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stringclasses 1
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|---|---|---|---|---|---|---|---|---|
2. Mark 10 points on a circle, how many different polygons can be formed using these points (all or part)? (Polygons are considered the same only if their vertices are exactly the same).
|
$$
\begin{array}{l}
\left.+C_{10}^{2}+\cdots+C_{10}^{0}+C_{10}^{10}\right]-\left[C_{10}^{0}+C_{10}^{\overrightarrow{1}}\right. \\
\left.+C_{10}^{2}\right]=(1+1)^{10}-(1+10+45)=968 \text{ different convex polygons.} \\
\end{array}
$$
For $3 \leqslant k \leqslant 10$, every selection of $k$ points can form a convex $k$-gon, i.e., $k$ points can be chosen in $C_{10}^{\mathrm{E}}$ ways,
|
968
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3 . Let $n$ be a positive integer, $d$ be a digit among the 10 decimal digits, and suppose $\frac{n}{810}=0 . d 25 d 25 d 25 \cdots$, find $n$.
|
$$
\begin{array}{c}
3 . \because \frac{n}{810}=0 . d 25 d 25 d 25 \cdots, \\
\therefore 1000 \cdot \frac{n}{810}=d 25 . d 25 d 25 \cdots \\
\therefore \frac{999 n}{810}=1000 \cdot \frac{n}{810}-\frac{n}{810} \\
=d 25=100 d+25, \\
\therefore \quad 999 n=810(100 d+25),
\end{array}
$$
And $(750,37)=1$, so $37 \mid (4d+1)$, and since $d$ is a single digit, hence $d=9$, then $n=750$.
|
750
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $a<b<c<d<e$ are consecutive positive integers, $b+c+d$ is a perfect square, and $a+b+c+d+e$ is a perfect cube, what is the minimum value of $c$?
|
$$
\begin{aligned}
4 . & b+c+d=3 c \\
& a+b+c+a+e=5 c
\end{aligned}
$$
and $b+c+d=m^{2}, a+b+c+d+e=n^{3}$
$$
\begin{aligned}
\therefore \quad 3 c & =m^{2}, \\
5 c & =n^{3} .
\end{aligned}
$$
$$
\begin{array}{l}
\therefore 3^{3} \mid c_{0} \text { and } 5\left|n_{0} \quad \therefore 5^{2}\right| c, \quad \therefore 25 \cdot 27 \mid c_{0} \\
\therefore \quad c=25 \cdot 27=675 .
\end{array}
$$
|
675
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Two Skaters
Allie and Bllie are located
at points $A$ and $B$ on
a smooth ice surface, with $A$ and $B$
being 100 meters apart. If Allie starts from $A$
and skates at 8 meters/second
along a line that forms a $60^{\circ}$
angle with $A B$, while Bllie starts from $B$ at 7 meters/second
along the shortest line to meet Allie, how far will Allie have skated when they meet?
|
6. Let the two people meet at point $C$ after $t$ seconds, then the sides of $\triangle A B C$ are $A B=100, A C=8 t, B C=7 t$. By the cosine rule, we get
$$
\begin{aligned}
(7 t)^{2}= & 100^{2}+(8 t)^{2}-2(100)(8 t) \cos 60^{\circ} \\
\therefore \quad 0 & =3 t^{2}-160 t+2000 \\
& =(3 t-100)(t-20) .
\end{aligned}
$$
$\therefore t=20$ or $t=\frac{100}{3}$. Since the meeting time should be as short as possible, hence $t=20$, then $A C=8 \cdot 20=160$ (meters).
|
160
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If the integer $k$ is added to $36,300,596$, respectively, the results are the squares of three consecutive terms in an arithmetic sequence, find the value of $k$.
|
7. Let $\left\{\begin{array}{l}36+k=(n-d)^{2} , \\ 300+k=n^{2} \\ 596+k=(n+d)^{2}\end{array}\right.$
(1)
From (3) - (1) we get $4 n d=560$,
$\therefore n d=140$.
From (2) $\times 2-(1)-$ (3) we get $2 d^{2}=32$,
$\therefore d= \pm 4$, then $n= \pm 35$.
From (2), $k=n^{2}-300=35^{2}-300=925$.
|
925
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Euler's conjecture was refuted by American mathematicians in 1960, who proved that there exists a positive integer $n$ such that $133^{5}+110^{5}+84^{5}+27^{5}=n^{5}$. Find $n$ when it is satisfied.
|
9. Clearly $n \geqslant 134$. Now we find the upper bound of $n$.
$$
\begin{aligned}
\because \quad n^{5}= & 133^{5}+110^{5}+84^{5}+27^{5}<133^{5} \\
& +110^{5}+(27+84)^{5}<3(133)^{5} \\
& <\frac{3125}{1024}(133)^{5}=\left(\frac{5}{4}\right)^{5}(133)^{5}, \\
\therefore \quad & n<\frac{5}{4}(133), \text { i.e., } n \leqslant 166 .
\end{aligned}
$$
When a positive integer is raised to the fifth power, the last digit remains unchanged,
$\therefore n$ has the same last digit as $133+110+84+27$. The last digit of the latter is 4,
$$
\begin{array}{l}
\therefore n \text { is one of 134, 144, 154, 164. } \\
\because \quad 133 \equiv 1(\bmod 3), 110 \equiv 2(\bmod 3), \\
84 \equiv 0(\bmod 3), 27 \equiv 0(\bmod 3), \\
\therefore n^{5}=133^{5}+110^{5}+84^{5}+27^{5} \text {. } \\
\equiv 1^{5}+2^{5} \equiv 0(\bmod 3) \text {. } \\
\therefore \quad n=144 \text {. } \\
\end{array}
$$
|
144
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given an integer array consisting of 121 integers, where each number is between 1 and 1000 (repetition is allowed), this array has a unique mode (i.e., the integer that appears most frequently). Let $D$ be the difference between this mode and the arithmetic mean of the array. When $D$ reaches its maximum value, what is $[D]$ equal to?
|
11. Let $M$ and $m$ be the "mode" and "arithmetic mean" respectively. Without loss of generality, assume $M \geqslant m$. To maximize $D$, then $M=1000$ (otherwise, if $M=1000-k$, then increasing $M$ by $k$ while $m$ increases by no more than $k$, $D$ is obviously non-decreasing). In the case of $M=1000$, we must make $m$ as small as possible. At this time, 1000 must appear at least twice in the array. If 1000 appears exactly twice, then the other numbers can only appear once. In this case, when the other (121-2) numbers are $1,2, \cdots, 119$,
If $M=1000$ appears 3 times, then the other numbers can appear at most 2 times. At this time, when the other 118 numbers are $1,1,2,2, \cdots, 59,59$, $m$ is the smallest, $m=\frac{\frac{118(59+1)}{2}+3000}{121}=\frac{6540}{121}$. If $M$ appears 4 times, then the smallest value of $m$ is $\frac{\frac{117(39+1)}{2}+4000}{121}=\frac{6340}{121}$. If $M$ appears 5 times, the smallest value of $m$ is $\frac{\frac{116(29+1)}{2}+5000}{121}=\frac{6740}{121}$. If $M$ appears 6 times, then the smallest value of $m$ is $\frac{115(23+1)+6000}{121}=\frac{7380}{121}$. If $M=1000$ appears $n$ times $(n \geqslant 7)$, then $m \geqslant \frac{1000 n}{121} \geqslant \frac{7000}{121}$.
$\therefore$ When $m=1000$ appears exactly 4 times, $m$ is the smallest, $m=\frac{6340}{121}=52+\frac{48}{121}$. The maximum value of $D$ is $1000-(52+\frac{48}{121})$, so $[D]=947$.
|
947
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. As shown in the figure, $ABCD$ is a tetrahedron, $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, $CD=13$. Let $d$ be the distance between the midpoints of $AB$ and $CD$. Find the value of $d^{2}$.
|
12. By the median formula, we get $m_{s}^{2}=\frac{1}{4}\left(2 b^{2}+2 c^{2}\right.$ $\left.-a^{2}\right)$, so $P C^{2}=\frac{1}{4}\left[2(A C)^{2}+2(B C)^{2}\right.$ $\left.-(A B)^{2}\right]=\frac{1009}{4}$.
$$
\begin{array}{l}
P D^{2}\left.=\frac{1}{4}\left[2(A D)^{2}+2(B D)^{2}\right]-(A B)^{2}\right] \\
=\frac{425}{4} . \\
\therefore \quad P Q^{2}=\frac{1}{4}\left[2(P C)^{2}+2(P D)^{2}\right. \\
\left.-(C D)^{2}\right]=137 . \quad \text { (to be continued) }
\end{array}
$$
(to be continued)
|
137
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5 . There are two piles of stones. If 100 stones are taken from the first pile and placed in the second pile, then the number of stones in the second pile will be twice that in the first pile. Conversely, if some stones are taken from the second pile and placed in the first pile, then the number of stones in the first pile will be six times that in the second pile. How many stones are there in the first pile at a minimum? And determine the number of stones in the second pile in this case.
|
Let the number of stones in the first and second piles be denoted as $x, y$. Let $z$ be the number of stones taken from the second pile and placed into the first pile. Thus, we obtain the equations
$$
\begin{array}{l}
2(x-100)=y+100, \\
x+z=6(y-z) .
\end{array}
$$
From (1), we get $y=2 x-300$, substituting this into (2) yields:
$$
4 x+7(x-z)=1800 \text{. }
$$
Therefore, $x-z$ must be divisible by 4. Let $x-z=4 t, t \in \mathbb{Z}$. From equation (3), we get $x=-7t+450$; hence $y=-14t+600, z=x-4t=-11t+450$. Since $x, y, z$ are natural numbers, they must satisfy the inequalities:
$$
\left\{\begin{array}{l}
-7 t+450>0, \\
-14 t+600>0, \\
-11 t+450>0 .
\end{array}\right.
$$
Solving this system of inequalities, we get $t \leqslant 40$. Since the value of $x$ decreases as $t$ increases, the minimum possible value of $x$ is achieved when $t$ is at its maximum, i.e., $t=40$. In this case, $x=170, y=40, z=10$. Upon verification, the values of $x, y, z$ satisfy the conditions of the problem.
|
170
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. The decimal representation of the natural number $A$ is $\overline{a_{n} a_{n}-\overline{1} \cdots a_{1} a_{0}}$, let
$f(A)=2^{\mathrm{n}} a_{0}+2^{\mathrm{n}-1} a_{1}+\cdots+2 a_{\mathrm{n}-1}^{\mathrm{n}}+a_{\mathrm{n}}$ denote $A_{1}=f(A), A_{\mathrm{i}+\mathrm{i}}=f\left(A_{\mathrm{i}}\right),(i=1,2, \cdots)$.
Prove:
(1) There must be a natural number $k$ such that $A_{\mathrm{k}+1}=A_{\mathrm{k} 3}$
(2) If $A=19^{86}$, what is the value of the above $A_{\mathrm{L}}$? And explain the reason.
(First National Mathematical Olympiad Training Team Selection Exam)
|
Prove $f(A)=2^{n} a_{0}+2^{n-1} a_{1}^{\gamma}+\cdots+2 a_{n-1}$
$$
\begin{array}{l}
+a_{n} \leqslant\left(2^{n}+2^{n-1}+\cdots+2+1\right)_{9} \\
=9\left(2^{n+1}-1\right) .
\end{array}
$$
While $A \geqslant 10^{\text { }} a_{\text {n }}$
Next, we prove that when $n \geqslant 2$, $A>f(A)$.
By $a_{n} \cdot 10^{n} \geqslant 10^{n}=10 \cdot 10^{n-1}$
$$
\begin{array}{l}
>9 \cdot 10^{n-1}>9 \cdot 8 \cdot 10^{n-2} \\
>9 \cdot 8 \cdot 2^{n-2}=9 \cdot 2^{n+1}>9\left(2^{n+1}-1\right) .
\end{array}
$$
Thus, $10^{\mathrm{n}} a_{\mathrm{n}}>9\left(2^{\mathrm{n}+1}-1\right)$.
That is, $A>f(A) . \quad(n \geqslant 2)$
This means that, for each operation on $A$, it decreases, thus
$$
A_{1}>A_{2}>A_{3}>\cdots,
$$
Since $A$ is a finite number, there must exist a natural number $k$ such that $A_{1}0$, for $A_{\mathrm{x}}$ to undergo another operation, making $A_{1+1}$ either equal to 19, in which case the problem is solved, or not equal to 19, in which case, since $A_{k+1}$ is a number less than 100, it will inevitably, after some operation, result in a single-digit number, denoted as $A_{\mathbf{k}^{\prime}}=a$, hence $A_{\mathbf{k}^{\prime}+1}=A_{\mathbf{k}^{\prime}}$, thereby solving the problem.
For (2), we first prove that if $10 x + y$ (where $x$ is a natural number, and $y$ is a non-negative integer) can be divided by 19, then $x + 2 y^{\prime}$ can also be divided by 19, and the converse is also true.
Because $10 x + y = 10 x + 20 y - 20 y + y$
$$
=10(x + 2 y) + 19 y \text {. }
$$
10 and 19 are coprime, thus
$$
19|10 x + y \Leftrightarrow \Rightarrow 19| x + 2 y \text {. }
$$
Since $A=\overline{a_{0} a_{n-1} \cdots a_{1} a_{0}}$,
if $19 \mid A$, then
$$
1 9 \longdiv { a _ { n } a _ { n - 1 } \cdots a _ { 1 } } + 2 a _ { 0 } \text {. }
$$
Similarly,
$$
\begin{array}{l}
19 \mid \overline{a_{\mathrm{n}} a_{\mathrm{n}-1} \cdots a_{2}} + 2\left(a_{1}^{-} + 2 a_{0}\right) . \\
19 \mid \overline{a_{\mathrm{n}} a_{\mathrm{n}-1} \cdots a_{2}} + 2 a_{1}^{-} + 2^{2} a_{0} .
\end{array}
$$
Continuing this process, we have
$$
\begin{array}{l}
19 \mid a_{\mathrm{n}} + 2 a_{\mathrm{n}-1} + \cdots + 2^{\mathrm{n}-1} a_{1} + 2^{n} a_{0} \\
=f(A) .
\end{array}
$$
Thus, from $A=19^{8}{ }^{8}$ and $19 \mid A$, we know that $A_{1}, A_{2}, A_{3}$, … can all be divided by 19. From (1)
$$
A_{1}>A_{2}>\cdots>A_{\mathrm{k}}
$$
and $f(19)=19$, we can conclude
$$
A_{\mathrm{k}}=19
$$
|
19
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
To 9. On a plane, there is a fixed point $P$, consider all possible equilateral triangles $ABC$, where $AP=3, BP=2$. What is the maximum length of $CP$? (1961 Autumn Competition)
|
Solve for Ling and Hui
$$
\begin{array}{l}
\angle A P B=\alpha . \\
\angle B A P=\beta,
\end{array}
$$
Given $A B^{2}=3^{2}+2^{2}$
$$
\begin{aligned}
- & 12 \cos \alpha, \\
\cos \beta & =\frac{3^{2}+A B^{2}-2^{2}}{6 A B}, \sin \beta=\frac{2 \sin \alpha}{A B} .
\end{aligned}
$$
From this, we get
$$
\begin{aligned}
\cos \left(\frac{\pi}{3}+\beta\right)= & \frac{1}{2} \cos \beta-\frac{\sqrt{3}}{2} \sin \beta \\
= & \frac{3^{2}+A B^{2}-2^{2}-2 \sqrt{3} \sin \alpha}{12 A B}=2 A B \\
= & \frac{3^{2}+3^{2}-12 \cos \alpha}{12 A B} . \\
& -\frac{2 \sqrt{3} \sin \alpha}{2 A B} .
