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 |
|---|---|---|---|---|---|---|---|---|---|
922. Find the fraction with the smallest possible denominator that lies between $\frac{7}{25}$ and $\frac{8}{25}$. List all solutions. | $\triangle$ Let the required fraction be $\frac{a}{b}$, where $a$ and $b$ are natural numbers. Then
$$
\frac{7}{25}<\frac{a}{b}<\frac{8}{25}
$$
Now let's consider several cases depending on $b$.
1) The cases $b=1$ and $b=2$ should be discarded, as the fraction $\frac{a}{b}$ would be too large in these cases.
2) Let ... | \frac{2}{7} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,318 |
925. What is the smallest number of members in a math club if the number of girls in it is more than $40 \%$, but less than $50 \%$? | Let in the club there be $a$ girls, and the total number of club members is $b$. Then the ratio of the number of girls to the total number of club members is $\frac{a}{b}$.
Since $40 \%$ is $\frac{2}{5}$, and $50 \% - \frac{1}{2}$ of the club members, then by the condition
$$
\frac{2}{5}<\frac{a}{b}<\frac{1}{2}
$$
W... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,319 |
928. Solve in integers $x$ and $y$ the system of inequalities
$$
\left\{\begin{array}{l}
\left|x^{2}-2 x\right|<y+0.5 \\
y+|x-1|<2
\end{array}\right.
$$ | $\triangle$ From the first inequality of the system, it is clear that $y \geqslant 0$, and from the second that $y<2$. Therefore, $0 \leqslant y<2$. Since only two integer values of $y$ satisfy the last inequality, $y=0$ and $y=1$, we will consider two cases.
1) Let $y=0$.
Then the system takes the form:
$$
\left\{\... | (0;0),(2;0),(1;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 45,320 |
930. How many solutions in integers $x$ and $y$ does the inequality
$$
|x|+|y|<10 ?
$$
have? | $\triangle$ From the condition, it is clear that
$$
|x|<10, \text { i.e. }-10<x<10
$$
Let's consider all cases.
If $x= \pm 9$, then $y=0$. We have two solutions: $(9 ; 0)$ and $(-9 ; 0)$.
If $x= \pm 8$, then $y=0, \pm 1$. This gives us six solutions: $(8 ; 0),(-8 ; 0),(8 ; 1),(-8 ; 1)$, $(8 ;-1)$ and $(-8 ;-1)$.
W... | 181 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 45,321 |
932. Find two natural numbers if their product is between 120 and 130, and their ratio is between 2 and 3. List all solutions. | Let's denote the required numbers by $k$ and $n$. Then
$$
\left\{\begin{array}{l}
120 < k n < 130 \\
2 < \frac{k}{n} < 3
\end{array}\right.
$$
We multiply these inequalities: $240 < k^2 < 390$. Additionally, we divide the inequalities, in the right part we need to divide 130 by 2 (but not by 3), and in the left part ... | 187 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,322 |
934. Three classmates bought 13 pies, with Kostya buying half as many as Tolya, and Volodya buying more than Kostya but less than Tolya. How many pies did each of them buy? | Let Kostya buy $x$ pies, and Volodya $-y$. Then Tolya bought $2 x$ pies. We get a mixed system, consisting of an equation and an inequality:
$$
\left\{\begin{array}{l}
3 x+y=13 \\
x<13
\end{array}\right.
$$
Solving this system of inequalities, we get:
$$
2 \frac{3}{5}<x<3 \frac{1}{4}
$$
Then $x=3, 2 x=6$. Let's fin... | 3,4,6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,323 |
938*. Two motor transport companies sent out cars to transport agricultural products in a total number of less than 18. The number of cars sent by the second company is less than twice the number of cars sent by the first company. If the first company had sent out 4 more cars, the number of cars sent by the first compa... | Let $x$ and $y$ be the number of cars sent by the first and second auto enterprises, respectively. We obtain the system of inequalities:
$$
\left\{\begin{array}{l}
x+y<18 \\
y<12 \\
x+4<y
\end{array}\right.
$$
Since $x+4<y$ and $y<18-x$ (from the first inequality of the system), we have
$$
x+4<18-x, \quad 2 x<14, \q... | 611 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,324 |
943. Solve in integers $x, y$ and $z$ the inequality
$$
x^{2}+y^{2}+z^{2}+3 < x y + 3 y + 2 z
$$ | $\triangle$ Let's gather all the terms of the inequality on the left side:
$$
x^{2}+y^{2}+z^{2}-x y-3 y-2 z+3<0
$$
Replace this inequality with the following:
$$
x^{2}+y^{2}+z^{2}-x y-3 y-2 z+4 \leqslant 0
$$
The obtained inequality is equivalent to the previous one for integer $x, y,$ and $z$ (think about why).
N... | (1;2;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 45,325 |
946. Find two natural numbers whose product does not exceed their sum. List all solutions. | Let's denote the required numbers by $x$ and $y$. Then
$$
x y \leqslant x+y .
$$
Let $x \leqslant y$. We obtain:
$$
x y \leqslant x+y \leqslant y+y=2 y
$$
From this,
$$
x y \leqslant 2 y, \quad x \leqslant 2
$$
Now let's consider two cases.
1) Suppose $x=1$.
The original inequality becomes:
$$
y \leqslant 1+y
... | one\\is\1,\the\other\is\any\natural\;\2,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,326 |
950. Solve the inequality in natural numbers
$$
x y < x + 2 y
$$ | $\triangle$ Here, the constraint of the type $x \leqslant y$ on the unknowns $x$ and $y$ (similar to what we applied when solving problem 946) should not be used, since the inequality is not symmetric with respect to $x$ and $y$.
Gather all terms of the inequality on the left side and add 2 to both sides of the result... | (x;1)(x\in\mathbb{N});(1;y)(y\in\mathbb{N},y\neq1);(2;y)(y\in\mathbb{N},y\neq1);(3;2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 45,327 |
954. Solve the inequality in natural numbers $x$ and $y$
$$
x^{2}+1<y^{2}<x^{2}+x+1
$$ | $\triangle$ The inequality holds
$$
x^{2}<x^{2}+1<y^{2}<x^{2}+x+1<x^{2}+2 x+1=(x+1)^{2}
$$
From this,
$$
x^{2}<y^{2}<(x+1)^{2}, \quad x<y<x+1
$$
But the last inequality has no solutions in natural numbers $x$ and $y$. Answer: no solutions.
Here we applied the "pinching" method, familiar to us from § 20. | nosolutions | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 45,329 |
958*. Solve the inequality in integers
$$
\frac{1}{\sqrt{x-2 y+z+1}}+\frac{2}{\sqrt{2 x-y+3 z-1}}+\frac{3}{\sqrt{3 y-3 x-4 z+3}}>x^{2}-4 x+3
$$ | $\triangle$ The numbers $x-2y+z+1, 2x-y+3z-1$, and $3y-3x-4z+3$ are not only integers but also positive. Their sum is equal to 3.
$$
(x-2y+z+1) + (2x-y+3z-1) + (3y-3x-4z+3) = 3
$$
But when is the sum of three natural numbers equal to 3? Only in one case - when each of them is equal to 1.
$$
\left\{\begin{array}{l}
x... | (3;1;-1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 45,330 |
960*. Solve the inequality in natural numbers
$$
3^{x}+4^{x}<5^{x}
$$ | $\triangle$ Calculations show that the inequality does not hold for $x=1$ and $x=2$, but it does hold for $x=3,4,5$.
