square marked with a number indicating the layer, starting from this square and counting cou... | Solution: (1) From the figure, we observe that the number of squares in each upper layer is twice the number of squares in the previous layer. Therefore, the fifth layer has 16 squares, and the $n$-th layer has $2^{n-1}$ squares.
(2) The sum of the areas of the two smaller squares on each branch equals the area of the ... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Figure 4 is a "sheep's head" pattern, the method of which is: starting from square (1), using one of its sides as the hypotenuse, an isosceles right triangle is constructed outward, and then using its legs as sides, squares (2) and (2)' are constructed outward, ... and so on. If the side length of square (1) ... | Solution: By the Pythagorean theorem, the side length of square (2) is $64 \times \frac{\sqrt{2}}{2}=32 \sqrt{2}(\mathrm{~cm})$,
the side length of square (3) is
$$
32 \sqrt{2} \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{2}=32(\mathrm{~cm}) \text {, }
$$
the side length of square (4) is
$$
32 \... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. A book has a total of 61 pages, sequentially numbered as 1, 2, ..., 61. Someone, while adding these numbers, mistakenly reversed the digits of two two-digit page numbers (a two-digit number of the form $\overline{a b}$ was treated as $\overline{b a}$), resulting in a total sum of 2008. Therefore, the maximum sum of ... | 3.68.
Notice that $1+2+\cdots+61=1891$,
$$
2008-1891=117 \text {. }
$$
Since the page numbers in the form of $\overline{a b}$ are read as $\overline{b a}$, the difference in the sum will be $9|a-b|$, because $a$ and $b$ can only take values from 1,2, $\cdots, 9$, $|a-b| \leqslant 8$, so,
$9|a-b| \leqslant 72$.
Since ... | 68 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. The largest integer not exceeding $(\sqrt{5}+\sqrt{3})^{6}$ is
$\qquad$ | 4.3903.
Notice that $(\sqrt{5}+\sqrt{3})^{6}=(8+2 \sqrt{15})^{3}$.
Let $8+2 \sqrt{15}=a, 8-2 \sqrt{15}=b$. Then we have
$$
a+b=16, ab=4 \text{. }
$$
We know that $a$ and $b$ are the roots of the equation $x^{2}-16 x+4=0$, so we have
$$
\begin{array}{l}
a^{2}=16 a-4, b^{2}=16 b-4 ; \\
a^{3}=16 a^{2}-4 a, b^{3}=16 b^{2... | 3903 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Let $a$ be an integer such that the equation $a x^{2}-(a+5) x+a+7=0$ has at least one rational root. Find all possible rational roots of the equation.
| When $a=0$, the rational root of the equation is $x=\frac{7}{5}$.
The following considers the case where $a \neq 0$. In this case, the original equation is a quadratic equation, and by the discriminant
$$
(a+5)^{2}-4 a(a+7) \geqslant 0,
$$
which simplifies to $3 a^{2}+18 a-25 \leqslant 0$.
Solving this, we get $-\frac... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) 120 people participated in a math competition, with the test consisting of 5 major questions. It is known that questions $1, 2, 3, 4, 5$ were correctly solved by 96, $83, 74, 66, 35$ people respectively. If at least 3 questions must be answered correctly to win an award, how many people won awards at... | Three, number these 120 people as $P_{1}, P_{2}, \cdots, P_{120}$, and consider them as 120 points on a number line. Let $A_{k}(k=1, 2,3,4,5)$ represent the group of people among these 120 who did not answer the $k$-th question correctly, and $\left|A_{k}\right|$ be the number of people in that group. Then,
$\left|A_{1... | 40 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+(1+x)^{3}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+b_{2} x^{2}+\cdots+b_{n} x^{n},
\end{array}
$$
and it satisfies $b_{1}+b_{2}+\cdots+b_{n}=26$. Then a possible value of the positive integer $n$ is $\qquad$ | Take $x=0$, we get $b_{0}=n$.
Take $x=1$, we get
$$
2+2^{2}+2^{3}+\cdots+2^{n}=b_{0}+b_{1}+b_{2}+\cdots+b_{n} \text {, }
$$
i.e., $\frac{2\left(1-2^{n}\right)}{1-2}=n+26$.
Thus, $2\left(2^{n}-1\right)=n+26$.
Therefore, $n=4$ is a possible value. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that the function $f(x)$ is defined on $\mathbf{R}$, and satisfies:
(1) $f(x)$ is an even function;
(2) For any $x \in \mathbf{R}$, $f(x+4) = f(x)$, and when $x \in [0,2]$, $f(x) = x + 2$.
Then the distance between the two closest points of intersection between the line $y=4$ and the graph of the function $f(... | 9.4.
Draw the graph of the function $f(x)$ (as shown in Figure 3).
From Figure 3, it is easy to see that the distance between the two closest intersection points of the line $y=4$ and the graph of the function $f(x)$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. If sets $A_{1}$ and $A_{2}$ satisfy $A_{1} \cup A_{2}=A$, then we denote $\left[A_{1}, A_{2}\right]$ as a pair of subset partitions of $A$. It is stipulated that: $\left[A_{1}, A_{2}\right]$ and $\left[A_{2}, A_{1}\right]$ are considered the same pair of subset partitions of $A$. Given the set $A=\{1,2,3\}$. Then, ... | 12. 14.
Divide into three categories.
First category: When at least one of $A_{1}$ and $A_{2}$ is $A$ (let's assume $A_{1}=A$), then $A_{2}$ can be any subset of $A$, for a total of 8 different subset decompositions.
Second category: $A_{1}$ and $A_{2}$ are one-element and two-element subsets, respectively, with the ... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. Let the function $f(x)=x^{2}+a x+b(a, b$ be real constants). It is known that the inequality $|f(x)| \leqslant 12 x^{2}+4 x-30|$ holds for any real number $x$. Define the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ as:
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, 2 a_{n}=f\left(a_{n-1}\right)+15(n=2,3, \c... | 15. (1) Let the two real roots of the equation $2 x^{2}+4 x-30=0$ be $\alpha, \beta$. Then $\alpha+\beta=-2, \alpha \beta=-15$.
Taking $x=\alpha$ in $|f(x)| \leqslant\left|2 x^{2}+4 x-30\right|$, we get $|f(x)| \leqslant 0$, hence $f(\alpha)=0$.
Similarly, $f(\beta)=0$.