\end{aligned}
$$
(Applying the Law of Cosines in $\triangle P A C$, we get
$$
\begin{aligned}
P C^{2} & =A C^{2}+3^{2}-6 A C \cos \left(\frac{\pi}{3}+?\right) \\
& =13-12 \cos \left(\frac{\pi}{3}+\alpha\right) .
\end{aligned}
$$
Clearly, when $a=\frac{2 \pi}{3}$, $P C$ reaches its maximum value of 5.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $\triangle A B C$, extend the three sides by 1, 2, 3 times respectively, to get $\triangle A^{\prime} B^{\prime} C^{\prime}$. Ask how many times the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is compared to the area of $\triangle A B C$.
|
Theorem: Let the area of $\triangle ABC$ be $S$. Extend the sides $AB, BC, CA$ of $\triangle ABC$ such that $BB'=\lambda_{1} AB$, $CC'=\lambda_{2} BC$, $AA'=\lambda_{3} CA$, and let $\triangle A' B' C'$ be the resulting triangle with area $S'$. Then,
$$
S'=1+\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{1} \lambda_{2}+\lambda_{2} \lambda_{3}+\lambda_{3} \lambda_{1}.
$$
The conclusion of the theorem is a formula for the area ratio of two triangles. The area of the new triangle is the sum of the areas of four smaller triangles. If we connect $A' B, B' C, C' A$, then as shown in the figure, connecting $A' B, B' C, C' A$ results in:
In $\triangle C' B' C$ and $\triangle C B' B$, we have
Similarly, we can obtain:
$$
\begin{array}{l}
\frac{S \wedge C' A'}{S}=\lambda_{2}+\lambda_{2} \lambda_{3}. \\
\frac{S \triangle A A' B'}{S}=\lambda_{3}+\lambda_{3} \lambda_{1}.
\end{array}
$$
Therefore,
$$
\begin{array}{l}
=\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{1} \lambda_{2}+\lambda_{2} \lambda_{3}+\lambda_{3} \lambda_{1} \\
\end{array}
$$
Thus,
$$
\begin{aligned}
\frac{S'}{S}= & 1+\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{1} \lambda_{2}+\lambda_{2} \lambda_{3} \\
& +\lambda_{3} \lambda_{1}.
\end{aligned}
$$
Using this theorem to solve the initial problem: $\lambda_{1}=1, \lambda_{2}=2, \lambda_{3}=3$, substituting into (*) we get
$$
\frac{S_{\triangle A' B' C'}}{S_{\triangle ABC}}=18.
$$
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. In the plane, 7 points are given, connect them with some line segments so that
(1) among any three points, at least two are connected,
(2) the number of line segments is minimized.
How many line segments are there? Provide such a graph.
|
17. The figure below shows that 9 line segments are sufficient.
Now prove that at least 9 line segments are needed.
If point $A$ is not an endpoint of any line segment, then due to (1), the other 6 points must connect at least $C_{6}^{2}>9$ line segments.
If point $A$ is the endpoint of only 1 line segment, then due to (1), the 5 points not connected to $A$ must connect at least $C_{5}^{2}>9$ line segments.
Assume each point is the endpoint of at least two line segments. Then if point $A$ is the endpoint of only two line segments $A B, A C$, by (1), the 4 points not connected to $A$ must connect at least $C_{4}^{2}=6$ line segments. At least one more line segment must be drawn from point $B$, so in this case, there are at least $2+6+1=9$ line segments.
If each point is the endpoint of at least 3 line segments, then there must be more than $\left[\frac{3 \times 7}{2}\right]=10$ line segments.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Arrange the positive rational numbers in the following sequence,
$$
\begin{array}{l}
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \\
\frac{2}{3}, \frac{1}{4}, \cdots,
\end{array}
$$
Then the position number of the number $\frac{1989}{1949}$ is $\qquad$
|
3. The number's position in the original
sequence is $(1+2+\cdots+3937)+$ $1949=7753902$,
|
7753902
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A hotel has 90 vacant rooms, each with a unique key. 100 guests arrive, and keys need to be distributed so that any 90 of them can stay in the 90 rooms, with each person getting one room (assuming there is no limit to the number of keys that can be issued for each room or the number of keys each person can receive). What is the minimum number of keys that need to be issued? $\qquad$
|
8. At least 990 keys should be prepared.
|
990
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, there are 6000 points inside a circle, and no three points are collinear:
(1) Can this circle be divided into 2000 parts, each containing exactly three points? How to divide it?
(2) If the three points in each part satisfy: the distance between any two points is an integer and does not exceed 9, then using the three points in each part as vertices to form triangles, how many of these triangles are of the same size at least?
|
(1) 6000 points inside a circle can determine $C_{6000}^{2}$ lines. Since $C_{6000}^{2}$ is a finite number, there must exist a tangent line to the circle that is not parallel to any of the $C_{5000}^{2}$ lines, denoted as $l$. Moving $l$ parallel within the circle, it is clear that the 6000 points will be crossed (if it crosses two points simultaneously, then the line connecting these two points must be parallel to $l$, which contradicts the choice of $l$). Specifically, when $l$ crosses 3 points and has not yet crossed the fourth point, a chord $l_{1}$ is drawn.
Similarly, when crossing the 4th, 5th, 6th points, chords $l_{2}, l_{3}, l_{4}, \cdots$ are drawn. In this way, 1999 chords can be drawn, dividing the circle into 2000 segments, each containing three points. Each set of three points forms a triangle, resulting in 2000 triangles.
(2) It can be calculated that the number of triangles with integer side lengths and the longest side not exceeding 9 is 95. Let the side lengths be $a \leqslant b \leqslant c \leqslant 9$, where $a, b, c$ are all integers.
When $c=9$, and $a+b>c$, we have
\begin{tabular}{|c|c|c|}
\hline Values of $b$ & Values $a$ can take & \begin{tabular}{l}
Number of triangles
\end{tabular} \\
\hline 5 & 5 & 1 \\
\hline 6 & $4,5,6$ & 3 \\
\hline 7 & $3,4,5,6,7$ & 5 \\
\hline 3 & $\mid 2,3,4,5, \cdots, 8$ & 7 \\
\hline 9 & $2,3,4,5, \cdots, 9$ & 9 \\
\hline
\end{tabular}
Thus, when $c=9$, 25 different triangles can be obtained. When $c=8$, and $a+b>c$, we have
Thus, when $c=8$, 20 different triangles can be obtained. Similarly, when $c=7,6,5,4,3$, 2,1, the number of different triangles are 16, 12, 9, 6, 4, 2, 1, respectively.
Therefore, the total number of different triangles with integer side lengths and the longest side not exceeding 9 is
$$
1+2+4+6+9+12+16+20+25=95
$$
Thus, among the 2000 triangles, the number of triangles that are exactly the same in size is at least
$$
\left[\frac{2000}{95}\right]+1=22 \text { triangles. }
$$
(Supplied by Chen Dingchang, Zhejiang)
|
22
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the integer solution to the equation
$$
\left(1+\frac{1}{m}\right)^{m+1}=\left(1+\frac{1}{1988}\right)^{1088}
$$
|
Answer $m=-1989$.
|
-1989
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. All vertices of a broken line lie on the faces of a cube with an edge length of 2, and each segment of the broken line is 3 units long. This broken line connects two farthest vertices of the cube. How many segments does such a broken line have at least?
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last part is a note about the translation request and is not part of the text to be translated. It has been translated here for completeness.
|
Solve: Consider the cube circumscribing a sphere with center $A$ and radius 3, intersecting the cube's faces at three arcs: $K \hat{L}, \widehat{L N}, \widehat{N K}$ (Figure 3). The points $K, L, N$ on the edges bisect these three edges. In fact, $A D_{1}=\sqrt{8}$, so $L D_{1}=\sqrt{A} L^{\overline{2}-\overline{D_{1}^{2}}}=1$, and similarly for points $K$ and $N$.
Let point $M$ be any point on the aforementioned arcs, for example, on the arc $K L$, and $M$ is different from any endpoint of the arc. We consider a sphere with center $M$ and radius 3, and point $A$ lies on this sphere. It is easy to prove that all other points of the cube are inside this sphere.
From this, it follows that point $M$ can only be connected to point $A$, and the length of the segment is 3. Therefore, the first node of the broken line can only be one of the points $L, K$, and $N$. From point $K$, only the distance to point $D$ is 3 (except for point $A$). Similarly, the points that can be reached from points $L$ and $N$ can only be points $B$ and $A_{1}$, so all nodes are vertices of the cube or adjacent to point $A$. Similarly, the next node of the broken line must be the midpoint of one of the cube's edges, and the next node will be one of the points $C, B_{1}$, $D_{1}$, and the next node will again be the midpoint of an edge, and the final endpoint of the sixth segment can fall on point $C_{1}$, the opposite vertex of point $A$ (Figure 4 is an example). Thus, the minimum number of segments in a broken line that satisfies the conditions of the problem is 6.
|
6
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. On the blackboard, there are numbers 1 and 2. It is stipulated that new numbers can be written according to the following method: If there are numbers $a$ and $b$ on the blackboard, then the number $a b + a + b$ can be written. Using this method, can the following numbers be obtained:
(a) Number 131213
(b) Number 12131
|
Let the new number $ab + a + b$ be $c$. This means $c + 1 = ab + a + b + 1 = (a + 1)(b + 1)$. This implies that if the number written on the blackboard is replaced by a number that is 1 greater, then each new number will be the product of two existing numbers. Starting with the numbers 2 and 3, after several multiplications, we will get a number of the form $2^n \cdot 3^m$, where $n, m$ are natural numbers. Clearly, all such numbers can be obtained. This means that in the original situation, only numbers of the form $2^n \cdot 3^n - 1$ can be obtained. The number 13121 is of this form, while 12131 is not.
|
13121
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A country has 21 cities, and some airlines can implement air transportation between these cities. Each airline connects pairs of cities with non-stop flights (and several airlines can operate flights between the same two cities at the same time). Every two cities are connected by at least one non-stop flight. How many airlines are needed at least to meet the above requirements?
|
To make this country form an aviation network that meets the conditions required by the problem, there must be at least 21 airlines, because the total number of non-stop routes is no less than $20+19+\cdots+3+2+1=210$, and each airline provides $4+3+2+1=10$ non-stop routes. Figure 5 is an example of a service route map for a country with 21 airlines (here, the cities are the vertices of a regular 21-sided polygon, and the non-stop routes are its sides and diagonals). The first airline provides services to cities 1, 3, 8, 9, and 12, while the other airlines provide services to cities obtained by rotating the above cities by $k \cdot \frac{360^{\circ}}{21}$ $(k=1,2, \cdots, 20)$. The satisfaction of the problem's conditions is derived as follows: the segments connecting the vertices numbered 1, 3, 8, 9, and 12, including the sides of the 21-sided polygon and all possible lengths of diagonals, appear exactly once.
|
21
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. (Wenzhou Junior High School Math Competition Question, 1987) Let the natural number $n$ have the following property: from 1, 2, ..., $n$, any 51 different numbers chosen will definitely have two numbers whose sum is 101. The largest such $n$ is $\qquad$
|
Consider $\{1,2, \cdots, n\}$ as the vertex set. When the sum of two numbers is 101, connect the corresponding two vertices to form a graph $G$. Clearly, the original problem is equivalent to finding the largest $n$ such that any selection of 51 vertices in $G$ must include two adjacent vertices.
i) When $51 \leqslant n \leqslant 100$, $G$ has exactly $n-50$ edges with no common vertices, and there are $50-(n-50)=100-n$ vertices of degree zero. Since the sum of the number of edges and the number of zero-degree vertices is $(n-50)+(100-n)=50$, taking 51 vertices must include two adjacent vertices.
ii) When $n>100$, $G$ has exactly 50 edges with no common vertices, and $n-100$ vertices of degree zero. Since $n>100$, we have $50+(n-100) \geqslant 51$. That is, in the set of 51 vertices formed by selecting one vertex from each of the 50 edges and adding any one of the zero-degree vertices, no two vertices are adjacent.
Combining i) and ii), the value of $n$ that meets the condition is 100.
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. (85 Provincial Six Autonomous Regions High School Mathematics Joint Competition Question) A football invitational tournament involves sixteen cities, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and teams from the same city do not play against each other. After several days of the tournament, a statistical review found that, except for Team A from City A, all other teams have played a different number of matches. How many matches has Team B from City A played? Please prove your conclusion.
|
Prove that if two teams have played against each other, then an edge is connected between the corresponding two vertices to form graph $G$. Let the team from City A be $v^{*}$, then the original problem is equivalent to: find the degree of the vertex corresponding to the other team from City A.
Obviously, $\max d\left(v_{i}\right) \leqslant 32-2=30$, i.e., $v_{1} \in V$
$V-\left\{u^{*}\right\}$ contains 31 vertices whose degrees can only take 0, 1, 2, ..., 31 values. Therefore, the degrees of the vertices in $V-\left\{v^{*}\right\}$ are exactly 0, 1, 2, ..., 30. Let the vertex with degree $i$ be $v_{i} (i=0,1,2, \cdots, 30)$, then $V=\left\{v_{0}, v_{1}, v_{2}, \cdots, v_{30}, v^{*}\right\}$. Clearly, the only vertex not adjacent to $v_{30}$ is $v_{0}$, so $v_{30}$ and $v_{0}$ are the two teams from the same city; the only vertices not adjacent to $v_{29}$ are $v_{0}$ and $v_{1}$, so $v_{29}$ and $v_{1}$ are the two teams from the same city; ..., and the set of vertices not adjacent to $v_{16}$ is $\left\{v_{0}, v_{1}, v_{2}, \cdots, v_{14}\right\}$, so $v_{16}$ and $v_{14}$ are the two teams from the same city.
Thus, the team from the same city as $U^{\prime \prime}$ can only be $v_{15}$, so the desired number of matches is 15.
Let a graph with $n$ vertices, where each pair of vertices is connected by exactly one edge, be denoted as $k_{n}$. Clearly, the number of edges in $k_{0}$ is $\frac{1}{2} n(n-1)$.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. (Hefei Mathematical Competition Question in 1983) A new station is opened, and several bus routes are planned to serve the community. Their wishes are: (1) to open as many routes as possible; (2) each route must have at least one bus stop; (3) ensure that each bus stop is served by at least two different routes. Under these conditions, what is the maximum number of routes they can open? How many stops should each route have at minimum?
|
Let $S$ be the number of lines that can be opened, and consider the lines as vertices to form a graph $K_{\mathrm{s}}$. Label the 1983 stations as $A_{1}, A_{2}, \cdots, A_{19}$. If two lines have a common station, color the edge between the corresponding two vertices with color $C$.
