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 |
|---|---|---|---|---|---|---|---|---|---|
2.20. (Jury, SRR, 77). Let numbers $a_{i}, \ldots, a_{n}, a_{n+1} \in \mathbf{N}$ be given, and for each value $i=1, \ldots, n$ the numbers $a_{i}$ and $a_{n+1}$ are coprime. Prove that the equation
$$
x_{1}^{a_{1}}+x_{2}^{a_{8}}+\ldots+x_{n}^{a_{n}}=x_{n+1}^{a_{n+1}}
$$
has infinitely many solutions in natural numbe... | 2.20. If $n=1$, then for each value of $z \in \mathbf{N}$, the pair of numbers
$$
x_{1}=z^{a_{3}}, x_{2}=2^{a_{1}}
$$
satisfies the equation. Now let $n>1$. Then, according to the Chinese Remainder Theorem (Theorem 23), there are infinitely many natural numbers $z$ that satisfy the two conditions:
$$
z=0\left(\bmod ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,829 |
2.22. (VNR, 78). Prove that for any values of $a, b \in \mathbf{Q}$ the equation $a x^{2}+b y^{2}=1$ in rational numbers either has no solutions or has infinitely many solutions. | 2.22. Let the pair of numbers $x_{0}, y_{0} \in \mathbb{Q}$ be a solution of the original equation, and let the number $k \in \mathbb{Q}$ satisfy the condition $a k^{2} + b \neq 0$ (there are infinitely many such values of $k$, since $a^{2} + b^{2} > 0$, which means that the equation $a k^{2} + b = 0$ can be satisfied ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,831 |
2.23. (Jury, SPP, 79). Prove that for any coprime numbers \(a, b \in \mathbf{Z}\) the equation \(a x^{2} + b y^{2} = z^{3}\) in integers has infinitely many solutions satisfying the condition \((x, y) = 1\). | 2.23. For any numbers $k, m, a, b$ the equality
$$
\left(a k^{2}+b m^{2}\right)^{3}=a\left(a k^{3}-3 b k m^{2}\right)^{2}+b\left(3 a k^{2} m-b m^{3}\right)^{2}
$$
holds. Therefore, if we set
$$
x=a k^{3}-3 b k m^{2}, \quad y=3 a k^{2} m-b m^{3}, \quad z=a k^{2}+b m^{2}
$$
then the equality
$$
a x^{2}+b y^{2}=z^{3}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,832 |
2.28*. (Jury, Canada, 82). Prove that the set of all values $n \in \mathbf{N}$ for which the equation
$$
\frac{1}{x}+\frac{1}{y}=\frac{3}{n}
$$
has no solutions in natural numbers cannot be represented as a union of a finite number of arithmetic progressions (both finite and infinite).
2.29*. (NRB, 79). Prove that t... | 2.28. Let's denote the set of numbers described in the problem by $M$ and assume that it can be represented as the union of a finite set of arithmetic progressions. We will prove that among these progressions, there is no infinite one. Indeed, suppose for some values $a, d \in \mathbf{N}$, the set $M$ contains all numb... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,834 |
3.1. (Netherlands, 82). What is larger: $(17091982!)^{2}$ or $17091982^{1701982} ?$ | 3.1. We will prove that for any value of $n \in \mathbf{N}$, greater than 2, the inequality
$$
(n!)^{2}>n^{n}
$$
holds. Indeed, we have
$$
n' \cdot n!=(1 \cdot 2 \ldots n)(n(n-1) \ldots .1)=(1 \cdot n)(2(n-1)) \ldots(n \cdot 1)>n^{n}
$$
since
$$
1 \cdot n=n \cdot 1=n \text { and } k(n-k+1)=(n-k)(k-1)+n>n
$$
for $... | (17091982!)^{2}>17091982^{17081982} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,835 |
3.2. (Yugoslavia, 74). Find all numbers $n \in \mathbf{N}$, for which for some value
$$
k \in\{1 ; 2 ; \ldots ; n-1\}
$$
the equality holds
$$
2 C_{n}^{k}=C_{n}^{k-1}+C_{n}^{k+1}
$$ | 3.2. Since for each value of $k=1,2$, ..., $n=1$ the relations
$$
C_{n}^{k-1}=C_{n}^{k} \cdot \frac{k}{n-k+1}, \quad C_{n}^{k+1}=C_{n}^{k} \cdot \frac{n-k}{k+1}, \quad C_{n}^{k} \neq 0
$$
are satisfied, the equality
$$
2 C_{n}^{k}=C_{n}^{k-1}+C_{n}^{k+1}
$$
is equivalent to the equality
$$
2=\frac{k}{n-k+1}+\frac{... | n=^2-2,where=3,4,\ldots | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,836 |
3.3. (Canada, 83). Solve the equation
$$
x!+y!+z!=u!
$$
in natural numbers. | 3.3. Let the numbers $x, y, z, u \in \mathbf{N}$ satisfy the original equation, and let $v$ denote the largest of the numbers $x, y, z$. Then $1 \leq v < u$ and
$$
u v! \leq u(u-1)! = u! = x! + y! + z! \leq 3 v!
$$
from which $u v! \leq 3 v!$, hence $u \leq 3$. For $u=3$, we have the equalities
$$
3! = 3 v! = x! + y... | x=y=z=2,u=3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,837 |
3.4. (USA, 82). Prove that for any value of $n \in \mathbf{N}$, the following equality holds:
$$
\sum_{k=0}^{n} \frac{(2 n)!}{(k!)^{2}((n-k)!)^{2}}=\left(C_{2 n}^{n}\right)^{2}
$$ | 3.4. From the identity
$$
(1+x)^{2 n}=(1+x)^{n}(1+x)^{n}
$$
using the binomial theorem, we get
$C_{2 n}^{0}+C_{2 n}^{1} x+\ldots+C_{2 n}^{2 n} x^{2 n}=$
$$
=\left(C_{n}^{0}+C_{n}^{1} x+\ldots+C_{n}^{n} x^{n}\right)\left(C_{n}^{0}+C_{n}^{1} x+\cdots+C_{n}^{n} x^{n}\right)
$$
Equating the coefficients of $x^{n}$ and... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,838 |
3.5. (Jury, Australia, 82). For a given value $m \in \mathbf{N}:$
a) prove that the number $\frac{1}{m+1} C_{2 m}^{m}$ is a natural number;
b) find the smallest value $k \in \mathbf{N}$, for which the number $\frac{k}{n+m+1} C_{2 n}^{n+m}$ is a natural number for each natural value $n \geqslant m$. | 3.5. a) The positive number
$$
\frac{1}{m+1} C_{2 m}^{m}=\left(1-\frac{m}{m+1}\right) C_{2 m}^{m}=C_{2 m}^{m}-\frac{(2 m)!}{(m-1)!(m+1)!}=C_{2 m}^{m}-C_{2 m}^{m-1}
$$
is an integer, since $C_{2 m}^{m}, C_{2 m}^{m-1} \in \mathbf{N}$ for $m \in \mathbf{N}$.
b) Let a number $m \in \mathbb{N}$ be given. Since for $n=m$ ... | 2m+1 | Combinatorics | proof | Yes | Yes | olympiads | false | 28,839 |
3.6. (New York, 74). Prove that for any natural values $n \geqslant k$ the greatest common divisor of the numbers $C_{n}^{k}, C_{n+1}^{k}, \ldots, C_{n+k}^{k}$ is 1. | 3.6. Let the numbers $C_{n}^{k}, C_{n+1}^{k}, \ldots, C_{n+k}^{k}$ have a common divisor $d \in \mathbf{N}$. Then the numbers
$$
C_{n}^{k-1}=C_{n+1}^{k}-C_{n}^{k}, C_{n+1}^{k-1}=C_{n+2}^{k}-C_{n+1}^{k}, \ldots, C_{n+k-1}^{k-1}=C_{n+k}^{k}-C_{n+k-1}^{k}
$$
also have a common divisor $d$. Similarly, we obtain that the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,840 |
3.7. (England, 81). Prove that for any values of $m, n \in \mathbf{N}$ the number
$$
S_{m, n}=1+\sum_{k=1}^{m}(-1)^{k} \frac{(n+k+1)!}{n!(n+k)}
$$
is divisible by $m l$, but for some values of $m, n \in \mathbf{N}$ the number $S_{m, n}$ is not divisible by $m l(n+1)$. | 3.7. For each value of $n \in N$, we will prove by induction on $m \in N$ the validity of the equality
$$
S_{m, n}=(-1)^{m} \frac{(n+m)!}{n!}
$$
For $m=1$, we have the correct statement
$$
S_{1, n}=1-\frac{(n+2)!}{n!(n+1)}=1-(n+2)=-\frac{(n+1)!}{n!}
$$
Suppose that for some value of $m \in \mathbf{N}$, the equality... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,841 |
3.8. (Yugoslavia, 70). Prove that if the number $p$ is prime, then the number $C_{2 p}^{p}-2$ is divisible by $p^{2}$. | 3.8. Consider a prime number $p>2$ (for $p=2$ the statement of the problem is true, since the number $C_{4}^{2}-2=4$ is divisible by $2^{2}=4$). First of all, we have the equalities
$$
C_{2 p}^{p}=\frac{(2 p)!}{(p))^{2}}=\frac{2 p(2 p-1)!}{p(p-1)!p!}=2 C_{2 p-1}^{p-1}
$$
From the relation
$$
(2 p-k)(p+k)=k(p-k)\left... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,842 |
3.10. (НРБ, 82; Australia, 83). Solve the equation
$$
(y+1)^{x}-1=y!