Therefore, $f(x)=(x-\alpha)(x-\beta)$
$$
=x^{2}-... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
16. In four-dimensional space, the distance between point $A\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ and point $B\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ is defined as
$$
A B=\sqrt{\sum_{i=1}^{4}\left(a_{i}-b_{i}\right)^{2}} .
$$
Consider the set of points
$I=\left\{P\left(c_{1}, c_{2}, c_{3}, c_{4}\right) \mid c_{i}=... | 16. Construct the following 8 points:
$$
\begin{array}{l}
P_{1}(0,0,0,0), P_{2}(0,1,0,0), P_{3}(0,0,0,1), \\
P_{4}(0,0,1,1), P_{5}(1,1,0,0), P_{6}(1,1,1,0), \\
P_{7}(1,1,1,1), P_{8}(1,0,1,1).
\end{array}
$$
By calculation, it is known that no three of these points can form an equilateral triangle, so, $n_{\min } \geqs... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $a$ is an integer, $14 a^{2}-12 a-27 \mid$ is a prime number. Then the sum of all possible values of $a$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. D.
From the problem, we know
$$
\left|4 a^{2}-12 a-27\right|=|(2 a+3)(2 a-9)|
$$
is a prime number.
Therefore, $2 a+3= \pm 1$ or $2 a-9= \pm 1$, which means $a=-1,-2$ or $a=5,4$.
Thus, the sum of all possible values of $a$ is 6. | 6 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given $a+\frac{1}{a+1}=b+\frac{1}{b-1}-2$, and $a-$ $b+2 \neq 0$. Then the value of $a b-a+b$ is $\qquad$ . | $=、 1.2$.
Let $a+1=x$,
$$
b-1=y \text {. }
$$
Then $x-y$
$$
\begin{array}{l}
=a-b+2 \\
\neq 0 .
\end{array}
$$
Substitute into the given equation,
we get $x+\frac{1}{x}=y+\frac{1}{y}$,
which means $(x-y)(x y-1)=0$.
Since $x-y \neq 0$, it follows that $x y=1$.
Therefore, $(a+1)(b-1)=1$, which means $a b-a+b=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $N=23x+92y$ be a perfect square, and $N$ does not exceed 2392. Then the number of all positive integer pairs $(x, y)$ that satisfy the above conditions is $\qquad$ pairs. | Solution: Since $N=23(x+4 y)$, and 23 is a prime number, there exists a positive integer $k$, such that $x+4 y=23 k^{2}$.
Also, because $23(x+4 y)=N \leqslant 2392$, we have $x+4 y \leqslant 104$.
Therefore, $23 k^{2}=x+4 y \leqslant 104$, which means $k^{2} \leqslant 4$.
Thus, $k^{2}=1,4$.
When $k^{2}=1$, $x+4 y=23$, ... | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let the perfect square $y^{2}$ be the sum of the squares of 11 consecutive integers. Then the minimum value of $|y|$ is $\qquad$ . | Solution: Let the middle number of 11 consecutive integers be $x$. Then
$$
\begin{aligned}
y^{2}= & (x-5)^{2}+(x-4)^{2}+\cdots+x^{2}+\cdots+ \\
& (x+4)^{2}+(x+5)^{2} \\
= & x^{2}+2\left(x^{2}+1^{2}\right)+2\left(x^{2}+2^{2}\right)+\cdots+2\left(x^{2}+5^{2}\right) \\
= & 11 x^{2}+2\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\rig... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 3. Given real numbers } a, b, c. \text { If } \\ \frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 c a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1, \\ \text { then }\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2008}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2008}+ \\ \left(\frac{a^{2}+b^{... | 3.3.
Notice
$$
\begin{array}{l}
\frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 a a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+1+\frac{(c-a)^{2}-b^{2}}{2 c a}+1+ \\
\frac{(a+b)^{2}-c^{2}}{2 a b}-1=1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+\frac{(c-a)^{2}-b^{2}}{2 ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given that $n$ is a positive integer greater than 10, and set $A$ contains $2n$ elements. If the family of sets
$$
\left\{A_{i} \subseteq A \mid i=1,2, \cdots, m\right\}
$$
satisfies the following two conditions, it is called "suitable":
(1) For any $i=1,2, \cdots, m, \operatorname{Card}\left(A_{i}\... | The maximum positive integer $m=4$.
Take two non-complementary $n$-element subsets $A_{1}, A_{2}$ of $A$, and let $A_{3}, A_{4}$ be the complements of $A_{1}, A_{2}$, respectively. From these four sets, any three sets will always have two sets that are complementary, hence the intersection of these three sets is the em... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example $4^{\circ}$ If a perfect square number has the last 3 digits the same and not 0, find the minimum value of this number.
保留源文本的换行和格式，直接输出翻译结果。 | Solution: Since the last two digits of a perfect square can only be even 0, even 1, even 4, even 9, $\overline{25}$, or odd 6, the last three digits of a perfect square can only be 444. Since 444 is not a perfect square, we have $n \geqslant 1444$. And $1444=38^{2}$ is a perfect square, so the smallest perfect square i... | 1444 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 There is a four-digit number
$$
N=\overline{(a+1) a(a+2)(a+3)} \text {, }
$$
which is a perfect square. Find $a$. | Solution: Note that the last digit of a perfect square can only be $0, 1, 4, 5, 6, 9$, thus
$$
a+3 \equiv 0,1,4,5,6,9(\bmod 10) .
$$
Also, $a \geqslant 0, a+3 \leqslant 9$, so $a=1,2,3,6$.
Therefore, the last two digits of $N$ are
$$
\overline{(a+2)(a+3)}=34,45,56,89 \text {. }
$$
Since the last two digits of a perfe... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 4, there are five circles that are externally tangent to each other in sequence, and all are tangent to lines $a$ and $b$. If the diameters of the smallest and largest circles are 18 and 32, respectively, then the diameter of $\odot \mathrm{O}_{3}$ is $\qquad$ | 8.24.
As shown in Figure 7, let the radii of the five circles $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}, \odot O_{5}$ be $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$, respectively. Draw perpendiculars from points $O_{1}, O_{2}, O_{3}$ to line $a$, with the feet of the perpendiculars being $A_{1}, A_{2}, A_{3}$, respe... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Given real numbers $a, b, c, d$, and $a \neq b, c \neq d$. If the equations: $a^{2}+a c=2, b^{2}+b c=2, c^{2}+a c=$ $4, d^{2}+a d=4$ all hold, then the value of $6 a+2 b+3 c+2 d$ is $\qquad$. | 9.0 .
$$
\begin{array}{l}
\text { Given }\left(a^{2}+a c\right)-\left(b^{2}+b c\right)=2-2=0, \\
\left(c^{2}+a c\right)-\left(d^{2}+a d\right)=4-4=0,
\end{array}
$$
we get
$$
\begin{array}{l}
(a-b)(a+b+c)=0, \\
(c-d)(a+c+d)=0 .