From (2), every edge of $K_{\mathrm{s}}$ can be colored;
and (3), the number of different colors used on any two different edges of $K_{\mathrm{s}}$ is
$$
\frac{1}{2} S(S-1) \leqslant 1983 \text {. }
$$
From (1), $S$ can only be 63. From the above solution process, it is easy to see that at most 63 lines can be opened; and because each vertex of $K_{:} 3$ is connected to 62 edges of different colors, each line must pass through at least 62 stations.
|
63
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10. (19th All-Soviet Union Middle School Olympiad Problem) A quadratic trinomial $a x^{2}+b x+c$, where $a>100$. How many different integer points can there be at which the absolute value of its value does not exceed 50?
|
Assume there are three different integer points satisfying the given condition, then there must be two points located on the same side of the axis of symmetry $x=-\frac{b}{2 a}$ of the quadratic function $y=a x^{2} +b x+c$, or one point is located on this axis of symmetry. Without loss of generality, assume
$-\frac{b}{2 a}1\left(x_{1} \neq x_{2}\right.$ are integers $)$, then we have
$$
\begin{aligned}
100 & \geqslant\left|a x_{2}^{2}+b x_{2}+c\right|+\left|a x_{1}^{2}+b x_{1}+c\right| \\
& >\left|\left(a x_{2}^{2}+b x_{2}+c\right) \cdots\left(1 x_{1}^{2}+b x_{1}+c\right)\right| \\
& =a\left(x_{2}-x_{1}\right)\left(x_{2}+x_{1}+\frac{b}{a}\right) \geqslant a>100,
\end{aligned}
$$
which is a contradiction. Therefore, the number of points satisfying the given condition is at most two.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Column 1. A tour group is selecting visit locations from $A, B, C, D, E$ under the following constraints:
(1) If visiting $A$, must also visit $B$,
(2) At least one of $D, E$ must be visited,
(3) Only one of $B, C$ can be visited,
(4) Both $C, D$ must be visited or neither,
(5) If visiting $E$, both $A, D$ must be visited.
Please explain the reasoning, which places can the tour group visit at most?
|
堔 We adopt the貑 deduction method.
(1)
(4)
(2)
$1^{\circ}$ If going to $A \Rightarrow$ must go to $B \Rightarrow$ do not go to $C \Rightarrow$ do not go to $D \Rightarrow$ must not go to $A$. Going to $A, D$. This leads to the same contradiction as $1^{\circ}$, so do not go to $B$.
(5)
$3^{\circ}$ If going to $E \Rightarrow$ must go to $A, D$. From $1^{\circ}$, we cannot go to $A$, so do not go to $E$.
(4)
$4^{\circ}$ If going to $C \Rightarrow$ must go to $D$, and all other conditions are met. Therefore, the maximum number of places to go is $C, D$ two places.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Find the largest integer $A$, such that for any permutation of all natural numbers from 1 to 100, there are 10 consecutive numbers whose sum is greater than or equal to $A$.
[22nd Polish Mathematical Olympiad]
|
This problem, although it is to find the maximum integer $A$, is actually an existence problem, i.e., there exists a maximum integer $A$ such that the sum of 10 consecutive numbers is not less than $A$.
Solution: Let $T=\left(a_{1}, a_{2}, \cdots, a_{100}\right)$ be a permutation of the natural numbers from 1 to 100.
Consider the sum of 10 consecutive terms
$$
\sum_{k=1}^{10} a_{a+k}, \quad n=1,2, \cdots, 90 .
$$
This is a finite set, and in a finite set, there must be a maximum number, denoted as
$$
A_{\mathrm{T}}=\max _{1 \leqslant n \leqslant 90}^{\text {io }} \sum_{\mathrm{D}=1} a_{n}+x_{0}
$$
Any other sum of 10 consecutive terms does not exceed $A_{x}$.
By the definition of $A_{\Gamma}$, we have
$$
\begin{array}{l}
A_{\mathrm{T}} \geqslant a_{1}+a_{2}+\cdots+a_{10} \\
A_{\mathrm{T}} \geqslant a_{11}+a_{12}+\cdots+a_{20}, \\
\quad \cdots \cdots \\
A_{\mathrm{X}} \geqslant a_{61}+a_{82}+\cdots+a_{100}
\end{array}
$$
Adding these up, we get
$$
10 A_{\mathrm{r}} \geqslant \sum_{1=1}^{100} a_{1}=5050,
$$
$\rightarrow \operatorname{mosers}$
$$
A_{\mathrm{T}}>505 \text {. }
$$
According to the problem, we need to find the smallest $A_{x}$ among all permutations $T$, i.e.,
$$
A=\min A_{\mathrm{r}}
$$
Below, we can find a permutation $T^{\prime}$ such that for the permutation $T^{\prime}$, its $A_{\mathrm{T}}{ }^{\prime} \leqslant 505$.
For example, we can arrange the numbers from 1 to 100 as follows:
$$
\begin{array}{c}
T^{\prime}=(100,1,99,2,98,3,97, \\
4, \cdots, 51,50),
\end{array}
$$
which satisfies
$$
\begin{array}{l}
a_{2 \Omega+1}=100-n, 0 \leqslant n \leqslant 49, \\
a_{2 \mathrm{n}}=n, 1 \leqslant n \leqslant 50 .
\end{array}
$$
At this point, we have
$$
\begin{array}{l}
a_{2 k}+a_{2 k+1}+\cdots+a_{2 k+2} \\
=\left(a_{2 k}+a_{2 k+2}+a_{2 k+4}+a_{2 k+0}\right. \\
\left.+a_{2 k+8}\right)+\left(a_{2 k+1}+a_{2 k+8}+a_{2 k+5}\right. \\
\left.+a_{2 k+1}+a_{2 k+0}\right) \\
=(k+k+1+k+2+k+3+k+4) \\
+[100-k+100-(k+1)+100-(k \\
+2)+100-(k+3)+100-(k+4)] \\
= 500 \\
a_{2 k+1}+a_{2 k+2}+\cdots+a_{2 k+10} \\
=\left(a_{2 k}+a_{2 k+1}\right)+\left(a_{2 k+2}+a_{2 k+8}\right)+\cdots
\end{array}
$$
$$
\begin{aligned}
& +\left(a_{2 k+8}+a_{2 k+2}\right)+a_{2 k+10}-a_{2 k} \\
= & 500+k+5-5 \\
= & 505 .
\end{aligned}
$$
Thus, $A \mathrm{x}^{\prime} \leqslant 505$.
From (1) and (2), we get $A=505$.
|
505
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In 1988, the Chinese Junior High School Mathematics League had the following problem: If natural numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$, what is the maximum value of $x_{5}$?
|
This problem is novel and unique, and some students find it difficult to start. Below, I will first discuss the solution to this problem.
Algorithm 1: Given the symmetry of the equation, we can assume without loss of generality that \( x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5} \), and then use the "bounding method":
\[
\begin{array}{l}
1=\frac{1}{x_{1} x_{2} x_{3} x_{4}}+\frac{1}{x_{1} x_{2} x_{3} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
+\frac{1}{x_{2} x_{3} x_{4} x_{5}} \leqslant \frac{1}{x_{4}}+\frac{1}{x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}} \\
+\frac{1}{x_{4} x_{5}}=\frac{3+x_{4}+x_{5}}{x_{4} x_{5}}
\end{array}
\]
Therefore,
\[
\begin{array}{l}
x_{4} x_{3} \leqslant 3+x_{4}+x_{5}, \\
\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4 .
\end{array}
\]
If \( x_{4} = 1 \), then \( x_{1} = x_{2} = x_{3} = x_{4} = 1 \), and the original equation becomes \( 4 + x_{5} = x_{5} \), which is a contradiction.
If \( x_{4} > 1 \), then \( x_{5} - 1 \leqslant 4 \), i.e., \( x_{5} \leqslant 5 \).
Thus, the maximum value of \( x_{5} \) is 5. It is easy to observe that one solution is \( (1,1,1,2,5) \).
Solution 2: When \( x_{5} \) is maximized, it should be no less than any \( x_{i} \) (for \( i=1,2,3,4 \)), so
\[
\begin{array}{l}
x_{5} < x_{2} + x_{2} + x_{3} + x_{4} + x_{5} \leqslant 5 x_{3}^{3}, \\
x_{3} < x_{1} x_{2} x_{3} x_{4} x_{5} \leqslant 5 x_{3}, \\
1 < x_{1} x_{2} x_{3} x_{4} \leqslant 5 .
\end{array}
\]
From the original equation, we get \( x_{5} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{x_{1} x_{2} x_{3} x_{4} - 1} \).
If \( x_{1} x_{2} x_{3} x_{4} = 2 \), then among \( x_{1}, x_{2}, x_{3}, x_{4} \), there is exactly one 2 and three 1s. Substituting, we get \( x_{5} = 5 \).
For \( x_{1} x_{2} x_{3} x_{4} = 3, 4, 5 \), similarly, \( x_{5} = 3 \) and \( x_{5} = 2 \).
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. The vertex of the parabola is at the origin, and the focus F is the center of the circle given by $x^{2}+y^{2}-4 x=0$. A line passing through point $F$ with a slope of 2 intersects the parabola at points $A$ and $D$, and intersects the circle at points $B$ and $C$.
Find $|A B|+|C D|$.
|
$$
\begin{array}{l}
\text { Given } F \text { as the pole, and the positive direction of the } x \text { axis as the polar axis, we establish the polar coordinate system. } \\
\left.\begin{array}{l}
\left.\therefore \text { The parabola equation is: } \rho=\frac{p}{1-\cos \theta}\right\} \Rightarrow \rho= \\
\because F(2,0), \therefore p=4 .
\end{array}\right\} \Rightarrow
\end{array}
$$
$$
\begin{array}{l}
1-\cos \theta \text {. } \\
\text { Also, } \because \operatorname{tg} \theta=2, \therefore \cos \theta=\frac{\sqrt{5}}{5}(0 \leqslant \theta<\pi) \text {. } \\
\because|A B|+|C D|=|A F|+|F D|-|B C| \text {, } \\
|A F|=\rho_{1}, \quad|F D|=\rho_{2}, \\
\therefore \rho_{1}+\rho_{2}=\frac{4}{1-\cos \theta}+\frac{4}{1-\cos (\theta+\pi)} \\
=\frac{8}{1-\cos ^{2} \theta}=\frac{1}{1-\frac{1}{5}}=10 \text {. } \\
\end{array}
$$
$$
\begin{array}{c}
\text { Also, } \because|B C|=2 R=4, \\
\therefore|A B|+|C D|=10-4=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
, Example 6. Find a four-digit number, which when multiplied by 4 is exactly equal to its reverse.
|
Assuming the four-digit number $x y z t(x \neq 0$, $1 \neq 0, y, z$ are integers) exists, at this point
$$
\text { 4. } x y z t=t z y x \text {. }
$$
Since $4 \cdot x y z t \geqslant 4 x \cdot 1000$,
$\overline{t z y x}7$. Thus, $t=8$ or $t=9$.
But $t=9$ is impossible, because if $t=9$, then the left-hand side of (*) is 6, while the right-hand side is $2$, making (*) invalid. Therefore, only $t=8$.
Substituting $x=2, t=8$ into (*) yields:
$$
\begin{array}{l}
4(2000+100 y+10 z+8)=8000+100 z \\
+10 y+2 .
\end{array}
$$
Simplifying gives $13 y+1=2 z$.
Given $z \leqslant 9$ and (**), we have $13 y \leqslant 17$, so $y=0$ or $y=1$.
When $y=0$, by (**) we get $z=\frac{1}{2}$, which is not an integer, so $y \neq 0$. Therefore, only $y=1$.
When $y=1$, substituting into (**) gives $z=7$. Thus, the number sought is 2178.
|
2178
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. (IMO-19-2) In a sequence of real numbers, the sum of any 7 consecutive terms is negative, while the sum of any 11 consecutive terms is positive. How many terms can such a sequence have at most?
保留源文本的换行和格式，翻译结果如下：
Example 7. (IMO-19-2) In a sequence of real numbers, the sum of any 7 consecutive terms is negative, while the sum of any 11 consecutive terms is positive. How many terms can such a sequence have at most?
|
Given that a junior high school student is not clear about the concept of "sequence", we can simply change the term "sequence" to "numbers arranged according to a certain rule".
Let the sequence be $a_{1}, a_{2}, \cdots a_{2}, \cdots, a_{0}$, i.e., $n \geq 17$. According to the problem:
$$
\begin{array}{l}
a_{k}+a_{k+1}+\cdots a_{k+0}0 . \quad(k \in N)
\end{array}
$$
(2) - (1) gives:
$$
a_{k+}+a_{k+1}+a_{k+0}+a_{k+10}>0 \text {. }
$$
From (3), we know that starting from the 8th term, the sum of any four consecutive terms is positive.
Then
$$
\begin{array}{l}
a_{8}+a_{8}+a_{10}+a_{11}>0, \\
a_{11}+a_{12}+a_{18}+a_{14}>0 .
\end{array}
$$
(4) + (5) gives: $a_{0}+a_{0}+a_{10}+2 a_{11}+a_{12}$
$$
+a_{13}+a_{14}>0 \text {. }
$$
Furthermore, from (1), we get: $a_{8}+a_{4}+a_{10}+a_{11}+a_{12}$
$$
+a_{13}+a_{14}0$.
Similarly, we get: $a_{12}>0, a_{18}>0$,
$$
a_{11}+a_{12}+a_{1 s}>0 \text {. }
$$
When $n \geqslant 17$, from (1) we get
$$
\begin{array}{l}
a_{11}+a_{13}+a_{18}+a_{14}+a_{17}+a_{10} \\
+a_{17}3$ and $a \in N$, sequences of the form:
$a, a,-\frac{5 a+1}{2}, a, a, a,-\frac{5 a+1}{2}$, $a, a,-\frac{5 a+1}{2}, a, a, a,-\frac{5 a+1}{2}, a, a$ are all sequences that satisfy the condition $n=16$.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $m, n$ be positive integers, prove that there exists a constant $\alpha>1$ independent of $m, n$, such that when $\frac{m}{n}<\sqrt{7}$, we have $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\alpha}{n^{2}}$. What is the largest value of $\alpha$?
|
Solve $\alpha=3$.
$$
\begin{array}{l}
\quad 7-\frac{m^{2}}{n^{2}} \geqslant \frac{a}{n^{2}} \Leftrightarrow 7 n^{2}-m^{2} \geqslant \alpha_{0} \\
\because \quad m+7 \equiv m(\bmod 7) \\
\therefore \quad(m+7)^{2} \equiv m^{2}(\bmod 7)
\end{array}
$$
$\therefore\left\{m^{2}(\bmod 7)\right\}$ is a purely periodic sequence with a period of 7, its first 7 terms are $1,4,2,2,4$,
$$
\begin{array}{l}
1, 0, \\
\therefore m^{2} \equiv 0,1,2,4(\bmod 7), \\
\therefore 7 n^{2}-m^{2} \equiv-m^{2} \\
=0,6,5,3(\bmod 7) . \\
\because \quad 7 n^{2}-m^{2}>0, \\
\therefore \quad 7 n^{2}-m^{2} \geqslant 3,
\end{array}
$$
When $n=1, m=2$, $7 n^{2}-m^{2}=3$, hence $\alpha_{\mathrm{m} x}=3$.
|
3
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Six The terms of the sequence $x_{1}, x_{2}, x_{3}, \cdots$ are non-zero real numbers, and satisfy $x_{\mathrm{a}+2}=\frac{x_{\square} \cdot x_{\mathrm{n}+1}}{2 x_{\mathrm{a}}-x_{\mathrm{a}+1}} \quad(n=1,2, \cdots)$,
(1) Find the necessary and sufficient conditions for $x_{1}$ and $x_{2}$ such that all $x_{0}$ are integers;
(2) If $x_{1}=\lg 5, x_{2}=\lg 2$, find the smallest integer $n$ such that $x_{\mathrm{n}}<\frac{\lg 5 \lg 2}{10+\lg 2}$ (1 \mathrm{~g} 2=0.3010).
|
Six, Solution (1) Hint: By mathematical induction, it is easy to prove
$$
x_{\mathrm{n}}=\frac{x_{1} x_{2}}{(n-1) x_{1}-(n-2) x_{2}},
$$
i.e., $x_{\mathrm{n}}=\frac{x_{1} x_{2}}{n\left(x_{1}-x_{2}\right)+\left(2 x_{2}-x_{1}\right)}$.
