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10. 1 Given? three quadratic trinomials; $P_{1}(x)=x^{2}+p_{1} x$ $+q_{1}, P_{2}(x)=x^{2}-p_{2} x+q_{2}$ and $P_{3}(x)=x^{2}+p_{3} x+q_{3}$. Prove: The equation $\left|P_{1}(x)\right|+\left|P_{2}(x)\right|=\left|P_{3}(x)\right|$ has at most 8 real roots.
|
10.1 Each root of the original equation should be a root of a quadratic trinomial of the form $\pm p_{1}: p_{2} \pm p_{3}$, and with different choices of signs, there are 8 such quadratic trinomials. Since the coefficient of $x^{2}$ has the form $\pm 1 \pm 1 \pm 1$, the coefficient of $x^{2}$ will never be 0 regardless of how the signs are chosen. Therefore, all 8 are genuine quadratic trinomials. However, the roots of two quadratic trinomials with coefficients that are opposite in sign are the same, so the roots of the equation $\left|P_{1}(x)\right|+\mid P_{2}(x) \mid = | P_{3}(x) \mid$ are included among the roots of 4 different quadratic equations. Thus, it has no more than 8 roots.
|
8
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a$ be the decimal part of $\sqrt{3}$, $b$ be the decimal part of $\sqrt{2}$: $\frac{a}{(a-b) b}$ has an integer part of $\qquad$
|
$$
\begin{array}{l}
\text { 3. } \frac{a}{(a-b) b}=\frac{\sqrt{3}-1}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{(\sqrt{3}-\sqrt{2})+(\sqrt{2}-1)}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}-\sqrt{2}} \\
=\sqrt{3}+2 \sqrt{2}+1 \\
\approx 5.56 .
\end{array}
$$
The integer part is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $p, q, \frac{2 p-1}{q}, \frac{2 q-1}{p}$ all be integers, and $p>1, q>$
1. Then $p+q=$ $\qquad$ .
|
6. $p+q=8$.
From $\frac{2 q-1}{p}, \frac{2 p-1}{q}$ both being positive integers, we know that one of them must be less than 2. Otherwise, $\frac{2 q-1}{p}>2, \frac{2 p-1}{q}>2$, leading to $2 p-1+2 q-1>2 p+2 q$, which is a contradiction. Assume $0<\frac{2 q-1}{p}<2$, then it must be that $2 q-1=p$. It is easy to see that $q=3, p=5$, thus $p+q=8$.
Note: The original problem statement has a typo in the final result, which should be $p+q=8$ instead of $p+q=3$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. As shown in the figure, in pentagon $A B C D E$, $\angle B=\angle A E D=$ $90^{\circ}, A B=C D=A E=B C+D E=1$. Find the area of this pentagon. (1992, Beijing Junior High School Grade 2 Mathematics Competition)
|
Analysis We immediately think of connecting $A C, A D$, because there are two right-angled triangles. But we also find that it is not easy to directly calculate the area of each triangle. Given the condition $B C+D E=1$, we wonder if we can join $B C$ to one end of $D E$, making $E F=B C$. Connect $A F$. Then we find $\triangle A B C \cong$ $\triangle A E F$. So $F D=1$, and $=A C \cdot \triangle E=A B, \triangle A D F$ are triangles with bases of 1, each with an area of $\frac{1}{2}$, and $\triangle A C D$ is congruent to $\triangle A D F$, thus the total area is 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. As shown in the figure, let $D$ and $E$ be on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BD$ and $CE$ intersect at $F$, $AE = EB$, $\frac{AD}{DC} = \frac{2}{3}$, and $S_{\triangle ABC} = 40$. Find $S_{\text{quadrilateral AEFD}}$. (6th National Junior High School Mathematics Correspondence Competition)
|
Analyzing, first connect $A F$ and then set:
$$
S_{\triangle A E P}=x, S_{\triangle A P D}=y, S_{\triangle C D F}=z, S_{\triangle B E P}=t\left(S_{\triangle B C F}\right.
$$
$=u$ ), to find $S_{\text {quadrilateral } A E F D}$, we only need to find the values of $x$ and $y$.
We easily get:
$$
\begin{array}{l}
x+y+z=20, \\
x+y+t=40 \times \frac{2}{5}=16, \\
x=t, \\
\frac{y}{z}=\frac{2}{3} .
\end{array}
$$
Solving these, $x=5, y=6$.
Thus, $S_{\text {quadrilateral } A E F D}=11$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ that satisfy: (1) $\left|A_{i}\right|=5, i=1,2, \cdots, k_{i}$ (2) $\mid A_{i} \cap$ $A_{j} \mid \leqslant 2, i, j=1,2, \cdots, k, i \neq j$. Find the maximum value of $k$. (6th test, 1st question, provided by Wu Chang)
|
First, prove that each element in $S$ belongs to at most 3 of the sets $A_{1}, A_{2}, \cdots, A_{k}$.
By contradiction, suppose there is an element in $S$ that belongs to at least 4 subsets. Without loss of generality, assume $1 \in A_{1}, A_{2}, A_{3}, A_{4}$.
(1) If each of the other elements in $S$ belongs to at most 2 of the sets $A_{1}, A_{2}, A_{3}, A_{4}$, each $A_{i} (i=1,2,3,4)$ has 5 positive integers. Removing 1, there are 4 positive integers left. Thus, excluding 1, $A_{1}, A_{2}, A_{3}, A_{4}$ should have $4 \times 4 = 16$ positive integers, some of which are repeated. However, $S - \{1\}$ consists of 9 different positive integers, so there should be $16 - 9 = 7$ repeated positive integers. Given the initial assumption, 7 positive integers belong to exactly 2 of these 4 sets. Therefore, using the problem's condition, we have
$$
2 C_{4}^{2} \geqslant \sum_{1<i<j<4}\left|A_{i} \cap A_{j}\right| \geqslant C_{4}^{2} + 7,
$$
but $2 C_{4}^{2} = 12$ and $C_{4}^{2} + 7 = 13$, which is a contradiction.
(2) If there is another element in $S$ that belongs to at least 3 of the sets $A_{1}, A_{2}, A_{3}, A_{4}$, without loss of generality, assume $2 \in A_{1} \cap A_{2} \cap A_{3}$. Since each of $A_{1}, A_{2}, A_{3}$ has 5 positive integers, removing 1 and 2, each $A_{i} (i=1,2,3)$ has 3 positive integers left. Thus, excluding 1 and 2, $A_{1}, A_{2}, A_{3}$ should have $3 \times 3 = 9$ positive integers, but $S - \{1, 2\}$ consists of 8 different positive integers. Therefore, $9 - 8 = 1$, meaning there is at least one positive integer, which is neither 1 nor 2, belonging to 2 of the sets $A_{1}, A_{2}, A_{3}$. Thus, we have
$$
6 = 2 C_{3}^{2} \geqslant \sum_{1<i<j<3}\left|A_{i} \cap A_{j}\right| \geqslant 2 \times 3 + 1 = 7,
$$
which is a contradiction.
Thus, $k \leqslant \frac{3 \times 10}{5} = 6$.
On the other hand, let
$$
\begin{array}{l}
A_{1}=\{1,2,3,4,5\}, \quad A_{2}=\{1,2,6,7,10\} \\
A_{3}=\{1,3,8,9,10\}, \quad A_{4}=\{2,4,5,8,9\} \\
A_{5}=\{3,5,6,7,8\}, \quad A_{6}=\{4,5,7,9,10\}.
\end{array}
$$
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ .
|
3. 1.
The formula yields
$$
\begin{aligned}
y & =(x-1)^{2}+x+\frac{1}{x}-1 \\
& =(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1 .
\end{aligned}
$$
When $x=1$, both $(x-1)^{2}$ and $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}$ simultaneously take the minimum value of 0, so the minimum value of $y=x^{2}-x+\frac{1}{x}$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The "Monkey Cycling" performance in the circus is carried out by 5 monkeys using 5 bicycles, with each monkey riding at least once, but no monkey can ride the same bicycle more than once. After the performance, the 5 monkeys rode the bicycles $2, 2, 3, 5, x$ times respectively, and the 5 bicycles were ridden $1, 1, 2, 4, y$ times respectively. Therefore, $x+y$ equals ( ).
(A) 5
(B) 6
(C) 7
(D) 8
|
,- 1 (B).
Every time the monkey rides a bike, the monkey and the bike each get one count, so the number of times the monkey rides a bike = the number of times the bike is ridden, i.e., $2+2+3+5+x=1+1+2+4+y$. Simplifying, we get $x+4=y$. From $x \geqslant 1$, we get $y \geqslant 5$. Also, from the problem, $y \leqslant 5$, so $y=5, x=1$. Therefore, $x+y=6$.
|
6
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. The ice numbers $a_{1}, u_{2}, \cdots, a_{9}$ can only take two different values, +1 or -1. Then, the maximum value of the expression $a_{1} a_{5} a_{9}-a_{1} a_{6} a_{8}+a_{2} a_{8} a_{7}-$ $a_{2} a_{4} a_{9}+a_{3} a_{4} a_{8}-a_{3} a_{5} a_{7}$ is $\qquad$.
|
4. 4 .
Since each term in the sum can only be 11 or -1, the sum is even (the parity of $a-b$ and $a+b$ is the same), so the maximum value of the sum is at most 6. It is easy to see that the sum cannot be 6, because in this case $a_{1} a_{5} a_{9}, a_{2} a_{6} a_{7}, a_{3} a_{6} a_{8}$ should all be +1, making their product equal to +1. However, $a_{1} a_{8} a_{8}, a_{2} a_{4} a_{9}, a_{3} a_{5} a_{8}$ should all be -1, making their product equal to -1.
But in fact, these two products should be equal, so the sum cannot be 6. However, the sum can be 4. For example, when $a_{1}=a_{2}=a_{3}=a_{5}=a_{8}=a_{0}=1, a_{4}=a_{6}=a_{7}=-1$, the sum is equal to 4. In conclusion, the maximum value of the sum is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $f(x)=a x^{2}+b x+c(a, b, c \in R, a \neq 0)$. If for $|x| \leqslant 1$, $|f(x)| \leqslant 1$, then for $|x| \leqslant 1$, the maximum value of $|2 a x+b|$ is $\qquad$
|
4. 4 .
Given $f(0)=c, f(1)=a+b+c, f(-1)=a-b+c$,
we solve to get $a=\frac{f(1)+f(-1)-2 f(0)}{2}, b=\frac{f(1)-f(-1)}{2}$.
For $|x| \leqslant 1$, $|f(x)| \leqslant 1$, we have
$$
\begin{array}{l}
|2 a x+b| \\
=\left|[f(1)+f(-1)-2 f(0)] x\right. \\
\left.+\frac{f(1)-f(-1)}{2} \right| \\
=\left|\left(x+\frac{1}{2}\right) f(1)+\left(x-\frac{1}{2}\right) f(-1)-2 x f(0)\right| \\
\leqslant\left|x+\frac{1}{2}\right||f(1)|+\left|x-\frac{1}{2}\right||f(-1)| \\
\quad \quad+|2 x||f(0)| \\
\leqslant\left|x+\frac{1}{2}\right|+\left|x-\frac{1}{2}\right|+2 \leqslant 4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(4) (Total 20 points) In the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $M$ is a point on $AB$, and $AM^2 + BM^2 + CM^2 = 2AM + 2BM + 2CM - 3$. If $P$ is a moving point on the segment $AC$, and $\odot O$ is the circle passing through points $P, M, C$, and a line through $P$ parallel to $AB$ intersects $\odot O$ at point $D$.
(1) (8 points) Prove that $M$ is the midpoint of $AB$;
(2) (12 points) Find the length of $PD$.
|
(1) From the given, we have
$$
(A M-1)^{2}+(B M-1)^{2}+(C M-1)^{2}=0,
$$
which implies $A M=B M=C M=1$, meaning $M$ is the midpoint of $A B$.
(2) From $\left\{\begin{array}{l}M A=M C \Rightarrow \angle A=\angle M C A, \\ P D / / A B: \angle A=\angle C P D\end{array}\right.$
$\Rightarrow \angle M C A=\angle C P D$, thus $\overparen{P} M=\overparen{C D}$, and $P C / / M D$. Since $A P \| A$, then $A P D M$ is a parallelogram, hence $P D=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In the table below, the average of any three adjacent small squares in the upper row is $1(x \neq 0)$, and the square of any four adjacent small squares in the lower row is also 1. Then the value of $\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}$ is $ـ$. $\qquad$
|
ii. 3 .
From the problem, we know that in the above row, the numbers in every two small squares separated by one are equal, thus we can deduce
$$
\frac{x+7-y}{3}=1 \text {. }
$$
Similarly, for the row below, we get
$$
\begin{array}{l}
\frac{z-x+9.5-1.5}{4}=1 . \\
\begin{array}{l}
\therefore y=x+4, z=x-4, \text { then } \\
\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}=\frac{x^{2}+\cdot(x+4)^{2}+(x-4)^{2}-32}{(x+4)(x-4)+16} \\
=\frac{3 x^{2}}{x^{2}}=3 .
