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6. A square piece of paper is divided into two parts by a straight line, one of which is then divided into two parts by another straight line, and then one of the three pieces is divided into two parts by another straight line, and so on. In the end, 19 96-sided polygons and some other polygons are obtained. Then the minimum number of cuts required is $\qquad$.
|
\%. 1766 . After each division, the sum of the angles of the resulting polygons increases by $2 \pi$. Thus, after $k$ divisions, $k+1$ polygons are obtained, the sum of whose interior angles is $2 \pi(k+1)$. Since among these $k+1$ polygons, there are 19 ninety-six-sided polygons, the sum of their interior angles is $19 \times 94 \pi$, and the remaining $k+1-19=k-18$ polygons, each of which has an interior angle sum of no less than $\pi$, have an interior angle sum of no less than $(k-18) \pi$. Therefore, the inequality $19 \times 94 \pi + (k-18) \pi \leqslant 2(k+1) \pi$ holds, which simplifies to $k \geqslant 19 \times 93 - 1 = 1766$, meaning the minimum number of divisions is 1766.
|
1766
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (1/3 points) There are 99 numbers $a_{1}, a_{2}, a_{3}, \cdots, a_{98}$, defined as: $S_{1} = a_{1}, S_{2} = a_{1} + a_{2}, \cdots, S_{k} = a_{1} + a_{2} + \cdots + a_{k} (1 \leqslant k \leqslant 99)$. There are 100 numbers $b_{1}, b_{2}, \cdots, b_{100}$, and $b_{1} = 1, b_{2} = 1 + a_{1}, b_{3} = 1 + a_{2}, \cdots, b_{k} = 1 + a_{k-2} (2 \leqslant k \leqslant 100)$, defined as: $S_{1}^{\prime} = b_{1}, S_{2}^{\prime} = b_{1} + b_{2}, \cdots, S_{k}^{\prime} = b_{1} + b_{2} + \cdots + b_{k} (1 \leqslant k \leqslant 100)$.
(1) Find: $S_{k}^{\prime} - S_{k-1} (2 \leqslant k \leqslant 100)$;
(2) If $A = \frac{S_{1} + S_{2} + \cdots + S_{90}}{99} = 50$, find the value of $B = \frac{S_{1}^{\prime} + S_{2}^{\prime} + \cdots + S_{100}^{\prime}}{100}$.
|
$\begin{array}{l}\text { Four, (1) } \because S_{k}^{\prime}=b_{1}+b_{2}+\cdots+b_{k} \\ ==?+\left(1+a_{1}\right)+\left(1+a_{2}\right)+\cdots+\left(1+a_{k-1}\right) \\ =k+\left(a_{0}+a_{2}+\cdots+a_{k-1}\right) \\ =k+S_{k-1} \text {. } \\ \therefore S^{\prime}{ }_{4}-S_{2-1}=k \text {. } \\ \text { (2) } \because A=\frac{S_{1}+S_{2}+\cdots+S_{99}}{99}=50 \text {, } \\ \therefore S_{1}+S_{2}+\cdots+S_{99}=50 \times 99 \text {. } \\ B=\frac{S_{2}{ }^{\prime}+S_{3}{ }^{\prime}+\cdots+S_{100}{ }^{\prime}}{100} \\ =\frac{1+\left(2+S_{1}\right)+\left(3+S_{2}\right)+\cdots+\left(100+S_{99}\right)}{100} \\ =\frac{100(100+1)}{2}+50 \times 99 \\ =\frac{101 \times 50+50 \times 99}{100} \\ =100 . \\\end{array}$
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initially 35. 4 distinct prime numbers, none of whose last digit is 1, have a sum of squares equal to \(4pq\) (where \(p, q\) are primes, \(p \neq q\)). \(A=2^n pq\) (where \(n\) is an integer greater than 6) is a 6-digit number. The sum of the two 3-digit numbers formed by the first 3 digits and the last 3 digits of \(A\) equals \(10pq\), and the 3-digit number formed by the first 3 digits of \(A\) is of the form \(2r\) (where \(r\) is a prime). Find the product of these 4 prime numbers.
|
Let
$$
2^{\prime \prime} p q=10^{5} a+10^{4} b+10^{3} c+10^{2} d+10 e+f,
$$
where $a, b, \cdots, f$ are all integers, $1 \leqslant a \leqslant 9,0 \leqslant b, c, \cdots, f \leqslant$
9. Then (1) can be written as
$$
\begin{aligned}
2^{2} p q= & 199(1002+10 s+c)+100(a+d) \\
& +10(b+e)+c+f \\
= & 3^{3} \cdot 37 \cdot 2 r+10 p q .
\end{aligned}
$$
That is, $\left(2^{x-1}-5\right) p q=3^{3} \cdot 37 r$.
Given that $r>50$. Suppose $p6$ leads to a contradiction.
Therefore, (2) and (3) are both impossible.
Let $w, x, y, z$ be prime numbers, satisfying
$$
w^{2}+x^{2}+y^{2}+z^{2}=444 .
$$
Assume $w444, \therefore z444$, thus, only one of these 4 primes can be $\geqslant 13$. Therefore,
$$
\begin{array}{l}
w=3, x=5, y=7, \\
z^{2}=444-\left(3^{2}+5^{2}+7^{2}\right)=361, z=19 .
\end{array}
$$
Thus, the product of these 4 primes is $3 \times 5 \times 7 \times 19=1995$.
Note: It is easy to prove that when $p=3, q=37$, the $A, r, n$ that satisfy the conditions in the problem exist, i.e., $A=454656, r=227, n=12$.
|
1995
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Second Pump Find all real numbers $p$ such that the cubic equation
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p
$$
has three roots that are all natural numbers.
|
The original problem is equivalent to
$$
5 x^{2}-5 p x+66 p-1=0
$$
having both roots as natural numbers.
Method 1: Let $u, v$ be the roots of equation (2), then
$$
\left\{\begin{array}{l}
u+v=p, \\
u v=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$
Eliminating the parameter $p$ yields
$$
\begin{array}{l}
5 u v=66 u+66 v-1, \\
u=\frac{66 v-1}{5 v-66}=\frac{66}{5}+\frac{4351}{5(5 v-66)}, \\
5 u=66+\frac{19 \times 229}{5 v-66} .
\end{array}
$$
Clearly, $5 v-66>0$, and $5 v-66$ is a positive divisor of $19 \times 229$ that is congruent to 4 modulo 5, i.e.,
$5 v-66=19$ or 229.
$$
\left\{\begin{array} { l }
{ u = 5 9 ; } \\
{ v = 1 7 ; }
\end{array} \text { or } \left\{\begin{array}{l}
u=17, \\
v=59 .
\end{array}\right.\right.
$$
Therefore, $p=u+v=76$.
Method 2: Clearly, $p$ should be a natural number, and the discriminant
$$
\Delta=25 p^{2}-20(66 p-1)=(5 p-132)^{2}-4351 \times 4
$$
is a perfect square. Let $A=5 p-132, \Delta=B^{2}(B \geqslant 0)$, then
$$
A^{2}-B^{2}=4 \times 19 \times 229 \text {. }
$$
$A, B$ have the same parity, and they can only both be even (if $A, B$ are both odd, then $8\left|\left(A^{2}-1\right), 8\right|\left(B^{2}-1\right)$; i.e., $8 \mid\left(A^{2}-B^{2}\right)$, which is a contradiction).
$$
\begin{array}{l}
\text { Let } A=2 A_{1}, B=2 B_{1} \text {, then } \\
A_{1}^{2}-B_{1}^{2}=19 \times 229 .
\end{array}
$$
From $A_{1}-B_{1} \leqslant A_{1}+B_{1}$, we get
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ A _ { 1 } - B _ { 1 } = - 4 3 5 1 , } \\
{ A _ { 1 } + B _ { 1 } = - 1 ; }
\end{array} \left\{\begin{array}{l}
A_{1}-B_{1}=-229, \\
A_{1}+B_{1}=-19 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ A _ { 1 } - B _ { 1 } = 1 , } \\
{ A _ { 1 } + B _ { 1 } = 4 3 5 1 ; }
\end{array} \left\{\begin{array}{l}
A_{1}-B_{1}=19, \\
A_{1}+\overline{8}=229 .
\end{array}\right.\right.
\end{array}
$$
Thus, $5 p-132==2, A_{1}=--432,-21 \hat{2}, 4352,240$.
The positive integer $\phi$ can only be 75.
|
76
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $x+\frac{1}{x}=3$, then $x^{4}+3 x^{3}-16 x^{2}+3 x-1$ ? The value equals $\qquad$ .
|
5. From $x^{2}+1=3 x$, we get $x^{4}=7 x^{2}-1, x^{3}=3 x^{2}-x$. Therefore, the original expression $=7 x^{2}-1+9 x^{2}-3 x-16 x^{2}+3 x-17$ $=-18$.
Another solution: From the given $x^{2}+1=3 x, x^{2}-3 x=-1$, so,
$$
\begin{array}{l}
x^{4}+3 x^{3}-16 x^{2}+3 x-17=\left(x^{4}+2 x^{2}+1\right)+3 x^{3}+3 x \\
-18 x^{2}-18=\left(x^{2}+1\right)^{2}+3 x\left(x^{2}+1\right)-18\left(x^{2}+1\right) \\
=(3 x)^{2}+(3 x)^{2}-18 \times 3 x=18\left(x^{2}-3 x\right)=-18 .
\end{array}
$$
|
-18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $x_{1}, x_{2}, \cdots, x_{19}$ be integers, and satisfy $x_{1}+x_{2}+\cdots$ $+x_{19}=95$. Then the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{19}^{2}$ is $\qquad$.
|
6. Let's assume $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant \cdots \leqslant x_{19}$. Note that when $1<x_{i}$ $\leqslant x_{j}$, $x_{i}^{2}+x_{j}^{2}<x_{i}^{2}+1-2 x_{i}+2 x_{j}+1+x_{j}^{2}=\left(x_{i}-\right.$ $1)^{2}+(x,+1 \cdot 1)^{2}$. Therefore, when $x_{1}=x_{2}=\cdots=x_{18}=1, x_{10}=$ 77, $x^{2} \cdot x_{2}^{2}+\cdots+x_{19}^{2}$ reaches its maximum value, which is $18+77^{2}=$ 5947.
|
5947
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (This question is worth 20 points)
In the border desert area, patrol vehicles travel 200 kilometers every day. Each vehicle departs from base $A$ at the same time, completes the mission, and then returns along the original route to the base (returning together). After vehicles A and B reach point $B$ on the way, they only keep enough gasoline for their own return to the base, and leave the extra gasoline for the other three vehicles to use. How far can the other three vehicles travel at most in kilometers?
|
When the car reaches point $B$ on the way, it takes $x$ days, and from $B$ to the farthest point, it takes $y$ days, then we have
$$
2[3(x+y)+2 x]=14 \times 5,
$$
which simplifies to $5 x+3 y=35$.
According to the problem, we need $x>0, y>0$, and
$$
14 \times 5-(5+2) x \leqslant 14 \times 3 \text {, }
$$
which simplifies to $x \geqslant 4$.
Thus, the essence of the problem is to find the maximum value of $y$ under the constraints
$$
\left\{\begin{array}{l}
5 x+3 y=35, \\
x \geqslant 4, \\
y>0
\end{array}\right.
$$
Clearly, $y=5$.
Therefore, $200 \times(4+5)=1800$ (km) is the farthest distance the other three cars can travel.
|
1800
|
Other
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (11 points) There are four numbers. Each time, select three of them, calculate their average, and then add the other number. Using this method, four calculations were made, resulting in the following four numbers: 86, 92, 100, 106. (1) Find the four numbers, (2) Find the average of these four numbers.
|
Three, let these four numbers be $a, b, c, d$. We can get
$$
\left\{\begin{array}{l}
\frac{a+b+c}{3}+d=86, \\
\frac{b+c+d}{3}+a=92, \\
\frac{c+d+a}{3}+b=100, \\
\frac{d+a+b}{3}+c=106 .
\end{array}\right.
$$
Solving it, we get $a=42, b=54, c=63, d=33$.
$$
\frac{1}{4}(a+b+c+d)=48 \text{. }
$$
Answer: These four numbers are $42,54,63,33$; the average is 48.
|
48
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11. There are 100 points on a plane, where the distance between any two points is no less than 3. Now, connect every two points whose distance is exactly 3 with a line segment. Prove: such line segments will not exceed 300.
(1984, Beijing Junior High School Mathematics Competition)
|
Given that the central angle subtended by an arc between two points is less than $60^{\circ}$, it means the distance between these two points is less than 3, which contradicts the problem's condition. Therefore, with $P$ as the endpoint, the maximum number of line segments that can be drawn to $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$, each with a distance of 3, is 6.
Since the 100 points on the plane are equally positioned, any point can be fixed. Considering that the two endpoints of any line segment are not ordered, removing the duplicates, we have $\frac{1}{2}(6 \times 100) = 300$. Therefore, the number of such line segments will not exceed 300.
Analyzing this problem, due to the large number of points involved, considering all at once can make it seem like a dead end. We might as well start with a simpler problem to find a solution path.
Consider a plane with three points $P, A_{1}, A_{2}$, where the distance between any two points is not less than 3. If we connect every two points that are exactly 3 units apart, there will be no more than 3 line segments.
Moving to a plane with four points $P, A_{1}, A_{2}, A_{3}$, with $P$ fixed, and the distance between any two points not less than 3, if we connect every two points that are exactly 3 units apart, there will be no more than 3 line segments. With $P$ as the center and a radius of 3 units, $A_{1}, A_{2}, A_{3}$ will not be inside the circle $\odot P$. Suppose $A_{1}, A_{2}, A_{3}$ are on $\odot P$ and are equidistant. We know that the distance from any point on $\odot P$ to $P$ is 3. However, whether there exists a point $A_{4}$ that is exactly 3 units away from $A_{1}, A_{2}, A_{3}$, and only one such point exists for $A_{1} A_{2}$. Similarly, for $A_{2}, A_{3}$, and $A_{3}, A_{2}$, there are also exactly one point $A_{5}, A_{6}$ (as shown in Figure 7). Thus, we find that on the plane, relative to a fixed point, there are at most 6 points that are 3 units away from this point.
Proof: As shown in Figure 7, suppose there are 7 points $P, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ on the plane, with $P$ as the fixed point. With $P$ as the center and a radius of 3 units, the circle will have at most 6 points that are 3 units away from $P$. This is because if there were 7 points, then
(Note: The last sentence is incomplete in the original text, so it is left as is in the translation.)
|
300
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Find all natural numbers with the following property: its unit digit is 6, and when its unit digit is moved to the front of the number, the resulting new number is four times the original number.
|
Let the required number be $a_{n} a_{n-1} \cdots \cdots a_{2} a_{1} 6$. According to the problem, we have $\overline{6 a_{n} a_{n-1} \cdots a_{2} a_{1}}=4 \overline{a_{n} a_{n-1} \cdots a_{2} a_{1} 6}$, where $0 \leqslant a_{i} \leqslant 9, a_{i} \in N$.
