problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
2. $42 N$ is the set of all positive integers. For a subset $S$ of $N$ and $n \in N$, define
$$S \oplus\{n\}=\{s+n \mid s \in S\}$$
Additionally, define the subset $S_{k}$ as follows:
$$S_{1}=\{1\}, S_{k}=\left\{S_{k-1} \oplus\{k\}\right\} \cup\{2 k-1\}, k=2,3,4 \cdots .$$
(1) Find $N-\bigcup_{k=1}^{\infty} S_{k}$.
(2... | [Solution] (1) Since
$$\begin{aligned}
C_{k+1}^{2}+(k+1)= & \frac{1}{2} k(k+1)+(k+1) \\
= & \frac{1}{2}(k+1)(k+2) \\
& =C_{k+2}^{2} \\
C_{k+2}^{2}-C_{k}^{2}= & \frac{1}{2}(k+1)(k+2)-\frac{1}{2} k(k-1)=2 k+1
\end{aligned}$$
By the problem statement, we have
$$\begin{array}{l}
S_{1}=\{1\} \\
S_{2}=\left\{S_{1} \oplus\{2... | 500 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,287 |
2. $44 \quad c \geqslant 1$ is a fixed positive integer. For each non-empty subset $A$ of the set $\{1,2, \cdots, n\}$, a positive integer $\omega(A)$ is assigned from the set $\{1,2, \cdots, c\}$, satisfying
$$\omega(A \cap B)=\min (\omega(A), \omega(B))$$
where $A, B$ are any two non-empty intersecting subsets of $\... | ［Solution］ When $c=1$, it is obvious that $a(n)=1$, thus we have
$$\lim _{n \rightarrow \infty} \sqrt[n]{a(n)}=1$$
Now consider the case when $c \geqslant 2$.
The set $\{1,2, \cdots, n\}$ has $2^{n}-1$ non-empty subsets, and these $2^{n}-1$ non-empty subsets form the domain of the function $\omega(A)$.
$$\text { Let }... | C | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,288 |
2.45 A monotonically increasing sequence of integers, if its first term is odd, the second term is even, the third term is odd, the fourth term is even, $\cdots$, and so on, is called an alternating sequence. The empty set is also considered an alternating sequence. Let $A(n)$ denote the number of alternating sequences... | [Solution] When $n=1$, $\phi$ and $\{1\}$ are two alternating sequences, so $A(1)=2$. When $n=2$, $\phi, \{1\}, \{1,2\}$ are three alternating sequences, so $A(2)=3$. If $\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{m}\right\}$ is a non-empty alternating sequence taken from the set $\{1,2, \cdots, n\}$, then when $a_{1}=1$,
... | 17711 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,289 |
2. 46 Insert "+" or "-" signs between the numbers $1,2,3 \cdots, 1989$, what is the smallest non-negative number that the sum can achieve? | [Solution] Except for 995, all numbers $1,2,3, \cdots, 1989$ can be divided into 994 pairs: (1, 1989), $(2,1988), \cdots,(994,996)$. Since the parity of the two numbers in each pair is the same, the result of the operation is always even regardless of how "+" or "-" signs are placed before each pair; and 995 is an odd ... | 1 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,290 |
2.47 If $a<b<c<d<e$ are consecutive positive integers, $b+c+d$ is a perfect square, and $a+b+c+d+e$ is a perfect cube, what is the minimum value of $c$? | [Solution] Since $a, b, c, d, e$ are consecutive positive integers, we have
$$\begin{array}{l}
b+c+d=3 c \\
a+b+c+d+e=5 c
\end{array}$$
According to the problem, we can set
$$\begin{array}{l}
3 c=m^{2} \quad(m \in N), \\
5 c=n^{3} \quad(n \in N),
\end{array}$$
From (1), $m$ is a multiple of 3, so $c$ is also a multip... | 675 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,291 |
2 - 49 Find the smallest natural number $n$ that satisfies the following properties:
(1) The unit digit of $n$ is 6;
(2) If the unit digit 6 of $n$ is moved to the front of the other digits, the resulting new number is 4 times $n$. | [Solution] Let the required natural number be
$$n=10 x+6,$$
where $x$ is a natural number.
Let the number of digits of $n$ be $m$. According to the problem, we have
$$6 \cdot 10^{m-1}+x=4(10 x+6)$$
which gives
$$\begin{array}{l}
13 x=2 \cdot 10^{m-1}-8 \\
2 \cdot 10^{m-1}=13 x+8
\end{array}$$
Thus, we can divide the... | 153846 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,292 |
2. $50 N$ is an integer, its base $b$ representation is 777, find the smallest positive integer $b$, such that $N$ is an integer to the fourth power. | [Solution] We need to find the smallest $b$ such that
$$7 b^{2}+7 b+7=x^{4}$$
has an integer solution for $x$. Since 7 is a prime number, 7 divides $x$, so $x=7 k$, substituting into (1) and simplifying, we get
$$b^{2}+b+1=7^{3} k^{4}$$
Let $k=1$, then the above equation becomes
$$b^{2}+b+1=7^{3}$$
which is $(b-18)(... | 18 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,293 |
1. 26 At the ends of a diameter of a circle, the numbers 1 are marked. Then, each resulting semicircle is bisected, and the sum of the numbers at the endpoints is written at the midpoint of each arc (first step). Next, the 4 arcs obtained are bisected, and the sum of the numbers at the endpoints is written at the midpo... | [Solution] After the first step, the sum of all numbers equals $6=2 \times 3$.
Let $S_{n}$ be the sum of all numbers after $n$ steps. It is not hard to prove that after $n+1$ steps, the sum of all numbers will be $2 S_{n}+S_{n}=3 S_{n}$.
Thus, the sum of the numbers triples with each step. Therefore, at the $n$-th ste... | 2 \times 3^{n} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,294 |
$2 \cdot 52$ some positive integer squared, its last three digits are the same non-zero digit, find the smallest positive integer with this property.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | ［Solution］Let $p=10 a \pm b(a, b \in \mathbb{Z}, 1 \leqslant b \leqslant 5)$, then
$$p^{2} \equiv 10 \cdot( \pm 2 a b)+b^{2}(\bmod 100)$$
It can be verified that when $b=1,3,4,5$, the tens and units digits of $p^{2}$ have opposite parity; only when $b=2$, the last two digits of $p^{2}$ have the same parity. Therefore,... | null | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,295 |
2・53 Let $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative real numbers, and define $M$ as the sum of all products $a_{i} a_{j}(i < j)$, that is,
$$M=a_{1}\left(a_{2}+a_{3}+\cdots+a_{n}\right)+a_{2}\left(a_{3}+a_{4}+\cdots+a_{n}\right)+\cdots+a_{n-1} a_{n},$$
Prove that at least one of the numbers $a_{1}, a_{2}, \cdots, a... | [Proof] Let $a$ be the smallest of $a_{1}, a_{2}, \cdots, a_{n}$, then $a_{i} a_{j} \geqslant a^{2}$.
Thus we have $M=\sum_{i<j} a_{i} a_{j} \geqslant C_{n}^{2} a^{2}$, hence
$$a^{2} \leqslant \frac{M}{C_{n}^{2}}=\frac{2 M}{n(n-1)}$$ | a^{2} \leqslant \frac{2 M}{n(n-1)} | Algebra | proof | Yes | Yes | inequalities | false | 735,296 |
2. 55 In decimal, find the smallest natural number: its square number starts with 19 and ends with 89 | [Solution] The last digit of the required number can only be 3 or 7. We write out the squares of all two-digit numbers ending in 3 and 7, among which only $17^{2}, 33^{2}, 67^{2}, 83^{2}$ end with 89. To make the first two digits of the natural number $x$ be 19, the inequality must hold:
$19 \leqslant x^{2} \cdot 10^{-... | 1383 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,298 |
$2 \cdot 56$ Find the smallest natural number, which when its last digit is moved to the front, it becomes 5 times larger.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | [Solution] Let $a_{1} a_{2} \cdots a_{n-1} a_{n}$ be the number we are looking for. According to the problem, we have
$$\overline{a_{n} a_{1} a_{2} \cdots a_{n-1}}=5 \cdot \overline{a_{1} a_{2} \cdots a_{n}}$$
Thus,
$$\begin{array}{l}
a_{n} \cdot 10^{n-1}+\overline{a_{1} a_{2} \cdots a_{n-1}}=5\left(\overline{a_{1} a_... | null | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,299 |
2.57 A quirky mathematician has a ladder with $n$ steps, and he climbs up and down the ladder, each time ascending $a$ steps or descending $b$ steps. Here, $a, b$ are fixed positive integers.
If he can start from the ground, climb to the very top step of the ladder, and then return to the ground, find the minimum valu... | [Solution] The minimum value is $a+b-(a, b)$, where $(a, b)$ denotes the greatest common divisor of $a$ and $b$.
Since the level the mathematician reaches each time is a multiple of $(a, b)$, we can ignore other levels, treating $(a, b)$ as the first level, and using $\frac{a}{(a, b)}, \frac{b}{(a, b)}$ to replace $a$... | a+b-(a, b) | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,300 |
$2 \cdot 62$ Given 1990 piles of stones, each consisting of $1, 2, \cdots, 1990$ stones, in each round, you are allowed to pick any number of piles and remove the same number of stones from these piles. How many rounds are needed at minimum to remove all the stones? | [Solution] After each round, we consider piles with the same number of stones as one group, and empty piles as another group. Suppose at a certain moment there are $n$ groups, and in the next round, the same number of stones is removed from some piles, which originally belonged to $k$ different groups. After the remova... | 11 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,302 |
$2 \cdot 63$ Insert "+" or "-" between $1^{2}, 2^{2}, 3^{2}, \cdots, 1989^{2}$, what is the smallest non-negative number that can be obtained from the resulting sum? | [Solution] Since there are 995 odd numbers in $1^{2}, 2^{2}, 3^{2}, \cdots, 1989^{2}$, the result of the expression obtained by adding "+" and "-" signs will always be odd, so the smallest non-negative value sought is no less than 1.