$$
in natural numbers. | 3.10. Let the pair of numbers $x, y \in \mathbf{N}$ satisfy the equation. Note that the number $p=y+1$ is prime. Indeed, if there exists a divisor $d$ of the number $p$ satisfying the condition $1<d<y+1$, then we have $y+1 \vdots d$ and
$$
1=(y+1)^{x}-y!\equiv 0(\bmod d)
$$
which is impossible. Therefore, we have the... | (1;1),(1;2),(2;4) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,843 |
3.11. (Austria, 73; SFRY, 77; GDR, 79). For a given value \( n \in \mathbf{N} \), greater than 1, it is denoted
\[
m_{k}=n l+k, \quad k \in \mathbf{N}
\]
Prove that for any value \( k \in \{1; \ldots; n\} \) there exists a prime number \( p \) that divides the number \( m_{k} \) and does not divide any of the other n... | 3.11. Let
$$
l_{k}=\frac{m_{k}}{k}=\frac{n!}{k}+1 \quad(k=1, \ldots, n)
$$
We will prove that if $p$ is a prime divisor of the number $l_{k}$, then the number $p$ does not divide any of the numbers $m_{j}, i \neq k$. Since in this case the number $p$ divides the number $m_{k}=l_{k} k$, it will be the desired one. Sup... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,844 |
3.12*. (NBR, 68). Prove that the number $C_{n}^{\prime 2}$ is odd if and only if the numbers $n, k \in \mathbf{N}$ satisfy the condition: if in some digit of the binary representation of the number $k$ there is a 1, then in the same digit of the binary representation of the number $n$ there is also a 1. | 3.12. The exponent to which two enters the prime factorization of the number $\|$ is equal to
$$
\left[\frac{l}{2}\right]+\left[\frac{l}{4}\right]+\left[\frac{l}{8}\right]+\cdots
$$
see Theorem 20). Therefore, the number $C_{n}^{k}=\frac{n!}{k!(n-k)!}$ is odd if and only if the highest power of two by which the numbe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,845 |
3.13*. (MMС, Luxembourg, 80). Prove that for any value of $n \in \mathbf{N}$ and a prime number $p$ the following conditions are equivalent:
a) none of the numbers $C_{n}^{k}$ for $k=0,1, \ldots, n$ is divisible by $p$.
b) $n=p^{s} m-1$, where $s \in \mathbf{Z}^{+}, m \in \mathbf{N}, m<p$. | 3.13. Let's prove that condition b) is equivalent to the following condition:
c) $n=p^{t} l+\left(p^{t}-1\right)$, where $t \in \mathbf{Z}^{+}, l \in \mathbf{N}, l>0$, then we set
$$
t=s, l=(m-1)>0
$$
If $m=1$, then $s>0$, because otherwise $n=p^{0} \cdot 1-1=0 \notin \mathbf{N}$. Therefore,
$$
n=p^{s-1} \cdot p-1=... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,846 |
3.14*. (Jury, USSR, 83). Let $h_{n}$ denote the last non-zero digit of the decimal representation of the number $n!$. Prove that the infinite decimal fraction
$$
0, h_{1} h_{2} h_{3} \ldots
$$
represents an irrational number.
## § 4. Numerical Sets
(see Appendix G: definitions 1, 2, 11, 12; theorems $1,2,10,13,18,2... | 3.14. Suppose the number
$$
0, h_{1} h_{2} h_{8} \ldots
$$
is rational. Then there exist numbers $N, T \in \mathbf{N}$, for which the equalities $h_{n+T}=h_{n}$ hold for all $n \geqslant N$. We will prove that there exists a number $T_{1}$, which is divisible by $T$ and has the last non-zero digit equal to 1. Indeed,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,847 |
4.1. (SRP, 78). Prove that for any partition of the set
$$
X=\{1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9\}
$$
into two subsets, at least one of the resulting subsets contains 3 numbers such that the sum of two of them is twice the third. | 4.1. Let $X=A \cup B$ and for definiteness $5 \in A$. Suppose that in neither of the sets $A$ and $B$ is there a triplet of numbers, the sum of two of which is equal to twice the third. If $3 \in A$, then $1 \in B, 4 \in B$ and $7 \in B$, which contradicts the assumption, since $1+7=2 \cdot 4$. Similarly, it can be sho... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,848 |
4.2. (Belgium, 79). Find the sum of all $7!$ numbers that can be obtained by all possible permutations of the digits in the number 1234567. | 4.2. For any values of $i, i \in\{1 ; \ldots ; 7\}$, the number of numbers in which the $i$-th place is occupied by the digit $j$ is 61. Therefore, the sum of all numbers is
$$
\begin{aligned}
& (61 \cdot 1+\ldots+61 \cdot 7)+(61 \cdot 1+\ldots+61 \cdot 7) 10+ \\
& \quad+(61 \cdot 1+\ldots+61 \cdot 7) 10^{2}+\ldots+(6... | 22399997760 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,849 |
4.3. (England, 66). Prove that from any 52 integers, it is always possible to select 2 numbers, the sum or difference of which is divisible by 100. | 4.3. Let's divide all possible remainders of division by 100 into the following groups:
$$
\{0\},\{1 ; 99\},\{2 ; 98\}, \ldots,\{49 ; 51\},\{50\}
$$
Since there are 51 groups in total and 52 numbers, by the Pigeonhole Principle (Theorem I), there must be 2 numbers among them whose remainders when divided by 100 fall ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,850 |
4.4. (England, 70). Prove that in any set of $n$ natural numbers, one can select a (non-empty) subset of numbers, the sum of which is divisible by $n$.
| 4.4. Let for some set consisting of numbers $a_{1}, a_{2}, \ldots, a_{n}$, the statement of the problem be false. Then none of the $n$ numbers
$$
S_{1}=a_{1}, S_{2}=a_{1}+a_{2}, \ldots, S_{n}=a_{1}+a_{2}+\ldots+a_{n}
$$
is divisible by $n$. Since the number of possible non-zero remainders upon division by $n$ is $n-1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,851 |
4.5. (PNR, 79). Natural numbers $a_{1}, a_{2}, \ldots, a_{n}$ when divided by some number $m \in \mathbf{N}$ give different remainders, and $n>m / 2$. Prove that for each number $k \in \mathbf{Z}$ there exist such indices
$$
i, j \in\{1 ; \ldots ; n\}
$$
(not necessarily different), that the number $a_{i}+a_{i}-k$ is... | 4.5. Consider $2 n$ numbers
$$
a_{1}, a_{2}, \ldots, a_{n}, k-a_{1}, k-a_{2}, \ldots, k-a_{n^{\prime}}
$$
Since $2 n>m$, at least two of them give the same remainder when divided by $m$. According to the problem's condition, the numbers $a_{1}, a_{2}, \ldots, a_{n}$ have different remainders when divided by $m$, from... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,852 |
4.7. (SFRY, 81). The set of numbers $1,2, \ldots, 100$ is partitioned into 7 subsets. Prove that in at least one of these subsets, there will be either 4 numbers $a, b, c, d$ such that $a+b=c+d$, or 3 numbers $e, f, g$ such that $e+f=2 g$. | 4.7. Note that in at least one of the 7 subsets, there are no fewer than 15 numbers (otherwise, together all subsets would contain no more than $7 \cdot 14=98$ numbers). For each pair of numbers $a>b$ in this subset, we associate the difference $a-b$. Then, in total, we get no fewer than
$$
C_{15}^{2}=15 \cdot 14 / 2=... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,854 |
4.8. (USA, 83). On the number line, an interval of length $1 / n (n \in \mathbb{N})$ is taken. Prove that this interval contains no more than $(n+1) / 2$ irreducible fractions of the form $p / q$, where $p, q \in \mathbb{Z}, 1 \leqslant q \leqslant n$. | 4.8. Suppose that, contrary to the statement of the problem, some interval of length $1 / n$ contains more than $(n+1) / 2$ irreducible fractions of the form $p / q$, where $q \in\{1 ; 2 ; \ldots ; n\}$. We will prove that among the denominators of these fractions, there will be two such that one divides the other. Ind... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,855 |
4.10. (NPR, 80). Prove that the number of ways to choose a sextet of different numbers from the set $1,2, \ldots, 49$ such that at least two of them are consecutive is
$$
C_{49}^{6}-C_{44}^{6}
$$ | 4.10. To each set of 6 different natural numbers from 1 to 49 (without loss of generality, we assume they are arranged in ascending order)
$$
a_{1}<a_{2}<a_{3}<a_{4}<a_{5}<a_{6}
$$
we correspond a set of the form
$$
a_{1}, a_{2}-1, a_{3}-2, a_{4}-3, a_{5}-4, a_{6}-5
$$
The numbers in the last set are distinct if an... | C_{49}^{6}-C_{44}^{6} | Combinatorics | proof | Yes | Yes | olympiads | false | 28,857 |
4.11. (Austria-Poland, 78). For a given positive rational value $c \neq 1$, prove that the set of natural numbers can be represented as the union of two non-intersecting subsets $A$ and $B$ such that the ratio of any two numbers from the set $A$, as well as the ratio of any two numbers from the set $B$, does not equal ... | 4.11. We will construct sets $A$ and $B$ by induction on $n \in \mathbf{N}$. Let $1 \in A$. Suppose the numbers $1, \ldots, n-1$ have already been distributed among the subsets $A$ and $B$. Consider the number $n$. If there exists a number
$$
k_{1} \in \{1; \ldots; n-1\}
$$
satisfying the equation $k_{1} / n = c$, th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,858 |
4.12. (Jury, Spain, 77). The sum of integers $a_{1}$, $a_{2}, \ldots, a_{n}$ is one. Prove that then among the numbers
$$
b_{i}=a_{i}+2 a_{i+1}+3 a_{i+2}+\cdots+(n-i+1) a_{n}+(n-i+2) a_{1}+
$$
$$
+(n-i+3) a_{2}+\ldots+n a_{i-1}(i=1,2, \ldots, n)
$$
there are no identical ones. | 4.12. Since
\[
\begin{aligned}
& b_{i}-b_{i+1}=(1-n) a_{i}+a_{i+1}+a_{i+2}+\ldots \\
& \ldots+a_{n}+a_{1}+a_{2}+\ldots+a_{i-1}=1-n a_{i} \equiv 1(\bmod n)
\end{aligned}
\]
for each value of \(i=1, \ldots, n-1\), then
\[
\begin{aligned}
& b_{n-1}=b_{n}+1(\bmod n), \\
& b_{n-2} \equiv b_{n-1}+1(\bmod n), \\
& b_{1}=b_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,859 |
4.13. (Jury, USSR, 82). Find all values of $n \in \mathbf{N}$ for each of which there exists a string of $2 n$ numbers with the following property: for any value $k=1, \ldots, n$ there are 2 numbers in the string equal to $k$, between which there are exactly $k$ numbers. | 4.13. Suppose that for some value of $n$ there exists a string of $2 n$ numbers with the required property. Let $m_{k}$ be the position of the first (leftmost) of the two numbers equal to $k$. Then the second of these numbers has the position $m_{k}+k+1$, and the sum of the positions of all $2 n$ numbers in the string ... | n=4l,n=4l-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,860 |
4.14. (Austria, 75). For a non-empty set $M \subset \mathbf{Q}$, the following two conditions are satisfied:
1) if $a \in M$ and $b \in M$, then $a+b \in M$ and $a b \in M$;
2) if $r \in \mathbf{Q}$, then exactly one of the following three statements is true: $r \in M, -r \in M, r=0$.