\end{array}
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In a tennis tournament, $n$ women and $2 n$ men participate, and each player plays against all other players exactly once. If there are no ties, the ratio of the number of games won by women to the number of games won by men is 7:5. Then $n=$
|
6. $n=3$.
Let $k$ represent the number of times the woman wins over Luo Ziyue.
Here, we set $\frac{k+\mathrm{C}_{0}^{2}}{\mathrm{C}_{\mathrm{jn}}^{2}}=\frac{7}{12}, k=\frac{7}{12} \mathrm{C}_{3 n}-\mathrm{C}_{n}^{2}$. From $k \leq 2 n^{2}$, we can solve to get $n \leq 3$.
When $n=1$, we get $k=\frac{7}{4}$; when $n=2$, $k=\frac{31}{4}$; when $n=3$, $k=18$. Thus, $n=3$.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at $D$, $P$ is the midpoint of $AD$, $BP$ intersects $AC$ at $E$, $EF \perp BC$ at $F$. Given $AE=3$, $EC=$ 12. Find the length of $EF$.
|
Analysis: Consider $\triangle A B C$ as its machine triangle, given that $E$, $P$ are the fixed ratio points of $A C, A D$ respectively, hence from (*) we get $\left(1+\frac{C D}{D B}\right) \cdot \frac{3}{12}=1$, i.e., $\frac{C D}{D B}=3$.
From the projection theorem, we can find $B D=\frac{5}{2} \sqrt{3}, D C$ $=\frac{15}{2} \sqrt{3}$, thus $A D=\frac{15}{2}$.
By $\triangle C A D C \sim \triangle C E F$ we get $E F=6$.
Using the above basic figure and (*) formula, prove the following.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Among the 35 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 35^{2}$, the numbers with an odd digit in the tens place are $\qquad$ in total.
|
3. 7. Among $1^{2}, 2^{2}, \cdots, 9^{2}$, those with an odd tens digit are only $4^{2}=16, 6^{2}=36$.
The square of a two-digit number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2} \text {. }
$$
It is evident that, when the unit digit is $A$ and $\hat{C}$, the tens digit of the square is odd. That is, only $4^{2}, 6^{2}, 14^{2}, 16^{2}, 24^{2}, 26^{2}, 34^{2}$ have
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\frac{a}{a^{3}+a^{2} b+6 b^{2}+b^{3}}+\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}} \\
+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$
|
\[
\begin{aligned}
=、 \text { Original expression } & =\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\
& +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\
= & \frac{a^{2}+b^{2} .}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\
& =0 .
\end{aligned}
\]
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the equation $2 x^{2}+2 k x-13 k+1=0$ has two real roots whose squares sum to 13. Then $k=$ $\qquad$ .
|
1.1. By the relationship between roots and coefficients, we solve to get $k=1$ or $k=$ -14. Substituting these values back into the original equation, we find that $k=-14$ does not satisfy the conditions, so it is discarded; $k=1$ satisfies the conditions.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the area of $\triangle A B C$ is $1, D$ is the midpoint of $B C, E$, $F$ are on $A C, A B$ respectively, and $S_{\triangle B D P}=\frac{1}{5}, S_{\triangle C D E}=\frac{1}{3}$. Then $S_{\triangle D S P}=$ $\qquad$
|
(7)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. You asked me to translate a text, but the text you provided is just a number in parentheses. If you need a translation, please provide the actual text you want translated.
For the given input, the translation would simply be:
(7)
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (18 points) In a round-robin football tournament with $n$ teams (i.e., each team must play a match against every other team), each match awards 2 points to the winning team, 1 point to each team in the case of a draw, and 0 points to the losing team. The result is that one team has more points than any other team, yet has fewer wins than any other team. Find the smallest possible value of $n$.
|
Let the team with the highest score be Team $A$, and assume Team $A$ wins $k$ games, draws $m$ games, and thus Team $A$'s total points are $2k + m$.
From the given conditions, every other team must win at least $k+1$ games, and their points must be no less than $2(k+1)$. Therefore, we have $2k + m > 2(k+1)$, which simplifies to $m > 2$. Hence, $m \geqslant 3$.
Thus, there is a team that draws with Team $A$, and this team's score is no less than $2(k+1) + 1$. Therefore,
$$
2k + m > 2(k+1) + 1, \quad m > 3.
$$
This implies $m \geqslant 4$.
From this, we derive $n \geqslant m + 1 = 5$.
If $n = 5$, then $m = 4$, $k = 0$, and Team $A$ scores 4 points. However, in this case, 5 teams play a total of 10 games, and the total points of all teams sum up to 20 points. According to the problem, Team $A$'s score $>\frac{20}{5} = 4$, leading to a contradiction.
Therefore, $n \geqslant 6$.
When $n = 6$, there exists a match that meets the conditions (see the table below), so the minimum possible value of $n$ is 6.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. On the diagonal $BD$ of square $ABCD$, take two points $E$ and $F$, such that the extension of $AE$ intersects side $BC$ at point $M$, and the extension of $AF$ intersects side $CD$ at point $N$, with $CM = CN$. If $BE = 3$, $EF = 4$, what is the length of the diagonal of this square?
|
3. From the condition, we get
$\triangle A^{\prime} B M I \triangle \triangle A C M$.
$\therefore \angle B E=\angle D A E$.
Also, $A B=A D$,
$\angle A B E=\angle A D F$,
$\therefore \triangle A B E \simeq \triangle A D F$.
Therefore, $D F=B E=3$.
$\therefore A C=B D=10$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3. 3 .
From $[\lg x] \leqslant \lg x$, we get $\lg ^{2} x-\lg x-2 \leqslant 0$, which means $-1 \leqslant \lg x \leqslant 2$.
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$. Substituting into the original equation, we get $\lg x= \pm 1$, but $\lg x=1$ is not valid, so $\lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \leqslant \lg x<1$, we have $[\lg x]=0$, substituting into the original equation, we get $\lg x$ $= \pm \sqrt{2}$, both of which are not valid.
When $1 \leqslant \lg x<2$, we have $[\lg x]=1$, substituting into the original equation, we get $\lg x$ $= \pm \sqrt{3}$. But $\lg x=-\sqrt{3}$ is not valid, so $\lg x=\sqrt{3}, x_{2}$ $=10 \sqrt{3}$.
When $\lg x=2$, we get $x_{3}=100$.
Therefore, the original equation has
3 real roots.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. 21 people participate in an exam, with the test paper containing 15 true/false questions. It is known that for any two people, there is at least one question that both of them answered correctly. What is the minimum number of people who answered the most questions correctly? Please explain your reasoning.
|
For the $i$-th problem, $a_{i}$ people answered correctly, thus there are exactly
$$
b_{i}=C_{s_{i}}^{z}
$$
pairs of people who answered the $i$-th problem correctly $(i=1,2, \cdots, 15)$.
Below, we focus on the sum $\sum_{i=1}^{15} b_{i}$.
Let $a=\max \left\{a_{1}, a_{2}, \cdots, a_{15}\right\}$, then we have
$$
15 C_{a}^{2} \geqslant \sum_{i=1}^{15} b_{i} \geqslant C_{21}^{2}, \quad a(a-1) \geqslant \frac{420}{15}=28,
$$
$a \geqslant 6$.
We will show that $a=6$ is impossible, hence, the problem with the most correct answers must have at least 7 people answering correctly.
Assume $a=6$, meaning no more than 6 people answered any question correctly. We will derive a contradiction. If someone answered only three questions correctly, then for each of these questions, they could only share the correct answer with at most 5 other people, meaning they could share correct answers with at most 15 other people, which contradicts the assumption. Therefore, each person must answer at least four questions correctly.
Since $21 \times 5 > 6 \times 15$,
it is impossible for everyone to answer five or more questions correctly, so there must be at least one person who answers exactly four questions correctly, and these four questions form four sets that have no common members except for this person, for example:
$$
\begin{array}{l}
S_{1}=\{1,2,3,4,5,6\}, \\
S_{2}=\{1,7,8, \cdots, 11\}, \\
S_{3}=\{1,12,13, \cdots, 16\}, \\
S_{4}=\{1,17,18, \cdots, 21\},
\end{array}
$$
On the other hand, among all 15 questions, at least 12 questions must have 6 people answering correctly, otherwise it would lead to a contradiction:
$$
C_{21}^{2} \leqslant \sum_{i=1}^{15} b_{i} \leqslant 11 C_{6}^{2}+4 C_{5}^{2}=205 .
$$
Excluding the aforementioned four questions, there are still eight questions with 6 people answering correctly. Consider the 6 people who answered one of these eight questions correctly. Among them, either more than three people belong to the same $S_{j}$, or two people belong to some $S_{j}$, and two others belong to $S_{k}(j, k \in\{1,2,3,4\}, j \neq k)$. In any case, when calculating $\sum_{i=1}^{15} b_{i}$, each of these eight questions will result in at least two duplicate counts. This leads to
$$
C_{21}^{2} \leqslant \sum_{i=1}^{15} b_{i}-8 \times 2 \leqslant 15 C_{6}^{2}-16=209 .
$$
The resulting contradiction shows that the case $a=6$ is impossible.
Based on the above discussion, we confirm that $a \geqslant 7$.
Next, we construct an example to show that the case $a=7$ is possible. For this, we first number the participants from 1 to 21 and define the following sets of participants:
$$
\begin{array}{l}
P_{i}=\{2 i-1,2 i\}, i=1, \cdots, 63 \\
P_{1}=\{13,14,15\}, \\
P_{8}=\{16,17,18\}, \\
P_{0}=\{19,20,21\},
\end{array}
$$
Using these notations, we construct the following table to indicate the participants who answered each question correctly. It is easy to verify that in the scenario shown in the table, every pair of the 21 participants has at least one question they both answered correctly, and no more than 7 people answered the same question correctly.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $S=\left\{A=\left(a_{1}, \cdots, a_{8}\right) \mid a_{i}=0\right.$ or $1, i=1, \cdots$, 8\}. For two elements $A=\left(a_{1}, \cdots, a_{8}\right)$ and $B=\left(b_{1}\right.$, $\cdots, b_{8}$ ) in $S$, denote
$$
d(A, B)=\sum_{i=1}^{\delta}\left|a_{i}-b_{i}\right|,
$$
and call it the distance between $A$ and $B$. What is the minimum number of elements that can be selected from $S$ such that the distance between any two of them is $\geqslant 5$?
|
Solution One
(I) First, we point out that the sum of the weights of any two codewords in $\mathscr{D}$ does not exceed 11; otherwise, if
$$
\omega(X)+\omega(Y) \geqslant 12,
$$
then because $12-8=4$, these two codewords must share at least four positions with 1s, and the distance between them
$$
a^{\prime}\left(X, y^{\prime}\right) \leq 8-4=4.
$$
Therefore, $\mathscr{D}$ can contain at most one codeword with a weight $\geqslant 6$. Furthermore, if $\mathscr{D}$ contains a codeword with a weight $\geqslant 7$, then $\mathscr{D}$ can only contain that codeword and the all-zero codeword.
(I) Noting that $5+5-8=2$, we conclude: if $\mathscr{D}$ contains two codewords $A$ and $B$ with a weight of 5, then these two codewords must share at least two positions with 1s. We can further conclude: these two codewords share exactly two positions with 1s. Otherwise, if they share three positions with 1s, then they must also share at least one position with 0s, and the distance between the two codewords
$$
d(A, B) \leqslant 4.
$$
Based on the above discussion, we can conclude: $\mathscr{D}$ can contain at most two codewords with a weight of 5 (otherwise, the third codeword with a weight of 5 must share at least three positions with 1s with either $A$ or $B$).
(1) In summary, $\mathscr{D}$ can contain at most four codewords, one with a weight of 0, one with a weight of 6, and two with a weight of 5. The following example shows that a $\mathscr{D}$ consisting of four codewords can be realized:
$$
\begin{array}{l}
(0,0,0,0,0,0,0,0) \\
(1,1,1,0,0,1,1,1) \\
(1,1,1,1,1,0,0,0) \\
(0,0,0,1,1,1,1,1).
\end{array}
$$
Solution Two First, consider an easier problem: what is the maximum number of codewords of length 7, where the distance between any two is $\geqslant 5$? The answer is: at most two. First, assume $(0,0,0,0,0,0,0)$ is one of the codewords. Then it is impossible to have two more codewords. Otherwise, each of these two codewords would have a weight $\geqslant 5$. However,
$$
5+5-7=3.
$$
So, these two codewords must share at least three positions with 1s. The distance between these two codewords is no more than $7-3=4$.
Now, let's return to the original problem. For codewords of length 8, the first bit can be 0 or 1. Codewords with the same first bit $(a \in\{0,1\})$, the remaining bits form a codeword of length 7, and thus there can be at most two such codewords with a distance $\geqslant 5$. Therefore, there can be at most four codewords of length 8, where the distance between any two is $\geqslant 5$.