We will prove that the inequality is valid for all natural $x \geqslant 3$. We will use the method of mathematical induction.
1) For $x=3$, the inequality is valid.
2) Assume that it is valid for some ... | allnaturalx,notthan3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 45,331 |
963. Alyosha says to Borya: «Think of a two-digit number. Multiply it by 2, add 5 to the product, and then multiply the resulting sum by 5. What did you get? 715? You thought of 69.» How did he guess? | $\triangle$ Let's denote the number thought of by Borey as $\overline{a b}$. We will perform all the specified actions.
$$
(\overline{a b} \cdot 2+5) \cdot 5=\overline{a b} \cdot 10+25
$$
To get to the number $\overline{a b}$, first, we discard the digit 5 from the obtained number; it will turn into the number $\over... | 69 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,332 |
968. Can 12 consecutive natural numbers be written in the cells of a rectangular table with 3 rows and 4 columns, arranging them so that:
a) the sum of the numbers in each row is divisible by 6;
b) the sum of the numbers in each column is divisible by 4;
c) the sum of the numbers in each column is divisible by 2, bu... | $\triangle$ a) Let's take natural numbers from 1 to 12 and arrange them in a $3 \times 4$ table as follows.
Since the sum of the numbers in each row is divisible by 6, it is possible to arrange them in the table in the required manner.
| 1 | 2 | 3 | 6 |
| :---: | :---: | :---: | :---: |
| 4 | 5 | 7 | 8 |
| 9 | 10 | 1... | )can;b)cannot;)cannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,333 |
972*. Is it possible to fill the cells of a $3 \times 3$ table with the numbers $1, -1$, and $0$ so that all six sums of the numbers in the rows and columns are distinct? | $\triangle$ The sums of the numbers in each row and in each column of the table in this case can only take integer values from -3 to 3. There are 7 integers from -3 to 3. But can all of them be values of such a sum?
Suppose that in the table there is a sum equal to -3, for example, in a row. It can only occur when eac... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 45,334 |
975. Natural numbers from 1 to 9 are placed in the cells of a third-order magic square. What number is in the center of the square? Where can 1 be placed? And 8? Restore at least one such magic square. | $\triangle$ First, let's find the sum of the magic square. It is equal to the sum of all natural numbers from 1 to 9, divided by 3.
$$
S=(1+2+3+\ldots+9): 3=45: 3=15
$$
Now let's find all the ways in which the number 15 can be represented as the sum of three different natural numbers from 1 to 9.
$$
\begin{gathered}... | \begin{pmatrix}\hline8&1&6\\\hline3&5&7\\\hline4&9&2\\\hline\end{pmatrix} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,335 |
979*. Prove that the sum of a magic square of the third order is divisible by 3. | $\triangle$ Let's denote the numbers in the cells of the magic square in sequential order from $a_{1}$ to $a_{9}$, and the sum of the square - through $S$.
We will consider the numbers $a_{1}, a_{2}, a_{4}$, and $S$ as known, and express the other numbers in the square in terms of them. Why? To express the same number... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,336 |
992*. Find all natural $n$ for which the fraction $\frac{n!}{n+1}$ is reducible. | $\triangle$ Consider two cases.
1) Let the number $n-$ be odd, greater than 1.
Then the fraction is reducible by 2.
2) Let $n$ be even.
If in this case the number $n+1-$ is prime: $n+1=p$, where $p-$ is a prime, then the fraction is irreducible.
If, however, $n+1-$ is composite: $n+1=a \cdot b$, where $a$ and $b$ ... | all\odd\n,\greater\than\1,\\all\even\n\neqp-1,\where\p\is\\ | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,337 |
994. A prime number $p$ has only two distinct divisors - 1 and $p$. When does a natural number have: a) 3; b) 4; c) 5 distinct divisors? List all such cases.
To solve such problems in more complex cases, it is useful to derive a formula for the number of divisors of a natural number.
$995^{\circ}$. Prove that the numb... | $\triangle$ Any divisor of a natural number $a$ has the form:
$$
d=p_{1}^{\beta_{1}} \cdot p_{2}^{\beta_{2}} \cdot \ldots \cdot p_{n}^{\beta_{n}}
$$
where $\beta_{1}, \beta_{2}, \ldots, \beta_{n}$ are non-negative integers not exceeding the respective numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$. (In particul... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,338 |
996. Find all cases when a natural number has 8 different divisors. | $\triangle$ Based on the statement of problem 995 (with the same notation), we obtain:
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{n}+1\right)=8
$$
On the other hand, let's find all factorizations of the number 8 into a product of natural numbers:
$$
8=1 \cdot 8=2 \cdot 4=2 \cdot 2 \cdo... | p^{7}, | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,339 |
998. Find a natural number that is divisible by 9 and 5 and has 14 different divisors. | $\triangle$ Let the desired number be denoted by
$$
a=3^{\alpha_{1}} \cdot 5^{\alpha_{2}} \cdot p_{3}^{\alpha_{3}} \ldots p_{n}^{\alpha_{n}}
$$
where $p_{3}, p_{4}, \ldots, p_{n}$ are distinct prime numbers (different from 3 and 5), $\alpha_{1}, \alpha_{2}$ are natural numbers, and $\alpha_{3}, \alpha_{4}, \ldots, \a... | 3645 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,340 |
1000. A certain natural number has only two prime divisors (in some powers), and its square has 35 different divisors. How many different divisors does the cube of this number have? | $\triangle$ Let the original number have the form:
$$
a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}
$$
where $p_{1}$ and $p_{2}$ are distinct prime numbers, and $\alpha_{1}$ and $\alpha_{2}$ are natural numbers. Then
$$
a^{2}=p_{1}^{2 \alpha_{1}} \cdot p_{2}^{2 \alpha_{2}}, \quad a^{3}=p_{1}^{3 \alpha_{1}} \cdot p_{... | 70 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,341 |
1003. A mathematician was asked: "How old are you?" He answered in a complicated way: "The number of my years has an odd number of distinct divisors, and the number of these divisors also has an odd number of distinct divisors." So how old is he? | $\triangle$ Based on the statement of problem 1001, the number of years of the mathematician must be a perfect square. Let's list all such numbers:
$$
4,9,16,25,36,49,64,81,100,121,144,169
$$
(it is probably unrealistic to go further). Let's count the number of divisors of each of these numbers, using the statement o... | 36or100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,342 |
1005. Prove that if natural numbers $a$ and $b$ are coprime, then
$$
\tau(a b)=\tau(a) \tau(b)
$$ | $\triangle$ Let's factorize the numbers $a$ and $b$ into prime factors:
$$
a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{m}^{\alpha_{m}}, \quad b=q_{1}^{\beta_{1}} \cdot q_{2}^{\beta_{2}} \cdot \ldots \cdot q_{n}^{\beta_{n}}
$$
where $p_{1}, p_{2}, \ldots, p_{m}, \quad q_{1}, q_{2}, \ldots, q_{n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,343 |
1007. From a natural number, the sum of its digits was subtracted, and from the resulting difference, the sum of its digits was subtracted again. If this process continues, with what number will the calculations end? | $\triangle$ The difference between a natural number and the sum of its digits is divisible by 9 (see supporting problem 120 from § 5), and this property is preserved at each step of the computation. But such a difference is non-negative and decreases with further continuation of the computations, so its last value is z... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,344 |
1008. Does there exist a natural number $n$ such that
$$
n+S(n)=125
$$
where $S(n)$ is the sum of the digits of $n ?$ | $\triangle$ The number $n$ can only be a three-digit number, and its first digit is equal to 1. Let $n=\overline{1 a b}$. We get:
$$
\overline{1 a b}+1+a+b=125, \quad 100+10 a+b+1+a+b=125, \quad 11 a+2 b=24
$$
The last equation has a unique solution in digits: $a=2, b=1$.