Prove that the set $M$ coincides... | 4.14. From condition 2) it follows that either $I \in M$, or $(-1) \in M$. But $(-1) \notin M$, because otherwise, by condition 1),
$$
(-1)(-1)=1 \in M
$$
which contradicts condition 2). Therefore, $1 \in M$. From condition 1) it follows that $1+1 \in M, 2+1 \in M$, etc., i.e., $M \supset \mathbf{N}$. If now $(-1 / m... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,861 |
4.15. (New York, 73). A finite set $B \subset \mathbf{R}$ is called a basis for a set $M \subset \mathbf{R}$ if each number in the set $M$ can be uniquely represented as a product of integer powers of numbers from the set $B$. Is it true that for any finite set of positive numbers, there exists a basis? | 4.15. We will prove that for a finite set $M$ of positive numbers, there exists a basis $B$. We will call a subset $S$ of positive numbers a superbasis for $M$ if each number in $M$ can be represented as a product
$$
\alpha_{1}^{i_{1}} \ldots \alpha_{m}^{i_{m}}, \text { where } \alpha_{i}, \ldots, \alpha_{m} \in S, i_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,862 |
4.16. (Austria-Poland, 80). Prove that for any value of $n \in \mathbf{N}$, the following equality holds:
$$
\sum_{1 \leqslant i_{1}<i_{2}<\ldots<i_{k} \leqslant n} \frac{1}{i_{1} i_{2} \ldots i_{k}}=n
$$
where the summation is taken over all possible sets of numbers
$$
i_{1}<i_{2}<\ldots<i_{k}, \quad k=1,2, \ldots,... | 4.16. First solution. Consider the polynomial
$$
P(x)=\left(x+\frac{1}{1}\right)\left(x+\frac{1}{2}\right) \ldots\left(x+\frac{1}{n}\right)
$$
which, by Vieta's theorem, has the form
$$
P(x)=x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n}
$$
where it is denoted
$$
a_{1}=\sum_{i_{1}=1}^{n} \frac{1}{i_{1}}, \quad a_{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,863 |
4.17*. (NRP, 83). Find all values of $n \in \mathbf{N}$ for each of which there exists a permutation
$$
\left(a_{1} ; a_{2} ; \ldots ; a_{n}\right)
$$
of the numbers $0,1, \ldots, n-1$, such that all numbers
$$
a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \ldots, a_{1} a_{2} \ldots a_{n}
$$
give different remainders when... | 4.17. For $n=1$ the permutation $\left(a_{1}\right)=(0)$ satisfies the condition of the problem. For $n=4$, the permutation
$$
\left(a_{1} ; a_{2} ; a_{8} ; a_{4}\right)=(1 ; 3 ; 2 ; 0)
$$
possesses the required property. Let $n$ be a prime number. According to the Chinese "remainder theorem" (Theorem 23), for each v... | 1,4, | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,864 |
4.18*. (Jury, Sweden, 83). Prove that if a number $n \in \mathbf{N}$ is not an integer power of a prime number, then there exists a permutation
$$
\left(i_{1} ; i_{2} ; \ldots ; i_{n}\right)
$$
of the numbers $1,2, \ldots, n$, for which the equality
$$
\sum_{k=1}^{n} k \cos \left(2 \pi i_{k} / n\right)=0
$$
holds. | 4.18. The number $n$ can be represented as the product $p-q$ of coprime numbers $p>1$ and $q>1$. For each $k=1,2, \ldots, n$, we will select numbers
$$
m \in\{0 ; 1 ; \ldots ; p-1\}, l \in\{1 ; 2 ; \ldots ; q\}
$$
such that the equality $k=m q+l$ holds, and we set $i_{k}=r+1$, where $r$ is the remainder of the divisi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,865 |
4.19*. (Jury, Finland, 82). Determine the sum of all natural numbers, each of which has digits in its decimal representation that form an increasing or decreasing sequence. | 4.19. Let $A$ and $B$ be the sets of natural numbers with strictly increasing and strictly decreasing digits, respectively. The sum of all numbers in a set $M$ will be denoted by $S(M)$. We will partition the set $B$ into two non-intersecting subsets $B_{0}$ and $B_{1}$, consisting of numbers that end in the digit 0 an... | \frac{80}{81}\cdot11^{10}-\frac{35}{81}\cdot2^{10}-45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,866 |
4.20*. (Jury, Finland, 79). A set $\left(a_{1} ; \ldots ; a_{n}\right)$ of natural numbers satisfying the equation
$$
a_{1}+2 a_{2}+\ldots+n a_{n}=1979
$$
is called even if the number $r$ is even, and is called odd if the number $n$ is odd. Prove that there are as many even sets as there are odd sets. | 4.20. To each set $(a_i; \ldots; a_n)$ there corresponds a set $b = (b_1; \ldots; b_n)$ of natural numbers defined by the equations
$$
b_i = a_i + a_{i+1} + \ldots + a_n \quad \text{for} \quad i = 1, \ldots, n.
$$
Thus, there is a one-to-one correspondence between the set of initial sets and the set $B$ of sets $b$ s... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,867 |
5.1. (ČSSR, 52). Prove that if positive numbers $a, b, c \in \mathbf{Q}$ satisfy the equation $\sqrt{a}+\sqrt{b}=c$, then $\sqrt{a}, \sqrt{b} \in \mathbf{Q}$. | 5.1. Since $a, b, c \in \mathbf{Q}$, the number
$$
d=\frac{a-b}{c}=\frac{(\sqrt{\bar{a}}+\sqrt{\bar{b}})(\sqrt{a}-\sqrt{\bar{b}})}{\sqrt{a}+\sqrt{\bar{b}}}=\sqrt{\bar{a}}-\sqrt{\bar{b}}
$$
is rational, therefore, the numbers
$$
\sqrt{\bar{a}}=(c+d) / 2, \quad \sqrt{\bar{b}}=(c-d) / 2
$$
are also rational. | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,869 |
5.2. (SRP, 75). Prove that there exist positive irrational numbers $a$ and $b$, for which the number $a^{b}$ is a natural number. | 5.2. Let's take positive numbers
$$
a=\sqrt{\overrightarrow{2}} \text { and } b=\log _{\sqrt{2}} 3_{2}
$$
then the number
$$
a^{b}=\sqrt{2}^{\log v^{2}}=3
$$
is a natural number. The number $a$ is irrational. We will prove that the number $b$ is also irrational. Indeed, otherwise we have $b=p / q$, where $p, q \in ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,870 |
5.4. (SIUSA, 78). Prove that any number $n \in \mathbf{N}$, greater than 32, can be represented as the sum of several natural numbers, the sum of the reciprocals of which equals 1.