\end{array}
$$
Since $a \neq b, c \neq d$, we have
$$
a+b+c=0, a+c+d=0 .
$$
Thus, $b=d=-(... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given that $m$ and $n$ are positive integers. If $1 \leqslant m \leqslant n$ $\leqslant 30$, and $mn$ is divisible by 21, then the number of pairs $(m, n)$ that satisfy the condition is $\qquad$ . | 10.57.
Given that positive integers $m, n$ satisfy $mn$ is divisible by 21, and $1 \leqslant m \leqslant n \leqslant 30$, therefore,
(1) If $m=21$, then $n=21,22, \cdots, 30$. Hence, there are 10 pairs $(m, n)$ that satisfy the condition.
(2) If $m \neq 21$,
(i) When $n=21$, $m=1,2, \cdots, 20$. There are 20 pairs $(m... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. Given that $a$ and $b$ are real numbers, and $a^{2}+a b+b^{2}=3$. If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$. | Three, 11. Let $a^{2}-a b+b^{2}=k$. From
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=3, \\
a^{2}-a b+b^{2}=k
\end{array} \Rightarrow a b=\frac{3-k}{2} .\right.
$$
Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$$
=3+\frac{3-k}{2}=\frac{9-k}{2} \text {. }
$$
Since $(a+b)^{2} \geqslant 0$, i.e., $\frac{9-k}{2} \geqs... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$. Then the value of $\frac{\tan \alpha}{\tan \beta}$ is | $$
\begin{array}{l}
\sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \\
= 3(\sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta), \\
\sin \alpha \cdot \cos \beta = 2 \cos \alpha \cdot \sin \beta. \\
\text{Therefore, } \frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cdot \cos \beta}{\cos \alpha \c... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. If $x^{5}+3 x^{3}+1=a_{0}+a_{1}(x-1)+$ $a_{2}(x-1)^{2}+\cdots+a_{5}(x-1)^{5}$ holds for any real number $x$, then the value of $a_{3}$ is $\qquad$ (answer with a number). | 12.13.
$$
\begin{array}{l}
\text { In } x^{5}+3 x^{3}+1 \\
=[(x-1)+1]^{5}+3[(x-1)+1]^{3}+1
\end{array}
$$
the coefficient of the $(x-1)^{3}$ term in the expanded form is $\mathrm{C}_{5}^{2}+3=$ 13, so, $a_{3}=13$. | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In the right-angled coordinate plane $x O y$, $\triangle A_{i} B_{i} A_{i+1}(i=1,2, \cdots)$ are equilateral triangles, and satisfy
$$
\boldsymbol{O A}_{1}=\left(-\frac{1}{4}, 0\right), \boldsymbol{A}_{i} \boldsymbol{A}_{i+1}=(2 i-1,0) \text {. }
$$
(1) Prove that the points $B_{1}, B_{2}, \cdots, B_{n}, \cdots$ lie on... | Let $B_{n}(x, y)$. Then
$$
\left\{\begin{array}{l}
x=-\frac{1}{4}+1+3+\cdots+(2 n-3)+\frac{2 n-1}{2}=\left(n-\frac{1}{2}\right)^{2}, \\
|y|=\frac{\sqrt{3}}{2}(2 n-1) .
\end{array}\right.
$$
Eliminating $n$ gives $y^{2}=3 x$.
Thus, the points $B_{1}, B_{2}, \cdots, B_{n}, \cdots$ lie on the same parabola $\Gamma: y^{2}... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Find the largest positive integer $n$, such that $4^{27}+4^{50}+4^{n}$ is a perfect square.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
Example 9 Find the largest positive integer $n$, such that $4^{27}+4^{50}+4^{n}$ is a perfect sq... | Solution: When $n>27$,
$$
A=4^{27}+4^{500}+4^{n}=4^{27}\left(1+4^{473}+4^{n-27}\right) \text {. }
$$
Since $4^{27}$ is a perfect square, for $A$ to be a perfect square, $1+4^{473}+4^{n-27}$ must be a perfect square.
Notice that
$$
\begin{array}{l}
1+4^{473}+4^{n-27}=1+2^{2 \times 473}+\left(2^{n-27}\right)^{2} \\
=1+2... | 972 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the sequence $\left\{b_{n}\right\}_{n \geqslant 1}$ satisfy
$$
\begin{array}{l}
b_{200 n}=2, b_{2008}=4, \\
b_{n+3}+b_{n+2} \\
=\frac{\left(b_{n+3}+b_{n+2}\right)\left(b_{n+1}+b_{n}\right)}{2}+1(n=1,2, \cdots)
\end{array}
$$
Then $b_{1}+b_{2}+\cdots+b_{1900}=$ | 5.1912 .
Let $a_{n}=b_{n+1}+b_{n}$. Then
$2 a_{n+2}-2=a_{n+2} a_{n}$
$$
\begin{array}{l}
\Rightarrow a_{n+2}=\frac{-2}{a_{n}-2}(n=1,2, \cdots) \\
\Rightarrow a_{n+4}=\frac{-2}{a_{n+2}-2}=\frac{-2}{\frac{-2}{a_{n}-2}-2}=\frac{a_{n}-2}{a_{n}-1} \\
\Rightarrow a_{n+x}=\frac{a_{n+4}-2}{a_{n+4}-1}=\frac{\frac{a_{n}-2}{a_{n... | 1912 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a, b \in \mathbf{N}_{+}$, and $a+b \sqrt{2}=(1+\sqrt{2})^{100}$. Then the last two digits of $a b$ are $\qquad$ . | 6.64.
Since $a+b \sqrt{2}=(1+\sqrt{2})^{100}$
$$
\begin{array}{l}
=(1+\sqrt{2})\left[\left((1+\sqrt{2})^{3}\right]^{33}\right. \\
=(1+\sqrt{2})(7+5 \sqrt{2})^{33},
\end{array}
$$
Therefore, by the binomial theorem we get
$$
\begin{array}{l}
a= \sum_{i=1}^{16} \mathrm{C}_{33}^{2 i} 7^{33-2 i}(5 \sqrt{2})^{2 i}+ \\
\s... | 64 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Let the four-digit number $\overline{a b c d}$ be a perfect square, and its arithmetic square root can be expressed as $\sqrt{\overline{a b c d}}=\overline{a b}+\sqrt{\overline{c d}}$. How many such four-digit numbers are there? | Given $\sqrt{a b c d}=\overline{a b}+\sqrt{\overline{c d}}$, we know
$$
\begin{array}{l}
(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}+\overline{c d} \\
=\overline{a b c d}=100 \overline{a b}+\overline{c d} .