Answer: It exists and is unique - 121. | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,345 |
1013. Find the two smallest consecutive natural numbers, each of which has a sum of digits divisible by 7. | $\triangle$ Obviously, this is possible only with "skipping" over the nine.
Let the smaller of the sought numbers be denoted by
$$
x=\overline{a b c \ldots k l 99 \ldots 9}
$$
$$
x+1=\ove... | 69999,70000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,346 |
1015*. Find a three-digit number such that its product by 6 is a three-digit number with the same sum of digits. List all such solutions. | $\triangle$ Let the desired number be denoted as $\overline{a b c}$, and the new number as $\overline{\mathrm{klm}}$.
Then
$$
\overline{a b c} \cdot 6=\overline{k l m}
$$
where, by the condition,
$$
a+b+c=k+l+m
$$
From this, it is clear that $\overline{a b c} \leqslant 166$, since the product $167 \cdot 6=1002$.
... | 117,135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,347 |
1017*. Prove that the sum of the digits of all numbers
$$
1,2,3, \ldots, 999 \ldots 9
$$
( $n$ nines) is $\frac{1}{2} 9 n \cdot 10^{n}$. | $\triangle$ The sum of the digits of two natural numbers $a$ and $99 \ldots 9-a$ ( $p$ nines), where $a<99 \ldots 9$, is equal to $9 n$. In particular, the sum of the digits of the numbers 1 and $99 \ldots 98$ ( $n-1$ nines) is
$$
1+9+9+\ldots+9+8=9 n
$$
The sum of the digits of the numbers 2 and $99 \ldots 97$ is al... | \frac{1}{2}9n\cdot10^{n} | Number Theory | proof | Yes | Yes | olympiads | false | 45,348 |
1021. Let's write down the digits from 0 to 9 in a row: \(0, 1, 2, 3, \ldots, 9\) and consider the sequence of sums of two adjacent digits:
\[
1 ; 3 ; 5 ; 7 ; 9 ; 11 ; 13 ; 15 ; 17
\]
In this sequence, each term, starting from the second, is 2 more than the previous one. Arrange the digits \(0, 1, 2, \ldots, 9\) so t... | Let's denote the desired sequence of digits as
$$
a_{1} ; a_{2} ; a_{3} ; \ldots ; a_{10}
$$
Then the new sequence of pairwise sums is
$$
a_{1}+a_{2} ; \quad a_{2}+a_{3} ; \quad a_{3}+a_{4} ; \quad a_{4}+a_{5} ; \ldots ; \quad a_{9}+a_{10}
$$
which is an arithmetic progression with a common difference of 1.
The su... | 0;5;1;6;2;7;3;8;4;9or5;0;6;1;7;2;8;3;9;4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,349 |
1022. Given the sequence:
$$
1 ; 1 ; 2 ; 3 ; 7 ; 22 ; \ldots
$$
where the first and second terms are 1, and each subsequent term, starting from the third, is the product of the two preceding terms plus 1. Prove that no term of this sequence is divisible by 4. | $\triangle$ In this sequence, after every two odd members, there comes one even member. Consider any five consecutive members
$$
a ; b ; c ; d ; e,
$$
such that the numbers $b$ and $e$ are even, and the numbers $a, c$ and $d$ are odd. By the definition of the sequence, we get:
$$
e=c d+1=(a b+1)(b c+1)+1=a b^{2} c+b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,350 |
1025. Prove that a member of the Fibonacci sequence is even if and only if its index is divisible by 3. | $\triangle$ Consider the remainders of the division of the terms of the sequence by 2. For the first two terms, they are 1, for the third - 0, for the fourth and fifth - again 1, for the sixth - 0, and so on. The remainders form the following sequence:
$$
1 ; 1 ; 0 ; 1 ; 1 ; 0 ; 1 ; 1 ; 0 ; \ldots .
$$
The divisibili... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,351 |
1026. Find the indices of all Fibonacci sequence members that are divisible by 3. | $\triangle$ Calculations show that the terms of the sequence divisible by 3 are those with indices 4, 8, and 12. It seems that this will always be the case if the index of the term is divisible by 4.
To prove this hypothesis, let's form a sequence of remainders from dividing the terms of the given sequence by 3:
$$
1... | allnaturaldivisible4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,352 |
1031*. Prove that any infinite arithmetic progression composed of natural numbers contains an infinite geometric progression. | $\triangle$ Let $a$ be any term of an arithmetic progression, and $d$ be its common difference. Then the number
$$
a + ad = a(1 + d)
$$
is also a term of this progression, since the $n$-th term of an arithmetic progression is calculated by the formula
$$
a_{n} = a_{1} + d(n-1) \text{.}
$$
It follows that the progre... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,353 |
1038. A piece of the book fell out. The first page of the piece has the number 163, and the number of the last page consists of the same digits but in a different order. How many sheets fell out of the book | $\triangle$ The last page of the piece has the number 316, as it must end with an even digit. Now we find the number of pages, and then the number of sheets in the piece:
$$
316-162=154, \quad 154: 2=77
$$
Answer: 77. | 77 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,354 |
1042. Does there exist a natural number, the product of whose digits equals $1560 ?$ | $\triangle$ Let's factorize the number 1560 into prime factors:
$$
1560=2^{3} \cdot 3 \cdot 5 \cdot 13
$$
The number 13 is not a digit, so such a natural number does not exist.
Answer: does not exist. | doesnotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,355 |
1048. Which is greater and by how much: the sum of all even two-digit numbers or the sum of all odd two-digit numbers? | $\triangle$ First, let's calculate the total number of two-digit numbers. There are 99 numbers in total from 1 to 99, including both single-digit and two-digit numbers. Among them, there are 9 single-digit numbers. Therefore, the number of two-digit numbers is $99-9=90$.
Out of the 90 two-digit numbers, 45 are even an... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,356 |
1053. A vendor was selling eggs. To the first customer, she sold half of all the eggs she had and another half-egg, to the second - half of the remainder and another half-egg, to the third - half of the new remainder and another half-egg, to the fourth - half of the last remainder and another half-egg. After this, she ... | $\triangle$ Let's solve the problem from the beginning.
After the third customer, the seller had one egg left.