Since the squares of 4 consecutive natural numbers: $k^{2},(k+1)^{2},(k+2)^{2},(k+3)^... | 1 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,303 |
$2 \cdot 66$ Write a number $+1$ or -1 on each card. You can point to 3 cards and ask: What is the product of the numbers on these three cards (but not tell what numbers are written on the cards)? When there are (1) 30 cards; (2) 31 cards; (3) 32 cards, what is the minimum number of questions needed to find out the pro... | [Solution] (1) Divide 30 numbers into 10 groups of 3 numbers and ask 10 questions to find the product of the 3 numbers in each group, which will allow us to know the product of all the numbers. On the other hand, since each number should be in one of the 3-number groups, at least 10 questions must be asked.
(2) Multip... | 10, 11, 12, 50 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,304 |
$2 \cdot 67$ There are 1000 tickets numbered $000, 001, \cdots, 999$ and 100 boxes numbered $00, 01, \cdots, 99$. If the number of a box can be obtained by deleting one digit from the number of a ticket, then the ticket can be placed in this box. Prove:
(1) All tickets can be distributed into 50 boxes.
(2) It is imposs... | [Proof] (1) We divide the 10 digits $0,1,2, \cdots, 9$ into two groups of 5 digits each, with the digits from 0 to 4 forming one group and the digits from 5 to 9 forming the other. Since in a three-digit number, there must be two digits from the same group, we only need to use the boxes with numbers formed by digits fr... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,305 |
$2 \cdot 70$ For a fixed $\theta \in\left(0, \frac{\pi}{2}\right)$, find the smallest positive number $a$ that satisfies the following two conditions:
(i) $\frac{\sqrt{a}}{\cos \theta}+\frac{\sqrt{a}}{\sin \theta}>1$;
(ii) There exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, su... | [Solution] From (i) we get
$$\sqrt{a}>\frac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}$$
Without loss of generality, assume $\quad \frac{a}{\sin ^{2} \theta}+\frac{a}{\cos ^{2} \theta} \leqslant 1$.
(ii) is equivalent to: there exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\righ... | \frac{\sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,308 |
$2 \cdot 71$ Let $M=\{2,3,4, \cdots, 1000\}$. Find the smallest natural number $n$, such that in any $n$-element subset of $M$, there exist 3 pairwise disjoint 4-element subsets $S, T, U$, satisfying the following conditions: $\square$
（1）For any two elements in $S$, the larger number is a multiple of the smaller numbe... | 【Solution】 Note that $999=37 \times 27$, we let
$$\left\{\begin{array}{l}
A=\{3,5, \cdots, 37\} \\
B=M-A
\end{array}\right.$$
Obviously, $|A|=18,|B|=981$.
We will prove that the subset $B$ of $M$ cannot satisfy conditions (1) and (3) simultaneously.
By contradiction.
If $B$ contains 3 mutually disjoint 4-element subse... | 982 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,310 |
2・76 Let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}, a_{2}, \cdots, a_{n}$ have the following property, for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
| [Solution] The minimum value of $n$ is 8.
First, we prove that each number in $s$ appears at least twice in the sequence $a_{1}, a_{2}, \cdots, a_{n}$.
In fact, if a number in $s$ appears only once in this sequence, since there are 3 subsets containing this number, but the number of adjacent pairs containing this numbe... | 8 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,311 |
$2 \cdot 77$ In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number, the number in the $i$-th row and $j$-th column is $x_{i j}(i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). However, the numbers in each column of Table 1 are rearranged in descending order from top ... | [Solution] The minimum value of $k$ is 97.
In fact, if we take
$$x_{i j}=\left\{\begin{array}{l}
0,4,(j-1)+1 \leqslant i \leqslant 4 j \\
\frac{1}{24} . \text { for the rest of } i
\end{array} \quad(j=1,2, \cdots, 25)\right.$$
Then, we have
$$\sum_{j=1}^{25} x_{i j}=0+24 \times \frac{1}{24}=1 \quad(i=1,2, \cdots, 100)... | 97 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,312 |
2. 7821 people participate in an exam, with the test paper containing 15 true/false questions. It is known that any two people have at least one question that they both answered correctly. What is the minimum number of people who answered the question with the most correct answers? Please explain your reasoning. | [Solution] Let $a_i$ be the number of people who answered the $i$-th question correctly. Then, the number of pairs who answered the $i$-th question correctly is
$$b_i = C_{a_i}^2$$
for $i=1,2, \cdots, 15$.
We will focus on the sum $\sum_{i=1}^{15} b_i$.
Let $a = \max \{a_1, a_2, \cdots, a_{15}\}$, then we have
$$\begin... | 7 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,313 |
2. 79 In a game, scoring is as follows: answering an easy question earns 3 points, and answering a difficult question earns 7 points. Among the integers that cannot be the total score of a player, find the maximum value. | [Solution] Let the set of natural numbers that cannot be the total score be $S$. Among the first 12 natural numbers, only $1,2,4,5,8,11$ belong to $S$.
Since $7=2 \times 3+1,3 \times 5=7 \times 2+1$, when the sum of several 3s and 7s $n \geqslant 12$, one 7 can be exchanged for two 3s, or five 3s can be exchanged for ... | 11 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,314 |
2.81 The sum of the following 7 numbers is exactly 19:
$$\begin{array}{l}
a_{1}=2.56, a_{2}=2.61, a_{3}=2.65, a_{4}=2.71, a_{5}=2.79, a_{6}= \\
2.82, a_{7}=2.86 .
\end{array}$$
To approximate $a_{i}$ with integers $A_{i}$ $(1 \leqslant i \leqslant 7)$, such that the sum of $A_{i}$ is still 19, and the maximum value $M... | [Solution] Since $21$.
To make $A_{1}+A_{2}+\cdots+A_{7}=19$, then $A_{1}, A_{2}, \cdots, A_{7}$ should consist of five 3s and two 2s.
To make $M$ as small as possible, the smallest two numbers $a_{1}$ and $a_{2}$ should have their approximations as $A_{1}=$ $A_{2}=2$, while the others are $A_{3}=A_{4}=A_{5}=A_{6}=A_... | 61 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,315 |
2 - 83 The sum of several positive integers is 1976. Find the maximum value of the product of these positive integers.
Several positive integers add up to 1976. Find the maximum value of the product of these positive integers. | [Solution] Let these numbers be $a_{1}, \cdots, a_{n}$, then $\square$
$$a_{1}+\cdots+a_{n}=1976$$
Assume without loss of generality that $a_{i}<4(1 \leqslant i \leqslant n)$, because when $a_{i} \geqslant 4$, $a_{i} \leqslant 2\left(a_{i}-2\right)$, so replacing $a_{i}$ with the two numbers 2 and $a_{i}-2$ will not d... | 2 \times 3^{658} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,316 |
$2 \cdot 84$ For any positive integer $q_{0}$, consider the sequence $q_{1}, q_{2}, \cdots, q_{n}$ defined by
$$q_{i}=\left(q_{i-1}-1\right)^{3}+3 \quad(i=1,2, \cdots, n)$$
If each $q_{i}(i=1,2, \cdots, n)$ is a power of a prime, find the largest possible value of $n$. | ［Solution］Since
$$m^{3}-m=m(m-1)(m+1) \equiv 0(\bmod 3)$$
Therefore $\square$
$$q_{i}=\left(q_{i-1}-1\right)^{3}+3 \equiv\left(q_{i-1}-1\right)^{3} \equiv q_{i-1}-1(\bmod 3)$$
Thus, among $q_{1}, q_{2}, q_{3}$, there must be one divisible by 3, and it should be a power of 3. However, when 3 । $((q-$ $\left.1)^{3}+3\r... | 2 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,317 |
2.85 Given that $a, b, c, d, e$ are real numbers satisfying
$$\begin{array}{l}
a+b+c+d+e=8 \\
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16
\end{array}$$
determine the maximum value of $e$. | ［Solution］ Using the Cauchy-Schwarz inequality.
Since
$$\begin{aligned}
(a+b+c+d)^{2} & =(a \cdot 1+b \cdot 1+c \cdot 1+d \cdot 1)^{2} \\
& \leqslant\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(1^{2}+1^{2}+1^{2}+1^{2}\right) \\
& =4\left(a^{2}+b^{2}+c^{2}+d^{2}\right)
\end{aligned}$$
Also, since
$$\begin{array}{c}
a+b+c+... | \frac{16}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,318 |
$2 \cdot 87$ Let $4^{27}+4^{500}+4^{n}$ be a perfect square (square of an integer), find the maximum value of the integer $n$.
| [Solution] Factorization yields
$$\begin{aligned}
& 4^{27}+4^{500}+4^{n} \\
= & 4^{27}\left(4^{n-27}+4^{473}+1\right) \\
& 4^{n-27}+4^{473}+1=\left(2^{n-27}\right)^{2}+2 \cdot 2^{945}+1
\end{aligned}$$
When $n-27=945$, i.e., $n=972$, (2) is a perfect square, and thus (1) is a perfect square.