An example of "at most four" being achievable is the same as in Solution One.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}(n>4)$ where all interior angles are integer multiples of $15^{\circ}$, and $\angle A_{1}+\angle A_{2}+\angle A_{3}=$ $285^{\circ}$. Then, $n=$
|
2. 10 (Hint: The sum of the $n-3$ interior angles other than the male one is $(n-2) \cdot$ $180^{\circ}-285^{\circ}$, it can be divided by $n-3$, and its quotient should also be an integer multiple of $15^{\circ}$.）
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. For real numbers $x, y$, define a new operation: $x * y=a x+b y+$ $c$, where $a, b, c$ are constants, and the right side of the equation is the usual addition and multiplication operations. It is known that $3 * 5=15,4 * 7=28$. Then, 1*1= $\qquad$
|
7. -11
Brief solution: According to the definition, we know that $3 * 5=3a+5b+c=15, 4 * 7$ $=4a+7b+c=28$, and $1 * 1=a+b+c=3(3a+5b+$c) $-2(4a+7b+c)=3 \times 15-2 \times 28=-11$.
|
-11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (12 points) Given points $P_{1}\left(x_{1}, 1994\right), P_{2}\left(x_{2}, 1994\right)$ are two points on the graph of the quadratic function $y=a x^{2}+b x+7(a \neq 0)$. Try to find the value of the quadratic function $y=x_{1}+x_{2}$.
|
Three, since $P_{1}, P_{2}$ are two points on the graph of a quadratic function, we have
$$
\begin{array}{l}
a x_{1}^{2}+b x_{1}+7=1994, \\
a x_{2}^{2}+b x_{2}+7=1994 .
\end{array}
$$
Subtracting the two equations and simplifying, we get
$$
\left(x_{1}-x_{2}\right)\left[a\left(x_{1}+x_{2}\right)+b\right]=0 .
$$
Since $P_{1}, P_{2}$ are two distinct points, i.e., $x_{1} \neq x_{2}$, we have
$$
x_{1}+x_{2}=-\frac{b}{a},
$$
Thus, when $x=x_{1}+x_{2}$, the value of $y$ is 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. (India) The sequence of positive integers $\left\{f_{n}\right\}_{m=1}^{m}$ is defined as follows:
$f(1)=1$, and for $n \geqslant 2$,
$$
f(n)=\left\{\begin{array}{l}
f(n-1)-n, \text { when } f(n-1)>n ; \\
f(n-1)+n, \text { when } f(n-1) \leqslant n .
\end{array}\right.
$$
Let $S=\{n \in \mathbb{N} \mid f(n)=1993\}$.
(1) Prove that $S$ is an infinite set;
(2) Find the smallest positive integer in $S$;
(3) If the elements of $S$ are arranged in increasing order as $n_{1}<n_{2}$ $<n_{3}<\cdots$, then $\lim _{k \rightarrow \infty} \frac{n_{k+1}}{n_{k}}=3$.
|
(i) We point out that if $f(n)=1$, then $f(3n+3)=1$. From $f(n)=1$, we can sequentially obtain according to the definition:
$$
\begin{array}{l}
f(n+1)=n+2, f(n+2)=2n+4, \\
f(n+3)=n+1, f(n+4)=2n+5, \\
f(n+5)=n, \quad f(n+6)=2n+6 \\
f(n+7)=n-1, \cdots
\end{array}
$$
It is evident that the sequence $f(n+1), f(n+3), f(n+5), \cdots, f(3n+3)$ has values exactly as the sequence $n+2, n+1, n, \cdots, 1$, and the sequence $f(n+2), f(n+4), f(n+6), \cdots, f(3n+2)$ has values exactly as the sequence $2n+4, 2n+5, 2n+6, \cdots, 3n+4$. Therefore, $f(3n+3)=1$ and when $n \geq 1993$, from the above argument, among all natural numbers $n$ such that $b_i < n \leq b_{i+1}$, there is exactly one that makes $f(n)=1993$. Thus, $S$ is an infinite set.
(iv) Similarly to (iii), when $i \geq 8$, we have
$$
f(b_i + 2j - 1) = b_i + 3 - j = 1993
$$
Solving for $j$ gives $j = b_i - 1990$. Therefore, the value of $n_i$ that makes $f(n_i) = 1393$ is
$$
\begin{aligned}
a_1 & = b_i + 2j - 1 = 3 \cdot 3^{i-2} - 3981 \\
& = \frac{9}{2}(5 \cdot 3^{i-2} - 1) - 3982; \quad i = 8, 9, \cdots.
\end{aligned}
$$
Thus, we have
$$
\lim_{k \to \infty} \frac{n_{k+1}}{n_k} = \lim_{k \to \infty} \frac{\frac{9}{2}(5 \cdot 3^{k-1} - 1) - 3981}{\frac{9}{2}(5 \cdot 3^{k-2} - 1) - 3981} = 3.
$$
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $[x]$ denotes the greatest integer not exceeding $x$, $a, b, c \in$ $R^{+}, a+b+c=1$, let $M=\sqrt{3 a+1}+\sqrt{3 b+1}+$ $\sqrt{3 c+1}$. Then the value of $[M]$ is $($.
|
5. B.
Given $a \in(0,1), b \in(0,1), c \in(0,1) \Rightarrow a^{2} & \sqrt{a^{2}+2 a+1}+\sqrt{b^{2}+2 b+1} \\
& +\sqrt{c^{2}+2 c+1} \\
= & (a+b+c)+3=4 .
\end{aligned}
$$
And $M=\sqrt{(3 a+1) \cdot 1}+\sqrt{(3 b+1) \cdot 1}$
$$
\begin{aligned}
& +\sqrt{(3 c+1) \cdot 1} \\
& \frac{(3 a+1)+1}{2}+\frac{(3 b+1)+1}{2}+\frac{(3 c+1)+1}{2} \\
= & 3+\frac{3}{2}(a+b+c)=4.5, \\
\therefore & 4<M<4.5,[M]=4 .
\end{aligned}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the functions $y=2 \cos \pi x(0 \leqslant x \leqslant 2)$ and $y=2(x \in$ $R$ ) whose graphs enclose a closed plane figure. Then the area of this figure is $\qquad$ .
|
2. As shown in the figure, by symmetry, the area of $CDE$ = the area of $AOD$ + the area of $BCF$, which means the shaded area is equal to the area of square $OABF$, so the answer is 4.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the natural numbers $a, x, y$ satisfy $\sqrt{a-2 \sqrt{6}}=\sqrt{x} -\sqrt{y}$, then the maximum value of $a$ is $\qquad$.
|
$$
=, 1, a_{\max }=7 \text {. }
$$
Squaring both sides of the known equation yields $a-2 \sqrt{6}=x+y-$ number, it must be that $x+y=\alpha$ and $x y=6$. From the known equation, we also know that $x>y$, so it can only be $x=6, y=1$ or $x=3, y=2$. Therefore, $a=7$ or 2, the maximum value is 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A and Person B start from the same point $A$ on a circular track at the same time and run in opposite directions. Person A's speed is $5 \mathrm{~m}$ per second, and Person B's speed is $7 \mathrm{~m}$ per second. They stop running when they meet again at point $A$ for the first time. During this period, they meet a total of $n$ times, so $n=$ $\qquad$
|
3. 12 .
Two $\lambda \mu A$ start from $A$ and meet once (not at point $A$). At the moment of their first meeting, the two have run a certain distance. From the ratio of their speeds, it can be deduced that person A has run $\frac{5}{12}$ of a lap, and person B has run $\frac{7}{12}$ of a lap. Therefore, after the $n$-th meeting, person A has run $\frac{5}{12} n$ laps, and person B has run $\frac{7}{12} n$ laps. When they meet again at point $A$, both person A and person B have run an integer number of laps, so $\frac{5}{12} n$ and $\frac{7}{12} n$ are both integers. The smallest value of $n$ is 12, thus they have met 12 times.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Consider the hyperbola \((x-2)^{2}-\frac{y^{2}}{2}=1\). A line \(l\) is drawn through the right focus of the hyperbola, intersecting the hyperbola at points \(A\) and \(B\). If \(|A B|=4\), then the number of such lines is \(\qquad\).
|
4. 3 lines.
The coordinates of the right focus are $(2+\sqrt{3}, 0)$. The length of the chord perpendicular to the real axis through the right focus is exactly 4. The length of the chord $d$ formed by passing through the right focus and intersecting both branches of the hyperbola has the range $d>2$, so there are two more chords of length 4, making a total of 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. Write the number $1234567802011 \cdots 19941995$ on the blackboard, forming the integer $N_{1}$. Erase the digits of $N_{1}$ that are in even positions, leaving the remaining digits to form the integer $N_{2}$. Remove the digits of $N_{2}$ that are in odd positions, leaving the remaining digits to form the integer $N_{3}$. Erase the digits of $N_{3}$ that are in even positions, leaving the remaining digits to form the integer $N_{4}$. This process continues until only one digit remains on the blackboard. Determine this digit. (Note: Count positions from left to right, for example, in 12345, 1 is in the first position, 2 is in the second position, and so on).
|
$$
=、 9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873 \text {. }
$$
The integer $N_{1}$ is composed of 6873 digits, which are sequentially numbered as $1,2,3, \cdots, 6873$. We still examine the set of numbers $\{1,2, \cdots, 6873\}$. Removing the digits at even positions from $N_{1}$ is equivalent to removing $\{2,4, \cdots, 6872\}$ from the set $\{1,2, \cdots, 6873\}$.
Similarly, from the remaining set $\{1,3,5, \cdots, 6873\}$, we remove $\{1,5,9, \cdots, 6873\}$, leaving $\{3,7,11, \cdots, 6871\}$. We can sequentially obtain the sets $\{3,3+8=11,11+8=19, \cdots\}$, $\{11,11+16=27, \cdots\}$, $\{11,11+32=43, \cdots\}$, $\{43,43+64=107, \cdots\}$, $\{43,43+128=371, \cdots\}$, $\{171,171+256=427, \cdots\}$, $\{171,171+512=683, \cdots\}$, $\{683,683+1024=1707, \cdots\}$, $\{683,683+2048=2731\}$, $\{2731,2731+4096=6827\}$. After the thirteenth step, only $\{2731\}$ remains.
The required digit is the 2731st digit from left to right in $N_{1}$.
$$
9 \times 1+90 \times 2<2731<9 \times 1+90 \times 2+900 \times 3 \text {, }
$$
This digit should be part of a three-digit number. The one-digit and two-digit numbers together occupy 189 positions, 2731 - 189 = 2542 = 3 \times 847 + 1, the 848th three-digit number is 947, and the digit to the left of this number is 9. The final remaining digit is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that the inverse function of $y=f(x)$ is $\varphi(x)$, and $\varphi(x)$ . $=\log _{m a_{0}{ }^{2}}\left(\frac{1996}{x}-\sin ^{2} 0\right), 0 \in\left(0, \frac{\pi}{2}\right)$. Then the solution to the equation $f(x)=$ 1996 is
|
$=, 1 .-1$.
Since $f(x)$ and $\varphi(x)$ are inverse functions of each other, solving $f(x) = 1996$ is equivalent to finding the value of $\varphi(1996)$.
$$
\varphi(1996)=\log _{\sec ^{2} \theta}\left(1-\sin ^{2} \theta\right)=-1 .
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
When the expression changes, the minimum value of the fraction $\frac{3 x^{2}+6 x-5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$. (1993, National Junior High School Competition)
|
Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$.
(*)
Rearranging (*), we get
$$
(y-6) x^{2}+(2 y-12) x+2 y-10=0 \text {. }
$$
Since $x$ is a real number, we have
$$
\Delta=(2 y-12)^{2}-4(y-6)(2 y-10) \geqslant 0 \text {, }
$$
which simplifies to $y^{2}-10 y+24 \leqslant 0$.
Solving this, we get $4 \leqslant y \leqslant 6$.
Substituting $y=4$ into (*) gives $x=-1$.
Thus, when $x=-1$, $y_{\text {min }}=4$.
Note: Generally, if a rational function $y=f(x)$ can be transformed into $p(y) x^{2}+q(y) x+r(y)=0$ after clearing the denominator, where $p, q, r$ are real functions of $y$, then the range of $y$ can be determined by $\Delta \geqslant 0$, and thus the extremum of $y$ can be found.
There is also a simpler solution for this problem:
$$
\begin{array}{l}
\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}=\frac{\left(6 x^{2}+12 x+12\right)-2}{x^{2}+2 x+2} \\
=6-\frac{2}{(x+1)^{2}+1} .
\end{array}
$$
Clearly, when $x=-1$, the original expression has a minimum value of 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ - (1995, National Junior High School Competition)
|
- $y=x^{2}-x+\frac{1}{x}$ $=(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1$.
From the above equation, it is known that when $x-1=0$ and $\sqrt{x}-\frac{1}{\sqrt{x}}$ $=0$, i.e., $x=1$, $y$ takes the minimum value of 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13. Let $a, b, c$ be distinct integers from 1 to 9. What is the largest possible value of $\frac{a+b+c}{a b c}$? (1992, 1st Dannevirke-Shanghai Friendship Correspondence Competition)
|
Let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a, b$ remain unchanged temporarily, and only let $c$ vary, where $c$ can take any integer from 1 to 9. Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that when $c=1$, $P$ reaches its maximum value. Therefore, $c=1$.