That is 6
$$
\begin{array}{l}
6 \cdot 10^{n}+a_{n} \cdot 10^{n-1}+\cdots+a_{2} \cdot 10+a_{1} \\
=4\left(a_{n} \cdot 10^{n}+a_{n-1} \cdot 10^{n-1}+\cdots+a_{2} \cdot 10^{2}\right. \\
\left.\quad+a_{1} \cdot 10+6\right) .
\end{array}
$$
Thus, $\left(6-4 a_{n}\right) \cdot 10^{n}+\left(a_{n}-4 a_{n}-1\right) \cdot 10^{n-1}+\cdots+$ $\left(a_{2}-4 a_{1}-2\right) \cdot 10+a_{1}-4=0$, and from $6-4 a_{n} > 0$ we know $a_{n}<\frac{3}{2}$. Therefore, $a_{n}=1$.
Hence, the required number must be 153846, 153846153846, $153846153846153846, \cdots$.
|
153846
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Let $S$ be a subset of the set of numbers $\{1,2,3, \cdots, 1989\}$, and the difference between any two numbers in $S$ is not equal to 4 or 7. How many numbers can $S$ contain at most?
|
Analysis Notice that the sum of 4 and 7 is 11. We first observe the first 11 natural numbers $1,2, \cdots, 11$, and find that $1,4,6,7,9$ satisfy the conditions of the problem. Then we observe the 11 numbers $12,13, \cdots, 22$, and find that $1+11,4+11,6+11,7+11,9+11$ also satisfy the conditions of the problem. And $1,4,6,7,9,12,15,17,18,20$ still satisfy the conditions of the problem. Thus, we have found a periodic number group with a period of 11. This makes the solution approach very clear.
Solution The difference between any two numbers in the array $(1,4,6,7,9)$ is not equal to 4 or 7, and the arrays $(11+1,11+4,11+6,11+7, 11+9), \cdots,(11 k+1,11 k+4,11 k+6,11 k+7,11 k+9), k \in N$, and $11 k+9 \leqslant 1989$, also have this property. On the other hand, the difference between any two numbers in the union of all these arrays is also not equal to 4 or 7. We form the set $S$ by combining all the numbers in these arrays. Noting that $1989=11 \times 180+9$, the number of numbers in $S$ is $5 \times 180+5=905$.
We then prove that $S$ cannot contain more numbers. If not, then among the arrays $(1,2, \cdots, 11),(12,13, \cdots, 22), \cdots,(1970,1971, \cdots, 1980),(1981,1982, \cdots, 1989)$, at least one array can be selected to contain six numbers such that the difference between any two numbers is not 4 or 7. Without loss of generality, consider the array $1,2, \cdots, 11$, and divide them into five groups $(4,7,11),(3,10),(2,6),(5,9),(1,8)$. At least one group must have two numbers selected. Clearly, none of the last four pairs can be selected simultaneously, so only the first group can select 4,7. Thus, in $(3,10)$, only 10 can be selected; in $(2,6)$, only 2 can be selected; in $(5,9)$, only 5 can be selected; and in $(1,8)$, neither number can be selected. In other words, it is impossible to select a sixth number, so $S$ can contain at most 905 numbers.
|
905
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the arithmetic sequence (1): $5,8,11, \cdots$, and the arithmetic sequence (2): $1,5,9, \cdots$, both having 1996 terms, then there are $\qquad$ numbers that appear in both sequences.
|
(Answer: 498)
|
498
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. What is the largest even integer that cannot be written as the sum of two odd composite numbers? (If a positive integer can be divided by a positive integer other than 1 and itself, then this positive integer is called a composite number).
(2nd American Invitational Mathematics Examination)
|
Solution: Because the last digit of a positive even number is $0, 2, 4, 6, 8$.
When the last digit is 0, there are even integers such as $0,10,20,30,40, \cdots$, among which $30=15+15$.
When the last digit is 2, there are even integers such as $2,12,22,32,42,52, \cdots$, among which $42=15+27,52=25+27, \cdots$. Therefore, even integers above 42 do not meet the criteria.
When the last digit is 4, there are even integers such as $4,14,24,34, \cdots$, among which $24=9+15,34=9+25, \cdots$. Therefore, even integers above 24 do not meet the criteria.
When the last digit is 6, there are even integers such as $6,16,26,36,46, \cdots$, among which $36=15+21,46=15+21$, therefore, even integers above 36 do not meet the criteria.
When the last digit is 8, there are even integers such as $8,18,28,38,48,58, \cdots$, among which $48=21+27,58=33+25, \cdots$. Therefore, even integers above 48 do not meet the criteria.
In summary, the largest even integer that cannot be written as the sum of two odd composite numbers can only be 38.
|
38
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, in a game played on a $1 \times 1000$ board, initially there are $n$ chips placed. The game is played by two players taking turns. The first player places a chip in any unoccupied cell, and the second player removes any number of chips occupying consecutive cells. If the first player can place all $n$ chips on the board, occupying consecutive cells, he wins.
1. If $n=98$, prove that he can always win.
2. What is the maximum value of $n$ for which he can always win?
|
Six, let the two players be $A$ and $B$.
1. On the board, arrange 96 chips in the pattern of 8 chips, 1 empty space, 8 chips, 1 empty space, $\cdots$. This is a winning position for $A$. Clearly, $A$ can achieve this winning position. After $B$ moves a certain number of consecutive chips, $A$ can return them to their original positions and continue to work towards the winning position. No matter how $B$ moves the consecutive chips, $A$ can always return them to their original positions. Then, with the remaining two groups of 17 chips, there are 8 groups in the middle, each with 8 chips. If $B$ moves a group of 8 chips, $A$ can return the moved group to its original position and then take 9 chips from one side to fill the 9 empty spaces, and $A$ wins. If $B$ moves a group of 17 chips, $A$ can fill the 8 empty spaces with the 17 chips and then place them consecutively, and $A$ wins.
2. Assuming $A$ can win, on the eve of $A$'s victory, without loss of generality, let $n$ chips be on the board, divided into several consecutive groups, with at least one empty space between each group. The two groups on the sides each have no more than 17 chips, otherwise $A$ would not win in the next step. Suppose there are $k$ groups in the middle, then there are $k+1$ empty spaces in the middle, and each group in the middle has no more than $17-k-1$ chips, otherwise $A$ would not win in the next step. In summary,
$$
n \leqslant 2 \times 17 + k(17-k-1) = 34 + k(16-k) \leqslant 98.
$$
The maximum value of $n$ for which $A$ can guarantee a win is 98.
|
98
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
II. Write the number $1234567802011 \cdots 19941995$ on the blackboard, forming the integer $N_{1}$. Erase the digits of $N_{1}$ that are in even positions, leaving the remaining digits to form the integer $N_{2}$. Remove the digits of $N_{2}$ that are in odd positions, leaving the remaining digits to form the integer $N_{3}$. Erase the digits of $N_{3}$ that are in even positions, leaving the remaining digits to form the integer $N_{4}$. This process continues until only one digit remains on the blackboard. Determine this digit. (Note: Count positions from left to right, for example, in 12345, 1 is in the first position, 2 is in the second position, and so on).
|
$$
=、 9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873 \text {. }
$$
The integer $N_{1}$ is composed of 6873 digits, which are sequentially numbered as $1,2,3, \cdots, 6873$. We still examine the set of numbers $\{1,2, \cdots, 6873\}$. Removing the digits at even positions from $N_{1}$ is equivalent to removing $\{2,4, \cdots, 6872\}$ from the set $\{1,2, \cdots, 6873\}$.
Similarly, from the remaining set $\{1,3,5, \cdots, 6873\}$, we remove $\{1,5,9, \cdots, 6873\}$, leaving $\{3,7,11, \cdots, 6871\}$. We can sequentially obtain the sets $\{3,3+8=11,11+8=19, \cdots\}$, $\{11,11+16=27, \cdots\}$, $\{11,11+32=43, \cdots\}$, $\{43,43+64=107, \cdots\}$, $\{43,43+128=371, \cdots\}$, $\{171,171+256=427, \cdots\}$, $\{171,171+512=683, \cdots\}$, $\{683,683+1024=1707, \cdots\}$, $\{683,683+2048=2731\}$, $\{2731,2731+4096=6827\}$. After the thirteenth step, only $\{2731\}$ remains.
The required digit is the 2731st digit from left to right in $N_{1}$.
$$
9 \times 1+90 \times 2<2731<9 \times 1+90 \times 2+900 \times 3 \text {, }
$$
This digit should be part of a three-digit number. The one-digit and two-digit numbers together occupy 189 positions, 2731 - 189 = 2542 = 3 \times 847 + 1, the 848th three-digit number is 947, and the digit to the left of this number is 9. The final remaining digit is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$, and $x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+$ $c\left(\frac{1}{a}+\frac{1}{b}\right)$. Then the algebraic expression $x^{18}-96 x y+y^{3}=$
|
$$
=\sqrt{ } .-316
$$
Given $-b c>0$, and $a+b+c=0$, we know that among $a, b, c$, there must be two negative numbers. Without loss of generality, assume $a>0$. Hence $x=-1$.
$$
\begin{array}{l}
\text { ini } \begin{array}{l}
y=\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b} \\
=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \\
=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-3 . \\
\therefore x^{2 v}-96 x y+y^{3}
\end{array} \\
\therefore=(-1)^{20}-96 \times(-1)(-3)+(-3)^{3} \\
\end{array}
$$
|
-316
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the parabola $y=x^{2}+k x+4-k$ intersects the $x$-axis at integer points $A, B$, and intersects the $y$-axis at point $C$. Then $S_{\triangle A B C}=$ $\qquad$ .
|
3.24.
Given the equation $x^{2}+k x+4-k=0$ has two distinct integer roots $x_{1}, x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-k, \\
x_{1} x_{2}=4-k .
\end{array}\right.
$$
(2) - (1) gives $x_{1}+x_{2}-x_{1} x_{2}+4=0$,
$$
\left(x_{1}-1\right)\left(x_{2}-1\right)=5 \text {. }
$$
Thus, $\left\{\begin{array}{l}x_{1}-1=1, \\ x_{2}-1=5\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}-1=5, \\ x_{2}-1=1\end{array}\right.$
or $\left\{\begin{array}{l}x_{1}-1=-1, \\ x_{2}-1=-5\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}-1=-5, \\ x_{2}-1=-1 .\end{array}\right.$
which means $x_{1}+x_{2}=8$ or $x_{1}+x_{2}=-4$.
Therefore, $k_{1}=-8$ or $k_{2}=4$.
When $k=-8$, from $x^{2}-8 x+12=0$, we get $x_{1}=2$, $x_{2}=6$. Thus, $A B=4$, and the coordinates of point $C$ are $(0,12)$.
$$
\therefore S_{\triangle A B C}=\frac{1}{2} \times 4 \times 12=24 \text {. }
$$
When $k=4$, from $x^{2}+4 x=0$, we get $x_{1}=0, x_{2}=-4$, then $A B=4$, and $C(0,0)$. In this case, points $A, B, C$ cannot form a triangle. Hence, $S_{\triangle A B C}=24$.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that the inverse function of $y=f(x)$ is $\varphi(x)$, and $\varphi(x)$ . $=\log _{m a_{0}{ }^{2}}\left(\frac{1996}{x}-\sin ^{2} 0\right), 0 \in\left(0, \frac{\pi}{2}\right)$. Then the solution to the equation $f(x)=$ 1996 is
|
$=, 1 .-1$.
Since $f(x)$ and $\varphi(x)$ are inverse functions of each other, solving $f(x) = 1996$ is equivalent to finding the value of $\varphi(1996)$.
$$
\varphi(1996)=\log _{\sec ^{2} \theta}\left(1-\sin ^{2} \theta\right)=-1 .
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10. As shown in Figure 6-1, it is a part of a city's street map, with five roads running both longitudinally and latitudinally. If one walks from point $A$ to point $B$ (only from north to south, and from west to east), how many different ways are there to do so?
|
Solution: Because the number of different ways from the top-left vertex of each small square to the endpoint equals the sum of the number of different ways from the two adjacent vertices to the endpoint, we can deduce the data at the top-left corner of each small square in Figure 6-4, which represents the number of different ways, and thus conclude that there are 70 different ways from $A$ to $B$.
Analysis: Let's first simplify the complex grid of routes to Figure 6-2. We can see that there are $1+1$ $=2$ ways to go from $E$. Then, expanding Figure 6-2 to Figure 6-3, we find that there are $3+3$ $=6$ ways to go from point $H$. Through observation and analysis, we can derive the following rule: the number of different ways from the top-left vertex of each small square to the endpoint equals the sum of the number of different ways from the two adjacent vertices to the endpoint.
|
70
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
When the expression changes, the minimum value of the fraction $\frac{3 x^{2}+6 x-5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$. (1993, National Junior High School Competition)
|
Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$.
(*)
Rearranging (*), we get
$$
(y-6) x^{2}+(2 y-12) x+2 y-10=0 \text {. }
$$
Since $x$ is a real number, we have
$$
\Delta=(2 y-12)^{2}-4(y-6)(2 y-10) \geqslant 0 \text {, }
$$
which simplifies to $y^{2}-10 y+24 \leqslant 0$.
Solving this, we get $4 \leqslant y \leqslant 6$.
Substituting $y=4$ into (*) gives $x=-1$.
Thus, when $x=-1$, $y_{\text {min }}=4$.
Note: Generally, if a rational function $y=f(x)$ can be transformed into $p(y) x^{2}+q(y) x+r(y)=0$ after clearing the denominator, where $p, q, r$ are real functions of $y$, then the range of $y$ can be determined by $\Delta \geqslant 0$, and thus the extremum of $y$ can be found.
There is also a simpler solution for this problem:
$$
\begin{array}{l}
\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}=\frac{\left(6 x^{2}+12 x+12\right)-2}{x^{2}+2 x+2} \\
=6-\frac{2}{(x+1)^{2}+1} .
\end{array}
$$
Clearly, when $x=-1$, the original expression has a minimum value of 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ - (1995, National Junior High School Competition)
|
- $y=x^{2}-x+\frac{1}{x}$ $=(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1$.
From the above equation, it is known that when $x-1=0$ and $\sqrt{x}-\frac{1}{\sqrt{x}}$ $=0$, i.e., $x=1$, $y$ takes the minimum value of 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12. $a, b, c$ are all integers, and $a b c=1990$. Find the minimum value of $a b+b c+c a$. (1990, Jinyun Cup Junior Competition)
|
$$
\begin{array}{l}
\text { Sol } \because 1990=1 \times 1 \times 1990 \\
=(-1) \times(-1) \times 1990 \\
= 1 \times(-1) \times(-1990) \\
= 1 \times 2 \times 995 \\
=(-1) \times(-2) \times 995 \\
= \cdots=2 \times 5 \times 199=1 \times 10 \times 199,
\end{array}
$$
When $a=-1, b=-1, c=1990$ (letters can be swapped), $ab+bc+ca=-1990-1990+1=-3979$ is the minimum.
Therefore, the minimum value of $ab+bc+ca$ is -3979.
|
-3979
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13. Let $a, b, c$ be distinct integers from 1 to 9. What is the largest possible value of $\frac{a+b+c}{a b c}$? (1992, 1st Dannevirke-Shanghai Friendship Correspondence Competition)
|
Let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a, b$ remain unchanged temporarily, and only let $c$ vary, where $c$ can take any integer from 1 to 9. Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that when $c=1$, $P$ reaches its maximum value. Therefore, $c=1$.