When $n>972$,
$$\begin{al... | 972 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,319 |
2.89 Let $s$ be a list composed of positive integers (the list can contain the same number), which includes the number 68, the arithmetic mean of the numbers in $s$ is 56, but if 68 is removed, the arithmetic mean of the remaining numbers drops to 55. What is the largest number that could appear in $s$? | [Solution] Let $n$ be the number of positive integers in $s$, and its $n$ numbers are $a_{1}, a_{2}, \cdots, a_{n-1}$, 68. According to the problem, we have
$$\begin{array}{l}
\frac{a_{1}+a_{2}+a_{3}+\cdots+a_{n-1}+68}{n}=56 \\
\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1}=55
\end{array}$$
From (1) and (2), we get $56 n-68=5... | 649 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,320 |
2. 90 Prove: For every integer $k$ greater than 1, there exists a multiple of $k$ that is less than $k^{4}$ and, in its decimal representation, contains at most 4 different digits (each digit can be used repeatedly).
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
2. 90 Prove: For every integer $k$ greater than 1, there exists a ... | ［Proof］ Suppose $n$ satisfies $\quad 2^{n-1} \leqslant k$,
Therefore, in $A_{n}$ there are two distinct elements $x, y$ that have the same remainder when divided by $k$, hence $k$ divides $p$ $=|x-y|$. Clearly, the digits of $p$ $\in\{0,1,8,9\}$.
Since $n \leqslant 1+\log _{2} k$, we have
$$\begin{aligned}
p & \leqsla... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,321 |
2.92 The largest even number that cannot be expressed as the sum of two odd composite numbers is?
Will the text be translated into English, keeping the original text's line breaks and format, and output the translation result directly.
Note: The note above is not part of the translation but is provided to clarify th... | [Solution] First, we prove: for even numbers not less than 40, they can definitely be expressed as the sum of two odd composite numbers.
Let $k$ be an even number not less than 40.
If the unit digit of $k$ is 2 (i.e., $42, 52, 62, \cdots$),
then $k=27+5n$ ($n=3, 5, 7, \cdots$);
If the unit digit of $k$ is 4 (i.e., $44... | 38 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,322 |
$2 \cdot 94$ Find all natural numbers $n$ such that
$$\min _{k \in N}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$$
where $\left[\frac{n}{k^{2}}\right]$ denotes the greatest integer not exceeding $\frac{n}{k^{2}}$, and $N$ is the set of natural numbers. | [Solution] The condition $\min _{k \in N}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$ is equivalent to the following two conditions:
(1) For any $k \in N$, $k^{2}+\frac{n}{k^{2}} \geqslant 1991$;
(2) There exists $k_{0} \in N$ such that $k_{0}^{2}+\frac{n}{k_{0}^{2}}<1992$. Conditions (1) and (2) are also resp... | 990208 \leqslant n \leqslant 991231 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,323 |
$2 \cdot 95$ A state issues license plate numbers consisting of 6 digits (made up of the digits $0 \sim 9$), and the state stipulates that any two plate numbers must differ in at least two places (thus, the plate numbers 027592 and 020592 cannot both be used). Determine the maximum number of license plate numbers possi... | [Proof] First, we use induction to prove that the maximum number of $n$-digit license plate numbers (where any two numbers differ in at least two digits) composed of digits $0 \sim 9$ is $10^{n-1}$.
In fact, when $n=1$, there can only be $1=10^{1-1}$ license plate numbers. Suppose the proposition holds for $n=k$, when... | 10^5 | Combinatorics | proof | Yes | Yes | inequalities | false | 735,324 |
2.96 Let $a_{1}, a_{2}, a_{3}, \cdots$ be a non-decreasing sequence of positive integers. For $m \geqslant 1$, define
$$b_{m}=\min \left\{n \mid a_{n} \geqslant m\right\}$$
i.e., $b_{m}$ is the smallest value of $n$ such that $a_{n} \geqslant m$.
If $a_{19}=85$, find the maximum value of $a_{1}+a_{2}+\cdots+a_{19}+b_{... | ［Solution］ In general, let $a_{q}=p$, we prove
$$a_{1}+a_{2}+\cdots+a_{q}+b_{1}+b_{2}+\cdots+b_{p}=p(q+1)$$
For this problem, when $q=19, p=85$,
$$p(q+1)=85(19+1)=1700 \text {. }$$
(i) If $a_{1}=a_{2}=\cdots=a_{q}=p$, then by definition
$$b_{1}=b_{2}=\cdots=b_{p}=1$$
Thus we have
$$a_{1}+a_{2}+\cdots+a_{q}+b_{1}+b_{2... | 1700 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,325 |
2.97 Let $a, b, c$ be pairwise coprime positive integers, prove: $2abc - ab - bc - ca$ is the largest integer that cannot be expressed in the form $xbc + yca + zab$ (where $x, y, z$ are non-negative numbers). | [Proof] If
$$2 a b c-b c-c a-a b=x b c+y c a+z a b,$$
where $x \geqslant 0, y \geqslant 0, z \geqslant 0$, then we have
$$2 a b c=b c(x+1)+c a(y+1)+a b(z+1).$$
Thus $a \mid (x+1) b c$ and since $(a, b c)=1$, it follows that $a \mid (x+1)$. Therefore, $a \leqslant x+1$.
Similarly, we have $b \leqslant y+1, c \leqslant... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,326 |
1.4 Proof: There exists a number divisible by $5^{1000}$ and whose decimal representation does not contain the digit 0.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
1.4 Proof: There exists a number divisible by $5^{1000}$ and whose decimal rep... | [Proof] The last digit of the number $5^{1000}$ is 5. If the representation of $5^{1000}$ does not contain 0, the proposition is proved. Otherwise, there exists an integer $k>1$ such that in the decimal notation of $5^{1000}$, the $k$-th digit from the last is 0, and all the digits from the 1st to the $(k-1)$-th positi... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,328 |
$1 \cdot 31$ Let $\alpha_{n}$ denote the integer closest to $\sqrt{n}$, find the sum $\frac{1}{\alpha_{1}}+\frac{1}{\alpha_{2}}+\cdots+\frac{1}{\alpha_{1980}}$ | [Sol] Since $\alpha_{n}=k$ is equivalent to
$$k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2}$$
which means $k^{2}-k<n \leqslant k^{2}+k$,
thus $\frac{1}{\alpha_{(k-1) k+1}}+\frac{1}{\alpha_{(k-1) k+2}}+\cdots+\frac{1}{\alpha_{k(k+1)}}=\frac{1}{k} \cdot 2 k=2$, hence $\left(\frac{1}{\alpha_{1}}+\frac{1}{\alpha_{2}}\right)+\left(... | 88 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,329 |
2. 102 Real numbers $u, v$ satisfy the relation
$$\begin{aligned}
\left(u+u^{2}+u^{3}+\cdots+u^{8}\right)+10 u^{9} & =\left(v+v^{2}+v^{3}+\cdots+v^{10}\right)+ \\
& 10 v^{11} \\
& =8 .
\end{aligned}$$
Determine which of $u$ and $v$ is larger, and prove your conclusion. | [Solution] From the given relation, we know that
$$u0, v>0$. Therefore,
$$00 \\
g(0)=-80
\end{array}$$
Thus, $u$ and $v$ are the unique roots of the equations $f(x)=0$ and $g(x)=0$ in the interval $(0,1)$, respectively.
From equation (1), we know that $u$ is the root of
$$f_{1}(x)=(10 x-9)\left(x^{9}-1\right)+x-1=0$$... | u < v | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,330 |
2. 103 Use 94 bricks, each of size $4 \times 10 \times 19$, to stack into a tower 94 bricks high, where each brick can be oriented to provide a height of 4, 10, or 19. If all 94 bricks are used, how many different heights can the tower be (in inches)? | [Solution] Since
$$10 \times 5 = 19 \times 2 + 4 \times 3,$$
this means that each of the 5 bricks can provide a height of 10, totaling a height of 50 for the tower, and even if none of the 5 bricks provide a height of 10, they can still provide a height of 50 for the tower. Therefore, we can assume that among the 94 b... | 465 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,331 |
2・104 Let $n \geqslant 2, x_{1}, x_{2}, \cdots, x_{n}$ be real numbers, and
$$\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}=1$$
For every fixed $k(k \in N, 1 \leqslant k \leqslant n)$, find the maximum value of $\left|x_{k}\right|$. | [Solution] Let $s=\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}$, then
$$\begin{aligned}
s= & \left(x_{1}+\frac{1}{2} x_{2}\right)^{2}+\frac{3}{4} x_{2}^{2}+\sum_{i=2}^{n-1} x_{i} x_{i+1}+\sum_{i=3}^{n} x_{i}^{2} \\
= & \left(x_{1}+\frac{1}{2} x_{2}\right)^{2}+\left(\sqrt{\frac{3}{4}} x_{2}+\frac{1}{2 \cdot \... | \max \left|x_{k}\right|=\sqrt{\frac{2 k(n-k+1)}{n+1}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,332 |
2・ 106 For each finite set of non-zero vectors $\cup$ in the plane, define $l(U)$ as the length of the sum vector of all vectors in $\cup$. Given a finite set of non-zero vectors $V$ in the plane, if for each non-empty subset $A$ of $V$, $l(B)$ is greater than or equal to $l(A)$, then the subset $B$ of $V$ is said to b... | [Solution] (a) For $n=4$, consider quadrilateral $ABCD$, where $AB=BC=CA=DB, AD=$ $DC$ (as shown in the figure). Take vectors $\overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{CD}, \overrightarrow{DA}$.