After the second $-2 \cdot 1+1=3$. Let's check: she sold half of the previous remainder plus half an egg to the third customer, i.e., $3 / 2+1 / 2=2$, and she was left with $3-2=1$ egg. After the first custo... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,357 |
1058. One four-digit number is composed of consecutive digits in ascending order, the second is from the same digits but in descending order, and the third four-digit number is written with the same digits. Find the first number if the sum of all three numbers is 14364. | $\triangle$ Let's consider several cases.
1) Suppose the original number is 1234. Then we need to restore the record:
$$
\begin{aligned}
& 1234 \\
& +4321 \\
& \overline{a b c d} \\
& 14364
\end{aligned}
$$
where $a, b, c, d$ are the same digits $1,2,3,4$, but arranged in a different order.
Obviously, there are no ... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,358 | |
1060. Among natural numbers from 1 to 10000, which are there more of: those divisible by 7 but not by 11, or those divisible by 11 but not by 7? | $\triangle$ First, let's calculate the number of natural numbers from 1 to 10000 that are divisible by 7, as well as the number of natural numbers within the same range that are divisible by 11. The first set forms the sequence
$$
7 ; \quad 14=7 \cdot 2 ; \quad 21=7 \cdot 3 ; \quad \ldots ; \quad 7 \cdot 1428
$$
(142... | there\\\\that\\divisible\\7\but\not\\11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,359 |
1061. There are candies in three boxes. In the first box, there are 8 candies less than in the other two together, and in the second box, there are 12 less than in the first and third together. How many candies are in the third box? | $\triangle$ Let the number of candies in the first, second, and third boxes be denoted by $a, b$, and $c$ respectively. We obtain the system of equations:
$$
\left\{\begin{array}{l}
a=b+c-8 \\
b=a+c-12
\end{array}\right.
$$
Adding these equations term by term:
$$
a+b=(a+b)+2 c-20
$$
Then
$$
2 c-20=0, \quad c=10
$$... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,360 |
1064. Several natural numbers, more than three, are written in a circle, the sum of which is 19. The sums of any three consecutive numbers are equal to each other. What numbers were written in the circle? | $\triangle$ Let's denote these numbers in sequential order as $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$, where $n-$ is the total number of numbers. Then
$$
a_{1}+a_{2}+a_{3}=a_{2}+a_{3}+a_{4}
$$
From this, $a_{1}=a_{4}$.
Similarly, from the equalities
$$
a_{2}+a_{3}+a_{4}=a_{3}+a_{4}+a_{5}, \quad a_{3}+a_{4}+a_{5}=a_{4}... | 19,eachofwhichisequalto1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,361 |
1067. Kolya had to square a certain natural number for his homework. Instead, he mistakenly doubled the number and got a two-digit number, written with the same digits as the square of the number, but in reverse order. What should the correct answer be? | $\triangle$ Let's denote the original number by $x$, its square - by $\overline{a b}$. Then
$$
\left\{\begin{array}{l}
x^{2}=10 a+b \\
2 x=10 b+a
\end{array}\right.
$$
From here, it is clear that the number $x$ is a single-digit number.
Let's add the equations of this system:
$$
x(x+2)=11(a+b)
$$
Then either $x$, ... | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,362 |
1071. Let $p$ be a prime number. How many natural numbers exist:
a) less than $p$ and coprime with $p$;
b) less than $p^{2}$ and coprime with $p^{2}$? | $\triangle$ a) The number of natural numbers less than $p$ and coprime with $p$ is $p-1$ - these are all natural numbers less than $p$.
b) First, let's count the number of natural numbers not exceeding $p^{2}$ and not coprime with $p^{2}$. These numbers are
$$
p, 2 p, 3 p, \ldots,(p-1) p, p^{2}
$$
Obviously, there a... | )p-1;b)p(p-1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,363 |
1074*. Find all such rational numbers $x$ and $y$ for which the numbers $x+\frac{1}{y}$ and $y+\frac{1}{x}$ are natural. | $\triangle$ Since the numbers
$$
x+\frac{1}{y}=\frac{x y+1}{y} \text { and } y+\frac{1}{x}=\frac{x y+1}{x}
$$
are positive, the number
$$
\frac{x y+1}{y} \cdot \frac{x y+1}{x}=\frac{(x y+1)^{2}}{x y}
$$
is also positive, which implies that $x y$ is positive. But then the number $x y+1$ is positive, and consequently... | 1,1;2,\frac{1}{2};\frac{1}{2},2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,364 |
1077*. Find all natural numbers $k$ and $n$ for which the number
$$
\frac{7 k+15 n-1}{3 k+4 n}
$$
is a natural number. | $\triangle$ For natural $k$ and $n$, the inequality holds:
$$
2<\frac{7 k+15 n-1}{3 k+4 n}<4
$$
(verify!). Therefore, this fraction can only be equal to $3$:
$$
\frac{7 k+15 n-1}{3 k+4 n}=3, \quad 7 k+15 n-1=9 k+12 n, \quad 3 n-2 k=1
$$
To solve the last equation, express $k$ from it:
$$
k=\frac{3 n-1}{2}
$$
From... | k=3a-2,n=2a-1(\inN) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,365 |
1082. How many terms of the sum
$$
1+2+3+\ldots+n
$$
are needed to get a three-digit number that is written with identical digits? | $\triangle$ Let's denote this three-digit number as $\overline{a a a}$. Applying the formula for the sum of $n$ terms of an arithmetic progression, we get:
$$
\frac{n(n+1)}{2}=\overline{a a a}=a \cdot 111, \quad n(n+1)=2 \cdot a \cdot 111=2 \cdot 3 \cdot 37 \cdot a
$$
From this, either $n$ or $n+1$ is divisible by 37... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,366 |
1085*. Several natural numbers form an arithmetic progression, starting with an even number. The sum of the odd terms of the progression is 33, and the sum of the even terms is 44. Find the progression and the number of its terms. List all solutions. | Let's denote the first term of the arithmetic progression by $2a$ (where $a \in \mathbb{N}$), the common difference by $d$, and the number of terms by $n$. We apply the formula for the sum of the first $n$ terms of an arithmetic progression:
$$
S_{n}=\frac{2a_{1}+d(n-1)}{2} n
$$
For the sum of the first $n$ terms of ... | 2;5;8;11;14;17;20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,367 |
1089. Given an infinite arithmetic progression $3 ; 16 ; 29 ; 42 ; \ldots$. Find:
a) the smallest term of the sequence that can be written using only sevens;
b) all such terms. | $\triangle$ a) To find the smallest term of the sequence with this property, one needs to divide the number, written with sevens, by the difference of the progression 13 until the first time the remainder is 3. Such a number is 7777:
$$
7777=13 \cdot 598+3
$$
b) Let's find the next term of the sequence with the same ... | 777\ldots7,wheretheofsevensis6k+4(k=0,1,2,\ldots) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,368 |
1092. Solve the equation
$$
a x^{2}+b x+b=0
$$
for all possible values of the parameters $a$ and $b$, given that all its roots are integers. | $\triangle$ Let's consider several cases.
1) Let $a \neq 0$ and $b \neq 0$.