When $x_{1} \neq x_{2}$ and $n$ is sufficiently large, there must be $x_{0}<1$. When $x_{1}=x_{2}$, $x_{0}=\frac{x_{1} x_{2}}{2 x_{2}-x_{1}}=x_{1}$ (constant).
Therefore, the necessary and sufficient condition for all $x_{0}$ to be integers is $x_{1}=x_{2}$ (integer).
(2) From $\frac{\lg 2 \cdot \lg 5}{n(\lg 5-\lg 2)+2 \lg 2-\lg 5}$ $<\frac{\lg 5 \lg 2}{10+\lg 2}$, the smallest integer solution is $n=27$.
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given a set $A$ composed of functions, there are 10 odd functions, 8 increasing functions defined on the interval $(-\infty,+\infty)$, and 12 functions whose graphs pass through the origin. Then the set $A$ can have at most $\qquad$ elements.
|
5. 14.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Insert “+” or “- -” between $1,2,3, \cdots, 1989$, what is the smallest non-negative number that the sum can achieve?
|
Except for 995, the numbers $1,2,3, \cdots, 1989$ can all be divided into 994 pairs: $(1,1989),(2,1988)$, $\cdots$, (994,996). Since the parity of the two numbers in each pair is the same, the result of the operation, regardless of how “+” or “-” signs are placed before each pair, can only be an even number. And 995 is an odd number. Therefore, no matter how “+” or “-” signs are placed before the numbers $1,2, \cdots, 1989$, the value of the sum is always odd. Thus, the smallest non-negative number sought is no less than 1. The number 1 can be obtained in the following way:
$$
\begin{aligned}
1 & =1+(2-3-4+5)+(6-7-8+9) \\
+\cdots+ & (1986-1987-1988+1989)
\end{aligned}
$$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Does a ten-digit number exist that is divisible by 11 and has all different digits on each of its positions?
翻译完成，保留了原文的换行和格式。
|
Solve Write a three-digit number that can be divided by 11 and has different digits on each place, such as 275, 396, 418. Using these three numbers, it is not difficult to construct a ten-digit number that can be divided by 11. For example:
$$
\begin{array}{l}
2753964180 \\
=275 \cdot 10^{7}+396 \cdot 10^{4}+418 \cdot 10 \\
=11 \cdot\left(25 \cdot 10^{7}+36 \cdot 10^{4}+38 \cdot 10\right) .
\end{array}
$$
Another solution Utilize the divisibility rule for 11, the number
$$
n=\overline{a_{1} a_{2} a_{3} \cdots a_{3}} \text { and } S(n)=a_{1}-a_{2}+a_{3}
$$
$-\cdots-a_{n}$ are simultaneously divisible or not divisible by 11.
Let $A$ be the sum of the numbers in $S(n)$ with a “+” sign, and $B$ be the sum of the numbers with a “-” sign. Clearly,
$A+B=1+2+\cdots+9+0=45$. At the same time, $A-B$ should be divisible by 11. We can take $A-B=11$. Then $A=28, B=17$. It is not difficult to select five different digits that sum to 17, such as $1+2+3+5+6=17$. Then $4+7+8+9=28$. Therefore, the number 1427385960 meets the conditions of the problem.
|
1427385960
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
*8. In decimal, find the smallest natural number: its square number starts with 19 and ends with 89.
|
The last digit of the required number can only be 3 or 7. We write out all two-digit numbers ending in 3 and 7, and we get $17, 33, 67, 83$. To ensure that the first two digits of the natural number $x$ are 19, the inequality $19 \leqslant x^{2} \cdot 10^{-N}<20$ must hold, where $N$ is a natural number. When $N = 2k$ and $N=2k+1$, we get the inequalities $19 \leqslant x^{2} \cdot 10^{-2k}<20$ and $190 \leqslant x^{2} \cdot 10^{-2k}<200$. Taking the square root, in the first case we get the inequality $4.3588989 \cdots \leqslant x \cdot 10^{-k}<4.4721359 \cdots$; and in the second case, $13.784048 \cdots \leqslant x \cdot 10^{-k}<14.142135 \cdots$.
It is not difficult to see that, due to the variation of $k$, there will be infinitely many possibilities. We examine $k=1$ and $k=2$. If $4.358 \leqslant x \cdot 10^{-1}<4.472$, then $x=44$; if $4.358 \leqslant x \cdot 10^{-2}<4.472$, then $436 \leqslant x \leqslant 447$. If $13.784 \cdots \leqslant x \cdot 10^{-1}<14.142 \cdots$, then $x=14$; if $13.784 \cdots \leqslant x \cdot 10^{-2}<14.142 \cdots$, then $138 \leqslant x \leqslant 141$. When $k=3$, we similarly examine the two possible cases, and so on. Among the $x$ values we obtain, the smallest number whose last two digits are $17, 33, 67, 83$ is $1383\left(1383^{2}=1912689\right)$.
|
1383
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Find the smallest natural number $x$, such that in decimal notation, the number $x^{3}$ has the first three digits and the last four digits as 1.
|
Solving the problem, the last four digits are only related to the last four digits of $x$.
Number $x^{3}$ ends with 1, so the last digit of number $x$ can only be 1. Cubing the numbers 01, 11, 21, ..., 91, we confirm that only $71^{3}=357911$. Therefore, the last two digits of number $x$ are 71. Cubing the numbers 071, 171, 271, ..., 971, we confirm that only the number $471^{3}=104+87111$. Therefore, the last three digits of $x$ are 471. Similarly, cubing the numbers 0471, 1471, 2471, ..., 9471, we find that the last four digits of number $x$ are 8471.
Let $n$ be the number of digits between 111 and 1111 in $x^{3}$. There are three possible scenarios.
$$
\begin{array}{l}
\text { 1) } n=3 k, k \geqslant 0, \text { then } \\
111 \cdot 10^{3 k+4}<x^{3}=111 \frac{\cdots \cdots 1}{3 k \text { digits }} 111 \\
\therefore 112 \cdot 10^{3 k+4} .
\end{array}
$$
Therefore,
$$
\begin{array}{l}
10.353988 \cdot 10^{k+1} \approx \sqrt[3]{1110} \cdot 10^{k}<x \\
<\sqrt[3]{1120} \cdot 10^{k} \approx 10.384988 \cdot 10^{k+1} .
\end{array}
$$
When $k=0$, it is easy to see that the required number $x$ does not exist. If $k \geqslant 1$, then the required number $x$ starts with 103, i.e., $x=103 \cdots 8471$. The smallest such number is 1038471. It is not difficult to verify that this number satisfies the required conditions.
$$
\begin{array}{l}
\text { 2) } n=3 k+1, k \geqslant 0 \text {, then } \\
111 \cdot 10^{3 k+5}<x^{3}=111 \ldots \ldots \ldots \cdots \cdot 1111 \\
3 k+1 \text { digits } \\
\end{array}
$$
$<112 \cdot 10^{3 k+5}$. Therefore, 22.306991 $10^{k+1}$ $\approx \sqrt[3]{11100} \cdot 10^{k+1}<x<\sqrt[3]{11200} \cdot 10^{k+1}$ $\approx 22.373779 \cdot 10^{k+1}$.
When $k=0$, the required number $x$ does not exist. When $k \geqslant 1$, the required number has the form $x=225 \cdots 8471$, and there is at least one digit in the middle position. However, any number of this form is clearly greater than the (seven-digit) number 1038471. Therefore, none of these numbers satisfy the condition.
3) $n=3 k+2, k \geqslant 0$. This case is discussed similarly to case 2). Similarly, we find numbers greater than 1038471. Therefore, the required number is 1038471.
|
1038471
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $p, q$ are both natural numbers, and the two roots of the equation $p x^{2}-$ $q x+1985=0$ are both prime numbers, then what is the value of $12 p^{2}+q$? (85 Beijing Mathematics Competition Question)
|
Since $1985=5 \times 397$, and the two roots $x_{1}, x_{2}$ are both prime numbers, and $x_{1} x_{2}=\frac{1985}{p}=\frac{5 \times 397}{p}$, thus $p=1$, then $x_{1}, x_{2}$ are 5 and $397, x_{1}+x_{2}=\frac{q}{p}$ $=q=402$, therefore $12 p^{2}+q=12+402=414$.
|
414
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. For the equation $(1989 x)^{2}-1988 \cdot 1990 x$ $-1=0$, the larger root is $r$. For the equation $x^{2}+1989 x-1990=0$, the smaller root is $s$. Find the value of $r-s$. (Adapted from the 1984 Beijing Mathematics Competition)
|
Solve the equation $(1989 x)^{2}-1988 \cdot 1990 x-1$ $=0$, we get
$(1989 x)^{2}-(1989-1)(1989+1) x-1=0$.
Obviously, $1989^{2}-(1989-1)(1989+1)-1$ $=0$, so the equation has one root $x_{1}=1$, then the other root is $\boldsymbol{x}_{2}=-\frac{1}{1989^{2}}$, thus the larger root $r=1$.
Similarly, for the equation $x^{2}+1989 x-1990=0$, the sum of the coefficients is 0, so it also has one root $x_{1}=1$. Then the other root is $x_{2}=-1990$. Therefore, $s=-1990$.
Hence $r-s=1-(-1990)=1991$.
|
1991
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. If $k$ is an integer, the quadratic equation $(k-1) x^{2}-p x+k=0$ has two positive integer roots, find $k^{k 0}\left(p^{p}+k^{k}\right)$. (84 Beijing Mathematics Competition Question)
|
Let the two roots be $x_{1}, x_{2}$, then $x_{1}, x_{2}$ are both positive integers. By Vieta's formulas, $x_{1} x_{2}=\frac{k}{k-1}$ is also a positive integer. Since $k$ is a positive integer, if $k-1 \neq 1$, then $k-1$ and $k$ are coprime. In this case, $\frac{k}{k-1}$ cannot be an integer, so $k-1=1$, i.e., $k=2$ is the only way for $\frac{k}{k-1}$ to be an integer. At this time, $x_{1} x_{2}=2=1 \times 2$, then $p=3$. Therefore, $k^{k p}\left(p^{p}+k^{k}\right)=2^{2} \times 3\left(3^{3}+2^{2}\right)=1984$.
|
1984
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given the equation $x^{2}+(a-6) x+a=0$ $(a \neq 0)$ has two integer roots, find the integer value of $a$.
---
The equation $x^{2}+(a-6) x+a=0$ $(a \neq 0)$ is known to have two integer roots. Try to find the integer value of $a$.
|
Let the two integer roots of the equation be \(x_{1}, x_{2}\), and \(x_{1} \geqslant x_{2}\). By Vieta's formulas, we have:
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=6-a, \\
x_{1} x_{2}=a .
\end{array}\right.
\]
(1)
\[
\begin{array}{l}
+ \text { (2): } x_{1} x_{2}+x_{1}+x_{2}=6, \\
\therefore \quad\left(x_{1}+1\right)\left(x_{2}+1\right)=7 .
\end{array}
\]
From this, solve for \(x_{1}, x_{2}\), and then combine with the problem statement to get \(a=16\).
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $x=\sqrt{19-8 \sqrt{3}}$, then the value of the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{3}-7 x^{2}+5 x+15}$ is $\qquad$
|
$5$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A person is walking along the tram route, and a tram catches up from behind every 12 minutes, while a tram comes from the opposite direction every 4 minutes. Assuming both the person and the trams are moving at a constant speed, then the trams are dispatched from the starting station every $\qquad$ minutes.
|
$6$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, as shown in the figure, in $\triangle A B C$, $\angle A=90^{\circ}$, $A D \perp B C$ at $D, P$ is the midpoint of $A D$, $B P$ intersects $A C$ at $E, E F \perp B C$ at $F, A E=3, E C=12$. Find the length of $E F$.
|
Extend $F E, B A$ to intersect at point $H$,
$\left.\begin{array}{l}A D \| H F \\ P \text { is the midpoint of } A D\end{array}\right\} \Rightarrow H E=E F$.
Also, $\angle H A C=90^{\circ}$ ?
$\left.\begin{array}{l}\angle E F C=90^{\circ} \\ \text { and on the same side of } H C\end{array}\right\} \Rightarrow H, A, F, C$ are concyclic
$$
\begin{array}{l}
\Rightarrow H E \cdot E F=A E \cdot E C, \\
\quad \therefore E F^{2}=A E \cdot E C=3 \times 12=36 \\
\Rightarrow E F=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Consider the following sequences:
$$
\begin{array}{l}
3,7,11,15,19,23,27,31, \cdots \\
2,5,8,11,14,17,20,23, \cdots
\end{array}
$$
The 20th pair of identical numbers in them is $\qquad$
|
10. 239 .
|
239
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let the integer part of $\frac{\sqrt{13}+3}{\sqrt{13}-3}$ be $m$, and the decimal part be $n$, then the value of $198 m+9 n+n^{2}+1$ is $\qquad$
|
3. 1790 ,
|
1790
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Find the largest odd number that cannot be expressed as the sum of three distinct composite numbers.
untranslated text:
4.求一个不能用三个不相等的合数之和表示的最大奇数。
|
The smallest sum of three unequal composite numbers is
$$
4+6+8=18 \text{.}
$$
We prove that 17 is the largest odd number that cannot be expressed as the sum of three unequal composite numbers.
In fact, it is only necessary to prove that any odd number $2k-1$ not less than 19 can always be expressed as the sum of three unequal composite numbers.
Since $2k-1=4+9+(2k-14)$.
For $k \geqslant 10$, we have $2k-14 \geqslant 6$, so 4, 9, and $2k-14$ are three unequal composite numbers.
Therefore, 17 is the largest odd number satisfying the condition.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, $E, F$ are
on the sides $B C$ and $C D$
of rectangle $A B C D$,
if the areas of $\triangle C E F$,
$\triangle A B E$, $\triangle A D F$
are 3,
4, 5 respectively. Find the area $S$ of $\triangle A E F$.
|
Let $AB = a, BC = b$, then $BE = \frac{8}{a}$,
$$
\begin{array}{l}
CE = b - \frac{8}{a}, DF = \frac{10}{b}, FC = a - \frac{10}{b}. \\
\left\{\begin{array}{l}
\frac{1}{2}\left(b - \frac{8}{a}\right) \times \left(a - \frac{10}{b}\right) = 3, \\
ab = 3 + 4 + 5 + S .
\end{array}\right. \\
S = \sqrt{144 - 80} = 8 .
\end{array}
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. When programming a computer to print out 10000 numbers, there is a bug, every time it prints the number 3, it prints “X” instead, how many numbers are printed incorrectly.
保留了源文本的换行和格式。
|
12. 3439 .
|
3439
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, as shown in the figure, $P$ is a point inside the square $ABCD$, $PA=5$, $PB=8$, $PC=13$. Find the area of square $ABCD$.