Hint. The values $n=33,34,35, \ldots, 73$ satisfy the required condition. | 5.4. Note that if a number $n$ satisfies the condition of the problem, then the numbers $2 n+2$ and $2 n+9$ also satisfy it. Indeed, if
$$
n=a_{1}+a_{2}+\ldots+a_{k}
$$
where
$$
a_{1}, a_{2}, \ldots, a_{k} \in \mathbf{N} \text { and } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{k}}=1
$$
Then
$$
2 n+2=2 a_{1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,872 |
5.5. (England, 82). The binary representation of a number \( n \in \mathbf{N} \), which is a multiple of 17, contains exactly 3 digits 1. Prove that this representation contains at least 6 digits 0, and if there are exactly 7, then the number \( n \) is even. | 5.5. Since the binary representation of the number $n$, divisible by 17, contains exactly 3 digits 1 (and the rest are 0), this number is represented as the sum
$$
n=2^{k}+2^{l}+2^{m}
$$
where the numbers $k, l, m \in \mathbf{Z}^{+}$ satisfy the inequalities $k<l<m$. If the considered binary representation contains f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,873 |
5.6. (Yugoslavia, 83). Find all values of $n \in \mathbf{N}$ that have the following property: if the numbers $n^{3}$ and $n^{4}$ are written next to each other (in the decimal system), then in the resulting record each of the 10 digits $0,1, \ldots, 9$ will appear exactly once. | 5.6. Let $f(m)$ denote the number of digits in the decimal representation of the number $m \in N$, then for the desired number $n$ we have
$$
f\left(n^{3}\right)+f\left(n^{4}\right)=10
$$
Moreover, $f\left(n^{9}\right) \geqslant 4$, because otherwise the inequalities $n^{3}4$, then $n>10, n^{4}>10 n^{3}$, from which
... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,874 |
5.7. (Yugoslavia, 77). Find all values of $n \in \mathbf{N}$ that have the following property: the fifth power of the sum of the digits in the decimal representation of the number $n$ is equal to $n^{2}$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
Note: The last sentence is a repetition of the instruction and should no... | 5.7. Let $n$ be the desired number, $s$ be the sum of the digits of its decimal representation, and $k$ be the number of these digits. Then the following relations hold:
$$
s^{6}=n^{2}, \quad s \leqslant 9 k, \quad n \geqslant 10^{k-1},
$$
from which we obtain that
$$
9^{5} k^{5} \geqslant s^{5}=n^{2} \geqslant 10^{... | n=1n=243 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,875 |
5.8. (England, 78). Prove that if the denominator of a proper fraction does not exceed 100, then in the decimal representation of this fraction, the three digits 1, 6, 7 cannot appear consecutively in the specified order. | 5.8. Suppose there exists a fraction $m / n$, where $m, n \in \mathbf{Z}$ and $m<n \leqslant 100$, such that the decimal representation of this fraction is
$$
m / n=0, a_{1} a_{2} \ldots a_{k} 167 a_{k+4} \ldots
$$
where $a_{1}, a_{2}, \ldots, a_{k}, a_{k+4}, \ldots$ are some digits. Then for the number $p=10^{k} m /... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,876 |
5.9. (New York, 78). Prove that any prime number of the form
$$
2^{2^{n}}+1(n \in \mathbf{N})
$$
cannot be represented as the difference of the fifth powers of two natural numbers. | 5.9. Let, contrary to the assertion of the problem, there exist numbers $n, m, k \in \mathbf{N}$, satisfying the equation
$$
2^{2^{n}}+1=m^{5}-k^{5}
$$
and the number
$$
m^{5}-k^{5}=(m-k)\left(m^{4}+m^{3} k+m^{2} k^{2}+m k^{3}+k^{4}\right)
$$
is prime. Then $m-k=1$ and
$$
2^{2^{n}}+1=(k+1)^{5}-k^{5}=5 k^{4}+10 k^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,877 |
5.10. (New York, 77). Does there exist a value $n \in \mathbb{N}$, for which the numbers
$$
2^{n+1}-1 \text { and } 2^{n-1}\left(2^{n}-1\right)
$$
are simultaneously cubes of integers? | 5.10. Suppose that for some value of $n \in \mathbb{N}$, the number
$$
2^{n-1}\left(2^{n}-1\right)
$$
is a cube of an integer. Note that the power of two in this product is $n-1$ (since the second factor $2^{n}-1$ is not divisible by 2), so $n-1=3k$, i.e., $n=3k+1$, where $k \in \mathbb{Z}^{+}$. But then the number
... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,878 |
5.13. (CRR, 80). Prove that for any values of $m, n \in \mathbf{N}$ there exists a number $k \in \mathbf{N}$, satisfying the equation
$$
(\sqrt{m}+\sqrt{m-1})^{n}=\sqrt{k}+\sqrt{k-1} .
$$ | 5.13. For given numbers $m, n \in \mathbf{N}$, using the binomial theorem, we get
$$
(\sqrt{m} \pm \sqrt{m-1})^{n}=\sum_{i=0}^{n} C_{n}^{i}(\sqrt{m})^{n-i}( \pm \sqrt{m-1})^{i}
$$
In the case $n=2j (j \in \mathbf{N})$, we have
$$
\begin{gathered}
(\sqrt{m} \pm \sqrt{m-1})^{n}= \\
=\sum_{i=0}^{j} C_{n}^{2 i}(\sqrt{m}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,880 |
5.14. (Jury, PNR, 77). Prove that for any values of $a, b \in \mathbf{R}$ and $\varepsilon>0$ there exist numbers $k, m \in \mathbf{Z}$ and $n \in \mathbf{N}$, satisfying the inequalities
$$
|n a-k|<\varepsilon \text { and }|n b-m|<\varepsilon .
$$ | 5.14. Let's choose an integer $N>1/$ and for each pair of values
$$
x, y \in[0 ; 1)
$$
we will associate a pair of numbers $u$, $v$, defined by the formulas
$$
u=[N x], \quad v=[N y] .
$$
Then, if two pairs $(x_{1} ; y_{1})$ and $(x_{2} ; y_{2})$ correspond to the same pair $(u ; v)$, we have
$$
\begin{aligned}
\l... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,881 |
5.15. (Jury, SRV, 76). Prove that there are infinitely many numbers of the form
$$
5^{n}(n \in \mathbf{N})
$$
in the decimal representation of each of which one can highlight at least 1976 consecutive zeros. | 5.15. We will prove that for any value of $k \in \mathbf{N}$, there exist infinitely many numbers $m \in \mathbf{N}$ satisfying the condition
$$
5^{m} \equiv 1\left(\bmod 2^{k}\right)
$$
Indeed, among the numbers
$$
5^{0}, 5^{1}, 5^{2}, \ldots, 5^{k}
$$
there will be two numbers $5^{p}$ and $5^{q}(p>q)$, giving the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,882 |
5.16*. (Jury, England, 77). Prove that for any value of $m \in \mathbf{N}$ there are infinitely many numbers of the form
$$
5^{n}(n \in \mathbb{N})
$$
such that each of the $m$ last digits of their decimal representation has a parity different from the parity of the digits adjacent to it. | 5.16. First, we will prove by induction on $1 \in \mathbf{Z}^{+}$ that the number $52^{j}-1$ is divisible by $2^{j+2}$ but not by $2^{j+3}$. For $i=0$ we have
$$
52^{0}-1=4,
$$
i.e., the statement is true. Next, suppose that for some $l \geqslant 0$ the number $52^{j}-1$ is divisible by $2^{l+2}$ but not by $2^{l+3}$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,883 |
5.17*. (SFRY, 73). Prove that if the lengths of the sides of a rectangle are odd numbers, then there is no point inside this rectangle such that the distance from it to any of the four vertices is an integer. | 5.17. Let $x_{1}$ and $x_{2}$ be the distances from some point inside a rectangle to two of its opposite sides, and $y_{1}$ and $y_{2}$ be the distances to the other two sides. Then, by the condition, the sides of the rectangle
$$
A=x_{1}+x_{2} \quad \text { and } \quad B=y_{1}+y_{2}
$$
are odd numbers. Suppose there... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,884 |
5.18*. (Jury, VNR, 79). Prove that there does not exist a regular quadrilateral pyramid in which the lengths of all edges, the total surface area, and the volume are integers. | 5.18. Suppose the pyramid mentioned in the problem exists. Let $g$ be the length of the side of the square base, and $h$ be the height of the pyramid. Then the length of the lateral edge of the pyramid, its total surface area, and volume are respectively
$$
f=\sqrt{h^{2}+2(g / 2)^{2}}, s=g^{2}+2 g \sqrt{h^{2}+(g / 2)^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,885 |
5.19*. (England, 81). For given distinct numbers $a_{1}, a_{2}, \ldots, a_{n} \in \mathbf{N}(n>1)$ and for each value $i=1,2, \ldots$ ..., $n$ denote
$$
p_{i}=\prod_{\substack{1 \leqslant j \leqslant n \\ j \neq i}}\left(a_{i}-a_{j}\right) .
$$
Prove that for any value $k \in \mathbf{N}$ the number
$$
\sum_{i=1}^{n}... | 5.19. We will first prove the required statement for the values
$$
k=0,1, \ldots, n-1
$$
We construct a polynomial of degree no higher than $(n-1)$, which at each point $a_{i}(i=1,2, \ldots, n)$ takes the value $a_{i}^{k}$. According to Lagrange's formula (see Theorem 62), this polynomial has the form
$$
P(x)=\sum_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,886 |
6.1. (New York, 78). Solve the equation $8^{x}(3 x+1)=4$.