Let $V=\{\overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{CD}, \overrightarrow{DA} \mid$
Then $V$ ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,333 |
2- 107 Let $n \geqslant 5$ be a natural number, and $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ different natural numbers with the following property: for any two different non-empty subsets $A$ and $B$ of the set
$$S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$$
the sum of the numbers in $A$ and the sum of the numbers in $B$... | ［Solution］ Without loss of generality, let $a_{1}C_{2}>\cdots>C_{n}, D_{k}=\sum_{i=1}^{k} a_{i}-\left(1+2+\cdots+2^{k-1}\right)$
$$=\sum_{i=1}^{k} a_{i}-\left(2^{k}-1\right) \geqslant 0$$
Thus, we have
$$\begin{aligned}
& 1+\frac{1}{2}+\cdots+\frac{1}{2^{n-1}}-\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n... | 2-\frac{1}{2^{n-1}} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,334 |
2・108 In a class of 30 students, each student has the same number of friends within the class. How many students at most can there be who are better than the majority of their friends in terms of academic performance (assuming that for any two students in the class, it can be determined who performs better)? | [Solution] Answer: 25 students.
In fact, students who perform better than the majority of their classmates are referred to as "good students." Let the class have $x$ "good students," and each student has $k$ friends in the class.
Obviously, the student with the best grades in the class is the better performer in $k$ "... | 25 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,335 |
$2 \cdot 109$ Let $S=\left\{A=\left(a_{1}, \cdots, a_{8}\right) \mid a_{i}=0\right.$ or $\left.1, i=1,2, \cdots, 8\right\}$. For two elements $A=\left(a_{1}, \cdots, a_{8}\right)$ and $B=\left(b_{1}, \cdots, b_{8}\right)$ in $S$, let
$$d(A, B)=\sum_{i=1}^{8}\left|a_{i}-b_{i}\right|,$$
and call it the distance between ... | [Solution] $S^{\prime}$ is a subset of $S$, and the distance between any two elements in $S^{\prime}$ is $\geqslant 5$. Let $A=\left(a_{1}, \cdots, a_{8}\right)$ and $B=\left(b_{1}, \cdots, b_{8}\right)$ be two elements in $S^{\prime}$, and let
$$\omega(A)=\sum_{i=1}^{8} a_{i}, \omega(B)=\sum_{i=1}^{8} b_{i}$$
Clearly... | 4 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,336 |
2-110 Let the sequence of positive real numbers $x_{0}, x_{1}, \cdots, x_{1995}$ satisfy the following two conditions:
(1) $x_{0}=x_{1995}$;
(2) $x_{i-1}+\frac{2}{x_{i-1}}=2 x_{i}+\frac{1}{x_{i}}, i=1,2, \cdots, 1995$.
Find the maximum value of $x_{0}$ for all sequences satisfying the above conditions. | 【Solution】 From condition (2), we know
$$\begin{array}{l}
x_{i}^{2}-\left(\frac{x_{i-1}}{2}+\frac{1}{x_{i-1}}\right) x_{i}+\frac{1}{2}=0, \\
\left(x_{i}-\frac{x_{i-1}}{2}\right)\left(x_{i}-\frac{1}{x_{i-1}}\right)=0, \\
x_{i}=\frac{x_{i-1}}{2} \text { or } x_{i}=\frac{1}{x_{i-1}} .
\end{array}$$
The transformation tha... | 2^{997} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,337 |
$1 \cdot 32$ Find all natural numbers $x$, such that $x^{2}$ is a twelve-digit number in the form of $x^{2}=$ $2525 * * * * * * 89$ (the six unknown digits represented by * do not have to be the same).
Translate the above text into English, please retain the original text's line breaks and format, and output the trans... | [Solution] Write down the squares of all two-digit numbers whose last digit is 3 or 7, and we can see that the last two digits of the number $x$ can be $17, 33, 67, 83$. Since the first four digits of the square of the number we are looking for are 2525. Therefore $\square$
$$2525 \times 10^{8}<x^{2}<2526 \times 10^{8}... | null | Geometry | math-word-problem | Yes | Yes | inequalities | false | 735,338 |
2・111 $n$ is a positive integer, $A$ is a set of subsets of the set $\{1,2, \cdots, n\}$, such that no element in $A$ contains another element in $A$, find the maximum number of elements in all elements of $A$.
untranslated text:
2・111 $n$ is a positive integer, $A$ is a set of subsets of the set $\{1,2, \cdots, n\}$... | [Solution] Consider the permutations of $n$ elements $1,2, \cdots, n$. Clearly, the number of such permutations is $n!$.
Let $A$ contain $f_{k}$ $k$-element subsets as elements, where $f_{k}(k=1,2, \cdots, n)$ are non-negative integers. Let $|A|$ denote the number of elements in $A$, then
$$|A|=\sum_{k=1}^{n} f_{k}$$
... | C_{n}^{\left[\frac{n}{2}\right]} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,339 |
3・1 Simplify:
$$\log _{a}\left[\left(\frac{m^{4} n^{-4}}{m^{-1} n}\right)^{-3} \div\left(\frac{m^{-2} n^{2}}{m n^{-1}}\right)^{5}\right]$$
Here $m$, $n$, and $a$ are all positive numbers, $a \neq 1$. | [Solution] $\begin{aligned} & \log _{a}\left[\left(\frac{m^{4} n^{-4}}{m^{-1} n}\right)^{-3} \div\left(\frac{m^{-2} n^{2}}{m n^{-1}}\right)^{5}\right] \\ = & \log _{a}\left[\left(m^{5} n^{-5}\right)^{-3} \div\left(m^{-3} n^{3}\right)^{5}\right] \\ = & \log _{a}\left(m^{-15} n^{15} \div m^{-15} n^{15}\right) \\ = & \log... | 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,340 |
$3 \cdot 2$ Find $1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\cdots+(n-1)(n-1)!+n \cdot n$ !, where $n!=n(n-1)(n-2) \cdots 2 \cdot 1$. | [Solution] Since $k \cdot k!=(k+1)!-k!\quad k=1,2,3, \cdots$
Therefore,
$$\begin{aligned}
& 1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\cdots+(n-1)(n-1)!+n \cdot n! \\
& =(2!-1!)+(3!-2!)+\cdots+[n!-(n-1)!]+[(n+1)!- \\
& n!] \\
= & (n+1)!-1 .
\end{aligned}$$ | (n+1)!-1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,341 |
$3 \cdot 5$ Simplify the fraction $\frac{a+a^{2}+a^{3}+\cdots+a^{m}}{a^{-1}+a^{-2}+a^{-3}+\cdots+a^{-m}}$. | $\begin{array}{l}\text { [Solution] Original expression }=\frac{a^{m} \cdot a\left(1+a+a^{2}+\cdots+a^{m-1}\right)}{a^{m}\left(a^{-1}+a^{-2}+a^{-3}+\cdots+a^{-m}\right)} \\ =\frac{a^{m+1}\left(1+a+a^{2}+\cdots+a^{m-1}\right)}{a^{m-1}+a^{m-2}+\cdots+a^{2}+a+1} \\ =a^{m+1} .\end{array}$ | a^{m+1} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,343 |
$\begin{array}{l}3 \cdot 7 \text { if } 0<x<1 \text {, simplify: } \\ \left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^{2}}+x-1}\right) \times\left(\sqrt{\frac{1}{x^{2}}-1}-\frac{1}{x}\right) .\end{array}$ | [Solution]
$$\begin{aligned}
\text { Original expression } & =\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} \cdot \frac{\sqrt{1-x^{2}}-1}{x} \\
& =\frac{(\sqrt{1+x})^{2}-(\sqrt{1-x})^{2}}{(\sqrt{1+x}-\sqrt{1-x})^{2}} \cdot \frac{\sqrt{1-x^{2}}-1}{x} \\
& =\frac{2 x}{2-2 \sqrt{1-x^{2}}} \cdot \frac{\sqrt{1-x^{2}}-... | -1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,344 |
3.9 Given numbers $\alpha$ and $\beta$ satisfy the following two equations: $\alpha^{3}-3 \alpha^{2}+5 \alpha=1, \beta^{3}-3 \beta^{2}+5 \beta=5$, try to find $\alpha+\beta$. | [Solution] Let $f(x)=x^{3}-3 x^{2}+5 x$
$$=(x-1)^{3}+2(x-1)+3 .$$
Then the left sides of the given equations are $f(\alpha)$ and $f(\beta)$, respectively. Let
$$\begin{array}{l}
g(y)=y^{3}+2 y \\
g(\alpha-1)=f(\alpha)-3=-2 \\
g(\beta-1)=f(\beta)-3=2
\end{array}$$
Since $g(y)$ is an odd function and also an increasing... | \alpha+\beta=2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,346 |
1.33 Find all natural numbers $n \in[1000,2000]$, such that $a_{n}=\sqrt{57121+35 n}$ is a natural number. | [Solution] According to the condition $1000 \leqslant n \leqslant 2000$, we get
$a_{n} \geqslant \sqrt{57121+35 \times 1000}=303.51441 \cdots$
and $a_{n} \leqslant \sqrt{57121+35 \times 2000}=356.54032 \cdots$,
which means $\quad 304 \leqslant a_{n} \leqslant 357$
Also, since $57121=1632 \times 35+1$,
we have $a_{n}=\s... | 1096, 1221, 1749, 1888, 1185, 1312, 1848, 1989 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,347 |
3. 11 Simplify: $\frac{1-x-x^{2}+x^{3}}{1-2|x|+x^{2}}$ | [Solution] Numerator $=(1-x)-x^{2}(1-x)=(1-x)^{2}(1+x)$.
