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3. The real number $x$ satisfies
$$
\sqrt{x^{2}+3 \sqrt{2} x+2}-\sqrt{x^{2}-\sqrt{2} x+2}=2 \sqrt{2} x \text {. }
$$
Then the value of the algebraic expression $x^{4}+x^{-4}$ is $\qquad$
|
3.14.
Obviously $x \neq 0$.
Let $\sqrt{x^{2}+3 \sqrt{2} x+2}=a$,
$$
\sqrt{x^{2}-\sqrt{2} x+2}=b \text {. }
$$
Then $a^{2}-b^{2}=(a+b)(a-b)=4 \sqrt{2} x$.
Also, $a-b=2 \sqrt{2} x$, so, $a+b=2$.
Thus, $a=\sqrt{2} x+1$.
Then $(\sqrt{2} x+1)^{2}=x^{2}+3 \sqrt{2} x+2$.
Rearranging gives $x^{2}-\sqrt{2} x-1=0$, which is $x-x^{-1}=\sqrt{2}$.
Squaring both sides gives $x^{2}-2+x^{-2}=2$, which is
$$
x^{2}+x^{-2}=4 \text {. }
$$
Squaring both sides of the above equation gives $x^{4}+2+x^{-4}=16$, which is $x^{4}+x^{-4}=14$.
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) A positive integer $M$ has 8 positive divisors. Xiao Feng, while calculating the sum of the 8 positive divisors of $M$, forgot to add one of the divisors, resulting in a sum of 2776. Find all possible values of the positive integer $M$.
|
Three, if $M$ has at least four different prime factors, then the number of positive divisors of $M$ is at least $(1+1)^{4}=16$, which contradicts the problem statement.
Therefore, $M$ has at most three different prime factors.
If all prime factors of $M$ are odd, then all positive divisors of $M$ are odd. In this case, the sum of the seven positive divisors of $M$ is odd, which contradicts the fact that the sum is 2776, an even number.
Therefore, one of the prime factors of $M$ is 2.
(1) When $M$ has only one prime factor, this prime factor is $2$, and $M$ is $2^{7}$. Let the missing positive divisor be $2^{n}(n \in \mathbf{N}, n \leqslant 7)$.
By the problem statement,
$1+2+2^{2}+\cdots+2^{7}-2^{n}=2776$.
Simplifying, we get $2^{n}=-2521$ (discard).
Therefore, $M \neq 2^{7}$.
Thus, $M$ cannot have only one prime factor.
(2) When $M$ has only two different prime factors, let these two different prime factors be $p$ and $q$. Then $M$ can be set as $p^{3} q$. The missing positive divisor is $p^{m} q^{n}(m, n \in \mathbf{N}, 0 \leqslant m \leqslant 3, 0 \leqslant n \leqslant 1$.
By the problem statement,
$$
\left(1+p+p^{2}+p^{3}\right)(1+q)-p^{m} q^{n}=2776 \text {. }
$$
(i) If $p=2$, then equation (1) becomes
$$
15 q-2^{m} \times q^{n}=2761 \text {. }
$$
After calculation, only when $(m, n)=(2,1)$, $q=251$ (a prime number) meets the problem's requirements.
At this point, $M=2^{3} \times 251=2008$.
(ii) If $q=2$, then equation (1) becomes
$3\left(p+p^{2}+p^{3}\right)-2^{n} \times p^{m}$
$$
=2773=47 \times 59 \text {. }
$$
When $m \geqslant 1$, the left side of equation (2) can be divided by $p$, then $p \mid 47 \times 59$.
Since $p$ is a prime factor of $M$, hence $p=47$ or 59.
Therefore, the left side of equation (1) $\geqslant p^{3} \geqslant 47^{3}>2776$.
So, $m \geqslant 1$ does not hold.
When $m=0$, equation (2) becomes
$3 p\left(1+p+p^{2}\right)=2773+2^{n}=2775$ or 2774.
Since $3 \nmid 2775, 3 \nmid 2774$, so,
$3 p\left(1+p+p^{2}\right)=2775$,
$p\left(1+p+p^{2}\right)=925=5^{2} \times 37$.
Therefore, $p \mid 5^{2} \times 37$, hence $p=5$ or 37.
When $p=5$ or 37, equation (3) does not hold.
So, $q \neq 2$.
(3) When $M$ has only three different prime factors, one of them is 2, and the other two are denoted as $p$ and $q(p \neq q)$.
By the problem statement, $M=2 p q$.
Let the missing positive divisor be $2^{m} p^{n} q^{t}(m, n, t$ are natural numbers not exceeding 1).
By the problem statement,
$$
(1+2)(1+p)(1+q)-2^{m} p^{n} q^{t}=2776,
$$
i.e., $3(1+p)(1+q)-2^{m} p^{n} q^{t}=2776$.
Since $p, q$ are odd prime numbers, therefore,
$4 \mid 3(1+p)(1+q)$.
Also, $4 \mid 2776$, by equation (4), $4 \mid 2^{m} p^{n} q^{t}$.
Hence, $4 \mid 2^{m}$.
Since $m \leqslant 1$, then $4 \nmid 2^{m}$. Therefore, equation (5) does not hold.
Thus, $M$ cannot have three different prime factors.
In summary, the value of $M$ is 2008.
(Xie Wenxiao, Huanggang High School, Hubei Province, 438000)
|
2008
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $\triangle A B C$ with side lengths $a=17, b=18, c$ $=19$, a point $O$ inside $\triangle A B C$ is drawn perpendiculars to the three sides of $\triangle A B C$, intersecting at points $D$, $E$, and $F$, such that $B D+C E+A F=27$. Then $B D+B F$ $=$ $\qquad$
|
2.18.
As shown in Figure 5, let \( BD = x \), \( CE = y \), \( AF = z \). Then
\[
\begin{array}{l}
CD = 17 - x, \\
AE = 18 - y, \\
BF = 19 - z.
\end{array}
\]
Connect \( OA \), \( OB \), \( OC \).
In right triangles \( \triangle OBD \), \( \triangle OCE \), \( \triangle OAF \),
\[
\left\{\begin{array}{l}
x^{2} + OD^{2} = (19 - z)^{2} + OF^{2}, \\
y^{2} + OE^{2} = (17 - x)^{2} + OD^{2}, \\
z^{2} + OF^{2} = (18 - y)^{2} + OE^{2}, \\
x + y + z = 27.
\end{array}\right.
\]
\[
\begin{array}{l}
\text{(1) } + \text{ (2) } + \text{ (3) gives } \\
x^{2} + y^{2} + z^{2} = (17 - x)^{2} + (18 - y)^{2} + (19 - z)^{2}. \\
\text{ Hence } \left\{\begin{array}{l}
17x + 18y + 19z = 487, \\
y = 27 - x - z.
\end{array}\right.
\end{array}
\]
Solving, we get \( x - z = -1 \).
Therefore, \( BD + BF = x + 19 - z = 18 \).
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A positive integer $M$, when added to 36, 300, and 596 respectively, results in the squares of three different positive integers. The sum of the smallest and the largest of these three different positive integers is twice the middle one. Then this positive integer $M=$
|
3.925 .
According to the problem, let $M+36=(k-m)^{2}$,
$M+300=k^{2}$,
$M+596=(k+m)^{2}(m>0)$.
Then $(M+36)+(M+596)=2\left(k^{2}+m^{2}\right)$.
Therefore, $M+316=M+300+16=k^{2}+m^{2}$.
Solving this, we get $m=4$.
$$
\begin{array}{l}
\text { Also, }(M+596)-(M+36) \\
=(k+m)^{2}-(k-m)^{2}=4 k m,
\end{array}
$$
Thus, $k=35$.
Therefore, $M=925$.
|
925
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $a-2$ is a positive integer and a divisor of $3 a^{2}-2 a+10$, then the sum of all possible values of $a$ is $\qquad$ .
|
2.51.
$$
\begin{array}{l}
\text { Given } \frac{3 a^{2}-2 a+10}{a-2} \\
=\frac{\left(3 a^{2}-6 a\right)+(4 a-8)+18}{a-2} \\
=3 a+4+\frac{18}{a-2}(a \neq 2),
\end{array}
$$
we know that $a-2$ must be a divisor of 18.
Thus, $a=3,4,5,8,11,20$.
Therefore, the sum of all possible values of $a$ is
$$
3+4+5+8+11+20=51 \text {. }
$$
|
51
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $x$ is a real number. Then the maximum value of $\sqrt{2000-x}+$ $\sqrt{x-2000}$ is $\qquad$ .
|
4.4.
$$
\begin{array}{l}
\text { Let } t_{1}=\sqrt{2008-x}+\sqrt{x-2000} \text {. Then } \\
t_{1}^{2}=8+2 \sqrt{(2008-x)(x-2000)} \\
\leqslant 8+8=16 .
\end{array}
$$
Therefore, $t \leqslant 4$, i.e., $t_{\max }=4$. At this point, $x=2004$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A ten-digit number, whose digits can only be 2 or 3, and there are no two adjacent 3s. How many such ten-digit numbers are there?
保留源文本的换行和格式，直接输出翻译结果。
|
5.144.
Solution 1: Consider all positive integers with the given property (using a recursive method).
The number of ten-digit numbers starting with 2 is the same as the number of nine-digit numbers;
Ten-digit numbers starting with 3 must start with 32, and their number is the same as the number of eight-digit numbers, i.e., the number of ten-digit numbers is the sum of the number of nine-digit numbers and the number of eight-digit numbers. This pattern is the same as the Fibonacci sequence.
There are 2 one-digit numbers, 3 two-digit numbers, so there are 5 three-digit numbers, 8 four-digit numbers, 13 five-digit numbers, 21 six-digit numbers, 34 seven-digit numbers, 55 eight-digit numbers, 89 nine-digit numbers, and 144 ten-digit numbers.
Solution 2: Since there cannot be more than two consecutive 3s, the ten digits of a ten-digit number can be divided into five pairs, so there can be at most 5 threes. Since there cannot be more than two consecutive 3s, it can be considered that after determining the number of 2s, 3s are inserted among them.
All positive integers with the given property can be divided into:
(1) If it has no digit 3, then there is only one possible number: 2222222222;
(2) If it has 1 three, then there are 9 twos, and there are $\mathrm{C}_{9+1}^{1}$ $=10$ cases;
(3) If it has 2 threes, then there are 8 twos, and there are $\mathrm{C}_{8+1}^{2}$ $=\frac{9!}{2!\times 7!}=36$ cases;
(4) If it has 3 threes, then there are 7 twos, and there are $\mathrm{C}_{7+1}^{3}$ $=\frac{8!}{3!\times 5!}=56$ cases;
(5) If it has 4 threes, then there are 6 twos, and there are $\mathrm{C}_{8+1}^{4}$ $=\frac{7!}{4!\times 3!}=35$ cases;
(6) If it has 5 threes, then there are 5 twos, and there are $\mathrm{C}_{5+1}^{5}$ $=\frac{6!}{5!}=6$ cases.
Therefore, there are $1+10+36+56+35+6=144$ ten-digit numbers.
|
144
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. There are $n(n>3)$ integers on a circle with a sum of 94, each of which is equal to the absolute value of the difference between the two numbers that follow it (in a clockwise direction). Then all possible values of $n$ are
|
6.141 .
Obviously, these $n$ integers are all non-negative.
Let $a$ be the largest one, and the next four numbers be $b, c, d, e$ respectively. Then $a=|b-c|$.
Thus, $b=a$ or $c=a$.
When $b=a$, $c=0$.
From $b=|c-d|$, we know $d=a$.
Also, from $c=|d-e|$, we get $d=e$.
Therefore, $n$ is a multiple of 3.
Assume $n=3m$. Then these $n$ numbers are $a, a, 0, a, a, 0, \cdots, a, a, 0$.
Since their sum is $2ma=94$, we have $m=47, a=1$ or $m=1, a=47$.
Thus, $n=141$ or 3 (not valid).
When $c=a$, $b=0$.
Thus, $d=a, e=0$, and these $n$ numbers are $a, 0, a, a, 0, a, \cdots, a, 0, a$.
Similarly, we get $n=141$ or 3 (not valid). Hence, $n$ can only be 141.
|
141
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that $M$ is a four-digit perfect square. If the thousand's digit of $M$ is reduced by 3 and the unit's digit is increased by 3, another four-digit perfect square can be obtained. Then the value of $M$ is $\qquad$
|
7.4761 .
Let $M=\overline{a b c d}=A^{2}$ and $M$'s thousands digit decreases by 3 and the units digit increases by 3 to equal $B^{2}$. Then
$$
\left\{\begin{array}{l}
A^{2}=1000 a+100 b+10 c+d, \\
B^{2}=1000(a-3)+100 b+10 c+(d+3) .
\end{array}\right.
$$
Therefore, $A^{2}-B^{2}=2997$.
Hence $(A-B)(A+B)=3^{4} \times 37$.
Since $A+B \leqslant 2 \times 99=198$, therefore,
$$
\left\{\begin{array}{l}
A-B=3^{3}, 37, \\
A+B=3 \times 37,3^{4} .
\end{array}\right.
$$
Solving yields $(A, B, M)=(69,42,4761)$ or $(59, 22,3481)$ (not valid).
|
4761
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 2, each segment of the broken line $A-B-C-D$ is parallel to the sides of the rectangle, and it divides the rectangle into two equal areas. Point $E$ is on the side of the rectangle such that segment $A E$ also bisects the area of the rectangle.
Given that segment $A B=30, B C=$
$24, C D=10$. Then $D E=$
$\qquad$
|
8.12.
Solution 1: Let line segment $A E$ intersect $B C$ at point $M$, and draw $E P \perp B C$, with the foot of the perpendicular being $P$.
Let $E D=x, B M=y$. Then $M P=24-x-y$.
By the problem, $S_{\triangle B B Y}=S_{\text {quadrilateral } U C D E}$, then
$15 y=120-5 y+5 x$.
Thus, $4 y=24+x$.
Also, $\triangle A M B \backsim \triangle E M P$, so $\frac{A B}{E P}=\frac{B M}{M P}$.
Therefore, $\frac{30}{10}=\frac{y}{24-x-y}$.
Hence, $4 y=72-3 x$.
From equations (1) and (2), we get $x=12$.
Solution 2: On both sides of the original rectangle, subtract equal-area smaller rectangles from each part so that $C D$ lies on the edge of the remaining rectangle (as shown in Figure 10). At this point, the broken line $A-B-C$ also divides this smaller rectangle into two equal-area parts. When both are further reduced by the area of the shaded part, they are still equal, i.e.,
$$
30 \times 24-10 \times 24=A G \times 40 \Rightarrow A G=12 .
$$
Since $A E$ also divides this smaller rectangle into two equal-area parts, we have $A G=E D=12$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given the function $f(x)=a x^{2}-c(a, c$ are real numbers). If $-4 \leqslant f(1) \leqslant-1,-1 \leqslant f(2) \leqslant 2$, then the maximum value of $f(8)$ is $\qquad$ .
|
9.122.
Since $f(1)=a-c, f(2)=4a-c$, therefore, $a=\frac{1}{3}(f(2)-f(1)), c=\frac{1}{3}(f(2)-4f(1))$.
Thus, $f(8)=64a-c=21f(2)-20f(1)$.
Hence $-1=21 \times(-1)-20 \times(-1) \leqslant f(8)$ $\leqslant 21 \times 2-20 \times(-4)=122$.
When $f(x)=2x^{2}-6$, it satisfies the given conditions, and $f(8)=122$. Therefore, the maximum value of $f(8)$ is 122.
|
122
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. The smallest positive integer $n$ for which $n^{2}-n+11$ has four prime factors (not necessarily distinct) is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
11.132.
From Table 1, it can be verified that for any positive integer $n, n^{2}-$ $n+11$ is not a multiple of $2,3,5,7$.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $\bmod 2$ & $\bmod 3$ & \multicolumn{3}{|c|}{$\bmod 5$} & \multicolumn{3}{|c|}{$\bmod 7$} \\
\hline$n$ & 0 & 1 & 0 & 1 & 2 & 0 & 1 & 2 & 3 & 4 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline$n^{2}$ & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 4 & 4 & 1 & 0 & 1 & 4 & 2 & 2 & 4 & 1 \\
\hline$n^{2}-n+11$ & 1 & 1 & 2 & 2 & 1 & 1 & 1 & 3 & 2 & 3 & 4 & 4 & 6 & 3 & 2 & 3 & 6 \\
\hline
\end{tabular}
Therefore, the smallest possible value is $n^{2}-n+11=11^{4}$.
Then $n(n-1)=2 \times 5 \times 7 \times 11 \times 19$.
However, this equation has no positive integer solutions.
When $n^{2}-n+11=13 \times 11^{3}$, we get
$n(n-1)=2 \times 2 \times 3 \times 11 \times 131$.
Solving this, we find $n=132$.
|
132
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. (40 points) If the simplest fraction $\frac{p}{q}$ is written in decimal form as $0 . a b a b a b a b \cdots$ (non-negative integers $a$ and $b$ can be equal, but at least one is non-zero), then, among the fractions that meet the condition, how many different numerators are there?
|
1. Decimal numbers in the form of 0.abababab $\cdots$ can all be written as $\frac{k}{99}$ $\left(k \in \mathbf{N}_{+}, k=1,2, \cdots, 98\right)$. Among the numbers $1,2, \cdots$, 99, those that are multiples of 3 or 11 are
$$
\left[\frac{99}{3}\right]+\left[\frac{99}{11}\right]-\left[\frac{99}{3 \times 11}\right]=39 \text {. }
$$
Therefore, the number of positive integers in $1,2, \cdots, 99$ that are coprime with 99 is $99-39=60$.
