problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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Fomin D:
How many pairs of natural numbers ( $m, n$ ), each not exceeding 1000, exist such that
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} ?
$$ | Consider all pairs $(m, n)$ of natural numbers for which $m+n+1=s \geq 3$. We have $\frac{1}{s}2>\sqrt{2}>1 / 2>1 / s$, and since the number $\sqrt{2}$ is irrational, it falls into exactly one of the specified intervals. Thus, among the pairs $(m, n)$ with a fixed value of $s$, there will be exactly one for which the i... | 1706 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
It seems there is no text in the provided image, only a mathematical expression. Here is the expression in LaTeX format:
```
\int_{0}^{1} \frac{x^4(1-x)^4}{1+x^2} \, dx = \frac{22}{7} - \pi
```
If you need the expression translated into a textual description, please let me know! | We can assume that the difference $d$ of the progression is positive and not a multiple of 10. Let $d$ be a $k$-digit number.
a) The numbers from A890...0 to A9...9 ($k+2$ nines) all contain nines. There are more than $d$ such numbers. The distance between two consecutive groups of such numbers is $89 \cdot 10^{k} \le... | 72 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
Find the maximum number $N$ for which there exist $N$ consecutive natural numbers such that the sum of the digits of the first number is divisible by 1, the sum of the digits of the second number is divisible by 2, the sum of the digits of the third number is divisible by 3, ..., and the sum of the digits... | Evaluation. Let's denote the sum of the digits of a number $a$ by $S(a)$. Suppose there exist 22 consecutive numbers $a_{1}, \ldots, a_{22}$, satisfying the conditions of the problem. Then the numbers $S\left(a_{2}\right), S\left(a_{12}\right)$, and $S\left(a_{22}\right)$ are even. On the other hand, if the second-to-l... | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest natural number $A$, satisfying the following conditions:
a) its notation ends with the digit 6;
b) when the digit 6 is moved from the end of the number to its beginning, the number is increased four times. | The desired number is written in the form $A=\overline{x_{1}} \overline{\overline{1}} \overline{-x} \bar{k} \overline{\bar{k}}$. Then $4 A=\overline{6 x_{1}} \bar{\ldots} \overline{x_{k}} \overline{\text {. }}$. To determine the digits of the number $4 A$, we will sequentially multiply the number $A$ by 4:
$6 \cdot 4=... | 153846 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Auto: : Agakhanov N. $\mathbf{H}$.
Find the free term of the polynomial $P(x)$ with integer coefficients, if it is known that it is less than a thousand in absolute value, and $P(19)=P(94)=1994$. | Let $a_{0}$ be the constant term of the polynomial $P(x)$. Then $P(x)=x Q(x)+a_{0}$, where $Q(x)$ is a polynomial with integer coefficients. Therefore, $P(19)=19 n+a_{0}$, and $P(94)=94 m+a_{0}$, where $m$ and $n$ are integers. From the condition, it follows that $19 n=94 m$, hence $n=94 k, m=19 k$. Thus, $19 \cdot 94 ... | 208 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Tokorevev. S.
Among 2000 indistinguishable balls, half are aluminum with a mass of 10 g, and the rest are duralumin with a mass of 9.9 g. It is required to separate the balls into two piles such that the masses of the piles are different, but the number of balls in them is the same. What is the smallest number of weig... | Let's compare the mass of 667 balls with the mass of another 667 balls. If the masses of these two piles are not equal, the required condition is achieved.
Suppose the specified masses are equal. Then the mass of 666 balls that did not participate in the weighing is not equal to the mass of any 666 balls lying on one ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Kozhesenikov P.A. }}$
A natural number $b$ is called lucky if for any natural number $a$, such that $a^{5}$ is divisible by $b^{2}$, the number $a^{2}$ is divisible by $b$.
Find the number of lucky natural numbers less than 2010. | Lemma. A number $b$ is lucky if and only if each prime number enters the prime factorization of $b$ with one of the following exponents: $0,1,2,3,4,6,8$.
Proof. Let's call a non-negative integer $k$ happy if there does not exist an integer $m$ such that $2 m < k < 2 m + 2$. If $k \leq 4$, then $k$ is happy, since we c... | 1961 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8,9,10 | |
What is the minimum number of weights needed to be able to weigh any number of grams from 1 to 100 on balance scales, if the weights can be placed on both pans of the scales? | When solving this problem, we need the following interesting property of the ternary numeral system: any natural number can be represented as the difference of two numbers, the ternary representation of which contains only 0 and 1.
To prove this, we need to write the original number in ternary notation and construct t... | 5 | Other | math-word-problem | Yes | Yes | olympiads | false |
Authors: Kanel-Belov A.Ya., Galochkin A.i.
For what largest $n$ can one invent two infinite in both directions sequences $A$ and $B$ such that any segment of sequence $B$ of length $n$ is contained in $A$, $A$ has a period of 1995, and $B$ does not have this property (is non-periodic or has a period of a different len... | An example where the period of sequence $A$ consists of one 1 and 1994 zeros, and $B$ is a non-periodic sequence of zeros and ones, with all ones at least 1994 positions apart, shows that the condition of all segments of length 1994 being identical is not sufficient for the periodicity of $B$.
The sequence is periodic ... | 1995 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Hramio }}$ D.
Find the largest natural number $N$, for which in any arrangement of different natural numbers from 1 to 400 in the cells of a $20 \times 20$ square table, there will be two numbers in the same row or column whose difference is at least $N$. | Example. Let's divide the table into two $20 \times 10$ rectangles vertically. In the first rectangle, we will place numbers from 1 to 200 in increasing order by rows (from 1 to 10 in the first row, from 11 to 20 in the second row, and so on). In the second rectangle, we will place numbers from 201 to 400 in the same m... | 209 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Golvanov A.C.