Thus, $P=\frac{a+b+1}{a b}$.
In equation (2), let $a$ remain unchanged temporarily, and only let $b$ vary, where $b$ can take any integer from 2 to 9. Then, from $P=\frac{a+b+1}{a b}=\frac{1}{a}+\frac{a+1}{a b}$, we know that when $b=2$, $P$ reaches its maximum value. Therefore, $b=2$.
Thus, $P=\frac{1}{2}+\frac{3}{2 a}$.
In equation (3), $P$ reaches its maximum value when $a$ takes its minimum value, and $a$ can take any integer from 3 to 9, i.e., $a=3$.
Therefore, when $a=3, b=2, c=1$ (the letters can be interchanged), $P$ reaches its maximum value of 1.
Regarding the application of function extremum problems, there is another important aspect, which is the application of function extremum. As long as we master the basic methods of finding the extremum of a function, such problems are not difficult to solve.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If 20 points divide a circle into 20 equal parts, then the number of regular polygons that can be formed with vertices only among these 20 points is ( ) .
(A) 4
(B) 8
(C) 12
(D) 24
|
6. (C).
Let the regular $k$-sided polygon satisfy the condition, then the 20 $-k$ points other than the $k$ vertices are evenly distributed on the minor arcs opposite to the sides of the regular $k$-sided polygon.
Thus, $\frac{20-k}{k}=\frac{20}{k}-1$ is an integer, so $k \mid 20$.
But $k \geqslant 3$,
$\therefore k=4$ or 5 or 10 or 20.
$\therefore$ The number of regular polygons sought is
$$
\frac{20}{4}+\frac{20}{5}+\frac{20}{10}+\frac{20}{20}=12(\uparrow)
$$
|
12
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given that $a, b, c$ are positive integers, and the parabola $y=$ $a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A, B$. If the distances from $A, B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Three, let the coordinates of $A, B$ be $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$ and $x_{1}<0, x_{2}>0$,
\[
\begin{array}{l}
\therefore x_{1} x_{2}=\frac{c}{a}<0, \\
\therefore a c<0, \text { i.e., } a>0, c<0 .
\end{array}
\]
\[
\begin{array}{l}
\because x_{1}+x_{2}=-\frac{b}{a}>0, \\
\therefore b<0 .
\end{array}
\]
\[
\begin{array}{l}
\because \text { the parabola opens upwards, and } x_{1} x_{2}<0, \\
\therefore \text { the parabola intersects the } x \text{-axis at two points, } \\
\therefore \Delta=b^{2}-4 a c>0, \\
\because x_{1} x_{2}<0, \text { get } b>2 \sqrt{a c} .
\end{array}
\]
\[
\begin{array}{l}
\because|O A|=|x_{1}|,|O B|=|x_{2}|, \\
\therefore|O A| \cdot|O B|=|x_{1} x_{2}|=\left|\frac{c}{a}\right| \geqslant 1, \\
\therefore|c| \geqslant|a|, \text { and } a>0, c<0, \\
\therefore c \leqslant -a .
\end{array}
\]
\[
\begin{array}{l}
\because \text { the parabola passes through the point } (-1,1), \\
\therefore a(-1)^{2}+b(-1)+c>0 . \\
\text { get } b<2 \sqrt{a c}+1 \Rightarrow(\sqrt{a}-\sqrt{c})^{2}>1 .
\end{array}
\]
\[
\begin{array}{l}
\text { From (2) we get } \sqrt{a}-\sqrt{c}>1 . \\
\therefore \sqrt{a}>\sqrt{c}+1 .
\end{array}
\]
\[
\begin{array}{l}
\text { i.e., } a>(\sqrt{c}+1)^{2} \geqslant(\sqrt{1}+1)^{2}=4, \\
\therefore a \geqslant 5 .
\end{array}
\]
\[
\begin{array}{l}
\text { Also, } b>2 \sqrt{a c} \geqslant 2 \sqrt{5 \times 1}>4, \\
\therefore b \geqslant 5 .
\end{array}
\]
\[
\begin{array}{l}
\text { Taking } a=5, b=5, c=1, \text { the parabola } y=5 x^{2}+5 x+1 \text { satisfies the given conditions. } \\
\text { Hence, the minimum value of } a+b+c \text { is } 5+5+1=11 .
\end{array}
\]
(Provided by Liu Yuxuan)
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) As shown in the figure, given that $AB, CD$ are perpendicular chords in a circle $\odot O$ with radius 5, intersecting at point $P$. $E$ is the midpoint of $AB$, $PD=AB$, and $OE=3$. Try to find the value of $CP + CE$.
---
The translation is provided as requested, maintaining the original text's format and line breaks.
|
By the intersecting chords theorem, we have
$$
C P \cdot P D=A P \cdot P B \text {. }
$$
Also, $A E=E B=\frac{1}{2} A B$,
$$
\begin{aligned}
\because C P \cdot P D & =\left(\frac{1}{2} A B+E P\right)\left(\frac{1}{2} A B-E P\right) \\
& =\frac{1}{4} A B^{2}-E P^{2} .
\end{aligned}
$$
And $P D=A B$,
$$
\therefore \quad C P=\frac{1}{4} A B-\frac{E P^{2}}{A B} \text {. }
$$
In the right triangle $\triangle C E P$,
$$
\begin{array}{l}
C E^{2}=C P^{2}+E P^{2}=\left(\frac{1}{4} A B-\frac{E P^{2}}{A B}\right)^{2}+E P^{2} \\
=\left(\frac{1}{4} A B+\frac{E P^{2}}{A B}\right)^{2}, \\
\therefore \quad C E=\frac{1}{4} A B+\frac{E P^{2}}{A B} .
\end{array}
$$
Thus, $C P+C E=\frac{1}{2} A B=A E=\sqrt{O A^{2}-O E^{2}}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The Dao sequence $\left\{\begin{array}{l}x_{1}=x_{2}=1, \\ x_{n+2}=a x_{n+1}+b x_{n}(n \in N) .\end{array}\right.$
If $T=1996$ is the smallest natural number such that $x_{T+1}=x_{T+2}=1$, then $\sum_{i=1}^{1006} x_{i}=$ $\qquad$ .
|
2. 0 .
From $T=1996$ being the smallest natural number that makes $x_{1+1}=x_{T+2}=1$, we know that $a+b \neq 1$. Otherwise, if $a+b=1$, then $x_{3}=a x_{2}+b x_{1}=a+b=1$. This would mean that when $T=1$, $x_{T+1}=x_{T+2}=1$, which contradicts the fact that $T=1996$ is the smallest value. Furthermore, we have
$$
\begin{array}{l}
x_{1}=x_{1997}=a x_{1006}+b x_{1905}, \\
x_{2}=x_{1998}=a x_{1997}+b x_{1996}=u x_{1}+b x_{1995}, \\
x_{3}=a x_{2}+b x_{1}, \\
x_{4}=a x_{3}+b x_{2}, \\
\cdots \cdots \\
x_{1096}=a x_{1995}+b x_{1994} .
\end{array}
$$
Summing up, we get $\quad \sum_{i=1}^{1004} x_{i}=a \sum_{i=1}^{1006} x_{i}+b \sum_{i=1}^{1096} x_{i}$,
which simplifies to $(a+b-1) \sum_{i=1}^{1098} x_{i}=0$.
Since $a+b-1 \neq 0$, it follows that $\sum^{1096} x_{i}=0$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the plane $\alpha$ there is a $\triangle A B C, \angle A B C=105^{\circ}$, $A C=2(\sqrt{6}+\sqrt{2})$. On both sides of the plane $\alpha$, there are points $S, T$, satisfying $S A=S B=S C=\sqrt{41}, T A=T B=T C=$ 5. Then $S T=$ $\qquad$.
|
3. 8 .
From Figure 4, since $S A=S B=S C$, we know that the projection $D$ of $S$ on plane $\alpha$ is the circumcenter of $\triangle A B C$. Similarly, the projection of $T$ on plane $\alpha$ is also the circumcenter of $\triangle A B C$. By the uniqueness of the circumcenter, we have $S T \perp a$.
Connect $A D$. By the Law of Sines, we have
T
$$
\begin{array}{l}
A D=\frac{A C}{\sin \angle A B \bar{C}} \\
=\frac{3(\sqrt{6}+\sqrt{2})}{2\left(\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}\right)}=4 .
\end{array}
$$
Thus,
$$
\begin{aligned}
S T & =S D+D T \\
& =\sqrt{S A^{2}-A D^{2}}+\sqrt{T A^{2}-A D^{2}} \\
& =\sqrt{41-16}+\sqrt{25-16}=5+3=8
\end{aligned}
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given functions $f(x)$ and $g(x)$ are defined on $R$, and $f(x-y)=f(x) g(y)-g(x) f(y)$. If $f(1)=f(2) \neq$ 0, then $g(1)+g(-1)=$ $\qquad$ .
|
4. 1 .
For $x \in R$, let $x=u-v$, then we have
$$
\begin{array}{l}
f(-x)=f(v-u) \\
=f(v) g(u)-g(v) f(u) \\
=-[f(u) g(v)-g(u) f(v)] \\
=-f(u-v)=-f(x) .
\end{array}
$$
Therefore, $f(x)$ is an odd function, and we have
$$
\begin{array}{l}
f(2)=f[1-(-1)] \\
=f(1) g(-1)-g(1) f(-1) \\
=f(1) g(-1)+g(1) f(1) .
\end{array}
$$
But $f(1)=f(2) \neq 0$, so $g(1)+g(-1)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ have side lengths $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, respectively, and their areas be $S$ and $S^{\prime}$. If for all values of $x$, $a x^{2}+b x+c=3\left(a^{\prime} x^{2}+b^{\prime} x+c^{\prime}\right)$ always holds, then $\frac{S}{S^{\prime}}$ $=$
|
II. 1. From the given, $\left(a-3 a^{\prime}\right) x^{2}+\left(b-3 b^{\prime}\right) x+(c$ $\left.-3 c^{\prime}\right)=0$ is an identity, hence $\left(a-3 a^{\prime}\right)=0,\left(b-3 b^{\prime}\right)=0$, $\left(c-3 c^{\prime}\right)=0$, which gives $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}=3$,
$\therefore \triangle A B C \sim \triangle A^{\prime} B^{\prime} C^{\prime}$, then $\frac{S}{S^{\prime}}=\left(\frac{a}{a^{\prime}}\right)^{2}=9$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In the interior angles of an $n(n \geqslant 3)$-sided polygon, the maximum number of acute angles is $\qquad$ .
|
II, 1.3.
Let a polygon with $n$ sides have $k$ acute angles, then
$$
(n-2) \times 180^{\circ}<k \times
$$
$90^{\circ}+(n-k) \times 180^{\circ}$.
Thus, $k<4$.
As shown in the figure, construct a $60^{\circ}$ sector $A_{1} A_{2} A_{1}$. When $n=3$, $\triangle A_{1} A_{2} A_{3}$ has three interior angles, all of which are part of $A_{1} A_{2} \cdots A_{n-1} A_{n}$, where $\angle A_{1} A_{2} A_{3}$, $\angle A_{n-1} A_{n} A_{1}$, and $\angle A_{n} A_{1} A_{2}$ are all acute angles, hence the maximum value of $k$ is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. There are two roads $O M, O N$ intersecting at a $30^{\circ}$ angle. Along the direction of road $O M$, 80 meters from $A$ is a primary school. When a tractor travels along the direction of $O N$, areas within 50 meters on both sides of the road will be affected by noise. Given that the speed of the tractor is 18 kilometers/hour. Then, the time during which the tractor traveling along $O N$ will cause noise impact to the primary school is $\qquad$ seconds.
|
3. 12 .
As shown in the figure, when the tractor reaches point $B$, it starts to have an impact, and when it reaches point $C$, the impact ceases. Thus, $A B = A C = 50$ meters. Draw $A D \perp$ $O N$ from $A$. Given that $\angle A O D = 30^{\circ}$, we know that $A D = \frac{1}{2} O A = 40$ meters. Therefore, $B D = \sqrt{A B^{2} - A D^{2}} = 30$ meters, and thus $B C = 60$ meters. Since the speed of the tractor is 18 kilometers/hour $= 5$ meters/second, the duration of the noise impact is $t = \frac{60}{5} = 12$ seconds.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let real numbers $x, y$ satisfy the equation $9 x^{2}+4 y^{2}-3 x+2 y=0$. Then the maximum value of $z=3 x+2 y$ is $\qquad$ .
|
From $9 x^{2}+4 y^{2}=3 x-2 y$ we know $3 x-2 y \geqslant 0$ and $\frac{9 x^{2}+4 y^{2}}{3 x-2 y}$ $=1$,
we have $z=3 x+2 y=\frac{(3 x)^{2}-(2 y)^{2}}{3 x-2 y} \leqslant \frac{(3 x)^{2}+(2 y)^{2}}{3 x-2 y}$ $=1$ (equality holds when $y=0$).
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 17. Given as shown, in quadrilateral $ABCD$, $AD=DC=1, \angle DAB=$ $\angle DCB=90^{\circ}, BC, AD$ extended intersect at $P$. Find the minimum value of $AB \cdot S_{\triangle PAB}$.