\end{array}
$$
Thus, $(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}=100 \overline{a b}$, whi... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The sum of all non-negative integer solutions to the inequality $|2 x-1|<6$ with respect to $x$ is $\qquad$ . | 2.1.6.
The original inequality is equivalent to $\left\{\begin{array}{l}2 x-1-6 .\end{array}\right.$
Solving it, we get $-\frac{5}{2}<x<\frac{7}{2}$.
Therefore, all non-negative integer solutions that satisfy the condition are $x=0,1,2,3$.
Thus, the sum of all non-negative integer solutions is 6. | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $x=\frac{1}{2+\sqrt{3}}, y=\frac{1}{2-\sqrt{3}}$. Then $x^{3}+12 x y+y^{3}=$ $\qquad$ | 2.64.
It is known that $x=2-\sqrt{3}, y=2+\sqrt{3}$. Therefore, $x+y=4$.
Then, $x^{3}+12 x y+y^{3}=x^{3}+y^{3}+3 x y(x+y)$ $=(x+y)^{3}=64$. | 64 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a$ and $b$ be positive integers, $1176a=b^{4}$. Find the minimum value of $a$.
untranslated text remains unchanged. | (Note: Observing that $b^{4}=1176 a=2^{3} \times 3 \times$ $7^{2} a$, then $a$ must contain the factor $2 \times 3^{3} \times 7^{2}$. Thus, the minimum value of $a$ is $2 \times 3^{3} \times 7^{2}=2646$.) | 2646 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (25 points) Let the real number $x$ satisfy
$$
\frac{3 x-1}{2}-\frac{4 x-2}{3} \geqslant \frac{6 x-3}{5}-\frac{13}{10} \text {. }
$$
Find the minimum value of $2|x-1|+|x+4|$. | Four, multiplying both sides of the original inequality by 30 gives
$$
15(3 x-1)-10(4 x-2) \geqslant 6(6 x-3)-39 \text {. }
$$
Solving this yields $x \leqslant 2$.
Let $y=2|x-1|+|x+4|$.
(1) When $x \leqslant-4$,
$$
y=-2(x-1)-(x+4)=-3 x-2 \text {. }
$$
Therefore, the minimum value of $y$ is $(-3) \times(-4)-2=$ 10, at... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Set
$$
A=\left\{x \left\lvert\, x=\left[\frac{5 k}{6}\right]\right., k \in \mathbf{Z}, 100 \leqslant k \leqslant 999\right\},
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then the number of elements in set $A$ is $\qquad$ | When $k=100$, $\left[\frac{5 k}{6}\right]=83$;
When $k=999$, $\left[\frac{5 k}{6}\right]=832$.
It is also easy to see that for $100 \leqslant k \leqslant 999$, we have
$$
0 \leqslant\left[\frac{5(k+1)}{6}\right]-\left[\frac{5 k}{6}\right] \leqslant 1 \text {. }
$$
Therefore, the elements in $A$ can cover all int... | 750 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=5, a_{n}=\frac{2 a_{n-1}-1}{a_{n-1}-2}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the sum of its first 100 terms is $\qquad$ | 8.400.
Since $a_{1}=5$, then $a_{2}=3, a_{3}=5, a_{4}=3$, so the sequence has a period of 2. Therefore, the sum of the first 100 terms is 400. | 400 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A tangent line is drawn from the left focus $F$ of the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ to the circle $x^{2}+y^{2}=9$, with the point of tangency being $T$. Extend $F T$ to intersect the right branch of the hyperbola at point $P$. If $M$ is the midpoint of segment $F P$, and $O$ is the origin, find th... | 14. Without loss of generality, place point $P$ in the first quadrant. As shown in Figure 2, let $F^{\prime}$ be the right focus of the hyperbola, and connect $P F^{\prime}$. Since $M$ and $O$ are the midpoints of $F P$ and $F F^{\prime}$ respectively, we have $|M O|=\frac{1}{2}\left|P F^{\prime}\right|$. By the defini... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the smallest positive integer $n$ such that there exists an $(n+1)$-term sequence $a_{0}, a_{1}, \cdots, a_{n}$, satisfying $a_{0}=0, a_{n}=2008$, and
$$
\left|a_{i}-a_{i-1}\right|=i^{2}(i=1,2, \cdots, n) .
$$ | If $n \leqslant 17$, then
$$
\begin{array}{l}
a_{n}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)+a_{0} \\
\leqslant \sum_{i=1}^{n}\left|a_{i}-a_{i-1}\right|=\frac{1}{6} n(n+1)(2 n+1) \\
\leqslant \frac{1}{6} \times 17 \times 18 \times 35<2008,
\end{array}
$$
a contradiction.
If $n=18$, then
$$
\begin{array}{l}
a_{n}=\sum_... | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9.5. On an infinitely large chessboard, the distance between any two squares is defined as the minimum number of steps a King needs to move from one square to another. Given three squares that are pairwise 100 steps apart, find the number of squares that are 50 steps away from each of these three squares. | 9.5. Let the side length of a small square be 1.
Represent a small square by the coordinates of its center. Let the absolute difference of the x-coordinates of two small squares be $x$, and the absolute difference of the y-coordinates be $y$.
It is not hard to see that a king can reach from one square to another in $... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Several people participated in a test, required to answer five questions, and it was stipulated that those who answered 3 or more questions correctly would be qualified. The test results are: the number of people who answered the first to fifth questions correctly are $81, 91, 85, 79, 74$, respectively, and 70 of th... | - 1.C.
Let the number of people participating in the test be $n$.
Among these $n$ people, the total number of questions answered incorrectly is:
$$
\begin{array}{l}
5 n-(81+91+85+79+74) \\
=5 n-410 \text{ (questions). }
\end{array}
$$
Since those who answer 3 or more questions incorrectly are considered不合格 (unqualifi... | 100 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
3. Given real numbers $a, b, c$ satisfy $2^{a}=5, 2^{b}=10, 2^{c}=80$. Then the value of the algebraic expression $2006 a-3344 b+1338 c$ is ( ).