Sasha wrote a non-zero digit on the board and keeps appending a non-zero digit to the right, until he writes a million digits. Prove that a perfect square was written on the board no more than 100 times. | Let's consider separately the numbers with an odd and even number of digits. Let $x_{1}^{2}, x_{2}^{2}, \ldots$ be the squares encountered on the board with an even number of digits, and their representations contain $2 n_{1}, 2 n_{2}, \ldots$ digits respectively. (If $n_{1}10^{2 n_{k}-1}$, from which it follows that $... | 100 | Number Theory | proof | Yes | Yes | olympiads | false |
[ Tasks with constraints $]$
The distance between numbers $a_{1}^{-} a_{2}^{2} a_{3}^{-} a_{4}^{-} a_{5}^{-}$ and $b_{1}^{-} b_{2}^{-} b_{3}^{-} b_{4}^{-} b_{5}^{-}$ is defined as the maximum $i$ for which $a_{i} \neq b_{i}$. All five-digit numbers are written one after another in some order. What is the minimum possi... | Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ denote the number of pairs of consecutive numbers with distances of $1, 2, 3, 4, 5$ respectively. Since there are only 10 last digits, they must change at least 9 times, so $x_{5} \geq 9$.
The number of pairs of the last two digits is 100, so similarly $x_{4}+x_{5} \geq 100-1=99... | 101105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$$
\begin{aligned}
& \text { [Tournaments and tournament tables] } \\
& \text { [Induction (other)] }
\end{aligned}
$$
In a tournament, 25 chess players are set to participate. They all play at different levels, and the stronger player always wins when they meet.
What is the minimum number of games required to determ... | Example. Let's organize a tournament according to the Olympic system in five rounds: first, 12 pairs will play, then 6, then 3. Four people will remain, two pairs will play, and then the winners will play against each other. Thus, the strongest will be determined. He played no more than five games.
Since 24 people wer... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Evoikinov M.A.
Polina has a deck of 36 cards (4 suits with 9 cards each). She chooses half of the cards she wants and gives them to Vasilisa, and keeps the other half for herself. Each turn, the players take turns opening one card of their choice (the opponent sees the suit and rank of the opened card), starting with ... | If Polina takes all the hearts, all the aces, all the kings, and all the queens, then Vasilisa will not be able to score points on the ace, king, and queen of hearts, i.e., she will score no more than 15 points.
Now let's prove that for any choice of Polina, Vasilisa can earn at least 15 points. Lay out the cards in t... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
KOVSHIN=BOTTLE+GLASS; $\quad$ TWO KOVSHINS=SEVEN GLASSES; BOTTLE=CUP+TWO GLASSES; $\quad$ BOTTLE=how many CUPS? | Try to replace the BOTTLE in the first line with its equivalent in CUPS and GLASSES (see the third line).
## Solution
Let's restate the conditions of the problem:
JUG = BOTTLE + GLASS; TWO JUGS = SEVEN GLASSES; BOTTLE = CUP + TWO GLASSES; BOTTLE = how many CUPS?
From the 1st and 3rd lines, it follows that the capac... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Along the running track, 12 flags are placed at equal distances from each other. The athlete starts at the first flag and runs at a constant speed. Already after 12 seconds, the athlete was at the 4th flag. How long will it take him to run the entire track? | Recall problem 4.
## Solution
Between 12 flags, there are 11 "distances." Between 4 flags, there are 3 "distances." It takes 4 seconds to cover one "distance." Therefore, it will take a total of 44 seconds.
## Answer
It takes 44 seconds. | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\underline{\text { Arithmetic. Mental calculation, etc. }}]} \\ {[\underline{\text { Evenness and oddness }}]}\end{array}\right.$
How many odd numbers are there between 300 and 700? | There are 399 numbers between 300 and 700, and there is 1 fewer even number than odd numbers. Therefore, there are 200 odd numbers.
## Answer
200 numbers. | 200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ] Common fractions
Half of half of a number is equal to half. What is this number?
# | Half of half is a quarter.
## Solution
If half of half (i.e., a quarter) of a given number is 0.5, then the number itself is $0.5 \times 4 = 2$.
## Answer
2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Sasha wrote down the numbers from one to one hundred, and Misha erased some of them. Among the remaining numbers, 20 have the digit one in their notation, 19 have the digit two, and 30 have neither the digit one nor the digit two. How many numbers did Misha erase? | Among the numbers from 1 to 100, the digit 1 appears exactly twenty times: the number 1 itself, ten numbers from 10 to 19, the numbers 21, 31, ..., 91 (eight of them), and the number 100. This means that none of these numbers were erased. Similarly, the digit 2 appears exactly nineteen times: the number 2 itself, ten n... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Raskina I.V.
Once a trader came to the city with umbrellas of three colors. He had half as many blue umbrellas as yellow and red ones, red umbrellas were one-third the number of yellow and blue ones, and there were 45 yellow umbrellas. How many blue and how many red umbrellas did the trader have? | There are half as many blue umbrellas as yellow and red ones. This means that blue umbrellas make up a third of the total. Similarly, red umbrellas make up a quarter of the total. Let's draw a segment representing all the umbrellas. To conveniently represent both a third and a quarter of the entire segment, we will mak... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Petya wrote a hundred-digit number $X$, in which there are no zeros. The fifty-digit number formed by the first fifty digits of the number $X$, Petya called the head of the number $X$. It turned out that the number $X$ is divisible by its head without a remainder. How many zeros are in the quotient? # | Let's append 50 zeros to the head of the number $X$ and subtract the resulting number from $X$. The result will be the "tail" of the number $X$, that is, the number formed by the last 50 digits of $X$. Since the difference between numbers that are divisible by the head is also divisible by the head, and the quotient ca... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Authors: Kazzczzna T.V., Frankinn B.R., Shapovalov A.V.
Out of 100 members of the Council of Two Tribes, some are elves, and the rest are dwarves. Each member wrote down two numbers: the number of elves in the Council and the number of dwarves in the Council. Each member correctly counted their own tribe but made an e... | Participants were required to find just one example. Let's show how to find them all.
Let the digit $A$ be written at least 222 times. Someone wrote $A$ at least three times (otherwise, the number of $A$s in the answers would not exceed $100 \cdot 2 = 200$). Therefore, Vasya wrote $A A$ and $B A$. Since the sum of the... | 66 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{ll}{[\text { Decimal numeral system }}\end{array}\right]$
From $A$ to $B 999$ km. Along the road, there are kilometer markers indicating the distances to $A$ and to $B$:
$0|999,1| 998, \ldots, 999 \mid 0$.