(1994, Sichuan Province Junior High School Mathematics League Competition)
|
Let $DP = x$, then $PC = \sqrt{x^2 - 1}$.
Since $\triangle PCD \sim \triangle PAB$,
$\therefore CD \perp AB = PC : PA$.
$\therefore AB = \frac{CD \cdot PA}{PC} = \frac{x + 1}{\sqrt{x^2 - 1}}$.
Let $y = AB \cdot S_{\triangle PAB}$, then
$$
y = \frac{1}{2} AB^2 \cdot PA = \frac{(x + 1)^3}{2(x^2 - 1)}.
$$
Eliminating the denominator and simplifying, we get
$$
x^2 - 2(1 - y)x + 1 + 2y = 0.
$$
Since $x$ is a real number, this equation has real roots, so
$$
\Delta = 4(1 - y)^2 - 4(1 + 2y) = 4y(y - 4) \geqslant 0.
$$
Since $y > 0$, solving gives $y \geqslant 4$,
i.e., the minimum value of $AB \cdot S_{\triangle PAB}$ is 4. E.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. If $a=\sqrt{17}-1$, find the value of $\left(a^{5}+2 a^{4}-17 a^{3}\right.$ $\left.-a^{2}+18 a-17\right)^{1993}$.
(Adapted from the 1987 Chongqing Junior High School Mathematics Invitational Competition)
|
Given $a+1=\sqrt{17}$, squaring both sides yields $a^{2}+2 a+1=17$.
Then
$$
\begin{array}{l}
\left(a^{5}+2 a^{4}-17 a^{3}-a^{2}+18 a-17\right)^{1993} \\
=\left[a^{5}+2 a^{4}-\left(a^{2}+2 a+1\right) a^{3}-a^{2}+\left(a^{2}\right.\right. \\
\left.+2 a+1+1) a-\left(a^{2}+2 a+1\right)\right]^{1993} \\
=(-1)^{1993}=-1 \text {. } \\
\end{array}
$$
Note: Here, we first constructed the expression $a^{2}+2 a+1$ with a value of 17, and then substituted $a^{2}+2 a+1$ for 17 in the expression to be evaluated.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. If $m^{2}=m+1, n^{2}=n+1$, then what is the value of $m^{5}+n^{5}$?
(1989, Jiangsu Province High School Mathematics Competition)
|
Let $S_{k}=m^{k}+n^{k}$, construct the recurrence relation for $S_{k}$.
$$
\begin{array}{l}
\because m^{2}=m+1, n^{2}=n+1, \\
\therefore m^{k}=m^{k-1}+m^{k-2}, n^{k}=n^{k-1}+n^{k-2} .
\end{array}
$$
Adding the two equations gives $S_{k}=S_{k-1}+S_{k-2}$.
$$
\begin{aligned}
\therefore S_{5} & =S_{4}+S_{3}=\left(S_{3}+S_{2}\right)+\left(S_{2}+S_{1}\right) \\
& =\cdots=3 S_{2}+2 S_{1} .
\end{aligned}
$$
From the given conditions, $S_{2}=m+n-2$, and $m, n$ are the roots of the quadratic equation $x^{2}-x-1=0$, so $m+n=1$, i.e., $S_{1}=1, S_{2}=1+2=3$.
Thus, $S_{5}=3 \times 3+2 \times 1=11$.
Note: Here, we first construct the recurrence relation for $S_{k}=m^{2}+n^{k}$, then express $S_{5}$ in terms of $S_{1}$ and $S_{2}$, and solve the problem. This is a general method for solving problems of the form $a^{n}+b^{n}$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10. Given $a^{2}+2 a-5=0, b^{2}+2 b-5=0$, and $a \neq b$. Then the value of $a b^{2}+a^{2} b$ is what?
(1989, Sichuan Province Junior High School Mathematics Competition)
|
Given that $a, b$ are the roots of the quadratic equation $x^{2}+$ $2 x-5=0$, then $a+b=-2, ab=-5$. Therefore, $ab^{2}+a^{2}b=ab(a+b)=10$. Note: Here, the definition of the roots is used to construct the equation.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a+b+c=0, a^{3}+b^{3}+c^{3}=0$. Find the value of $a^{15}+b^{15}+c^{15}$.
|
2. From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right.$ $-a b-b c-a c)$, we have $a b c=0$. Then at least one of $a, b, c$ is zero, for example, $c=0$, then it is known that $a, b$ are opposites, i.e., $a^{15}+b^{15}+c^{15}$ $=0$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Find a natural number $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square.
(2nd All-Russian High School Mathematics Olympiad)
|
Let $2^{4}=x$, then $2^{8}=x^{2}, 2^{11}=2^{7} \cdot x$. Therefore, $2^{8}+2^{11}+2^{n}=x^{2}+2^{7} x+2^{n}$.
According to the condition for a quadratic trinomial to be a perfect square, we get $\Delta=\left(2^{7}\right)^{2}-4 \times 2^{n}=0$.
Solving for $n$ yields $n=12$.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $\frac{x}{m}+\frac{y}{n}+\frac{z}{p}=1, \frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$. Calculate the value of $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}$.
|
6. From $\frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$, we have $\frac{m y z+n x z+p x y}{x y z}=0$.
Since $x, y, z$ are not zero, then $m y z+n x z+p x y=0$. Also,
$$
\begin{aligned}
& \frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}} \\
= & \left(\frac{x}{m}+\frac{y}{n}+\frac{z}{p}\right)^{2}-2\left(\frac{x y}{m n}+\frac{x z}{m p}+\frac{y z}{n p}\right) \\
= & 1-2 \cdot \frac{1}{m n p}(p x y+n x z+m y z)=1 .
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that $x, y, z$ are three non-negative rational numbers, and satisfy $3 x$ $+2 y+z=5, x+y-z=2$. If $S=2 x+y-z$, then what is the sum of the maximum and minimum values of $S$?
|
9. $\left\{\begin{array}{l}3 x+y+z=5, \\ x+y-z=2, \\ 2 x+y-z=S .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}x=S-2, \\ y=\frac{15-4 S}{3}, \\ z=\frac{3-S}{3} .\end{array}\right.$ Since $x, y, z$ are all non-negative, we have
$$
\left\{\begin{array} { l }
{ S - 2 \geqslant 0 , } \\
{ \frac { 1 5 - 4 S } { 3 } \geqslant 0 , } \\
{ \frac { 3 - S } { 3 } \geqslant 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
S \geqslant 2, \\
S \leqslant \frac{15}{4}, \\
S \leqslant 3 .
\end{array}\right.\right.
$$
Therefore, $2 \leqslant S \leqslant 3$. The sum of the maximum and minimum values of $S$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In $\triangle A B C$, $G$ is the centroid, and $I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$. If $I G / / B C$, and $B C=5$, then $A B+A C$ $=$ . $\qquad$
|
3. Connect $A G, B G$ and $C G$, and extend $A G$ to intersect $B C$ at $M$, then $S_{\triangle G B C}: S_{\triangle A B C}=G M: A M=1: 3$.
Connect $A I$, then by $I G$ $/ / B C$, we know
$$
\begin{array}{l}
S_{\triangle I B C}=S_{\triangle G B C}=\frac{S_{\triangle A B C}}{3}, \\
\therefore S_{\triangle I A B}+S_{\triangle I A C}=2 S_{\triangle I B C} .
\end{array}
$$
$\because I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$ in $\triangle A B C$,
$\therefore I$ is equidistant from the three sides of $\triangle A B C$.
Thus, $A B+A C=2 B C=10$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points, Question (1) 8 points, Question (2) 12 points) As shown in the figure, there is a cube-shaped wire frame, and the midpoints of its sides $I, J, K, L$ are also connected with wire.
(1) There is an ant that wants to crawl along the wire from point A to point G. How many shortest routes are there? Represent these routes using letters (use the letters of the connection points passed through, for example, if the ant starts from point A, passes through point I, point L, and finally reaches point H, this route is represented as $A I L H$).
(2) Is it possible for the ant to start from point A, pass through each connection point exactly once along the wire, and finally reach point G? If possible, find one such route; if not, explain why.
|
(1) "The shortest path" means that the ant can neither move left nor down, otherwise it would be taking a "detour" rather than the "shortest" path. The ant can crawl on face $ABCGFE$ or on $ADCGHE$. The number of "shortest paths" on these two faces is the same, and the possible paths can be represented by the following diagram:
Therefore, there are a total of 12 shortest paths, which are:
$A B C K G, A B J K G, A B J F G, A I E F G, A I J K G$,
$A I J F G, A I E H G, A I L H G, A I L K G, A D C K G$,
$A D L K G, A D L H G$.
(2) Using the application of integer properties - coloring method or labeling method
Solution. The answer sought in the problem is impossible.
As shown in the figure, all 12 connection points are colored with black and white, and each adjacent pair of points is colored with different colors. The ant starts from the black point $A$, and must follow a route pattern like: black $\rightarrow$ white $\rightarrow$ black $\rightarrow$ white $\rightarrow$ $\cdots$, and by the 12th point $G$ it should be $\rightarrow$ white $\rightarrow$ black $(G)$, but according to the route pattern, the 12th point should be: $\rightarrow$ black $\rightarrow$ white. Therefore, it is impossible for the ant to start from point $A$, pass through each connection point exactly once, and finally reach point $G$.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers $1,2, \cdots, n$. Let $f(n)$ be the number of such permutations such that
(i) $a_{1}=1$;
(ii) $\left|a_{i}-a_{i+1}\right| \leqslant 2 . i=1,2, \cdots, n-1$.
Determine whether $f(1996)$ is divisible by 3.
|
3. Verify that $f(1)=f(2)=1$ and $f(3)=2$.
Let $n \geqslant 4$. Then it must be that $a_{1}=1, a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$, because by deleting the first term and reducing all subsequent terms by 1, we can establish a one-to-one correspondence of sequences.
If $a_{2}=3$, then $a_{3}=2$. In this case, $a_{4}=4$, and the number of such permutations is $f(n-3)$.
If $a_{3} \neq 2$, then 2 must come after 4. This implies that all odd numbers are in ascending order and all even numbers are in descending order. Therefore, $f(n)=f(n-1)+f(n-3)+1$.
For $n \geqslant 1$, let $r(n)$ be the remainder when $f(n)$ is divided by 3. Then we have
$$
\begin{array}{l}
r(1)=r(2)=1, r(3)=2, \\
r(n)=r(n-1)+r(n-3)+1 .
\end{array}
$$
The sequence of remainders forms a periodic sequence with a period of 8:
$$
\{1,1,2,1,0,0,2,0\} \text {. }
$$
Since $1994=4(\bmod 8)$, and $r(4)=1$,
$f(1996)$ is not divisible by 3.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $360^{x}=3, 360^{y}=5$, then $72^{\frac{1-2 x-y}{3(1-y)}}=$
|
\begin{array}{l}=1.2. \\ \because 72=\frac{360}{5}=\frac{360}{360^{y}}=360^{1-y}, \\ \therefore \text { original expression }=\left(360^{1-y}\right)^{\frac{1}{3(2 x-y)}} \\ =(360)^{\frac{1}{3}(1-2 x-y)}-\left(360^{1-2 x-y}\right)^{\frac{1}{3}} \\ =\left(\frac{360}{\left(360^{x}\right)^{2} \cdot 360^{5}}\right)^{\frac{1}{3}}=\sqrt[3]{\frac{360}{9 \times 5}}=2 . \\\end{array}
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $x=\frac{1}{2}-\frac{1}{4 x}$, then $1-2 x+2^{2} x^{2}-2^{3} x^{3}+2^{4} x^{4}$ $-\cdots-2^{1995} x^{1995}$ is. $\qquad$.
|
2. 1 .
From $x=\frac{1}{2}-\frac{1}{4 x}$ we can get $1-2 x+(2 x)^{2}=0$.
$$
\begin{array}{c}
\therefore \text { the original expression }=1-2 x\left[1-2 x+(2 x)^{2}\right]+(2 x)^{4}[1-2 x \\
\left.+(2 x)^{2}\right]-\cdots-(2 x)^{1993}\left[1-2 x+(2 x)^{2}\right]=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $x=\sqrt{19-8 \sqrt{3}}$, then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$ $\qquad$
|
Solve: From $x=\sqrt{19-2 \sqrt{48}}=4-\sqrt{3}$, we get $x-4=-\sqrt{3}$. Squaring both sides and rearranging, we obtain
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore,
$$
\begin{aligned}
\text { Original expression } & =\frac{\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)+10}{\left(x^{2}-8 x+13\right)+2} . \\
& =5 .
\end{aligned}
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $a, b$ be positive integers, and $a+b \sqrt{2}$ $=(1+\sqrt{2})^{100}$. Then the units digit of $a b$ is $\qquad$
|
6.4.