Additionally, there are 3 multiples of 27 (since $\frac{27}{99}=\frac{3}{11}, \frac{54}{99}=\frac{6}{11}$, $\frac{81}{99}=\frac{9}{11}$, we need to add back the numbers $3, 6, 9$), so the total number of different numerators is $60+3=63$.
|
63
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. (40 points) A rectangular box with dimensions $a_{1} \times b_{1} \times c_{1}$ can fit into another rectangular box with dimensions $a_{2} \times b_{2} \times c_{2}$ if and only if $a_{1} \leqslant a_{2} 、 b_{1} \leqslant b_{2} 、 c_{1} \leqslant c_{2}$. Therefore, among the rectangular boxes with dimensions $a \times b \times c$ (where $a 、 b 、 c$ are integers and $1 \leqslant a \leqslant b \leqslant c \leqslant 5$), what is the maximum number of boxes that can be selected such that no box can fit into another?
|
4. Among the boxes that meet the conditions, there are 5 types of boxes that are cubes, and 20 types of boxes that are square-based but not cubic prisms, because they have five different heights and four different choices for the square side length.
In addition, there are 10 types of boxes where the length, width, and height are all different, because three can be chosen from five lengths, resulting in $\frac{5 \times 4 \times 3}{3 \times 2}$ types. Table 2 categorizes these 35 types of boxes based on their length + width + height values. Within the same category, boxes with smaller length + width + height values can fit into boxes with larger length + width + height values.
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline Length+Width+Height & First Class Second Class Third Class Fourth Class Fifth Class \\
\hline 15 & 555 & & & & \\
\hline 14 & 455 & & & & \\
\hline 13 & 355 & 445 & & & \\
\hline 12 & 255 & 345 & & 444 & \\
\hline 11 & 155 & 245 & 335 & 344 & \\
\hline 10 & 145 & 244 & 235 & 334 & \\
\hline 9 & 135 & 144 & 225 & 234 & 333 \\
\hline 8 & & 134 & 125 & 224 & 233 \\
\hline 7 & & 124 & 115 & 223 & 133 \\
\hline 6 & & & 114 & 222 & 123 \\
\hline 5 & & & 113 & & 122 \\
\hline 4 & & & & & 112 \\
\hline 3 & & & & & 111 \\
\hline
\end{tabular}
One box with a length + width + height value of 9 can be selected from each category, i.e., $(1,3,5)$, $(1,4,4)$, $(2,2,5)$, $(2,3,4)$, and $(3,3,3)$. Each of these boxes has at least one dimension that is larger than the others, making it impossible for them to fit inside each other. Therefore, the maximum number of boxes that can be selected is 5.
Note: Giving an example with 4 boxes earns 10 points, giving an example with 5 boxes earns 20 points. Simply stating 5 boxes without giving an example earns 0 points. Proving the maximum of 5 boxes earns 20 points.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. (40 points) On the blackboard, the numbers 1 and 4 are originally written. For the numbers on the blackboard, an operation can be performed: each time, any two numbers can be chosen from the blackboard, and a new number \( c = ab + a + b \) can be added to the blackboard. After several operations, what is the smallest number not less than 2008 that appears on the blackboard?
|
6. From $c_{1}=a b+a+b=(a+1)(b+1)-1$, we have
$$
c_{1}+1=(a+1)(b+1) \text {. }
$$
Taking numbers $a$ and $c_{1}$, we get
$$
c_{2}+1=(a+1)\left(c_{1}+1\right)=(a+1)^{2}(b+1) \text {; }
$$
Taking numbers $b$ and $c_{1}$, we get
$$
c_{2}+1=(b+1)\left(c_{1}+1\right)=(a+1)(b+1)^{2} \text {; }
$$
$\qquad$
Suppose after several operations, the new number added to the blackboard is $x$. Then $x$ can always be expressed as
$$
x=(a+1)^{m}(b+1)^{n}-1\left(m, n \in \mathbf{N}_{+}\right) \text {. }
$$
Let after several operations, the numbers on the blackboard be
$$
h=(a+1)^{\rho}(b+1)^{q}-1
$$
and $k=(a+1)^{r}(b+1)^{s}-1$.
Then after one more operation, a new number is obtained as
$$
\begin{array}{l}
(h+1)(k+1)-1 \\
=(a+1)^{p+r}(b+1)^{q+s}-1 .
\end{array}
$$
When $a=1, b=4$, we have $x+1=2^{m} \times 5^{n}$, i.e., the numbers on the blackboard are all of the form $2^{m} \times 5^{n}-1$.
From Table 3, we know that the smallest number of this form and not less than 2008 is
$$
2^{11}-1=2048-1=2047 .
$$
Table 3
\begin{tabular}{|c|r|r|r|r|r|}
\hline $2^{m} \times 5^{n}-1$ & $n=0$ & $n=1$ & $n=2$ & $n=3$ & $n=4$ \\
\hline$m=0$ & 0 & 4 & 24 & 124 & 624 \\
\hline$m=1$ & 1 & 9 & 49 & 249 & 1249 \\
\hline$m=2$ & 3 & 19 & 99 & 499 & 2499 \\
\hline$m=3$ & 7 & 39 & 199 & 999 & \\
\hline$m=4$ & 15 & 79 & 399 & 1999 & \\
\hline$m=5$ & 31 & 159 & 799 & 3999 & \\
\hline$m=6$ & 63 & 319 & 1599 & & \\
\hline$m=7$ & 127 & 639 & 3199 & & \\
\hline$m=8$ & 255 & 1279 & & & \\
\hline$m=9$ & 511 & 2559 & & & \\
\hline$m=10$ & 1023 & & & & \\
\hline$m=11$ & 2047 & & & & \\
\hline
\end{tabular}
|
2047
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.(40 points) As shown in Figure 7, quadrilateral $ABCD$ is inscribed in a circle, $AB=AD$, and its diagonals intersect at point $E$. Point $F$ lies on segment $AC$ such that $\angle BFC = \angle BAD$. If $\angle BAD = 2 \angle DFC$, find the value of $\frac{BE}{DE}$.
保留源文本的换行和格式，直接输出翻译结果。
|
7. From $AB=AD$, we know $\angle ABD=\angle ADB=\theta$. By the property of equal arcs subtending equal angles at the circumference, we have
$$
\angle ACD=\angle ACB=\theta.
$$
Let $\angle DFC=\varphi$. Then
$$
\angle BAD=\angle BFC=2\varphi.
$$
Therefore, $\angle ABD + \angle ADB + \angle BAD$
$$
=\theta + \theta + 2\varphi = 180^{\circ}.
$$
Thus, $\theta + \varphi = 90^{\circ}$, and $\angle CDF = 90^{\circ}$.
On the other hand, from
$$
\begin{array}{l}
\angle FBC = 180^{\circ} - \theta - 2\varphi = \theta = \angle FCB \\
\Rightarrow FB = FC.
\end{array}
$$
Let $M$ be the midpoint of side $BC$, and connect $FM$. It is easy to see that $\triangle FCD \cong \triangle FBM$, and $BC = 2CD$.
Since $AC$ is the angle bisector of $\angle BCD$, by the Angle Bisector Theorem, we have $\frac{BE}{DE} = \frac{BC}{CD} = 2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A sequence of numbers, the first three numbers are $1, 9, 9$, and each subsequent number is the remainder of the sum of the three preceding numbers divided by 3. What is the 1999th number in this sequence?
|
(Tip: Apart from the first three numbers $1,9,9$, this sequence repeats every 13 terms (i.e., $1,1,2,1,1,1,0,2,0,2,1,0,0$). Since $1999-3=13 \times 153+7$, the 1999th number is the 7th number in the 154th cycle, which is exactly 0.)
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Observe the array:
(1),(3,5),(7,9,11),(13,15,17,19), .....
In which group is 2003?
|
(Tip: Use the trial method. The first 45 groups have a total of $1+2+\cdots$ $+45=1035$ numbers. The last number of the 45th group is 2069, and the first number is 1981, so, 2003 is in the 45th group.)
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Several 1s and 2s are arranged in a row
$$
1,2,1,2,2,1,2,2,2,1,2, \cdots
$$
The rule is: the 1st number is 1, the 2nd number is 2, the 3rd number is 1, ... Generally, first write a row of 1s, then insert $k$ 2s between the $k$th 1 and the $(k+1)$th 1 ($k=1$, $2, \cdots$). Try to answer:
(1) Is the 2005th number 1 or 2?
|
Explanation: Clearly, the position of 1 is somewhat special. For the convenience of calculation, we might as well divide this sequence of numbers into $n$ groups: the 1st group has 1 number, the 2nd group has 2 numbers, the 3rd group has 3 numbers, ..., the $n$th group has $n$ numbers, and the last number of each group is 1, while the rest are 2, i.e.,
(1), $(2,1),(2,2,1), \cdots,(\underbrace{2,2, \cdots, 2,1}_{n-1})$.
Thus, to determine whether the 2005th number is 1 or 2, we just need to use the Gauss summation formula to find out which group the 2005th number is in.
Below, we will use the trial method to estimate.
These $n$ groups have a total of $1+2+\cdots+n=\frac{n(n+1)}{2}$ numbers.
When $n=62$, there are 1953 numbers;
When $n=63$, there are 2016 numbers.
From this, we can see that the 2005th number is in the 63rd group, and it is not the last number.
Therefore, the 2005th number is 2.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If $a, b$ are integers, and the equation $a x^{2}+b x-$ $2008=0$ has two distinct prime roots, then the value of $3 a+b$ is ( ).
(A) 100
(B) 400
(C) 700
(D) 1000
|
6.D.
Let the roots of the equation be $x_{1}, x_{2}\left(x_{1}<x_{2}\right)$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, \\
x_{1} x_{2}=-\frac{2008}{a},
\end{array}\right.
$$
which means
$$
\left\{\begin{array}{l}
b=-a\left(x_{1}+x_{2}\right), \\
a x_{1} x_{2}=-2008=-2 \times 2 \times 2 \times 251.
\end{array}\right.
$$
Since $x_{1}, x_{2}$ are prime numbers, we can take $x_{1}=2$, $x_{2}=251$, from which we get $a=-4$.
And $b=4(2+251)=1012$, thus
$$
3 a+b=1000.
$$
|
1000
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $a>0, b>0$, satisfying $a+\sqrt{a}=2008, b^{2}+b=2008$. Then the value of $a+b$ is $\qquad$
|
$=, 1.2008$
For the convenience of calculation, let $2008=m$.
From $a+\sqrt{a}-m=0$, we get
$$
\sqrt{a}=\frac{-1+\sqrt{1+4 m}}{2} \text{, }
$$
which means $a=\frac{1+2 m-\sqrt{1+4 m}}{2}$.
From $b^{2}+b-m=0$, we get
$$
b=\frac{-1+\sqrt{1+4 m}}{2} \text{. }
$$
Therefore, $a+b=m=2008$.
|
2008
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Arrange positive integers starting from 1 according to the pattern shown in Figure 2, where 2 is at the first turn, 3 is at the second turn, 5 is at the third turn, 7 is at the fourth turn, … then the number at the 2007th turn is $\qquad$
|
To explore the relationship between the number of turns and the numbers at the turns, let $a_{n}$ represent the number corresponding to the $n$-th turn. If we observe in the order of natural numbers, then
$$
\begin{array}{l}
a_{1}=2, a_{2}=3, a_{3}=5, a_{4}=7, a_{5}=10, \\
a_{6}=13, a_{7}=17, a_{8}=21, \cdots \cdots
\end{array}
$$
The relationship between the number of turns $n$ and $a_{n}$ is relatively complex. If we skip some numbers (such as observing in the order of odd numbers), then
$$
a_{3}=a_{1}+3, a_{5}=a_{3}+5, a_{7}=a_{5}+7, \cdots \cdots
$$
That is, the next turning number = the previous turning number + the next turning number.
Thus, $a_{2007}=a_{2005}+2007$
$$
\begin{array}{l}
=a_{2003}+2005+2007=\cdots \\
=a_{3}+5+7+\cdots+2007 \\
=a_{1}+3+5+\cdots+2007 \\
=2+3+5+\cdots+2007 \\
=1+1+3+5+\cdots+2007 \\
=1+\frac{(1+2007) \times 1004}{2}
\end{array}
$$
$$
=1004^{2}+1=1008017 \text {. }
$$
Therefore, the number at the 2007th turn is 1008017.
|
1008017
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11.2. Given that the weight ratio of any two out of $N$ weights belongs to $\left[\frac{4}{5}, \frac{5}{4}\right]$, and these $N$ weights can be divided into 10 groups of equal weight, as well as 11 groups of equal weight. Find the minimum possible value of $N$.
|
11.2. $N_{\min }=50$.
An example for $N=50$.
In fact, 20 weights of $50 \mathrm{~g}$ and 30 weights of $40 \mathrm{~g}$ satisfy the requirement. These 50 weights can be divided into 10 equal groups, each consisting of 2 weights of $50 \mathrm{~g}$ and 3 weights of $40 \mathrm{~g}$. They can also be divided into 11 equal groups (5 groups each consisting of 4 weights of $50 \mathrm{~g}$, and 6 groups each consisting of 5 weights of $40 \mathrm{~g}$).
Now assume $N<50$.
Lemma: Suppose the ratio of the weights of any two weights is in $\left[\frac{4}{5}, \frac{5}{4}\right]$. If two groups of weights, one consisting of $k$ weights and the other of $l (k<l)$ weights, are equal in weight, then $k \geqslant 4$. When $k=4$, $l=5$ and each group consists of weights of the same weight.
Proof of the lemma: Let the weight of each group be $M$, and let $m$ be the weight of the lightest weight in the group of $l$ weights. Then the weight of each weight in both groups does not exceed $\frac{5}{4} m$. This gives us
$$
\frac{5}{4} k m \geqslant M \geqslant l m \geqslant(k+1) m,
$$
which implies $k \geqslant 4$.
If $k=4$, all inequalities in the above estimate become equalities, so $l=k+1=5$.
In the group of 5 weights, all weights are of weight $m=4 x$, and in the group of 4 weights, all weights are of weight $5 x$.
Returning to the original problem.
First, assume that among the 10 equal groups of weights, there are 2 groups with a different number of weights. Then, by the lemma and $N<50$, some groups consist of 4 weights each of weight $5 x$, and the remaining groups consist of 5 weights each of weight $4 x$. In this case, the weight of each group is $20 x$, so the total weight of the weights is $200 x$.
Since 200 is not a multiple of 11, these weights cannot be divided into 11 equal groups. This is a contradiction.
Therefore, the 10 equal groups of weights must have the same number of weights in each group. This implies $10 \mid N$.
If among the 11 equal groups, there are 2 groups with a different number of weights, then by the lemma, each group must have at least 4 weights, which implies $N \geqslant 44$. Since $N$ is a multiple of 10, we get $N \geqslant 50$, a contradiction.
If among the 11 equal groups, each group has the same number of weights, then $11 \mid N$, which implies $110 \mid N$. This contradicts the assumption $N<50$.
|
50
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $\sqrt{m^{2}+1203}$ is an integer, then the sum of all positive integers $m$ that satisfy the condition is ( ).
(A)401
(B) 800
(C) 601
(D) 1203
|
4.B.
Let $\sqrt{m^{2}+1203}=n\left(n \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
n^{2}-m^{2}=1203 \\
\Rightarrow(n+m)(n-m)=401 \times 3=1203 \times 1 .
\end{array}
$$
Therefore, $\left\{\begin{array}{l}n+m=401, \\ n-m=3\end{array}\right.$ or $\left\{\begin{array}{l}n+m=1203, \\ n-m=1 .\end{array}\right.$
Solving these, we get $m=199$ or $m=601$. Hence, the sum is 800.
|
800
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure $2, \odot O$ is internally tangent to $\odot O^{\prime}$ at point $P, \odot O$'s chord $A B$ is tangent to $\odot O^{\prime}$ at point $C$, and $A B / / O O^{\prime}$. If the area of the shaded part is $4 \pi$, then the length of $A B$ is $\qquad$
|
4.4.
As shown in Figure 10, connect $O^{\prime} C$ and $O A$. Draw $O D \perp A B$ at $D$. Then quadrilateral $O O^{\prime} C D$ is a rectangle.
Therefore, $O^{\prime} C=O D$.
Thus, $S_{\text {fill }}$
$=\pi O A^{2}-\pi O^{\prime} C^{2}$ $=\pi\left(O A^{2}-O D^{2}\right)=\pi A D^{2}=4 \pi$.
So, $A D=2$.
By the Perpendicular Diameter Theorem, $A B=2 A D=2 \times 2=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The digit at the 2007th position after the decimal point of the irrational number $0.2342343423434342343434342 \cdots$ is $\qquad$ .
|
(Observation: Note the position of the digit 2 after the decimal point. The $n$th 2 is at the $n^{2}$th position after the decimal point. Since $44^{2}<2007<45^{2}$, the digit at the 2007th position after the decimal point is between the 44th and 45th 2, with 44 instances of 34 in between. The digit 3 is in the odd positions, and $2007-1936=71$, so the digit at the 2007th position is 3.)