$$\begin{array}{c}
\text { Denominator }=\left\{\begin{array}{ll}
1-2 x+x^{2}=(1-x)^{2}, & (\text { when } x \geqslant 0 \text { ) } \\
1+2 x+x^{2}=(1+x)^{2} . & (\text { when } x<0 \text { ) }
\end{array}\right. \\
\text { Original expression }=\left\{\begin{arr... | 1+x \text{ (when } x \geqslant 0 \text{)}, \frac{(1-x)^{2}}{1+x} \text{ (when } x < 0 \text{)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,348 |
3.12 Try to find the sum:
$$S=m!+\frac{(m+1)!}{1!}+\frac{(m+2)!}{2!}+\cdots+\frac{(m+n)!}{n!}$$ | [Solution] $S=m!+\frac{(m+1)!}{1!}+\frac{(m+2)!}{2!}+\cdots+\frac{(m+n)!}{n!}$
$$\begin{aligned}
= & m!\left[1+\frac{m+1}{1!}+\frac{(m+2)(m+1)}{2!}+\cdots+\right. \\
& \left.+\frac{(m+n)(m+n-1) \cdots(m+2)(m+1)}{n!}\right] \\
= & m!\left(C_{m}^{0}+C_{m+1}^{1}+C_{m+2}^{2}+\cdots+C_{m+n}^{n}\right)
\end{aligned}$$
Using... | \frac{(m+n+1)!}{(m+1) \cdot n!} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,349 |
3. 13 Prove: If $x, y, z$ are distinct integers, and $n$ is a non-negative integer, then $\frac{x^{n}}{(x-y)(x-z)}+\frac{y^{n}}{(y-x)(y-z)}+\frac{z^{n}}{(z-x)(z-y)}$ is an integer. | [Proof] After making the original expression have a common denominator, it becomes
$$\frac{x^{n}(y-z)+y^{n}(z-x)+z^{n}(x-y)}{(x-y)(x-z)(y-z)}$$
When $n=0,1$, the numerator of (1) is 0.
When $n \geqslant 2$, the numerator of (1) can be transformed into
$$\begin{aligned}
& \left(x^{n}-z^{n}\right)(y-z)+\left(y^{n}-z^{n}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,350 |
3.14 If \( x + \frac{1}{x} = a \), where \( a \) is a known number, find: \( x^{13} + \frac{1}{x^{13}} \). | [Solution] Since
$$\left(x^{m}+\frac{1}{x^{m}}\right)\left(x^{n}+\frac{1}{x_{n}}\right)=\left(x^{m+n}+\frac{1}{x^{m+n}}\right)+\left(x^{m-n}+\frac{1}{x^{m-n}}\right),$$
Therefore,
$$x^{m+n}+\frac{1}{x^{m+n}}=\left(x^{m}+\frac{1}{x^{m}}\right)\left(x^{n}+\frac{1}{x_{n}}\right)-\left(x^{m-n}+\frac{1}{x^{m-n}}\right) .$$... | a^{13}-13 a^{11}+65 a^{9}-156 a^{7}+182 a^{5}-91 a^{3}+13 a | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,351 |
$3 \cdot 15$ Simplify: $\frac{\csc \alpha}{\sqrt{1+\operatorname{ctg}^{2} \alpha}}-\frac{\cos \alpha}{\sqrt{1-\sin ^{2} \alpha}}$. | [Solution] Original expression $=\frac{\csc \alpha}{|\csc \alpha|}-\frac{\cos \alpha}{|\cos \alpha|}$ $=\left\{\begin{array}{l}0, \text { when } 2 k \pi<\alpha<\left(2 k+\frac{1}{2}\right) \pi \text {; } \\ 2, \text { when }\left(2 k+\frac{1}{2}\right) \pi<\alpha<(2 k+1) \pi \text {; } \\ 0, \text { when }(2 k+1) \pi<\... | \begin{cases}
0, & \text{when } 2 k \pi < \alpha < \left(2 k + | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,352 |
3・17 (1) Simplify $\frac{1-a^{2}}{(1+a x)^{2}-(a+x)^{2}}$;
(2) When $x=0.44$, find the value of $\sqrt{1-x-x^{2}+x^{3}}$. | [Solution]
$$\text { (1) } \begin{aligned}
\text { Original expression } & =\frac{(1+a)(1-a)}{(1+a x+a+x)(1+a x-a-x)} \\
& =\frac{(1+a)(1-a)}{(1+a)(1+x)(1-a)(1-x)} \\
& =\frac{1}{(1+x)(1-x)} .
\end{aligned}$$
(2)
$$\begin{aligned}
& \sqrt{1-x-x^{2}+x^{3}}=\sqrt{(1-x)-x^{2}(1-x)} \\
= & \sqrt{(1-x)\left(1-x^{2}\right)}=... | 0.672 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,354 |
$$3 \cdot 19 \text { Let } \frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}+\frac{1}{1}}}=\frac{m}{n} \text { , }$$
where $m$ and $n$ are coprime natural numbers, and the left side of the equation contains 1988 fraction bars, try to calculate the value of $m^{2}+m n-n^{2}$ | [Solution] Let the value of the above complex fraction with $k$ fraction lines be $\frac{m_{k}}{n_{k}}\left(m_{k} 、 n_{k}\right.$ being coprime natural numbers), then
$$\frac{m_{k+1}}{n_{k+1}}=\frac{1}{1+\frac{m_{k}}{n_{k}}}=\frac{n_{k}}{m_{k}+n_{k}}$$
Notice that
$$\begin{array}{l}
\frac{m_{1}}{n_{1}}=\frac{1}{1}, m_... | -1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,355 |
3. 20 Let $[x]$ denote the greatest integer not exceeding $x$. Try to compute the sum $\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right]$ for any positive integer $n$. | [Solution] First, we prove that for all real numbers $x$, the equation
$$\left[x+\frac{1}{2}\right]=[2 x]-[x]$$
holds.
In fact, let $x=[x]+\alpha$, then $0 \leqslant \alpha<1$,
so $\left[x+\frac{1}{2}\right]=\left[[x]+\alpha+\frac{1}{2}\right]$
$$=[x]+\left[\alpha+\frac{1}{2}\right]$$
$$\begin{aligned}
{[2 x]-[x] } & ... | n | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,356 |
$3 \cdot 21 x$ takes what real value(s) when the equation
(a) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{2}$,
(b) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=1$,
(c) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=2$.
all hold (here the square root takes non-negative values)? | [Solution] Construct the auxiliary function
$$f(x)=\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}$$
Then the domain of $f(x)$ is $\left\{x \left\lvert\, x \geqslant \frac{1}{2}\right.\right\}$.
Notice that $(\sqrt{2 x-1}+1)^{2}=2(x+\sqrt{2 x-1})$
and $(\sqrt{2 x-1}-1)^{2}=2(x-\sqrt{2 x-1})$,
so we have $\sqrt{x+\sqrt{2 x... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,357 |
3. 22 Let $p(x)$ be the product of the digits of the decimal number $x$. Try to find all positive numbers $x$ that satisfy $p(x)=x^{2}-10 x-22$. | [Solution] Let $x$ be an $n$-digit number, then
$$p(x) \leqslant 9^{n}, \quad x^{2} \geqslant 10^{n-1} \cdot x$$
If $n>2$, then
$$\begin{aligned}
p(x)=x^{2}-10 x-22 & \geqslant x\left(10^{n-1}-10\right)-22 \\
& \geqslant 10^{n-1} \cdot 90-22=9 \cdot 10^{n}-22 \\
& >9^{n}, \text { contradiction. }
\end{aligned}$$
$$\th... | 12 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,358 |
3. 23 Given the numbers $1,2,2^{2}, \cdots, 2^{n-1}$, for any permutation $\sigma=\left(x_{1}\right.$, $\left.x_{2}, \cdots, x_{n}\right)$ of them, define $S_{1}(\sigma)=x_{1}, S_{2}(\sigma)=x_{1}+x_{2}, S_{3}(\sigma)=x_{1}+x_{2}+$ $x_{3}, \cdots, S_{n}(\sigma)=x_{1}+x_{2}+\cdots+x_{n}$, and let $Q(\sigma)=S_{1}(\sigma... | [Solution] We prove a more general conclusion:
For any $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, the sum $\sum \frac{1}{Q(\sigma)}=$ $S\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\frac{1}{a_{1} a_{2} \cdots a_{n}}$.
We use induction on $n$.
When $n=1$, $S\left(a_{1}\right)=\sum \frac{1}{Q(\sigma)}=\frac{1}{a_{1... | 2^{-\frac{n(n-1)}{2}} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,359 |
$3 \cdot 25$ For each positive integer $n$, let
$$\begin{array}{l}
S_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} \\
T_{n}=S_{1}+S_{2}+S_{3}+\cdots+S_{n} \\
U_{n}=\frac{1}{2} T_{1}+\frac{1}{3} T_{2}+\frac{1}{4} T_{3}+\cdots+\frac{1}{n+1} T_{n}
\end{array}$$
Find integers $0<a, b, c, d<1000000$, such that
$$\begin{... | [Solution] From the given information,
$$\begin{aligned}
T_{n}= & S_{1}+S_{2}+S_{3}+\cdots+S_{n} \\
= & 1+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{2}+\frac{1}{3}\right)+\cdots+\left(1+\frac{1}{2}+\frac{1}{3}+\right. \\
& \left.\cdots+\frac{1}{n}\right) \\
= & n+\frac{1}{2}(n-1)+\frac{1}{3}(n-2)+\cdots+\frac{1}{n}[n-... | T_{1988}=1989 S_{1989}-1989, \quad U_{1988}=1990 S_{1989}-3978 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,361 |
$3 \cdot 27$ When the natural number $n \geqslant 2$ is the smallest, find integers $a_{1}, a_{2}, \cdots, a_{n}$, such that the following equation $a_{1}+a_{2}+\cdots+a_{n}=a_{1} \cdot a_{2} \cdot \cdots a_{n}=1990$ holds. | [Solution] First, we prove that the minimum value of $n$ is 5. The number 1990 can only be divided by 2 but not by 4, so there is only one even number in $a_{1} \cdot a_{2} \cdots a_{n}$. Therefore, in the sum $a_{1}+a_{2}+\cdots+a_{n} = 1990$, there must be an even number of odd addends, which means $n$ is odd.