Thus, $P=\frac{a+b+1}{a b}$.
In equation (2), let $a$ remain unchanged temporarily, and only let $b$ vary, where $b$ can take any integer from 2 to 9. Then, from $P=\frac{a+b+1}{a b}=\frac{1}{a}+\frac{a+1}{a b}$, we know that when $b=2$, $P$ reaches its maximum value. Therefore, $b=2$.
Thus, $P=\frac{1}{2}+\frac{3}{2 a}$.
In equation (3), $P$ reaches its maximum value when $a$ takes its minimum value, and $a$ can take any integer from 3 to 9, i.e., $a=3$.
Therefore, when $a=3, b=2, c=1$ (the letters can be interchanged), $P$ reaches its maximum value of 1.
Regarding the application of function extremum problems, there is another important aspect, which is the application of function extremum. As long as we master the basic methods of finding the extremum of a function, such problems are not difficult to solve.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. There are 120 equally distributed spheres inside a regular tetrahedron $A-B C D$. How many spheres are placed at the bottom of this regular tetrahedron?
|
Analysis (1) The base of the regular tetrahedron $A-BCD$ is the equilateral $\triangle BCD$. Assuming the number of spheres closest to the edge $BC$ is $n$, then the number of rows of spheres tangent to the base $\triangle BCD$ is also $n$, with the number of spheres in each row being $n, n-1, \cdots, 3,2,1$. Thus, the total number of spheres tangent to the base is
$$
1+2+\cdots+n=\frac{n(n+1)}{2} \uparrow .
$$
The number of spheres in each layer is
$$
\begin{array}{l}
1+2+\cdots+(n-1)=\frac{(n-1) n}{2} . \\
1+2+\cdots+(n-2)=\frac{(n-2)(n-1)}{2} . \\
\cdots \cdots \\
1+2=\frac{2 \times 3}{2} . \\
1=\frac{1 \times 2}{2} . \\
\therefore \frac{n(n+1)}{2}+\frac{(n-1) n}{2}+\cdots+\frac{1 \times 2}{2}=120 .
\end{array}
$$
That is, $\frac{1}{6} n(n+1)(n+2)=120$.
$$
\begin{array}{l}
(n-8)\left(n^{2}+11 n+90\right)=0 . \\
\therefore n=8 .
\end{array}
$$
Therefore, a total of 8 small spheres are placed inside the regular tetrahedron, and the number of spheres at the bottom is $\frac{8 \times 9}{2}=36$.
（3）The problem of densely packing small spheres inside a regular tetrahedron: From top to bottom, the 1st layer has 1 sphere, the 2nd layer has 3 spheres, the 3rd layer has 6 spheres, $\cdots$, the $n$-th layer has $\frac{1}{2} n(n+1)$ spheres. The number of spheres placed in each layer is equal to half the product of the layer number and the next layer number.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If 20 points divide a circle into 20 equal parts, then the number of regular polygons that can be formed with vertices only among these 20 points is ( ) .
(A) 4
(B) 8
(C) 12
(D) 24
|
6. (C).
Let the regular $k$-sided polygon satisfy the condition, then the 20 $-k$ points other than the $k$ vertices are evenly distributed on the minor arcs opposite to the sides of the regular $k$-sided polygon.
Thus, $\frac{20-k}{k}=\frac{20}{k}-1$ is an integer, so $k \mid 20$.
But $k \geqslant 3$,
$\therefore k=4$ or 5 or 10 or 20.
$\therefore$ The number of regular polygons sought is
$$
\frac{20}{4}+\frac{20}{5}+\frac{20}{10}+\frac{20}{20}=12(\uparrow)
$$
|
12
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. When $|x+1| \leqslant 6$, the maximum value of the function $y=x|x|$ $-2 x+1$ is $\qquad$. (1994, National Junior High School Competition)
|
Solve: From $|x+1| \leqslant 6$ we get $-7 \leqslant x \leqslant 5$.
When $0 \leqslant x \leqslant 5$,
$$
y=x^{2}-2 x+1=(x-1)^{2} \text {. }
$$
Since $x=1 \in[0,5]$, and $1>\frac{-7+0}{2}, \\
\therefore y_{\text {max }}=f(5)=(5-1)^{2}=16; \\
y_{\text {min }}=f(1)=(1-1)^{2}=0 .
$$
When $-7 \leqslant x < 0$,
$$
y=x^{2}+2 x+1=(x+1)^{2} \text {. }
$$
Since $x=-1 \in[-7,0]$, and $-1>\frac{-7+0}{2}, \\
\therefore y_{\text {min }}=f(-1)=(-1+1)^{2}=0; \\
y_{\text {max }}=f(-7)=(-7+1)^{2}=36 .
$$
Comprehensive analysis shows that when $x=5$, $y$ reaches $16$.
(2) The maximum and minimum values of a quadratic function on a moving interval
Since $x=-\frac{b}{2 a}$ is fixed, such problems can be discussed in cases based on whether the moving interval $t \leqslant x \leqslant t^{\prime}$ includes $x=-\frac{b}{2 a}$.
$1^{\circ}$ The moving interval includes $x=-\frac{b}{2 a}$;
$2^{\circ}$ The moving interval does not include $x=-\frac{b}{2 a}$, and is to the left of $x=-\frac{b}{2 a}$;
$3^{\circ}$ The moving interval does not include $x=-\frac{b}{2 a}$, and is to the right of $x=-\frac{b}{2 a}$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given that $a, b, c$ are positive integers, and the parabola $y=$ $a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A, B$. If the distances from $A, B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Three, let the coordinates of $A, B$ be $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$ and $x_{1}<0, x_{2}>0$,
\[
\begin{array}{l}
\therefore x_{1} x_{2}=\frac{c}{a}<0, \\
\therefore a c<0, \text { i.e., } a>0, c<0 .
\end{array}
\]
\[
\begin{array}{l}
\because x_{1}+x_{2}=-\frac{b}{a}>0, \\
\therefore b<0 .
\end{array}
\]
\[
\begin{array}{l}
\because \text { the parabola opens upwards, and } x_{1} x_{2}<0, \\
\therefore \text { the parabola intersects the } x \text{-axis at two points, } \\
\therefore \Delta=b^{2}-4 a c>0, \\
\because x_{1} x_{2}<0, \text { get } b>2 \sqrt{a c} .
\end{array}
\]
\[
\begin{array}{l}
\because|O A|=|x_{1}|,|O B|=|x_{2}|, \\
\therefore|O A| \cdot|O B|=|x_{1} x_{2}|=\left|\frac{c}{a}\right| \geqslant 1, \\
\therefore|c| \geqslant|a|, \text { and } a>0, c<0, \\
\therefore c \leqslant -a .
\end{array}
\]
\[
\begin{array}{l}
\because \text { the parabola passes through the point } (-1,1), \\
\therefore a(-1)^{2}+b(-1)+c>0 . \\
\text { get } b<2 \sqrt{a c}+1 \Rightarrow(\sqrt{a}-\sqrt{c})^{2}>1 .
\end{array}
\]
\[
\begin{array}{l}
\text { From (2) we get } \sqrt{a}-\sqrt{c}>1 . \\
\therefore \sqrt{a}>\sqrt{c}+1 .
\end{array}
\]
\[
\begin{array}{l}
\text { i.e., } a>(\sqrt{c}+1)^{2} \geqslant(\sqrt{1}+1)^{2}=4, \\
\therefore a \geqslant 5 .
\end{array}
\]
\[
\begin{array}{l}
\text { Also, } b>2 \sqrt{a c} \geqslant 2 \sqrt{5 \times 1}>4, \\
\therefore b \geqslant 5 .
\end{array}
\]
\[
\begin{array}{l}
\text { Taking } a=5, b=5, c=1, \text { the parabola } y=5 x^{2}+5 x+1 \text { satisfies the given conditions. } \\
\text { Hence, the minimum value of } a+b+c \text { is } 5+5+1=11 .
\end{array}
\]
(Provided by Liu Yuxuan)
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the product of four real numbers is 1, and the sum of any one of the numbers with the product of the other three is 1000. Then the sum of these four numbers is
|
$=、 1.2000$.
These four real numbers are all non-zero and not all equal, denoted as $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$. Taking any one of these numbers $x_{i} (i=1,2,3,4)$, the other three numbers are $x_{j}$, $x_{k}$, $x_{r}$, then we have
$$
\left\{\begin{array}{l}
x_{i}+x_{j} x_{k} x_{r}=1000, \\
x_{i} x_{j} x_{k} x_{r}=1 .
\end{array}\right.
$$
Since $x_{i}$ is any one of the four numbers $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$,
Therefore, $x_{i}$, $x_{j}$, $x_{k}$, $x_{r}$ are all roots of the equation $x^{2}-1000 x+1=0$.
Since the four numbers are not all equal, without loss of generality, let $x_{i} > x_{j}$, we have
$$
\begin{array}{r}
x_{i}=\frac{1000+\sqrt{1000^{2}-4}}{2}, \\
x_{j}=\frac{1000-\sqrt{1000^{2}-4}}{2} .
\end{array}
$$
Similarly, $x_{4}=\frac{1000+\sqrt{1000^{2}-4}}{2}$,
$$
x_{r}=\frac{1000-\sqrt{1000^{2}-4}}{2} \text {. }
$$
Therefore, the sum of the four real numbers is 2000.
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) As shown in the figure, given that $AB, CD$ are perpendicular chords in a circle $\odot O$ with radius 5, intersecting at point $P$. $E$ is the midpoint of $AB$, $PD=AB$, and $OE=3$. Try to find the value of $CP + CE$.
---
The translation is provided as requested, maintaining the original text's format and line breaks.
|
By the intersecting chords theorem, we have
$$
C P \cdot P D=A P \cdot P B \text {. }
$$
Also, $A E=E B=\frac{1}{2} A B$,
$$
\begin{aligned}
\because C P \cdot P D & =\left(\frac{1}{2} A B+E P\right)\left(\frac{1}{2} A B-E P\right) \\
& =\frac{1}{4} A B^{2}-E P^{2} .
\end{aligned}
$$
And $P D=A B$,
$$
\therefore \quad C P=\frac{1}{4} A B-\frac{E P^{2}}{A B} \text {. }
$$
In the right triangle $\triangle C E P$,
$$
\begin{array}{l}
C E^{2}=C P^{2}+E P^{2}=\left(\frac{1}{4} A B-\frac{E P^{2}}{A B}\right)^{2}+E P^{2} \\
=\left(\frac{1}{4} A B+\frac{E P^{2}}{A B}\right)^{2}, \\
\therefore \quad C E=\frac{1}{4} A B+\frac{E P^{2}}{A B} .
\end{array}
$$
Thus, $C P+C E=\frac{1}{2} A B=A E=\sqrt{O A^{2}-O E^{2}}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8. If the polynomial $P=2a^{2}-8ab+17b^{2}$ $-16a-4b+1990$, what is the minimum value of $P$? (1990, Junior High School Correspondence Contest in Some Provinces and Cities)
|
$$
\begin{array}{r}
\text { Sol } P=2 a^{2}-8 a b+17 b^{2}-16 a-4 b \\
+1990=2(a-2 b-4)^{2}+9(b-2)^{2}+1922 .
\end{array}
$$
From this, for all real numbers $a, b, P \geqslant 1922$. The equality holds only when $a-2 b-4=0, b-2=0$, i.e., $a=$ $8, b=2$. Therefore, the minimum value of $P$ is $1922$.
|
1922
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The Dao sequence $\left\{\begin{array}{l}x_{1}=x_{2}=1, \\ x_{n+2}=a x_{n+1}+b x_{n}(n \in N) .\end{array}\right.$
If $T=1996$ is the smallest natural number such that $x_{T+1}=x_{T+2}=1$, then $\sum_{i=1}^{1006} x_{i}=$ $\qquad$ .
|
2. 0 .
From $T=1996$ being the smallest natural number that makes $x_{1+1}=x_{T+2}=1$, we know that $a+b \neq 1$. Otherwise, if $a+b=1$, then $x_{3}=a x_{2}+b x_{1}=a+b=1$. This would mean that when $T=1$, $x_{T+1}=x_{T+2}=1$, which contradicts the fact that $T=1996$ is the smallest value. Furthermore, we have
$$
\begin{array}{l}
x_{1}=x_{1997}=a x_{1006}+b x_{1905}, \\
x_{2}=x_{1998}=a x_{1997}+b x_{1996}=u x_{1}+b x_{1995}, \\
x_{3}=a x_{2}+b x_{1}, \\
x_{4}=a x_{3}+b x_{2}, \\
\cdots \cdots \\
x_{1096}=a x_{1995}+b x_{1994} .
\end{array}
$$
Summing up, we get $\quad \sum_{i=1}^{1004} x_{i}=a \sum_{i=1}^{1006} x_{i}+b \sum_{i=1}^{1096} x_{i}$,
which simplifies to $(a+b-1) \sum_{i=1}^{1098} x_{i}=0$.
Since $a+b-1 \neq 0$, it follows that $\sum^{1096} x_{i}=0$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the plane $\alpha$ there is a $\triangle A B C, \angle A B C=105^{\circ}$, $A C=2(\sqrt{6}+\sqrt{2})$. On both sides of the plane $\alpha$, there are points $S, T$, satisfying $S A=S B=S C=\sqrt{41}, T A=T B=T C=$ 5. Then $S T=$ $\qquad$.
|
3. 8 .
From Figure 4, since $S A=S B=S C$, we know that the projection $D$ of $S$ on plane $\alpha$ is the circumcenter of $\triangle A B C$. Similarly, the projection of $T$ on plane $\alpha$ is also the circumcenter of $\triangle A B C$. By the uniqueness of the circumcenter, we have $S T \perp a$.
Connect $A D$. By the Law of Sines, we have
T
$$
\begin{array}{l}
A D=\frac{A C}{\sin \angle A B \bar{C}} \\
=\frac{3(\sqrt{6}+\sqrt{2})}{2\left(\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}\right)}=4 .
\end{array}
$$
Thus,
$$
\begin{aligned}
S T & =S D+D T \\
& =\sqrt{S A^{2}-A D^{2}}+\sqrt{T A^{2}-A D^{2}} \\
& =\sqrt{41-16}+\sqrt{25-16}=5+3=8
\end{aligned}
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given functions $f(x)$ and $g(x)$ are defined on $R$, and $f(x-y)=f(x) g(y)-g(x) f(y)$. If $f(1)=f(2) \neq$ 0, then $g(1)+g(-1)=$ $\qquad$ .
|
4. 1 .
For $x \in R$, let $x=u-v$, then we have
$$
\begin{array}{l}
f(-x)=f(v-u) \\
=f(v) g(u)-g(v) f(u) \\
=-[f(u) g(v)-g(u) f(v)] \\
=-f(u-v)=-f(x) .
\end{array}
$$
Therefore, $f(x)$ is an odd function, and we have
$$
\begin{array}{l}
f(2)=f[1-(-1)] \\
=f(1) g(-1)-g(1) f(-1) \\
=f(1) g(-1)+g(1) f(1) .