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given a height of 241, find the minimum value of the function
$$
S_{n}=|n-1|+2|n-2|+\cdots+100|n-100|
$$
$\left(n \in \mathbf{N}_{+}\right)$.
|
Solution: When $n \geqslant 100$,
$$
\begin{array}{l}
S_{n}=(n-1)+2(n-2)+\cdots+100(n-100) \\
=n(1+2+\cdots+100)-\left(1^{2}+2^{2}+\cdots+10^{2}\right) \\
\geqslant S_{100} .
\end{array}
$$
Therefore, the minimum value of $S_{n}$ can only be obtained when $1 \leqslant n \leqslant 100$.
When $2 \leqslant n \leqslant 100$,
$$
\begin{array}{l}
S_{n}-S_{n-1} \\
=|n-1|+|n-2|+\cdots+ \\
|n-100|-100|n-101| \\
=|n-1|+|n-2|+\cdots+ \\
|n-100|-100(101-n) \\
=|n-1|+|n-2|+\cdots+ \\
|n-100|-100 \times 101+100 n .
\end{array}
$$
Let the above expression be $f(n)$.
Thus, when $3 \leqslant n \leqslant 100$,
$$
\begin{array}{l}
f(n)-f(n-1)=2 n-2>0 . \\
\text { Hence, } f(2) < 0 .
\end{array}
$$
Notice that
$$
\begin{array}{c}
f(71)=70+69+\cdots+1+0+1+ \\
2+\cdots+29-100 \times 30 \\
=-800 .
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } f(2) < 0, \text{ and } S_{2} > S_{3} > \cdots > S_{70} > S_{71}, \\
S_{71} < S_{72} < \cdots < S_{99} < S_{100} .
\end{array}
$$
In summary, the minimum value of $S_{n}$ is obtained if and only if $n=71$.
Now, let's find the minimum value.
$$
\begin{array}{l}
S_{71}=|71-1|+2|71-2|+\cdots+ \\
71|71-71|+72|71-72|+\cdots+ \\
100|71-100| \\
=(1+2+\cdots+70) \times 71- \\
\left(1^{2}+2^{2}+\cdots+70^{2}\right)+ \\
\left(72^{2}+73^{2}+\cdots+100^{2}\right)- \\
71(72+73+\cdots+100) \\
=\frac{70 \times 71}{2} \times 71-\frac{70 \times 71 \times 141}{6} \times 2-71^{2}+ \\
\frac{100 \times 101 \times 201}{6}-\frac{172 \times 29}{2} \times 71 \\
=176435-233590-5041+ \\
338350 - 177074 \\
=99080 .
\end{array}
$$
|
99080
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find four distinct natural numbers such that the sum of any two of them can be divided by their difference. If the sum of the largest and smallest of these four numbers is to be minimized, what is the sum of the middle two numbers?
(3rd Hua Luogeng Cup)
|
Analysis: Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four numbers that meet the conditions, and $a_{1}8$, which contradicts the condition that $a_{1}+a_{4}$ is the smallest.
In summary, $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(2,3,4,6)$, the sum of the middle two numbers is 7. .
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If a natural number $N$ is appended to the right of any natural number, the resulting number can be divided by $N$ (for example, 2 appended to 35 results in 352, which is divisible by 2), then $N$ is called a "magic number". Among the natural numbers less than 130, how many magic numbers are there?
|
Analysis: To calculate how many magic numbers there are, we first need to clarify what constitutes a magic number. Although the problem provides the definition of a magic number, it is not convenient to directly use this definition to determine whether a number is a magic number. Therefore, we need to explore a simplified and equivalent condition.
Obviously, the key to simplification is how to accurately describe in mathematical language "a number appended to the right of a natural number." Imagine $P$ appended to the right of $Q$; what is the resulting number? How can this number be expressed in terms of $P$ and $Q$? This clearly depends on how many digits $P$ has. If $P$ is an $m$-digit number, then the number obtained by appending $P$ to the right of $Q$ is $Q \cdot 10^{m} + P$. Using this, we can derive a simplified and equivalent condition based on the definition of a magic number.
Solution: Let $P$ be a magic number, and assume it is an $m$-digit number.
According to the condition, for any natural number $Q$, we have
$$
P \mid\left(Q \cdot 10^{m} + P\right),
$$
which implies $P \mid Q \cdot 10^{m}$ (for any $Q \in \mathbf{N}$).
Taking $Q=1$, we get $P \mid 10^{m}$.
Conversely, if $P \mid 10^{m}$, then for any $Q \in \mathbf{N}$, we have $P \mid Q \cdot 10^{m}$.
Thus, $P \mid\left(Q \cdot 10^{m} + P\right)$.
Therefore, $P$ is an $m$-digit magic number if and only if $P \mid 10^{m}$.
(1) When $m=1$, $P \mid 10$, and $P$ is a one-digit number, so $P=1,2,5$.
(2) When $m=2$, $P \mid 10^{2}$, i.e., $P \mid\left(2^{2} \times 5^{2}\right)$, and $P$ is a two-digit number, so,
$$
P=2^{0} \times 5^{2}, 2^{1} \times 5^{1}, 2^{1} \times 5^{2}, 2^{2} \times 5^{1}.
$$
(3) When $m=3$, $P \mid 10^{3}$, i.e., $P \mid\left(2^{3} \times 5^{3}\right)$, and $P$ is a three-digit number, $P<130$, so,
$$
P=2^{0} \times 5^{3}, 2^{2} \times 5^{2}.
$$
In summary, there are $3+4+2=9$ magic numbers that meet the conditions.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initially 243 Hexagon $A B C D E F$ is inscribed in $\odot O$, $A F // D C, F E // C B, E D // B A, A B+B C=$ $2 C D$. Constructing squares on each of the six sides, the sum of the areas of these six squares is 2008. Find the perimeter of hexagon $A B C D E F$.
|
Solution: As shown in Figure 3, from $A F$
$/ / D C, F E / / C B, E D / /$
$B A$, we know that $A D, B E, C F$ are
all diameters of $\odot O$, thus
$$
\begin{array}{l}
D E=A B, \\
E F=B C, \\
F A=C D .
\end{array}
$$
Let $C D=x, A B=x-d$. Then
$$
\begin{array}{l}
B C=x+d, D E=x-d, \\
A F=x, E F=x+d .
\end{array}
$$
According to the problem, we have
$$
2(x-d)^{2}+2 x^{2}+2(x+d)^{2}=2008,
$$
which simplifies to $(x-d)^{2}+x^{2}+(x+d)^{2}=1004$.
Thus, $3 x^{2}+2 d^{2}=1004$.
Clearly, $x^{2}=\frac{1004-2 d^{2}}{3}1004$, so $x^{2}>200$. Therefore, $x \geqslant 15$.
Noting that $x$ is an even number, we have $x=16$ or 18.
If $x=16$, then $d^{2}=\frac{1004-3 x^{2}}{2}=118$, which clearly has no solution.
If $x=18$, then $d^{2}=\frac{1004-3 x^{2}}{2}=16$, so $d=4$. In this case, $A B=x-d=14, C D=x=18, B C$ $=x+d=22$.
Therefore, the perimeter of the hexagon $A B C D E F$ is
$$
\begin{array}{l}
A B+B C+C D+D E+E F+F A \\
=2(14+18+22)=108 .
\end{array}
$$
|
108
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
244 Given real numbers $a, b, c, d$ satisfy
$$
a+b+c+d=ab+ac+ad+bc+bd+cd=3 \text{. }
$$
Find the maximum real number $k$, such that the inequality
$$
a+b+c+2ab+2bc+2ca \geqslant k d
$$
always holds.
|
Given:
$$
\begin{aligned}
& (a+b+c+d)^{2}=3^{2} \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}+2 a b+2 a c+2 a d+ \\
& 2 b c+2 b d+2 c d=9 \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}=3 .
\end{aligned}
$$
Let \( y=(x-a)^{2}+(x-b)^{2}+(x-c)^{2} \).
Then \( y=3 x^{2}-2(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right) \).
Since \( y \geqslant 0 \) and the coefficient of \( x^{2} \) is \( 3 > 0 \), we have:
$$
\Delta=[-2(a+b+c)]^{2}-4 \times 3\left(a^{2}+b^{2}+c^{2}\right)
$$
\(\leqslant 0\).
Thus, \( 4(3-d)^{2}-4 \times 3\left(3-d^{2}\right) \leqslant 0 \), which simplifies to:
$$
2 d^{2}-3 d \leqslant 0 \text {. }
$$
Solving this, we get \( 0 \leqslant d \leqslant \frac{3}{2} \) (when \( a=b=c=1 \), \( d=0 \); when \( a=b=c=\frac{1}{2} \), \( d=\frac{3}{2} \)).
From the given, \( a+b+c=3-d \),
$$
\begin{array}{l}
a b+b c+c a=3-d(a+b+c) \\
=3-d(3-d)=3-3 d+d^{2} .
\end{array}
$$
Then \( a+b+c+2 a b+2 b c+2 c a \)
$$
\begin{array}{l}
=3-d+2\left(3-3 d+d^{2}\right):=2 d^{2}-7 d+9 \\
=2\left(\frac{3}{2}-d\right)^{2}+3\left(\frac{3}{2}-d\right)+2 d
\end{array}
$$
Since \( 0 \leqslant d \leqslant \frac{3}{2} \), we have:
$$
2\left(\frac{3}{2}-d\right)^{2}+3\left(\frac{3}{2}-d\right) \geqslant 0 \text {. }
$$
Therefore, \( a+b+c+2 a b+2 b c+2 c a \geqslant 2 d \).
When \( a=b=c=\frac{1}{2}, d=\frac{3}{2} \), the equality holds.
Thus, the largest real number \( k \) such that the inequality
$$
a+b+c+2 a b+2 b c+2 c a \geqslant k d
$$
always holds is \( k=2 \).
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given the function $f(x)=x+\frac{t}{x}(t>0)$ and the point $P(1,0)$, draw two tangent lines $P M$ and $P N$ from $P$ to the curve $y=f(x)$, with the points of tangency being $M$ and $N$.
(1) Let $|M N|=g(t)$, find the expression for the function $g(t)$.
(2) Does there exist a $t$ such that $M$, $N$, and $A(0,1)$ are collinear? If so, find the value of $t$; if not, explain the reason.
(3) Under the condition in (1), if for any positive integer $n$, there exist $m+1$ real numbers $a_{1}, a_{2}, \cdots, a_{m}, a_{m+1}$ in the interval $\left[2, n+\frac{64}{n}\right]$ such that the inequality
$$
g\left(a_{1}\right)+g\left(a_{2}\right)+\cdots+g\left(a_{m}\right)<g\left(a_{m+1}\right)
$$
holds, find the maximum value of $m$.
|
Solution: (1) The equation of the chord of tangents $MN$ passing through the external point $P(1,0)$ of the curve $y=x+\frac{t}{x}(t>0)$ is
$$
y=2 x+2 t \text {. }
$$
Substituting into $y=x+\frac{t}{x}(t>0)$, we get $x^{2}+2 t x-t=0$.
Thus, $x_{1}+x_{2}=-2 t, x_{1} x_{2}=-t \ldots$
Therefore, $g(t)=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}+\frac{t}{x_{1}}-x_{2}-\frac{t}{x_{2}}\right)^{2}}$
$$
=\sqrt{5}\left|x_{1}-x_{2}\right|=2 \sqrt{5 t^{2}+5 t}(t>0) \text {. }
$$
(2) When $M, N$ and $A(0,1)$ are collinear, substituting $A(0,1)$ into $y=2 x+2 t$, we get $t=\frac{1}{2}$.
(3) It is easy to see that $g(t)$ is an increasing function on the interval $\left[2, n+\frac{64}{n}\right]$, then
$$
\begin{array}{l}
m g(2) \leqslant g\left(a_{1}\right)+g\left(a_{2}\right)+\cdots+g\left(a_{m}\right) \\
<g\left(a_{m+1}\right) \leqslant g\left(n+\frac{64}{n}\right)
\end{array}
$$
holds for all positive integers $n$, i.e.,
$$
m<\sqrt{\frac{1}{6}\left[\left(n+\frac{64}{n}\right)^{2}+\left(n+\frac{64}{n}\right)\right]}
$$
holds for all positive integers $n$.
Since $n+\frac{64}{n} \geqslant 16$, we have
$$
\sqrt{\frac{1}{6}\left[\left(n+\frac{64}{n}\right)^{2}+\left(n+\frac{64}{n}\right)\right]} \geqslant \sqrt{45 \frac{1}{3}} \text {. }
$$
Since $m$ is a positive integer, we take $m=\left[\sqrt{45 \frac{1}{3}}\right]=6$.
When $m=6$, there exist
$$
a_{1}=a_{2}=\cdots=a_{m}=2, a_{m+1}=16 \text {, }
$$
satisfying the condition for all $n$, thus, the maximum value of $m$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
110 players participate in a table tennis tournament, with each pair of players playing one match. If player $i$ beats player $j$, player $j$ beats player $k$, and player $k$ beats player $i$, this is called a "triangle." Let $W_{i}$ and $L_{i}$ represent the number of wins and losses, respectively, for the $i$-th player. If player $i$ beats player $j$, then $L_{i} + W_{j} \geqslant 8$. Prove that this tournament has exactly 40 triangles.
(1994, Hong Kong Mathematical Olympiad)
|
Solution: Let the points $-v_{1}, v_{2}, \cdots, v_{10}$ represent the 10 players. If $v_{i}$ beats $v_{j}$, then draw a directed edge between the corresponding points: $v_{i} \rightarrow v_{j}$. Thus, we obtain a 10-vertex directed graph $G$, then
$$
\text { and } \begin{aligned}
d^{+}\left(v_{i}\right) & =W_{i}, d^{-}\left(v_{i}\right)=L_{i}, \\
& =9(i=1,2, \cdots, 10) .
\end{aligned}
$$
Assume $W_{1} \geqslant W_{2} \geqslant \cdots \geqslant W_{10}$. Then
$L_{1} \leqslant L_{2} \leqslant \cdots \leqslant L_{10}$.
Since graph $G$ has 45 edges, each edge produces one "out-degree" and one "in-degree", hence
$\sum_{t=1}^{10} W_{i}=\sum_{i=1}^{10} L_{i}=45$.
Thus, $W_{1} \geqslant 5, W_{10} \leqslant 4, L_{1} \leqslant 4, L_{10} \geqslant 5$.
By the problem statement, when $v_{i}$ beats $v_{j}$, we have $L_{i}+W_{j} \geqslant 8$. Then $W_{i}+L_{j} \leqslant 10$.
We will prove: $W_{1}, W_{2}, \cdots, W_{10}$ only take values from 5 and 4.
(1) Prove by contradiction: $W_{1}=5$.
Suppose $W_{1} \geqslant 6$, then $L_{1} \leqslant 3$. If $v_{j}$ is any person defeated by $v_{1}$, by $L_{1}+W_{j} \geqslant 8$, we get $W_{j} \geqslant 5$.
Also, by $W_{10} \leqslant 4$, we know $v_{10}$ does not belong to the people defeated by $v_{1}$. Thus, $v_{10}$ beats $v_{1}$. Therefore, the players defeated by $v_{1}$ (at least 6 people) belong to the set $\left\{v_{2}, v_{3}, \cdots, v_{9}\right\}$. Again, by $W_{10} \leqslant 4, L_{10} \geqslant 5$, we know the people who beat $v_{10}$ (at least 5 people) must be in the set $\left\{v_{2}, v_{3}, \cdots, v_{9}\right\}$, and there must be one person $v_{r}$ who is both defeated by $v_{1}$ and beats $v_{10}$.
By “$v_{1}$ beats $v_{r}$” we know $W_{r} \geqslant 5$, hence $L_{r} \leqslant 4$.
Also, by “$v_{r}$ beats $v_{10}$” we know $L_{r}+W_{10} \geqslant 8$, thus
$W_{10} \geqslant 4$.
Combining with $W_{10} \leqslant 4$ we get $W_{10}=4$.
Thus, the values of $W_{1}, W_{2}, \cdots, w_{10}$ are $W_{1} \geqslant 6,\left\{W_{2}, w_{3}, \cdots, w_{9}\right\}$ with at least six numbers greater than or equal to 5, and the rest greater than or equal to 4.
Since $W_{10}=4$, we have $\sum_{i=1}^{10} W_{i} \geqslant 48$, which is a contradiction.
Therefore, the assumption is false.
Thus, $w_{1} \leqslant 5$.
We have already proved $W_{1} \geqslant 5$, so $W_{1}=5$.
Using the same method, we get $L_{10}=5$, thus $W_{10}=4$.
Notice that $5=W_{1} \geqslant W_{2} \geqslant \cdots \geqslant W_{10}=4$.
Assume there are $k$ 5s and $10-k$ 4s.
By $45=5 k+4(10-k)$ we get $k=5$. Therefore,
$W_{1}=W_{2}=\cdots=W_{5}=5$,
$W_{6}=W_{7}=\cdots=W_{10}=4$.
Thus, $L_{1}=L_{2}=\cdots=L_{5}=4$,
$L_{6}=L_{7}=\cdots=L_{0}=5$.
(2) Call the triangle in the problem statement a “cyclic triangle”.
Notice in graph $G$,
the number of cyclic triangles +
the number of non-cyclic triangles
$=\mathrm{C}_{10}^{3}=120$.
First, find the number of non-cyclic triangles.