---
The translation maintains the original text's line breaks and format.
|
Four, Hint: Like tears, draw $P E \perp$ $A B, P F \perp B C$, and let the side length of the square be a, $\quad P_{E}=x, \quad P F$ $=y$, solve to get the area of square $A B C D$ is approximately 153.
|
153
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. There is a car, the circumference of its front wheel is $5 \frac{5}{12}$ meters, and the circumference of its rear wheel is $6 \frac{1}{3}$ meters. Then it must travel $\qquad$ meters to make the number of revolutions of the front wheel 99 more than that of the rear wheel.
|
2. 3705 ;
|
3705
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. (Addition to the problem $1^{*}$ ) $n$ countries, each country's team of 3 representatives, form $m$ representative meetings $A_{0}(1), A_{0}(2), \cdots, A_{0}(m)$ called a meeting circle. If
1 each meeting has $n$ representatives, exactly 1,
() no two meetings have the same representatives;
(3) $A_{0}(i)$ and $A_{0}(i+1)$ have no common representatives, $i=1,2, \cdots, m$. It is agreed that $A_{\mathrm{n}}(n l+1)=A_{\mathrm{o}}(1)$,
(1) if $1<|i-j|<m-1$, then $A_{\mathrm{D}}$ (1) and $A_{0}$ (j) have at least one common representative.
Is there a meeting circle of 11 countries with 1990 meetings?
|
We use $(m, n)$ to denote a conference circle with $n$ countries and $m$ meetings, and use $\left(A_{\mathrm{n}}(i), j\right)$ to denote a concept formed by all the representatives and the $j$-th representative of the $n+1$-th country in the meeting $A_{n}(i)$. If $A,(0), \cdots, A \sim(m)$ is an $(m, n)$ assembly, then when $m$ is an even number, $(4 n(1), 1), 4 n(2), 2), \cdots$, $(A 0(m), 1),\left(A_{0}(1), 2\right),\left(A_{0}(2), 1\right), \cdots$ $(A n(m), 2)$ is a $(2 m, n+1)$ conference circle. In this case, $\left(A_{\mathrm{a}}(1), 3\right),\left(A_{\mathrm{a}}(2), 1\right),\left(A_{\mathrm{a}}(3), 2\right)$, $\left(A_{n}(4), 1\right),\left(A_{0}(5), 2\right), \cdots,\left(A_{n}(k-2), 1\right.$, ; $\left(A_{n}(k-1), 2\right),\left(A_{0}(k), 3\right),\left(A_{n}(k-1), 1\right)$, $(A n(k-2), 2), \cdots,\left(A_{0}(5), 1\right),\left(A_{n}(4), 2\right)$, $\left(A_{n}(3), 1\right),\left(A_{n}(2), 2\right)$. . . a $(2(k-1), 1)$ conference circle, which satisfies an even number not greater than $m$, and except for the first sub-concept corresponding to $A_{n}(1), A_{0}(k)$ modulo 3, all other corresponding second sub-concepts are updated according to the method of constructing new conference circles. Thus, we can sequentially obtain the full circles (6,2), (10,3), (18,4), (34,5), (66, 6), $(130,7),(258,8),(514,9),(1026$, 10), and by taking $k=996$, we get the conference circle $(1990,11)$.
|
1990
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In a dark room, a drawer contains socks of two colors and two sizes, 4 pairs in total (one pair of each color and size). How many socks must be taken out to ensure that there are two pairs of different colors and different sizes?
|
New $A_{1}, A, B, B$ each two. Period out 6 as $A ., A_{1}, 1, B_{1}, B_{1}, B_{2}$, does not meet the question's requirements. But when any 7 are taken out, there must be one color (let's say A) with 4 pieces all taken out, and among the 3 pieces of another color, there must be two.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For the set $\{00,01, \cdots, 98,99\}$, a subset $X$ satisfies: in any infinite sequence of digits, there are two adjacent digits that form an element of $X$. What is the minimum number of elements that $X$ should contain?
|
4. For any $i, j \in\{0,1, \cdots, 9\}, \lambda$ should include $i j$ or $i i$ (otherwise the sequence $i j i j i \ldots$ would satisfy the problem's requirements). There are $10+C_{10}^{2}=55$ such unordered pairs $(i, j)$, hence $|X| \geqslant 55$.
On the other hand, if we take $X=\{i j: 0 \leqslant i \leqslant j \leqslant 9\}$, then $|X|=55$, and for any non-decreasing sequence, let $i$ be the smallest digit it contains, and $j$ be the digit following $i$, then $i j \in X$.
Therefore, $X$ must contain at least 55 elements.
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The street plan of a city is a $5 \times 5$ square grid, (as shown in the figure) and there is a sweeper at point $A$. i. Find the total length of all streets.
保留了原文的换行和格式，翻译结果如上。
|
4. This graph has 16 odd vertices (the four inner points on each boundary) and can be divided into eight pairs of adjacent ones. Therefore, the length of the shortest cleaning route is $60+8=68$.
|
68
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. i will arrange $1 \times 1$ square tiles into a strip with a diagonal length of 100. Find the approximate minimum value of the total length of the cuts, not exceeding 2.
|
For $a>$. $a^{2}+b^{2}=100^{2}, ab=1$,
$$
\therefore c+b=\sqrt{100^{2}+2}>100 .
$$
The perimeter of the rectangle $2(a+b)$ is due to the original square's perimeter.
Given its two uses, $2 L+4 \geqslant 2(a+b)$, $L \geqslant a+b-2>98$.
On the other hand,
Any two parallelograms with the same base and height can be:
tightly arranged in the manner shown in the right figure,
taking $ABCD$ as a parallelogram with length $a$ and height $b$, then one length is $a+b$, so $L \leqslant a+b<100.01$.
Therefore, $L \approx 99$, with an error of no more than 1.01.
|
99
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The sequence $2,3,5,6,7,10,11, \cdots$ contains all positive integers that are neither perfect squares nor perfect cubes, find the 500th term of this sequence.
|
1. Answer: 528.
|
528
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. $n$ is the smallest positive integer satisfying the following condition:
(2) $n$ has exactly 75 positive divisors (including 1 and itself). Find $\frac{n}{75}$.
|
5. To ensure that $n$ is divisible by 75 and the obtained $n$ is the smallest, we can set $n=2^{\gamma_{13}} r_{2} \gamma_{3}$ and
$(r_{1}+1)(r_{2}+1)(r_{3}+1)=75$ $(r_{2} \geqslant 1, r_{3} \geqslant 2)$.
It is easy to prove that when $r_{1}=r_{2}=4, r_{3}=2$, $n$ has the minimum value. At this time, $\frac{n}{75}=2^{4} \cdot 3^{3}=432$.
|
432
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A biologist wants to estimate the number of fish in a lake. On May 1, he randomly caught 60 fish and marked them, then released them back into the lake. On September 1, he found that some of the fish were no longer in the lake (due to death or migration), and on September 1, 40% of the fish in the lake were not there on May 1 (due to new births or recent migration into the lake). How many fish were in the lake on September 1?
|
6. Let the number of fish in the lake on May 1 be $x$, and the number of fish in the lake on September 1 be $y$. According to the problem, we have
$$
y=0.75 x+0.40 y \text {. }
$$
On September 1, the number of tagged fish in the lake is $0.75 \times 60 = 45$. Assuming that the tagged fish on September 1 can represent the whole, we have
$$
\begin{aligned}
3 & =\frac{45}{y} . \\
\therefore y & =1050, x=\frac{0.60 y}{0.75}=840 .
\end{aligned}
$$
|
1050
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The vertices of a triangle are $P(-8$, 5), $Q(-15,-19), R(1,-7), \angle P$ bisector can be expressed as $a x+2 y+c=0$, find $a+c$,
|
7. Extend $P R$ to $T$, such that $P Q=P T$. Since $P Q=25, P R=15$, then the coordinates of $T$ are $(7,-15)$, and the angle bisector intersects $Q T$ at the midpoint $(-4,-17)$.
Thus, the slope of the angle bisector is $-\frac{11}{2}$, and the equation is
$$
\begin{array}{r}
11 x+2 y+78=0 . \\
\text { Hence } a+c=11+78=89 .
\end{array}
$$
|
89
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Find $i \div j$.
Translate the text above into English, keep the original text's line breaks and format, and output the translation result directly.
|
9. Tossing a coin 10 times has $2^{10}$ outcomes. Let $A(n)$ be the number of outcomes where no heads appear consecutively. By enumeration, it is easy to see that $A(1)=2, A(2)=3, A(3)=5$. It can also be proven by induction that there is a recurrence relation:
$$
A(n+2)=A(n+1) + A(n) \text {. }
$$
This is because $A(n+2)$ includes two types of sequences: one where no heads are consecutive and ends with a tail, of which there are $A(n+1)$; and another where no heads are consecutive but ends with a head, of which there are $A(n)$. Thus, $A(n)$ is a Fibonacci number. Therefore, $A(10)=144$.
Thus, the probability $P=\frac{144}{1024}=\frac{9}{64}$, so $i+1=73$.
|
73
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. We notice that $6!=8 \cdot 9 \cdot 10$. Try to find the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers.
|
11. If $n!$ can be expressed as the product of $(n-3)$ consecutive integers, then there exists an integer $k$, such that
$$
\begin{array}{l}
n!=(n+k)(n+k-1) \cdots(k+4) \\
=-(n+k)! \\
(k+3)!
\end{array}
$$
Originally,
$$
\begin{array}{l}
\frac{n+k}{k+3} \cdot \frac{n+k-1}{k+2} \cdot \ldots \cdot \frac{n+2}{5} \cdot \frac{n+1}{4!} \\
=1
\end{array}
$$
It is evident that $n>23$ is impossible.
On the other hand, $n=23, k=1$, is clearly one of its solutions. Therefore, the value of $n$ that satisfies the problem's requirements is 23.
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
The sum of all side lengths and diagonal lengths of a 12-sided polygon can be written in the form $a+b \sqrt{2}+c \sqrt{3}+d \sqrt{6}$, where $a, b, c$, $d$ are positive integers. Find $a+b+c+d$.
The text above is translated into English, preserving the original text's line breaks and format.
|
Coinciding with the origin, one vertex is at $(12,0)$, and the coordinates of the other vertices are $(12 \cos k x, 12 \sin k x)$, where $x=30^{\circ}$, $k=1,2, \cdots, 11$. The length of the line segment connecting $(12,0)$ and $(12 \cos k x, 12 \sin k x)$ is $24 \sin \frac{k x}{2}$. Therefore, from this,
$$
\begin{array}{l}
S=24\left(\sin 15^{\circ}+\sin 30^{\circ}+\cdots+\sin 165^{\circ}\right) \\
=48\left(\frac{1}{2} \sqrt{6}+\frac{1}{2}+\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\right)+24 \\
=48+24 \sqrt{2}+24 \sqrt{3}+24 \sqrt{6} .
\end{array}
$$
For the other points, similarly, noting that each line segment is counted twice, therefore,
$$
\begin{array}{l}
\text { Total length }=\frac{1}{2}(12 S) \\
=288+144 \sqrt{2}+144 \sqrt{3}+144 \sqrt{6} . \\
\text { Thus, } a+b+c+d=720 .
\end{array}
$$
|
720
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let $T=\left\{9^{4} \mid k\right.$ be an integer, $0 \leqslant k$ $\leqslant 4000\}$. It is known that $9^{1000}$ has 3817 digits, and its most significant digit is 9. How many elements in $T$ have 9 as their most significant digit?
|
13. Notice that $9^{k}$ has one more digit than $9^{k-1}$, unless $9^{1}$ starts with the digit 9. In the latter case, by long division, $9^{k-1}$ starts with the digit 1, and $9^{k}$ has the same number of digits. Therefore, from $9^{0}$ to $9^{4000}$, there are 3816 digit increases, so there must be 184 times when the number of digits from $9^{k-1}$ to $9^{k}$ does not increase.
$\because 9^{\circ}=1$, the leading digit is not 9,
$\therefore$ when the number of digits does not increase from $9^{\mathrm{k}-1}$ to $9^{\text { }}$, $9^{\mathbf{k}}(1 \leqslant k$ $\leqslant 4000$ ) has 9 as the leading digit.
Thus, there are 184 numbers that have 9 as the leftmost digit.
|
184
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. $A B C D$ is a rectangle, $A B=12 \sqrt{3}, B C=$ $13 \sqrt{3}$, diagonals $A C, B D$ intersect at $P$. If the triangle $A B P$ is cut off and $A P, B P$ are joined, find its volume.
|
14. As shown in the figure, $N$ is the intersection of the altitude from $P$ to $BCD$.
It is easy to prove that $N$ is the circumcenter of $\angle BCD$, and
\[
\begin{array}{l}
r=\frac{n^{2}}{\sqrt{4 n^{2}-m^{2}}} . \\
\because P B=\frac{1}{2} \sqrt{m^{2}+n^{2}}, \text { by the Pythagorean theorem, we have } \\
P N=\frac{m}{2} \sqrt{\frac{3 n^{2}-m^{2}}{4 n^{2}-m^{2}}}, \\
\therefore V_{\mathrm{PBCD}}=\frac{1}{3} P N \cdot S \triangle \mathrm{BCD} \\
\quad=\frac{m^{2}}{24} \sqrt{3 n^{2}-m^{2} .} \\
\because \quad m^{2}=432, n^{2}=507, \\
\therefore \quad V=594 .
\end{array}
\]
|
594
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
f_{u:-1}(k)=f_{1}\left(f_{u}(k)\right) .
$$
Find $f_{1991}\left(2^{1000}\right)$.
|
Let $k$ be much smaller, and we can estimate the value range of $f_{1}(k)$ as follows:
Suppose the positive integer $a$ has $m$ digits, then by changing all the digits of $a$ to 9, we have
$$
f_{1}(a) \leqslant 9^{2} m^{2}=81 m^{2}.
$$
If $a \leqslant b$, then $m$ is no greater than the integer part of $\lg b$ plus 1, thus
$$
f_{1}(a) \leqslant 81(1 + \lg b)^{2} < \left(4 \log _{2} 16 b\right)^{2}.
$$
Using the above inequality, we have
$$
\begin{array}{l}
f_{1}\left(2^{1990}\right) < 2^{4} 1994^{2} < 2^{20}, \\
f_{2}\left(2^{1900}\right) < (4 \cdot 30)^{2} = 14400.
\end{array}
$$
Therefore, the sum of the digits of $f_{2}\left(2^{1990}\right)$ is no more than $4 \cdot 9$, so
$$
\begin{array}{l}
f_{3}\left(2^{1990}\right) < 36^{2} = 1296, \\
f_{4}\left(2^{1800}\right) < (9 + 9 + 9)^{2} = 729, \\
f_{5}\left(2^{1000}\right) < (6 + 9 + 9)^{2} = 24^{2}.
\end{array}
$$
On the other hand, we have
$$
f_{1}(k) = k^{2} (\bmod 9).
$$
Since $12^{8} \equiv 1 (\bmod 9)$, we get
$$
\begin{aligned}
f_{1}\left(2^{1900}\right) & \equiv \left(2^{1990}\right)^{2} \equiv \left(2^{4}\right)^{2} \\
& \equiv 4 (\bmod 9), \\
f_{2}\left(2^{1900}\right) & \equiv -2 (\bmod 9), \\
f_{3}\left(2^{1900}\right) & \equiv 4 (\bmod 9).
\end{aligned}
$$
By analogy, generally, we have
$$
f_{0}\left(2^{1900}\right) \equiv \left\{\begin{array}{l}
4 (\bmod 9), 2 \mid n_{3} \\
-2 (\bmod 9), 2 \mid n_{0}.