\end{array}
$$
But $f(1)=f(2) \neq 0$, so $g(1)+g(-1)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ have side lengths $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, respectively, and their areas be $S$ and $S^{\prime}$. If for all values of $x$, $a x^{2}+b x+c=3\left(a^{\prime} x^{2}+b^{\prime} x+c^{\prime}\right)$ always holds, then $\frac{S}{S^{\prime}}$ $=$
|
II. 1. From the given, $\left(a-3 a^{\prime}\right) x^{2}+\left(b-3 b^{\prime}\right) x+(c$ $\left.-3 c^{\prime}\right)=0$ is an identity, hence $\left(a-3 a^{\prime}\right)=0,\left(b-3 b^{\prime}\right)=0$, $\left(c-3 c^{\prime}\right)=0$, which gives $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}=3$,
$\therefore \triangle A B C \sim \triangle A^{\prime} B^{\prime} C^{\prime}$, then $\frac{S}{S^{\prime}}=\left(\frac{a}{a^{\prime}}\right)^{2}=9$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $a_{1}, a_{2}, \cdots, a_{k}$ be $k$ distinct positive integers, and $a_{1}+a_{\varepsilon}+\cdots+a_{k}=1995$. Then, the maximum value of $k$ is $\qquad$
|
2. $\because a_{1}, a_{2}, \cdots, a_{4}$ are all distinct, without loss of generality, let $a_{1}<a_{2}<\cdots<a_{k}$. Since $a_{1}, a_{2}, \cdots, a_{k}$ are all positive integers, we have $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{k} \geqslant k$, thus $1+2+\cdots+k \leqslant a_{1}+a_{2}+\cdots+a_{k}=$ 1995, which means $\frac{k(k+1)}{2} \leqslant 1995$, solving this yields $1 \leqslant k \leqslant 62$. On the other hand, when $a_{1}=1, a_{2}=2, \cdots, a_{61}=61, a_{62}=104$, such $a_{1}$, $a_{2}, \cdots, a_{4}$ indeed satisfy the conditions of the problem, therefore the maximum value of $k$ is 62.
|
62
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If $\sqrt{a^{2}-1996}$ is an integer, then the minimum value of the integer $a$ is
|
5. Consider the case of natural number $a$: Let $\sqrt{a^{2}-1996} = m$, then $a^{2}-m^{2}=1996$, $(a+m)(a-m)=1996$. Since $a+m$ and $a-m$ are both even numbers, and $1996=2 \times 998$ is the only way to split it into the product of two even numbers, we have $a+m=998, a-m=2$, solving for $a=500, m=498$. Since the largest natural number satisfying $\sqrt{a^{2}-1996}$ being an integer is 500, the smallest integer satisfying $\sqrt{a^{2}-1996}$ being an integer is -500.
|
-500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 14 points) If the sum of the digits of a natural number and the product of its digits add up to exactly the natural number itself, we call it a "lucky number." Try to find the sum of all "lucky numbers."
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let the $n$-digit natural number be $N=\overline{a_{1} a_{2} \cdots a_{n}}$, then
\[ N=a_{1} \times 10^{n-1}+a_{2} \times 10^{n-2}+\cdots+a_{n-1} \times 10+a_{n} \]
(Where $a_{1}, a_{2}, \cdots, a_{n}$ are the digits of $N$).
If $N$ is a "lucky number", then
\[
\begin{array}{l}
a_{1} \times 10^{n-1}+a_{2} \times 10^{n-2}+\cdots+a_{n-1} \times 10+a_{n} \\
=\left(a_{1}+a_{2}+\cdots+a_{n}\right)+a_{1} \times a_{2} \times \cdots \times a_{n},
\end{array}
\]
Rearranging gives
\[
\begin{array}{l}
a_{1}\left(10^{n-1}-1\right)+a_{2}\left(10^{n-2}-1\right)+\cdots+a_{n-1}(10-1) \\
=a_{1} \times a_{2} \times \cdots \times a_{n},
\end{array}
\]
That is,
\[ a_{1}(\underbrace{99 \cdots 9}_{n-14}-a_{2} \times a_{3} \times \cdots \times a_{n})+A=0, \]
where $A=a_{2}\left(10^{n-1}-1\right)+\cdots+a_{n-1}(10-1)$ is a non-negative number.
When $n \geqslant 3$, $\underbrace{99 \cdots 9}_{n-14}>9^{n-1} \geqslant a_{2} \times a_{3} \times \cdots \times a_{n}$, hence (1) does not hold, so there are no "lucky numbers" with 3 or more digits;
When $n=2$, from (1) we get $a_{1}\left(9-a_{2}\right)=0$,
Since $a_{1} \neq 0, \therefore 9-a_{2}=0$, thus $a_{2}=9$.
Therefore, two-digit numbers ending in 9 are "lucky numbers" (it can be verified that no single-digit numbers are "lucky numbers"), they are $19,29,39,49$, $59,69,79,89,99$ and their sum is 531.
|
531
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of diagonals in a convex nonagon is $\qquad$.
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 blank space represented by $\qquad$ is kept as is in the translation.
|
3. 27. (Hint: Each vertex leads to $(9-3)=6$ diagonals, there are 9 vertices in total, and each diagonal passes through two vertices, so the total number of diagonals is $6 \times 9 \div 2=27$.)
|
27
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that the six interior angles of a regular hexagon are all $120^{\circ}$, and the lengths of four consecutive sides are $1, 9, 9, 5$ cm respectively. Then, the perimeter of this hexagon is $\qquad$ cm.
|
7. 42. (Hint: Extend the three pairs of opposite sides of the hexagon, the result is an equilateral triangle, it is easy to know that the side length of the equilateral triangle is 19 cm, thus find the other two side lengths of the hexagon to be 13 cm and 5 cm.)
|
42
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In the interior angles of an $n(n \geqslant 3)$-sided polygon, the maximum number of acute angles is $\qquad$ .
|
II, 1.3.
Let a polygon with $n$ sides have $k$ acute angles, then
$$
(n-2) \times 180^{\circ}<k \times
$$
$90^{\circ}+(n-k) \times 180^{\circ}$.
Thus, $k<4$.
As shown in the figure, construct a $60^{\circ}$ sector $A_{1} A_{2} A_{1}$. When $n=3$, $\triangle A_{1} A_{2} A_{3}$ has three interior angles, all of which are part of $A_{1} A_{2} \cdots A_{n-1} A_{n}$, where $\angle A_{1} A_{2} A_{3}$, $\angle A_{n-1} A_{n} A_{1}$, and $\angle A_{n} A_{1} A_{2}$ are all acute angles, hence the maximum value of $k$ is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. There are two roads $O M, O N$ intersecting at a $30^{\circ}$ angle. Along the direction of road $O M$, 80 meters from $A$ is a primary school. When a tractor travels along the direction of $O N$, areas within 50 meters on both sides of the road will be affected by noise. Given that the speed of the tractor is 18 kilometers/hour. Then, the time during which the tractor traveling along $O N$ will cause noise impact to the primary school is $\qquad$ seconds.
|
3. 12 .
As shown in the figure, when the tractor reaches point $B$, it starts to have an impact, and when it reaches point $C$, the impact ceases. Thus, $A B = A C = 50$ meters. Draw $A D \perp$ $O N$ from $A$. Given that $\angle A O D = 30^{\circ}$, we know that $A D = \frac{1}{2} O A = 40$ meters. Therefore, $B D = \sqrt{A B^{2} - A D^{2}} = 30$ meters, and thus $B C = 60$ meters. Since the speed of the tractor is 18 kilometers/hour $= 5$ meters/second, the duration of the noise impact is $t = \frac{60}{5} = 12$ seconds.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let real numbers $x, y$ satisfy the equation $9 x^{2}+4 y^{2}-3 x+2 y=0$. Then the maximum value of $z=3 x+2 y$ is $\qquad$ .
|
From $9 x^{2}+4 y^{2}=3 x-2 y$ we know $3 x-2 y \geqslant 0$ and $\frac{9 x^{2}+4 y^{2}}{3 x-2 y}$ $=1$,
we have $z=3 x+2 y=\frac{(3 x)^{2}-(2 y)^{2}}{3 x-2 y} \leqslant \frac{(3 x)^{2}+(2 y)^{2}}{3 x-2 y}$ $=1$ (equality holds when $y=0$).
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 17. Given as shown, in quadrilateral $ABCD$, $AD=DC=1, \angle DAB=$ $\angle DCB=90^{\circ}, BC, AD$ extended intersect at $P$. Find the minimum value of $AB \cdot S_{\triangle PAB}$.
(1994, Sichuan Province Junior High School Mathematics League Competition)
|
Let $DP = x$, then $PC = \sqrt{x^2 - 1}$.
Since $\triangle PCD \sim \triangle PAB$,
$\therefore CD \perp AB = PC : PA$.
$\therefore AB = \frac{CD \cdot PA}{PC} = \frac{x + 1}{\sqrt{x^2 - 1}}$.
Let $y = AB \cdot S_{\triangle PAB}$, then
$$
y = \frac{1}{2} AB^2 \cdot PA = \frac{(x + 1)^3}{2(x^2 - 1)}.
$$
Eliminating the denominator and simplifying, we get
$$
x^2 - 2(1 - y)x + 1 + 2y = 0.
$$
Since $x$ is a real number, this equation has real roots, so
$$
\Delta = 4(1 - y)^2 - 4(1 + 2y) = 4y(y - 4) \geqslant 0.
$$
Since $y > 0$, solving gives $y \geqslant 4$,
i.e., the minimum value of $AB \cdot S_{\triangle PAB}$ is 4. E.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Write $(\sqrt{7}-\sqrt{6})^{6}$ in the form $\sqrt{N+1}-\sqrt{N}$, where $N$ is a natural number. Then $N=$ $\qquad$
|
5. $N=76545000$.
Let $(\sqrt{7}-\sqrt{6})^{6}=\sqrt{N+1}-\sqrt{N}$, then $(\sqrt{7}-\sqrt{6})^{6}+\sqrt{N}=\sqrt{N+1}$.
Square it to get
$$
\begin{array}{l}
(\sqrt{7}-\sqrt{6})^{12}+2 \sqrt{N}(\sqrt{7}-\sqrt{6})^{6}+N \\
= N+1 \\
2 \sqrt{N}=\frac{1-(\sqrt{7}-\sqrt{6})^{12}}{(\sqrt{7}-\sqrt{6})^{6}} \\
=\frac{(7-6)^{6}-(\sqrt{7}-\sqrt{6})^{12}}{(\sqrt{7}-\sqrt{6})^{6}} \\
=(\sqrt{7}+\sqrt{6})^{6}-(\sqrt{7}-\sqrt{6})^{6} \\
= 2\left[C_{6}^{1}(\sqrt{7})^{5} \cdot \sqrt{6}+C_{6}^{3}(\sqrt{7})^{3}(\sqrt{6})^{3}\right. \\
\left.+C_{6}^{5} \cdot \sqrt{7} \cdot(\sqrt{6})^{5}\right] \\
= 2 \sqrt{7} \cdot \sqrt{6} \cdot 6 \cdot 225,
\end{array}
$$
Thus, $N=7 \cdot 2^{3} \cdot 3^{7} \cdot 5^{4}=76545 \times 10^{3}$.
|
76545000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
*6. From $1,2, \cdots, 1996$, select $k$ numbers such that the sum of any two numbers cannot be divisible by their difference. Then the maximum possible value of $k$ is $\qquad$ .
|
6. 666 .
Classify $1,2, \cdots, 1996$ by their remainder when divided by 3:
$$
\begin{array}{l}
3 k+1: 1,4,7, \cdots, 1996, \text { a total of 666; } \\
3 k+2: 2,5,8, \cdots, 1994, \text { a total of 665; } \\
3 k: 3,6,9, \cdots, 1995, \text { a total of 665. }
\end{array}
$$
Select all $3 k+1$ numbers, a total of 666, where the sum of any two is $3 k+2$, which cannot be divided by the difference $3 k$, meeting the problem's requirements.
If 667 numbers can be selected, consider $(1,2,3),(4,5,6), \cdots$, $(1993,1994,1995),(1996)$ as 666 drawers. There must be two numbers in the same drawer. Their difference $x-b$ is less than 3.
If $a-b=1$, then $a-b \mid a+b$;
If $a-b=2$, then $a$ and $b$ have the same parity.
Thus, $a-b \mid a+b$, so the maximum number that can be selected is 666.
|
666
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For any natural number $n$, connect the origin $O$ and the point $A_{n}(n, n+3)$. Let $f(n)$ denote the number of integer points on the line segment $O A_{n}$, excluding the endpoints. Try to find: $f(1)+f(2)+\cdots+f(1996)$.
|
The equation of the line segment $O A_{n}$ is $y=\frac{n+3}{n} x(0 \leqslant x \leqslant n)$. Therefore, $f(x)$ is the number of integer solutions of this equation in the interval $(0, n)$. When $n=3 k$ (where $k$ is a positive integer), the equation becomes
$$
y=\frac{k+1}{k} x(0<x<3 k) \text {. }
$$
It has two sets of integer solutions
$$
\left\{\begin{array} { l }
{ x = k , } \\
{ y = k + 1 }
\end{array} \text { and } \left\{\begin{array}{l}
x=2 k, \\
y=2(k+1) .
\end{array}\right.\right.
$$
When $n=3 k \pm 1$ (where $k$ is a non-negative integer), it is easy to see that $n$ and $n+3$ are coprime, so the equation has no integer solutions satisfying $0<x<n$.
From the above discussion, we get
$$
f(n)=\left\{\begin{array}{ll}
0 & \text { when } n=3 k \pm 1, \\
2 & \text { when } n=3 k.
\end{array}\right.
$$
Therefore,
$$
\begin{array}{l}
f(1)+f(2)+\cdots+f(1996) \\
=2 \times\left[\frac{1996}{3}\right]=2 \times 665=1330 .
\end{array}
$$
(Shi Li Lu, Henan Provincial Traffic School, 450052)
|
1330
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In $\triangle A B C$, $A B=37, A C=58$. With $A$ as the center and $A B$ as the radius, an arc is drawn intersecting $B C$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $B D$ and $D C$ are both integers, find the length of $B C$.
|
Solve as shown in the figure, first prove
$$
\begin{array}{l}
B C \cdot C D \\
=A C^{2}-A B^{2} .
\end{array}
$$
Draw $A H \perp B C$ at $H$, then
$$
\begin{array}{l}
A C^{2}=A H^{2}+C H^{2}, \\
A B^{2}=A H^{2}+B H^{2},
\end{array}
$$
we have
$$
\begin{array}{l}
A C^{2}-A B^{2}=C H^{2}-B H^{2} \\
\quad=(C H+B H)(C H-B H)=B C \cdot C D .
\end{array}
$$
Also, $\because A B=37, A C=58$.
$$
\therefore B C \cdot C D=58^{2}-37^{2}=3 \times 5 \times 7 \times 19 \text {. }
$$
Since $A C - A B < B C, D$ is not between $B, C$, so it should be discarded, hence we should take $B C=57$, at this time $C D$ $=35, B D=24$.
|
57
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. If $a=\sqrt{17}-1$, find the value of $\left(a^{5}+2 a^{4}-17 a^{3}\right.$ $\left.-a^{2}+18 a-17\right)^{1993}$.