If $a... | 5 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,362 |
$3 \cdot 28$ Try to express $\sum_{k=0}^{n} \frac{(-1)^{k} C_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}$ in the form $\frac{p(n)}{q(n)}$, where $p(n)$ and $q(n)$ are two polynomials with integer coefficients. | [Solution] Factorization yields
$$k^{3}+9 k^{2}+26 k+24=(k+2)(k+3)(k+4)$$
Let
$$S(n)=\sum_{k=0}^{n} \frac{(-1)^{k} \cdot C_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}$$
Then $\square$
Let
$$\begin{aligned}
S(n)= & \sum_{k=0}^{n} \frac{(-1)^{k} \cdot n!}{k!(n-k)!(k+2)(k+3)(k+4)} \\
= & \sum_{k=0}^{n}\left[\frac{(-1)^{k}(n+4)!}{(... | \frac{1}{2(n+3)(n+4)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,363 |
3. 29 For a natural number $n \geqslant 3$. In the plane, a line $l$ is given, on which there are $n$ distinct points $p_{1}, p_{2}, \cdots, p_{n}$ in sequence. The product of the distances from point $p_{1}$ to the other $n-1$ points is denoted as $d_{i}(i=1,2, \cdots, n)$. There is also a point $Q$ in the plane not o... | [Solution] Without loss of generality, let these $n$ points be on the real axis, with coordinates $p_{i}\left(x_{i}, 0\right)(i=1,2, \cdots, n), x_{1}3 \text { when } .
\end{array}\right.$$
Substituting into (1), we get
$$S_{n}=\left(\alpha^{2}+\beta^{2}\right) T_{0}-2 \alpha T_{1}+T_{2}=T_{2}=\left\{\begin{array}{l}
... | S_{n} = \left\{\begin{array}{l} 1, n=3 \\ 0, n>3 \end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,364 |
1.35 A four-digit number is composed of four consecutive digits in sequence. If the first two digits from the left are swapped, the resulting number is a perfect square. Find the four-digit number.
保留源文本的换行和格式，直接输出翻译结果。 | [Solution] Let the thousand's digit of the required four-digit number be $a$, then the hundred's digit, ten's digit, and unit's digit are $a+1, a+2$, and $a+3 (1 \leqslant a \leqslant 6)$, or $a-1, a-2$, and $a-3 (3 \leqslant a \leqslant 9)$, respectively.
From the problem, we have
$$(a+1) \times 1000 + a \times 100 +... | 3456 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,365 |
$3 \cdot 30$ (1) Let $n$ be a prime number greater than 3. Find the value of $\left(1+2 \cos \frac{2 \pi}{n}\right)$ $\left(1+2 \cos \frac{4 \pi}{n}\right)\left(1+2 \cos \frac{6 \pi}{n}\right) \cdots \cdot\left(1+2 \cos \frac{2 n \pi}{n}\right)$;
(2) Let $n$ be a natural number greater than 3, find the value of $\left(... | [Solution] (1) Let $w=e^{\frac{2 \pi i}{n}}$. Clearly, $w^{n}=1, w^{-\frac{n}{2}}=-1, w^{k}+w^{-k}=2 \cos \frac{2 k \pi}{n}$. Therefore, we have
$$\begin{aligned}
& \prod_{k=1}^{n}\left(1+2 \cos \frac{2 k \pi}{n}\right)=\prod_{k=1}^{n}\left(1+w^{k}+w^{-k}\right) \\
= & \prod_{k=1}^{n} w^{-k}\left(w^{k}+w^{2 k}+1\right)... | 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,366 |
3. 31 Given real numbers $a, b, c$. It is known that the complex numbers $z_{1}, z_{2}, z_{3}$ satisfy:
$$\left\{\begin{array}{l}
\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1, \\
\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{3}}+\frac{z_{3}}{z_{1}}=1 .
\end{array}\right.$$
Find the value of $\left|a z_{1}+b z_{2}+... | [Solution] Let $e^{i \theta}=\cos \theta+i \sin \theta$. Suppose
$$\frac{z_{1}}{z_{2}}=e^{i \theta}, \frac{z_{2}}{z_{3}}=e^{i \varphi},$$
then $\quad \frac{z_{3}}{z_{1}}=e^{-i(\theta+\varphi)}$.
From the given condition, we have
$$e^{i \theta}+e^{i \varphi}+e^{-i(\theta+\varphi)}=1$$
Taking the imaginary part on both... | \sqrt{(a+b)^{2}+c^{2}} \text{ or } \sqrt{a^{2}+(b+c)^{2}} \text{ or } \sqrt{(c+a)^{2}+b^{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,367 |
3.35 Prove:
$$\begin{aligned}
& (1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right) \cdots\left(1+x^{2^{k-1}}\right) \\
= & 1+x+x^{2}+x^{3}+\cdots+x^{2^{k}-1}
\end{aligned}$$ | [Proof] (By mathematical induction) When $k=1$, the equation obviously holds. If the equation holds for $k \geqslant 1$, i.e., equation (1), then multiplying both sides of equation (1) by $1+x^{2^{k}}$, we get
$$\begin{aligned}
& (1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right) \cdots\left(1+x^{2^{k-1}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,368 |
3・36 Given $x+y+z=0$, prove:
$$\left(\frac{x^{2}+y^{2}+z^{2}}{2}\right)\left(\frac{x^{5}+y^{5}+z^{5}}{5}\right)=\left(\frac{x^{7}+y^{7}+z^{7}}{7}\right) .$$ | $$\begin{aligned}
{[\text { Proof }] } & \because x+y+z=0, z=-(x+y) \\
\text { Left side }= & \frac{1}{2}\left[x^{2}+y^{2}+(x+y)^{2}\right] \cdot \frac{1}{5}\left[x^{5}+y^{5}-(x+y)^{5}\right] \\
= & \frac{1}{2}\left[2\left(x^{2}+y^{2}+x y\right)\right] \cdot \frac{1}{5}\left\{\left[-5\left(x^{4} y+2 x^{3} y^{2}+2 x^{2}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,369 |
3-38 Proof: For any natural number $n$, the following equation holds:
$$\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}}_{n \text { square roots }}=2 \cdot \cos \frac{\pi}{2^{n+1}} \text {. }$$ | [Proof] When $n=1$, the conclusion holds.
Assume that when $n=k$, the conclusion holds. Then, using the identity $\cos \alpha=2 \cdot \cos ^{2} \frac{\alpha}{2}-1$ and the induction hypothesis, we get
$$\begin{aligned}
& \underbrace{}_{k+1 \text { terms }} \sqrt{2+\sqrt{2+\cdots+\sqrt{2}}} \\
= & \sqrt{2+2 \cdot \cos \... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,370 |
$3 \cdot 39$ Ten athletes participate in a table tennis round-robin tournament, where each pair of them plays exactly one match. During the tournament, the first player wins $x_{1}$ matches and loses $y_{1}$ matches, the second player wins $x_{2}$ matches and loses $y_{2}$ matches, and so on. Prove:
$$x_{1}^{2}+x_{2}^{... | [Proof] Since each player plays 9 games, we have
$$x_{1}+y_{1}=x_{2}+y_{2}=\cdots=x_{10}+y_{10}=9 \text {, }$$
Also, since each game results in exactly one win and one loss, we have
$$x_{1}+x_{2}+\cdots+x_{10}=y_{1}+y_{2}+\cdots+y_{10}$$
Thus, we have
$$\begin{aligned}
& x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}-y_{1}^{2... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,371 |
1-36 Find all such four-digit numbers, when 400 is written to their left, the result is a perfect square.
Find all such four-digit numbers, when 400 is written to their left, the result is a perfect square. | [Solution] Let $\overline{a b c d}$ be the required four-digit number. Since $\overline{400 a b c d}$ is a seven-digit number and a perfect square, we have
$$\overline{400 a b c d}=(20 l m)^{2}$$
Rewriting (1) as
$$400 \times 10^{4}+\overline{a b c d}=\left(20 \times 10^{2}+10 l+m\right)^{2}$$
Expanding and simplifyi... | 4001 \text{ or } 8004 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,372 |
$3 \cdot 40$ Divide $1,2,3, \cdots, 2 n-1,2 n$ into two groups, each containing $n$ numbers. Let $a_{1}b_{2}>\cdots>b_{n}$ be the second group of numbers written in decreasing order. Prove:
$$\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\cdots+\left|a_{n}-b_{n}\right|=n^{2}$$ | [Proof] Each term $\left|a_{k}-b_{k}\right|(k=1,2, \cdots, n)$ is exactly the absolute value of the difference between a number greater than $n$ and a number not exceeding $n$. Therefore,
$$\begin{aligned}
& \left|a_{1}-b_{1}\right|+\cdots+\left|a_{n}-b_{n}\right| \\
= & (n+1)+(n+2)+\cdots+2 n-(1+2+\cdots+n) \\
= & n^{... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,373 |
3. 41 Proof: The identity $\frac{2}{x^{2}-1}+\frac{4}{x^{2}-4}+\frac{6}{x^{2}-9}+\cdots+\frac{20}{x^{2}-100}=$
$$11\left(\frac{1}{(x-1)(x+10)}+\frac{1}{(x-2)(x+9)}+\cdots+\frac{1}{(x-10)(x+1)}\right) .$$ | [Proof] Since $\frac{2 k}{x^{2}-k^{2}}=\frac{1}{x-k}-\frac{1}{x+k},(k=1,2, \cdots, 10)$ and $\frac{11}{(x-11+k)(x+k)}=\frac{1}{x-11+k}-\frac{1}{x+k}(k=1,2, \cdots, 10)$.