Since each pair of out-degrees (or in-degrees) from a point determines a non-cyclic triangle, and for any point $P$ in graph $G$, it either has 5 out-degrees and 4 in-degrees (or 5 in-degrees and 4 out-degrees), forming $\mathrm{C}_{5}^{2}$ pairs of "out-degrees", $\mathrm{C}_{4}^{2}$ pairs of "in-degrees" (or $\mathrm{C}_{5}^{2}$ pairs of in-degrees, $\mathrm{C}_{4}^{2}$ pairs of out-degrees), totaling $C_{5}^{2}+C_{4}^{2}=16$ non-cyclic triangles with $P$ as a vertex. If we do not count repetitions, 10 points produce 160 non-cyclic triangles.
Since each non-cyclic triangle contains one pair of out-degrees and one pair of in-degrees, it is counted twice: thus, there are $\frac{160}{2}=80$ non-cyclic triangles. Therefore, the number of cyclic triangles is $120-80=40$.
|
40
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Choose $n$ numbers from $1,2, \cdots, 9$, among which there must be several numbers (at least one, or all), the sum of which can be divisible by 10. Find the minimum value of $n$.
(2008, National Junior High School Mathematics Competition)
|
Solution: When $n \leqslant 4$, since no combination of $n$ numbers from $1,3,5,8$ can sum to a multiple of 10, we have $n \geqslant 5$.
Below, we prove that $n=5$ meets the requirement, i.e., we prove that from $1,2, \cdots, 9$, any 5 numbers chosen will always include some numbers whose sum is a multiple of 10.
We use proof by contradiction.
Assume that 5 numbers can be chosen such that no combination of these numbers sums to a multiple of 10. Divide $1,2, \cdots, 9$ into the following 5 groups:
$$
\{1,9\},\{2,8\},\{3,7\},\{4,6\},\{5\},
$$
Then the chosen numbers can only contain one number from each group, thus 5 must be chosen.
(1) If 1 is chosen from the first group, then 9 is not chosen. Since $1+5+4=10$, 4 is not chosen, thus 6 is chosen. Since $1+6+3=10$, 3 is not chosen, thus 7 is chosen. Since $1+7+2=10$, 2 is not chosen, thus 8 is chosen. But $5+7+8=20$ is a multiple of 10, a contradiction. (2) If 9 is chosen from the first group, then 1 is not chosen. Since $9+5+6=20$, 6 is not chosen, thus 4 is chosen. Since $9+4+7=20$, 7 is not chosen, thus 3 is chosen. Since $9+3+8=20$, 8 is not chosen, thus 2 is chosen. But $5+3+2=10$ is a multiple of 10, a contradiction. In summary, the minimum value of $n$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. As shown in Figure 5, in quadrilateral $A B C D$, $\angle A=$ $\angle B C D=90^{\circ}, B C=$ $C D, E$ is a point on the extension of $A D$. If $D E=A B$ $=3, C E=4 \sqrt{2}$, then the length of $A D$ is
|
11.5.
In Figure 5, connect $A C$. It is easy to prove that
$\triangle C D E \cong \triangle C B A, \angle A C E=90^{\circ}$.
Since $C A=C E=4 \sqrt{2}$, therefore, $A E=8$.
Thus, $A D=5$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Xiao Wang walks along the street at a constant speed, and he notices that a No. 18 bus passes him from behind every $6 \mathrm{~min}$, and a No. 18 bus comes towards him every $3 \mathrm{~min}$. Assuming that each No. 18 bus travels at the same speed, and the No. 18 bus terminal dispatches a bus at fixed intervals, then, the interval between dispatches is $\qquad$ min.
|
13.4 .
Let the speed of bus No. 18 be $x \mathrm{~m} / \mathrm{min}$, the walking speed of Xiao Wang be $y \mathrm{~m} / \mathrm{min}$, and the distance between two consecutive buses traveling in the same direction be $s \mathrm{~m}$. From the problem, we have
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s .
\end{array}\right.
$$
Solving these equations, we get $s=4 x$, i.e., $\frac{s}{x}=4$.
Therefore, the interval time for bus No. 18 to depart from the terminal is $4 \mathrm{~min}$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. There are 6 seats arranged in a row, and three people are to be seated, with exactly two adjacent empty seats. The number of different seating arrangements is ( ).
(A) 48
(B) 60
(C) 72
(D) 96
|
7.C.
Arrange the seats in order as $1,2, \cdots, 6$.
If the adjacent empty seats are 1 and 2 or 5 and 6, there are $\mathrm{C}_{3}^{1} \cdot \mathrm{A}_{3}^{3}$ arrangements each; if the adjacent empty seats are 2 and 3, 3 and 4, or 4 and 5, there are $\mathrm{C}_{2}^{1} \cdot \mathrm{A}_{3}^{3}$ arrangements each.
Therefore, the number of different arrangements where exactly two empty seats are adjacent is $2 \mathrm{C}_{3}^{1} \cdot \mathrm{A}_{3}^{3}+3 \mathrm{C}_{2}^{1} \cdot \mathrm{A}_{3}^{3}=72$.
|
72
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given 99 | $\overline{62 x y 427}$. Find $950 x+24 y+1$. (1997, Hope Cup Mathematics Invitational)
Find the value of $950 x+24 y+1$ if 99 divides $\overline{62 x y 427}$. (1997, Hope Cup Mathematics Invitational)
|
(提示: From $91 \overline{62 x y 427}$, we get $91(x+y+21)$. Therefore, $9 \mid(x+y+3)$. Noting that $3 \leqslant x+y+3 \leqslant$ 21, hence $x+y=6$ or 15. From $11 \overline{62 x y 427}$, we get $x-$ $y=-2$ or 9. Also note that $x+y$ and $x-y$ have the same parity, solving gives $(x, y)=(2,4)$. Therefore, $950 x+24 y+1=$ 1997.)
|
1997
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. For a real number $x$, $[x]$ denotes the greatest integer not exceeding the real number $x$. It is known that the sequence of positive numbers $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right),
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then
$$
\left[\frac{1}{S}+\frac{1}{S_{2}}+\cdots+\frac{1}{S_{100}}\right]=(\quad) \text {. }
$$
(A) 17
(B) 18
(C) 19
(D) 20
|
10. B.
Notice
$$
\begin{array}{l}
S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)=\frac{1}{2}\left[\left(S_{n}-S_{n-1}\right)+\frac{1}{S_{n}-S_{n-1}}\right], \\
S_{n}+S_{n-1}=\frac{1}{S_{n}-S_{n-1}}, \\
S_{n}^{2}=S_{n-1}^{2}+1 .
\end{array}
$$
Since $S_{1}=a_{1}=1$, we have $S_{n}^{2}=n, S_{n}=\sqrt{n}$. Therefore, $\sqrt{n}+\sqrt{n-1}\sqrt{101}-1>9 \Rightarrow S>18 .
\end{array}
$$
Also, since $S_{1}=a_{1}=1$, we have
$$
\begin{array}{l}
\frac{S}{2}-\frac{1}{2 S_{1}}=\frac{1}{2 S_{2}}+\frac{1}{2 S_{3}}+\cdots+\frac{1}{2 S_{100}} \\
<\sqrt{100}-1=9,
\end{array}
$$
which means $S<2\left(9+\frac{1}{2}\right)=19$.
Thus, $[S]=18$.
|
18
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. For a positive integer $n$, its decimal representation consists only of the digits 0 and 1, and it is divisible by 225. Find the minimum value of $n$.
|
(Tip: From $25 \mid n$, we know that the last two digits of $n$ are multiples of 25. But since all the digits of $n$ are 0 and 1, the last two digits of $n$ must be 00. From $91n$, we know that the sum of the digits of $n$ is a multiple of 9, so $n$ must contain at least 9 ones. Therefore, the smallest value of $n$ is 11111111100.)
|
11111111100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$1 . \underbrace{66 \cdots}_{100 \uparrow} \underbrace{77 \cdots 7}_{100 \uparrow} \underbrace{88 \cdots 8}_{100 \uparrow} 8$ divided by 72 has a remainder of
|
$$
\begin{array}{l}
\underbrace{66 \cdots 6}_{100 \uparrow} \underbrace{77 \cdots}_{100 \uparrow} \underbrace{88 \cdots 8}_{100 \uparrow} \\
=\underbrace{66 \cdots 677 \cdots}_{10 \uparrow} \underbrace{77}_{100 \uparrow} \underbrace{88 \cdots 82000}_{96 \uparrow}+72 \times 95+48 . \\
\underbrace{66 \cdots}_{100 \uparrow} \underbrace{77 \cdots}_{100 \uparrow} \underbrace{78 \cdots}_{96} 82000=a .
\end{array}
$$
Since the sum of the digits of $a$
$$
100 \times 6 + 100 \times 7 + 96 \times 8 + 2 = 2070
$$
is divisible by 9, thus $a$ is divisible by 9; also, the last three digits of $a$ are divisible by 8, so $a$ is also divisible by 8. Since 8 and 9 are coprime, therefore, $a$ is divisible by $8 \times 9 = 72$. Hence, the remainder of the original number when divided by 72 is 48.
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In trapezoid $A B C D$, $A D / / B C, A B=A C$, $B C=B D=(\sqrt{2}+1) C D$. Then the degree measure of $\angle B A C+\angle B D C$ is $\qquad$ .
|
2.180 .
As shown in Figure 5, draw $A E \perp B C$ at $E$ and $D F \perp B C$ at $F$. Then
$$
\begin{array}{l}
B E=E C=\frac{1}{2} B C, \\
\angle C A E=\frac{1}{2} \angle B A C,
\end{array}
$$
and quadrilateral $A E F D$ is a rectangle.
|
180
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $n$ is an integer, and the quadratic equation in $x$
$$
(n-1)^{2} x^{2}-5 n(n-1) x+\left(6 n^{2}-n-1\right)=0
$$
has at least one integer root. Then the sum of all possible values of $n$ is $\qquad$
|
$$
[(n-1) x-(2 n-1)][(n-1) x-(3 n+1)]=0 .
$$
Since $n \neq 1$, then
$$
\begin{array}{l}
x_{1}=\frac{2 n-1}{n-1}=2+\frac{1}{n-1}, \\
x_{2}=\frac{3 n+1}{n-1}=3+\frac{4}{n-1} .
\end{array}
$$
Because the original equation has at least one integer root, and $n$ is an integer, so, $n-1= \pm 1, \pm 2, \pm 4$.
Solving gives $n=2,0,3,-1,5,-3$.
Therefore, the sum of all $n$ values is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) In any permutation of $1,2, \cdots, 200$, there always exists a sum of 20 consecutive numbers that is not less than $a$. Find the maximum value of $a$.
|
Three, take any permutation, and divide it into 10 groups, each containing 20 numbers. Let $A_{1}=\left(b_{1}, b_{2}, \cdots, b_{20}\right), A_{2}=\left(b_{21}, b_{2}\right.$, $\left.\cdots, b_{40}\right), \cdots, A_{10}=\left(b_{181}, b_{182}, \cdots, b_{200}\right)$.
Take the average of the 10 groups of numbers
$$
\bar{A}=\frac{1+2+\cdots+200}{10}=2010 \text {. }
$$
Obviously, at least one of $A_{1}, A_{2}, \cdots, A_{10}$ is not less than $\bar{A}$. At this point, it is always possible to find a sum of 20 consecutive numbers that is not less than 2010.
Below, construct a permutation of $1, 2, \cdots, 200$ such that the sum of any 20 consecutive numbers is not greater than 2010.
Construct a $10 \times 20$ number table: place 200, 199, $\cdots, 101$ sequentially in the first $10 \times 10$ number table (large numbers on the left (top) side of smaller numbers), and place $1, 2, \cdots, 100$ sequentially in the second $10 \times 10$ number table (small numbers on the right (top) side of larger numbers).
Arrange the numbers in this table from left to right and from top to bottom into a sequence, which is the permutation that meets the requirements of the problem (the sum of any 20 consecutive numbers is one of $2010-10 i(i=0,1, \cdots, 10)$).
Thus, the maximum value of $a$ is 2010.
Note: The constructive part of the proof in this problem was provided by Mr. Song Qiang from our editorial department.
|
2010
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9}$, $x \in[0,1]$. Provide the following conclusions:
(1) $f(x)>0$;
(2) $f(x)<0$;
(3) There exists $x_{0} \in[0,1)$, such that $f\left(x_{0}\right)=0$;
(4) There exists $x_{0} \in[0,1]$, such that $f\left(x_{0}\right)<0$.
Among them, the correct conclusion numbers are $\qquad$.
|
2. (1)
$$
\begin{array}{l}
f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9} \\
=x^{3}(1-x)+3 x^{3}-3 x^{2}+(x-1)^{2}+\frac{4}{9} \\
=x^{3}(1-x)+(x-1)^{2}+\frac{3}{2} x^{3}+\frac{3}{2} x^{3}+\frac{4}{9}-3 x^{2} \\
\geqslant x^{3}(1-x)+(x-1)^{2}+3 \sqrt[3]{\frac{3}{2} x^{3} \cdot \frac{3}{2} x^{3} \cdot \frac{4}{9}}-3 x^{2} \\
=x^{3}(1-x)+(x-1)^{2} \geqslant 0 .
\end{array}
$$
The first equality holds when $x=\frac{2}{3}$, and the second equality holds when $x=1$.
Therefore, the equalities cannot hold simultaneously.
Thus, $f(x)>0$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $x$ is a
four-digit number, the sum of its digits is $y$. If the value of $\frac{x}{y}$ is minimized, then $x=$ $\qquad$
|
4.1099.
Let $x=\overline{a_{1} a_{2} a_{3} a_{4}}$, where $a_{1} \in \mathbf{N}_{+}, a_{2}, a_{3}, a_{4} \in \mathbf{N}$. Therefore,
$$
\begin{array}{l}
\frac{x}{y}=\frac{1000 a_{1}+100 a_{2}+10 a_{3}+a_{4}}{a_{1}+a_{2}+a_{3}+a_{4}} . \\
=1+9 \times \frac{111 a_{1}+11 a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}+a_{4}} .
\end{array}
$$
To make $\frac{x}{y}$ the smallest, then $a_{4}=9$, at this time,
$$
\begin{array}{l}
\frac{x}{y}=1+9 \times \frac{111 a_{1}+11 a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}+9} \\
=10+9 \times \frac{110 a_{1}+10 a_{2}-9}{a_{1}+a_{2}+a_{3}+9} ;
\end{array}
$$
To make $\frac{x}{y}$ the smallest, then $a_{3}=9$, at this time,
$$
\begin{array}{l}
\frac{x}{y}=10+9 \times \frac{110 a_{1}+10 a_{2}-9}{a_{1}+a_{2}+18} \\
=100+9 \times \frac{100 a_{1}-189}{a_{1}+a_{2}+18} .
\end{array}
$$
If $a_{1}=1$, then $\frac{x}{y}=100-9 \times \frac{89}{a_{2}+19}$.
To make $\frac{x}{y}$ the smallest, then $a_{2}=0$, at this time,
$$
x=1099, \frac{x}{y}=\frac{1099}{19} \text {. }
$$
If $a_{2}>1$, to make $\frac{x}{y}$ the smallest, then $a_{2}=9$, at this time,
$$
\begin{array}{l}
\frac{x}{y}=100+9 \times \frac{100 a_{1}-189}{a_{1}+27} \\
=1000-9 \times \frac{9 \times 3 \times 107}{a_{1}+27} .
\end{array}
$$
To make $\frac{x}{y}$ the smallest, then $a_{1}=2$, at this time,
$$
x=2999, \frac{x}{y}=\frac{2999}{29} \text {. }
$$
Since $\frac{2999}{29}>\frac{1099}{19}$, therefore, $x=1099$.
|
1099
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. As shown in Figure 3, there are 16 points, and the distance between any two adjacent points, whether left and right or up and down, is equal to 1. If these points are used as the vertices of triangles, then, a total of right-angled triangles can be obtained.
|
5.200.
As shown in Figure 6, the number of right-angled triangles with $A$ as the right-angle vertex is $C_{3}^{1} \cdot C_{3}^{1}=9$;
the number of right-angled triangles with $B$ as the right-angle vertex is
$$
C_{3}^{1} \cdot C_{3}^{1}+C_{2}^{1}+1=12 \text {; }
$$
the number of right-angled triangles with $C$ as the right-angle vertex is
$$
C_{3}^{1} \cdot C_{3}^{1}+C_{2}^{1} \cdot C_{3}^{1}+2=17 \text {; }
$$
By symmetry, the total number of right-angled triangles is
$$
9 \times 4+12 \times 8+17 \times 4=200
$$
right-angled triangles.
|
200
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $f(x)=x^{2}+2 x+1$, there exists a real number $t$ such that when $x \in[1, m]$, $f(x+t) \leqslant x$ always holds, then the maximum value of $m$ is $\qquad$ .
|
2.4.
Translate the graph of $f(x)$ to the right by $-t$ units, and from the graphical analysis, the maximum value of $m$ is the larger of the x-coordinates of the two intersection points of $\left\{\begin{array}{l}y=x, \\ y=f(x+t)\end{array}\right.$
From $f(1+t)=1$, we get $t=-1, t=-3$.
Then, from $f(x-3)=x$, we have
$x=1$ (discard), $x=4$.