(Adapted from the 1987 Chongqing Junior High School Mathematics Invitational Competition)
|
Given $a+1=\sqrt{17}$, squaring both sides yields $a^{2}+2 a+1=17$.
Then
$$
\begin{array}{l}
\left(a^{5}+2 a^{4}-17 a^{3}-a^{2}+18 a-17\right)^{1993} \\
=\left[a^{5}+2 a^{4}-\left(a^{2}+2 a+1\right) a^{3}-a^{2}+\left(a^{2}\right.\right. \\
\left.+2 a+1+1) a-\left(a^{2}+2 a+1\right)\right]^{1993} \\
=(-1)^{1993}=-1 \text {. } \\
\end{array}
$$
Note: Here, we first constructed the expression $a^{2}+2 a+1$ with a value of 17, and then substituted $a^{2}+2 a+1$ for 17 in the expression to be evaluated.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. If $m^{2}=m+1, n^{2}=n+1$, then what is the value of $m^{5}+n^{5}$?
(1989, Jiangsu Province High School Mathematics Competition)
|
Let $S_{k}=m^{k}+n^{k}$, construct the recurrence relation for $S_{k}$.
$$
\begin{array}{l}
\because m^{2}=m+1, n^{2}=n+1, \\
\therefore m^{k}=m^{k-1}+m^{k-2}, n^{k}=n^{k-1}+n^{k-2} .
\end{array}
$$
Adding the two equations gives $S_{k}=S_{k-1}+S_{k-2}$.
$$
\begin{aligned}
\therefore S_{5} & =S_{4}+S_{3}=\left(S_{3}+S_{2}\right)+\left(S_{2}+S_{1}\right) \\
& =\cdots=3 S_{2}+2 S_{1} .
\end{aligned}
$$
From the given conditions, $S_{2}=m+n-2$, and $m, n$ are the roots of the quadratic equation $x^{2}-x-1=0$, so $m+n=1$, i.e., $S_{1}=1, S_{2}=1+2=3$.
Thus, $S_{5}=3 \times 3+2 \times 1=11$.
Note: Here, we first construct the recurrence relation for $S_{k}=m^{2}+n^{k}$, then express $S_{5}$ in terms of $S_{1}$ and $S_{2}$, and solve the problem. This is a general method for solving problems of the form $a^{n}+b^{n}$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Let real numbers $a, b, x$ and $y$ satisfy $a x + b y = 3, a x^2 + b y^2 = 7, a x^3 + b y^3 = 16, a x^4 + b y^4 = 42$. Find the value of $a x^5 + b y^5$.
(Eighth American Invitational Mathematics Examination)
|
Let $S_{\mathrm{n}}=a x^{n}+b y^{\star}, A=x+y, B=x y$, then $x, y$ are the two roots of the equation $u^{2}-A u+B=0$. Therefore,
$$
\begin{array}{l}
x^{2}-A x+B=0, y^{2}-A y+B=0 . \\
\therefore a x^{n}-A\left(a x^{n-1}\right)+B\left(a x^{n-2}\right)=0, \\
b y^{n}-A\left(b y^{n-1}\right)+B b y^{n-2}=0 .
\end{array}
$$
Adding the two equations, we get
$$
\begin{array}{l}
S_{n}=A S_{n-1}-B S_{n-2} . \\
\therefore S_{5}=A S_{4}-B S_{3}=42 A-16 B .
\end{array}
$$
Also, since $S_{4}=A S_{3}-B S_{2}, S_{3}=A S_{2}-B S_{1}$,
$\therefore$ from $S_{1}=3, S_{2}=7, S_{3}=16, S_{4}=42$, we have
$$
42=16 A-7 B, 16=7 A-3 B \text {, }
$$
Solving these, we get $A=-14, B=-38$.
Thus, $S_{5}=42 \times(-14)-16 \times(-38)=20$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. Find the integer part of $(\sqrt{3}+1)^{6}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
$$
\begin{array}{l}
\text{Solve for the dual number } B = (\sqrt{3}-1)^{6} \text{ of the number } A = (\sqrt{3}+1)^{6}, \text{ and examine} \\
A+B=(\sqrt{3}+1)^{6}+(\sqrt{3}-1)^{6} \\
= \left[(\sqrt{3}+1)^{3}\right]^{2}+\left[(\sqrt{3}-1)^{3}\right]^{2} \\
= \left[(\sqrt{3}+1)^{3}-(\sqrt{3}-1)^{3}\right] \\
+2[(\sqrt{3}+1)(\sqrt{3}-1)]^{3} \\
=\{[(\sqrt{3}+1)-(\sqrt{3}-1)][(\sqrt{3} \\
+1)^{2}+(\sqrt{3}+1)(\sqrt{3}-1) \\
\left.\left. +(\sqrt{3}-1)^{2}\right]\right\}+16 \\
= 2^{2}(4+2 \sqrt{3}+2+4-2 \sqrt{3})^{2}+16 \\
= 416 . \\
\because 0<\sqrt{3}-1<1, \\
\therefore 0<(\sqrt{3}-1)^{6}<1,
\end{array}
$$
Thus, the integer part of $(\sqrt{3}+1)^{6}$ is 415. Note, the dual form of $(a+b)^{n}$ is $(a-b)^{n}$.
|
415
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10. Given $a^{2}+2 a-5=0, b^{2}+2 b-5=0$, and $a \neq b$. Then the value of $a b^{2}+a^{2} b$ is what?
(1989, Sichuan Province Junior High School Mathematics Competition)
|
Given that $a, b$ are the roots of the quadratic equation $x^{2}+$ $2 x-5=0$, then $a+b=-2, ab=-5$. Therefore, $ab^{2}+a^{2}b=ab(a+b)=10$. Note: Here, the definition of the roots is used to construct the equation.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a+b+c=0, a^{3}+b^{3}+c^{3}=0$. Find the value of $a^{15}+b^{15}+c^{15}$.
|
2. From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right.$ $-a b-b c-a c)$, we have $a b c=0$. Then at least one of $a, b, c$ is zero, for example, $c=0$, then it is known that $a, b$ are opposites, i.e., $a^{15}+b^{15}+c^{15}$ $=0$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Find a natural number $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square.
(2nd All-Russian High School Mathematics Olympiad)
|
Let $2^{4}=x$, then $2^{8}=x^{2}, 2^{11}=2^{7} \cdot x$. Therefore, $2^{8}+2^{11}+2^{n}=x^{2}+2^{7} x+2^{n}$.
According to the condition for a quadratic trinomial to be a perfect square, we get $\Delta=\left(2^{7}\right)^{2}-4 \times 2^{n}=0$.
Solving for $n$ yields $n=12$.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $\frac{x}{m}+\frac{y}{n}+\frac{z}{p}=1, \frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$. Calculate the value of $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}$.
|
6. From $\frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$, we have $\frac{m y z+n x z+p x y}{x y z}=0$.
Since $x, y, z$ are not zero, then $m y z+n x z+p x y=0$. Also,
$$
\begin{aligned}
& \frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}} \\
= & \left(\frac{x}{m}+\frac{y}{n}+\frac{z}{p}\right)^{2}-2\left(\frac{x y}{m n}+\frac{x z}{m p}+\frac{y z}{n p}\right) \\
= & 1-2 \cdot \frac{1}{m n p}(p x y+n x z+m y z)=1 .
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Find the minimum value of $|x-1|+|x-2|+|x-3|+\cdots$ $+|x-1996|$.
|
8. When $998 \leqslant x \leqslant 999$, the original expression has a minimum value, the minimum value is:
$$
\begin{array}{l}
(1996-1)+(1995-2)+\cdots+(999-998) \\
=1995+1993+\cdots+1 \\
=\frac{(1+1995) \times 998}{2}=998^{2}=996004 .
\end{array}
$$
|
996004
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that $x, y, z$ are three non-negative rational numbers, and satisfy $3 x$ $+2 y+z=5, x+y-z=2$. If $S=2 x+y-z$, then what is the sum of the maximum and minimum values of $S$?
|
9. $\left\{\begin{array}{l}3 x+y+z=5, \\ x+y-z=2, \\ 2 x+y-z=S .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}x=S-2, \\ y=\frac{15-4 S}{3}, \\ z=\frac{3-S}{3} .\end{array}\right.$ Since $x, y, z$ are all non-negative, we have
$$
\left\{\begin{array} { l }
{ S - 2 \geqslant 0 , } \\
{ \frac { 1 5 - 4 S } { 3 } \geqslant 0 , } \\
{ \frac { 3 - S } { 3 } \geqslant 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
S \geqslant 2, \\
S \leqslant \frac{15}{4}, \\
S \leqslant 3 .
\end{array}\right.\right.
$$
Therefore, $2 \leqslant S \leqslant 3$. The sum of the maximum and minimum values of $S$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Among the students in the second year of junior high school, 32 students participated in the math competition, 27 students participated in the English competition, and 22 students participated in the Chinese competition. Among them, 12 students participated in both math and English, 14 students participated in both English and Chinese, and 10 students participated in both math and Chinese. It is known that $\frac{5}{8}$ of the students who participated in the competitions won awards. How many students won awards?
|
10. Let the number of people who participate in all three subjects be $x$, and the total number of participants be $y$, then
$$
\begin{aligned}
y & =32+27+22-12-14-10+x \\
& =45+x .
\end{aligned}
$$
Given $2 \leqslant x \leqslant 10$, hence $47 \leqslant y \leqslant 55$.
Only when $y=48$, it can be divisible by 8, so the number of award winners is $48 \times \frac{5}{8}=30$ (people).
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (16 points) A plot of land can be covered by $n$ identical square tiles. If smaller identical square tiles are used, then $n+76$ such tiles are needed to cover the plot. It is known that $n$ and the side lengths of the tiles are integers. Find $n$.
|
Let the side length of the larger tile be $x$, and the side length of the smaller tile be $r$. Then $x, y$ are both natural numbers, and
$$
n x^{2}=(n+76) y^{2} \text {. }
$$
If $(x, y)=d$, let $x_{1}=\frac{x}{d}, y_{1}=\frac{y}{d}$, then $x_{1}, y_{1}$ are both natural numbers, and $\left(x_{1}, y_{1}\right)=1$. Thus, the previous equation can be transformed into
$$
n x_{1}^{2}=(n+76) y_{1}^{2} \text {. }
$$
We have $\frac{x_{1}^{2}}{y_{1}^{2}}=\frac{n+76}{n}, \frac{x_{1}^{2}-y_{1}^{2}}{y_{1}^{2}}=\frac{76}{n}$.
Since $\left(x_{1}^{2}-y_{1}^{2}, y_{1}^{2}\right)=1$, $\left(x_{1}-y_{1}\right)\left(x_{1}+y_{1}\right)$ is a divisor of 76 $=$ $2^{2} \times 19$. Noting that $x_{1}+y_{1}, x_{1}-y_{1}$ have the same parity, and $x_{1}+y_{1}>x_{1}-y_{1}$, we have
$$
\left\{\begin{array} { l }
{ x _ { 1 } + y _ { 1 } = 1 9 , } \\
{ x _ { 1 } - y _ { 2 } = 1 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}+y_{1}=2 \times 19, \\
x_{1}-y_{1}=2 .
\end{array}\right.\right.
$$
Since $\left(x_{1}, y_{1}\right)=1$, we can only have $\left\{\begin{array}{l}x_{1}=10 \\ y_{1}=9 .\end{array}\right.$
Thus, $\frac{76}{n}=\frac{19}{81}, n=324$.
|
324
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four points are such that no three are collinear, and no two are coincident. Through each point, draw the perpendiculars to the lines joining the other three points. If we do not count the original four points, how many different intersection points are there among these perpendiculars? Prove your conclusion.
|
For each point, draw perpendicular lines to the lines connecting it to the other three points. Since there are 3 such lines for each point, and there are 4 points, there are a total of $4 \times 3 = 12$ perpendicular lines.
If every two perpendicular lines intersect, there would be $\frac{12 \times 11}{2} = 66$ intersection points.
Among the four points, any two points can be connected by a line, and the other two points can draw perpendicular lines to this line. The two perpendicular lines are parallel and do not intersect, and there are 6 such lines, so 6 points need to be deducted.
Among the four points, any three points can form a triangle, and the three perpendicular lines drawn from the three vertices to the opposite sides intersect at one point. These points were counted three times in the previous calculation. There are four triangles formed by any three points among the four points. Therefore, a total of $4 \times 2 = 8$ points need to be deducted.
Each point should draw three perpendicular lines to the three lines connecting it to the other three points, which should result in 3 intersection points, but these 3 points coincide with the original points, so $4 \times 3 = 12$ points need to be deducted.
As stated above, the number of intersection points should be $66 - (6 + 8 + 12) = 40$.
|
40
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. In $\triangle A B C$, $G$ is the centroid, and $I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$. If $I G / / B C$, and $B C=5$, then $A B+A C$ $=$ . $\qquad$
|
3. Connect $A G, B G$ and $C G$, and extend $A G$ to intersect $B C$ at $M$, then $S_{\triangle G B C}: S_{\triangle A B C}=G M: A M=1: 3$.
Connect $A I$, then by $I G$ $/ / B C$, we know
$$
\begin{array}{l}
S_{\triangle I B C}=S_{\triangle G B C}=\frac{S_{\triangle A B C}}{3}, \\
\therefore S_{\triangle I A B}+S_{\triangle I A C}=2 S_{\triangle I B C} .
\end{array}
$$
$\because I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$ in $\triangle A B C$,
$\therefore I$ is equidistant from the three sides of $\triangle A B C$.
Thus, $A B+A C=2 B C=10$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $x_{1}, x_{2}, \cdots, x_{51}$ be natural numbers, $x_{1}<x_{2}$ $<\cdots<x_{51}$, and $x_{1}+x_{2}+\cdots+x_{51}=1995$. When $x_{26}$ reaches its maximum value, the maximum value that $x_{51}$ can take is
|
5. To maximize $x_{26}$, then $x_{1}, x_{2}, \cdots, x_{25}$ must be minimized, thus we have
$$
\begin{array}{l}
x_{26}+x_{27}+\cdots+x_{51} \\
=1995-(1+2+\cdots+25)=1670 .
\end{array}
$$
To maximize $x_{51}$, then $x_{26}, x_{27}, \cdots, x_{50}$ must be minimized, thus we have
$$
\begin{array}{l}
x_{26}+\left(x_{26}+1\right)-1-\cdots+\left(x_{26}+24\right)+x_{01} \\
=1670\left(x_{11}>x_{23}+\% 4\right),
\end{array}
$$
i.e. $\square$
$$
\begin{array}{l}
25 x_{26}+-x_{51}=1670-(1+2+\cdots+24) \\
=1370=25 \times 51+95 .
\end{array}
$$
Therefore, the maximum value that $x_{51}$ can achieve is 95.
|
95
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. There is a four-digit number. Dividing it
in the middle into two parts, we get the front and back two $\uparrow=$ digit numbers. By adding a 0 to the end of the front two-digit number, and then adding the product of the front and back two-digit numbers, we get exactly the original four-digit number. It is also known that the unit digit of the original number is 5. Therefore, this four-digit number is $ـ$. $\qquad$
|
3. 1995
|
1995
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points, Question (1) 8 points, Question (2) 12 points) As shown in the figure, there is a cube-shaped wire frame, and the midpoints of its sides $I, J, K, L$ are also connected with wire.