(A) 2007
(B) 2008
(C) 2009
(D) 2010 | 3. B.
From $2^{b-a}=2^{b} \div 2^{a}=2^{1}$, we get
$$
b=a+1 \text {; }
$$
From $2^{c-a}=2^{c} \div 2^{a}=80 \div 5=2^{4}$, we get $c=a+4$.
Thus, $2006 a-3344 b+1338 c$
$$
\begin{array}{l}
=2006 a-3344(a+1)+1338(a+4) \\
=2008 .
\end{array}
$$ | 2008 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. The number of simplest proper fractions with 2088 as the denominator is
$\qquad$.
| II.1.672.
Since $2088=2^{3} \times 3^{2} \times 29$, the number of integers from $1,2, \cdots$, 2088 that contain at least one of the factors $2,3,29$ is
$$
\begin{array}{l}
\left(\left[\frac{2088}{2}\right]+\left[\frac{2088}{3}\right]+\left[\frac{2008}{29}\right]\right)- \\
\left(\left[\frac{2088}{2 \times 3}\right]+\... | 672 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $x$, $m$, and $n$ are positive integers, $m+n=5$, and $x^{2}+m$ and $\left|x^{2}-n\right|$ are both prime numbers. Then the number of possible values of $x$ is $\qquad$ . | 3.2.
From the problem, $m$ can take the values $1,2,3,4$, and accordingly, $n$ can be $4,3,2,1$, with $m$ and $n$ being one odd and one even.
Thus, $x^{2}+m$ and $\left|x^{2}-n\right|$ are one odd and one even.
Since $x^{2}+m$ and $\left|x^{2}-n\right|$ are both prime numbers, it follows that $x^{2}+m=2$ or $\left|x^{... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) A toy factory has a total of 273 labor hours and 243 units of raw materials for producing a batch of teddy bears and kittens. Producing a teddy bear requires 9 labor hours and 12 units of raw materials, with a profit of 144 yuan; producing a kitten requires 6 labor hours and 3 units of raw materials, w... | Let the number of bears and cats produced be $x$ and $y$, respectively, with the total profit being $z$ yuan, requiring $m$ labor hours and $n$ units of raw materials. Then,
$$
\left\{\begin{array}{l}
m \leqslant 273, n \leqslant 243, \\
9 x+6 y=m, \\
12 x+3 y=n .
\end{array}\right.
$$
From equations (1) and (2), we g... | 3978 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) As shown in Figure 4, the incircle $\odot I$ of $\triangle ABC$ touches $AB$ and $AC$ at points $D$ and $E$, respectively. Extend $DI$ to $M$ and $EI$ to $N$ such that $IM = IN = 9IE$. Points $B$, $I$, $C$, $M$, and $N$ are concyclic. The ratio of the perimeter of $\triangle ABC$ to side $BC$ is the sim... | As shown in the figure,
10, let $A I$ intersect the circumcircle of $\triangle A B C$ at point $P, Q$ be the midpoint of side $B C$, and draw $P F \perp D M$ at point $F$, connecting $P C$. Then
$P Q \perp B C, C Q=\frac{1}{2} B C$,
Rt $\triangle A D I \sim$ Rt $\triangle P F I$,
Rt $\triangle C Q P \sim \mathrm{Rt} \t... | 2009 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
238 Test: Can 2007 be expressed in the form
$$
a_{1}^{x_{1}}+a_{2}^{x_{2}}+\cdots+a_{m}^{x_{n}}-b_{1}^{y_{1}}-b_{2}^{y_{2}}-\cdots-b_{n}^{y_{n}}
$$
where $m, n$ are both positive integers greater than 130 and less than 140 (allowing $m$ to equal $n$), $a_{1}, a_{2}, \cdots, a_{m}, b_{1}, b_{2}, \cdots, b_{n}$ are all ... | Solution: It can be done. The reasons are as follows:
Let $m=n=132$,
$$
\begin{array}{l}
a_{i}=\frac{16(i+2)^{2}+(2 i+3)^{2}}{(2 i+4)(2 i+3)}(i=1,2, \cdots, 125), \\
b_{i}=\frac{16(i+2)^{2}-(2 i+3)^{2}}{(2 i+4)(2 i+3)}(i=1,2, \cdots, 125), \\
a_{k}=\frac{(2 k+4)^{2}+(2 k+3)^{2}}{2(2 k+4)(2 k+3)}(k=126,127, \cdots, 132)... | 2007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $f(x)=a x+b(a, b$ be real numbers),
$$
\begin{array}{l}
f_{1}(x)=f(x), \\
f_{n+1}(x)=f\left(f_{n}(x)\right)(n=1,2, \cdots) .
\end{array}
$$
If $f_{7}(x)=128 x+381$, then $a+b=$ | Ni.7.5.
From the problem, we know
$$
\begin{array}{l}
f_{n}(x)=a^{n} x+\left(a^{n-1}+a^{n-2}+\cdots+a+1\right) b \\
=a^{n} x+\frac{a^{n}-1}{a-1} \cdot b .
\end{array}
$$
From $f_{7}(x)=128 x+381$, we get
$$
a^{7}=128, \frac{a^{7}-1}{a-1} \cdot b=381 \text {. }
$$
Therefore, $a=2, b=3, a+b=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. As shown in Figure $1, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$. | 15. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), and \( C(0, c) \), and assume \( b > c \).
The equation of line \( PB \) is \( y - b = \frac{y_{0} - b}{x_{0}} x \).
Simplifying, we get \( \left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0 \).
The distance from the circle center \((1,0)\) to \( PB \) is 1, so
\[ ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (20 points) Find the positive integer $n$ such that
$$
\left[\log _{3} 1\right]+\left[\log _{3} 2\right]+\cdots+\left[\log _{3} n\right]=2007 \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. | $$
\begin{array}{l}
{\left[\log _{3} 1\right]+\left[\log _{3} 2\right]=0,} \\
{\left[\log _{3} 3\right]+\left[\log _{3} 4\right]+\cdots+\left[\log _{3} 8\right]} \\
=6 \times 1=6, \\
{\left[\log _{3} 9\right]+\left[\log _{3} 10\right]+\cdots+\left[\log _{3} 26\right]} \\
=18 \times 2=36, \\
{\left[\log _{3} 27\right]+\... | 473 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. From $1,2, \cdots, 2008$, at least $\qquad$ even numbers must be taken to ensure that there are definitely two even numbers whose sum is 2012. | 2.504.