Therefore, the maximum value of $m$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $n$ be a positive integer. If
$$
A(n)=1986^{n}+1987^{n}+1988^{n}+1989^{n}
$$
is not divisible by 5, find $n$.
|
(Tip: Only consider the unit digit $G(A(n))$ of $A(n)$. When $n=4 k\left(k \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
G(A(n))=G\left(6^{n}+7^{n}+8^{n}+9^{n}\right) \\
=G\left(6^{4}+7^{4}+8^{4}+9^{4}\right) \\
=G(6+1+6+1)=4,
\end{array}
$$
at this time, 5 does not divide $A(n)$;
$$
\begin{array}{l}
\text { When } n=4 k+r(k \in \mathbf{N}, 1 \leqslant r \leqslant 3) \text {, } \\
G(A(n))=G\left(6^{n}+7^{n}+8^{n}+9^{n}\right) \\
=G\left(6^{r}+7^{\prime}+8^{r}+9^{r}\right)=0,
\end{array}
$$
contradiction.)
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) On an infinite grid paper, some cells are colored red, and the rest are colored blue. In every $2 \times 3$ rectangle of six cells, there are exactly two red cells. How many red cells are there in a $9 \times 11$ rectangle of 99 cells?
|
II. 33 Red Squares.
As shown in Figure 3, take any red square $K_{0}$ as the center of a $3 \times 3$ square; it is not allowed to color $K$ red. If $K$ is colored red, then in the $2 \times 3$ rectangles $A F H D$, $A B S T$, and $M N C D$, there will be two red squares each. To ensure that the shape $B C G E$ contains two red squares, no matter where the red square is placed, it will result in one of the aforementioned three rectangles having 3 red squares. This means that the red square cannot share a side with $K_{0}$, but must be on its diagonal. This is feasible overall (as shown in Figure 4).
Therefore, each $3 \times 3$ square contains exactly 3 red squares.
Furthermore, in a $9 \times 11$ rectangle, it can be divided into nine $3 \times 3$ squares and three $2 \times 3$ rectangles, so there are a total of $9 \times 3 + 3 \times 2 = 33$ red squares.
|
33
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. There are 10 positive integers arranged from smallest to largest: $1, 4, 8$, $10, 16, 19, 21, 25, 30, 43$. How many groups of consecutive numbers have a sum that is divisible by 11?
|
Let 10 positive integers be $a_{1}, a_{2}$, $\cdots, a_{10}$, and let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Then
$$
a_{i+1}+a_{i+2}+\cdots+a_{j}=S_{j}-S_{i} .
$$
The sequence $a_{n}(n=1,2, \cdots, 10)$ has remainders modulo 11 of $1,4,-3,-1,5,-3,-1,3,-3,-1$; the sum of the first $n$ terms $S_{n}$ has remainders modulo 11 of $1,5,2$, $1,6,3,2,5,2,1$. Therefore,
$$
\begin{array}{l}
S_{1} \equiv S_{4} \equiv S_{10} \equiv 1(\bmod 11), \\
S_{2} \equiv S_{8} \equiv 5(\bmod 11), \\
S_{3} \equiv S_{7} \equiv S_{9} \equiv 2(\bmod 11) .
\end{array}
$$
Thus, there are $3+1+3=7$ pairs $(i, j)$ such that $11 \mid \left(S_{j}-S_{i}\right)$, i.e., $11 \mid \left(a_{i+1}+a_{i+2}+\cdots+a_{j}\right)$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If $p$ and $q$ are both prime numbers, and $7p + q$ and $pq + 11$ are also prime numbers, find the value of $p^q + q^p$.
(1997, Jingzhou City, Hubei Province, Junior High School Mathematics Competition)
|
(Tip: Since $7 p+q$ is a prime number, and $7 p+q>7$, therefore, $7 p+q$ is odd. Thus, $p, q$ do not share the same parity.
If $p$ is even, then $p=2$. According to the problem, $14+$ $q, 2 q+11$ are both prime numbers. Therefore,
$$
14+q \neq \equiv 0(\bmod 3), 2 q+11 \not \equiv 0(\bmod 3),
$$
which means $q \neq 1(\bmod 3), q \neq 2(\bmod 3)$.
Hence, $q \equiv 0(\bmod 3)$. Thus, $q=3$.
At this point, $p^{q}+q^{p}=2^{3}+3^{2}=17$.
If $q$ is even, then $q=2$. According to the problem, $7 p+$ $2,2 p+11$ are both prime numbers, so,
$$
7 p+2 \neq 0(\bmod 3), 2 p+11 \neq \equiv 0(\bmod 3) \text {, }
$$
which means $p \neq 1(\bmod 3), p \neq 2(\bmod 3)$.
Hence, $p \equiv 0(\bmod 3)$. Thus, $p=3$.
At this point, $p^{q}+q^{p}=2^{3}+3^{2}=17$.)
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If
$$
\frac{4^{5}+4^{5}+4^{5}+4^{5}}{3^{5}+3^{5}+3^{5}} \times \frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5}}{2^{5}+2^{5}}=2^{n},
$$
then, $n=$ . $\qquad$
|
$-1.12$
The left side of the original equation can be transformed into
$$
\frac{4 \times 4^{5}}{3 \times 3^{5}} \times \frac{6 \times 6^{5}}{2 \times 2^{5}}=\frac{4^{6}}{3^{6}} \times \frac{6^{6}}{2^{6}}=4^{6}=2^{12} .
$$
Thus, $2^{12}=2^{n}$. Therefore, $n=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let the integer $a$ divided by 7 leave a remainder of 3, and the integer $b$ divided by 7 leave a remainder of 5. If $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7.
(1994, Tianjin City Junior High School Mathematics Competition)
|
Solution: Let $a=7 m+3, b=7 n+5$. Then
$$
\begin{array}{l}
a^{2}-4 b=(7 m+3)^{2}-4(7 n+5) \\
=49 m^{2}+42 m+9-28 n-20 \\
=7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{array}
$$
Therefore, the remainder when $a^{2}-4 b$ is divided by 7 is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $p$ is a prime number, and the equation
$$
x^{2}+p x-444 p=0
$$
has two integer roots. Then $p=$ $\qquad$
|
3.37.
According to
$$
x^{2}=p(444-x)
$$
we know that $p(444-x)$ is a perfect square.
Since $p$ is a prime number, it follows that $p\left|x^{2} \Rightarrow p\right| x$.
Let $x=n p(n \in \mathbf{Z})$ and substitute into equation (1) to get
$$
(n p)^{2}=p(444-n p) .
$$
Since $p \neq 0$, we have $n^{2} p=444-n p$, which is
$$
n(n+1) p=2^{2} \times 3 \times 37 \text {. }
$$
Therefore, $p=37$.
|
37
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $a, b, c$ simultaneously satisfy $a-7b+8c=4$ and $8a+4b-c=7$. Then, $a^{2}-b^{2}+c^{2}=$ $\qquad$ .
|
4.1.
According to the conditions $a+8c=4+7b, 8a-c=7-4b$. Squaring both sides of the two equations and then adding them yields
$$
\begin{array}{l}
(a+8c)^{2}+(8a-c)^{2} \\
=(7+4b)^{2}+(7-4b)^{2} .
\end{array}
$$
Simplifying and organizing, we get $65\left(a^{2}+c^{2}\right)=65\left(1+b^{2}\right)$.
Therefore, $a^{2}-b^{2}+c^{2}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In trapezoid $A B C D$, $A B / / C D$, the base angles $\angle D A B=36^{\circ}, \angle C B A=54^{\circ}, M$ and $N$ are the midpoints of sides $A B$ and $C D$, respectively. If the lower base $A B$ is exactly 2008 units longer than the upper base $C D$, then the line segment $M N=$ $\qquad$
|
5.1004.
As shown in Figure 4, draw $N S / / A D$ and $N T / / C B$ through point $N$.
Then, from $\square A S N D$ and $\square B T N C$, we get
$D N=A S, N C=T B$,
and
$$
\begin{array}{l}
\angle N S T=\angle D A B=36^{\circ}, \\
\angle N T S=\angle C B A=54^{\circ} . \\
\text { Hence, } \angle S N T=180^{\circ}-(\angle N S T+\angle N T S) \\
=180^{\circ}-\left(36^{\circ}+54^{\circ}\right)=90^{\circ}
\end{array}
$$
We can determine that $\triangle N S T$ is a right triangle.
Noting that $A S=D N=N C=T B$, we get
$$
\begin{array}{l}
S T=A B-(A S+T B) \\
=A B-(D N+N C) \\
=A B-D C=2008 .
\end{array}
$$
Therefore, $M N=\frac{1}{2} S T=1004$.
|
1004
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Calculate: $1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+$ $\cdots+\frac{1}{20}(1+2+\cdots+20)$.
|
$\begin{array}{l}\text { 1. Original expression }=1+\frac{1}{2} \times \frac{2 \times 3}{2}+\frac{1}{3} \times \frac{3 \times 4}{2}+ \\ \cdots+\frac{1}{20} \times \frac{20 \times 21}{2} \\ =\frac{1}{2}(2+3+4+\cdots+21) \\ =\frac{1}{2} \times \frac{23 \times 20}{2}=115 .\end{array}$
|
115
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let real numbers $x, y, z$ simultaneously satisfy
$$
\begin{array}{l}
x^{3}+y=3 x+4, \\
2 y^{3}+z=6 y+6, \\
3 z^{3}+x=9 z+8 .
\end{array}
$$
Try to find the value of $2008(x-1)^{2}+2009(y-1)^{2}+$ $2010(z-1)^{2}$.
|
4. Transform to get
$$
\begin{array}{l}
y-2=-x^{3}+3 x+2=-(x-2)(x+1)^{2}, \\
z-2=-2 y^{3}+6 y+4=-2(y-2)(y+1)^{2}, \\
x-2=-3 z^{3}+9 z+6=-3(z-2)(z+1)^{2} .
\end{array}
$$
Multiplying the above three equations yields
$$
\begin{array}{l}
(x-2)(y-2)(z-2) \\
=-6(x-2)(y-2)(z-2) . \\
(x+1)^{2}(y+1)^{2}(z+1)^{2} .
\end{array}
$$
It can only be that $(x-2)(y-2)(z-2)=0$.
Without loss of generality, let $x=2$. Then we get $x=y=z=2$.
Therefore, $2008(x-1)^{2}+2009(y-1)^{2}+2010(z-1)^{2}$ $=6027$.
|
6027
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In an equilateral triangle $\triangle A B C$ with side length $2 \sqrt{7}$, $M$ is the midpoint of side $B C$, and $P$ is a point on $A C$.
(1) For what value of $P C$ is $B P+P M$ minimized?
(2) Find the minimum value of $B P+P M$.
|
7. As shown in Figure 8, connect $A M$, and flip $\triangle A B C$ along the axis $A C$ by $180^{\circ}$ to get $\triangle A D C$. Let $N$ be the symmetric point of $M$ (about $A C$). At this point, we always have $P M = P N$. Since $B$ and $N$ are fixed points, $B P + P N \geqslant B N$, with equality holding if and only if point $P$ lies on the side $B N$.
To find the value of $P C$, extend $B C$ to $E$ such that $C E = C N$, forming an equilateral $\triangle C E N$. By the construction, we have
$$
P C \parallel N E, \frac{P C}{N E} = \frac{B C}{B E}, \text{ i.e., } \frac{P C}{\sqrt{7}} = \frac{2 \sqrt{7}}{3 \sqrt{7}}.
$$
Thus, when $P C = \frac{2 \sqrt{7}}{3}$, $B P + P M$ achieves its minimum value.
Draw $N F \perp C E$, with $F$ as the foot of the perpendicular. Then $N F = \frac{\sqrt{21}}{2}$, $B F = \frac{5 \sqrt{7}}{2}$,
$$
B N = \sqrt{B F^{2} + F N^{2}} = \sqrt{\frac{175}{4} + \frac{21}{4}} = 7
$$
Therefore, the minimum value of $B P + P M$ is 7.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 There is a sequence of numbers: $1,3,4,7,11,18, \cdots$, starting from the third number, each number is the sum of the two preceding numbers.
(1) What is the remainder when the 1991st number is divided by 6?
(2) Group the above sequence as follows:
$(1),(3,4),(7,11,18), \cdots$,
where the $n$-th group has exactly $n$ numbers. What is the remainder when the sum of the numbers in the 1991st group is divided by 6?
(6th Spring Festival Cup Mathematics Competition)
|
Solution: Let the $n$-th number of the sequence be $a_{n}$. Then
$$
a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3) .
$$
Through experimentation, it is known that the remainders of $a_{1}, a_{2}, \cdots, a_{26}$ when divided by 6 are
$$
\begin{array}{l}
1,3,4,1,5,0,5,5,4,3,1,4,5, \\
3,2,5,1,0,1,1,2,3,5,2,1,3 .
\end{array}
$$
Notice that $a_{25} \equiv a_{1}(\bmod 6), a_{26} \equiv a_{2}(\bmod 6)$, and using $a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$, we get
$$
a_{27}=a_{26}+a_{25} \equiv a_{2}+a_{1}=a_{3}(\bmod 6) .
$$
Continuing this way, when $n>24$, we have
$$
a_{n} \equiv a_{n-24}(\bmod 6) \text {. }
$$
(1) Since $1991=24 \times 82+23$, we have
$$
a_{1991} \equiv a_{23} \equiv 5(\bmod 6) \text {. }
$$
Thus, the remainder when the 1991st number is divided by 6 is 5.
(2) When $n>24$, $a_{n} \equiv a_{n-24}(\bmod 6)$, so for any natural number $n$, we have
$$
\begin{array}{l}
a_{n+1}+a_{n+2}+\cdots+a_{n+x} \equiv a_{1}+a_{2}+\cdots+a_{21} \\
\equiv 1+3+4+1+5+0+5+5+4+ \\
3+1+4+5+3+2+5+1+0+ \\
1+1+2+3+5+2 \\
=66 \equiv 0(\bmod 6) .
\end{array}
$$
Since the first 1990 groups contain a total of $1+2+\cdots+1990$ $=1981045$ (let this be $m$) numbers, the sum of the numbers in the 1991st group is
$$
\begin{array}{l}
a_{m+1}+a_{m+2}+\cdots+a_{m+1991} \\
\equiv\left(a_{m+1}+a_{m+2}+\cdots+a_{m+1991}+a_{m+192}\right)- \\
\quad a_{m+192} \\
\equiv 83 \times 0-a_{m+1922} \equiv-a_{13} \equiv-5 \equiv 1(\bmod 6) .
\end{array}
$$
Thus, the remainder when the sum of the numbers in the 1991st group is divided by 6 is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (15 points) The function $f(x)$ defined on the interval $[0,1]$ satisfies $f(0)=f(1)=0$, and for any $x_{1}, x_{2} \in [0,1]$, we have
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant f\left(x_{1}\right)+f\left(x_{2}\right) \text {. }
$$
(1) Prove: For any $x \in [0,1]$, we have
$$
f(x) \geqslant 0 \text {; }
$$
(2) Find the value of $f\left(\frac{3}{4}\right)$;
(3) Calculate $f\left(\frac{1}{2}\right)+f\left(\frac{1}{4}\right)+\cdots+f\left(\frac{1}{2^{k}}\right)+$
$$
\cdots+f\left(\frac{1}{2^{2 \alpha B}}\right) .
$$
|
(1) Take $x_{1}=x_{2}=x \in[0,1]$, then
$$
f\left(\frac{2 x}{2}\right) \leqslant f(x)+f(x),
$$
i.e., $f(x) \leqslant 2 f(x)$.
Thus, $f(x) \geqslant 0$.
Therefore, for any $x \in[0,1]$, $f(x) \geqslant 0$.
(2) From $f(0)=f(1)=0$, we get
$$
f\left(\frac{0+1}{2}\right) \leqslant f(0)+f(1)=0+0=0.
$$
Thus, $f\left(\frac{1}{2}\right) \leqslant 0$.
But from (1), we know $f\left(\frac{1}{2}\right) \geqslant 0$, so $f\left(\frac{1}{2}\right)=0$.
From $f\left(\frac{1}{2}\right)=0, f(1)=0$, we get
$$
f\left(\frac{\frac{1}{2}+1}{2}\right) \leqslant f\left(\frac{1}{2}\right)+f(1)=0+0=0.
$$
Thus, $f\left(\frac{3}{4}\right) \leqslant 0$.
From (1), we know $f\left(\frac{3}{4}\right) \geqslant 0$, so $f\left(\frac{3}{4}\right)=0$.
(3) From $f(0)=0, f\left(\frac{1}{2}\right)=0$, we get
$$
f\left(\frac{0+\frac{1}{2}}{2}\right) \leqslant f(0)+f\left(\frac{1}{2}\right)=0+0=0.
$$
Thus, $f\left(\frac{1}{4}\right) \leqslant 0$.
But from (1), we know $f\left(\frac{1}{4}\right) \geqslant 0$, so
$$
f\left(\frac{1}{4}\right)=f\left(\frac{1}{2^{2}}\right)=0.
$$
Continuing this process, we get
$$
f\left(\frac{1}{2^{k}}\right)=0(k=1,2, \cdots, 2008).
$$
Therefore, $f\left(\frac{1}{2}\right)+f\left(\frac{1}{4}\right)+\cdots+f\left(\frac{1}{2^{k}}\right)+\cdots$
$$
+f\left(\frac{1}{2^{2008}}\right)=0.
$$
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 A positive integer, if it can be expressed as the difference of squares of two positive integers, is called a "wise number". Arrange all the wise numbers in ascending order. Find the wise number at the 2009th position.
|
Solution: Let $n$ be any positive integer.