(1) There is an ant that wants to crawl along the wire from point A to point G. How many shortest routes are there? Represent these routes using letters (use the letters of the connection points passed through, for example, if the ant starts from point A, passes through point I, point L, and finally reaches point H, this route is represented as $A I L H$).
(2) Is it possible for the ant to start from point A, pass through each connection point exactly once along the wire, and finally reach point G? If possible, find one such route; if not, explain why.
|
(1) "The shortest path" means that the ant can neither move left nor down, otherwise it would be taking a "detour" rather than the "shortest" path. The ant can crawl on face $ABCGFE$ or on $ADCGHE$. The number of "shortest paths" on these two faces is the same, and the possible paths can be represented by the following diagram:
Therefore, there are a total of 12 shortest paths, which are:
$A B C K G, A B J K G, A B J F G, A I E F G, A I J K G$,
$A I J F G, A I E H G, A I L H G, A I L K G, A D C K G$,
$A D L K G, A D L H G$.
(2) Using the application of integer properties - coloring method or labeling method
Solution. The answer sought in the problem is impossible.
As shown in the figure, all 12 connection points are colored with black and white, and each adjacent pair of points is colored with different colors. The ant starts from the black point $A$, and must follow a route pattern like: black $\rightarrow$ white $\rightarrow$ black $\rightarrow$ white $\rightarrow$ $\cdots$, and by the 12th point $G$ it should be $\rightarrow$ white $\rightarrow$ black $(G)$, but according to the route pattern, the 12th point should be: $\rightarrow$ black $\rightarrow$ white. Therefore, it is impossible for the ant to start from point $A$, pass through each connection point exactly once, and finally reach point $G$.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. Given $x, y, z \in R^{+}$, and $x+y+z=1$. Prove: $\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant 36$.
(1990, Japan IMO Team Selection Test)
|
Prove that for $x=\sin ^{2} \alpha \cos ^{2} \beta, y=\cos ^{2} \alpha \cos ^{2} \beta$, $z=\sin ^{2} \beta$, where $\alpha, \beta$ are acute angles, we have
$$
\begin{array}{l}
\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \\
=\left(1+\operatorname{ctg}^{2} \alpha\right)\left(1+\operatorname{tg}^{2} \beta\right)+4\left(1+\operatorname{tg}^{2} \alpha\right) \\
\text { - }\left(1+\operatorname{tg}^{2} \beta\right)+9\left(1+\operatorname{ctg}^{2} \beta\right) \\
=14+\left(4 \operatorname{tg}^{2} \alpha+\operatorname{ctg}^{2} \alpha\right)+\left(4 \operatorname{tg}^{2} \alpha+\operatorname{ctg}^{2} \alpha\right. \\
+5) \operatorname{tg}^{2} \beta+9 \operatorname{ctg}^{2} \beta \\
\geqslant 14+4+9 \operatorname{tg}^{2} \beta+9 \operatorname{ctg}^{2} \beta \\
\geqslant 18+18=36 \text {. } \\
\end{array}
$$
|
36
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers $1,2, \cdots, n$. Let $f(n)$ be the number of such permutations such that
(i) $a_{1}=1$;
(ii) $\left|a_{i}-a_{i+1}\right| \leqslant 2 . i=1,2, \cdots, n-1$.
Determine whether $f(1996)$ is divisible by 3.
|
3. Verify that $f(1)=f(2)=1$ and $f(3)=2$.
Let $n \geqslant 4$. Then it must be that $a_{1}=1, a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$, because by deleting the first term and reducing all subsequent terms by 1, we can establish a one-to-one correspondence of sequences.
If $a_{2}=3$, then $a_{3}=2$. In this case, $a_{4}=4$, and the number of such permutations is $f(n-3)$.
If $a_{3} \neq 2$, then 2 must come after 4. This implies that all odd numbers are in ascending order and all even numbers are in descending order. Therefore, $f(n)=f(n-1)+f(n-3)+1$.
For $n \geqslant 1$, let $r(n)$ be the remainder when $f(n)$ is divided by 3. Then we have
$$
\begin{array}{l}
r(1)=r(2)=1, r(3)=2, \\
r(n)=r(n-1)+r(n-3)+1 .
\end{array}
$$
The sequence of remainders forms a periodic sequence with a period of 8:
$$
\{1,1,2,1,0,0,2,0\} \text {. }
$$
Since $1994=4(\bmod 8)$, and $r(4)=1$,
$f(1996)$ is not divisible by 3.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $360^{x}=3, 360^{y}=5$, then $72^{\frac{1-2 x-y}{3(1-y)}}=$
|
\begin{array}{l}=1.2. \\ \because 72=\frac{360}{5}=\frac{360}{360^{y}}=360^{1-y}, \\ \therefore \text { original expression }=\left(360^{1-y}\right)^{\frac{1}{3(2 x-y)}} \\ =(360)^{\frac{1}{3}(1-2 x-y)}-\left(360^{1-2 x-y}\right)^{\frac{1}{3}} \\ =\left(\frac{360}{\left(360^{x}\right)^{2} \cdot 360^{5}}\right)^{\frac{1}{3}}=\sqrt[3]{\frac{360}{9 \times 5}}=2 . \\\end{array}
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $x=\frac{1}{2}-\frac{1}{4 x}$, then $1-2 x+2^{2} x^{2}-2^{3} x^{3}+2^{4} x^{4}$ $-\cdots-2^{1995} x^{1995}$ is. $\qquad$.
|
2. 1 .
From $x=\frac{1}{2}-\frac{1}{4 x}$ we can get $1-2 x+(2 x)^{2}=0$.
$$
\begin{array}{c}
\therefore \text { the original expression }=1-2 x\left[1-2 x+(2 x)^{2}\right]+(2 x)^{4}[1-2 x \\
\left.+(2 x)^{2}\right]-\cdots-(2 x)^{1993}\left[1-2 x+(2 x)^{2}\right]=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. From .1 to Э these nine natural numbers, select two, separately as the logarithm's number and base, to get different logarithmic values ( ).
(A) 52
(B) 53
(C) 57
(D) 72
|
2. (B).
Since the base is not 1, the logarithm and the base cannot take the same value, so only 64 different logarithmic forms can be taken. Among them, there are 8 cases where the true value is 1, all taking the value 0. In addition,
$$
\begin{array}{ll}
\log _{2} 3=\log _{4} 9 & \log _{2} 4=\log _{3} 9, \\
\log _{3} 2=\log _{9} 4 & \log _{4} 2=\log _{9} 3 .
\end{array}
$$
Therefore, the total number of different logarithmic values is $64-7-4=53$.
|
53
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $P=\{$ natural numbers no less than 3 $\}$. Define the function $f$ on $P$ as follows: if $n \in P, f(n)$ represents the smallest natural number that is not a divisor of $n$, then $f(360360)=$ $\qquad$ .
|
4. 16 .
Since $360360=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$, we know that the divisors of 360360 in ascending order are $1,2,3,4,5,6,7,8,9,10$, $11,12,13,14,15,18,20, \cdots$, the smallest natural number that is not a divisor of 360360 is 16. Therefore, $f(360360)=16$.
|
16
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$5 . n$ is a positive integer not exceeding 1996. If there is a 0 such that $(\sin \theta+i \cos \theta)^{n}=\sin n \theta+i \cos n 0$ holds, then the number of $n$ values satisfying the above condition is $\qquad$.
|
5. 498 .
$$
\begin{array}{l}
\because(\sin \theta+i \cos \theta)^{n}=[i(\cos \theta-i \sin \theta)]^{n} \\
\quad=i^{n}(\cos n \theta-\sin n \theta)=i^{n-1}(\sin n \theta+i \cos n \theta), \\
\text { and }(\sin \theta+i \cos \theta)^{n}=\sin n \theta+i \cos n \theta, \\
\therefore i^{n-1}(\sin n \theta+i \cos n \theta)=\sin n \theta+i \cos n \theta. \\
\because \sin \theta+i \cos \theta \neq 0 . \\
\therefore i^{n-1}=1, n=4k+1 .
\end{array}
$$
That is, all positive integers that leave a remainder of 1 when divided by 1996 meet the requirement, and there are $\left[\frac{1996-1}{4}\right]=498$ such numbers.
|
498
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the sequence of natural numbers starting from 1, certain numbers are colored red according to the following rules. First, color 1; then color two even numbers 2, 4; then color the three consecutive odd numbers closest to 4, which are $5, 7, 9$; then color the four consecutive even numbers closest to 9, which are $10, 12, 14, 16$; then color the five consecutive odd numbers closest to 16, which are $17, 19, 21, 23, 25$. Continue this pattern, resulting in a red subsequence $1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, \cdots$ - The 1996th number in the red subsequence, starting from 1, is $\qquad$
|
6. 3929 .
The first time, color one number red: $1,1=1^{2}$;
The second time, color 2 numbers red: $2,4,4=2^{2}$;
The third time, color 3 numbers red: $5,7,9,9=3^{2}$;
Guessing, the last number colored red in the $k$-th time is $k^{2}$. Then the $k+1$ numbers colored red in the $(k+1)$-th time are:
$$
k^{2}+1, k^{2}+3, k^{2}+5, \cdots, k^{2}+2 k-1, k^{2}+2 k+1 \text {. }
$$
The last number is $(k+1)^{2}$. According to this rule, the first $k$ segments have colored a total of
$$
\begin{array}{l}
1+2+3+\cdots+(k+1)+k=\frac{k(k+1)}{2} \text { numbers. } \\
\text { Solve } \frac{k(k+1)}{2} \leqslant 1996, \text { we get } k \leqslant 62 .
\end{array}
$$
At this point, the first 62 segments have colored a total of $\frac{62 \times 63}{2}=1953$ numbers. The last number in the 62nd segment is $62^{2}=3844$, which is the 1953rd red number. The 1954th red number is 3845. And $1996-1953=$ 43, so the 1996th red number is
|
3929
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $x=\sqrt{19-8 \sqrt{3}}$, then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$ $\qquad$
|
Solve: From $x=\sqrt{19-2 \sqrt{48}}=4-\sqrt{3}$, we get $x-4=-\sqrt{3}$. Squaring both sides and rearranging, we obtain
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore,
$$
\begin{aligned}
\text { Original expression } & =\frac{\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)+10}{\left(x^{2}-8 x+13\right)+2} . \\
& =5 .
\end{aligned}
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $a, b$ be positive integers, and $a+b \sqrt{2}$ $=(1+\sqrt{2})^{100}$. Then the units digit of $a b$ is $\qquad$
|
6.4.
By the binomial theorem, we have
$$
a-b \sqrt{2}=(1-\sqrt{2})^{100} \text {. }
$$
Therefore, $a=\frac{1}{2}\left((1+\sqrt{2})^{100}+(1-\sqrt{2})^{100}\right)$,
$$
\begin{array}{l}
b=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{100}-(1-\sqrt{2})^{100}\right) . \\
\text { Hence } \left.a b=\frac{1}{4 \sqrt{2}}(1+\sqrt{2})^{200}-(1-\sqrt{2})^{200}\right) \\
=\frac{1}{4 \sqrt{2}}\left((3+2 \sqrt{2})^{100}-(3-2 \sqrt{2})^{100}\right) . \\
\text { Let } x_{n}=\frac{1}{4 \sqrt{2}}\left((3+2 \sqrt{2})^{n}\right. \\
\left.-(3-2 \sqrt{2})^{n}\right)(n=1,2,3,4, \cdots),
\end{array}
$$
then $x_{1}=1, x_{2}=6$.
By the identity $a^{n}-b^{n}=(a+b)\left(a^{n-1}-b^{n-1}\right)-a b$ $\left(a^{n-2}-b^{n-2}\right)$, we can obtain the recurrence relation for the sequence $\left\{x_{n}\right\}$
$$
x_{n}=6 x_{n-1}-x_{n-2} \cdot(n=3,4,5, \cdots)
$$
Thus, the last digit of $x_{n}$ is as follows:
$$
\begin{array}{l}
x_{1}=1, x_{2} \equiv 6, x_{3} \equiv 5, x_{4} \equiv 4, x_{5} \equiv 9, x_{6}=0, \\
x_{7}=1, x_{8} \equiv 6, x_{9}=5, x_{10} \equiv 4(\bmod 10) .
\end{array}
$$
From this, we can see that $x_{n+2}=x_{n+8}(\bmod 10)(n=1,2,3$, $\cdots$.
Therefore, $x_{100}=x_{6 \times 16+4} \equiv x_{4}=4(\bmod 10)$, so the last digit of $a b=x_{100}$ is 4.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
46. Given that $\overline{a b}$ is a two-digit number, and $\overline{a b c d e}$ is a five-digit number. If $(\overline{a b})^{3}=\overline{a b c d e}$, find $\overline{a b c d e}$.
|
Solve $(\overline{a b})^{3}=\overline{a b c d e}=1000 \cdot \overline{a b}+\overline{c d e}$.
We have $\left[(\overline{a b})^{2}-1000\right] \cdot \overline{a b}=\overline{c d e}$.
Thus, $(\overline{a b})^{2}>1000$, which means $\overline{a b}>\sqrt{1000}>31$.
Therefore, $\bar{a} \bar{v} \geqslant 32$.
Also, $(\overline{a b})^{2}-1000=\frac{\overline{c d e}}{\overline{a b}} \leqslant \frac{999}{32}<32$,
Therefore, $(\overline{a b})^{2}<1032$, which means $\overline{a b}<\sqrt{1032}<33$.
Hence, $\overline{a b}=32,32^{3}=32768$.
Therefore, the required $\overline{a b c d e}$ is 32768.
|
32768
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13. Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0, x y z \neq 0$. Then the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$ is equal to
|
Solve the system of equations by treating $\approx$ as a constant,
$$
\left\{\begin{array}{l}
4 x-3 y=6 z, \\
x-2 y=7 z,
\end{array}\right.
$$
we get
$$
\begin{array}{l}
x=3 z, y=2 z . \\
\text { Therefore, the original expression }=\frac{2 \cdot(3 z)^{2}+3 \cdot(2 z)^{2}+6 z^{2}}{(3 z)^{2}+5 \cdot(2 z)^{2}+7 z^{2}}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 14. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq$ $n$, then $m^{5}+n^{5}=$ $\qquad$
|
$$
\begin{array}{l}
\text { Sol } \because(a-b)\left(a^{n}+b^{n-1}\right) \\
=a^{n}+b^{n}+a b\left(a^{n-2}+b^{n-2}\right), \\
\therefore a^{n}+b^{n}=(a+b)\left(a^{n-1}+b^{n-1}\right) \\
-a b\left(a^{n-2}+b^{n-2}\right) \text {. } \\
\end{array}
$$
Let $S_{n}=a^{n}-b^{n}$, we get the recursive formula
$$
S_{n}=(a+b) S_{n-1}-a b S_{n-2} .(n=2,3, \cdots)
$$
From the given, $m, n$ are the unequal real roots of $x^{2}-x-1=0$, hence
$$
\begin{array}{l}
m+n=1, m n=-1, m^{2}+n^{2}=3 . \\
\text { Therefore, } S_{n}=S_{n-1}+S_{n-2} . \\
\text { Thus, } S_{3}=S_{2}+S_{1}=3+1=4, \\
S_{4}=S_{3}+S_{2}=4+3=7, \\
S_{5}=S_{4}+S_{3}=7+4=11 .