Denote the roots of the quadratic equation by $x_{1}$ and $x_{2}$. By Vieta's formulas, we get:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a} \\
x_{1} x_{2}=\frac{b}{a}
\end{array}\right.
$$
Add the equations of this system:
$$
x_{1}... | if\\neq0\\b=4a,\then\x_{1}=x_{2}=-2;\if\\neq0\\b\neq4a\or\=b=0,\then\there\\no\solutions;\if\=0,\b\neq0,\then\-1;\if\\neq | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,369 |
1096. Let $2a, a+b$ and $c$ be integers. Prove that the quadratic trinomial
$$
f(x)=a x^{2}+b x+c
$$
takes an integer value for any integer $x$. | $\triangle$ It is sufficient to prove that the expression $a x^{2}+b x$ takes only integer values. Let's transform it as follows:
$$
a x^{2}+b x=\left(a x^{2}-a x\right)+(a x+b x)=2 a \frac{x(x-1)}{2}+(a+b) x
$$
Since $2 a, \frac{x(x-1)}{2}, a+b$ and $x$ take integer values, then $a x^{2}+b x$ also takes only integer... | proof | Algebra | proof | Yes | Yes | olympiads | false | 45,370 |
1101*. Find all natural $n$ for which in the expression
$$
* 1 * 2 * 3 * \ldots * n=n+1
$$
it is possible to replace the asterisks with plus and minus signs so that the resulting equation is true. | $\triangle$ First, let's replace all the stars with plus signs. We get:
$$
1+2+3+\ldots+n=\frac{n(n+1)}{2}
$$
When replacing some of the plus signs with minus signs, the parity of the sum does not change. Therefore, the number $\frac{n(n+1)}{2}$ must have the same parity as the number $n+1$. But then the difference b... | n=4k+2orn=4k+3(k=0,1,2,\ldots) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,371 |
1108*. Two natural numbers are guessed. Mathematician A was told their sum, and Mathematician B was told the sum of their squares. If they combined their information, they would, of course, find both numbers. However, they discovered these numbers as a result of the following conversation:
B: - I don't know what these... | $\triangle$ Let the sought natural numbers be denoted by $x$ and $y$.
Since Mathematician V initially did not know what these numbers were, there exists another pair of natural numbers $a$ and $b$ such that
$$
x^{2}+y^{2}=a^{2}+b^{2}
$$
These pairs differ in that
$$
x+y \geqslant 10, \quad a+b<10
$$
Then $(x+y)^{2... | 74 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,372 |
1109*. A natural number $n$ is such that the number $n^{2}+1$ is a ten-digit number. Prove that the decimal representation of this ten-digit number contains at least two identical digits.
$1110^{*}$. Prove that the number
$$
1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}
$$
is never an integer for any natural number $... | $\triangle$ Let's take the term of the sum with the highest power of two in the denominator. Such a term is unique, since for any given natural number $n>1$ there exists a unique natural number $k$ that satisfies the inequality
$$
2^{k} \leqslant n<2^{k+1}
$$
After bringing the sum to a common denominator, the denomi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,373 |
1113*. The number
$$
1010101 \ldots 0101
$$
is divisible by 9999. How many ones can be in the notation of this number? List all cases. | $\triangle$ Let's factorize the number 9999 into prime factors:
$$
9999=3^{2} \cdot 11 \cdot 101
$$
According to the divisibility rules for 9 and 11, the number of ones in the original number's representation must be divisible by 9 and 11.
A number is divisible by 101 if and only if, disregarding the first three dig... | 198k(k\inN) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,374 |
1114*. Find all prime numbers that can be divisors of numbers of the form $111 \ldots 1$. | $\triangle$ Numbers of the form $11 \ldots 1$ are not divisible by 2 or 5. Let $p$ be any prime number different from 2 and 5. We will write a sequence of $p+1$ numbers
$$
1,11,111, \ldots, 111 \ldots 1
$$
(the last number is written with $p+1$ ones).
By the pigeonhole principle, among these numbers, there will be t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,375 |
## PROBLEM 1. CHESS TOURNAMENT
In New Pork, there are more chess masters than in the rest of the United States. A chess tournament is planned to include all masters. It has been decided that the tournament will be held in a place where the total travel distance between the cities of the participants is minimized. The ... | Solution. New Yorkers are right! Let's denote New York masters by the letters $N_{1}, N_{2}, \ldots, N_{k}$, and the other players by the letters $O_{1}, O_{2}, \ldots, O_{t}$. Since more than half of the players are in New York, then $k>t$. If we consider pairs of players $\left(N_{1}, O_{1}\right),\left(N_{2}, O_{2}\... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,376 |
## PROBLEM 2. ORDERED PARTITIONS OF A NUMBER $n$
The number 3 can be expressed as the sum of one or more natural numbers in four ways, if the order of the terms is taken into account:
$$
3,1+2,2+1,1+1+1
$$
In how many ways can the number $n$ be expressed as a sum? | Solution. Consider a row of $n$ ones placed in a line. Any arrangement of $n-1$ or fewer dividing dashes in the gaps 8
between the ones corresponds to a partition of the number into summands and vice versa:
$$
\begin{gathered}
11|1| 11|1| 11 \ldots 111 \\
n=2+3+1+n-6 .
\end{gathered}
$$
Since we can either insert a d... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 45,377 |
## PROBLEM 3. REGIONS IN A CIRCLE
Place $n$ points on a circle and connect them pairwise with chords. Assume that no three chords have a common point inside the circle. Into how many regions will the circle be divided by these chords? | Solution. Suppose we draw the chords one by one sequentially. A new chord divides different regions, increasing the number of parts (Fig. 2). The additional number of regions is equal to the number of segments into which the new chord is divided by the chords it intersects. Therefore, this number is one more than the n... | C_{n}^{2}+C_{n}^{4}+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 45,378 |
## PROBLEM 4. FERRIES
Two ferries shuttle across a river at constant speeds, turning at the banks without losing time. They depart simultaneously from opposite banks and meet for the first time 700 feet from one of the banks, continue to the banks, return, and meet for the second time 400 feet from the opposite bank. ... | Solution. The total distance covered by the ferries at the time of their first meeting is exactly equal to the width of the river (Fig. 5). However, it may be surprising that by the time they meet again, the total distance they have covered is three times the width of the river. Since the speeds are constant, the secon... | 1700 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,379 |
## PROBLEM 5. PROTRUDING SEMICIRCLE
Chord $A B$ of a circle with radius 1 and center at $O$ is the diameter of semicircle $A C B$, located outside the first circle (Fig. 6). It is clear that
^2 = \left(\sqrt{a} + \sqrt{1-a^2}\right) = a + 2\sqrt{a(1-a)} + 1 - a$. We get $OC^2 = 1 + 2\sqrt{a(1-a)}$. 12
This expression is maximized when the value of $a(1-a)$ is maximized. But since
... | AB=\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,380 |
## PROBLEM 7. SCREENS IN THE CORNER
A corner of a rectangular room is partitioned off by two four-foot screens. At what position will the enclosed area be the largest? | Solution. The solution is based on the double application of the following well-known result.