How many of them have only two different digits? | If the number on the left is composed of digits $a, b, c$, then the number on the left is composed of digits $a_{1}=9-a, b_{1}=9-b$ and $c_{1}=9$ $-c$. If $a=b=c$, then the required condition is satisfied. There are exactly 10 such columns.
Now let's assume that among the digits $a, b, c$ there are exactly two differe... | 40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10,11
The number $N$ is a perfect square and does not end in zero. After erasing the last two digits of this number, another perfect square is obtained. Find the largest number $N$ with this property. | Let $a$ be a number such that the square of $a$ becomes the number $N$ after erasing its last two digits.
Then $N>100 a^{2}$, hence $\sqrt{N}>10 a$. By the condition, the number $\sqrt{N}$ is an integer. Therefore, $\sqrt{N} \geq 10 a+1$, which means $N \geq (10 a+1)^{2}=100 a^{2}+20 a+1$. On the other hand, $N<1700$,... | 1681 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A natural number A, when divided by 1981, gave a remainder of 35, and when divided by 1982, it also gave a remainder of 35. What is the remainder when the number $A$ is divided by 14?
# | Let $A=1981 k+35=1982 l+35$. Then $1981 k=1982 l$; in particular, $k$ is even. Therefore, the number $1981 k$ is divisible by 14. Hence, when divided by 14, the number A gives the same remainder as 35.
## Answer
7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{ll}{\left[\begin{array}{l}\text { Irrational Equations } \\ \text { [Completing the Square. Sums of Squares] }\end{array}\right]}\end{array}\right]$
Solve the equation
$$
\left(x^{2}+x\right)^{2}+\sqrt{x^{2}-1}=0
$$ | Since the numbers $\left(x^{2}+x\right)^{2}$ and $\sqrt{x^{2}-1}$ are non-negative, and their sum is zero, then both these numbers are equal to zero. On the other hand, if both these numbers are equal to zero, then their sum is zero. Therefore, the original equation is equivalent to the following system:
$$
\left\{\be... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Sequences ]
Continue the sequence of numbers: 1, 11, 21, 1112, 3112, 211213, 312213, 212223, 114213...
# | In each subsequent number, it is recorded how many ones are in the previous number, then how many twos, how many threes, and so on. Thus, in the fourth number, it is recorded that the third number contains three ones and one two. Thus, in the last ninth number, it is recorded that the eighth number contains one one, fo... | 31121314 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Divisibility of numbers. General properties ]
On a bus rode Andrey
To the circle and back home,
Paying 115 rubles,
He bought himself a pass.
In January he didn't get it,
And therefore for several days
He bought a ticket from the driver
For 15 rubles for himself.
And on another day the conductor
Took only 11... | The amount of rubles spent by Andrey on the days when he bought a ticket from the driver is divisible by 5; the total amount of rubles spent by him in January is also divisible by 5. Therefore, the total amount of money spent on other days is also divisible by 5. Hence, the number of days when Andrey bought a ticket fr... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Investigation of a quadratic trinomial ] [ Methods for solving problems with parameters ]
For the quadratic trinomial $f(x)=a x^{2}-a x+1$, it is known that $|f(x)| \leq 1$ for $0 \leq x \leq 1$. Find the greatest possible value of $a$. | Since $f(0)=f(1)=1$, the graph of the quadratic polynomial is a parabola symmetric with respect to the line $x=0.5$. From the condition $|f(x)| \leq 1$ for
$0 \leq x \leq 1$, it follows that the branches of the parabola are directed upwards. The minimum value of $f(x)$ is $f(0.5)=1-a / 4$.
The maximum possible value ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find the largest four-digit number, all digits of which are different and which is divisible by 2, 5, 9, and 11.
# | The numbers $2, 5, 9$ and 11 have no common divisors, so if a number is divisible by each of them, it is also divisible by their product. That is, the number we are looking for is divisible by $2 \cdot 5 \cdot 9 \cdot 11=990$. Let's list all four-digit numbers that are divisible by 990: 1980, 2970, 3960, 4950, 5940, 69... | 8910 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6,7,8 |
In 2 seconds, the kangaroo mother makes three jumps, and the joey makes five jumps. The length of the mother kangaroo's jump is 6 meters, while the length of the joey's jump is three times less. The mother and joey are playing tag: the joey jumps ahead by 12 jumps, after which the mother starts to chase him, a... | In two seconds, the mother kangaroo makes 3 jumps, which means she jumps 9 jumps of the joey. Therefore, in two seconds, the distance between the mother and the joey decreases by 4 jumps of the joey. Initially, there were 12 jumps of the joey between them, so it will take the mother 6 seconds to catch up.
## Answer
I... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Help Neznaika restore the division example of two numbers, if it is known that the quotient is five times smaller than the dividend and seven times larger than the divisor.
# | Since the quotient is five times smaller than the dividend, the divisor is 5. Since the quotient is seven times larger than the divisor, it is equal to \(5 \cdot 7 = 35\).
## Problem
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result di... | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Carlson was given a bag of candies: chocolate and caramels. In the first 10 minutes, Carlson ate 20% of all the candies, and 25% of them were caramels. After that, Carlson ate three more chocolate candies, and the proportion of caramels among the candies Carlson had eaten decreased to 20%. How many candies were in the ... | In the first portion of candies eaten, there were 3 times more chocolate ones than caramels, and then their number became 4 times more. This means that the number of additional chocolate candies eaten equals the number of caramels. Thus, at the beginning, Karlson ate 3 caramels and 9 chocolate candies, and these 12 can... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Two dozen lemons cost as many rubles as lemons can be bought for 500 rubles. How much does a dozen lemons cost?
# | Let's set up a proportion according to the problem: 20 lemons - x rubles; x lemons - 500 rubles.
Thus, $20: x = x: 500$, which means $x^2 = 10000; x = 100$. Two dozen lemons cost 100 rubles, so one dozen costs 50 rubles.