From $1,2, \cdots, 2008$ select two even numbers, the sum of which is 2012, there are a total of 501 pairs, that is
$$
4+2008,6+2006, \cdots, 1004+1008 \text {. }
$$
Since the sum of 2 or 1006 and any one of the even numbers above does not equal 2012, therefore, at least $501+2+$ $1=504$ even numbers must be t... | 504 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the function $y=f(x)\left(x, y \in \mathbf{N}_{+}\right)$ satisfies:
(1) For any $a, b \in \mathbf{N}_{+}, a \neq b$, it holds that $a f(a)+b f(b)>a f(b)+b f(a)$;
(2) For any $n \in \mathbf{N}_{+}$, it holds that $f(f(n))=3 n$. Then the value of $f(5)+f(12)$ is $(\quad$.
(A) 17
(B) 21
(C) 25
(D) 29 | 5.D.
For any $n \in \mathbf{N}_{+}$, from (1) we get
$$
\begin{array}{l}
(n+1) f(n+1)+n f(n) \\
>(n+1) f(n)+n f(n+1),
\end{array}
$$
which means $f(n+1)>f(n)$.
Therefore, $f(x)$ is a monotonically increasing function on $\mathbf{N}_{+}$.
For any $n \in \mathbf{N}_{+}$, from (2) we get
$$
f(3 n)=f(f(f(n)))=3 f(n) \tex... | 29 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. As shown in Table 1, in the use of computers, codes are arranged according to certain rules, and they are infinite from left to right and from top to bottom. In Table 1, the code 100 appears $\qquad$ times.
\begin{tabular}{|c|c|c|c|c|c|}
\hline 1 & 1 & 1 & 1 & 1 & $\cdots$ \\
\hline 1 & 2 & 3 & 4 & 5 & $\cdots$ \\
\... | From Table 1, we know that the $m$-th row is an arithmetic sequence with the first term 1 and common difference $m-1$. Therefore, the $n$-th number in the $m$-th row is
$$
\begin{array}{l}
a_{m n}=1+(n-1)(m-1) . \\
\text { Let } a_{m n}=100 \text {. Then } \\
(m-1)(n-1)=99=3^{2} \times 11 \text {. }
\end{array}
$$
The... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A positive integer $n$, if it does not contain any perfect square factor greater than 1, is called a "simple number". Then, among the 1000 numbers $1,2, \cdots, 1000$, the number of simple numbers is $\qquad$. | 6.508.
Notice that $1000<33^{2}$. If a number in the set $M=\{1,2, \cdots, 1000 \}$ is not a simple number, then it must contain one or more factors from the set
$$
A=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}, 19^{2}, 23^{2}, 29^{2}, 31^{2}\right\}.
$$
Let $P(a)$ denote the number of multiples of $a^{... | 508 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (40 points) Among nine visually identical gold coins, one weighs $a$, seven weigh $b$, and the last one weighs $c$, and $a < b < c$. Using a balance scale, find the minimum number of weighings required to identify the coin weighing $a$ and the coin weighing $c$.
To solve this problem, we need to devise a strategy t... | 9. Solution 1: First, label the nine coins as
$A, B, C, D, E, F, G, H, I$.
Weigh $(A, B, C, D)$ against $(E, F, G, H)$. If they balance, then $a+c=2b$. If they do not balance, assume without loss of generality that $(A, B, C, D) > (E, F, G, H)$.
For the second weighing, weigh $(A, B)$ against $(C, D)$ and $(E, F)$ aga... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
10. (40 points) Given that $n$ is a positive integer, such that
$$
1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}
$$
$=2^{k}$ ( $k$ is a positive integer).
Find the sum of all possible values of $n$. | 10. Since
$$
\begin{array}{l}
1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6} \\
=\frac{(n+1)\left(n^{2}-n+6\right)}{6},
\end{array}
$$
Therefore, $n+1$ is a power of 2 or 3 times a power of 2.
(1) If $n+1=2^{m}\left(m \in \mathbf{N}_{+}\right)$, then
$$
n^{2}-n+6=2^{2 m}-3 \times 2^{m}+8
$$
is 3 times a power of 2.
When ... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. The number of values of $n$ $\left(n \in \mathbf{N}_{+}\right)$ for which the equation $x^{2}-6 x-2^{n}=0$ has integer solutions is $\qquad$ | 2.1.
$x=3 \pm \sqrt{3^{2}+2^{n}}$, where $3^{2}+2^{n}$ is a perfect square. Clearly, $n \geqslant 2$.
When $n \geqslant 2$, we can assume
$$
2^{n}+3^{2}=(2 k+1)^{2}\left(k \in \mathbf{N}_{+}, k \geqslant 2\right),
$$
i.e., $2^{n-2}=(k+2)(k-1)$.
It is evident that $k-1=1, k=2, n=4$.
The number of values of $n$ that mak... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. If a number can be expressed as the sum of the digits of some multiple of 91, then this number is called a "harmonious number". Then, among $1,2, \cdots, 2008$, the number of harmonious numbers is $\qquad$ | 5.2007 .
Notice that $91=7 \times 13$.
Numbers with a digit sum of 1 are not multiples of 91.
$1001, 10101, 10011001, 101011001$,
$100110011001, 1010110011001, \cdots$
are all multiples of 91, and their digit sums are 2, 3, $4, 5, 6, 7, \cdots$ respectively. Therefore, among $1, 2, \cdots, 2008$, the number of numbers... | 2007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. It is known that the selling price of a certain model of car is 230,000 yuan per unit. A factory's total cost for producing this model of car in a year consists of fixed costs and production costs. The fixed cost for one year is 70,000,000 yuan. When producing $x$ units of this car in a year, the production cost for... | 6.318.
If the sales revenue of this type of car produced by the factory in a year is not less than the total cost, then
$$
\begin{array}{l}
23 x-\left[7000+\frac{70-\sqrt{x}}{3 \sqrt{x}} \cdot x\right] \geqslant 0 \\
\Rightarrow x-\sqrt{x}-300 \geqslant 0 \Rightarrow \sqrt{x} \geqslant \frac{1+\sqrt{1201}}{2} \\
\Righ... | 318 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Select 1004 numbers from $1,2, \cdots, 2008$ such that their total sum is 1009000, and the sum of any two of these 1004 numbers is not equal to 2009. Then the sum of the squares of these 1004 numbers is $\qquad$
Reference formula:
$$
1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1) \text {. }
$$ | 10.1351373940.