By the binomial theorem, we have
$$
a-b \sqrt{2}=(1-\sqrt{2})^{100} \text {. }
$$
Therefore, $a=\frac{1}{2}\left((1+\sqrt{2})^{100}+(1-\sqrt{2})^{100}\right)$,
$$
\begin{array}{l}
b=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{100}-(1-\sqrt{2})^{100}\right) . \\
\text { Hence } \left.a b=\frac{1}{4 \sqrt{2}}(1+\sqrt{2})^{200}-(1-\sqrt{2})^{200}\right) \\
=\frac{1}{4 \sqrt{2}}\left((3+2 \sqrt{2})^{100}-(3-2 \sqrt{2})^{100}\right) . \\
\text { Let } x_{n}=\frac{1}{4 \sqrt{2}}\left((3+2 \sqrt{2})^{n}\right. \\
\left.-(3-2 \sqrt{2})^{n}\right)(n=1,2,3,4, \cdots),
\end{array}
$$
then $x_{1}=1, x_{2}=6$.
By the identity $a^{n}-b^{n}=(a+b)\left(a^{n-1}-b^{n-1}\right)-a b$ $\left(a^{n-2}-b^{n-2}\right)$, we can obtain the recurrence relation for the sequence $\left\{x_{n}\right\}$
$$
x_{n}=6 x_{n-1}-x_{n-2} \cdot(n=3,4,5, \cdots)
$$
Thus, the last digit of $x_{n}$ is as follows:
$$
\begin{array}{l}
x_{1}=1, x_{2} \equiv 6, x_{3} \equiv 5, x_{4} \equiv 4, x_{5} \equiv 9, x_{6}=0, \\
x_{7}=1, x_{8} \equiv 6, x_{9}=5, x_{10} \equiv 4(\bmod 10) .
\end{array}
$$
From this, we can see that $x_{n+2}=x_{n+8}(\bmod 10)(n=1,2,3$, $\cdots$.
Therefore, $x_{100}=x_{6 \times 16+4} \equiv x_{4}=4(\bmod 10)$, so the last digit of $a b=x_{100}$ is 4.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13. Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0, x y z \neq 0$. Then the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$ is equal to
|
Solve the system of equations by treating $\approx$ as a constant,
$$
\left\{\begin{array}{l}
4 x-3 y=6 z, \\
x-2 y=7 z,
\end{array}\right.
$$
we get
$$
\begin{array}{l}
x=3 z, y=2 z . \\
\text { Therefore, the original expression }=\frac{2 \cdot(3 z)^{2}+3 \cdot(2 z)^{2}+6 z^{2}}{(3 z)^{2}+5 \cdot(2 z)^{2}+7 z^{2}}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 14. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq$ $n$, then $m^{5}+n^{5}=$ $\qquad$
|
$$
\begin{array}{l}
\text { Sol } \because(a-b)\left(a^{n}+b^{n-1}\right) \\
=a^{n}+b^{n}+a b\left(a^{n-2}+b^{n-2}\right), \\
\therefore a^{n}+b^{n}=(a+b)\left(a^{n-1}+b^{n-1}\right) \\
-a b\left(a^{n-2}+b^{n-2}\right) \text {. } \\
\end{array}
$$
Let $S_{n}=a^{n}-b^{n}$, we get the recursive formula
$$
S_{n}=(a+b) S_{n-1}-a b S_{n-2} .(n=2,3, \cdots)
$$
From the given, $m, n$ are the unequal real roots of $x^{2}-x-1=0$, hence
$$
\begin{array}{l}
m+n=1, m n=-1, m^{2}+n^{2}=3 . \\
\text { Therefore, } S_{n}=S_{n-1}+S_{n-2} . \\
\text { Thus, } S_{3}=S_{2}+S_{1}=3+1=4, \\
S_{4}=S_{3}+S_{2}=4+3=7, \\
S_{5}=S_{4}+S_{3}=7+4=11 .
\end{array}
$$
Therefore, $m^{5}+n^{5}=11$.
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given the equation $a x^{2}+b x+c=0(a \neq 0)$, the sum of the roots is $s_{1}$, the sum of the squares of the roots is $s_{2}$, and the sum of the cubes of the roots is $s_{3}$. Then the value of $a s_{3}+$ $\left\langle s_{2}\right.$ $+c s_{1}$ is . $\qquad$
|
(Tip: Let the two roots of the equation be $x_{1}, x_{2}$.
Then by definition, we have $a x_{1}^{2}+b x_{1}+c=0, a x_{2}^{2}+b x_{2}+c=0$. The original expression $=0$.)
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. If $a$ is a root of $x^{2}-3 x+1=0$, try to find the value of $\frac{2 a^{5}-5 a^{4}+2 a^{3}-8 a^{2}}{a^{2}+1}$.
|
From the definition of the root, we know that $a^{2}-3 a+1=0$.
Thus, $a^{2}+1=3 a$ or $a^{2}-3 a=-1$ or
$$
\begin{array}{l}
a^{3}=3 a^{2}-a \\
\therefore \text { the original expression }=\frac{a\left[2 a^{2}\left(a^{2}+1\right)-5 a^{3}-8 a\right]}{3 a} \\
=\frac{1}{3}\left(a^{3}-8 a\right)=\frac{1}{3}\left(3 a^{2}-9 a\right) \\
=a^{2}-3 a=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 8. Let } x>\frac{1}{4} \text {. Simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$
|
Let $a$
$$
\begin{array}{l}
=x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}, b=x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1} . \text { Original } \\
\text { expression }=1 .)
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. The value of $x$ that satisfies the following equations is
$$
\begin{array}{l}
(123456789) x+9=987654321, \\
(12345678) x+8=98765432 . \\
(1234567) x+7=9876543 . \\
\cdots \cdots .
\end{array}
$$
|
Observe the numerical changes on both sides of each equation, it is easy to know that the last equation should be $x+1=9$, i.e., $x=8$. Upon verification, $x=8$ is the solution.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. Let $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=0$ and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is $\qquad$.
|
From the given, we have $\left(\frac{1}{a}\right)^{2}-\frac{2}{a}-1=0$, $\left(b^{2}\right)^{2}-2 b^{2}-1=0$. By the definition of roots, $\frac{1}{a}, b^{2}$ are the roots of the equation $x^{2}-2 x-1=0$. Then
$$
\begin{aligned}
\frac{1}{a}+b^{2} & =2, \frac{1}{a} \cdot b^{2}=-1 . \\
\therefore \text { the original expression } & =\left(\frac{1}{a}+b^{2}+\frac{1}{a} \cdot b^{2}\right)^{1990} \\
& =(2-1)^{1990}=1 .
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Glue the bases of two given congruent regular tetrahedra together, precisely to form a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
|
4. 3.
As shown in the figure, construct $CE \perp AD$, connect $EF$, it is easy to prove that $EF \perp AD$. Therefore, $\angle CEF$ is the plane angle of the dihedral angle formed by plane $ADF$ and plane $ACD$.
Let $G$ be the midpoint of $CD$. Similarly, $\angle AGB$ is the plane angle of the dihedral angle formed by plane $ACD$ and plane $BCD$.
D
Given that $\angle CEF = \angle AGB$.
Let the side length of the base $\triangle CDF$ be $2a$, and the length of the side $AD$ be $b$. In $\triangle ACD$, $CE \cdot b = AG \cdot 2a$.
Thus, $CE = \frac{AG \cdot 2a}{b} = \frac{\sqrt{b^2 - a^2} \cdot 2a}{b}$.
In $\triangle ABC$, it is easy to find that
$AB = 2 \sqrt{b^2 - \left(\frac{2}{3} \sqrt{3} a\right)^2} = 2 \sqrt{b^2 - \frac{4}{3} a^2}$.
Since $\triangle CEF \sim \triangle AGB$, we have $\frac{AB}{CF} = \frac{AG}{CE}$,
which means $\frac{2 \sqrt{b^2 - \frac{4}{3} a^2}}{2a} = \frac{\sqrt{b^2 - a^2}}{\frac{\sqrt{b^2 - a^2} \cdot 2a}{b}}$.
Thus, we find that $b = \frac{4}{3} a$.
Therefore, $b = 2$, and $2a = 3$.
Hence, the distance between the farthest two vertices is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$ .
|
6. 4 .
Let $A(x, y)$ be an integer point on circle $O$. As shown in the figure, the equation of circle $O$ is $y^{2}+(x-199)^{2}=$ $199^{2}$.
$$
\begin{array}{l}
\text { Clearly, } x=0, y=0 ; \\
x=199, y=199 ; \\
x=199, y=-199 ; x=389, y=0
\end{array}
$$
These are 4 solutions to the equation. However, when $y \neq 0, \pm 199$, $y$ is coprime with 199, so 199 can be expressed as the sum of the squares of two positive integers, i.e., $199=m^{2}+n^{2}$. Since $199=4 \times 49+3$, we can set $m=2 k, n=2 l+1$, then.
$$
199=4 k^{2}+4 l^{2}+4 l+1=4\left(k^{2}+l^{2}+l\right)+1 .
$$
The integer points are:
$$
(0,0),(199,199),(389,0),(199,-199) \text {. }
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8. Calculate
$$
\begin{array}{l}
\sqrt{3633 \times 3635 \times 3639 \times 3641+36} \\
-3636 \times 3638=
\end{array}
$$
|
Let $3637=a$. Then
the original expression $=$
$$
\begin{array}{l}
\sqrt{(a-4)(a-2)(a+2)(a+4)+36} \\
-(a+1)(a-1) \\
=\sqrt{\left(a^{2}-10\right)^{2}}-\left(a^{2}-1\right) \\
=a^{2}-10-a^{2}+1=-9 . \\
\end{array}
$$
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $a, b$ be unequal real numbers, and $a^{2}+2 a=$ $5 \cdot b^{2}+2 b=5$. Then $a^{2} b+a b^{2}=$
|
4. (10).
It is known that $a, b$ are two unequal real numbers which are exactly the two real roots of the equation $x^{2}+2 x-5=0$. Therefore, $a+b=-2, ab=-5$.
Thus, $a^{2} b+ab^{2}=ab(a+b)=(-2) \cdot(-5)=$
10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If $m=1996^{3}-1995^{3}+1994^{3}-1993^{3}$ $+\cdots+4^{3}-3^{3}+2^{3}-1^{3}$, then the last digit of $m$ is
|
5. (0).
$$
2^{3}-1^{3}=7,4^{3}-3^{3}=37,6^{3}-5^{3}=91,8^{3}-7^{3}=
$$
$169,10^{3}-9^{3}=271$. Therefore, the sum of the last digits of the first 10 numbers $10^{3}-9^{3}+8^{3}-7^{3}$ $+6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}$ is $1+9+1+$ $7+7=25$, and the last digit of 25 is 5. The last digit of the algebraic sum of every ten consecutive cubic numbers from the unit digit 1 to the unit digit 0 is always 5. For $1990^{3}$, there are a total of 199 fives, and the last digit remains 5. Additionally, $1996^{3}-1995^{3}+1994^{3}-1093^{3}+1992^{3}-1591^{3}$ also has a last digit of 5. Therefore, the last digit of $m$ is ).
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$. To make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National Competition)
|
Solution: According to the problem, all logarithms are positive. Therefore, from (*) we get
$$
\begin{aligned}
\text { LHS } & =\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \\
& \geqslant \frac{(1+1+1)^{2}}{\log _{1993}\left(\frac{x_{0}}{x_{1}} \cdot \frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}}\right)} \\
& =9\left(\log _{1993} \frac{x_{0}}{x_{3}}\right)^{-1} .
\end{aligned}
$$
It is clear that the maximum value of $k$ is 9.
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.3. $x, y$ are natural numbers. $k$ is a natural number of size 1. Find all natural numbers $n$ that satisfy: $3^{n}=x^{k}+y^{k}$, and provide a proof.
|
10. 3. Answer: $n=2$.
At the time: Let $3^{n}=x^{k}+y^{k}$, where $x$ and $y$ are prime (assuming $x > y$), $k>1$, and $n$ is a natural number. Of course, neither $x$ nor $y$ can be divisible by 3.
If $k$ is even, then $x^{k}$ and $y^{k}$ leave a remainder of 1 when divided by 3. Thus, the sum of $x^{k}$ and $y^{k}$ leaves a remainder of 2 when divided by 3, which is not a power of 3. This leads to a contradiction, so $k$ cannot be even.
If $k$ is odd and $k>1$, then
$$
3^{n}=(x+y)\left(x^{k-1}-\cdots+y^{k-1}\right).
$$
Thus, $x+y=3^{m}, m \geqslant 1$.
We will now prove that $n \geqslant 2 m$.
Since $k$ is divisible by 3 (see 9.3. Answer), let $x_{1}=x^{\frac{k}{3}}$, $y_{1}=y^{\frac{k}{3}}$ and substitute, so we can assume $k=3$.
Thus, $x^{3}+y^{3}=3^{n}, x+y=3^{m}$.
To prove $n \geqslant 2 m$, we need to use $x^{2}+y^{3} \geqslant(x+y)^{2}$, which means proving $x^{2}-x y+y^{2} \geqslant x y$.