Lemma. Among the set of triangles with a given base $b$ and the angle $\theta$ opposite to it, the isosceles triangle has the largest area. (It is clear that this set is contained within a segment of a circle, and the height ... | \angleOAP=67.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,381 |
## PROBLEM 8. COLORING THE PLANE
Suppose that each point on the plane is colored either red or blue. Show that there exists a rectangle, all of whose vertices are colored the same color. | Solution. Out of seven points, at least four must be of the same color. Let's select four points $P_{1}, P_{2}, P_{3}, P_{4}$ from the seven points on a line, all colored red, for example. Consider two more lines parallel to the initial one and two sets of four points $\left(Q_{1}, Q_{2}, Q_{3}, Q_{4}\right)$ and $\lef... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 45,382 |
## PROBLEM 10. $\cos 17 x=f(\cos x)$
Let $f$ be the function of $\cos x$ that expresses $\cos 17 x$, i.e.
$$
\cos 17 x=f(\cos x)
$$
Show that the same function $f$ expresses $\sin 17 x$ in terms of $\sin x: \sin 17 x=f(\sin x)$. | Solution. Let $x=\frac{\pi}{2}-y$. Then $\sin x=\cos y$ and
$$
\begin{aligned}
& \sin 17 x=\sin {\left[17\left(\frac{\pi}{2}-y\right)\right]=\sin \left(8 \pi+\frac{\pi}{2}-17 y\right)=} \\
&=\sin \left(\frac{\pi}{2}-17 y\right)=\cos 17 y=f(\cos y)=f(\sin x)
\end{aligned}
$$
Note that the number 17 can be replaced by ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 45,383 |
## PROBLEM 11. SQUARE ON A GRID
An arbitrary square $S$ of size $n \times n$ on the coordinate plane covers $(n+1)^{2}$ integer points (points where both coordinates $x$ and $y$ are integers) if one of its corners is at an integer point and its sides are parallel to the coordinate axes. Intuitively prove the obvious r... | Solution. Consider $S$ in an arbitrary position on the plane. Suppose now that the boundary of the square $S$ is made of a circular rubber band, and a nail is hammered into the plane at each integer point. Let the rubber band shrink, surrounding the nails located at the lattice points that fall within the square $S$. T... | (n+1)^{2} | Geometry | proof | Yes | Yes | olympiads | false | 45,384 |
## PROBLEM 12. OPAQUE SQUARE
A set of straight line segments inside or on the boundary of a square with side length 1 is called "opaque" if any straight line that intersects the square also intersects at least one of these segments. For example, the two diagonals form an opaque set
 has a length of
$$
2+\frac{\sqrt{2}}{2}=2+\frac{1.42}{2}=2.71<1+\sqrt{3}
$$
Sides $AB$ and $BC$ have a common point with any line intersecting triangle $ABC$. If the entire triangle $ABC$ is consider... | \frac{\sqrt{2}}{2}(2+\sqrt{3})\approx2.64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,385 |
## PROBLEM 13. TIC-TAC-TOE
Consider the game of "tic-tac-toe" on a three-dimensional cube $8 \times 8 \times 8$. How many straight lines can be indicated on which 8 symbols lie in a row? | Solution. This is not a difficult problem; it can be solved by simple counting. However, there is a brilliant solution proposed by Leo Moser. It involves considering a $10 \times 10 \times 10$ cube, obtained by covering the given $8 \times 8 \times 8$ cube with a layer of unit thickness. A winning line in the inner $8 ... | 244 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 45,386 |
## PROBLEM 14. AN AMAZING PROPERTY OF RIGHT TRIANGLES
Prove that if the legs are rotated around their vertices so that they lie on the hypotenuse, the length of their common part is equal to the diameter of the inscribed circle (Fig. 18).
 - (length of the hypotenuse). This can be easily shown as follows. Proje... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,387 |
## PROBLEM 17. 0 - $k$-CLOUDS
Circles of unit radius have no common interior points and lie within a strip $S$, formed by two parallel lines, spaced apart by a distance $w$.
Fig. 21
We wi... | Solution. Draw a line $m$ through point $O$ - the center of some circle $C$ from the 2-cloud, perpendicular to the lines bounding the strip $S$ (Fig. 22). Then the line $m$ must intersect another circle $A$. Let $Q$ -
$ from the set of numbers ( $1,2,3, \ldots, n$ ), allowing the numbers $a_{i}$ to repeat, then there will be $n^{k}$ such sets. For each such set, we mark the smallest of the $a_{i}$. Prove the unexpected result that... | Solution. This is a beautiful solution based on a very simple fact that is so obvious it usually cannot help in solving other problems. When adding a sequence of natural numbers $r_{i}$, one can distribute their members $\geqslant 1$, members $\geqslant 2$, and so on. Then $\sum r_{i}=$ (number of members $\left.\geqsl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 45,390 |
## PROBLEM 20. ROLLING DICE
A regular die has the numbers $1,2,3,4,5,6$ on its faces. It is rolled randomly until the sum of the points that come up exceeds the number 12.
What is the most likely total sum of points? | Solution. Consider the penultimate roll. After it, the total sum should be either $12,11,10,9,8$,
or 7. If it is 12, the overall result will with equal probability take the values $13,14,15$, $16,17,18$. Similarly, with a sum of 11, the final result will with equal probability take the values 13, 14, $15,16,17$, and so... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 45,391 |
## PROBLEM 21. PIERCING THE CUBE
A solid cube $C$ of size $20 \times 20 \times 20$ is constructed from 2000 bricks of size $2 \times 2 \times 1$. Prove that this cube can be pierced by a straight line perpendicular to a face of the cube, passing through the cube and not piercing any of the bricks. | Solution. Consider 8000 unit cubes in $C$. On each face of the cube $C$, their edges form a $20 \times 20$ grid. Inside the face, there are $19 \cdot 19=361$ nodes of this grid.
Lines drawn through these 361 nodes perpendicular to the face of the cube pass through the edges of the unit cubes to the corresponding nodes... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,392 |
## PROBLEM 22. DOUBLE SEQUENCE
For some natural numbers $n$, a sequence can be constructed in which each of the numbers $1,2,3, \ldots, n$ appears twice, and the second appearance of each number $r$ occurs $r$ places after its first appearance. For example, for $n=4$
$$
4,2,3,2,4,3,1,1
$$
For $n=5$ we have
$$
3,5,2... | Solution. Let's number the positions in the sequence by $1,2,3, \ldots, 2 n$. Suppose the first occurrence of the number 1 happens at the position numbered $p_{1}$, the first occurrence of the number 2 - at the position numbered $p_{2}$, and so on. Then the second occurrence of the number 1 will happen at the position ... | 2^{n-1} | Combinatorics | proof | Yes | Yes | olympiads | false | 45,393 |
PROBLEM 26. $a^{b}$ and $b^{a}$
It is often not difficult to determine which is greater: $a^{b}$ or $b^{a}$. It is obvious that
$$
2^{3} < 4^{3}
$$
However, when considering two numbers, one of which is between 2 and 3, and the other between 3 and 4, this question can be challenging. Which is greater:
$$
e^{\pi} \t... | Solution. For positive $x$ we have
$$
e^{x}=1+x+\frac{x^{2}}{2}+\ldots>1+x
$$
Since $\pi>e, \pi / e>1$ and $x=\pi / e-1>0$. Thus,
$$
\begin{gathered}
e^{\pi / e-1}>1+\left(\frac{\pi}{e}-1\right), \quad \frac{e^{\pi / e}}{e}>\frac{\pi}{e} \\
e^{\pi / e}>\pi, \quad e^{\pi}>\pi^{e} .