Divide $1,2, \cdots, 2008$ into 1004 groups:
$$
\{1,2008\},\{2,2007\}, \cdots,\{1004,1005\} \text {. }
$$
By the problem's setup, exactly one number is taken from each group. Replace 2, 4, $\cdots, 2008$ with 1004, 1006, 1008, 1010 by their counterparts in the same group, 1005, 1003, 1001, 999, respect... | 1351373940 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Divide $1,2, \cdots, 9$ into three groups, each containing three numbers, such that the sum of the numbers in each group is a prime number.
(1) Prove that there must be two groups with equal sums;
(2) Find the number of all different ways to divide them.
| Three, (1) Since the sum of three different numbers in $1,2, \cdots, 9$ is between 6 and 24, the prime numbers among them are only 7, 11, 13, 17, 19, 23, these six. Now, these six numbers are divided into two categories based on their remainders when divided by 3:
$A=\{7,13,19\}$, where each number leaves a remainder o... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a$, $b$, $c$ satisfy $(a+b)(b+c)(c+a)=0$ and $abc<0$. Then the value of the algebraic expression $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}$ is | $=.1 .1$.
From $(a+b)(b+c)(c+a)=0$, we know that at least two of $a, b, c$ are opposite numbers. Also, since $abc<0$, it follows that among $a, b, c$, there must be two positive and one negative.
Thus, the value of $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}$ is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In Rt $\triangle A B C$, $F$ is the midpoint of the hypotenuse $A B$, and $D, E$ are points on sides $C A, C B$ respectively, such that $\angle D F E=$ $90^{\circ}$. If $A D=3, B E=4$, then the length of segment $D E$ is $\qquad$ | 3.5 .
As shown in Figure 6, extend $D F$ to point $G$ such that $D F = F G$, and connect $G B$ and $G E$.
Given $A F = F B$, we have
$\triangle A D F \cong \triangle B G F$
$\Rightarrow B G = A D = 3$
$\Rightarrow \angle A D F = \angle B G F$
$\Rightarrow A D \parallel G B$
$\Rightarrow \angle G B E + \angle A C B = 1... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Divide the positive integers $1,2, \cdots, 10$ into two groups, $A$ and $B$, where group $A$: $a_{1}, a_{2}, \cdots, a_{m}$; group $B$: $b_{1}, b_{2}, \cdots$, $b_{n}$. Now, take one number from each of the groups $A$ and $B$, and multiply the two numbers taken. Then the maximum value of the sum of all different pro... | 4.756 .
From the condition, the sum of all different products of two numbers is $S=\left(a_{1}+\cdots+a_{m}\right)\left(b_{1}+\cdots+b_{n}\right)$.
$$
\begin{array}{l}
\text { Let } x=a_{1}+\cdots+a_{m}, y=b_{1}+\cdots+b_{n} . \text { Then } \\
x+y=1+2+\cdots+10=55, \\
S=x y=\frac{1}{4}\left[(x+y)^{2}-(x-y)^{2}\right]... | 756 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (25 points) As shown in Figure 2, in the right trapezoid $ABCD$, $\angle ABC = \angle BAD = 90^{\circ}$, $AB = 16$. The diagonals $AC$ and $BD$ intersect at point $E$. A line $EF \perp AB$ is drawn through $E$ at point $F$, and $O$ is the midpoint of side $AB$, with $FE + EO = 8$. Find the value of $AD + BC$. | Let $O F=x$.
Then $F B=8-x, F A=8+x$.
Given that $D A / / E F / / C B$, we have $\frac{F E}{A D}=\frac{F B}{A B}$, which means
$$
A D=\frac{16}{8-x} E F \text {. }
$$
Similarly, $B C=\frac{16}{8+x} E F$.
Thus, $A D+B C=\left(\frac{16}{8-x}+\frac{16}{8+x}\right) E F$
$$
=\frac{16 \times 16}{8^{2}-x^{2}} E F \text {. }
... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (25 points) In the expression “ $\square 1 \square 2 \square 3 \square 4 \square 5 \square 6 \square 7$ $\square 8 \square 9$ ”, fill in the small squares with “+” or “-” signs. If the algebraic sum can be $n$, then the number $n$ is called a "representable number"; otherwise, it is called an "unrepresentable num... | (1) Since $+1-2-3+4+5-6+7-8$ $+9=7$, therefore, 7 is a number that can be represented.
$$
\text { Also, }+1+2+3+4+5+6+7+8+9=45
$$
is an odd number, and for any two integers $a$ and $b$, $a+b$ and $a-b$ have the same parity. Therefore, no matter how the “ $+\cdots \times$ - ” signs are filled, the algebraic sum must be ... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Given $a-b=1, a^{2}-b^{2}=-1$. Then $a^{2008}-b^{2008}=$ | 12. -1 .
Given $a^{2}-b^{2}=(a+b)(a-b)=-1$, and $a-b=1$, then $a+b=-1$.
Therefore, $\left\{\begin{array}{l}a+b=-1 \\ a-b=1 .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}a=0, \\ b=-1 .\end{array}\right.$ Hence, $a^{2 \alpha R}-b^{20 R}=0^{2008}-(-1)^{2008}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Given that $a$ and $b$ are real numbers, and $ab=1, a \neq 1$, let $M=\frac{a}{a+1}+\frac{b}{b+1}, N=\frac{1}{a+1}+\frac{1}{b+1}$. Then the value of $M-N$ is $\qquad$. | 15.0.
Given $a b=1, a \neq 1$, so,
$$
\begin{array}{l}
M=\frac{a}{a+1}+\frac{b}{b+1}=\frac{a}{a+a b}+\frac{b}{b+a b} \\
=\frac{a}{a(1+b)}+\frac{b}{b(1+a)}=\frac{1}{a+1}+\frac{1}{b+1}=N .
\end{array}
$$
Thus, $M-N=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $x\left(y-\frac{1}{x}\right)=2008$, where $x$ and $y$ are positive integers. Then the sum of the maximum and minimum values of $x+y$ is ( ).
(A) 2010
(B) 2100
(C) 2008
(D) 2000 | -.1.B.
From the given, we have
$$
\begin{array}{l}
2008=x\left(y-\frac{1}{x}\right) \\
\Rightarrow x y=2009=1 \times 2009=7 \times 7 \times 41 .