## Answer
50 rubles. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Folklore
What values can the expression $(x-y)(y-z)(z-x)$ take if it is known that $\sqrt{x-y+z}=\sqrt{x}-\sqrt{y}+\sqrt{z}$? | Let $\sqrt{x}=a, \sqrt{y}=b, \sqrt{z}=c, \sqrt{x-y+z}=d$, we get the system $a-b+c=d$ $a^{2}-b^{2}+c^{2}=d^{2}$, or
$a-b=d-c$
$a^{2}-b^{2}=d^{2}-c^{2}$.
If $a=b$, then $x=y$. Otherwise, dividing the second equation by the first, we get $a+b=d+c$.
Subtracting the first equation from this, we get $b=c$, that is, $y=z... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On the table, 100 cards were lying with the white side up, each having one white side and one black side. Kostya flipped 50 cards, then Tanya flipped 60 cards, and after that, Olya - 70 cards. As a result, all 100 cards ended up lying with the black side up. How many cards were flipped three times? | Since all the cards ended up being flipped, each of them was flipped either once or three times. In total, there were 180 flips: 100 of them were needed to flip each card once; the remaining 80 were to flip some cards two more times. Therefore, 40 cards were flipped three times.
## Answer
40 cards. | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$2+$ |
| [ | ]
For the angles of triangle $ABC$, it is known that $\sin A + \cos B = \sqrt{2}$ and $\cos A + \sin B = \sqrt{2}$. Find the measure of angle $C$. | The first method. We square both equalities and add them. We get $4=2+2(\sin A \cos B+\cos A \sin B)=2+2 \sin (A+B)$. Therefore,
$\sin (A+B)=1$, that is, $A+B=90^{\circ}$.
The second method. Adding the original equalities, we get $\sin \left(A+45^{\circ}\right)+\sin \left(B+45^{\circ}\right)=2$, from which $\sin \lef... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Raskina I.V.
The cake is packed in a box with a square base. The height of the box is half the side of this square. A ribbon of length 156 cm can be used to tie the box and make a bow on top (as shown in the left image). To tie it with the same bow on the side (as shown in the right image), a ribbon of length 178 cm i... | For the first method of tying, the ribbon encircles the box twice the length, twice the width, and four times the height, meaning its length is equal to six sides of the base plus the bow. For the second method of tying, the ribbon encircles the box twice the length, four times the width, and twice the height, meaning ... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{ll}{[\text { Arithmetic. Mental calculation, etc. }]} \\ {[\text { Word problems (miscellaneous) }]}\end{array}\right]$
The length of the crocodile from head to tail is three times less than ten ken, and from tail to head it is three ken and two shaku. It is known that one shaku is equal to 30 cm.... | According to the condition, the tripled length of the crocodile is 9 ken and 6 shaku or 10 ken. Therefore, 1 ken = 6 shaku = 180 cm. The length of the crocodile is 3 ken 2 shaku, which is 600 cm.
## Answer
6 m. | 6 | Other | math-word-problem | Yes | Yes | olympiads | false |
[ [Decimal number system ] $[$ Problems with constraints $]$
Find the number of five-digit numbers in decimal notation that contain at least one digit 8.
# | Find the number of all five-digit numbers and five-digit numbers that do not contain the digit 8.
## Solution
There are a total of 90000 five-digit numbers (see the solution to problem 60336). Let's find the number of five-digit numbers that do not contain the digit 8. The first digit of such a number cannot be 0 or ... | 37512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic progression ]
[ Arithmetic. Mental calculation, etc. ]
What is the sum of the digits of all numbers from one to a billion? | Add zero to these numbers and form 500 million pairs: (0, 999999 999), (1, 999999 998), and so on. In each pair, the sum of the digits is 81, and besides, we forgot the number 1000000 000; therefore, the total sum is $500000000 \times 81+1=40500000001$.
## Answer
40500000001. | 40500000001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Mathematical logic (miscellaneous).] [ Arithmetic. Mental arithmetic, etc. ]
This ancient problem was known even in Ancient Rome.
A wealthy senator, on his deathbed, left his wife pregnant. After the senator's death, it was discovered that his estate, worth 210 talents, was left with the following will: "In the eve... | Since it is impossible to fulfill all the testator's requirements, only a part of them will have to be fulfilled. Depending on which specific part we fulfill, one or another method of division will be adopted.
## Solution
Since it is impossible to fulfill all the testator's requirements, we will have to fulfill only ... | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
35298 topics: [ Equations in integers ] [Products and factorials]
Find all natural integer solutions of the equation $(n+2)!-(n+1)!-n!=n^{2}+n^{4}$.
# | The equation can be written as $n!((n+2)(n+1)-(n+1)-1)=n^{2}\left(n^{2}+1\right)$, or $n!(n+2) n=n^{2}\left(n^{2}+1\right)$.
Checking shows that $n=1$ is not a solution.
Let $n>1$. Write the equation as $(n-2)!(n+2)(n-1)=\left(n^{2}+1\right)$. From this, $n^{2}+1 \geq(n+2)(n-1)$.
Expanding the brackets, we get $n \l... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Algebra and arithmetic (miscellaneous).] $[$ Sequences (miscellaneous) ]
Find the value of the expression $1!* 3-2!* 4+3!* 5-4!* 6+\ldots-2000!* 2002+2001!$. | Start simplifying the given expression from the end.
## Solution
Let's denote the given expression by S and start transforming it from the end: -2000!*2002 + 2001! = 2000! * (-2002 + 2001) = -2000!. Thus, S = 1!*3 - 2!*4 + 3!*5 - 4!*6 + ... + 1999!*2001 - 2000!. Next, 1999!*2001 - 2000! = 1999!(2001 - 2000) = 1999!. ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}[\text { Modular arithmetic (other) })] \\ {[\text { Word problems (other). }]}\end{array}\right]$
In a store, there were 6 boxes weighing 15, 16, 18, 19, 20, and 31 kilograms, respectively. Two companies bought five boxes, with one company taking twice as much by weight as the other. Which box ... | The mass of the purchased apples is divisible by 3.