When $n=2 k+1\left(k \in \mathbf{N}_{+}\right)$, we have $n=2 k+1=(k+1)^{2}-k^{2}$.
Thus, every odd number greater than 1 is a wise number.
When $n=4 k\left(k \in \mathbf{N}_{+}, k \geqslant 2\right)$, we have
$$
n=4 k=(k+1)^{2}-(k-1)^{2} \text {. }
$$
Thus, every number greater than 4 and divisible by 4 is a wise number.
When $n=4 k-2\left(k \in \mathbf{N}_{+}\right)$, let $n=p^{2}-q^{2}$.
Then $2(2 k-1)=(p+q)(p-q)$.
Thus, $(p+q)(p-q)$ is even.
Since $p+q$ and $p-q$ have the same parity, both $p+q$ and $p-q$ are even. Therefore, $4 \mid (p+q)(p-q)$. Hence, $4 \mid 2(2 k-1)$, which implies $2 \mid (2 k-1)$, a contradiction.
Therefore, every number of the form $4 k-2\left(k \in \mathbf{N}_{+}\right)$ is not a wise number.
If all wise numbers are arranged in ascending order and, except for the first wise number, the rest are grouped into sets of 3:
$$
3,(5,7,8),(9,11,12), \cdots
$$
According to the above discussion, the last number in each set is a multiple of 4, so the $k$-th set is
$$
(4 k+1,4 k+3,4 k+4) \text {. }
$$
Since $2009-1=3 \times 669+1$, the 2009th position in the sequence is the first number of the 670th set, which is $4 \times 670+1=2681$.
|
2681
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Calculate:
(1) $\sqrt{627953481}+\sqrt{672935481}=$ $\qquad$
(2) $\sqrt{\sqrt{254817369}-\sqrt{152843769}}=$
$\qquad$ .
|
2. (1) $51000(2) 60$
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Find the smallest positive integer $n$, such that the last three digits of $n^{3}$ are 888.
|
Solution: Since the unit digit of $n^{3}$ is 8, then the unit digit of $n$ is 2. Let $n=10 k+2$. Then
$$
\begin{array}{l}
n^{3}=(10 k+2)^{3}=1000 k^{3}+600 k^{2}+120 k+8 \\
=1000 k^{3}+100 k(6 k+1)+20 k+8 .
\end{array}
$$
According to the problem, we know that the last three digits of $100\left(6 k^{2}+k\right)+20 k+8$ are 888, so the last two digits of $20 k+8$ are 88.
Therefore, $20 k=100 m+80$, which means $k=5 m+4$.
Thus, $100\left(6 k^{2}+k\right)+20 k+8$
$$
\begin{array}{l}
=100(5 m+4)(30 m+25)+20(5 m+4)+8 \\
=15000 m^{2}+24600 m+10088 .
\end{array}
$$
Therefore, the unit digit of $6 m$ is 8. Hence, the smallest value of $m$ is 3. Consequently, the smallest value of $k$ is 19. Therefore, the smallest value of $n$ is 192.
|
192
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Sure, here is the translated text:
```
II. (20 points) Find all positive integers $n$ such that
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{4}\right]+\left[\frac{n}{5}\right]=69 \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
```
|
$$
\begin{array}{l}
\frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}-4 \\
<\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{4}\right]+\left[\frac{n}{5}\right] \\
\leqslant \frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5} . \\
\text { Therefore, } \frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}-4 \\
<69 \leqslant \frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}
\end{array}
$$
Thus, $53<n<57$. Hence, $n=54,55,56$. Upon inspection, $n=55$ satisfies the condition.
|
55
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) For a set $S\left(S \subseteq \mathbf{N}_{+}\right)$, if for any $x \in S, \dot{x}$ cannot divide the sum of elements of any non-empty subset of $S \backslash\{x\}$, then $S$ is called a "good set" ($S \backslash\{x\}$ represents the set $S$ after removing the element $x$).
(1) If $\{3,4, n\}$ is a good set, find the minimum value $n_{0}$ of $n$;
(2) Prove: the set $\left\{3,4, n_{0}, m\right\}$ is not a good set, where $m \in \mathbf{N}_{+}, m \neq 3,4, n_{0}$.
|
Three, (1) Obviously, $n>4$.
If $n=5$, then $4 \mid (3+5)$; if $n=6$, then $3 \mid 6$; if $n=7$, then $7 \mid (3+4)$; if $n=8$, then $4 \mid 8$; if $n=9$, then $3 \mid 9$.
When $n=10$, it is verified that $\{3,4,10\}$ is a good set.
Therefore, the minimum value of $n$ is $n_{0}=10$.
(2) If $\{3,4,10, m\}$ is a good set, then $m$ is not a multiple of 3, and 3 cannot divide $10+m$, i.e., $m$ is not of the form $3k+2$. 3 also cannot divide $10+4+m$, i.e., $m$ is not of the form $3k+1$. Therefore, for any positive integer $m (m \neq 3,4,10)$, the set $\{3,4,10, m\}$ is not a good set.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Reverse the order of the four digits of a four-digit number, and add the resulting number to the original number. If the sum obtained is divisible by 35, then this four-digit number is called a "good number". Then, among all four-digit numbers, the number of good numbers is ( ).
(A) 234
(B) 252
(C) 270
(D) 369
|
3. A.
Let the four-digit number be $\overline{a_{1} a_{2} a_{3} a_{4}}$. Then
$$
\begin{array}{l}
A=\overline{a_{1} a_{2} a_{3} a_{4}}+\overline{a_{4} a_{3} a_{2} a_{1}} \\
=1001\left(a_{1}+a_{4}\right)+110\left(a_{2}+a_{3}\right) .
\end{array}
$$
Since $1001=7 \times 143$, we have
$$
\begin{array}{l}
7 \mid A \Leftrightarrow 7 \mid 110\left(a_{2}+a_{3}\right) \Leftrightarrow 7 \mid \left(a_{2}+a_{3}\right), \\
5 \mid A \Leftrightarrow 5 \mid \left(a_{1}+a_{4}\right) . \\
\text { Then }\left(a_{1}, a_{4}\right) \\
=(1,4),(1,9),(2,3),(2,8),(3,2) \text {, } \\
(3,7),(4,1),(4,6),(5,0),(5,5) \text {, } \\
(6,4),(6,9),(7,3),(7,8),(8,2) \text {, } \\
(8,7),(9,1),(9,6) \text {. } \\
\left(a_{2}, a_{3}\right) \\
=(0,7),(1,6),(2,5),(3,4),(4,3) \text {, } \\
(5,2),(5,9),(6,1),(6,8),(7,0) \text {, } \\
(7,7),(8,6),(9,5) \text {. } \\
\end{array}
$$
Therefore, the number of numbers that satisfy the conditions is $18 \times 13=234$.
|
234
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Given Theorem: "If three prime numbers $a, b, c$ greater than 3 satisfy the equation $2a + 5b = c$, then $a + b + c$ is a multiple of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion.
(1997, National Junior High School Mathematics League)
|
Solution: Let $a=3 k_{1}+r_{1}, b=3 k_{2}+r_{2}$. Then
$$
\begin{array}{l}
a+b+c=3(a+2 b) \\
=3\left(3 k_{1}+r_{1}+6 k_{2}+2 r_{2}\right) \\
=9\left(k_{1}+2 k_{2}\right)+3\left(r_{1}+2 r_{2}\right) .
\end{array}
$$
Since $a, b$ are both primes greater than 3, we have
$$
r_{1} r_{2} \neq 0.
$$
If $r_{1} \neq r_{2}$, then $r_{1}=1, r_{2}=2$, or $r_{1}=2, r_{2}=1$. In this case,
$$
\begin{aligned}
c & =2 a+5 b=6 k_{1}+2 r_{1}+15 k_{2}+5 r_{2} \\
& =3\left(2 k_{1}+5 k_{2}+r_{2}\right)+2\left(r_{1}+r_{2}\right) \\
& =3\left(2 k_{1}+5 k_{2}+r_{2}\right)+6,
\end{aligned}
$$
which contradicts the fact that $c$ is a prime. Therefore, $r_{1}=r_{2}$.
$$
\begin{array}{l}
\text { Hence } a+b+c=9\left(k_{1}+2 k_{2}\right)+3\left(r_{1}+2 r_{2}\right) \\
=9\left(k_{1}+2 k_{2}+r_{1}\right) .
\end{array}
$$
Thus, $a+b+c$ is a multiple of 9, i.e., $n=9$ when the conclusion holds.
Next, we prove: $n \leqslant 9$.
Take $a=11, b=5$, then $c=2 a+5 b=47$. In this case, $a+b+c=63$, so $n \mid 63$.
Take $a=13, b=7$, then $c=2 a+5 b=61$. In this case, $a+b+c=81$, so $n \mid 81$.
Therefore, $n$ is a common divisor of 63 and 81.
Hence $n \leqslant(63,81)=9$.
In conclusion, the maximum value of $n$ is 9.
|
9
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. Polynomial
$$
\begin{array}{l}
|x+1|+|x-2|+|x+3|+\cdots+ \\
|x+2007|+|x-2008|+|x+2009|
\end{array}
$$
The minimum value of the above expression is ( ).
(A) 2019044
(B) 2017035
(C) 2009
(D) 0
|
4. A.
$$
\begin{array}{l}
| x+ 1|+| x-2|+| x+3 |+\cdots+ \\
|x+2007|+|x-2008|+|x+2009| \\
=|x+1|+(|x-2|+|x+3|)+\cdots+ \\
(|x-2008|+|x+2009|) \\
\geqslant|x+1|+|(x-2)-(x+3)|+\cdots+ \\
|(x-2008)-(x+2009)| \\
=|x+1|+2+3+\cdots+2009 \\
\geqslant 0+2+3+\cdots+2009 \\
= \frac{2009 \times 2010}{2}-1 \\
= 2019044 .
\end{array}
$$
The equality holds if and only if \( x = -1 \).
|
2019044
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $n$ is a natural number, $n^{2}+4 n+2009$ can be expressed as the sum of the squares of four consecutive natural numbers. Then the sum of all $n$ that satisfy this condition is $\qquad$ .
|
2.1.712.
Let $n^{2}+4 n+2009$
$$
=(m-1)^{2}+m^{2}+(m+1)^{2}+(m+2)^{2} \text {, }
$$
where $m$ is a positive integer. Then
$$
(n+2)^{2}+2005=(2 m+1)^{2}+5 \text {, }
$$
which implies $(2 m+1)^{2}-(n+2)^{2}=2000$.
Factoring, we get
$$
(2 m+n+3)(2 m-n-1)=2000 \text {. }
$$
Since $2 m+n+3$ and $2 m-n-1$ have the same parity, and $2 m+n+3>2 m-n-1$, we have
$$
\begin{array}{l}
(2 m+n+3,2 m-n-1) \\
=(1000,2),(500,4),(250,8), \\
(200,10),(100,20),(50,40) . \\
\text { Hence }(m, n) \\
=(250,497),\left(\frac{251}{2}, 246\right),(64,119), \\
(52,93),\left(\frac{59}{2}, 38\right),(22,3) .
\end{array}
$$
Therefore, the sum of all $n$ that satisfy the condition is
$$
497+119+93+3=712 \text {. }
$$
|
712
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Find the remainder when $47^{37^{2}}$ is divided by 7.
Try to find the remainder of $47^{37^{2}}$ when divided by 7.
|
(Note that $47^{6} \equiv(-2)^{6} \equiv 2^{6} \equiv 8^{2} \equiv 1^{2}$ $\equiv 1(\bmod 7)$. Also, $37^{23} \equiv 1^{23} \equiv 1(\bmod 6)$, let $37^{23}$ $=6k+1$. Then $47^{37^{23}}=47^{k+1}=\left(47^{6}\right)^{k} \times 47 \equiv$ $\left.1^{k} \times 47 \equiv 47 \equiv 5(\bmod 7).\right)$
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2009}\right]$ when divided by 7 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
7.6.
Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$. Then
$$
-1<\beta<0,1<\alpha<2 \text {, }
$$
and $\alpha+\beta=1, \alpha \beta=-1$.
Therefore, $\alpha, \beta$ are the two distinct real roots of $x^{2}-x-1=0$.
Let $A_{n}=\alpha^{n}+\beta^{n}$. Then $A_{n+2}=A_{n+1}+A_{n}$.
By $A_{1}=\alpha+\beta=1$,
$$
A_{2}=\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3 \text {, }
$$
we know that $\left\{A_{n}\right\}$ is an integer sequence, and the sequence of remainders when divided by 7 is 1, $3,4,0,4,4,1,5,6,4,3,0,3,3,6,2,1,3,4, \cdots$, which is a periodic sequence with a period of 16.
Since $2009=16 \times 125+9$, we know that the remainder when $A_{2000}$ is divided by 7 is 6.
$$
\text { Since }-1<\left(\frac{1-\sqrt{5}}{2}\right)^{200}<0 \text {, hence }\left[\left(\frac{1+\sqrt{5}}{2}\right)^{209}\right]
$$
is 6 when divided by 7.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that the three non-zero real roots of the equation $x^{3}+a x^{2}+b x+c$ $=0$ form a geometric progression. Then $a^{3} c-b^{3}$ $=$ . $\qquad$
|
8.0.
Let the three roots be $d$, $d q$, and $d q^{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
d+d q+d q^{2}=-a, \\
d^{2} q+d^{2} q^{2}+d^{2} q^{3}=b, \\
d^{3} q^{3}=-c .
\end{array}\right.
$$
Dividing (2) by (1) gives $d q=-\frac{b}{a}$.
Substituting into (3) yields $\left(-\frac{b}{a}\right)^{3}=-c$. Therefore, $a^{3} c-b^{3}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (15 points) Given two lines
$$
\begin{array}{l}
l_{1}: 3 x+4 y-25=0, \\
l_{2}: 117 x-44 y-175=0,
\end{array}
$$
Point $A$ has projections $B$ and $C$ on $l_{1}$ and $l_{2}$, respectively.
(1) Find the locus curve $\Gamma$ of point $A$ such that $S_{\triangle A B C}=\frac{1728}{625}$;
(2) If $\odot T:\left(x-\frac{39}{5}\right)^{2}+\left(y-\frac{27}{5}\right)^{2}=r^{2}$ intersects the curve $\Gamma$ at exactly 7 points, find the value of $r$.
|
10. (1) Let $A(x, y)$. It is easy to know
$$
\begin{array}{l}
A B=\frac{|3 x+4 y-25|}{\sqrt{3^{2}+4^{2}}}=\frac{|3(x-3)+4(y-4)|}{5}, \\
A C=\frac{|117 x-44 y-175|}{\sqrt{117^{2}+44^{2}}} \\
=\frac{|117(x-3)-44(y-4)|}{125} .
\end{array}
$$
Let the angles of inclination of $l_{1}$ and $l_{2}$ be $\dot{\alpha}_{1}$ and $\alpha_{2}$, respectively, and the angle between $l_{1}$ and $l_{2}$ be $\beta=\alpha_{1}-\alpha_{2}$. Then $\sin \beta=\frac{24}{25}$. Therefore,
$$
\begin{aligned}
& S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \sin \beta \\
= & \frac{12}{5}\left|351(x-3)^{2}-176(y-4)^{2}+336(x-3)(y-4)\right| \\
= & \frac{1728}{625} \\
\Rightarrow & \mid 351(x-3)^{2}-176(y-4)^{2}+ \\
& 336(x-3)(y-4) \mid=3600 .
\end{aligned}
$$
Equation (1) is the trajectory equation of point $A$.
(2) It is easy to know $T\left(\frac{39}{5}, \frac{27}{5}\right)$, and the intersection point of $l_{1}$ and $l_{2}$ is $D(3,4)$.
Translate $\odot T$ and curve $\Gamma$ so that point $D$ is moved to the origin, and then perform a rotation transformation centered at point $D^{\prime}$ to make the angle bisector of $l_{1}$ and $l_{2}$ parallel to the coordinate axes, obtaining $\odot T^{\prime}$ and curve $\Gamma^{\prime}$, i.e.,
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{\prime}=\frac{24}{25}(x-3)+\frac{7}{25}(y-4), \\
y^{\prime}=\frac{24}{25}(y-4)-\frac{7}{25}(x-3)
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x-3=\frac{24 x^{\prime}-7 y^{\prime}}{25}, \\
y-4=\frac{7 x^{\prime}+24 y^{\prime}}{25} .
\end{array}\right.
\end{array}
$$
Substitute into equation (1) and simplify to get
$$
\left|\left(4 x^{\prime}\right)^{2}-\left(3 y^{\prime}\right)^{2}\right|=144 \text {. }
$$
Then curve $\Gamma^{\prime}$ is two sets of hyperbolas:
$$
\Gamma_{1}: \frac{x^{2}}{9}-\frac{y^{2}}{16}=1, \Gamma_{2}: \frac{x^{2}}{9}-\frac{y^{2}}{16}=-1 \text {. }
$$
Also, $T^{\prime}(5,0)$, so $\odot T^{\prime}:(x-5)^{2}+y^{2}=r^{2}$.