\end{array}
$$
Thus, $(x+y)_{\text {max }}=1+2009=2010$,
$$
(x+y)_{\text {min }}=49+41=90 \text {. }
$$
Therefore, the sum of the maximum and minimum values of $x+y$ is 2100... | 2100 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then
$$
\begin{array}{l}
{\left[\frac{2009 \times 83}{2009}\right]+\left[\frac{2010 \times 83}{2009}\right]+\cdots+\left[\frac{4017 \times 83}{2009}\right]} \\
=(\quad) .
\end{array}
$$
(A) 249075
(B) 250958
(C) 174696
(D) 25... | 2.A.
Original expression
$$
\begin{aligned}
= & {\left[\frac{(2009+0) \times 83}{2009}\right]+\left[\frac{(2009+1) \times 83}{2009}\right]+} \\
& \cdots+\left[\frac{(2009+2008) \times 83}{2009}\right] \\
= & 83+\left[\frac{0 \times 83}{2009}\right]+83+\left[\frac{1 \times 83}{2009}\right]+\cdots+ \\
& 83+\left[\frac{2... | 249075 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
2. Let $a$ and $b$ be integers, and the equation
$$
a x^{2}+b x+1=0
$$
has two distinct positive roots both less than 1. Then the minimum value of $a$ is | 2. 5 .
Let the two roots of the equation be $x_{1}, x_{2}$.
From $x_{1} x_{2}=\frac{1}{a}>0$, we know $a>0$.
Also, $f(0)=1>0$, so according to the problem, we have
$$
\left\{\begin{array}{l}
\Delta=b^{2}-4 a>0, \\
00 .
\end{array}\right.
$$
Since $a$ is a positive integer, from equations (2) and (3), we get
$-(a+1)<b... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) There are 2009 athletes, numbered $1, 2, \cdots, 2009$. Some of them are selected to participate in the honor guard, but it must be ensured that among the remaining athletes, no one's number is equal to the product of the numbers of any two other people. How many athletes must be selected for the hon... | Three, since $45^{2}=2025>2009$, if both people's numbers are greater than or equal to 45, then their product is greater than 2009 and cannot be another person's number. If one of the two numbers is 1, then the product of 1 and the other number equals that number, which does not meet the requirement. Therefore, if the ... | 43 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a, b, c, d, e$ be distinct positive odd numbers. If the equation
$$
(x-a)(x-b)(x-c)(x-d)(x-e)=2009
$$
has an integer root $x$, then the last digit of $a+b+c+d+e$ is ( ).
(A) 1
(B) 3
(C) 7
(D) 9 | 5.I).
For any integer $x, x-a, x-b, x-c, x-$ $d, x-e$ are $I$ distinct integers. And expressing 2009 as the product of 7 $I$ distinct integers has only one unique form:
$$
2009=1 \times(-1) \times 7 \times(-7) \times 41.
$$
From this, $x$ is even.
Let $x=2 m$. Then
$$
\begin{array}{l}
\{2 m-a, 2 m-b, 2 m-c, 2 m-d, 2 ... | 9 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given $S=1^{2}-2^{2}+3^{2}-4^{2}+\cdots-100^{2}+$ $101^{2}$. Then the remainder when $S$ is divided by 103 is | $=.1 .1$.
Notice that
$$
\begin{array}{l}
S=1+\left(3^{2}-2^{2}\right)+\left(5^{2}-4^{2}\right)+\cdots+\left(101^{2}-100^{2}\right) \\
=1+2+3+\cdots+100+101 \\
=\frac{101 \times 102}{2}=5151=50 \times 103+1 .
\end{array}
$$
Therefore, the required remainder is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$ | 2. -1 .
Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$. Since $\triangle ABC$ is a right triangle, it is known that $x_{1}$ and $x_{2}$ must have opposite signs, so $x_{1} x_{2}=\frac{c}{a}<0$. By the projection theorem, we know that $|O C|^{2}=|A O| \cdot|B O|$, which means $c^{2}=\left|x_{1}\right| \cdot\... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y, z$ satisfy
$$
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \leqslant 60 \text {. }
$$
Then the value of $x+y-z$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | -.1.D.
From the given equation, we have
$$
\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \leqslant 60 \text {. }
$$
Therefore, $x=-2, y=5, z=3$.
Thus, $x+y-z=-2+5-3=0$. | 0 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Given that $m, n$ can each take one of $1,2, \cdots, 2009$. Then the number of pairs $(m, n)$ that make the equation $x^{2}-m x+n=0$ have real roots is ( ).
(A)676 211570
(B)676 211571
(C)676 211572
(D)676211578 | 2.A.
From the problem, we have
$$
\Delta=(-m)^{2}-4 n \geqslant 0 \text {, }
$$
which means $n \leqslant \frac{1}{4} m^{2}$.
(1) If $m$ is odd, let $m=2 k-1(k=1,2$, $\cdots, 1005$ ). Then
$$
n \leqslant \frac{1}{4}(2 k-1)^{2}=k^{2}-k+\frac{1}{4} \text {. }
$$
Thus, $n$ can take $1,2, \cdots, k^{2}-k$.
So $n$ can tak... | 676211570 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Given integers $a, b, c, d$ satisfy
$$
27\left(3^{a}+3^{b}+3^{c}+3^{d}\right)=6888 \text {. }
$$
Then the value of $a+b+c+d$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 4.C.
Assume $a \leqslant b \leqslant c \leqslant d$. From the given equation, we have
$$
3^{a+3}\left(1+3^{b-a}+3^{c-a}+3^{d-a}\right)=3 \times 2296 \text {. }
$$
(1) If $3^{a+3}=1$, i.e., $a=-3$, then
$$
31\left(1+3^{b+3}+3^{c+3}+3^{d+3}\right)=3 \times 2296 \text {. }
$$
Since $3^{b+3} \leqslant 3^{c+3} \leqslant 3... | 6 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. When $x=\frac{\sqrt{21}-5}{2}$, the value of the algebraic expression
$$
x(x+1)(x+2)(x+3)(x+4)(x+5)
$$
is $\qquad$ . | $$
\text { II. 1. }-15 \text {. }
$$
From $2 x+5=\sqrt{2 \mathrm{i}}$, after rearrangement, we get
$$
\begin{array}{l}
x^{2}+5 x+1=0 . \\
\text { Therefore, } x(x+5)=-1 . \\
\text { Also, } (x+1)(x+4)=x^{2}+5 x+4=3, \\
(x+2)(x+3)=x^{2}+5 x+6=5 .
\end{array}
$$
Thus, multiplying the equations, the value sought is -15. | -15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.