## Solution
The total mass of all purchased apples is three times the mass of apples purchased from the first company, meaning it is divisible by 3. The total mass of all boxes gives the same remainder when divided by 3 as the sum of $0+1+0+1+2+1$ = 5. Therefore, th... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a chess tournament, sports masters and candidates for masters are participating. What is the smallest number of people that can participate in this tournament if it is known that among them, masters are less than half, but more than $45 \%$.
# | Let $n$ chess players participate in a tournament, and $k$ of them are masters. According to the condition, $0.9 n2 k$ $-n>0.2 k-n$ is an integer, so it is at least 1. Therefore, $0.1 n>1, n>10$.
The case $n=11$ works: in the tournament, 5 masters and 6 candidates can play.
## Answer
11. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Dirichlet's Principle (miscellaneous). ] $[\quad$ Tables and tournaments (miscellaneous). $\quad]$
On each cell of a $9 \times 9$ board, there is a beetle. When a whistle is blown, each beetle moves to one of the diagonally adjacent cells. As a result, some cells may end up with more than one beetle, while some cel... | Paint the vertical lines of the board in black and white, alternating. Then, from a black cell, the beetle moves to a white one, and from a white one - to a black one.
## Solution
We will paint the vertical lines of the board in black and white, alternating. As a result, $5 \times 9$ $=45$ cells will be painted black... | 9 | Combinatorics | proof | Yes | Yes | olympiads | false |
[ Examples and counterexamples. Constructions ]
In the cells of a $3 \times 3$ table, the digits from 1 to 9 were placed. Then, the sums of the digits in each row were found. What is the maximum number of these sums that can be perfect squares?
# | The smallest possible sum of digits in a row is $1+2+3=6$, and the largest is $7+8+9=24$. Therefore, if the sum is a perfect square, it must be 9 or 16. Suppose that in each row, the sum is 9 or 16, which means it gives a remainder of 2 when divided by 7. Then the sum of all digits in the table, when divided by 7, will... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kazzytsina t.v.
Petrov booked an apartment in a new building with five identical entrances. Initially, the entrances were numbered from left to right, and Petrov's apartment had the number 636. Then the developer changed the numbering to the opposite direction (from right to left, see the figure). As a result, Petrov'... | From the condition, it follows that the number of entrances to the left and right of Petrov's entrance differ by 0.2 or 4, that is, by an even number. At the same time, the difference between the apartment numbers of Petrov under the two numbering options shows how many more apartments are in the entrances to the left ... | 985 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tokkymov M.A.
Find the smallest natural number that starts with 2016 (in decimal notation) and is divisible by 2017.
# | Let the number $n$ have $k+4$ digits. Then $2016 \cdot 10^{k} \leq n < 2017 \cdot 10^{k}$. Since $n$ is divisible by 2017, then $n \leq 2017 \cdot 10^{k} - 2017$. Therefore, $2017 \leq (2017 - 2016) \cdot 10^{k} = 10^{k}$, which means $k \geq 4$. Therefore, the smallest such number is $20170000 - 4 \cdot 2017$.
## Ans... | 20161932 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Konyagin S. $\mathbf{\text { K }}$.
In a family album, there are ten photographs. In each of them, three people are depicted: a man stands in the center, to the left of the man is his son, and to the right is his brother. What is the smallest number of different people who can be depicted in these photographs, given t... | Let's call the ten men standing in the center of the photographs the main faces. We will divide all men in the photographs into levels. We will assign level 0 to those who do not have fathers in the photographs, and level $k+1$ ($k=0,1$, $2, \ldots$) to men who have fathers at level $k$. Let $r_{k}$ denote the number o... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9,10,11 |
| | Product Rule | |
| | Cooperative Algorithms | |
| | Evaluation + Example | |
Authors: $\underline{\text { Knop K.A., }}$ Leontyeva O.
A magician and an assistant are going to perform the following trick. A spectator writes a sequence of $N$ digits on a board. The assistant of the magician covers t... | Suppose that for some value of $N$ the trick is possible. Then for each variant of the sequence with two closed digits (let their number be $k_{1}$), the magician can restore the original; that is, to each sequence with two closed digits, the magician can uniquely correspond the restored sequence of $N$ digits (let the... | 101 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
For an arbitrary number $x$, consider the sum $Q(x)=\lfloor x \rfloor + \left\lfloor \frac{x}{10000} \right\rfloor$.
Find the difference $Q(2023) - Q(2022)$. (Here $\lfloor x \rfloor$ denotes... | 
\$ \left\lfloor \left| \frac{\mathrm{x}-1}{\mathrm{k}} \right| \right\rfloor < \left\lfloor \left| \frac{\mathrm{x}}{\mathrm{k}} \right| \right\rfloor = m \$ means that \$ \mathrm{x}-1 < \mathr... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Authors: Shapovalov A.V., Kupakov A.
There is a set of weights, the masses of which in grams are: $1, 2, 4, \ldots, 512$ (consecutive powers of two) - one weight of each mass. It is allowed to weigh a load using this set, placing the weights on both pans of the scales.
a) Prove that no load can be weighed using these... | Let $K_{n}(P)$ be the number of ways to weigh a weight $P$ using weights of $1,2, \ldots, 2^{n}$, and $K_{n}=\max _{P} K_{n}(P)$ (the maximum number of ways to weigh any weight using these weights). Obviously, $K_{0}=1, K_{1}=2$.
a) Our task is to prove that $K_{9} \leq 89$. We will prove that $K_{n+1} \leq K_{n}+K_{n-... | 89 | Combinatorics | proof | Yes | Yes | olympiads | false |
frankin 5.
In the city of Udoyev, mayoral elections are held as follows. If in a given round of voting no candidate receives more than half of the votes, then a subsequent round is held with the participation of all candidates except the one with the fewest votes. (No two candidates ever receive the same number of vot... | a) Ostap could not take the last, 2002nd place in the first round, as otherwise he would have been immediately eliminated from the candidates. Therefore, $k \leq 2001$.
Suppose all candidates in the first round received almost the same number of votes, Ostap took the second-to-last place, and in each subsequent round,... | 2001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.I.