Solve the equation $8^{x}(3 x+1)=4$. | 6.1. The number $x=1 / 3$ is a solution to the given equation. Let's prove that there are no other solutions. For $x>-1 / 3$, the functions $y_{1}(x)=8^{x}$ and $y_{2}(x)=3 x+1$ take positive values and increase, so their product (the left side of the equation) is also an increasing function. Therefore, on the interval... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,887 |
6.2. (Austria, 74). Prove that for any values of $a, b, c \in \mathbb{R}$ the equation
$$
(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
$$
has at least one solution. | 6.2. Let
$$
f(x)=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)
$$
Without loss of generality, we can assume that $a \leqslant b \leqslant c$. If $a=b$ or $b=c$, then $f(b)=(b-a)(b-c)=0$. If, however, $a < b < c$, then $f(a)=(a-b)(a-c)>0$ and $f(c)=(c-a)(c-b)>0$. Since the function $f(x)$ is continuous, there exists a number $x_{0... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,888 |
6.3. (GDR, 83). Prove that the equation
$$
x^{4}+5 x^{3}+6 x^{2}-4 x-16=0
$$
has exactly two solutions. | 6.3. We will prove that the function
$$
f(x)=x^{4}+5 x^{3}+6 x^{2}-4 x-16
$$
takes the value 0 at exactly two points. To do this, we will investigate the derivative of this function
$$
f^{\prime}(x)=4 x^{3}+15 x^{2}+12 x-4=(x+2)^{2}(4 x-1)
$$
For $x < 1 / 4$, the inequality $f^{\prime}(x) < 0$ holds, and for $x > 1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,889 |
6.4. (New York, 80). Find all pairs of numbers \(a > 1\), \(b > 0\), for which the equation \(a^x = x^b\) has exactly one positive solution. Indicate this solution for each pair of found values \(a, b\). | 6.4. The equation $a^{x}=x^{b}$ is equivalent to the equation $c^{x}=x$, where $c=a^{1 / b}>1$. Since the derivative of the function $f(x)=c^{x}-x$, which is $f^{\prime}(x)=c^{x} \ln c-1$, is positive when $c^{x}>\log _{c} e$ and negative when $c^{x}<\log _{c} e$, there exists a unique $x_{0}>0$ that satisfies the equa... | =,b=e\ln,e | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,890 |
6.5. (SFRY, 72). For each value of $a \in \mathbb{R}$, solve the equation
$$
(a-1)\left(\frac{1}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x \cos x}\right)=2
$$ | 6.5. Transforming the left side of the equation, we get
$$
\begin{aligned}
& (a-1)\left(\frac{1}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x \cos x}\right)=(a-1)\left(\frac{\cos x+\sin x+1}{\sin x \cos x}\right)= \\
& =(a-1)\left(\frac{\sqrt{2} \sin (x+\pi / 4)+1}{\sin ^{2}(x+\pi / 4)-\frac{1}{2}}\right), \text { i.e., }... | (-1)^{n}\arcsin(\sqrt{2}/2)-\pi/4+\pin | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,891 |
6.6. (New York, 73). Find all values of $x \in [0, \pi / 2]$, satisfying the equation $\cos ^{8} x + \sin ^{8} x = 97 / 128$. | 6.6. Transforming the left side of the equation, we get
$$
\begin{aligned}
& \cos ^{8} x+\sin ^{8} x=\left(\frac{1+\cos 2 x}{2}\right)^{4}+\left(\frac{1-\cos 2 x}{2}\right)^{4}= \\
& =\frac{1}{16}\left(2+12 \cos ^{2} 2 x+2 \cos ^{4} 2 x\right)=\frac{1}{32}\left(4+12(1+\cos 4 x)+(1+\cos 4 x)^{2}\right)= \\
& =\frac{1}{... | \pi/12,5\pi/12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,892 |
6.7. (New York, 78). Find all pairs of natural numbers $A \neq B$ for which the system
$$
\left\{\begin{array}{l}
\cos A x+\cos B x=0 \\
A \sin A x+B \sin B x=0
\end{array}\right.
$$
has a solution. | 6.7. Let the given system have a solution \( x \) for some sought values of \( A, B \). From the first equation of the system, we obtain that \( \cos B x = -\cos A x \), hence \( \sin B x = \varepsilon \sin A x \), where \( \varepsilon \in \{-1 ; 1\} \). Then from the second equation of the system, we get
\[
(A + \var... | A=2^p,B=2^q | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,893 |
6.8. (CSSR, 56). Find all pairs of numbers \( x, y \in (0, \pi / 2) \) that satisfy the system
\[
\left\{\begin{array}{l}
\frac{\cos x}{\cos y}=2 \cos ^{2} y \\
\frac{\sin x}{\sin y}=2 \sin ^{2} y
\end{array}\right.
\] | 6.8. By adding the equations of the original system, we get for any solution of the system
$$
\frac{\cos x \sin y + \sin x \cos y}{\cos y \sin y} = 2
$$
i.e.,
$$
\frac{\sin (x+y)}{\sin 2 y} = 1
$$
Therefore,
$$
\sin \left(\frac{x-y}{2}\right) \cos \left(\frac{x+3 y}{2}\right) = 0
$$
Since \( x, y \in (0 ; \pi / 2... | x=y=\pi/4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,894 |
6.9. (GDR, 66). Solve the equation
$$
(\sin (x-y)+1)(2 \cos (2 x-y)+1)=6
$$ | 6.9. Due to the estimates $-1 \leqslant \sin \alpha \leqslant 1, -1 \leqslant \cos \alpha \leqslant 1$, we have the inequality
$$
(\sin (x-y)+1)(2 \cos (2 x-y)+1)<6
$$
in which equality is possible only in the case when
$$
\sin (x-y)=1 \quad \text { and } \quad \cos (2 x-y)=1 .
$$
Therefore, the equation is equival... | 2\pik-\frac{\pi}{2},\,(2+1)\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,895 |
6.10. (SRP, 78). For each value of $n \in \mathbf{N}$, solve the equation
$$
\sin x \sin 2 x \ldots \sin n x + \cos x \cos 2 x \ldots \cos n x = 1
$$ | 6.10. Let $n=1$, then
$$
1=\sin x+\cos x=\sqrt{2} \cos (x-\pi / 4)
$$
i.e.
$$
x=2 \pi m \quad \text { or } \quad x=\pi / 2+2 \pi k \quad(m, k \in \mathbf{Z})
$$
Let $n=2$, then
$$
1=\sin x \sin 2 x+\cos x \cos 2 x=\cos (2 x-x)=\cos x
$$
i.e.
$$
x=2 \pi m \quad(m \in \mathbb{Z})
$$
Finally, let $n>2$, then
$1=\... | \begin{pmatrix}ifn=1,&2\pior\pi/2+2\pik\(,k\in{Z}),\\ifn=4-2orn=4+1(\in\mathbb{N}),&2\pi\(\in\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,896 |
6.11. (GDR, 66). For each value of $n \in \mathbf{N}$, solve the equation $(x+y)^{n}=x^{n}+y^{n}$. | 6.11. If $n=1$, then $x$ and $y$ can be any values. Further, for any $n \in N$ we have: if $x=0$, then the equation holds for any $y$, and if $y=0$, then the equation holds for any $x$. Pairs $(x ; y)$, for which $x y \neq 0$ and $x+y=0$, satisfy the equation for odd values of $n$ (and only for them). We will prove tha... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,897 |
6.12. (GDR, 78). Solve the system
$$
\left\{\begin{array}{l}
x + xy + y = 2 + 3 \sqrt{2} \\
x^{2} + y^{2} = 6
\end{array}\right.
$$ | 6.12. Multiplying the first equation by 2 and adding it to the second, we get for any solution of the system
$$
x^{2}+2 x y+y^{2}+2(x+y)=(x+y)^{2}+2(x+y)=10+6 \sqrt{2},
$$
i.e., $(x+y+1)^{2}=(3+\sqrt{2})^{2}$, hence $x+y+1= \pm(3+\sqrt{2})$. If $x+y=-4-\sqrt{2}$, then $x y=6+4 \sqrt{2}$ and $(x-y)^{2}=(x+y)^{2}-4 x y... | x_{1}=2,y_{1}=\sqrt{2}x_{2}=\sqrt{2},y_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,898 |
6.13. (England, 75). Prove that for any value of $n \in \mathbf{N}$ there exists a unique set of numbers $x_{1}, \ldots, x_{u}$, satisfying the equation
$$
\left(1-x_{1}\right)^{2}+\left(x_{1}-x_{2}\right)^{2}+\ldots+\left(x_{n-1}-x_{n}\right)^{2}+x_{n}^{2}=1 /(n+1)
$$ | 6.13. By the mean value theorem (Theorem 6) we have
$$
\begin{aligned}
& \left(1-x_{1}\right)^{2}+\left(x_{1}-x_{2}\right)^{2}+\ldots+\left(x_{n-1}-x_{n}\right)^{2}+x_{n}^{2} \geqslant \\
& \quad \geqslant \frac{1}{n+1}\left(\left(1-x_{1}\right)+\left(x_{1}-x_{2}\right)+\ldots+\left(x_{n-1}-x_{n}\right)+x_{n}\right)^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,899 |
6.14. (NPR, 68). Find all values of $n \in \mathbf{N}$ for which there exists a set of positive numbers $x_{1}, \ldots, x_{\mu}$, satisfying the system
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\ldots+x_{n}=9 \\
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}=1
\end{array}\right.