\end{array}\right.
$$
Thus, since $f_{5}\left(2^{1580}\right) < 24^{2}$, $f_{5}\left(2^{1900}\right) \equiv 4 (\bmod 9)$, and noting that $f_{n}$ is a perfect square, we can deduce
$$
f_{5}\left(2^{1980}\right) \in \{4, 49, 121, 256, 400\}.
$$
For these 5 numbers, we find the square of the sum of their digits, yielding $f_{8}\left(2^{1900}\right) \in \{16, 169\}$.
Continuing to find the square of the sum of the digits, we get
$$
\begin{array}{l}
f_{7}\left(2^{1900}\right) \in \{49, 256\}; \\
f_{8}\left(2^{1090}\right) = 169.
\end{array}
$$
Therefore, when $n \geqslant 8$, we have
$$
f_{n}\left(2^{1000}\right) = \left\{\begin{array}{l}
169, 2 \mid n_{3} \\
256, 2 \nmid n_{0}.
\end{array}\right.
$$
In particular, we have $f_{1001}\left(2^{1000}\right) = 256$.
Solution 2: To quickly obtain the desired result for $f_{0+1}\left(2^{1990}\right)$, we simultaneously consider the remainder of $f_{0}\left(2^{1800}\right)$ modulo 9.
Since $2^{1000} < 8^{700} < 10^{700}$, $2^{1200}$ has at most 700 digits. Thus,
$$
\begin{array}{l}
f_{1}\left(2^{1990}\right) < (9 \cdot 700)^{2} < 4 \cdot 10^{7}, \\
f_{2}\left(2^{1990}\right) < (3 + 9 \cdot 7)^{2} < 4900, \\
f_{3}\left(2^{1990}\right) < (3 + 9 \cdot 3)^{2} = 30^{2}.
\end{array}
$$
Since $f_{3}\left(2^{1900}\right)$ is a perfect square, let $f_{s}\left(2^{1900}\right) = a^{2}$, then $a \equiv f_{2}\left(2^{1090}\right) \equiv 7 (\bmod 9)$. Therefore,
$$
\begin{aligned}
f_{3}\left(2^{1900}\right) & \in \{7^{2}, 16^{2}, 25^{2}\} \\
& = \{49, 256, 625\}.
\end{aligned}
$$
Noting that the sum of the digits of 49, 256, and 625 is 13, we have
$$
f_{4}\left(2^{1900}\right) = 13^{2} = 169.
$$
Continuing, we get
$$
\begin{array}{l}
f_{5}\left(2^{1990}\right) = 16^{2} = 256, \\
f_{0}\left(2^{1990}\right) = 13^{2} = 169.
\end{array}
$$
And so on. Finally, we have
$$
f_{1801}\left(2^{1800}\right) = 256.
$$
|
256
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $d_{1}, d_{2}, \cdots, d_{k}$ be all the divisors of the positive integer $n$, $1=d_{1}<d_{2}<d_{3}<\cdots<d_{k}=n$. Find all $n$ such that $k \geqslant 4$ and $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}=n$.
|
Number, but the sum of the squares of four odd numbers is even, which cannot equal $n$. Contradiction.
Thus, $n$ is even, $d_{2}=2$.
If $n$ is a multiple of 4, then $4 \in\left\{d_{3}, d_{4}\right\}$. Among the squares of the 4 factors, there are already two $\left(2^{2}\right.$ and $\left.4^{2}\right)$ that are multiples of 4, and one is 1. When another factor is odd, its square is congruent to 1 modulo 4; when even, its square is divisible by 4. Therefore, the sum of the squares of these 4 factors is not a multiple of 4, leading to a contradiction again.
Thus, $n=2 m, m$ is odd. $d_{3}$ is the smallest prime factor of $m$. Since $1^{2}+2^{2}+d_{3}^{2}+d_{4}^{2}=n$ is even, $d_{4}$ must be even, so $d_{4}=2 d_{3}$.
$$
n=1+2^{2}+d_{3}^{2}+4 d_{3}^{2} \Rightarrow 5 \text { is a divisor of } n.
$$
Excluding $d ; 4$, so $d \geqslant 6$, thus $d_{3}=5, d_{4}=10, n=1+2^{2}+5^{2}+10^{2}=130$.
|
130
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A natural number $N$ has exactly 12 divisors (including 1 and $N$), and these divisors are numbered in increasing order: $d_{1}<$ $d_{2}<\cdots<d_{12}$. The divisor with the index $d{ }_{4}-1$ equals $\left(d_{1}+d_{2}+d_{1}\right) \times d_{8}$, find $N$.
|
4. First, use the relationship
$$
d_{d_{4}-1}=\left(d_{1}+d_{2}+d_{4}\right) \times d_{8} \geqslant d_{5} \times d_{8}
$$
$=N$, we get $d_{4}=13, d_{5}=d_{2}+14, \quad N$ $=\left(d_{2}+14\right) \times d_{8}$. Then, using the enumeration method to take $d_{2}=2,3$, $5, 7, 11$ to get the only possible solution $d_{2}=3, d_{3}=9$, thus $N=9 \times 13 \times 17=1989$.
|
1989
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If a $23 \times 23$ square is composed of 1000 squares of sizes $1 \times 1, 2 \times 2, 3 \times 3$, how many $1 \times 1$ squares are needed at minimum?
|
4. At least 1 $1 \times 1$ square is needed. First, it needs to be shown that a $23 \times 23$ square can be composed of the following: $2 \times 2, 3 \times 3$ squares, and exactly 1 $1 \times 1$ square. Then, it also needs to be proven that a $1 \times 1$ square is indispensable.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. As shown in the figure,
$2, P$ is a point inside the triangle,
where the areas of the four
small triangles are marked in the figure. Find the area of $A B C$. (Third A1ME)
Note: The translation keeps the original text's line breaks and format as requested. However, the mathematical notation and the figure reference (like "As shown in the figure") are kept in their original form, as they are standard across languages. If the figure is not actually present, the reference to it might need to be adjusted.
|
Solve $-\frac{S \triangle \triangle P D}{S \triangle C D}=\frac{B D}{C D}$,
X
$\frac{S \therefore \operatorname{LA} 1}{S \triangle \mathrm{CAD}}=\frac{B D}{C D}$.
Let $S \triangle \mathrm{BPF}=x, S: A P E=y$,
Then
$$
\begin{array}{l}
S \frac{\mathrm{BAD}}{S-\mathrm{CAD}}=-\frac{S P C D}{S \triangle \mathrm{CPD}}=\frac{B D}{D C}, \\
\frac{84+x+40}{y+35+30}=\frac{40}{30} . \\
\frac{S-C B D}{S-A B R}=\frac{S \triangle C P R}{S}=\frac{C E}{E A}, \\
\frac{40+30+35}{x+84+y}=\frac{35}{y} .
\end{array}
$$
Solve the system of equations $\frac{84+x+40}{y+35+30}=\frac{40}{30}$,
$$
\frac{40+30+35}{x+84+y}=\frac{35}{y}
$$
Get $\left\{\begin{array}{l}x=70, \\ y=56 .\end{array}\right.$
Therefore, $S \triangle \triangle B C=315$.
|
315
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Calculate: $\sqrt{31 \cdot 30 \cdot 29 \cdot 28+1}$. (7th American Invitational Mathematics Examination)
|
```
\begin{array}{l}
\because \sqrt{ }(x+1) x(x-1)(x-2)+1 \\
= \sqrt{ }\left(x^{2}-x-2\right)\left(x^{2}-x\right)+1 \\
= \sqrt{ }\left(x^{2}-x-1\right)^{2} \\
=\left|x^{2}-x-1\right| . \\
\therefore \text { when } x=30 \text {, the original expression }=\left|30^{2}-30-1\right| \\
=869 .
\end{array}
```
|
869
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Take a point $P$ inside $\triangle ABC$, and draw three lines through $P$ parallel to the three sides of $\triangle ABC$,
thus forming three triangles
$t_{1}, t_{2}, t_{3}$ with areas
4, 9, 49 (as
shown in Figure 3).
Find the area of $\triangle ABC$.
(From the 2nd $\triangle I M E$)
|
$$
\begin{array}{l}
\text{Given that } \triangle t_{1}, \triangle t_{2}, \triangle t_{3} \text{ and } \triangle A B C \text{ are all similar triangles.} \\
\text{Let the areas of } \triangle A B C, \triangle t_{1}, \triangle t_{2}, \triangle t_{3} \text{ be } S, S_{1}, S_{2}, S_{3} \text{, respectively.} \\
\text{It is easy to know,} \\
\frac{D A}{C A}=\frac{P I}{C A}=\frac{\sqrt{S_{1}}}{\sqrt{S}}, \frac{C G}{C A}=\frac{H P}{C A} \\
=\frac{\sqrt{S_{2}}}{\sqrt{S}}, \quad G D=\sqrt{S_{3}} . \\
\because D A+C G+G D=C A, \\
\therefore \quad \frac{\sqrt{S_{1}}}{\sqrt{S}}+\frac{\sqrt{S_{2}}}{\sqrt{S}}+\frac{\sqrt{S_{3}}}{\sqrt{S}}=1. \\
\text{Thus,} \quad \sqrt{S}=\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S_{3}} \\
=2+3+7=12. \\
\end{array}
$$
Therefore, $S=144$.
|
144
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. A football invitational tournament has 16 cities participating, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and the two teams from the same city do not play against each other. If, after two days of the tournament, it is found that except for Team A of City $A$, all other teams have played a different number of matches. How many matches has Team B of City $A$ played? Please prove your conclusion. (1985, National High School League)
|
Solve the problem with a general discussion. Suppose there are $n$ cities participating in the competition, all of which meet the conditions of the problem. Let the number of matches played by Team B of City $A$ be $a_{a}$, obviously $a_{1}=0$.
In the case of $n$ cities, according to the competition rules, each team can play at most $2(n-1)$ matches. By the conditions of the problem, the number of matches played by the $2n-1$ teams, excluding Team A of City $A$, should be $0,1,2, \cdots, 2(n-1)$. For definiteness, let Team A of City $B$ have played $2(n-1)$ matches, meaning it has completed all its matches. Thus, except for Team B of City $B$, all other teams have played at least one match. Therefore, the number of matches played by Team B of City $B$ is 0. Now, remove the two teams from City $B$ and consider the remaining $n-1$ cities. At this point, excluding Team A of City $A$, the number of matches played by each team is $1-1=0, 2-1=1, \cdots, (2n-3)-1=2(n-2)$ (each reduced by one match played against Team A of City $B$). Since in the case of $n-1$ cities, the number of matches played by Team B of City $A$ is $a_{n-1}$, it follows that $a_{n}=a_{n-1}+1, n \geqslant 2$. This shows that the sequence $\left\{a_{0}\right\}$ is an arithmetic sequence with the first term $a_{1}=0$ and common difference $d=1$, thus
$$
a_{n}=0+(n-1) \cdot 1=n-1.
$$
Substituting $n=16$ into the above formula, we get that Team B of City $A$ has played 15 matches.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 - What is the maximum number of knights that can be placed on an $8 \times 8$ chessboard so that no two knights attack each other (assuming there are enough knights)?
|
We will alternately color the chessboard in black and white, so there will be 32 black squares and 32 white squares. According to the knight's move (see Figure 1), a knight on a black square can only capture a knight on a white square. Therefore, placing knights on all black squares means they will not capture each other. This means we can place 32 knights, and they will not capture each other. Now, we need to prove that placing 33 knights will inevitably result in some being captured.
In fact, dividing the chessboard
into 8 smaller $2 \times 4$
boards (as shown in Figure 6),
at least one of these smaller
boards will have to contain
5 knights. The possible
placements are: either
one row has 1 knight and the other has 4; or one row has 2 knights and the other has 3. Clearly, both of these placements will inevitably result in knights capturing each other.
Therefore, the maximum number of knights that can be placed so that they do not capture each other is 32.
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Given the sequence $\left\{x_{0}\right\}: x_{n+1}=$ $\frac{x_{\mathrm{n}}+(2-\sqrt{3})}{1-x_{n}(2-\sqrt{3})}$. Find the value of $x_{1001}-x_{401}$.
|
Given the shape of the recurrence relation, we can set $x_{\mathrm{n}}=\operatorname{tg} \alpha_{\mathrm{n}}$, and since $\operatorname{tg} \frac{\pi}{12}=2-\sqrt{3}$, we know that $x_{n+1}=\operatorname{tg} \alpha_{n+1}$
$$
=\frac{\operatorname{tg} \alpha_{\mathrm{a}}+\operatorname{tg} \frac{\pi}{12}}{1-\operatorname{tg} \alpha_{\mathrm{n}} \operatorname{tg} \frac{\pi}{12}}=\operatorname{tg}\left(\alpha_{\mathrm{n}}+\frac{\pi}{12}\right) \text {. }
$$
Therefore, $x_{\mathrm{n}+12}=x_{\mathrm{n}}$.
$$
\text { Hence } x_{1001}-x_{401}=x_{5}-x_{3}=0\left(\left\{x_{0}\right\}\right. \text { is a }
$$
periodic sequence).
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. As shown in Figure 5, in $\triangle ABC$, $AB=2$, $AC=3$, I, II, and III respectively denote the squares constructed on $AB$, $BC$, and $CA$. What is the maximum value of the sum of the areas of the three shaded regions? (1988, National Junior High School League)
---
The translation preserves the original text's line breaks and formatting.
|
Solve in $\triangle A D E$ and $\triangle A B C$ !! $\angle B A C + \angle D A E = 180^{\circ}$.
Also, $A C = A D, A B = A E$,
thus $S \triangle A D B = S \triangle A B C$.
Similarly, the areas of the other two shaded triangles are also equal to $S \triangle \triangle B C$.
$$
\begin{array}{r}
\therefore S \text { shaded } = 3 S \triangle \mathrm{ABC} = 3 \cdot \frac{1}{2} A B \cdot A C \\
\cdot \sin \angle B A C \leqslant 3 \cdot \frac{1}{2} \cdot 2 \cdot 3 \cdot \sin 90^{\circ} = 9 .
\end{array}
$$
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. A person earns a monthly salary of 200 yuan and receives a bonus of 20 yuan at the end of the month. After receiving the salary each month, they arrange the monthly expenses according to $\frac{4}{5}$ of the current savings. How much will be the remaining balance in the twelfth month?
|
Let the balance at the end of the $n$-th month be $a_{\mathrm{n}}$ yuan. According to the problem,
$$
\begin{array}{l}
a_{1}=200 \times\left(1-\frac{4}{5}\right)+20=60, \\
a_{2}=260 \times\left(1-\frac{4}{5}\right)+20=72 .
\end{array}
$$
In general, $a_{0}=\left(a_{0-1}+200\right) \times\left(1-\frac{4}{5}\right)+20$, which means $a_{\mathrm{n}}=\frac{1}{5} a_{n-1}+60$.
Let $a_{1}=60, \quad a_{n}=\frac{1}{5} a_{1-1}+60$
From this, we get
$$
a_{n}=75-15\left(\frac{1}{5}\right)^{n-1} \text {. }
$$
Therefore, $a_{12}=75-15\left(\frac{1}{5}\right)^{11} \approx 75$ (yuan).
The balance at the end of the 12th month is approximately 75.
|
75
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. The lengths of the three sides $a, b, c$ of a triangle are all integers, and $a \leqslant b \leqslant c$. If $b=10$, then the number of such triangles is ( ).
(A) 10.
(B) 55.
(C) $10^{2}$.
(D) Infinitely many.