\end{array}
$$
Therefore, $m^{5}+n^{5}=11$.
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given the equation $a x^{2}+b x+c=0(a \neq 0)$, the sum of the roots is $s_{1}$, the sum of the squares of the roots is $s_{2}$, and the sum of the cubes of the roots is $s_{3}$. Then the value of $a s_{3}+$ $\left\langle s_{2}\right.$ $+c s_{1}$ is . $\qquad$
|
(Tip: Let the two roots of the equation be $x_{1}, x_{2}$.
Then by definition, we have $a x_{1}^{2}+b x_{1}+c=0, a x_{2}^{2}+b x_{2}+c=0$. The original expression $=0$.)
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. If $a$ is a root of $x^{2}-3 x+1=0$, try to find the value of $\frac{2 a^{5}-5 a^{4}+2 a^{3}-8 a^{2}}{a^{2}+1}$.
|
From the definition of the root, we know that $a^{2}-3 a+1=0$.
Thus, $a^{2}+1=3 a$ or $a^{2}-3 a=-1$ or
$$
\begin{array}{l}
a^{3}=3 a^{2}-a \\
\therefore \text { the original expression }=\frac{a\left[2 a^{2}\left(a^{2}+1\right)-5 a^{3}-8 a\right]}{3 a} \\
=\frac{1}{3}\left(a^{3}-8 a\right)=\frac{1}{3}\left(3 a^{2}-9 a\right) \\
=a^{2}-3 a=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 8. Let } x>\frac{1}{4} \text {. Simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$
|
Let $a$
$$
\begin{array}{l}
=x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}, b=x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1} . \text { Original } \\
\text { expression }=1 .)
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given
$$
\left\{\begin{array}{l}
1988(x-y)+1989(y-z)+1990(z-x)=0, \\
1988^{2}(x-y)+1989^{2}(y-z)+1990^{2}(z-x)=1989 .
\end{array}\right.
$$
Find the value of $y-z$.
|
$y-z=-1989$
|
-1989
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. The value of $x$ that satisfies the following equations is
$$
\begin{array}{l}
(123456789) x+9=987654321, \\
(12345678) x+8=98765432 . \\
(1234567) x+7=9876543 . \\
\cdots \cdots .
\end{array}
$$
|
Observe the numerical changes on both sides of each equation, it is easy to know that the last equation should be $x+1=9$, i.e., $x=8$. Upon verification, $x=8$ is the solution.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Let $G$ be a simple graph of order 10 and does not contain a cycle of length 4, $C_{4}$. Then the maximum number of edges in $G$ is 16.
|
Proof: Let $f(n)$ be the maximum number of edges in an $n$-order simple graph without $C_{4}$. Clearly, $f(n)=4$. Now we prove that $f(5)=6$. First, if $G$ is two triangles sharing exactly one common vertex, then $G$ is a 5-order graph with 6 edges and no $C_{4}$. Second, if a 5-order graph $G$ has 7 edges, then there must be a vertex with a degree no greater than 2. Removing this vertex and the edges associated with it, we get a 4-order graph $G^{\prime}$. $G^{\prime}$ still has at least 5 edges, which must contain a quadrilateral. Now, we will prove that $f(10)=16$.
Let $G$ be a 10-order graph without $C_{4}$, and let $A$ be the vertex with the highest degree in $G$. Let $S=\{x \mid x$ is adjacent to $A\}$, $T=\{x \mid x$ is not adjacent to $A\}$, and assume $|S|=k$, then $|T|=9-k$, where $1 \leqslant k \leqslant 9$. For any point $x$ in $S \cup T$, it can be adjacent to at most one point in $S$, otherwise, adding $A$ would form a $C_{4}$. Thus, $S$ can have at most $\left[\frac{k}{2}\right]$ edges. Also, $S$ and $T$ can have at most $9-k$ edges between them, and $T$ can have at most $f(9-k)$ edges. Therefore,
$$
\begin{array}{l}
f(10) \leqslant k+\left[\frac{k}{2}\right]+(9-k)+f(9-k) . \\
\text { Let } g(k)=9+\left[\frac{k}{2}\right]+f(9-k) . \text { Then } \\
f(10) \leqslant g(k) .
\end{array}
$$
The relevant data is calculated as follows:
\begin{tabular}{c|c|c|c|c|c|c}
\hline$k$ & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$f(9-k)$ & 6 & 4 & 3 & 1 & 0 & 0 \\
\hline$g(k)$ & 17 & 15 & 15 & 13 & 13 & 13 \\
\hline
\end{tabular}
From the table, we can see that when $k \geqslant 5$, $g(k) \leqslant 15$, so $f(10) \leqslant 15$; also, when $k \leqslant 3$, $f(10) \leqslant \frac{3 \times 10}{2}=15$.
Thus, we only need to prove that when $k=4$, $f(10) \leqslant 16$. Since $T$ does not contain $C_{4}$ and has 5 vertices, from the initial discussion, $f(5)=6$. If there is a vertex with degree 1, removing this vertex would leave a 4-order graph with 5 edges, which must contain a $C_{4}$. Therefore, each vertex must have a degree of at least 2. If there is a vertex $A$ with degree 3, let it be adjacent to $B, C, D$ but not to $E$. Since $E$ has a degree of at least 2, $E$ must be adjacent to two of $B, C, D$, which would form a $C_{4}$. Therefore, each vertex must have degrees of 4, 2, 2, 2, 2, meaning $T$ forms two triangles sharing exactly one common vertex. Since $|S|=4$, if $S$ and $T$ are connected by at least 5 edges, then there must be a vertex in $S$ that is connected to two vertices in $T$, forming a $C_{4}$. Thus,
$$
\begin{array}{l}
\quad f(10) \leqslant 4+\left[\frac{4}{2}\right]+4 \\
+f(5)=16 .
\end{array}
$$
Finally, the graph on the right is a 10-order graph without $C_{4}$ and with 16 edges, so $f(10)=16$.
For $n \leqslant 21$, the values of $f(n)$ are as follows:
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline$n$ & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline$f(n)$ & 4 & 6 & 7 & 9 & 11 & 13 & 16 & 18 & 21 \\
\hline
\end{tabular}
\begin{tabular}{r|l|l|l|l|l|l|l|l}
\hline 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 \\
\hline 24 & 27 & 30 & 33 & 36 & 39 & 42 & 46 & 50 \\
\hline
\end{tabular}
Where $f(8)=11$ is the 5th problem of the 1992 CMO.
For a general $n \in \mathbb{N}$, what is the expression for $f(n)$? This remains an open problem, and people have only found an upper bound for $f(n)$, which is given below.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. Let $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=0$ and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is $\qquad$.
|
From the given, we have $\left(\frac{1}{a}\right)^{2}-\frac{2}{a}-1=0$, $\left(b^{2}\right)^{2}-2 b^{2}-1=0$. By the definition of roots, $\frac{1}{a}, b^{2}$ are the roots of the equation $x^{2}-2 x-1=0$. Then
$$
\begin{aligned}
\frac{1}{a}+b^{2} & =2, \frac{1}{a} \cdot b^{2}=-1 . \\
\therefore \text { the original expression } & =\left(\frac{1}{a}+b^{2}+\frac{1}{a} \cdot b^{2}\right)^{1990} \\
& =(2-1)^{1990}=1 .
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Glue the bases of two given congruent regular tetrahedra together, precisely to form a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
|
4. 3.
As shown in the figure, construct $CE \perp AD$, connect $EF$, it is easy to prove that $EF \perp AD$. Therefore, $\angle CEF$ is the plane angle of the dihedral angle formed by plane $ADF$ and plane $ACD$.
Let $G$ be the midpoint of $CD$. Similarly, $\angle AGB$ is the plane angle of the dihedral angle formed by plane $ACD$ and plane $BCD$.
D
Given that $\angle CEF = \angle AGB$.
Let the side length of the base $\triangle CDF$ be $2a$, and the length of the side $AD$ be $b$. In $\triangle ACD$, $CE \cdot b = AG \cdot 2a$.
Thus, $CE = \frac{AG \cdot 2a}{b} = \frac{\sqrt{b^2 - a^2} \cdot 2a}{b}$.
In $\triangle ABC$, it is easy to find that
$AB = 2 \sqrt{b^2 - \left(\frac{2}{3} \sqrt{3} a\right)^2} = 2 \sqrt{b^2 - \frac{4}{3} a^2}$.
Since $\triangle CEF \sim \triangle AGB$, we have $\frac{AB}{CF} = \frac{AG}{CE}$,
which means $\frac{2 \sqrt{b^2 - \frac{4}{3} a^2}}{2a} = \frac{\sqrt{b^2 - a^2}}{\frac{\sqrt{b^2 - a^2} \cdot 2a}{b}}$.
Thus, we find that $b = \frac{4}{3} a$.
Therefore, $b = 2$, and $2a = 3$.
Hence, the distance between the farthest two vertices is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Choose several colors from the given six different colors to color the six faces of a cube, with each face being colored with exactly one color, and any two faces sharing a common edge must be colored differently. Then, the number of different coloring schemes is $\qquad$. (Note: If we color two identical cubes and can rotate one of them so that the colors of the corresponding top, bottom, left, right, front, and back faces of the two cubes are the same, then we say that the two cubes have the same coloring scheme.)
|
5. 230 ways.
(1) Using 6 colors, the division is $\frac{6}{5 \times} \times \frac{4 \times 3 \times 2 \times 1}{6 \times 4}=$ 30 ways;
(3) Using 4 colors, the division is $C_{6}^{4} \cdot C_{4}^{2} \cdot \frac{2 \times 1}{2}=90$ ways;
(4) Using 3 colors, the painting method is $C_{6}^{3}=20$ ways; therefore, the total number of coloring schemes is 230 ways.
|
230
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$ .
|
6. 4 .
Let $A(x, y)$ be an integer point on circle $O$. As shown in the figure, the equation of circle $O$ is $y^{2}+(x-199)^{2}=$ $199^{2}$.
$$
\begin{array}{l}
\text { Clearly, } x=0, y=0 ; \\
x=199, y=199 ; \\
x=199, y=-199 ; x=389, y=0
\end{array}
$$
These are 4 solutions to the equation. However, when $y \neq 0, \pm 199$, $y$ is coprime with 199, so 199 can be expressed as the sum of the squares of two positive integers, i.e., $199=m^{2}+n^{2}$. Since $199=4 \times 49+3$, we can set $m=2 k, n=2 l+1$, then.
$$
199=4 k^{2}+4 l^{2}+4 l+1=4\left(k^{2}+l^{2}+l\right)+1 .
$$
The integer points are:
$$
(0,0),(199,199),(389,0),(199,-199) \text {. }
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8. Calculate
$$
\begin{array}{l}
\sqrt{3633 \times 3635 \times 3639 \times 3641+36} \\
-3636 \times 3638=
\end{array}
$$
|
Let $3637=a$. Then
the original expression $=$
$$
\begin{array}{l}
\sqrt{(a-4)(a-2)(a+2)(a+4)+36} \\
-(a+1)(a-1) \\
=\sqrt{\left(a^{2}-10\right)^{2}}-\left(a^{2}-1\right) \\
=a^{2}-10-a^{2}+1=-9 . \\
\end{array}
$$
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. (Ireland) A conference is attended by $12 k$ people, each of whom has greeted exactly $3 k+6$ others. For any two people, the number of people who have greeted both of them is the same. How many people attended the conference?
|
For any two people, let the number of other people who have greeted these two people be $n$. By the problem, $n$ is fixed. For a specific person $a$, let $\mathrm{B}$ be the set of all people who have greeted $a$, and $\mathrm{C}$ be the set of people who have not greeted $a$. Then, $\mathbf{B}$ contains $3 k+6$ people, and $\mathrm{C}$ contains $9 k-7$ people. For any person $b$ in $\mathbf{B}$, the people who have greeted both $a$ and $b$ must be in $\mathbf{B}$. Therefore, $b$ has greeted $n$ people in $\mathbf{B}$ and $3 k+5-n$ people in $\mathrm{C}$. For any person $c$ in $\mathrm{C}$, the people who have greeted both $a$ and $c$ must also be in $\mathbf{B}$. Thus, $c$ has greeted $n$ people in $\mathbf{B}$. The total number of greetings between $\mathbf{B}$ and $\mathbf{C}$ is
$(3 k+6)(3 k+5-n)=(9 k-7) n$.
Simplifying, we get $9 k^{2}-12 k n+33 k+n+30=0$. From this, we know that $n=3 m$, where $m$ is a positive integer, and $4 m=k+3+\frac{9 k+43}{12 k-1}$. When $k \geqslant 15$, $12 k-1>9 k+43$, and $4 m$ is not an integer. For $1 \leqslant k \leqslant 14$, only when $k=3$, $\frac{9 k+43}{12 k-1}$ is an integer. Therefore, the only possibility is that 36 people attended the meeting. The following construction shows that the solution to the original problem indeed exists.
\begin{tabular}{ll}
ROYGBV & $R:$ Red \\
VROYGB & $O:$ Orange \\
BVROYG & $Y$ : Yellow \\
GBVROY & $G$ : Green \\
YGBVRO & $B$ : Blue \\
OYGBVR & $V$ : Violet
\end{tabular}
People of the same color. Clearly, each person knows exactly 15 people. Let $P, Q$ be any two attendees. If they are in the same row, the people who know both of them are the other 4 people in that row, and the person in the column of $P$ with the color of $Q$, and the person in the column of $Q$ with the color of $P$. When $P, Q$ are in the same column or have the same color, similar analysis applies. Assume $P, Q$ are in different rows, different columns, and have different colors. Then the 6 people who know both of them are: the person in the row of $P$ and the column of $Q$, the person in the row of $P$ with the color of $Q$, the person in the column of $P$ and the row of $Q$, the person in the column of $P$ with the color of $Q$, the person with the color of $P$ in the row of $Q$, or the column of $Q$.