$$
Indicate all such sets for eac... | 6.14. Let $R=\sqrt[n]{x_{1} \ldots x_{n}}$, where $\left(x_{1} ; \ldots ; x_{n}\right)$ is a solution to the system. Applying the theorem of means, we get
$$
1=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}} \geqslant \frac{n}{R}, \quad 9=x_{1}+x_{2}+\ldots+x_{n} \geqslant n R,
$$
from which $R \geqslant n$ an... | 2,3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,900 |
6.15*. (CSSR, 82). For each pair of values $n, k \in \mathbf{N}$, find all sets of non-negative numbers $x_{1}, x_{2}, \ldots, x_{n}$, satisfying the system
$$
\left\{\begin{array}{l}
x_{1}^{k}+x_{2}^{k}+\ldots+x_{n}^{k}=1 \\
\left(1+x_{1}\right)\left(1+x_{2}\right) \ldots\left(1+x_{n}\right)=2
\end{array}\right.
$$ | 6.15. Note that for any number $i \in\{1 ; \ldots ; n\}$ the set of unknowns
$$
x_{i}=1, \quad x_{1}=\ldots=x_{i-1}=x_{i+1}=\ldots=x_{n}=0
$$
satisfies the original system. We will prove that there are no other sets of $n$ non-negative numbers that satisfy this system. Note that if $x_{i} \geqslant 0$ for $i=1,2, \ld... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,901 |
6.16*. (GDR, 80). Solve the system
$$
\left\{\begin{array}{l}
2 x+x^{2} y=y \\
2 y+y^{2} z=z \\
2 z+z^{2} x=x
\end{array}\right.
$$ | 6.16. Let the set ( $x ; y ; z$ ) satisfy the original system. If we denote $x=\operatorname{tg} a$, where $a \in(-\pi / 2 ; \pi / 2)$, then from the first equation of the system we get
$$
y=2 x^{\prime}\left(1-x^{2}\right)=\operatorname{tg} 2 a
$$
from the second-
$$
z=\operatorname{tg} 4 a
$$
and from the third-
... | \operatorname{tg}(\pik/7),\quad\operatorname{tg}(2\pik/7),\quad\operatorname{tg}(4\pik/7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,902 |
6.17*. (Austria-Poland, 79). For any values of $n \in \mathbf{N}$ and $a \in \mathbb{R}$, solve the system
$$
\left\{\begin{array}{c}
x_{1}+x_{2}+\ldots+x_{n}=a \\
x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=a^{2} \\
\cdot \cdot \cdot \cdot \cdot \cdot \cdot \\
x_{1}^{n}+x_{2}^{n}+\ldots+x_{n}^{n}=a^{n}
\end{array}\right.
$$... | 6.17. Note that for any number $i \in\{1 ; \ldots ; n\}$ the set of unknowns
$$
x_{i}=a, \quad x_{1}=\ldots=x_{i-1}=x_{i+1}=\ldots=x_{n}=0
$$
satisfies the original system. Let's prove that this system has no other solutions.
1) For $n=1$ we get the single condition $x_{1}=a$.
2) For $n=2$ we have the system
$$
\le... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,903 |
7.1. (GDR, 74). What is greater: $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}$ or 0? | 7.1. Using Theorem 4, we get
$$
\begin{aligned}
& \sqrt{4+\sqrt{7}}=\sqrt{4-\sqrt{7}}-\sqrt{\overline{2}}= \\
& =\left(\sqrt{\frac{4+\sqrt{16-7}}{2}}+\sqrt{\frac{4-\sqrt{16-7}}{2}}\right)- \\
& -\left(\sqrt{\frac{4+\sqrt{16-7}}{2}}-\sqrt{\frac{4-\sqrt{16-7}}{2}}\right)-\sqrt{2}= \\
& =\sqrt{\frac{7}{2}}+\sqrt{\frac{1}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,904 |
7.2. (Belgium, 79). Arrange the numbers
$$
\begin{gathered}
x=(a+b)(c+d) \\
y=(a+c)(b+d) \\
z=(a+d)(b+c)
\end{gathered}
$$
in ascending order, given that $a<b<c<d$. | 7.2. Since $a0
$$
and
$$
z-y=a c+b d-a d-b c=(a-b)(c-d)>0
$$
Thus, $x<y<z$. | x<y<z | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,905 |
7.3. (Yugoslavia, 76). Prove that if the product of three numbers is 1, and their sum is greater than the sum of their reciprocals, then exactly one of these numbers is greater than 1. | 7.3. Let the numbers $a, b, c$ satisfy the conditions of the problem
$$
a b c=1 \text { and } a+b+c>\frac{1}{a}+\frac{1}{b}+\frac{1}{c} .
$$
Then
$(a-1)(b-1)(c-1)=a b c-a b-a c-b c+a+b+c-1=$
$$
\begin{aligned}
&=(a b c-1)+(a+b+c)-a b c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)= \\
&=(a+b+c)-\left(\frac{1}{a}+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,906 |
7.5. (YFR, 76). Prove that for any numbers $a_{\text {, }}$ $b, c$, greater than 1, the following inequality holds
$$
2\left(\frac{\log _{b} a}{a+b}+\frac{\log _{c} b}{b+c}+\frac{\log _{a} c}{c+a}\right) \geqslant \frac{9}{a+b+c}
$$ | 7.5. Applying the mean value theorem and considering that the numbers $\log _{b} a$, $\log _{c} b, \log _{a} c$ are positive and their product equals one, we get
$\frac{\log _{b} a}{a+b}+\frac{\log _{c} b}{b+c}+\frac{\log _{a} c}{a+c} \geqslant 3 \sqrt[3]{\frac{\log _{b} a \log _{c} b \log _{a} c}{(a+b)(b+c)(a+c)}}=$ ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,908 |
7.6. (Austria, 71). Prove that for any positive numbers $a, b, c$ the inequality
$$
a^{2}(b+c-a)+b^{2}(a+c-b)+c^{2}(a+b-c) \leqslant 3 a b c
$$
holds. | 7.6. Without loss of generality, we can assume that $a \geqslant b \geqslant c>0$. Transform the difference between the right and left sides of the inequality as follows:
$$
\begin{aligned}
3 a b c + a^{2}(a-b-c) + b^{2}(b-a-c) + c^{2}(c-a-b) = \\
= 3 a b c + a^{3} + b^{3} + c^{3} - a^{2} b - b^{2} a - a^{2} c - b^{2}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,909 |
7.7. (USA, 80). Prove that for any numbers $a, b, c \in [0; 1]$ the inequality
$$
\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leqslant 1
$$
holds. | 7.7. If $a=b=c=0$, then the inequality holds. Let $s=a+b+c>0$, then
$$
\begin{aligned}
& \frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c)= \\
& =\frac{a}{s}-\frac{a(1-a)}{s(b+c+1)}+\frac{b}{s}-\frac{b(1-b)}{s(a+c+1)}+\frac{c}{s}-\frac{c(1-c)}{s(a+b+1)}+ \\
& +(1-a)(1-b)(1-c)\left(\frac{a}{s}+\frac{b}{s}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,910 |
7.8. (CSPR, 59). Prove that if the numbers $a, b, c \in \mathbf{R}$ satisfy the inequalities
$$
a+b+c>0, \quad ab+bc+ca>0, \quad abc>0
$$
then these numbers are positive. | 7.8. Let $p=a+b+c>0, \quad q=a b+b c+a c>0, \quad r=a b c>0$. Then the polynomial $P(x)=x^{3}-p x^{2}+q x-r$ takes negative values for $x \leqslant 0$. Therefore, all roots of this polynomial, which by the inverse of Vieta's theorem are equal to the numbers $a, b, c$, are positive. | proof | Algebra | proof | Yes | Yes | olympiads | false | 28,911 |
7.9. (Belgium, 76). Prove that for all $\alpha \in \mathbb{R}$ the inequality
$$
\sin (\cos \alpha)<\cos (\sin \alpha)
$$
holds. | 7.9. Let $\alpha \in[0 ; \pi / 2)$. First, substituting $x=\cos \alpha$ into the inequality $\sin x > 0$, we obtain the estimate $\sin (\cos \alpha) < \cos \alpha$. Second, from the inequality $\alpha \geqslant \sin \alpha$, we have the estimate $\cos \alpha \leq \cos (\sin \alpha)$, since the function $\cos x$ is decr... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,912 |
7.10. (Balkaniad, 84). Prove that for any positive numbers $\alpha_{1}, \ldots, \alpha_{n}(n \geqslant 2)$, the sum of which is 1, the following inequality holds:
$$
\sum_{i=1}^{n} \frac{\alpha_{i}}{2-\alpha_{i}} \geqslant \frac{n}{2 n-1}
$$ | 7.10. Let $\beta_{i}=2-\alpha_{i}>0(i=1, \ldots, n)$. Then, considering the equality $\sum_{i=1}^{n} \beta_{i}=2 n-1$ and the mean value theorem, we obtain
$$
\begin{aligned}
\sum_{i=1}^{n} \frac{\alpha_{i}}{\beta_{i}} & =\sum_{l=1}^{n}\left(\frac{\alpha_{i}}{\beta_{i}}+1\right)-n=\sum_{i=1}^{n} \frac{\alpha_{i}+\beta... | \frac{n}{2n-1} | Inequalities | proof | Yes | Yes | olympiads | false | 28,913 |
7.11. (GDR, 67; England, 76). Prove that for any positive numbers \(a_{1}, \ldots, a_{n}(n \geqslant 2)\) the inequality
$$
\sum_{i=1}^{n} \frac{a_{i}}{s-a_{i}} \geqslant \frac{n}{n-1}, \quad \text { where } s=\sum_{i=1}^{n} a_{i} \text { is denoted. }
$$ | 7.11. Let $b_{i}=s-a_{i}>0(l=1, \ldots, n)$. Then, considering the equality $\sum_{i=1}^{n} b_{i}=(n-1) s$ and the theorem of means, we obtain
$$
\begin{aligned}
\sum_{i=1}^{n} \frac{a_{i}}{s-a_{i}} & =\sum_{i=1}^{n}\left(\frac{a_{i}}{b_{i}}+1\right)-n=\sum_{i=1}^{n} \frac{s}{b_{i}}-n=s \sum_{i=1}^{n} \frac{1}{b_{i}}-... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,914 |
7.12. (New York, 75). Is it true that for any positive numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $a_{n+1}=a_{1}$, the inequality
$$
\sum_{i=1}^{n}\left(\frac{a_{i}}{a_{i+1}}\right)^{n} \geqslant \sum_{i=1}^{n} \frac{a_{i+1}}{a_{i}} ?