(1990, Suzhou High School Competition)
|
When $b=n$, take $a=k(1 \leqslant k \leqslant n)$, and $b \leqslant c<a+b$, then $n \leqslant c<n+k$, at this time the value of $c$ has exactly $k$ possibilities, i.e., $c=n, n+1, \cdots, n+k-1$. The table is as follows:
\begin{tabular}{c|c|c|c}
\hline$a$ & $b$ & $c$ & Number of triangles \\
\hline 1 & $n$ & $n$ & 1 \\
\hline 2 & $n$ & $n, n+1$ & 2 \\
\hline 3 & $n$ & $n, n+1, n+2$ & 3 \\
\hline$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
\hline$k$ & $n$ & $n, n+1, \cdots, n+k-1$ & $k$ \\
\hline$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
\hline$n$ & $n$ & $n, n+1, n+2, \cdots, 2 n-2$, & $n$ \\
\hline
\end{tabular}
Therefore, when $b=n$, the total number of triangles that meet the conditions is $1+2+\cdots+n=\frac{1}{2} n(n+1)$. Taking $n=10$, substituting into the above formula, we get
$\frac{1}{2} \cdot 10 \cdot(10+1)=55($ triangles), so the answer is $(B)$.
|
55
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. (1987, National Junior High School League) Among the triangles with three consecutive natural numbers as sides and a perimeter not exceeding 100, the number of acute triangles is $\qquad$
|
(29)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
29
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. (1986, Shanghai) The three sides of a triangle are all positive integers, one of which has a length of 4, but it is not the shortest side. How many different triangles are there? ...
The translation is provided while retaining the original text's line breaks and format.
|
(8).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. A pile of toothpicks 1000 in number, two people take turns to pick any number from it, but the number of toothpicks taken each time must not exceed 7. The one who gets the last toothpick loses. How many toothpicks should the first player take on the first turn to ensure victory? (New York Math Competition)
|
From $1000=125 \times 8$, we know that one should first dare to take 7 moves, so that the latter can achieve a balanced state: $124 \times(7+1)+1$.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. On a $4 \times n$ chessboard, if any cells are colored red, white, or blue, then the necessary and sufficient condition for the existence of a monochromatic rectangle is $n=19$.
|
If $n=19$, then in a $4 \times 19$ chessboard, there must be $\left[\frac{4 \times 19-1}{3}\right]+1=26$ small squares of the same color, let's assume they are red. At this point,
$$
\begin{array}{l}
m=4, n=19, a=26, q=1, r=7, \\
r C_{4+1}^{2}+(n-r) C_{4}^{2}=7, \\
C_{\mathrm{m}}^{2}=6,
\end{array}
$$
The conditions of Theorem 2 are satisfied, so there must exist a red monochromatic rectangle.
If $n=18$, then
$$
r C_{q+1}^{2}+(n-r) C_{\mathrm{q}}^{2}=C_{\mathrm{m}}^{2} \text { . }
$$
If we color 24 small squares red, 24 white, and 24 blue on this chessboard, and ensure that the small squares of each color are evenly distributed across the 18 columns, then there will be no monochromatic rectangles on this chessboard. As shown in Figure 3. $R, W, B$ represent red, white, and blue, respectively.
In summary, the necessary and sufficient condition for the existence of a monochromatic rectangle is $n=19$.
|
19
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. In $\triangle A B C$, $D, E, F$ are the midpoints of $B C$, $C A, A B$ respectively, and $G$ is the centroid. For each value of $\angle B A C$, how many non-similar $\triangle A B C$ are there such that $A E G F$ is a cyclic quadrilateral?
|
1. From $A, E, G, F$ being concyclic, we get
$$
\begin{aligned}
\angle C G_{E} & =\angle B A C \\
= & \angle C E D .
\end{aligned}
$$
Let $C G$ intersect $D E$ at $M$. Then from (1),
it is easy to derive
$$
C M \times C G=C E^{2},
$$
i.e., $\frac{1}{2} m_{\mathrm{e}} \times \frac{2}{3} m_{\mathrm{c}}=\left(\frac{1}{2} b\right)^{2}$.
Here, the median
$$
m_{c}^{2}=\frac{1}{2} a^{2}+\frac{1}{2} b^{2}-\frac{1}{4} c^{2} .
$$
From (2) and (3), we get
$$
2 a^{2}=b^{2}+c^{2} \text {. }
$$
Thus, by the cosine rule and (4), we have
$$
b^{2}+c^{2}=4 b c \cos \angle B A C \text {. }
$$
Let $\frac{b}{c}=\lambda$, then
$$
\lambda^{2}+1=4 \lambda \cos \angle B A C .
$$
In (6), there are real solutions if and only if $\angle B A C \leqslant 60^{\circ}$. The solutions $\lambda_{1}, \lambda_{2}$ are reciprocals of each other, thus generating similar triangles. Therefore, for $\angle B A C \leqslant 60^{\circ}$, there is only one non-similar $\triangle A B C$ that satisfies the conditions. When $\angle B A C > 60^{\circ}$, there are no $\triangle A B C$ that satisfy the conditions.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4.199J individuals are divided into several mutually disjoint subsets, such that
(a) in each subset, no one knows everyone else.
(b) in each subset, among any three people, at least two do not know each other.
(c) in each subset, for any two people who do not know each other, there is exactly one person in the subset who knows both of them.
(i) Prove that in each subset, each person knows the same number of people.
(ii) Find the maximum number of such subsets.
Note: If $A$ knows $B$, then $B$ knows $A$, and everyone knows themselves.
|
4. (i) Consider only one group. Let $y_{1}$ and $y_{2}$ be unknown to each other within the same group. By (c), there exists $x$ who knows both $y_{1}$ and $y_{2}$. Suppose, apart from $x$ and themselves, $y_{1}$ knows $z_{11}, z_{12}, \cdots, z_{11}$. $y_{2}$ knows $z_{21}$, $z_{22}, \cdots, z_{2 \mathrm{k}}$.
By (b), $x$ and $z_{11}$ do not know each other. By (c), $z_{11}$, $z_{12}, \cdots, z_{1}$ do not know $y_{2}$. Therefore, $z_{11}$, $z_{12}, \cdots, z_{11}, z_{21}, z_{22}, \cdots, z_{2 \mathrm{k}}$ are all distinct, and they are also distinct from $x, y_{1}, y_{2}$.
By (c), $z_{11}$ and $y_{2}$ have a common acquaintance, let's say $z_{21}$. $z_{12}$ and $y_{2}$ also have a common acquaintance, which cannot be $z_{21}$ (otherwise, $y_{1}$ and $z_{21}$ would have two common acquaintances, one of which is $z_{2}$). Continuing this process, we find that $h \leqslant k$ (the number of acquaintances of $z_{11}$, and these acquaintances are all distinct). By symmetry, we also have $k=h$. Therefore, $k=h$, and any two people who do not know each other have the same number of acquaintances. From the above argument, we know that $x$ and $z_{11}$ have the same number of acquaintances, and $z_{11}$ and $y_{2}$ have the same number of acquaintances. Thus, in this group, everyone has $h+1$ acquaintances.
(ii) In a group, $x$ has an unknown person $y$, and $x$ and $y$ have a common acquaintance $z$. $z$ has an unknown person $u$. $u$ might know both $x$ and $y$ (otherwise, if $u$ does not know $x$, then $u$ and $x$ have a common acquaintance $v$, and $v$ also knows $y$ and $z$. Thus, there are at least 5 people in this group who are all acquainted with each other.
Therefore, the maximum number of groups is $\frac{1990}{5}=398$.
|
398
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10: There are 100 points on a plane, where the distance between any two points is no less than 3. Now, connect every two points that are exactly 3 units apart with a line. Prove: the number of these line segments will not exceed 300. (1984, Beijing Junior High School Competition)
|
First, consider the maximum number of points whose distance from a certain point $A$ is: 3. Let $B_{1} A=B_{2} A=\cdots=B_{1} A$
$$
\begin{array}{l}
=3, \text { and } B_{1} B_{2} \geqslant 3, B_{2} B_{3} \geqslant 3, \cdots, B_{\mathrm{k}-1} B_{1} \\
\geqslant 3, B_{\mathrm{k}} B_{1} \geqslant 3 . \text { Then } \angle B_{1} A B_{2} \geqslant 60^{\circ}, \\
\angle B_{2} A B_{3} \geqslant 60^{\circ}, \cdots, \angle B_{1} A B_{1} \geqslant 60^{\circ},
\end{array}
$$
we get
$$
\begin{aligned}
360^{\circ} & =\angle B_{1} A B_{2}+\angle B_{2} A B_{3}+\cdots \\
& +\angle B_{k} A B_{1} \geqslant k \cdot 60^{\circ}, k \leqslant 6 .
\end{aligned}
$$
That is, on the circumference of a circle with $A$ as the center and 3 as the radius, there are at most 6 points. Now, there are 100 points, each of which can be considered as $A$, so the number of line segments with a length of 3 will not exceed $6 \times 100$. However, in this calculation, each point as the endpoint of a line segment is counted twice, so the number of line segments with a length of 3 will not exceed $\frac{6 \times 100}{2}=300$.
|
300
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. The decimal representation of the natural number $A$ is $\overline{a_{n} a_{n-1} \cdots a_{0}}$. Let $f(A)=2^{\mathrm{n}} a_{0}+2^{\mathrm{n}-1} a_{1}+\cdots+2 a_{\mathrm{n}-1}+a_{\mathrm{n}}$, and denote $A_{1}=f(A), A_{1+1}=f\left(A_{1}\right) . \quad(i=1,2$, $\cdots, n$ ) Prove:
i) There must be a $k \in \mathbb{N}$, such that $A_{\mathrm{k}+1}=A_{\mathrm{k}}$;
ii) If $A=19^{38}$, what is the value of the above $A_{\mathrm{k}}$?
|
Proof (i) When $n=0$. For any $k$, we have $A_{\mathrm{s}}=A$.
$$
\begin{array}{l}
\text { When } n=1 \text {, } \\
A-f(A)-10 a_{1}+a_{0}-2 a_{0}-a_{1} \\
=9 a_{1}-a_{0} \geqslant 9 \times 1-9=0 . \\
\text { When } n \geqslant 2 \text {, } \\
A-f(A) \\
\geqslant 10^{\mathrm{n}}-\left(2^{\mathrm{n}}+2^{\mathrm{n}+1}+\cdots+2+1\right) \times 9 . \\
>10^{\mathrm{n}}\left(10^{\mathrm{n}-1}-2^{\mathrm{n}+1}+1\right) \\
>10\left(2^{\mathrm{na}-3}-2^{\mathrm{n}+1}+1\right) \\
>0 .
\end{array}
$$
From the above, it is easy to see that $\left\{A_{1}\right\}$ is decreasing. Also, for any $i \in N$, we have $A_{1} \in N$. Therefore, there exists $k \in N$ such that $A_{\mathrm{k}+1}=A_{\mathfrak{k}}$.
$$
\text { ii) } \begin{aligned}
& \because 10^{n} f(A)-A \\
& =\left(20^{n}-1\right) a_{0} \div(20-1-1) \times 10 a_{1} \\
& +\cdots+(20-1) \times 10^{n-1} a_{n-1}
\end{aligned}
$$
Each term above is divisible by 19,
$$
\therefore 19 \mid\left(10^{n} f(A)-A\right) \text {. }
$$
Thus, if $19|A$, then $19 \mid f(A)$.
Since $19 \mid 19^{86}$, then $19 \mid A_{\mathrm{k}}$.
From the above, when $n=2$, $f(A)<A$.
When $n=1$, only when $a_{1}=1, a_{0}=9$ do we have
$$
A=f(A) \text {. }
$$
Therefore, $A_{\mathrm{k}}=19$.
|
19
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
In the May 1989 issue of this journal, there was a problem from the Zhejiang Province High School Mathematics Competition as follows:
Arrange the positive rational numbers: $\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, \cdots$, then the position number of the number $\frac{1989}{1949}$ is $\qquad$.
|
The original answer is $(1+2+\cdots+3937)+1949$ $=7753902$.
In fact, the original sequence is formed by segments of fractions with the sum of the numerator and denominator being 2, 3, 4, ․ respectively. The fraction 1989/1949 has a sum of numerator and denominator of 3938, making it the 1949th number in the 3937th segment. Therefore, the sequence number is $(1+2+\cdots+3936)+$ $1949=7749965$.
|
7749965
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
On a plane, there is a convex quadrilateral $A B C D$.
(1) If there exists a point $P$ on the plane such that the areas of $\triangle A B P, \triangle B C P, \triangle C D P, \triangle D A P$ are all equal, what condition should the quadrilateral $A B C D$ satisfy?
(2) How many such points $P$ can there be on the plane at most? Prove your conclusion.
|
. Solving (i) First consider the case where $P$ is inside the quadrilateral.
If points $A, P, C$ are collinear, and points $B, P, D$ are also collinear, then quadrilateral $ABCD$ is a parallelogram, and $P$ is the intersection of the two diagonals.
If points $A, P, C$ are not collinear, by the fact that $\triangle PAB \cong \triangle PAC$ in area, line $AP$ must pass through the midpoint of diagonal $BD$. Similarly, line $CP$ must also pass through the midpoint of $BD$. Therefore, $P$ is the midpoint of $BD$. Clearly, in this case, $\triangle ABD$ and $\triangle BCD$ have equal areas. This means that if there is a point $P$ inside the quadrilateral that meets the requirements, then the quadrilateral $ABCD$ is divided into two equal areas by one of its diagonals. It is easy to see that this condition is also sufficient.
(ii) Next, consider the case where point $P$ is outside the quadrilateral.
Since $\triangle APB \cong \triangle APD$ in area, $AP \parallel BD$. Similarly, $AC \parallel PD$. Therefore, quadrilateral $AEDP$ is a parallelogram. At this time, $S_{\triangle BVP} = S_{\triangle DVP} = \frac{1}{2} S_{\triangle BCD}$. Additionally, it is easy to see that any point $Q$ in the angle domain formed by extending two adjacent sides of the quadrilateral, such as extending $BA$ and $DA$, does not meet the requirements. This means that if a point $P$ outside the quadrilateral meets the requirements, then one of the four triangles formed by the two diagonals of the quadrilateral must have an area equal to half the area of the quadrilateral. We will now prove that this condition is also sufficient.
Assume $S_{\triangle BDP} = \frac{1}{2} S_{\triangle BCD}$. Draw a line through $A$ parallel to $BD$ and a line through $D$ parallel to $CA$, and let these two lines intersect at point $P$. Then quadrilateral $AEDP$ is a parallelogram and $S_{\triangle PAD} = \frac{1}{2} S_{ABCD}$. Since $S_{\triangle BE} < S_{\triangle ED}$, it follows that $BE < ED$. Therefore, $PB$ is inside the pentagon $PABCD$ and will not be outside the quadrilateral. Similarly, $PC$ is also inside the quadrilateral. Thus, $S_{\triangle APB} = S_{\triangle PBD} = S_{\triangle PDC}$. Consequently, $S_{\triangle PBC} = S_{PABCD} - S_{\triangle PAB} - S_{\triangle PCD} = S_{\triangle PAD}$. This proves that point $P$ meets the requirements of the problem.
(iii) From (i) and (ii), we know that there is at most one point $P$ outside the quadrilateral $ABCD$ that meets the requirements, and the interior and exterior of the quadrilateral cannot simultaneously have points that meet the requirements. Furthermore, if both diagonals of the quadrilateral divide the quadrilateral into equal areas, it is easy to prove that the quadrilateral is a parallelogram, so there is at most one point $P$ inside the quadrilateral that meets the requirements. In summary, the point $P$ that satisfies (1) can have at most one point.