\end{gathered}
$$
As an easy exerc... | e^{\pi}>\pi^{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,395 |
## PROBLEM 27. MATHEMATICAL JOKE
A person bought several one-cent stamps at the post office, as many two-cent stamps so that $3 / 4$ of their quantity equals the number of one-cent stamps, as many five-cent stamps so that $3 / 4$ of their quantity equals the number of two-cent stamps, and five eight-cent stamps. He pa... | Solution. Suppose a person bought $y$ one-cent stamps. Then he bought $3 y / 4$ two-cent stamps and $9 y / 16$ five-cent stamps. Since $9 y / 16$ is an integer, $y$ is divisible by 16. Then for some integer $16 x=y$ and the purchase consisted of 16 one-cent, $12 x$ two-cent, $9 x$ five-cent, and 5 eight-cent stamps. Su... | 18816 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 45,396 |
## PROBLEM 28. MAPS ON A SPHERE
Consider a geographical map $M$ on a sphere. Suppose that at each vertex of the map $M$, exactly three countries meet, and that no country border is a loop (i.e., each border contains at least two vertices). The map is colored with colors $A, B, C, D$ such that adjacent countries are co... | Solution. This question seems very difficult because, even knowing that the country is colored with color $A$, we cannot attribute it to the considered set until we know whether it is even or odd.
\end{aligned}
$$ | Solution. Since $3 a b c$ is a positive number, then $a^{3}$ must be greater than the sum of $b^{3}$ and $c^{3}$; thus, 52
we have
$$
b<a \quad \text { and } \quad c<a
$$
Adding these inequalities, we get that $b+c<2 a$, therefore, $2(b+c)<4 a$. From the second equation, we get that
$$
a^{2}<4 a \quad \text { and } ... | 2,1,1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,399 |
## PROBLEM 31. REFLECTED TANGENTS
Let $A$ and $B$ be two circles lying on the same side of a line $m$. Construct a tangent to circle $A$ that, after being reflected off line $m$, also touches circle $B$ (see Fig. 37 on p. 54). | Solution. Symmetrically reflect circle $B$ with respect to line $m$ to position $B^{\prime}$. Then the common tangents to circles $A$ and $B^{\prime}$ provide four possible solutions (the tangents to circle $B^{\prime}$, when reflected from line $m$, become tangents to circle $B$) (Fig. 38). | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,400 |
## PROBLEM 33. SNOWBALLS
A boy made two snowballs, one of which has a diameter twice as large as the other. He brought them into a warm room and let them melt. Since only the surface of the snowball is exposed to the warm air, assume that the melting rate is proportional to the surface area. When half of the volume of... | Solution. We will prove that the assumption that the melting rate of a snowball is proportional to the area of its surface leads to a surprising result, namely that the radius decreases at a constant rate regardless of the size of the snowball. Therefore, both radii will decrease by the same amount.
The volume and sur... | 40500000001 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,402 |
## PROBLEM 36. A DIOPHANTINE EQUATION
Let $a, b, c ; d$ be integers and $a \neq 0$. Prove that if $b c - a d \neq 0$, then the equation
$$
a x y + b x + c y + d = 0
$$
has only a finite number of solutions in integers. | Solution. Multiplying the equation by $a$, we get
$$
a^{2} x y+a b x+a c y+a d=0
$$
and
$$
(a x+c)(a y+b)=b c-a d
$$
This means that each of the factors $a x+c \mathbf{x}$ $a y+b$ must be a divisor of the number $b c-a d$. Thus, if
5. Ionetarsos.
$b c-a d \neq 0$, then it has only a finite number of divisors, and ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,404 |
## PROBLEM 39. DIVIDED INTEGER POINTS
A segment with endpoints at points $A(p, 0)$ and $B(0, p)$ passes through $p-1$ integer points: $(1, p-1),(2, p-2), \ldots,(p-1,1)$. If segments are drawn through these points, connecting them to the origin - point $O$, there will be $p-1$ such segments, and they will divide $\tri... | Solution. Let $C(a, b)$ be an integer point inside $\triangle O A B$. The tangent of the angle of inclination of the line $O C$ is $b / a$. If the point $C$ lay on a separating segment, say, on the segment leading to the point ( $i, p-i$ ) $(1 \leqslant i \leqslant p-1)$, then the tangent of the angle of inclination of... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,406 |
## PROBLEM 40. PERFECT NUMBERS
The ancient Greeks discovered that some natural numbers $n$ have a remarkable property: the sum of the divisors of the number $n$ equals the number $n$ itself (the number itself is not considered a divisor). For example, $n=28$ gives
$$
1+2+4+7+14=28
$$
Such numbers were called "perfec... | Solution. First, let us assume that $n$ is an odd perfect number. Then $\sigma(n)=2n$, where 2 and $n$ are coprime, and we get
$$
\sigma[\sigma(n)]=\sigma(2n)=\sigma(2)\sigma(n)=3\sigma(n)=3 \cdot 2n=6n
$$
Clearly, $6n$ is even, and if it is also perfect, then for some prime number $p$ we must have
$$
6n=2^{p-1}(2^{... | 6 | Number Theory | proof | Yes | Yes | olympiads | false | 45,407 |
## PROBLEM 41. SIDES OF A QUADRILATERAL
Prove that if the lengths of all sides of a quadrilateral are integers (relative to some unit of length) and if the length of each side is a divisor of the sum of the other three sides, then some two of the sides are equal. | Solution. Suppose no two sides are equal. Then let the side lengths be $s_{1}>s_{2}>s_{3}>s_{4}$, and let $p$ denote the perimeter. We get that $p-s_{6}$ is divisible by $s_{i}$ for $i=1,2,3,4$. In this case, we also have that $p$ is divisible by $s_{i}$. But under what conditions is $p$ divisible by $s_{i}$?
Consider... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,408 |
## PROBLEM 42. PRIME NUMBERS IN AN ARITHMETIC PROGRESSION
Show that in any arithmetic progression of natural numbers with a difference less than 2000, there cannot be 12 consecutive terms that are prime numbers.