$$ | 7.12. Let $a_{i} / a_{i+1}=b_{i}(i=1, \ldots, n)$ and set $b_{n+i}=1$. Then we have $\prod_{i=1}^{n+1} b_{i}=1$ and, using the theorem of means, we obtain the relations
$$
\sum_{i=1}^{n+1} \frac{1}{b_{i}}=\sum_{i=1}^{n+1} \prod_{j \neq i} b_{j}<\sum_{i=1}^{n+1}\left(\frac{1}{n} \sum_{i \neq i} b_{j}^{n}\right)=\sum_{i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,915 |
7.13. (Jury, Canada, 82). Prove that for any positive numbers $x \neq 1, a<1$ the inequality
$$
\frac{1-x^{a}}{1-x}<(1+x)^{a-1}
$$
holds. | 7.13. Let's introduce the notation $1-a=b$. First, consider the case when $0 < x < 1$. We need to show that
$$
x^{a} + b x^{a+1} > (1+b) x.
$$
Now we have
$$
\begin{aligned}
(1+x)^{-b} - \frac{1-x^{a}}{1-x} & \geqslant \frac{1}{1+b x} - \frac{1-x^{a}}{1-x}= \\
& = \frac{1-x + (x^{a}-1)(1+b x)}{(1+b x)(1-x)} = \frac{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,916 |
7.14. (Jury, USSR, 82). Prove that for any value of $\alpha \leqslant 1$ and any numbers $x_{1}, \ldots, x_{n}$, satisfying the conditions $1 \geqslant x_{1} \geqslant x_{2} \geqslant \ldots \geqslant x_{n}>0$, the inequality
$\left(1+x_{i}+x_{2}+\ldots+x_{n}\right)^{\alpha} \leqslant 1+1^{\alpha-1} x_{1}^{\alpha}+2^{... | 7.14. We will prove this by induction on $n \in \mathbb{Z}^{+}$. For $n=0$, we obtain the true inequality $1 \alpha \leqslant 1$. Assume that the inequality holds for the number $n$, and we will prove it for the number $n+1$. We have
$$
\begin{aligned}
& \left(1+x_{1}+\ldots+x_{n}+x_{n+1}\right)^{\alpha}-\left(1+x_{1}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,917 |
7.15. (NBR, 82). Prove that for any numbers $a_{1}, \ldots, a_{n} \in[0 ; 2](n \geqslant 2)$ the inequality
$$
\sum_{i=1}^{n} \sum_{i=1}^{n}\left|a_{i}-a_{j}\right| \leqslant n^{2}
$$
holds. Determine for which values of $a_{1}, \ldots, a_{n}$ equality is achieved. | 7.15. Without loss of generality, assume that $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{n}$. Then
$$
\begin{array}{r}
S=\sum_{i=1}^{n} \sum_{j=1}^{n}\left|a_{i}-a_{j}\right|=\sum_{1 \leqslant i<j \leqslant n}\left(a_{j}-a_{i}\right)=\sum_{1 \leqslant i<j \leqslant n} a_{j}-\sum_{1 \leqslant i<j \leqslant n}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,918 |
7.16. (SFRY, 72). Prove that if the number $M$ and the set of numbers
$$
\begin{aligned}
& a_{11}, a_{12}, \ldots, a_{1 n} \\
& a_{21}, a_{22}, \ldots, a_{2 n}, \\
& a_{n 1}, a_{n 2}, \ldots, a_{n n}
\end{aligned}
$$
satisfy the inequalities
$$
\sum_{j=1}^{n}\left|a_{j 1} x_{1}+a_{j 2} x_{2}+\ldots+a_{i n} x_{n}\rig... | 7.16. Since
$$
\sum_{j=1}^{n}\left|a_{j 1} x_{1}+\cdots+a_{j n} x_{n}\right|<M
$$
for any set of numbers \( x_{1}, \ldots, x_{n} \in\{-1 ; 1\} \), we have the inequality
$$
\frac{1}{2^{n}} \sum_{\left\{x_{1} ; \ldots ; x_{n}\right\}}\left(\sum_{l=1}^{n}\left|a_{j 1} x_{1}+\ldots+a_{j n} x_{n}\right|\right) \leqslant... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,919 |
7.17. (Jury, USA, 82). Prove that for any set of numbers $a_{i}, \ldots, a_{n} \in \mathbb{R}$, one can choose a value $k \in\{1 ; \ldots ; n\}$ such that any non-negative numbers $b_{1} \geqslant b_{2} \geqslant \ldots \geqslant b_{n}$, not exceeding 1, satisfy the inequality
$$
\left|\sum_{i=1}^{n} b_{i} a_{i}\right... | 7.17. Note that the sum $S=\sum_{i=1}^{n} b_{i} a_{i}$ can be represented as $S=b_{1} a_{1}+\sum_{i=2}^{n} b_{i}\left(a_{1}+\ldots+a_{i}\right)-\sum_{i=2}^{n} b_{i}\left(a_{1}+\ldots+a_{i-1}\right)=$ $=\sum_{i=1}^{n}\left(a_{1}+\ldots+a_{i}\right) b_{i}-\sum_{i=1}^{n-1}\left(a_{1}+\ldots+a_{i}\right) b_{i+1}=$ $=\sum_{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,920 |
7.18*. (NRA, 84). Prove that for any values of $m, n \in \mathbf{N}$ and any numbers $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n} \in[0 ; 1]$, satisfying the conditions $x_{i}+y_{i}=1$ for $i=1, \ldots, n$,
$$
\left(1-x_{1} \ldots x_{n}\right)^{m}+\left(1-y_{1}^{m}\right) \ldots\left(1-y_{n}^{m}\right) \geqslant 1
$$ | 7.18. We will prove the statement by induction on $n \in \mathbf{N}$. Since
$$
\left(1-x_{1}\right)^{m}+\left(1-y_{1}^{m}\right)=y_{1}^{m}+\left(1-y_{1}^{m}\right)=1
$$
the statement is true for $n=1$. Suppose it is true for the number $n-1$, then we have
$$
\begin{aligned}
& \left(1-x_{1} \ldots x_{n}\right)^{m}+\l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,921 |
7.19*. (USA, 77). Prove that for any positive numbers $a \leqslant b \leqslant c \leqslant d$ the inequality
$$
a^{b} b^{c} c^{d} d^{a} \geqslant b^{a} c^{b} d^{c} a^{d}
$$
holds. | 7.19. Let's make the substitution $b=a x, c=a y, d=a z$, then due to the condition of the problem we have $1 \leqslant x \leqslant y \leqslant z$, and the required inequality can be written as follows:
$$
a^{a x}(a x)^{a y}(a y)^{a z}(a z)^{a} \geqslant(a x)^{a}(a y)^{a x}(a z)^{a y} a^{a z}
$$
Further, after canceli... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,922 |
7.20*. (GDR, 80). Prove that for any numbers $n, k \in \mathbf{N}$, greater than 1, the inequality
$$
\sum_{j=2}^{n^{k}} \frac{1}{j}>k \sum_{j=2}^{n} \frac{1}{j}
$$
holds. | 7.20. Transform the left part of the inequality to the form
$$
\begin{aligned}
& \sum_{i=2}^{n^{k}} \frac{1}{j}=\left(\frac{1}{1+1}+\ldots+\frac{1}{n}\right)+\left(\frac{1}{n+1}+\ldots+\frac{1}{n^{2}}\right)+\ldots \\
& \quad \cdots+\left(\frac{1}{n^{k-1}+1}+\cdots+\frac{1}{n^{k}}\right)=\sum_{i=1}^{k}\left(\frac{1}{n... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,923 |
7.21*. (Jury, France, 82). Prove that for any positive numbers $a_{1}, \ldots, a_{n}$ the following inequality holds:
$$
\sum_{k=1}^{n} \sqrt[k]{a_{1} \ldots a_{k}} \leqslant e \sum_{k=1}^{n} a_{k}
$$
where $e$ is the base of the natural logarithms. | 7.21. Let us denote
$$
b_{k}=(k+1)^{k} / k^{k-1}=k(1+1 / k)^{k} \quad(k=1, \ldots, n)
$$
Considering the relations $b_{k} \leqslant k e$ (see Theorem 7) and
$$
b_{1} \ldots b_{k}=(k+1-1)^{k}
$$
as well as the theorem of means, we have
$$
\begin{aligned}
\sqrt[k]{a_{1} \ldots a_{k}}= & \frac{1}{k+1} \sqrt[k]{\left(... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,924 |
7.22*. (USA, 77). Prove that if two positive numbers $p<q$ are fixed, then for any numbers $\alpha, \beta, \gamma, \delta, \varepsilon \in[p ; q]$ the following inequality holds:
$(\alpha+\beta+\gamma+\delta+\varepsilon)\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}+\frac{1}{\varepsilon}\righ... | 7.22. We will consider that the quintet of numbers $\alpha, \beta, \gamma, \delta, \varepsilon$ coincides (up to order) with the quintet of numbers $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{1} \leqslant x_{1}$. Then
$$
\begin{aligned}
& \left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)\left(\frac{1}{x_{1}}+\frac{1}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,925 |
7.23*. (GDR, 70). Prove that for any positive numbers $a, b, c, d$ the following inequality holds:
$$
\sqrt[3]{\frac{a b c + a b d + a c d + b c d}{4}} \leqslant \sqrt{\frac{a b + a c + a d + b c + b d + c d}{6}}
$$
Determine for which values of $a, b, c, d$ equality is achieved.