Therefore,
$$\begin{aligned}
\sum_{k=1}^{10} \frac{2 k}{x^{2}-k^{2}} & =\sum_{k=1}^{10} \frac{1}{x-k}-\sum_{k=1}^{10} \frac{1}{x+k} \\
& =\sum_{k=1}^... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,374 |
3. 43 If $x=\left(1+\frac{1}{n}\right)^{n}, y=\left(1+\frac{1}{n}\right)^{n+1}$, prove: $y^{x}=x^{y}$.
| [Proof]
$$\begin{aligned}
y^{x}=[ & \left.\left(1+\frac{1}{n}\right)^{n+1}\right]^{\left(1+\frac{1}{n}\right)^{n}} \\
& =\left[\left(1+\frac{1}{n}\right)^{n\left(1+\frac{1}{n}\right)}\right]^{\left(1+\frac{1}{n}\right)^{n}} \\
& =\left[\left(1+\frac{1}{n}\right)^{n}\right]^{\left(1+\frac{1}{n}\right)^{n+1}} \\
& =x^{y}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,375 |
3.44 Let $p_{n}(k)$ denote the number of permutations of $n$ elements in which $k$ elements are fixed. Prove that:
$$\sum_{k=0}^{n} k p_{n}(k)=n!$$ | [Proof] Since $p_{n}(k)=c_{n}^{k} p_{n-k}(0)=\frac{n!}{k!(n-k)!} p_{n-k}(0)$, we have
$$\begin{aligned}
\sum_{k=0}^{n} k p_{n}(k) & =\sum_{k=1}^{n} k \frac{n!}{k!(n-k)!} p_{n-k}(0) \\
& =n \sum_{k=0}^{n-1} \frac{(n-1)!}{k!(n-k-1)!} p_{n-1-k}(0) \\
& =n \sum_{k=0}^{n-1} c_{n-1}^{k} p_{n-1-k}(0) \\
& =n \sum_{k=0}^{n-1} ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,376 |
3.45 Proof: $\frac{1}{\cos 0^{\circ} \cos 1^{\circ}}+\frac{1}{\cos 1^{\circ} \cos 2^{\circ}}+\cdots+\frac{1}{\cos 88^{\circ} \cos 89^{\circ}}=\frac{\cos 1^{\circ}}{\sin ^{2} 1^{\circ}}$. | [Proof] For the convenience of writing, we agree that the angles in the following trigonometric functions are in degrees, i.e., $\sin 1$ is $\sin 1^{\circ}, \sin (x+1)$ refers to $\sin (x+1)^{\circ}$. Thus,
$$\text { the right side of the original equation }=\frac{\cos 1}{\sin ^{2} 1}=\frac{\operatorname{ctg} 1}{\sin 1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,377 |
3. 48 If $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{A+B+C}$, then for any odd number $n$,
$$\frac{1}{A^{n}}+\frac{1}{B^{n}}+\frac{1}{C^{n}}=\frac{1}{A^{n}+B^{n}+C^{n}}$$
Prove it. | [Solution] Transform the original equation into:
$$\begin{aligned}
& \frac{1}{A}+\frac{1}{B}+\frac{1}{C}-\frac{1}{A+B+C} \\
\equiv & \frac{(A+B)(B+C)(C+A)}{A B C(A+B+C)}=0
\end{aligned}$$
Thus, $A, B, C$ should at least satisfy one of the following three equations:
$$A=-B, B=-C, C=-A .$$
Since $n$ is any odd number, ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,378 |
3.49 Positive numbers $u_{1}, u_{2}, \cdots, u_{n}$ form an arithmetic sequence, prove:
$$\begin{aligned}
t_{n} & =\frac{1}{\sqrt{u_{1}}+\sqrt{u_{2}}}+\frac{1}{\sqrt{u_{2}}+\sqrt{u_{3}}}+\cdots+\frac{1}{\sqrt{u_{n-1}}+\sqrt{u_{n}}} \\
& =\frac{n-1}{\sqrt{u_{1}}+\sqrt{u_{n}}} .
\end{aligned}$$ | [Proof] Let $d=u_{k}-u_{k-1}$ be the common difference of the arithmetic sequence, then
$$\begin{aligned}
t_{n} & =\frac{\sqrt{u_{2}}-\sqrt{u_{1}}}{u_{2}-u_{1}}+\frac{\sqrt{u_{3}}-\sqrt{u_{2}}}{u_{3}-u_{2}}+\cdots+\frac{\sqrt{u_{n}}-\sqrt{u_{n-1}}}{u_{n}-u_{n-1}} \\
& =\frac{\sqrt{u_{n}}-\sqrt{u_{1}}}{d} . \\
& =\frac{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,379 |
1.37 Four three-digit numbers with the same first digit are distinct and have the property: the sum of these numbers is divisible by any 3 of them. Find these 4 numbers.
The text above is translated into English, preserving the original text's line breaks and format. | [Solution] Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the numbers we are looking for, $S$ be their sum, and $a$ be their common first digit. Thus, we have
$$100 a \leqslant x_{i} \leqslant 100(a+1), \quad (i=1,2,3,4)$$
Using these inequalities, we get
$$x_{i} + 300 a \leqslant S < x_{i} + 300(a+1)$$
From this, we have
$$1 +... | 108, 135, 180, 117 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,380 |
3・50 Prove:
$$\begin{aligned}
& \frac{1}{1 \times 2}+\frac{1}{3 \times 4}+\frac{1}{5 \times 6}+\cdots+\frac{1}{(2 n-1) \times 2 n} \\
= & \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2 n} .
\end{aligned}$$ | [Proof] It is easy to see that
$$\begin{array}{l}
\frac{1}{1 \times 2}=\frac{1}{1}+\frac{1}{2}-\frac{1}{1} \\
\frac{1}{3 \times 4}=\frac{1}{3}+\frac{1}{4}-\frac{1}{2} \\
\cdots \cdots \cdots \cdots \\
\frac{1}{(2 n-1) 2 n}=\frac{1}{2 n-1}+\frac{1}{2 n}-\frac{1}{n} .
\end{array}$$
Adding both sides of the above $n$ equ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,381 |
3. $51 n, r$ are non-negative integers, the symbol $\binom{n}{r}$ represents the number of combinations of selecting $r$ objects from $n$ objects, with the convention $\binom{n}{0}=1$, and when $n<r$, $\binom{n}{r}=0$. Prove: For all integers $n, r, 1 \leqslant r \leqslant n$, the equation
$$\sum_{d=1}^{\infty}\binom{n... | [Proof] Compare both sides of the following equation:
$$\begin{aligned}
(1+x)^{n} & =(1+x)^{n-r+1}(1+x)^{r-1} \\
& =\left[\sum_{i=0}^{n-r+1}\binom{n-r+1}{i} x^{i}\right]\left[\sum_{j=0}^{r-1}\binom{r-1}{j} x^{j}\right]
\end{aligned}$$
the coefficient of \(x^{r}\), we get
$$\begin{aligned}
\binom{n}{r} & =\sum_{d=0}^{r}... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,382 |
3.52 Prove or show that it is incorrect: There exist prime numbers $a, b, c, d$, and $a<b<c<d$, satisfying:
$$\frac{1}{a}+\frac{1}{d}=\frac{1}{b}+\frac{1}{c} .$$ | ［Solution］ Suppose the conclusion is correct, then
$$\begin{aligned}
& \frac{1}{a}+\frac{1}{d}=\frac{1}{b}+\frac{1}{c} \Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d} \Rightarrow c d(b-a) \\
= & a b(d-c)
\end{aligned}$$
Thus, $b$ divides $c d(b-a)$.
On the other hand, since $b, c, d$ are distinct primes, $... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,383 |
3. 53 Given that $a$ is a positive integer, $r=\sqrt{a+1}+\sqrt{a}$, prove that for each positive integer $n$, there exists a positive integer $a_{n}$ such that
(1) $r^{2 n}+r^{-2 n}=4 a_{n}+2$;
(2) $r^{n}=\sqrt{a_{n}+1}+\sqrt{a_{n}}$. | [Proof] First, prove (2)
$$\begin{aligned}
r^{n}= & (\sqrt{a+1}+\sqrt{a})^{n} \\
= & (\sqrt{a+1})^{n}+C_{n}^{1}(\sqrt{a+1})^{n-1} \sqrt{a}+C_{n}^{2}(\sqrt{a+1})^{n-2} a \\
& +\cdots+(\sqrt{a})^{n}, \\
r^{-n}= & \left(r^{-1}\right)^{n}=\left(\frac{1}{\sqrt{a+1}+\sqrt{a}}\right)^{n}=\left(\frac{\sqrt{a+1}-\sqrt{a}}{1}\ri... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,384 |
3. 54 Prove that for any natural number $n(\neq 0)$ and any real number $x \neq \frac{N \pi}{2^{k}}(k=0$, $1,2, \cdots, n ; N$ is any integer), we have
$$\frac{1}{\sin 2 x}+\frac{1}{\sin 4 x}+\cdots \cdots+\frac{1}{\sin 2^{n} x}=\operatorname{ctg} x-\operatorname{ctg} 2^{n} x$$ | $$\text { [Proof] } \begin{aligned}
& \operatorname{ctg} 2^{k} x-\frac{1}{\sin 2^{k+1} x} \\
= & \frac{1}{\sin 2^{k+1} x}\left(2 \cos ^{2} 2^{k} x-1\right) \\
= & \frac{\cos 2^{k+1} x}{\sin 2^{k+1} x} \\
= & \operatorname{ctg} 2^{k+1} x,
\end{aligned}$$
Taking $k=0,1,2, \cdots, n-1$, and adding them up yields the resu... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,385 |
3.56 Prove the following theorem: Let the sides of a triangle be $a, b, c$, and the corresponding angles be $\alpha, \beta, \gamma$, and satisfy the condition
$$a+b=\operatorname{tg} \frac{\gamma}{2}(a \operatorname{tg} \alpha+b \operatorname{tg} \beta) .$$
Then the triangle is isosceles. | [Proof] $a+b=\operatorname{tg} \frac{\gamma}{2}(a \operatorname{tg} \alpha+b \operatorname{tg} \beta)$,
$$\therefore \quad(a+b) \operatorname{tg} \frac{\alpha+\beta}{2}=a \operatorname{tg} \alpha+b \operatorname{tg} \beta,$$
i.e. $\square$
$$\begin{array}{l}
a\left(\operatorname{tg} \frac{\alpha+\beta}{2}-\operatornam... | proof | Geometry | proof | Yes | Yes | inequalities | false | 735,386 |
$3 \cdot 58$ (a) Prove: $\operatorname{tg} \frac{\pi}{12}=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}$.