Since $\odot T$ intersects curve $\Gamma$ at exactly 7 points, i.e., $\odot T^{\prime}$ intersects $\Gamma^{\prime}$ at exactly 7 points, and both $\odot T$ and $\Gamma^{\prime}$ are symmetric about the $x$-axis, $\odot T$ and $\Gamma^{\prime}$ must have one intersection point on the $x$-axis, i.e., $(-3,0)$ or $(3,0)$ is on $\odot T^{\prime}$. Therefore,
$r=2$ or 8.
If $r=2$, since the distance from $T^{\prime}(5,0)$ to the asymptotes of $\Gamma^{\prime}$ is
$$
\frac{|4 \times 5 \pm 3 \times 0|}{\sqrt{4^{2}+3^{2}}}=4>r,
$$
so, $\odot T^{\prime}$ does not intersect $\Gamma_{2}$. And $\odot T^{\prime}$ intersects $\Gamma_{1}$ at most 4 points, which contradicts the fact that $\odot T^{\prime}$ intersects $\Gamma^{\prime}$ at 7 points. Therefore, $r=8$.
It is easy to see that when $r=8$, the conditions are satisfied.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Arrange $1, 2, 3$ in a cyclic order from left to right, forming a 2009-digit number: $x=123123 \cdots 12312$. Find the remainder when $x$ is divided by 101.
|
(Tip: Since $100 \equiv -1 \pmod{101}$, we have
$$
\begin{array}{l}
123123 = 123 \times 100 + 123 \\
\equiv -123 + 123 \equiv 0 \pmod{101}.
\end{array}
$$
Therefore, $123123 \cdots 12312 = 123123 \cdots 12300 + 12$
$$
\equiv 0 + 12 \equiv 12 \pmod{101}.
$$
)
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $p$ and $8 p^{2}+1$ both be prime numbers. Prove that $8 p^{2}-p+2$ is also a prime number.
|
(Tip: If $p \equiv 0(\bmod 3)$, then $p=3$. In this case, $8 p^{2}-p+2=72-3+2=71$ is a prime number.
If $p \equiv \pm 1(\bmod 3), 8 p^{2}+1 \equiv 8+1 \equiv$ $0(\bmod 3)$, i.e., $3 \mid\left(8 p^{2}+1\right)$, which is a contradiction.)
|
71
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $p$ be a prime number, and $q=4^{p}+p^{4}+4$ is also a prime number. Find the value of $p+q$.
The text above is translated into English, keeping the original text's line breaks and format.
|
(Tip: Mod 3 we get $q=4^{p}+p^{4}+4 \equiv 1^{p}+p^{4}+$ $1 \equiv p^{4}+2(\bmod 3)$. Also, $p^{2} \equiv 0,1(\bmod 3)$, so $p^{4}$ $\equiv 0,1(\bmod 3)$.
If $p^{4} \equiv 1(\bmod 3)$, then $q \equiv p^{4}+2 \equiv 1+2 \equiv$ $0(\bmod 3)$. But $q=4^{p}+p^{4}+4>3$, which contradicts the fact that $q$ is a prime number.
If $p^{4} \equiv 0(\bmod 3)$, then $p=3$. In this case, $q=4^{p}+$ $p^{4}+4=149$ is a prime number, which meets the requirement. Therefore, $p+q=$ 152.)
|
152
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find the unit digit of the natural number $2^{100}+3^{101}+4^{100}$.
|
Solution: Notice that
$$
\begin{array}{l}
2^{100} \equiv 2^{1 \times 25} \equiv\left(2^{4}\right)^{25} \equiv 16^{25} \\
\equiv 6^{25} \equiv 6(\bmod 10), \\
3^{101} \equiv 3^{4 \times 25+1} \equiv\left(3^{4}\right)^{25} \times 3^{1} \\
\equiv 1^{25} \times 3^{1} \equiv 3(\bmod 10), \\
4^{102} \equiv 4^{2 \times 51} \equiv\left(4^{2}\right)^{51} \equiv 16^{51} \equiv 6(\bmod 10) . \\
\text { Therefore, } 2^{100}+3^{101}+4^{102} \equiv 6+3+6 \equiv 5(\bmod 10),
\end{array}
$$
Thus, the unit digit of the natural number $2^{100}+3^{101}+4^{102}$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The inequality $x^{2}+|2 x-6| \geqslant a$ holds for all real numbers $x$. Then the maximum value of the real number $a$ is $\qquad$
|
2.5.
When $x \geqslant 3$,
left side $=x^{2}+2 x-6=(x+1)^{2}-7 \geqslant 9$;
When $x<3$,
left side $=x^{2}-2 x+6=(x-1)^{2}+5 \geqslant 5$.
Therefore, for the given inequality to hold for all real numbers $x$, the maximum value of $a$ should be 5.
|
5
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{n}$ denote the last digit of the number $n^{4}$. Then $a_{1}+a_{2}+\cdots+a_{2008}=$ $\qquad$
|
3.6632 .
Obviously, the unit digit of $(10+n)^{4}$ is the same as that of $n^{4}$, that is,
$$
1^{4}, 2^{4}, 3^{4}, 4^{4}, 5^{4}, 6^{4}, 7^{4}, 8^{4}, 9^{4}, 10^{4}
$$
the unit digits are
$$
\begin{array}{l}
1,6,1,6,5,6,1,6,1,0 \text {. } \\
\text { Then } a_{1}+a_{2}+\cdots+a_{10} \\
=1+6+1+6+5+6+1+6+1+0=33 \text {. } \\
\text { Therefore, } a_{1}+a_{2}+\cdots+a_{2008} \\
=200 \times 33+1+6+1+6+5+6+1+6 \\
=6632 \text {. }
\end{array}
$$
|
6632
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For the positive integer $n$, it is defined that $n!=1 \times 2 \times \cdots \times n$. Then, among all the divisors of the product $1! \times 2! \times \cdots \times 9!$, the number of divisors that are perfect squares is $\qquad$.
|
7.672.
Given $1!\times 2!\times \cdots \times 9!=2^{30} \times 3^{13} \times 5^{5} \times 7^{3}$, then the divisors of its perfect square must be in the form of $2^{2 a} \times 3^{2 b} \times$ $5^{2 c} \times 7^{2 d}$, where $a, b, c, d$ are non-negative integers, and $0 \leqslant a \leqslant 15,0 \leqslant b \leqslant 6,0 \leqslant c \leqslant 2,0 \leqslant d \leqslant 1$. Therefore, such divisors total $16 \times 7 \times 3 \times 2=672$.
|
672
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $k$ is a positive integer not exceeding 2008, such that the equation $x^{2}-x-k=0$ has two integer roots. Then the sum of all such positive integers $k$ is $\qquad$ .
|
8.30360
Since the roots of $x^{2}-x-k=0$ are $x=\frac{1 \pm \sqrt{4 k+1}}{2}$, therefore,
$x^{2}-x-k=0$ has two integer roots $\Leftrightarrow 4 k+1$ is a square of an odd number.
Let $1+4 k=(2 a+1)^{2}\left(a \in \mathbf{N}_{+}\right)$. Then $k=a(a+1)$.
Also, $k \leqslant 2008$, so $a \leqslant 44$.
Therefore, the sum of all such $k$ is
$$
\begin{array}{l}
1 \times 2+2 \times 3+\cdots+44 \times 45 \\
= \frac{1}{3}(1 \times 2 \times 3-0 \times 1 \times 2)+ \\
\frac{1}{3}(2 \times 3 \times 4-1 \times 2 \times 3)+\cdots+ \\
\frac{1}{3}(44 \times 45 \times 46-43 \times 44 \times 45) \\
= \frac{1}{3} \times 44 \times 45 \times 46=30360 .
\end{array}
$$
|
30360
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (15 points) Find the largest positive integer $n$ that satisfies the inequality
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{11}\right]+\left[\frac{n}{13}\right]<n
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
|
Five, Proof: For any integers $x, k$, we have
$$
\left[\frac{x}{k}\right] \geqslant \frac{x-k+1}{k} \text {. }
$$
Fact .. By the division algorithm, there exist integers $q, r$ such that $x=k q+r(0 \leqslant r \leqslant k-1)$. Thus,
$$
\begin{array}{l}
{\left[\frac{x}{k}\right]=\left[\frac{k q+r}{k}\right]=q+\left[\frac{r}{k}\right]=q \text {, }} \\
\frac{x-k+1}{k}=q+\frac{r-(k-1)}{k} \leqslant q=\left[\frac{x}{k}\right] \text {. } \\
\end{array}
$$
Let the positive integer $n$ satisfy the inequality
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{11}\right]+\left[\frac{n}{13}\right]<n \text {. }
$$
Then $n-1 \geqslant\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{11}\right]+\left[\frac{n}{13}\right]$
$$
\begin{array}{l}
\geqslant \frac{n-1}{2}+\frac{n-2}{3}+\frac{n-10}{11}+\frac{n-12}{13} \\
=\frac{859}{858} n-\frac{2573}{858} .
\end{array}
$$
Thus, $n \leqslant 1715$.
Furthermore, when $n=1715$,
$$
\begin{array}{l}
{\left[\frac{1715}{2}\right]+\left[\frac{1715}{3}\right]+\left[\frac{1715}{11}\right]+\left[\frac{1715}{13}\right]} \\
=857+571+155+131=1714<1715 .
\end{array}
$$
Therefore, the largest positive integer $n$ is 1715.
|
1715
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Find the largest positive number $\lambda$, such that for any real numbers $x, y, z$ satisfying $x^{2}+y^{2}+z^{2}=1$, the inequality
$|\lambda x y+y z| \leqslant \frac{\sqrt{5}}{2}$ holds.
(Zhang Zhengjie)
|
5. Notice that
$$
\begin{array}{l}
1=x^{2}+y^{2}+z^{2}=x^{2}+\frac{\lambda^{2}}{1+\lambda^{2}} y^{2}+\frac{1}{1+\lambda^{2}} y^{2}+z^{2} \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(\lambda|x y|+|y z|) \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(|\lambda x y+y z|),
\end{array}
$$
and when $y=\frac{\sqrt{2}}{2}, x=\frac{\sqrt{2} \lambda}{2 \sqrt{\lambda^{2}+1}}, z=\frac{\sqrt{2}}{2 \sqrt{\lambda^{2}+1}}$, both equalities can be achieved simultaneously.
Therefore, $\frac{\sqrt{1+\lambda^{2}}}{2}$ is the maximum value of $|\lambda x y+y z|$.
Let $\frac{\sqrt{1+\lambda^{2}}}{2}=\frac{\sqrt{5}}{2}$, then $\lambda=2$.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $n$ be a positive integer, and let $f(n)$ denote the number of $n$-digit numbers (called wave numbers) $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfy the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and $a_{i} \neq$ $a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, the signs of $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ $(i=1,2, \cdots)$ are opposite.
(1) Find the value of $f(10)$;
(2) Determine the remainder when $f(2008)$ is divided by 13.
|
And the $n$-digit waveform number $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfies $a_{1}>a_{2}$ is called a "B-type number". According to symmetry, when $n \geqslant 2$, the number of B-type numbers is also $g(n)$. Therefore, $f(n)=2 g(n)$.
Next, we find $g(n)$: Let $m_{k}(i)$ represent the number of $k$-digit A-type waveform numbers with the last digit being $i$ $(i=1,2,3,4)$. Then $g(n)=\sum_{i=1}^{4} m_{n}(i)$. Since $a_{2 k-1}a_{2 k+1}$, then 1) when $k$ is even,
$m_{k+1}(4)=0, m_{k+1}(3)=m_{k}(4)$,
$m_{k+1}(2)=m_{k}(4)+m_{k}(3)$,
$m_{k+1}(1)=m_{k}(4)+m_{k}(3)+m_{k}(2)$.
2) When $k$ is odd,
$m_{k+1}(1)=0, m_{k+1}(2)=m_{k}(1)$,
$m_{k+1}(3)=m_{k}(1)+m_{k}(2)$,
$m_{k+1}(4)=m_{k}(1)+m_{k}(2)+m_{k}(3)$. (2)
It is easy to see that $m_{2}(1)=0, m_{2}(2)=1, m_{2}(3)=2$, $m_{2}(4)=3$, then $g(2)=6$. Therefore,
$$
\begin{array}{l}
m_{3}(1)=m_{2}(2)+m_{2}(3)+m_{2}(4)=6, \\
m_{3}(2)=m_{2}(3)+m_{2}(4)=5, \\
m_{3}(3)=m_{2}(4)=3, \\
m_{3}(4)=0 .
\end{array}
$$
So, $g(3)=\sum_{i=1}^{4} m_{3}(i)=14$.
And by $m_{4}(1)=0, m_{4}(2)=m_{3}(1)=6$,
$m_{4}(3)=m_{3}(1)+m_{3}(2)=11$,
$$
m_{4}(4)=m_{3}(1)+m_{3}(2)+m_{3}(3)=14,
$$
So, $g(4)=\sum_{i=1}^{4} m_{4}(i)=31$.
Similarly, we get
$g(5)=70, g(6)=157, g(7)=353$,
$g(8)=793, \cdots$.
Generally, when $n \geqslant 5$,
$g(n)=2 g(n-1)+g(n-2)-g(n-3)$.
Now we prove formula (3). By induction on $n$.
When $n=5,6,7,8$, it has been verified.
Assume formula (3) holds up to $n$. Consider the case for $n+1$.
When $n$ is even, according to 1) and 2), we have
$m_{n+1}(4)=0$,
$m_{n+1}(3)=m_{n}(4)$,
$m_{n+1}(2)=m_{n}(4)+m_{n}(3)$,
$m_{n+1}(1)=m_{n}(4)+m_{n}(3)+m_{n}(2)$.
And $m_{n}(1)=0$, then
$g(n+1)=\sum_{i=1}^{4} m_{n+1}(i)$
$=2\left(\sum_{i=1}^{4} m_{n}(i)\right)+m_{n}(4)-m_{n}(2)$
$=2 g(n)+m_{n}(4)-m_{n}(2)$.
Also, $m_{n}(4)$
$=m_{n-1}(1)+m_{n-1}(2)+m_{n-1}(3)+0$
$=\sum_{i=1}^{4} m_{n-1}(i)=g(n-1)$,
$m_{n}(2)=m_{n-1}(1)$
$=m_{n-2}(4)+m_{n-2}(3)+m_{n-2}(2)+0$
$=g(n-2)$,
At this point, we have
$g(n+1)=2 g(n)+g(n-1)-g(n-2)$.
When $n$ is odd, $g(n+1)=\sum_{i=1}^{4} m_{n+1}(i)$.
And $m_{n+1}(1)=0, m_{n+1}(2)=m_{n}(1)$,
$m_{n+1}(3)=m_{n}(1)+m_{n}(2)$,
$m_{n+1}(4)=m_{n}(1)+m_{n}(2)+m_{n}(3)$,
Then $g(n+1)=\sum_{i=1}^{4} m_{n+1}(i)$
$=2 \sum_{i=1}^{4} m_{n}(i)+m_{n}(1)-m_{n}(3)$
$=2 g(n)+m_{n}(1)-m_{n}(3)$.
Also, $m_{n}(1)$
$=m_{n-1}(4)+m_{n-1}(3)+m_{n-1}(2)+0$
$=g(n-1)$,
$m_{n}(3)=m_{n-1}(4)$
$=m_{n-2}(1)+m_{n-2}(2)+m_{n-2}(3)+0$
$=g(n-2)$,
At this point, we also have
$g(n+1)=2 g(n)+g(n-1)-g(n-2)$.
Therefore, by induction, for all $n \geqslant 5$, formula (3) holds.
According to formula (3),
$g(9)=2 g(8)+g(7)-g(6)=1782$,
$g(10)=2 g(9)+g(8)-g(7)=4004$,
So, $f(10)=2 g(10)=8008$.
Now consider the sequence $\{g(n) \mid$ modulo 13.
Using formula (3), it is easy to calculate that when $n=2,3, \cdots, 14,15$,
$16, \cdots$, the remainders of $g(n)$ when divided by 13 are:
$6,1,5,5,1,2,0,1,0,1,1,3 ; 6,1,5,5, \cdots$
Therefore, when $n \geqslant 2$, the sequence of remainders of $\{g(n)\}$ when divided by 13 is a periodic sequence, with the smallest period being 12.
Since $2008=12 \times 167+4$, then
$g(2008) \equiv 5(\bmod 13)$.
Therefore, $f(2008) \equiv 10(\bmod 13)$.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Given that positive integers $p, q$ are both prime numbers, and $7p+q, pq+11$ are also prime numbers. Find $p+q$.
保留源文本的换行和格式，直接输出翻译结果。
|
When $p \equiv q \equiv 1(\bmod 2)$,
$$
7 p+q \equiv 7+1 \equiv 0(\bmod 2) \text{. }
$$
But $7 p+q \geqslant 7>2$, which contradicts the fact that $7 p+q$ is a prime number. Therefore, one of $p, q$ must be even.
Since $p, q$ are both primes, then $p=2$ or $q=2$.
(1) When $p=2$, $14+q$ and $2 q+11$ are both primes.
If $q \equiv 1(\bmod 3)$, then
$$
14+q \equiv 14+1 \equiv 0(\bmod 3) ; \quad \div i
$$
Contradiction.
If $q \equiv-1(\bmod 3)$, then
$$
2 q+11 \equiv-2+11 \equiv 0(\bmod 3),
$$
Contradiction.
Therefore, $q \equiv 0(\bmod 3)$.
But $q$ is a prime, so $q=3$.