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (Flowing Orchid) For $x \geqslant 1$, let $p(x)$ be the smallest prime number that does not divide $x$, and $q(x)$ be the product of all primes less than $p(x)$. Specifically, $p(1)=2$. If some $x$ makes $p(x)=2$, then define $q(x)=1$. The sequence $x_{0}, x_{1}, x_{2}, \cdots$ is defined by the following formula, with $x_{0}=1$.
$$
x_{n+1}=\frac{x_{n} p\left(x_{n}\right)}{q\left(x_{n}\right)},
$$
where $n \geqslant 0$. Find all integers $n$ such that $x_{n}=1995$.
|
Obviously, from the definitions of $p(x)$ and $q(x)$, it can be derived that for any $x$, $q(x)$ divides $x$. Therefore,
$$
x_{n=1}=\frac{x_{i}}{q\left(x_{x}\right)} \cdot p(x ;)
$$
Moreover, it is easy to prove by induction that for all $n$, $x_{n}$ has no square factors. Thus, a unique encoding can be assigned to $x$ based on whether it is divisible by a certain prime number, as follows: Let $p_{0}=2, p_{1}=3, p_{2}=5, \cdots$ be the sequence of all prime numbers in ascending order. Let $x>1$ be any square-free number, and let $p_{m}$ be the largest prime dividing $x$. Then the encoding of $x$ is $\left(1, s_{m-1}, s_{m-1}, \cdots, s_{1}, s_{0}\right)$, where $s_{i}=1$ if $p_{i}$ divides $x$, and $s_{i}=0$ otherwise, for $0 \leqslant i \leqslant m-1$. Define $f(x)=\frac{x p(x)}{q(x)}$. If the last digit of $x$'s encoding is 0, then $x$ is odd, $p(x)=2, q(x)=1, f(x)=2 x$, and the encoding of $f(x)$ is the same as that of $x$ except that the last 0 is replaced by 1. If the last several digits of $x$'s encoding are $011 \cdots 1$, then the last several digits of $f(x)$'s encoding are $100 \cdots 0$. If the encoding traverses all binary numbers, then the encoding of $f(x)$ can be obtained by adding 1 to the encoding of $x$. Given $x_{1}=2$ and for $n \geqslant 2$, $x_{n+1}=f\left(x_{n}\right)$, the encoding of $x_{n}$ can directly equal the binary representation of $n$. Therefore, when $x_{n}=1995=3 \cdot 5 \cdot 7 \cdot 19$, there exists a unique $n$, and since the encoding of $x_{n}$ is 10001110, it follows that $n=142$ is the decimal representation of this encoding.
|
142
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. (India) Let $N$ be the set of all positive integers. Prove that there exists a unique function $f: \mathrm{N} \rightarrow \mathrm{N}$, satisfying
$$
f(m+f(n))=n+f(m+95),
$$
where $m, n$ are any elements of $N$. And calculate the value of $\sum_{k=1}^{19} f(k)$.
|
Prove that for all $n \geqslant 1$, let $F(n)=f(n)-95$. By substituting $k$ for $m+95$, the given condition becomes
$$
F(k+F(n))=n+F(k),
$$
where $n \geqslant 1, k \geqslant 96$. In (1), substitute $m$ for $k$, then add $k$ to both sides, and apply the function $F$, we get $F(k+n+F(m))=F(k+F(m+F(n)))$. Using (1) again, we can derive
$$
F(k+n)=F(k)+F(n),
$$
where $n \geqslant 1, k \geqslant 96$. We can assert that for all $q \geqslant 1$,
$$
F(96 q)=q F(96).
$$
Indeed, when $q=1$, (3) naturally holds. Using (2) and mathematical induction, (3) can be immediately proven. For any $m$ and set $F(m)=96 q+r, 0 \leqslant r \leqslant 95$. For $n \geqslant 1$, using (1), (2), and (3), we get
$$
\begin{array}{l}
m+F(n)=F(n+F(m)) \\
=F(n+96 q+r)=F(n+r)+F(96 q) \\
=F(n+r)+q F(96).
\end{array}
$$
If $1 \leqslant n \leqslant 96-r$, then $1+r \leqslant n+r \leqslant 96$. If $97-r \leqslant n \leqslant 96$, where $r \geqslant 1$, then $1 \leqslant n+r-96 \leqslant r$. Using (2) and (4), we get (continued on page 37)
$$
\begin{array}{l}
m+F(n)=F(n+r-96+96)+q F(96) \\
=F(n+r-96)+(q+1) F(96).
\end{array}
$$
Sum (4) from $n=1$ to $n=96-r$. If $r \geqslant 1$, sum (5) from $n=97-r$ to $n=96$, then cancel $F(1)+F(2)+\cdots+F(96)$ from both sides, and finally get
$$
\begin{aligned}
96 m & =F(96)\{q(96-r)+(q+1) r\} \\
& =F(96) F(m).
\end{aligned}
$$
In (6), let $m=96$, we get $96^2=[F(96)]^2$. Since $F(96)>0$, we have $F(96)=96$. Using (6), we can deduce $F(m)=m$, i.e., $f(m)=m+95$, where $m \geqslant 1$. Therefore, the required sum is $1+2+\cdots+19+19 \cdot 95=1995$.
|
1995
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. $\sqrt{1991 \cdot 1993 \cdot 1995 \cdot 1997+16}=$
|
Let $x=1994$, then
$$
\begin{array}{l}
\sqrt{1991 \cdot 1993 \cdot 1995 \cdot 1997+16} \\
=\sqrt{(x-3)(x-1)(x+1)(x+3)+16} \\
=\sqrt{\left(x^{2}-1\right)\left(x^{2}-9\right)+16}=\sqrt{x^{4}-10 x^{2}+25} \\
=\sqrt{\left(x^{2}-5\right)^{2}}=1994^{2}-5 \\
=3976031 .
\end{array}
$$
|
3976031
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $a, b$ be unequal real numbers, and $a^{2}+2 a=$ $5 \cdot b^{2}+2 b=5$. Then $a^{2} b+a b^{2}=$
|
4. (10).
It is known that $a, b$ are two unequal real numbers which are exactly the two real roots of the equation $x^{2}+2 x-5=0$. Therefore, $a+b=-2, ab=-5$.
Thus, $a^{2} b+ab^{2}=ab(a+b)=(-2) \cdot(-5)=$
10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If $m=1996^{3}-1995^{3}+1994^{3}-1993^{3}$ $+\cdots+4^{3}-3^{3}+2^{3}-1^{3}$, then the last digit of $m$ is
|
5. (0).
$$
2^{3}-1^{3}=7,4^{3}-3^{3}=37,6^{3}-5^{3}=91,8^{3}-7^{3}=
$$
$169,10^{3}-9^{3}=271$. Therefore, the sum of the last digits of the first 10 numbers $10^{3}-9^{3}+8^{3}-7^{3}$ $+6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}$ is $1+9+1+$ $7+7=25$, and the last digit of 25 is 5. The last digit of the algebraic sum of every ten consecutive cubic numbers from the unit digit 1 to the unit digit 0 is always 5. For $1990^{3}$, there are a total of 199 fives, and the last digit remains 5. Additionally, $1996^{3}-1995^{3}+1994^{3}-1093^{3}+1992^{3}-1591^{3}$ also has a last digit of 5. Therefore, the last digit of $m$ is ).
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (Full marks 20 points) Given that positive integers $p, q$ are both prime numbers, and $7 p+q$ and $p q+11$ are also prime numbers. Calculate the value of $\left(p^{2}+q^{p}\right)\left(q^{2}+p^{q}\right)$.
|
Given that $p q+11$ is a prime number, we know that $p q+11$ must be odd. Therefore, $p q$ is even. So, at least one of $p, q$ must be even. But since $p, q$ are both primes, one of $p, q$ must be 2. If $p=q=2$, then $p q+11=15$ is not a prime number. Therefore, $p, q$ cannot both be 2. So, one and only one of $p, q$ is 2.
If $p=2$, then $14+q$ and $2 q+11$ are both primes. If $q$ leaves a remainder of 1 when divided by 3, then $q+14$ is divisible by 3. If $q$ leaves a remainder of 2 when divided by 3, then $2 q+11$ is divisible by 3, which contradicts the fact that $14+q$ and $2 q+11$ are primes. Therefore, $q=3$. Checking, we find that $p=2, q=3$ meets the requirements.
If $q=2$, then $7 p+2$ and $2 p+11$ are both primes. If $p$ leaves a remainder of 1 when divided by 3, then $7 p+2$ is divisible by 3. If $p$ leaves a remainder of 2 when divided by 3, then $2 p+11$ is divisible by 3, which contradicts the fact that $7 p+2$ and $2 p+11$ are primes. Therefore, $p=3$. Checking, we find that $p=3, q=2$ also meets the requirements.
Whether $p=2, q=3$ or $p=3, q=2$,
we have $\left(p^{2}+q^{p}\right)\left(q^{2}+p^{q}\right)=\left(3^{2}+2^{3}\right)\left(2^{2}+3^{2}\right)$
$$
=17 \times 13=221 .
$$
|
221
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$. To make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National Competition)
|
Solution: According to the problem, all logarithms are positive. Therefore, from (*) we get
$$
\begin{aligned}
\text { LHS } & =\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \\
& \geqslant \frac{(1+1+1)^{2}}{\log _{1993}\left(\frac{x_{0}}{x_{1}} \cdot \frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}}\right)} \\
& =9\left(\log _{1993} \frac{x_{0}}{x_{3}}\right)^{-1} .
\end{aligned}
$$
It is clear that the maximum value of $k$ is 9.
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (This question is worth 16 points) Given the set $\{1,2,3,4,5, 6,7,8,9,10\}$. Find the number of subsets of this set that have the following property: each subset contains at least 2 elements, and the absolute difference between any two elements in each subset is greater than 1.
|
Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ that have the given property.
The set $\{1,2, \cdots, n, n+1, n+2\}$ has subsets with the given property divided into two categories: the first category of subsets contains the element $n+2$, and there are $a_{n}+n$ such subsets (i.e., the union of each subset of $\{1, 2, \cdots, n\}$ with $\{n+2\}$, as well as $\{1, n+2\},\{2, n+2\}, \cdots,\{n, n+2\}$); the second category of subsets does not contain $n+2$, and there are $a_{n+1}$ such subsets.
Thus, we have $a_{n+2}=a_{n}+a_{n+1}+n$.
Clearly, $a_{3}=1, \quad a_{4}=3$ (i.e., $\{1,3\},\{2,4\}$, $\{1,4\}$ )
$\therefore a_{5}=7, \quad a_{6}=14, \quad a_{7}=26, \quad a_{8}=46$, $a_{9}=79, \quad a_{10}=133$
|
133
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.3. $x, y$ are natural numbers. $k$ is a natural number of size 1. Find all natural numbers $n$ that satisfy: $3^{n}=x^{k}+y^{k}$, and provide a proof.
|
10. 3. Answer: $n=2$.
At the time: Let $3^{n}=x^{k}+y^{k}$, where $x$ and $y$ are prime (assuming $x > y$), $k>1$, and $n$ is a natural number. Of course, neither $x$ nor $y$ can be divisible by 3.
If $k$ is even, then $x^{k}$ and $y^{k}$ leave a remainder of 1 when divided by 3. Thus, the sum of $x^{k}$ and $y^{k}$ leaves a remainder of 2 when divided by 3, which is not a power of 3. This leads to a contradiction, so $k$ cannot be even.
If $k$ is odd and $k>1$, then
$$
3^{n}=(x+y)\left(x^{k-1}-\cdots+y^{k-1}\right).
$$
Thus, $x+y=3^{m}, m \geqslant 1$.
We will now prove that $n \geqslant 2 m$.
Since $k$ is divisible by 3 (see 9.3. Answer), let $x_{1}=x^{\frac{k}{3}}$, $y_{1}=y^{\frac{k}{3}}$ and substitute, so we can assume $k=3$.
Thus, $x^{3}+y^{3}=3^{n}, x+y=3^{m}$.
To prove $n \geqslant 2 m$, we need to use $x^{2}+y^{3} \geqslant(x+y)^{2}$, which means proving $x^{2}-x y+y^{2} \geqslant x y$.
Since $x \geqslant y+1$, then $x^{2}-x=x(x-1) \geqslant x y$.
$$
\left(x^{2}-x-x y\right)+\left(y^{2}-y\right) \geqslant 0.
$$
The inequality $n \geqslant 2 m$ is proven:
From the equation $(x+y)^{3}-\left(x^{3}+y^{3}\right)=3 x y(x+y)$, we get:
$$
3^{2 m-1}-3^{n-m-1}=x y,
$$
and $2 m-1 \geqslant 1$,
and $n-m-1 \geqslant n-2 m \geqslant 0$.
Therefore, if at least one of the inequalities in (**) is a strict inequality, then the left side of (*) is divisible by 3, but the right side is not. This leads to a contradiction.
If $n-m-1=n-2 m=0$, then $m=1, n=2$ and $3^{2}=2^{3}+1^{3}$. Hence, $n=2$.
|
2
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) On the first day of operation, the factory's production does not exceed 20 units, and the daily production increases thereafter, but the increase in production each time does not exceed 20 units. When the daily production reaches 1996 units, find the minimum value of the total production of the factory.
When the daily production reaches 1996 units, find the minimum value of the total production of the factory.
|
Three, suppose the production on the $n$-th day after the start of work reaches 1996 pieces, and the production on the first day is $a_{1}$ pieces, with the increase in production on the $i$-th day being $a_{i}$ pieces $(i=2,3, \cdots, n)$, then we have
$$
a_{1}+a_{2}+a_{3}+\cdots+a_{n}=1996,0<a_{i} \leqslant 20 .
$$
The total production over these $n$ days is
$$
S=n a_{1}+(n-1) a_{2}+(n-2) a_{3}+\cdots+a_{n} \text {. }
$$
The problem is to find the minimum value of (2) under condition (1).
In $a_{1}, a_{2}, \cdots, a_{k}, \cdots, a_{n}$, counting from the end, let the first number less than 20 be $a_{k}(1<k \leqslant 20)$, at this time, make the following adjustments to these $n$ numbers, let
$$
\begin{array}{l}
a_{1}^{\prime}=a_{1}-1, a_{2}^{\prime}=a_{2}, a_{3}^{\prime}=a_{3}, \cdots, a_{k-1}^{\prime}=a_{k-1}, a_{k}^{\prime}=a_{k} \\
+1, a_{k+1}^{\prime}=a_{k+2}^{\prime}=\cdots=a_{n}^{\prime}=20 .
\end{array}
$$
We still have $a_{1}^{\prime}+a_{2}^{\prime}+a_{3}^{\prime}+\cdots+a_{n}^{\prime}=1996$.
However, the corresponding $S^{\prime}$ is smaller than the original $S$, i.e.,
$$
S-S^{\prime}=k-1 \text {. }
$$
It can be seen that the minimum value of $S$ is obtained when $0<a_{1} \leqslant 20$, and $a_{2}$
$$
\begin{array}{l}
=a_{3}=\cdots=a_{n}=20 \text {, since } \\
1996=99 \times 20+16 \text {, } \\
\text { hence when } a_{1}=16, n=100 \text {, } \\
S_{\text {min }}=100 \times 16+(1+2+\cdots+99) \times 20 \\
=100600 \text {. } \\
\end{array}
$$
|
100600
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The smallest natural number $a$ that makes the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$ $<a-1995 \frac{1}{3}$ hold for all natural numbers $n$ is $\qquad$.
|
6. 1997.
Let $\dot{f}(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$, then $f(n+1)-f(n)=\frac{-1}{(2 n+3)(2 n+2)}<0$, which means $f(n)$ is a decreasing function when $n \geq 0$. Therefore, $f(1)$ is the maximum value. To ensure the inequality holds for all $n \in N$, the maximum value of $f(n)$ must be less than: $a-1995 \frac{1}{3}$, i.e.,
$$
a-1995 \frac{1}{3}>f(1)=\frac{5}{6} \text {. }
$$
Thus, $a>1996 \frac{1}{6}$, and the smallest natural number $a$ is 1997.
|
1997
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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