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are there on the circle with the most birds, at minimum?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are there on the circle with the most birds, at minimum?
|
We use 10 points to represent 10 birds.
(1) Among the 10 points, there must be 5 points that are concyclic. If not, then any 5 points among the 10 points are not concyclic. The 10 known points can form $C_{10}^{5}=252$ groups of 5 points. According to the given, each group of 5 points has four points that are concyclic (hereinafter referred to as a four-point circle), which can form 252 four-point circles (including repeated counts). Each four-point circle belongs to 6 different groups of 5 points, so there should be 42 different four-point circles. Each four-point circle has four known points, totaling 168 points. Since there are 10 known points, there is one point $A$ on 17 different four-point circles. On these 17 four-point circles, apart from $A$, there are 3 other known points, totaling 51 points. Apart from $A$, there are 9 known points, so there is a point $B$ different from $A$, passing through $A$: 17 circles at least 6 pass through $B$, meaning there are at least two different four-point circles passing through $A$ and $B$, which have 12 points and they do not overlap, which is impossible.
(2) Suppose 5 known points $A_{1}, A_{2}, A_{3}, A_{4}$, $A_{8}$ are all on circle $S_{1}$, then the other 5 points can have at most one point not on $S_{1}$. If not, let $B, C$ not be on $S_{1}$.
Consider the 5-point group $A_{1}, A_{2}, A_{3}, B, C$. According to the given, 4 points among the 5 points must be concyclic on $S_{2}$. Since $B, C$ are not on $S_{1}$, $A_{1}, A_{2}, A_{3}$ cannot be on $S_{2}$, otherwise $S_{2}$ and $S_{1}$ would coincide. Assume $A_{1}, A_{2}$ are on $S_{2}$, then $A_{3}, A_{4}, A_{8}$ are not on $S_{2}$.
Consider the 5-point group $A_{3}, A_{4}, A_{5}, B, C$. Assume $A_{3}, A_{4}, B, C$ are concyclic on $S_{3}$, then $A_{1}, A_{2}, A_{5}$ are not on $S_{3}$.
Finally, consider the 5-point group $A_{1}, A_{3}, A_{8}, B, C$. Since $A_{1}, A_{3}, A_{8}$ are on $S_{1}$ and $B, C$ are not on $S_{1}$, $A_{1}, A_{3}, A_{8}, B$ and $A_{1}, A_{3}, A_{8}, C$ are not concyclic. Since $A_{1}, B, C$ are on $S_{2}$ and $A_{3}, A_{5}$ are not on $S_{2}$, $A_{1}, A_{3}, B, C$ and $A_{1}, A_{5}, B, C$ are not concyclic; similarly, it can be proven that $A_{3}, A_{5}, B, C$ are not concyclic, meaning any 4 points in this 5-point group are not concyclic, which contradicts the given.
In summary, the circle containing the most points must have at least 9 known points. Additionally, if 9 known points are on one circle and the other point is not on this circle, these 10 points clearly meet the requirements of the problem. Therefore, the circle with the most birds must have at least 9 birds.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. If from the numbers $1,2, \cdots, 141$, we sequentially select $a_{1}, a_{2}, a_{2}$, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$, then how many different ways of selection are there that meet the above requirements?
|
Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1, 2, \cdots, 10\}$. Suppose $\left\{a_{1}, a_{2}, a_{3}\right\}$ is a subset of $S$ and satisfies $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-2, a_{3}^{\prime}=a_{3}-4$, i.e., the following mapping is established:
$$
\begin{array}{c}
\left(a_{1}, a_{2}, a_{3}\right) \rightarrow\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right) \\
=\left(a_{1}, a_{2}-2, a_{3}-4\right) .
\end{array}
$$
The number of ways to choose such subsets is equal to the number of ways to choose any 3 different numbers from $S^{\prime}$, i.e., $C_{10}^{3}=120$.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. (Shanghai $\cdot$ Senior High School) If positive integers $p, q, r$ make the quadratic equation $p x^{2}-{ }_{y} x+r=0$ have two distinct real roots in the open interval $(0,1)$, find the minimum value of $p$.
|
Let $\alpha, \beta$ be the two roots of the quadratic equation $p x^{2}-q x+\gamma=0$, $\alpha \neq \beta, 0<\alpha, \beta<1$. Then
$$
\alpha \beta(1-\alpha)(1-\beta) = \frac{r}{p} \left[1 - \frac{q}{p} + \frac{r}{p}\right] = \frac{r}{p^{2}}(p-q+r).
$$
Therefore, $p^{2} > 16 r(p-q+r)$.
Since the quadratic term coefficient $p > 0$ of $f(x) = p x^{2} - q x + r$, and its roots are within $(0,1)$, we have $f(1) = (p - q + r) = 0$, and $r(p - q + r) > 0$. Given that $p, q, r$ are positive integers, we have $r(p - q + r) \geq 1$. Thus, from the inequality, we get
$$
p^{2} > 16 r(p - q + r) = 16.
$$
Therefore, $p > 1$, i.e., $p > 5$.
When $p = 5$, the quadratic equation $5 x^{2} - 5 x + 1 = 0$ has two roots $\alpha = \frac{5 + \sqrt{5}}{10}, \beta = \frac{5 - \sqrt{5}}{10}$, both of which are within $(0,1)$. Hence, $p_{10} = 5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. (First State "Killing Hope Cup" - Grade 10) Let the function $f(n)=k$, where $n$ is a natural number, and $k$ is the $n$-th digit after the decimal point of the irrational number $\pi=3.1415926535 \cdots$. It is also defined that $f(0)=3$.
$$
\text { Let } F(n)=\frac{f(f(f(f \cdots(f(n)) \cdots)))}{10 \uparrow \bar{f}} \text {. }
$$
Prove: $F[f(1990)+f(5)+f(13)]$
$$
=F[f(1990) f(3) f(25)] .
$$
|
It can be easily proven that for non-negative integers $n$, $F(n)=1$ always holds.
Below $f(1990)+f(5)+f(13)$ and $f(1990) f(3)$
- $f(25)$ are all non-negative integers, so $F(f(1990)+f(5)$
$$
+f(13) \equiv 1 \equiv F[(1990) f(3) f(25)] .
$$
|
1
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. (14th All-Russian Mathematics Competition) Let
$$
\frac{1}{1+\frac{1}{1+\frac{1}{1+}}+\frac{1}{1}}=-\frac{m}{n},
$$
where $m$ and $n$ are coprime natural numbers, and the left side of the equation has 1988 fraction lines. Calculate $m^{2}+m n-n^{2}$.
|
Let the value of the complex fraction with $k$ layers of fraction lines be $\frac{m_{k}}{n_{k}},\left(m_{k}, n_{k}^{*}\right)=1$, then
$$
\frac{m_{\mathrm{k}+1}}{n_{\mathrm{k}+1}}=\frac{1}{1+\frac{m_{\mathrm{k}}}{n_{\mathrm{k}}}}=\frac{n_{\mathrm{k}}}{m_{\triangle}+n_{\mathrm{k}}},
$$
i.e., $m_{\mathrm{k}+1}=n_{\mathrm{k}}, n_{\mathrm{k}+1}=m_{\mathrm{k}}+n_{\mathrm{b}}$.
Notice that $\frac{m_{1}}{n_{1}}=\frac{1}{1}, m_{1}=1, n_{1}=1$, comparing it with the Fibonacci sequence: $f_{1}=1, f_{2}=1, f_{k+1}$ $=f_{k}+f_{k+1}(k \geqslant 1)$, we can see that,
$$
m_{1}=f_{\mathrm{k}}, \quad n_{\mathrm{k}}=f_{\mathrm{k}+1} \text {. }
$$
Thus, $m^{2}+m n-n^{2}$
$$
\begin{array}{l}
=f_{1988}^{2}+f_{1988} f_{1989}-f_{1989}^{2} \\
=f_{1988}^{2}+f_{1988}\left(f_{1988}+f_{1987}\right) \\
-\left(f_{1987}^{2}+2 f_{1988} f_{1987}+f_{1988}^{2}\right) \\
=-\left(f_{1937}^{2}+f_{1987} f_{1988}-f_{1988}\right) \\
=f_{1980}^{2}+f_{1986} f_{1087}-f_{1987}^{2} \\
=\cdots \cdots
\end{array}
$$
$$
\begin{array}{l}
=f_{2}^{2}+f_{2} f_{3}-f_{3}^{2} \\
=1^{2}+1 \cdot 2-2^{2}=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
19. (Wuhu City)
In a square plate $ABCD$ with a side length of 8 units, there is an equilateral triangular colored plate $OEF$ (vertex $O$ is painted red, $E, F$ are white) placed as shown (E coincides with D, and $EF$ coincides with $AD$). Now, the triangular colored plate rotates to the left with $F$ as the pivot until $O$ touches the $AB$ side, completing one "left move". Then, with $O$ touching the $AB$ side as the pivot, the triangular colored plate rotates to the left again, making the second "left move". This continues until the 1990th "left move" is completed and the plate stops. How far is the red vertex $O$ from point $A$ at this time? Explain your reasoning.
|
The "left advance" rule for the triangular color board is:
?.
(1) Every 3 left advances, the red vertex is inside the disk;
(2) Every 16 left advances, the triangular board returns to its original starting position, but the red vertex moves to the point F on the AD side;
(3) Every 48 left advances, the triangular board and its red vertex completely return to their original positions.
Since $1990=48 \times 41+22$,
1990 left advances are equivalent to 22 left advances. After 16 left advances, O moves to the F position. After another 6 left advances, the vertex moves to the midpoint P on the AB side, at which point $PA=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. (Lokni Army) Let $A$ be a finite set, $N=\{1, 2, 3, \cdots\}$. If there exists a function $f: N \rightarrow A$ with the following property: if $|i-j|$ is a prime number, then $f(i) \neq f(j)$. Try to find the minimum number of elements in $A$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
Consider in $N$ taking $i=1,3,6,8$. The absolute value of the difference between any two of these numbers is a prime number, so their images $f(1), f(3), f(6), f(8)$ are the same, meaning that the set $A$ must have at least 4 elements.
Below, we prove that the minimum number of elements in set $A$ is 4. For this, we construct a function $f$ that satisfies the given conditions for a four-element set.
$$
\begin{array}{c}
\text { Let } A=\left\{a_{0}, a_{1}, a_{2}, a_{3}\right\} \text { and } N_{k} \\
=\{x \mid x=k+4 n, n \in Z\}(k=0,1,2,3) .
\end{array}
$$
For any $x \in N$, define $f(x)=a_{k}$ if $x \in N_{1}$. In this case, for any $i, j \in N$, if $f(i) = f(j)$, by the definition of the function, $i$ and $j$ must be elements of the same set $N_{1}$, so $i-j$ is divisible by 4, and thus $|i-j|$ cannot be a prime number. This means that for a four-element set $A$, a function $f$ that satisfies the conditions exists. In summary, the minimum number of elements in set $A$ is 4.
|
4
|
Logic and Puzzles
|
other
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x y+x+y=71, x^{2} y+x y^{2}$
$=830 . x, y$ are positive integers, find $x^{2}+y^{2}$.
|
1. Answer: 146. (Hint: Let $a=x+y$, $b=x y$ )
|
146
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Expand $(1+0.2)^{1000}$ using the binomial theorem, i.e. let $C_{1000}^{0}(0.2)^{0}+C_{1000}^{1}(0.2)^{1}+\cdots+$ $C_{1000}(0.2)^{1000}=A_{0}+A_{1}+\cdots+A_{1000}$. Let $A_{h}=C_{1000}^{k}(0.2)^{h}, k=0,1,2, \cdots$, 1000 . Find the value of $k$ for which $A_{k}$ is the largest.
|
3. For $1 \leqslant k \leqslant 1000$,
$$
\begin{array}{l}
\frac{A_{\mathrm{s}}}{A_{\mathrm{k}-1}}=\frac{\frac{1000!}{k!(1000-k)!}(0.2)^{k}}{\frac{1000!}{(k-1)!(1001-k)!}(0.2)^{k}} \\
=\frac{1001-k}{k}(0.2) \text {. } \\
\end{array}
$$
When and only when $1001-k>5k$, this ratio is greater than 1. That is, when $k \leqslant 166$, the above ratio is greater than 1.
Therefore, $A_{0}A_{108}>\cdots>A_{1000 \text { . }}$
And $A_{1 \in \theta}>A_{107}$,
which means when $\dot{x}=1$, $A$ is the largest.
|
166
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The number of real solutions to $\frac{1}{5} \log _{2} x=\sin (5 \pi x)$ is?
|
Consider only those $x$ that satisfy $\left|\frac{1}{5} \log _{2} x\right| \leqslant 1$. Solving this, we get $\frac{1}{32} \leqslant x \leqslant 32$.
First, consider $\frac{1}{32} \leqslant x<1$, in this case $-1 \leqslant \frac{1}{5} \log _{2} x <0$. When $\frac{1}{5} \leqslant x \leqslant \frac{2}{5}, \frac{3}{5} \leqslant x \leqslant \frac{4}{5}$, $\sin 5 \pi x \leqslant 0$.
The graph of $y=\frac{1}{5} \log _{2} x$ intersects at 4 points within $\frac{1}{32} \leqslant x<1$ (as shown in the figure).
When $1<x \leqslant 32$, $0<\frac{1}{5} \log _{2} x \leqslant 1$.
If $\frac{2 k}{5} \leqslant x \leqslant$
$$
\frac{2 k+1}{5}(k=3,4, \cdots,
$$
79), $\sin 5 \pi x=0$.
On these 77 intervals, the graphs of the two functions intersect at $2 \times 74=154$ points. $(1<x \leqslant 32)$
At $x=1$, the graphs of the two functions also intersect at one point.
Therefore, there are a total of $4+154+1=159$ solutions.
|
159
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. For a given rational number, represent it as a reduced fraction. Then find the product of its numerator and denominator. How many rational numbers between 0 and 1 have a product of their numerator and denominator equal to 20! ?
|
5. 20! has 8 prime factors: $2, 3, 5, 7, 11, 13, 17, 19$. A prime factor appearing in the denominator cannot appear in the numerator (and vice versa). For each prime factor, it can be in the numerator or in the denominator, so there are $2^8 = 256$ choices. However, not all of them are less than 1. In fact, the numbers can be paired as reciprocals of each other. Therefore, only $\frac{256}{2} = 128$ satisfy the requirement.
|
128
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Find the value of $r$ such that
$$
\begin{array}{l}
{\left[r+\frac{19}{100}\right]+\left[r+\frac{20}{100}\right]+\cdots} \\
+\left[r+\frac{91}{100}\right]=546 .
\end{array}
$$
Find $[100 r]$. (Where $[x]$ denotes the greatest integer not greater than $x$)
|
6. The left side of the equation has 73 terms, each being $[r]$ or $[r+1]$. This is because $\frac{19}{100}, \frac{20}{100}, \cdots, \frac{91}{100}$ are all less than 1, and $73 \times 7 < 546 < 73 \times 8$. To make the sum 546, $[r]$ must be 7.
Assume $\left[r+\frac{k}{100}\right]=7(19 \leqslant k \leqslant m),[r+$ $\left.\frac{k}{100}\right]=8(m+1 \leqslant k \leqslant 91)$. Then $7(m-18)+$ $8(91-m)=546$, yielding $m=5.6$, so $\left[r+\frac{56}{1.00}\right]=7$, $\left[r+\frac{5 ?}{100}\right]=8$, thus $7.43 \leqslant r \approx 7.44$. Therefore, $[100 r]=743$.
|
743
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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