## § 8. Problems with the Integer Pa... | 7.23. We will consider that the quadruple of numbers $a, b, c, d$ coincides (up to order) with the quadruple of numbers $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4}$. Let's find the derivative of the polynomial
$$
P(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right) .
$$
On ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,926 |
8.2. (England, 75). Solve the equation
$$
[\sqrt[3]{1}]+[\sqrt[3]{2}]+\ldots+\left[\sqrt[3]{x^{3}-1}\right]=400
$$
in natural numbers. | 8.2. Note that the relation $[\sqrt[3]{m}]=k$, where $m, k \in \mathrm{N}$, is equivalent to the inequality $k^{3} \leqslant m \leqslant(k+1)^{3}-1$. The number of natural numbers $m$ satisfying this condition (for a fixed $k$) is $(k+1)^{3}-k^{3}=3 k^{2}+3 k+1$. Therefore, the left side of the equation is $\sum_{k=1}^... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,928 |
8.3. (Canada, 81). Prove that the equation
$$
[x]+[2 x]+[4 x]+[8 x]+[16 x]+[32 x]=12345
$$
has no solutions. | 8.3. We will prove that for any $k \in \mathbf{N}$ the following estimates hold:
$$
k[x] < [k x] \leqslant k[x] + k - 1
$$
Indeed, let $m = [x]$, $\alpha = \{x\}$, then we have $x = m + \alpha$, $m \in \mathbf{Z}$, $0 \leqslant \alpha < 1$ and
$$
k m = [k m] \leqslant [k x] = [k m + k \alpha] \leqslant k m + k \alph... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,929 |
8.4. (Sweden, 82). For each value of $n \in \mathbb{N}$, determine how many solutions the equation $x^{2}-\left[x^{2}\right]=\{x\}^{2}$ has on the interval $[1 ; n]$. | 8.4. Let $m=[x], \alpha=\{x\}$. Then $x=m+\alpha$ and the original equation will have the form
$$
(m+\alpha)^{2}-\left[m^{2}+2 m \alpha+\alpha^{2}\right]=\alpha^{2},
$$
from which
$$
m^{2}+2 m \alpha=\left[m^{2}+2 m \alpha+\alpha^{2}\right]
$$
But $m^{2} \in Z$, therefore
$$
2 m \alpha=\left[2 m \alpha+\alpha^{2}\... | n^2-n+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 28,930 |
8.5. (Austria, 74). Prove that for any value of $n \in \mathbb{N}$ the following equality holds:
$$
[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4 n+2}]
$$ | 8.5. We will prove that for any $n \in \mathbf{N}$, the inequality $\sqrt{n}+\sqrt{n+1}<\sqrt{4 n+2}$ holds. Indeed, since $4 n(n+1)<(2 n+1)^{2}$, it follows that $2 \sqrt{n(n+1)}<2 n+1$. Therefore,
$$
(\sqrt{n}+\sqrt{n+1})^{2}=2 n+1+2 \sqrt{n(n+1)}<4 n+2
$$
from which the required inequality follows, and hence the i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,931 |
8.6. (Jury, Belgium, 79). Which natural numbers cannot be represented in the form $[n+\sqrt{n}+1 / 2]$, where $n \in \mathbf{N}$? | 8.6. Let
$$
f(n)=[n+\sqrt{} \bar{n}+1 / 2], \quad n \in N
$$
then the difference
$$
\begin{aligned}
f(n+1)-f(n)=[n+1+\sqrt{n+1} & +1 / 2]-[n+\sqrt{n}+1 / 2]= \\
& =1+[\sqrt{n+1}+1 / 2]-[\sqrt{n}+1 / 2]
\end{aligned}
$$
is greater than one if and only if
$$
[\sqrt{n+1}+1 / 2]>[\sqrt{n}+1 / 2]
$$
i.e., if $\sqrt{n}... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,932 |
8.7. (Yugoslavia, 83). Prove that among the members of the sequence $\left\{a_{n}\right\}$, defined by the relations
$$
a_{1}=2, a_{n+1}=\left[(3 / 2) a_{n}\right] \text { for } n \in \mathbf{N} \text {, }
$$
there are infinitely many even and infinitely many odd numbers. | 8.7. Suppose that among the members of the sequence $\{a\}$ there is only a finite set of odd numbers. Then, by choosing an odd number $a_{m}$ with the largest index $m$, we obtain that all numbers $a_{m+n}, n \in N$, are even. Let $a_{m+1}=2^{p} q$, where $q$ is odd and then
$$
a_{m+2}=\left[\frac{3}{2} a_{m+1}\right... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,933 |
8.8. (Austria - PDR, 79). For each value of $n \in \mathbf{N}$, find the greatest value of $k \in \mathbf{Z}^{+}$, for which the number $\left[(3+\sqrt{11})^{2 n-1}\right]$ is divisible by $2^{k}$. | 8.8. Let
$$
a_{n}=(3+\sqrt{11})^{n}+(3-\sqrt{11})^{n}
$$
then the following equality holds:
$$
a_{n+2}=6 a_{n+1}+2 a_{n}, \quad n \in \mathbf{Z}^{+}
$$
Indeed, let $\alpha=(3+\sqrt{11})^{n}, \beta=(3-\sqrt{11})^{n}$, then
$$
\begin{gathered}
a_{n}=\alpha+\beta \\
a_{n+1}=(3+\sqrt{11}) \alpha+(3-\sqrt{11}) \beta \\... | n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 28,934 |
8.9. (Jury, RPP, 79). Prove that for any value of $n \in \mathbf{N}$, the estimate
$$
\{n \sqrt{2}\}>1 /(2 n \sqrt{2})
$$
holds, and for any $\varepsilon>0$ there exists a number $n \in N$ satisfying the inequality
$$
\{n \sqrt{2}\}<(1+\varepsilon) /(2 n \sqrt{2})
$$ | 8.9. For a given value $n \in \mathbb{N}$, denote $m = [n \sqrt{2}]$. Since $m \neq n \sqrt{2}$ (otherwise, the number $\sqrt{2} = m / n$ would be rational), we have $m < n \sqrt{2}$. Let $\varepsilon = n \sqrt{2} - m > 0$. Consider the sequences $\{n_i\}$ and $\{m_i\}$ defined as follows: $n_1 = m_1 = 1$
$$
n_{i+1} =... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,935 |
8.10*. (USA, 75). a) Prove that for any non-negative numbers $x, y$ the inequality
$$
[5 x]+[5 y] \geqslant[3 x+y]+[3 y+x]
$$
holds.
b) Prove that the number
$$
\frac{(5 m)!(5 n)!}{m!n!(3 m+n)!(3 n+m)!}
$$
is an integer for all values of $m, n \in \mathbf{N}$. | 8.10. a) Consider the function
$$
f(x, y)=[5 x]+[5 y]-[3 x+y]-[3 y+x]-[x]-[y]
$$
and prove that $f(x, y) \geqslant 0$ for $x, y \in \mathbf{R}$. Suppose for some values $x, y \in[0 ; 1)$ the inequality
$$
f(x, y)=[5 x]+[5 y]-[3 x+y]-[3 y+x](5 x-1)+(5 y-1)-(3 x+y)-(3 y+x)=x+y-2
$$
holds, i.e., $x+y-2 \geqslant 2 / 5... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28,936 |
8.12*. (NPR, 83). Prove that if the numbers $a, b, c$ satisfy the equation $[n a] + [n b] = [n c]$ for each value of $n \in \mathbf{N}$, then at least one of the numbers $a, b$ is an integer.
## Chapter 3 PLANIMETRY
## § 9. Triangles
(see Appendix G: definitions $36,40,41$; theorems $64,67,74,75,77,82,83$ ) | 8.12. Note that from the conditions of the problem, the equality $c = a + b$ follows, since otherwise, for sufficiently large values of $n \in \mathbb{N}$, the equalities
$$
[an] + [bn] = [cn]
$$
cannot hold. Indeed, if $c > a + b$ and $n \geq 1 / (c - a - b)$, then
$$
[n c] > n c - 1 \geq n a + n b \geq [n a] + [n ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 28,938 |
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