In a class of 30 students, each has the same number of friends among their classmates. What is the maximum possible number of students who study better than the majority of their friends? (For any two students in the class, it can be said who studies better; if $A$ studies better than $B$, and $B$ studies... | Students who perform better than most of their friends are called good. Let $n$ be the number of good students, and $k$ be the number of friends each student has. We will prove that $n \leq 25$. For this, we will consider two cases.
1) $k \geq 8$. Then the five worst students in the class are not good.
2) $k \leq 7$. ... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
How many three-digit numbers exist in which the digits 1, 2, 3 appear exactly once each?
# | We can place any of the three digits in the first position, any of the two remaining digits in the second position, and the last remaining digit in the third position. Thus, we get a total of 6 numbers.
## Answer
$3!=6$ numbers. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the box, there are blue, red, and green pencils. In total, there are 20 pieces. Blue pencils are 6 times more than green ones, and there are fewer red pencils than blue ones.
How many red pencils are in the box?
# | The total number of blue and green pencils is divisible by 7. Therefore, they are either 7 or 14. In the first case, the number of red pencils is 13 more than the blue ones. Therefore, only the second option is possible: 6 red pencils.
## Answer
6 red pencils. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
There are nuts in three boxes. In the first box, there are 6 kg fewer nuts than in the other two together. And in the second box, there are 10 kg fewer nuts than in the other two together. How many nuts are in the third box? | Let's combine both given conditions and obtain the following statement: "In the first and second boxes, there are $6+10$ kg fewer nuts than in the first, second, and two-thirds of the third box." From this, it follows that in two-thirds of the third box, there are 16 kg of nuts, which means that in the third box, there... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ $[$ Divisibility of numbers. General properties $]$
Vovochka had more nuts than Pavlik. If Vovochka had given Pavlik as many nuts as Pavlik had, both boys would have had an equal number of nuts. But instead, Vovochka gave Pavlik a few nuts (no more than five), and divided the rest equally among three squirrels. How ... | Initially, Volodya had 3 times more nuts than Pavlik.
## Solution
From the condition, it follows that Volodya had 3 times more nuts than Pavlik. After giving some nuts to Pavlik, the number of his nuts remained a multiple of 3—otherwise, he wouldn't have been able to divide them among three squirrels. The only number... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How, without any measuring tools, can you measure 50 cm from a string that is 2/3 of a meter long?
# | Try cutting off a quarter of the shoelace.
## Solution
If you cut off a quarter of the shoelace (for this, you need to fold it in half twice), exactly 50 cm will remain. Indeed, $3 / 4 \cdot 2 / 3=1 / 2$. | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Append the same digit to the left and right of the number 10 so that the resulting four-digit number is divisible by 12.
# | Let the digit $a$ be appended. Then the resulting number will be written in the form $a 10 a$. Since this number is divisible by 12, it must be divisible by both 4 and 3. This, in turn, means that $a$ is divisible by 4, and $2a + 1$ is divisible by 3. This is only possible when $a=4$.
## Answer
4104. | 4104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bootin D. ..
Several guard brigades of the same size slept for the same number of nights. Each guard slept more nights than there are guards in a brigade, but fewer than the number of brigades. How many guards are in a brigade if all the guards together slept for 1001 person-nights? | $1001=7 \cdot 11 \cdot 13$.
## Solution
Let $s$ be the number of guards in a squad, $b$ be the number of squads, and $n$ be the number of nights one guard slept. Then $2 b n=1001$.
But $1001=7 \cdot 11 \cdot 13$, and the numbers $7,11,13$ are prime. Considering that $s<n<b$, we get $s=7$.
Answer
7 guards.
Submit ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2 [ Inequalities problems. Case analysis ]
Sasha invited Petya to visit, saying that he lives in the 10th entrance in apartment No. 333, but forgot to mention the floor. Approaching the building, Petya noticed that the building is nine-story. Which floor should he go to? (The number of apartments on each floor is the ... | If there are no more than three apartments on a floor, then in ten entrances, there will be no more than $10 \cdot 9 \cdot 3=270$ apartments, which means apartment № 333 will not be in the 10th entrance. If there are no fewer than five apartments on a floor, then already in nine entrances, there will be no fewer than $... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}\text { [Inclusion-Exclusion Principle]} \\ {[\quad \text { Word Problems (Miscellaneous). }}\end{array}\right]$
In the garden, Anya and Vitya had 2006 rose bushes. Vitya watered half of all the bushes, and Anya watered half of all the bushes. It turned out that exactly three bushes, the most be... | Vitya watered 1003 bushes, of which 1000 he watered alone, and three - together with Anya. Similarly, Anya watered 1003 bushes, of which 1000 she watered alone, and three - with Vitya. Therefore, together they watered $1000+1000+3=2003$ bushes. Thus, $2006-2003=3$ rose bushes remained unwatered.
## Answer
3 bushes. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kolya and his sister Masha went to visit someone. After walking a quarter of the way, Kolya remembered that they had forgotten the gift at home and turned back, while Masha continued on. Masha arrived at the visit 20 minutes after leaving the house. How many minutes later did Kolya arrive at the visit, given that they ... | Kolya walked an "extra" half of the distance. This means the delay time is equal to half of the time spent on the entire journey, which is 10 minutes.
## Answer
10 minutes. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2 [Chess boards and chess pieces]
During a chess game, at some point, Hedgehog had twice fewer pieces on the board than Bear, and they were five times fewer than the free squares on the board. How many of Bear's pieces had been captured by this point?
# | According to the condition, the number of cells on the board (64) is $1+2+5=8$ times more than the number of pieces the Hedgehog has. Therefore, the Hedgehog has 8 pieces left, and the Teddy Bear has 16.
## Answer
None.
Author: $\underline{\text { Folklore }}$
Find the smallest number divisible by 45, whose decimal... | 1111111110 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
"And this is for you to see too early," - said Baba Yaga to her 33 students and commanded: "Close your eyes!" The right eye was closed by all the boys and a third of the girls. The left eye was closed by all the girls and a third of the boys. How many students still saw what they were not supposed to s... | What is yet to be seen, two thirds of the girls saw with their right eye, and two thirds of the boys with their left. In total, then, one eye was not closed by two thirds of all the students - 22 people.