(b) Given any 13 different real numbers, prove that there exist at least two, say $x$ and $y$, satisfying the inequality
$$0<\frac{x-y}{1+x y}<\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}$$ | [Proof] (a) Let $\theta=\frac{\pi}{12}$, then $\operatorname{tg} 2 \theta=\frac{1}{\sqrt{3}}$,
i.e., $\frac{2 \operatorname{tg} \theta}{1-\operatorname{tg}^{2} \theta}=\frac{1}{\sqrt{3}}$,
solving this equation for $\operatorname{tg} \theta$ yields $\operatorname{tg} \theta=2-\sqrt{3}$,
i.e., $\operatorname{tg} \theta=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,388 |
3. 59 Let $A$, $B$, and $C$ be the three interior angles of a triangle, and $x$ satisfies
$$\cos (x+A) \cdot \cos (x+B) \cdot \cos (x+C) + \cos^3 x = 0.$$
(1) Prove: $\tan x = \cot A + \cot B + \cot C$.
(2) Prove: $\sec^2 x = \csc^2 A + \csc^2 B + \csc^2 C$. | [Proof] Dividing both sides of the known equation by $\cos ^{3} x \cdot \sin A \cdot \sin B \cdot \sin C$, we get
$$(\operatorname{tg} x-\operatorname{ctg} A)(\operatorname{tg} x-\operatorname{ctg} B)(\operatorname{tg} x-\operatorname{ctg} C)-\frac{1}{\sin A \sin B \sin C}=0 .$$
Let $S=\operatorname{ctg} A+\operatorna... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,389 |
$1 \cdot 38$ Find all such natural numbers: each of them is equal to the square of the number of all its divisors.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] Obviously, the natural number 1 is a solution to this problem.
Let the number $n=m^{2}$ have $m(>1)$ divisors. Since the number of divisors of a perfect square is odd, $m$ is odd. Let $m=2 k+1$, then $n$ has $k$ divisors less than $m$. But $n$ is odd, so all divisors of $n$ are odd, thus $n$ can be divided b... | null | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,390 |
3.61 Given three angles $A, B, C$ satisfying $\cos A+\cos B+\cos C=\sin A+$ $\sin B+\sin C=0$. Prove: $\cos ^{2} A+\cos ^{2} B+\cos ^{2} C$ is a constant, and find this constant. | [Solution] From the given, we have $(\cos A+\cos B+\cos C)^{2}=0$,
which means
$$\begin{aligned}
& \cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2(\cos A \cos B+\cos B \cos C+\cos A \cos C) \\
= & 0
\end{aligned}$$
$$\text { and } \begin{aligned}
& 2(\cos A \cos B+\cos B \cos C+\cos C \cos A) \\
= & \cos (A+B)+\cos (A-B)+\cos (B... | not found | Algebra | proof | Yes | Yes | inequalities | false | 735,391 |
3.62 Let $a, b$ be the legs of a right triangle, and $c$ be the hypotenuse. Prove: $\log _{b+c} a+\log _{c-b} a=2 \log _{c+b} a \cdot \log _{c-b} a$. | $$\begin{array}{l}
\text { [Proof] Left side }=\frac{1}{\log _{a}(b+c)}+\frac{1}{\log _{a}(c-b)} \\
=\frac{\log _{a}\left(c^{2}-b^{2}\right)}{\log _{a}(c+b) \cdot \log _{a}(c-b)} \\
=\frac{2}{\log _{a}(c+b) \cdot \log _{a}(c-b)} \\
=2 \log _{c+b}a \cdot \log _{c-b}a \\
=\text { Right side } \text {. }
\end{array}$$
Pr... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,392 |
3.63 $x, y, z$ are pairwise distinct integers, prove: $(x-y)^{5}+(y-z)^{5}+$ $(z-x)^{5}$ is divisible by $5(y-z)(z-x)(x-y)$. | ［Proof］ Let $x-y=u, y-z=v$, then $z-x=-(u+v)$. Using the binomial formula, we get
i.e. $\square$
$$\begin{array}{ll}
& (u+v)^{5}=u^{5}+5 u^{4} v+10 u^{3} v^{2}+10 u^{2} v^{3}+5 u v^{4}+v^{5} \\
\text { i.e. } \quad(u+v)^{5}=u^{5}+v^{5}+5 u v(u+v)\left(u^{2}+u v+v^{2}\right) \\
& u^{5}+v^{5}+\left[-(u+v)^{5}\right]=-5... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,393 |
3.65 As shown in the figure, the engine connecting rod $AP=b$, crank $OA=a$, $\angle AOP=\alpha$, $\angle APO=\beta$.
(1) Prove that $a \cdot \sin \alpha=b \cdot \sin \beta$;
(2) Find the maximum value of $\sin \beta$;
(3) If $BQ=b$, $PQ=x$,
prove that $x=a(1-\cos \alpha)+b(1-\cos \beta)$. | [Solution] (1) Draw $A C \perp O B$, with $C$ as the foot of the perpendicular.
Since $A C=O A \cdot \sin \alpha=a \cdot \sin \alpha$,
and $A C=A P \cdot \sin \beta=b \cdot \sin \beta$,
thus $a \cdot \sin \alpha=b \cdot \sin \beta$.
(2) Since $\sin \beta=\frac{a}{b} \sin \alpha$ (from (1)),
when $\alpha=\frac{\pi}{2}$,... | proof | Geometry | proof | Yes | Yes | inequalities | false | 735,395 |
3.66 $n(>3)$ is an integer, $a_{0}, a_{1}, \cdots, a_{n}$ are integers satisfying $1 \leqslant a_{0}<a_{1}<\cdots<$ $a_{n} \leqslant 2 \mathcal{k}-3$, prove that there exist different integers $i, j, k, l, m$, such that
$$a_{i}+a_{j}=a_{k}+a_{l}=a_{m}$$ | [Proof] We prove a stronger conclusion: among integers satisfying $1 \leqslant a_{0}<a_{1}<\cdots<a_{n} \leqslant 2 n$ -3, there must exist $i, j, k, l$, all not equal to $n$, such that
$$a_{i}+a_{j}=a_{k}+a_{l}=a_{n}.$$
We use proof by contradiction. Assume that there do not exist $i, j, k, l$ satisfying the above eq... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,396 |
3.67 Consider the polynomial determined by the identity
$$a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n} \equiv\left(x+2 x^{2}+\cdots+n x^{n}\right)^{2}$$
Prove: $\sum_{k=n+1}^{2 n} \cdot a_{k}=\frac{1}{24} n(n+1)\left(5 n^{2}+5 n+2\right)$. | [Proof] Clearly, $a_{0}=a_{1}=0, a_{2}=1$.
When $3 \leqslant k \leqslant n$,
$$\begin{aligned}
a_{k} & =1 \cdot(k-1)+2 \cdot(k-2)+\cdots+(k-1) \cdot 1 \\
& =k(1+2+\cdots+k-1)-\left[1^{2}+2^{2}+\cdots+(k-1)^{2}\right] \\
& =\frac{1}{2} k^{2}(k-1)-\frac{1}{6}(k-1) k(2 k-1) \\
& =\frac{1}{6}(k-1) k(k+1) \\
& =C_{k+1}^{3} ... | \frac{1}{24} n(n+1)\left(5 n^{2}+5 n+2\right) | Algebra | proof | Yes | Yes | inequalities | false | 735,397 |
3・68 Given natural numbers $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{m}$, the two sums $x_{1}+x_{2}+\cdots+x_{n}$ and $y_{1}+y_{2}+\cdots+y_{m}$ are equal to each other and less than $mn$. Prove: In the equation $x_{1}+x_{2}+\cdots+x_{n}=y_{1}+y_{2}+\cdots+y_{m}$, some addends can be deleted so that t... | [Proof] Induction on $m+n$.
When $m+n=4$, $m=2, n=2$. Let $S=x_{1}+x_{2}+\cdots+x_{n}=y_{1}+y_{2}+\cdots+y_{m}$, and by $S \geqslant m$, we have $S \geqslant 2$. Also, by $Sy_{1}$, let $x^{\prime}=x_{1}-y_{1}$,
$$S^{\prime}=\left(x_{1}-y_{1}\right)+x_{2}+\cdots+x_{n}=y_{2}+\cdots+y_{m}$$
Thus,
$$S^{\prime}=x^{\prime}{... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,398 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.