(2) When $q=2$, $7 p+2$ and $2 p+11$ are both primes.
If $p \equiv 1(\bmod 3)$, then
$$
7 p+2 \equiv 7+2 \equiv 0(\bmod 3),
$$
Contradiction.
If $p \equiv-1(\bmod 3)$, then
$$
2 p+11 \equiv-2+11 \equiv 0(\bmod 3),
$$
Contradiction.
Therefore, $p \equiv 0(\bmod 3)$.
But $p$ is a prime, so $p=3$.
Upon verification, $p=2, q=3$ or $p=3, q=2$ satisfy the conditions. Hence $p+q=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Consider the set $S=\{1,2, \cdots, 10\}$ and all its non-empty subsets. If the number of even numbers in a non-empty subset is no less than the number of odd numbers, then this subset is called a "good subset". Therefore, the number of good subsets is ( ) .
(A) 631
(B) 633
(C) 635
(D) 637
|
5.D.
Suppose a good subset contains $i(i=1,2,3,4,5)$ even numbers. Then the number of odd numbers can be $j(j=0,1, \cdots, i)$.
Therefore, the number of good subsets is
$$
\begin{array}{l}
\sum_{i=1}^{5}\left(C_{5}^{i} \sum_{j=0}^{i} C_{5}^{j}\right) \\
= C_{5}^{1}\left(C_{5}^{0}+C_{5}^{1}\right)+C_{5}^{2}\left(C_{5}^{0}+C_{5}^{1}+C_{5}^{2}\right)+ \\
C_{5}^{3}\left(C_{5}^{0}+C_{5}^{1}+C_{5}^{2}+C_{5}^{3}\right)+ \\
C_{5}^{4}\left(C_{5}^{0}+C_{5}^{1}+C_{5}^{2}+C_{5}^{3}+C_{5}^{4}\right)+ \\
C_{5}^{5}\left(C_{5}^{0}+C_{5}^{1}+C_{5}^{2}+C_{5}^{3}+C_{5}^{4}+C_{5}^{5}\right) \\
= 637 .
\end{array}
$$
|
637
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
8. If the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is $\qquad$ .
|
8. -1 .
Let $z$ be a root satisfying the condition. Then the original equation is equivalent to
$$
z^{2008}(z+1)=-1 \text {. }
$$
Taking the modulus on both sides, we get $|z+1|=1$.
Since $|z|=1$, all roots with modulus 1 can only be the complex numbers corresponding to the intersection points of the circle I: with center $(-1,0)$ and radius 1, and the circle with center $(0,0)$ and radius 1 in the complex plane. Therefore, $z=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$.
: Upon verification, both roots are roots of the original equation. Thus, the sum of all roots with modulus 1 is -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Consider the 25 grid points in a $4 \times 4$ square grid. The number of different lines passing through at least 3 grid points is $\qquad$ .
|
9.32.
There are 10 horizontal and vertical lines, 10 lines parallel to the two diagonals, and 12 other lines that meet the conditions. In total, there are 32 lines.
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the value of $\sum_{k=1}^{2008}\left[\frac{2008 k}{2009}\right]$ is $\qquad$ .
|
10.2015028.
For $k=1,2, \cdots, 2008$, since $\frac{2008 k}{2009}$ is not an integer, so,
$$
\begin{array}{l}
{\left[\frac{2008 k}{2009}\right]+\left[\frac{2008(2009-k)}{2009}\right]} \\
=\left[\frac{2008 k}{2009}\right]+\left[2008-\frac{2008 k}{2009}\right]=2007 . \\
\text { Therefore, } \sum_{k=1}^{2008}\left[\frac{2008 k}{2009}\right] \\
=\frac{1}{2} \sum_{k=1}^{2008}\left(\left[\frac{2008 k}{2009}\right]+\left[\frac{2008(2009-k)}{2009}\right]\right) \\
=\frac{1}{2} \times 2008 \times 2007=2015028 .
\end{array}
$$
|
2015028
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given the rectangular cuboid $A B C D-A_{1} B_{1} C_{1} D_{1}$ satisfies $A A_{1}=2, A D=3, A B=251$, the plane $A_{1} B D$ intersects $C C_{1} 、 C_{1} B_{1} 、 C_{1} D_{1}$ at points $L 、 M 、 N$. Then the volume of the tetrahedron $C_{1} L M N$ is $\qquad$ .
|
11.2008.
Since $B D / / M N, A_{1} D / / L M, A_{1} B / / L N$, therefore, $A_{1} 、 B 、 D$ are the midpoints of $M N 、 L M 、 L N$ respectively. Thus,
$$
\begin{array}{l}
C_{1} L=2 C C_{1}=4, C_{1} M=2 C_{1} B_{1}=6, \\
C_{1} N=2 C_{1} D_{1}=502 .
\end{array}
$$
Therefore, the volume of the tetrahedron $C_{1} L M N$ is
$$
\frac{1}{6} C_{1} L \cdot C_{1} M \cdot C_{1} N=2008 .
$$
|
2008
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. There are 10 players $A_{1}, A_{2}, \cdots, A_{10}$, whose points are $9,8,7,6,5,4,3,2,1,0$, and their rankings are 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th. Now a round-robin tournament (i.e., each pair of players plays exactly one match) is held, and each match must have a winner. If the player with a higher ranking beats the player with a lower ranking, the winner gets 1 point, and the loser gets 0 points; if the player with a lower ranking beats the player with a higher ranking, the winner gets 2 points, and the loser gets 0 points. After all the matches, the total points of each player (i.e., the sum of the points obtained in this round-robin tournament and the previous points) are calculated, and the players are re-ranked based on the total points. Find the minimum value of the new champion's total points (ties in ranking are allowed).
|
15. The minimum cumulative score of the new champion is 12.
If the new champion's score does not exceed 11 points, then
$A_{1}$ can win at most 2 games; $A_{2}$ can win at most 3 games;
$A_{3}$ can win at most 4 games; $A_{4}$ can win at most 5 games.
$A_{5}$ can increase by at most 6 points, but there are only 5 players with fewer points than him at the start. Therefore, if he increases by 6 points, he must win at least 1 game against players ranked higher than him, meaning he can win at most 4 games against players ranked lower. Thus, he can win at most 5 games.
$A_{6}$ can increase by at most 7 points, but there are only 4 players with fewer points than him at the start. Therefore, if he increases by 7 points, he must win at least 2 games against players ranked higher than him, meaning he can win at most 3 games against players ranked lower. Thus, he can win at most 5 games.
$A_{7}$ can increase by at most 8 points, but there are only 3 players with fewer points than him at the start. Therefore, if he increases by 8 points, he must win at least 3 games against players ranked higher than him, meaning he can win at most 2 games against players ranked lower. Thus, he can win at most 5 games.
$A_{8}$ can increase by at most 9 points, but there are only 2 players with fewer points than him at the start. Therefore, if he increases by 9 points, he must win at least 4 games against players ranked higher than him, meaning he can win at most 1 game against players ranked lower. Thus, he can win at most 5 games.
$A_{9}$ can increase by at most 10 points, but there is only 1 player with fewer points than him at the start. Therefore, if he increases by 10 points, he must win at least 5 games against players ranked higher than him. Thus, he can win at most 5 games.
$A_{10}$ can increase by at most 11 points, and he can win at most 5 games against players ranked higher than him. Thus, he can win at most 5 games.
In summary, the maximum number of games won by all players is $2+3+4+5 \times 7=44$, but each match between two players results in one win, totaling $\mathrm{C}_{10}^{2}=45$ wins, which is a contradiction.
The following example shows that the new champion's cumulative score can be 12 points.
$A_{1}$ wins against $A_{2}, A_{3}, A_{4}$, loses to $A_{5}, A_{6}, A_{7}, A_{8}, A_{9}, A_{10}$, with a cumulative score of $9+3=12$;
$A_{2}$ wins against $A_{3}, A_{4}, A_{5}, A_{6}$, loses to $A_{7}, A_{8}, A_{9}, A_{10}$, with a cumulative score of $8+4=12$;
$A_{3}$ wins against $A_{4}, A_{5}, A_{6}, A_{7}$, loses to $A_{8}, A_{9}, A_{10}$, with a cumulative score of $7+4=11$;
$A_{4}$ wins against $A_{5}, A_{6}, A_{7}, A_{8}$, loses to $A_{9}, A_{10}$, with a cumulative score of $6+4=10$;
$A_{5}$ wins against $A_{6}, A_{7}, A_{8}, A_{9}$, loses to $A_{10}$, with a cumulative score of $5+2+4=11$;
$A_{6}$ wins against $A_{7}, A_{8}, A_{9}, A_{10}$, with a cumulative score of $4+2+4=10$;
$A_{7}$ wins against $A_{8}, A_{9}, A_{10}$, with a cumulative score of $3+2 \times 2+3=10$;
$A_{8}$ wins against $A_{9}, A_{10}$, with a cumulative score of $2+2 \times 3+2=10$;
$A_{9}$ wins against $A_{10}$, with a cumulative score of $1+2 \times 4+1=10$; $A_{10}$ has a cumulative score of $0+2 \times 5=10$.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If a positive integer cannot be expressed as the difference of squares of two positive integers, then this positive integer is called a "non-wise number". If these non-wise numbers are arranged in ascending order, then the 2009th non-wise number is $\qquad$
|
ニ、1.8026.
1 cannot be expressed as the difference of squares of two positive integers, so 1 is the 1st non-wise number.
Odd numbers greater than 1, $2 n+1\left(n \in N_{+}\right)$, can be expressed as $(n+1)^{2}-n^{2}$.
Even numbers divisible by 4, $4 m$, can be expressed as $(m+1)^{2}-(m-1)^{2}(m>1)$. However, 4 cannot be expressed as the difference of squares of two positive integers. Even numbers divisible by 2 but not by 4, $4 m+2(m \in N)$, cannot be expressed as the difference of squares of two positive integers. Otherwise,
$$
x^{2}-y^{2}=(x+y)(x-y)=4 m+2 \text {. }
$$
Since $x+y$ and $x-y$ have the same parity, and $4 m+2$ is even, therefore, $x+y$ and $x-y$ are both even.
Thus, $2|(x+y), 2|(x-y)$.
Hence, $4 \mid\left(x^{2}-y^{2}\right)$.
Therefore, $4 \mid(4 m+2)$, which contradicts $4 \nmid(4 m+2)$.
So, $4 m+2$ is a non-wise number.
Thus, the non-wise numbers arranged in ascending order are:
$1,2,4,6,10,14, \cdots$
When $n>3$, the $n$-th non-wise number is $4 n-10$.
Therefore, the 2009th non-wise number is
$$
4 \times 2009-10=8026 \text {. }
$$
|
8026
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $x$ and $y$ are integers, $y=\sqrt{x+2003}-$ $\sqrt{x-2009}$. Then the minimum value of $y$ is $\qquad$ .
|
3.2 .
Let $\sqrt{x+2003}=a, \sqrt{x-2009}=b$.
Since $x, y$ are integers,
$y=a-b=\frac{a^{2}-b^{2}}{a+b}=\frac{4012}{a+b} \in \mathbf{Z}_{+}$.
Thus, $a-b \in \mathbf{Q}, a+b \in \mathbf{Q}$
$\Rightarrow a \in \mathbf{Q}, b \in \mathbf{Q} \Rightarrow a, b \in \mathbf{Z}$.
Then $(a+b)(a-b)=4012=2006 \times 2$.
Since $a+b$ and $a-b$ have the same parity, the minimum value of $a-b$ is 2.
Therefore, the minimum value of $y$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In Rt $\triangle A B C$, $\angle A C B=90^{\circ}$, on the hypotenuse $A B$ respectively intercept $A D=$
$$
A C, B E=B C, D E=6 \text{, }
$$
$O$ is the circumcenter of $\triangle C D E$, as shown in Figure 2. Then the sum of the distances from $O$ to the three sides of $\triangle A B C$ is . $\qquad$
|
4.9.
As shown in Figure 8, connect
$$
O A, O B, O C, O D \text {, and }
$$
$O E$.
Since $O C=O D$,
we know that point $O$ lies on the perpendicular bisector of segment $C D$.
Also, $A D=A C$, so $O A$ bisects $\angle C A D$.
Similarly, $O B$ bisects $\angle C B D$.
Therefore, $O$ is the incenter of $\triangle A B C$.
Thus, the distances from point $O$ to the three sides of $\triangle A B C$ are all equal to the radius of the incircle of the right triangle $\triangle A B C$. This radius is
$$
\begin{array}{l}
\frac{1}{2}(A C+B C-A B) \\
=\frac{1}{2}(A D+B E-A B)=\frac{1}{2} D E=3 .
\end{array}
$$
Therefore, the required sum is $3 \times 3=9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the sum of $2 n+1\left(n \in \mathbf{N}_{+}\right)$ consecutive positive integers is $a$, and the difference between the sum of the squares of the last $n$ numbers and the sum of the squares of the first $n$ numbers is $b$. If $\frac{a}{b}=\frac{11}{60}$, then the value of $n$ is
|
2.5
Let the $(n+1)$-th number be $m$. Then
$$
\begin{array}{l}
a=(2 n+1) m, \\
b=\sum_{i=1}^{n}(m+i)^{2}-\sum_{i=1}^{n}(m-i)^{2} \\
=2 m \sum_{i=1}^{n} 2 i=2 m n(n+1) . \\
\text { Hence } \frac{a}{b}=\frac{2 n+1}{2 n(n+1)}=\frac{11}{60} .
\end{array}
$$
Solving gives $n=5$ or $-\frac{6}{11}$ (discard).
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. For a right-angled triangle with a hypotenuse of 2009, if the two legs are also integers, then its area is $\qquad$ .
|
$-.1 .432180$.
Let the two legs be $x, y$. Then $x^{2}+y^{2}=2009^{2}$. According to Pythagorean triples and the symmetry of $x, y$, there exist positive integers $m, n, k (m>n)$, such that
$$
\begin{array}{l}
x=2 m n k, y=\left(m^{2}-n^{2}\right) k, \\
2009=\left(m^{2}+n^{2}\right) k .
\end{array}
$$
Since $2009=41 \times 7^{2}$, and $1,7,7^{2}, 7 \times 41$ cannot be expressed as the sum of squares of two positive integers, we have
$$
k=49, m^{2}+n^{2}=41 \text{. }
$$
Solving gives $m=5, n=4$.
Thus, $x=1960, y=441$.
Then $S=\frac{1}{2} x y=432180$.
|
432180
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. From the arithmetic sequence $2,5,8,11, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$
|
2.8 .
First, take $2,5,8,11,20,41,110,1640$, it is easy to see that the sum of their reciprocals is 1, i.e., $k=8$ satisfies the requirement.
Second, suppose we take $x_{1}, x_{2}, \cdots, x_{k}$ from the sequence, such that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1$.
Let $y_{i}=\frac{x_{1} x_{2} \cdots x_{k}}{x_{i}}$. Then
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$
Since $x_{n} \equiv 2(\bmod 3)$, taking the modulus 3 of both sides of equation (1) gives
$k \cdot 2^{k-1} \equiv 2^{k}(\bmod 3)$, i.e., $k \equiv 2(\bmod 3)$.
When $k=2$, $\frac{1}{x_{1}}+\frac{1}{x_{2}} \leqslant \frac{1}{2}+\frac{1}{5}<1$;
When $k=5$,
$\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{5}} \leqslant \frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}<1$.
Therefore, $k \geqslant 8$. Hence, the minimum value of $k$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$ are any two points (which can coincide) on the graph of the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{2 x}{1-2 x}, & x \neq \frac{1}{2} \\
-1, & x=\frac{1}{2}
\end{array}\right.
$$
Point $M$ lies on the line $x=\frac{1}{2}$, and $\overrightarrow{A M}=\overrightarrow{M B}$. Then the value of $y_{1}+y_{2}$ is
|
4. -2 .
Given that point $M$ is on the line $x=\frac{1}{2}$, let $M\left(\frac{1}{2}, y_{M}\right)$.
Also, $\overrightarrow{A M}=\overrightarrow{M B}$, that is,
$$
\begin{array}{l}
\overrightarrow{A M}=\left(\frac{1}{2}-x_{1}, y_{M}-y_{1}\right), \\
\overrightarrow{M B}=\left(x_{2}-\frac{1}{2}, y_{2}-y_{M}\right) .
\end{array}
$$
Thus, $x_{1}+x_{2}=1$.
(1) When $x_{1}=\frac{1}{2}$, $x_{2}=\frac{1}{2}$, then
$$
y_{1}+y_{2}=f\left(x_{1}\right)+f\left(x_{2}\right)=-1-1=-2 \text {; }
$$
(2) When $x_{1} \neq \frac{1}{2}$, $x_{2} \neq \frac{1}{2}$, then
$$
\begin{array}{l}
y_{1}+y_{2}=\frac{2 x_{1}}{1-2 x_{1}}+\frac{2 x_{2}}{1-2 x_{2}} \\
=\frac{2 x_{1}\left(1-2 x_{2}\right)+2 x_{2}\left(1-2 x_{1}\right)}{\left(1-2 x_{1}\right)\left(1-2 x_{2}\right)} \\
=\frac{2\left(x_{1}+x_{2}\right)-8 x_{1} x_{2}}{1-2\left(x_{1}+x_{2}\right)+4 x_{1} x_{2}}=-2 .
\end{array}
$$
In summary, $y_{1}+y_{2}=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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