## Answer
22 students. | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Find the smallest natural value of $n$ for which the number $n!$ is divisible by 990. | The number $11!=1 \cdot 2 \cdot \ldots \cdot 9 \cdot 10 \cdot 11$ is divisible by $9 \cdot 10 \cdot 11=990$.
For $n!<11$, the number $n!$ is not even divisible by 11.
## Answer
$n=11$. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
\left.\begin{array}{l}{[\text { Arithmetic. Mental calculation, etc. }]} \\ {[\text { Arithmetic progression }}\end{array}\right]
When little Clive approached his grandfather's cuckoo clock, it was 12:05.
Clive started turning the minute hand until the hour hand returned to its original position. How many "cuckoos" d... | Since the clock strikes every hour, Grandpa counted 1+2+3+...+12 = 78 "cuckoo".
## Answer
Grandpa counted 78 "cuckoo". | 78 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Kazitsvna t.v.
Four mice: White, Gray, Fat, and Thin were dividing a piece of cheese. They cut it into 4 visually identical slices. Some slices had more holes, so Thin's slice weighed 20 grams less than Fat's, and White's slice weighed 8 grams less than Gray's. However, White was not upset because his slice weighed ex... | Notice that now the slice of Gray weighs as much as the slice of White, i.e., exactly a quarter of the total cheese mass. This means that both White and Gray have already received their share, and all 28 grams should be divided between Fat and Thin. Currently, they have an equal amount of cheese, so they should receive... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Nine identical sparrows peck at less than 1001 grains, while ten such sparrows peck at more than 1100 grains. How many grains does each sparrow peck?
# | A sparrow can peck only an integer number of seeds.
## Solution
Since 10 sparrows peck more than 1100 seeds, 9 sparrows will peck more than (1100 : 10) $\cdot$ 9 = 990 seeds. At the same time, it is known that 9 sparrows peck fewer than 1001 seeds.
The only number divisible by 9 in the range from 991 to 1000 is 999.... | 111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Sosukhin 0.
A sequence of two different numbers was continued in two ways: to form a geometric progression and to form an arithmetic progression. The third term of the geometric progression coincided with the tenth term of the arithmetic progression. With which term of the arithmetic progression did the fourth term of... | Let $a$ be the first of two numbers in the original sequence, $d$ be the difference of the arithmetic progression, and $q$ be the ratio of the geometric progression. Then, according to the problem, $a+d=a q, a+9 d=a q^{2}$. Therefore, $a(q-1)=d$ and $a(q-1)(q+1)=a\left(q^{2}-1\right)=9 d=9 a(q-1)$. Since $q \neq 1$, fr... | 74 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Can $n!$ end with exactly five zeros?
# | 24! ends with four zeros. By adding the factor 25, two more zeros will be added (since the number 25 adds two fives to the prime factorization of the factorial). Therefore, 25! ends with six zeros.
## Answer
How many zeros does the number $100!$ end with?
## Hint
It is sufficient to find how many fives are in the p... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2+ |
| :---: | :---: |
Find the remainder when $6^{100}$ is divided by 7. | $6^{100} \equiv(-1)^{100}=1(\bmod 7)$
## Answer
1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Factorization ] [ Equations in integers ]
In how many ways can the number 1979 be represented as the difference of squares of natural numbers? | Let $1979=x^{2}-y^{2}=(x-y)(x+y)$. Since $1979-$ is a prime number, there is a unique representation of it as a product of two natural numbers $1979=1 \cdot 1979$. Since $x-y<x+y$, then $x-y=$ $1, x+y=1979$, from which $x=990, y=989$.
## Answer
One way. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many twos will be in the prime factorization of the number 1984! ?
# | Among the numbers from 1 to 1984, there are 992 even numbers. Each of them contributes at least one factor of 2 to the prime factorization of 1984!. Two factors of 2 will be contributed by numbers divisible by 4 (there are 496 such numbers). Further, 3, 4, 5, 6, 7, 8, 9, and 10 factors of 2 will be contributed by 248, ... | 1979 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ [Properties of numbers. General properties]
In a row, numbers divisible by $9$ are written in ascending order: $9, 18, 27, 36, \ldots$. Under each number in this row, its sum of digits is written.
a) At which position will the number 81 first appear in the second row?
b) Which will appear earlier: the number 27 fo... | a) The number 81 will first appear in the second row under the number 999999999, so its position will be $999999999: 9=111111111$.
b) The number 36 will first appear in the second row under the number 9999, the number 27 will appear four times in a row earlier, for example under the numbers $3969,3978,3987,3996$.
## ... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Factorization } \\ {[\underline{\text { Modular Arithmetic (other) })]}]}\end{array}\right.$
What remainders can be obtained when dividing $n^{3}+3$ by $n+1$ for a natural number $n>2$? | Find a number close to $n^{3}+3$ that is divisible by $n+1$.
## Solution
$n^{3}+3=(n+1)\left(n^{2}-n+1\right)+2$.
## Answer
Only 2. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Every day at noon, a scheduled steamboat departs from Moscow to Astrakhan and from Astrakhan to Moscow. A steamboat departing from Moscow takes exactly four days to reach Astrakhan, then stays for two days, and at noon, two days after its arrival in Astrakhan, it departs for Moscow. A steamboat departing from Astrakhan... | From the condition, it is clear that the steamboat, which left from Moscow, will leave from it again exactly $4+5+2+2=13$ days later.
## Oтвет
13 steamboats.
## Problem | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On the Island of Misfortune, only knights, who always tell the truth, and liars, who always lie, live. In the Parliament of the island, there are 101 deputies. To reduce the budget, it was decided to reduce the Parliament by one deputy. However, each deputy stated that if they were removed from the Parliament, the majo... | Write two inequalities corresponding to the statements of the deputy liar and the deputy knight.
## Solution
Let $P$ be the number of knights in the Duma, and $L$ be the number of liars ($P+L=101$). According to the statement of the deputy knight, $P-1<50$. Therefore, $P<51$ (from this, in particular, it follows that... | 51 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Tournaments and tournament tables ] [ Problems on percentages and ratios ]
In a volleyball tournament, which was held in a round-robin format, $20 \%$ of all teams did not win a single game. How many teams were there
# | Can it happen that two teams did not win a single game?
## Solution
Note that there cannot be two teams that did not win a single game. Indeed, in the match between these teams, one of them won. Thus, $20 \%$ (that is, $1 / 5$) of the total number of teams is one team.