76 | Solution. Suppose that all $n$ consecutive terms of an arithmetic progression
$$
a, a+d, a+2 d, \ldots, a+(n-1) d
$$
are prime numbers, $n \geqslant 3$. If $a<n$, then one of these terms will be $a+a d=a(1+d)$, which is not a prime number since both $a$ and $1+d$ are greater than 1. Therefore, $a \geqslant n \geqslan... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 45,409 |
## PROBLEM 43. ON CEVIANS
Suppose that $B C$ is the largest side of triangle $A B C$. Let point $O$ be chosen somewhere inside the triangle and lines $A O, B O, C O$ intersect the opposite sides at points $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Prove that
$$
O A^{\prime}+O B^{\prime}+O C^{\prime}<B C
$$ | Solution. A segment passing through a vertex of a triangle to the opposite side is called a "cevian." It is clear that a cevian has a smaller length than the larger of the two sides emanating from the vertex from which it is drawn. Accordingly, the largest side of the triangle exceeds all its cevians. Therefore, $BC$ i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,410 |
## PROBLEM 44. COWS AND SHEEP
Two men jointly owned $x$ cows, which they sold for $x$ dollars a head. With the money they received, they bought sheep at 12 dollars a head. Since the proceeds from the sale of the cows did not divide by 12, they used the remaining money to buy a lamb. Then they divided the flock so that... | Solution. The revenue from sales equals $x^{2}$ dollars. If $x$ were divisible by 6, then $x^{2}$ would be divisible by 36, and consequently, by 12. Since this is not the case, $x$ is not divisible by 6. In this case, $x=12 k+r$, where $|r|=1,2,3,4$ or 5. Accordingly,
$$
\frac{x^{2}}{12}=\frac{(12 k+r)^{2}}{12}=\frac{... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 45,411 |
## PROBLEM 45. SEQUENCE OF SQUARES
Show that each term of the sequence
$$
49,4499,444889,44448889, \ldots, \underbrace{44 \ldots 4}_{n} \underbrace{88 \ldots 89}_{n}
$$
is a perfect square. | Solution. The general term is
$$
\begin{aligned}
T= & \underbrace{44 \ldots 48}_{n}, \underbrace{8 \ldots 89}_{n}=9+8 \cdot 10+8 \cdot 10^{2}+\ldots \\
& \ldots+8 \cdot 10^{n}+4 \cdot 10^{n+1}+4 \cdot 10^{n+2}+\ldots+4 \cdot 10^{2 n+1}
\end{aligned}
$$
Representing 9 as $1+4+4$, and each 8 as $4+4$, we have
$$
\begi... | (\frac{2\cdot10^{n+1}+1}{3})^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 45,412 |
## PROBLEM 46. INSCRIBED DECAGON
The problem of inscribing a regular decagon in a circle has aroused keen interest among mathematicians since ancient Greece. Euclid wrote about it and, in particular, gave a very beautiful method for inscribing a decagon. First, we will show that the side $x$ of a regular decagon inscr... | Solution. It is not difficult to see that the radius $x$ of the circle inscribed in an equilateral triangle is one third of the height (which is also the median and the bisector). Referring to the figure, we see that
$$
O B=\frac{2}{3} \text { of the height }=2 x
$$
 | Geometry | proof | Yes | Yes | olympiads | false | 45,413 |
## PROBLEM 47. RED AND BLUE POINTS
Consider points arranged in a square of 20 columns and 20 rows, each of which is colored either red or blue. Whenever two points of the same color are adjacent in some row or column, they are connected by a segment of the same color. Adjacent points of different colors are connected ... | Solution. Each row contains 19 segments, so we get $19 \cdot 20=380$ horizontal segments. There are as many vertical segments, so their total number is 760. Since 237 of them are black, the remaining 523 are blue or red.
Let $k$ be the number of red segments and let's count how many times red points are endpoints of s... | 223 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 45,414 |
## PROBLEM 49. ON THE FUNCTION $\pi(n)$
Prime numbers have long been of particular interest to mathematicians. One of the most interesting functions related to prime numbers is the function $\pi(n)$ - the number of prime numbers not exceeding the number $n$.
Problem. Prove that $\pi(n) \geqslant \frac{\ln n}{\ln 4}$. | Solution. The outstanding Hungarian mathematician Paul Erdős found the following elementary solution to this problem.
Let $m$ be some natural number. Suppose that $k^{2}$ is the largest square that is a divisor of the number $m$, i.e., $m=k^{2} v$. Then $v$ cannot have repeated factors, otherwise a full square larger ... | \pi(n)\geqslant\frac{\lnn}{\ln4} | Number Theory | proof | Yes | Yes | olympiads | false | 45,416 |
## PROBLEM 50. CONSTANT CHORD
Suppose that two circles $Q$ and $R$ intersect at points $A$ and $B$ (Fig. 61). Point $P$ lies on the arc of circle $Q$ that is outside circle $R$. It is projected through points $A$ and $B$, defining chord $C D$ on circle $R$. Prove that regardless of where point $P$ is taken on the spec... | Solution. Let $P$ and $P^{\prime}$ denote the positions of point $P$ corresponding to the chords $C D$ and $C^{\prime} D^{\prime}$ (Fig. 62). We have
$$
\begin{aligned}
& \angle P A P^{\prime}=\angle P B P^{\prime} \\
& \angle P A P^{\prime}=\angle C A C^{\prime} \\
& \angle P B P^{\prime}=\angle D B D^{\prime}
\end{a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,417 |
## PROBLEM 51. NUMBER OF INTERNAL DIAGONALS
A simple polygon is one that does not intersect itself. However, such a polygon can be far from convex, and many of its diagonals may lie entirely or partially outside it. Prove that nevertheless, every simple $n$-gon has at least $n-3$ diagonals that lie entirely inside it. | Solution. It is clear that this statement is true for quadrilaterals (Fig. 63). Suppose the statement is true for $k$-gons $(k=4,5, \ldots, n)$. Consider a simple $(n+1)$-gon $P$. Note that it is impossible for a polygon to have no diagonals located inside it. This result is proven in my
![](https://cdn.mathpix.com/cro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 45,418 |
4. List the properties of a line that you know. Indicate no fewer than five properties. | 4. A straight line is not closed, is infinite, is a line of constant curvature, divides the plane in which it lies into two parts, and is determined by any two of its points. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,422 |
6. What is described in the following statement: a property of a line or a property of a plane?
"If two points of a line belong to a plane, then every point of this line belongs to the same plane."
And in this statement?
"Any line lying in a plane divides this plane into two convex regions." | 6. Both properties are equally properties of a line and a plane. More precisely: they do not describe properties of lines and planes themselves, but the r elationships between these concepts.
In the language of ordinary grammar: if a sentence has a subject(s) and an object(s), then the predicate expresses not a proper... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,426 |
7. Indicate the properties inherent to all triangles (essential or necessary properties); only some triangles (separable properties); properties that do not belong to any triangle (contradictory properties). | 7. In any triangle, the sum of the angles is $180^{\circ}$. Only some triangles have equal sides. A triangle cannot have two right angles. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,427 |
11. List the properties common to all rectangles; properties common to only some of them. | 11. Equality of diagonals, presence of two axes of symmetry - these are properties of all rectangles. Mutual perpendicularity of diagonals, presence of four axes of symmetry, equality of all sides - these are properties of only some rectangles. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,431 |
12. For a parallelepiped, all faces are parallelograms. Is this property unique to parallelepipeds? | 12. No, it is not. There are polyhedra other than parallelepipeds that have all faces as parallelograms. The polyhedron shown in Figure 14 is one of such polyhedra. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,432 |
19. Indicate the properties common to a rectangle and a rhombus. Compare these properties with those of a parallelogram. | 19. Example: equality of all sides and equality of diagonals. An unequal-angled rhombus has the first property but not the second; an unequal-sided rectangle has the second property but not the first; a square has both properties. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,439 |
27. For all polygons, are the equality of sides and the equality of angles independent properties? | 27. For a pentagon, all the listed properties are pairwise independent, but from each pair of them, the third one already follows. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,447 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.