Since $x \geqslant y+1$, then $x^{2}-x=x(x-1) \geqslant x y$.
$$
\left(x^{2}-x-x y\right)+\left(y^{2}-y\right) \geqslant 0.
$$
The inequality $n \geqslant 2 m$ is proven:
From the equation $(x+y)^{3}-\left(x^{3}+y^{3}\right)=3 x y(x+y)$, we get:
$$
3^{2 m-1}-3^{n-m-1}=x y,
$$
and $2 m-1 \geqslant 1$,
and $n-m-1 \geqslant n-2 m \geqslant 0$.
Therefore, if at least one of the inequalities in (**) is a strict inequality, then the left side of (*) is divisible by 3, but the right side is not. This leads to a contradiction.
If $n-m-1=n-2 m=0$, then $m=1, n=2$ and $3^{2}=2^{3}+1^{3}$. Hence, $n=2$.
|
2
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find the positive integer root of the equation
$$
\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2}=\frac{133}{60}
$$
(1990, Shanghai Junior High School Mathematics Competition)
|
Solution: Since $n$ is a natural number, by the properties of natural numbers and
$$
\begin{array}{l}
\frac{1}{n}>\frac{1}{n+1}>\frac{1}{n+2} \\
\therefore \frac{2+3+4}{n+2}<\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2} \\
<\frac{2+3+4}{n} .
\end{array}
$$
That is, $\frac{9}{n+2}<\frac{133}{60}<\frac{9}{n}$.
Solving this, we get $2 \frac{8}{133}<n<4 \frac{8}{133}$.
Since $n$ is a natural number, $n$ can only be 3 or 4. Upon verification, $n=3$ is the positive integer solution of the equation.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Solve the equation
\[
\begin{array}{l}
\sqrt{x^{2}+5 x-14}+\sqrt{x+7}+\sqrt{2-x}+x-5 \\
=0 .
\end{array}
\]
|
Solution: By the properties of quadratic radicals, we get
$$
\left\{\begin{array}{l}
x^{2}+5 x-14 \geqslant 0, \\
x+7 \geqslant 0, \\
2-x \geqslant 0 .
\end{array}\right.
$$
Solving, we get $-7 \leqslant x \leqslant-7$ or $2 \leqslant x \leqslant 2$.
That is, $x=-7$ or $x=2$.
Upon verification, $x=2$ is a root of the original equation.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 For a finite set $A$, function $f: N \rightarrow A$ has only the following property: if $i, j \in N, |H| i-j |$ is a prime number, then $f(i) \neq f(j)$. How many elements does set $A$ have at least?
|
Solution: Let $|A|$ denote the number of elements in the finite set $A$, and estimate the lower bound of $|A|$.
Since the absolute value of the difference between any two numbers among $1, 3, 6, 8$ is a prime number, by the problem's condition: $f(1)$, $f(3)$, $f(6)$, $f(8)$ are four distinct elements in $A$. Therefore, $|A| \geqslant 4$.
Construct the set $A=\{0,1,2,3\}$, then the function $f: N \rightarrow A$ has the following correspondence:
. If $x \in N, x=4 k+r$, then $f(x)=r$. That is, $k \in N \bigcup\{0\}, r=0,1,2,3$.
Next, we prove that $f$ satisfies the conditions of the problem:
For any $x, y \in N$, if $|x-y|$ is a prime number, then $f(x) \neq f(y)$. Otherwise, if $f(x)=f(y)$, then
$x \equiv y(\bmod 4)$.
Thus, $4 \mid |x-y|$, which contradicts the fact that $|x-y|$ is a prime number.
From the above, we conclude that $|A|_{\text {min }}=4$.
The proof of the inequality involves various techniques, such as constructing counterexamples, constructing sequences, constructing propositions, etc. Please list them.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, let $A=\{1,2,3, \cdots, 17\}$. For any function $f: A \rightarrow A$, denote
$$
f^{[1]}(x)=f(x), f^{[k+1]}(x)=f\left(f^{[k]}(x)\right)
$$
$(k \in \mathbb{N})$. Find the natural number $M$ such that:
$$
\begin{array}{l}
\text { (1) When } m<M, 1 \leqslant i \leqslant 16, \text { we have } \\
f^{[m]}(i+1)-f^{[m]}(i) \neq \pm 1(\bmod 17), \\
f^{[m]}(1)-f^{[m]}(17) \neq \pm 1(\bmod 17) ;
\end{array}
$$
(2) When $1 \leqslant i \leqslant 16$, we have
$$
\begin{array}{l}
f^{[M]}(i+1)-f^{[M]}(i) \\
\quad \equiv 1 \text { or }-1(\bmod 17), \\
\quad f^{[M]}(1)-f^{[M]}(17) \equiv 1 \text { or }-1(\bmod 17) .
\end{array}
$$
Determine the maximum possible value of $M$ for all functions $f$ that satisfy the above conditions, and prove your conclusion.
|
Five, the required $M_{0}=8$.
First, prove $M_{0} \geqslant 8$.
In fact, we can define the mapping $f(i) \equiv 3 i-2(\bmod 17)$, where $i \in A, f(i) \in A$.
If $f(i) \equiv f(j)(\bmod 17)$,
then $3 i-2 \equiv 3 j-2(\bmod 17)$,
we have $i \equiv j(\bmod 17)$,
$$
\therefore i=j \text {. }
$$
The mapping $f$ is a mapping from $A$ to $A$.
From the definition of the mapping $f$, we know
$$
\begin{array}{l}
f^{[n]}(i) \equiv 3^{n} \cdot i-3^{n}+1(\bmod 17) \text {. } \\
\text { If }\left\{\begin{array}{l}
f^{[M]}(i+1)-f^{[M]}(i) \equiv 1 \text { or }-1(\bmod 17), \\
f^{[M]}(1)-f^{[M]}(17) \equiv 1 \text { or }-1(\bmod 17),
\end{array}\right. \\
\text { then }\left\{\begin{array}{l}
{\left[3^{M}(i+1)-3^{M}+1\right]-\left[3^{M} \cdot i-3^{M}+1\right]} \\
\equiv+1 \text { or }-1(\bmod 17), \\
1 \quad\left[3^{M} \times 17-3^{M}+1\right]=+1 \text { or }-1(\bmod 17) .
\end{array}\right. \\
\end{array}
$$
Then $3^{M}=+1$ or $-1(\bmod 17)$
$$
\begin{aligned}
\text { Since } 3^{1} \equiv 3,3^{2} \equiv 9,3^{3} \equiv 10,3^{4}=13,3^{5}=5,3^{6} \equiv 15, \\
\left.3^{7} \equiv 11,3^{4} \equiv-1 \text { (mod } 17\right),
\end{aligned}
$$
Thus $M_{0} \geqslant 8$.
Proof: $M_{0} \leqslant 8$.
Consider a 17-sided polygon $A_{1} A_{2} \cdots A_{17}$, denoted as $G$. We first define:
When $i+1=18$, take $i+1$ as 1; when $i-1=0$, take $i-1$ as 17. Then, according to the rule, connect the segments: if $i \leqslant mp>q>0$, such that
$$
\begin{array}{l}
f_{[f]}^{[f]}(i)=f^{[\varphi]}(j), \\
f_{[f]}^{[f]}(i+1)=f^{[q]}(j+1) \\
f^{[p]}(i+1)=f^{[q]}(j-1) .
\end{array}
$$
Thus $f^{[\rho]}(i+1)=f^{[q]}(j-1)$.
From this, we have
$$
\left\{\begin{array}{l}
f^{[p-q]}(i)=j, \\
f^{[p-q]}(i+1)=j+1 \text { or } f^{[p-q]}(i+1)=j-1,
\end{array}\right.
$$
Therefore, the connected segments do not overlap.
Since $G$ has $17 \times 7$ diagonals.
Thus, $17 \times\left(M_{0}, 1\right) \leqslant 17 \times 7$.
Thus $M=8$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. Solve the equation
$$
\begin{array}{l}
\sqrt{x-1}+\sqrt{2 x-3}+\sqrt{3 x-5} \\
+\sqrt{4 x-7}=5 x-6 .
\end{array}
$$
(First Yangtze Cup Correspondence Competition for Junior High School Students)
|
Solution: By applying the method of completing the square, the original equation can be transformed into
$$
\begin{array}{l}
(\sqrt{x-1}-1)^{2}+(\sqrt{2 x-3}-1)^{2} \\
+(\sqrt{3 x-5}-1)^{2}+(\sqrt{4 x-7}-1)^{2}=0 . \\
\therefore \sqrt{x-1}-1=\sqrt{2 x-3}-1 \\
\quad=\sqrt{3 x-5}-1=\sqrt{4 x-7}-1=0 .
\end{array}
$$
Solving this, we get $x=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
24. (15 points) Given the parabola $y^{2}=\operatorname{tar}(0<a<1)$ with focus $\vec{F}$, a semicircle is constructed above the $x$-axis with center $A(a+4,0)$ and radius $|A F|$, intersecting the parabola at two distinct points $M$ and $N$. Let $P$ be the midpoint of segment $M N$.
(1) Find the value of $|M F|+|N F|$;
(2) Does there exist a value of $a$ such that $|M F|$, $|P F|$, $|N F|$ form an arithmetic sequence? If so, find the value of $a$; if not, explain the reason.
|
24. (1) From the given, we have $F(a, 0)$, and the semicircle is
$$
[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0) \text {. }
$$
Substituting $y^{2}=4 a x$ into the equation, we get
$$
x^{2}-2(4-a) x+a^{2}+8 a=0 \text {. }
$$
Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then, by the definition of the parabola, we have
$$
\begin{aligned}
|M F|+|N F| & =x_{1}+x_{2}+2 a \\
& =2(4-a)+2 a=8 .
\end{aligned}
$$
(2) If $|M F|,|P F|,|N F|$ form an arithmetic sequence, then
$$
2|P F|=|M F|+|N F| \text {. }
$$
On the other hand, let the projections of $M, P, N$ on the directrix of the parabola be $M^{\prime}, P^{\prime}, N^{\prime}$. Then, in the right trapezoid $M^{\prime} M N N^{\prime}$, $P^{\prime} P$ is the midline, and we have
$$
\begin{aligned}
2\left|P^{\prime} P\right| & =\left|M^{\prime} M\right|+\left|N^{\prime} N\right| \\
& =|F M|+|F N|,
\end{aligned}
$$
thus $|P F|=\left|P^{\prime} P\right|$. This indicates that point $P$ should be on the parabola.
However, it is known that $P$ is the midpoint of segment $M N$, i.e., $P$ is not on the parabola.
$\therefore$ There does not exist a value of $a$ such that $|M F|,|P F|,|N F|$ form an arithmetic sequence.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given that $x, y, z$ are two non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $S=2x+y-z$, then what is the sum of the maximum and minimum values of $S$?
(1996, Tianjin City Junior High School Mathematics Competition)
|
Analysis: By representing the algebraic expression $S$ with one letter and determining the range of this letter, we can find the maximum or minimum value of $S$.
From the given, we solve $y=\frac{7-4 x}{3}, z=\frac{1-x}{3}$, and.
$$
\left\{\begin{array}{l}
\frac{7-4 x}{3} \geqslant 0, \\
\frac{1-x}{3} \geqslant 0, \\
x \geqslant 0.
\end{array} \text { Thus, } \left\{\begin{array}{l}
x \leqslant \frac{7}{4}, \\
x \leqslant 1, \\
x \geqslant 0,
\end{array}\right.\right.
$$
That is, $0 \leqslant x \leqslant 1$.
And $S=2 x+\frac{7-4 x}{3}-\frac{1-x}{3}=x+2$, obviously 2
$\leqslant S \leqslant 3$. Therefore, the sum of the maximum and minimum values of $S$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
$(1996$, National Junior High School Mathematics League)
|
$$
\begin{array}{l}
\text{Analysis: Let } x_{1} \text{ and } x_{2} \text{ be the x-coordinates of the points where the parabola intersects the x-axis, then } x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}. \text{ It is easy to know that } -\frac{b}{a}<0, \text{ and } -116 c(a-b+c). \\
\because a-b+c>0, c \geqslant 1, \\
\therefore a^{2}>16, a>4 .
\end{array}
$$
$>20 c$. Thus, the minimum value of $b$ is $5$, the minimum value of $c$ is $1$, and the parabola is $y=5 x^{2}+5 x+1$, intersecting the x-axis at $\left(\frac{-5+\sqrt{5}}{10}, 0\right),\left(\frac{-5-\sqrt{5}}{10}, 0\right)$, which meets the requirements.
Therefore, $(a+b+c)_{\text{min}}=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Natural numbers $m, n$ satisfy $8 m+10 n>9 m n$. Then
$$
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4}=
$$
$\qquad$
|
3. 2 .
It is known that $\frac{8}{n}+\frac{10}{m}>9$. It is easy to see that one of the two natural numbers $m, n$ must be 1 (otherwise $\frac{8}{m}+\frac{10}{m} \leqslant \frac{8}{2}+\frac{10}{2}=9$ which is a contradiction). Without loss of generality, let $m=1$, then
$$
\begin{array}{l}
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4} \\
=1+n^{2}-n^{2}+1+n^{4}-n^{4}=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $(-1)^{n} 2^{n} \equiv b_{n}(\bmod 9), 0 \leq b_{n} \leq 8$. Then the range of values for $b_{n}$ is \{ $\qquad$ \}.
|
$\begin{array}{l}\text { II.1.1. } \\ \text { \| }(-1)^{n} 2^{3 n}=(-8)^{n}=(1-9)^{n} \\ \quad=C_{n}^{n}-C_{n}^{1} 9+C_{n}^{2} 9^{2}+\cdots+(-1)^{n} C_{n}^{n} 9^{n}, \\ \text { know }(-1)^{n} 2^{3 n} \equiv 1(\bmod 9) .\end{array}$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
The struggle of the land, by 1996, the forest coverage rate of the whole county had reached 30% (becoming an oasis). From 1997 onwards, each year will see such changes: 16% of the original desert area will be converted, and 4% of it will be eroded, turning back into desert.
(1) Assuming the total area of the county is 1, in 1996 the oasis area is $a_{1}=\frac{3}{10}$, after $n$ years the oasis area becomes $a_{n+1}$. Prove:
$$
a_{n+1}=\frac{4}{5} a_{n}+\frac{4}{25} \text {. }
$$
(2) How many years at least are needed to make the forest coverage rate of the whole county exceed 60% (round to the nearest whole number)?
|
Let the accumulated area be $b_{n+1}$. Then
$$
a_{1}+b_{1}=1, \quad a_{n}+b_{n}=1 .
$$
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} \cdot a_{n}$, which is
$$
a_{n}-\frac{4}{100} a_{n}=\frac{96}{100} a_{n} .
$$
The other part is the newly greened area $\frac{16}{100} b_{n}$. Therefore,
$$
\begin{aligned}
a_{n+1} & =\frac{96}{100} a_{n}+\frac{16}{100} b_{n}=\frac{96}{100} a_{n}+\frac{16}{100}\left(1-a_{n}\right) \\
& =\frac{4}{5} a_{n}+\frac{4}{25} .
\end{aligned}
$$
(2) From (1), we have
$$
\begin{aligned}
a_{n+1}-\frac{4}{5} & =\frac{4}{5}\left(a_{n}-\frac{4}{5}\right)=\left(\frac{4}{5}\right)^{2}\left(a_{n-1}-\frac{4}{5}\right) \\
& =\cdots=\left(\frac{4}{5}\right)^{n}\left(a_{1}-\frac{4}{5}\right) \\
& =-\frac{1}{2}\left(\frac{4}{5}\right)^{n} . \\
\therefore a_{n+1} & =\frac{4}{5}-\frac{1}{2}\left(\frac{4}{5}\right)^{n} .
\end{aligned}
$$
According to the problem, $a_{n+1} \geqslant 60 \%$, i.e.,
$$
\frac{4}{5}-\frac{1}{2}\left(\frac{4}{5}\right)^{n}>\frac{3}{5} \text {. }
$$
This simplifies to $\left(\frac{4}{5}\right)^{n}<\frac{2}{5}$,
or $(1-0.2)^{n}<0.4$.
To find the integer value, we use an upper approximation for $(1-0.2)^{n}$
$$
(1-0.2)^{n}<1-C_{n}^{1} \times 0.2+C_{n}^{2} \times 0.04 \leqslant 0.4,
$$
which simplifies to $n^{2}-11 n+30 \leqslant 0$.
Solving this, we get $5 \leqslant n \leqslant 6$.
Taking the smallest integer $n=5$, we verify that $a_{5}<0.6<a_{6}$, so after 5 years of effort, the greened area can exceed $60 \%$.
Here we used an approximation. We can also solve $a_{n} \leqslant 0.6<a_{n+1}$ to get $n=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initial 52. Given real numbers $a, b, x, y$ satisfy $a x^{n}+b y^{n}=1+2^{n+1}$ for any natural number $n$. Find the value of $x^{a}+y^{b}$.
|
Let $S_{n}=a x^{\prime \prime}+b y^{\prime \prime}$. Then
$$
S_{1}=a x+b y=5 \text {, }
$$
$$
\begin{array}{l}
S_{0}=a x^{2}+b y^{2}=9 . \\
S=a x^{3}+b y^{3}=17 . \\
S_{1}=a x^{4}+b y^{\prime}=33 .
\end{array}
$$
Let $x+y=A, x y=B$. Then $x, y$ are the roots of the quadratic equation $t^{2}-A t+B=0$. Therefore,
$$
\begin{array}{l}
x^{2}-A x+B=0, \\
y^{2}-A y+B=0 .
\end{array}
$$
a. $x^{n-2} \cdot$ (3) $+b y^{n-2} \cdot$ (1), we get
$$
S_{n}=A S_{n-1}-B S_{n-2} \text {. }
$$
From the problem, we have $\left\{\begin{array}{l}S_{3}=A S_{2}-B S_{1}, \\ S_{4}=A S_{3}-B S_{2},\end{array}\right.$
which gives $\left\{\begin{array}{l}17=9 A-5 B, \\ 33=17 A-9 B .\end{array}\right.$
Solving these, we get $\left\{\begin{array}{l}A=3, \\ B=2 .\end{array}\right.$.
Thus, $\left\{\begin{array}{l}x+y=3 \\ x y=2\end{array}\right.$
Substituting $x=1, y=2$ into (1) and (2), we get $a=1, b=2$.
Similarly, when $x=2, y=1$, we get $a=2, b=1$.
(1) When $x=1, y=2, a=1, b=2$,
$$
x^{4}+y^{b}=1^{1}+2^{2}=5 \text {. }
$$
(2) When $x=2, y=1, a=2, b=1$,
$$
x^{2}+y^{3}=1^{1}+2^{2}=5 \text {. }
$$
（From Yu Yuanzheng, "Zhonghua Yinghao School, Conghua, Guangzhou, 510960）
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{c}
\text { Example } 1 \text { Calculate } \frac{a}{a^{3}+a^{2} b+a b^{2}+b^{3}}+ \\
\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}}+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$
(1995, Tianjin City, Grade 7 "Mathematics Competition)
|
$\begin{array}{l}\text { Solution: Original expression }=\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\ +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\ =\frac{a^{2}+b^{2}}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\ =\frac{1}{a^{2}-b^{2}}-\frac{1}{a^{2}-b^{2}}=0 . \\\end{array}$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If $a+b+c=a b c \neq 0$, find the value of $\frac{\left(1-a^{2}\right)\left(1-b^{2}\right)}{a b}+\frac{\left(1-b^{2}\right)\left(1-c^{2}\right)}{b c}+$ $\frac{\left(1-c^{2}\right)\left(1-a^{2}\right)}{a c}$.
(1990, Wuhan City Mathematics Competition)
|
$\begin{array}{l}\text { Solution: Original expression }=\frac{1-a^{2}-b^{2}+a^{2} b^{2}}{a b} \\ +\frac{1-b^{2}-c^{2}+b^{2} c^{2}}{b c}+\frac{1-a^{2}-c^{2}+a^{2} c^{2}}{a c} \\ =\frac{1}{a b}-\frac{a}{b}-\frac{b}{a}+a b+\frac{1}{b c}-\frac{b}{c}-\frac{c}{b}+b c \\ +\frac{1}{a c}-\frac{a}{c}-\frac{c}{a}+a c \\ =\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right)-\frac{b+c}{a}-\frac{a+c}{b}-\frac{a+b}{c} \\ +a b+b c+c a \\ =\frac{a+b+c}{a b c}-(b c-1)-(a c-1) \\ -(a b-1)+a b+b c+c a \\ =4 \text {. } \\\end{array}$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 20 points) Given the equation $x^{2}+m x-m+1=0$ (where $m$ is an integer) has two distinct positive integer roots. Find the value of $m$.
---
The translation maintains the original format and line breaks as requested.
|
Three, let the two distinct positive integer roots be $\alpha, \beta(\alpha<\beta)$. By Vieta's formulas, we have $\left\{\begin{array}{l}\alpha+\beta=-m, \\ \alpha \beta=-m+1 .\end{array}\right.$ Eliminating $m$, we get $\alpha \beta-\alpha-\beta=1$.
That is, $(\alpha-1)(\beta-1)=2$.
Then $\left\{\begin{array}{l}\alpha-1=1, \\ \beta-1=2\end{array} \Rightarrow \alpha=2, \beta=3\right.$.
Therefore, $m=-(\alpha+\beta)=-5$.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a, b, c$ are distinct. Find the value of $\frac{2 a-b-c}{(a-b)(a-c)}$ $+\frac{2 b-c-a}{(b-c)(b-a)}+\frac{2 c-a-b}{(c-a)(c-b)}$.
|
(Answer: 0)
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$. If $x=$
$$
\begin{array}{l}
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}\right. \\
\left.+\frac{1}{b}\right) \text {. Then, } x+2 y+3 x y=
\end{array}
$$
.
|
1. 2
The above text has been translated into English, maintaining the original text's line breaks and format.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $X=\{-1,0,1\}, Y=\{-2,-1,0,1,2\}$, and for all elements $x$ in $X$, $x+f(x)$ is even. Then the number of mappings $f$ from $X$ to $Y$ is ( ).
(A) 7
(B) 10
(C) 12
(D) 15.
|
4. (C).
$x$ and $f(x)$ are both even, or both odd.
When $x=0$, there are three possibilities: $-2, 0, 2$;
When $x=-1, 1$, there are two possibilities each for -1 and 1, making 4 possibilities.
Therefore, the total is $3 \times 4=12$.
|
12
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 1. } \lim _{n \rightarrow \infty} \frac{1}{\sqrt[3]{n}}\left(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+ \\ \frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+ \\ \left.\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}}\right)=\end{array}$
|
$$
\approx .1 .1 \text {. }
$$
Since
$$
\begin{array}{l}
\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+\cdots+ \\
\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}} \\
=\frac{\sqrt[3]{2}-1}{2-1}+\cdots+\frac{\sqrt[3]{n}-\sqrt[3]{n-1}}{n-(n-1)} . \\
=\sqrt[3]{n}-1,
\end{array}
$$
Therefore, the required limit is $\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n}-1}{\sqrt[3]{n}}=1$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Simplify $\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}$ $+\frac{(x+a)(x+b)}{(c-a)(c-b)}$.
|
(Answer: 1).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $x=b y+c z, y=c z+a x, z=a x$ $+b y$. Find the value of $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$.
|
Solution: $\frac{a}{a+1}=\frac{a x}{a x+x}=\frac{a x}{a x+b y+c z}$.
Similarly, $\frac{b}{b+1}=\frac{b y}{a x+b y+c z}$,
$$
\frac{c}{c+1}=\frac{c z}{a x+b y+c z} \text {. }
$$
Adding them up, the value of the desired expression is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If $\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=$ $\frac{t}{x+y+z}$, let $f=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}$. Prove: $f$ is an integer.
(1990, Hungarian Mathematical Competition)
|
Proof: If $x+y+z+t \neq 0$, by the property of proportion, we have
$$
\begin{array}{l}
\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y} \\
=\frac{t}{x+y+z}=\frac{x+y+z+t}{3(x+y+z+t)}=\frac{1}{3} . \\
\text { Then we have }\left\{\begin{array}{l}
y+z+t=3 x, \\
z+t+x=3 y, \\
t+x+y=3 z, \\
x+y+z=3 t .
\end{array}\right.
\end{array}
$$
Solving, we get $x=y=z=t, f=4$ is an integer.
If $x+y+z+t=0$, it is easy to see that $f=-4$ is also an integer.
|
-4
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0(x y z \neq 0)$. Find the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$.
(1992, Sichuan Province Junior High School Mathematics League Preliminary)
|
Solution: Let $y=k_{1} x, z=k_{2} x$. Then
$$
\left\{\begin{array} { l }
{ 4 - 3 k _ { 1 } - 6 k _ { 2 } = 0 , } \\
{ 1 + 2 k _ { 1 } - 7 k _ { 2 } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
k_{1}=\frac{2}{3}, \\
k_{2}=\frac{1}{3} .
\end{array}\right.\right.
$$
Then $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}=\frac{2+3 k_{1}^{2}+6 k_{2}^{2}}{1+5 k_{1}^{2}+7 k_{2}^{2}}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given the theorem: "If three prime numbers greater than 3, $a, b$, and $c$, satisfy the equation $2a + 5b = c$, then $a + b + c$ is a deficient number of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion.
|
Three, the maximum possible value of $n$ is 9.
First, prove that $a+b+c$ can be divided by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$, so $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, then $r_{a} \neq 0, r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, then $r_{a}=1, r_{b}=2$ or $r_{a}=2, r_{b}=1$. In this case, $2a+5b$ must be a multiple of 3, i.e., $c$ is a composite number, which is a contradiction.
Therefore, $r_{a}=r_{b}$, then $r_{a}=r_{b}=1$ or $r_{a}=r_{b}=2$,
In this case, $a+2b$ must be a multiple of 3, thus $a+b+c$ is a multiple of 9.
Next, prove that 9 is the largest.
$$
\because 2 \times 11+5 \times 5=47 \text {, and } 11+5+47=63 \text {, }
$$
While in $2 \times 13+5 \times 7=61$, $13+7+61=81$, and $(63, 81)=9$, hence 9 is the largest possible value.
|
9
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
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