## Answer
5 teams. | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
The set of five-digit numbers $\{N_1, \dots, N_k\}$ is such that any five-digit number, all of whose digits are in increasing order, coincides in at least one digit with at least one of the numbers $N_1, \dots, N_k$. Find the smallest possible value of $k$. | Let's prove that the condition is satisfied by the set consisting of a single number $13579$. Indeed, let
$\overline{a_1a_2a_3a_4a_5}$ be a five-digit number, the digits of which satisfy the inequalities $a_1 < a_2 < a_3 < a_4 < a_5$. Then, if $a_1 \ne 1$, we have $2 \leq a_1 < a_2$. If at the same time $a_2 \ne 3$, t... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Badzian A.I.
In a $10 \times 10$ square, numbers from 1 to 100 are arranged: in the first row - from 1 to 10 from left to right, in the second row - from 11 to 20 from left to right, and so on. Andrey plans to cut the square into dominoes $1 \times 2$, calculate the product of the numbers in each domino, and sum the r... | Let's number the dominoes in the partition. Let the numbers in the $i$-th domino be $a_{i}$ and $b_{i}$. Notice that $a_{i} b_{i}=\frac{a_{i}^{2}+b_{i}^{2}}{2}-\frac{\left(a_{i}-b_{i}\right)^{2}}{2}$. Summing these equalities over all dominoes, we get that the sum of all 50 products is $\frac{a_{1}^{2}+\ldots+a_{100}^{... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
For different positive numbers $a$ and $b$, the equality $\frac{1}{1+a}+\frac{1}{1+b}=\frac{2}{1+\sqrt{a b}}$ holds. Prove that $a$ and $b$ are reciprocal numbers. | $0=\frac{1}{1+a}-\frac{1}{1+\sqrt{a b}}+\frac{1}{1+b}-\frac{1}{1+\sqrt{a b}}=\frac{1}{1+\sqrt{a b}}\left(\frac{\sqrt{a b}-a}{1+a}-\frac{\sqrt{a b}-b}{1+b}\right)=$
$\frac{\sqrt{b}-\sqrt{a}}{1+\sqrt{a b}}\left(\frac{\sqrt{a}}{1+a}-\frac{\sqrt{b}}{1+b}\right)$
Therefore, $0=\frac{\sqrt{a}}{1+a}-\frac{\sqrt{b}}{1+b}=\fr... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
$\underline{\text { Markelov S.V. }}$
Can we, by applying the functions sin, cos, tan, cot, arcsin, arccos, arctan, arccot to the number 1 in some order, obtain the number 2010? (Each function can be used any number of times.) | Let $f(x)=\operatorname{ctg}(\operatorname{arctg} x)=1 / x, g(x)=\sin (\operatorname{arctg} x)=\frac{x}{\sqrt{1+x^{2}}}$. By induction, we check that $g^{n}(x)=g\left(g^{n-1}(x)\right)=$ $\frac{x}{\sqrt{1+n x^{2}}}$
Taking $n=2010^{2}-1$, we get that $g^{n}(1)=1 / 2010$, from which $f\left(g^{n}(1)\right)=2010$.
## A... | 2010 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Antipov $M$.
In the cells of an $8 \times 8$ board, the numbers 1 and -1 are placed (one number per cell). Consider all possible placements of the figure $\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). We will call such a placement unsuccessful if the sum of... | Evaluation. Let's show that in each "cross" of five cells on the board, there will be at least one unsuccessful placement. Let the numbers in the outer cells of the cross be \(a, b, c, d\), and in the central cell be \(e\); denote by \(S\) the sum of all these five numbers. Suppose all placements in the cross are succe... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Fermat's Little Theorem ]
Find the remainder when $3^{102}$ is divided by 101.
# | Since $101$ is a prime number, then $3^{100} \equiv 1(\bmod 101)$. Therefore, $3^{102}=9 \cdot 3^{100} \equiv 9(\bmod 101)$.
Send a comment | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Trigonometric substitutions $]$
How many roots does the equation $8 x\left(1-2 x^{2}\right)\left(8 x^{4}-8 x^{2}+1\right)=1$ have on the interval $[0,1]$? | Note that $8 x^{4}-8 x^{2}+1=2\left(2 x^{2}-1\right)^{2}-1$. By making the substitution $x=\cos \varphi$, we get $8 \cos \varphi \cos 2 \varphi \cos 4 \varphi=-1$. Multiplying by $\sin \varphi$, we obtain $\sin 8 \varphi=-\sin \varphi$, from which $8 \varphi=-\varphi+2 k \pi$ or $8 \varphi=\pi+\varphi+2 k \pi$, that is... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Pooolish A.
Chichikov is playing with Nozdryov. First, Nozdryov distributes 1001 nuts into three boxes. After looking at the distribution, Chichikov names any integer \( N \) from 1 to 1001. Then Nozdryov must, if necessary, move one or several nuts to an empty fourth box and present Chichikov with one or several boxe... | Upper Bound. By placing 143,286 = 2 * 143 and 572 = 4 * 143 nuts in the boxes, Nozdryov can, for any N, move no more than 71 nuts. Indeed, N can be represented as 143k + r, where 0 ≤ k ≤ 7, and -71 ≤ r < 71. The number 143k can be collected with one or several boxes without moving anything. If r < 0, then 1001 - N = 14... | 71 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Inequalities problems. Case analysis ] [Examples and counterexamples. Constructions]
Authors: Khachatryan A.v., Raskina I.v.
Thirty-three bogatyrs (Russian epic heroes) hired themselves to guard Lukomorye for 240 coins. The cunning Chernomor can divide the bogatyrs into squads of any size (or record all in one squa... | From a detachment of $N$ bogatyrs, Chernomor will receive at most $N-1$ coins, since the remainder is less than the divisor. Therefore, he will receive no more than
$33-K$ coins, where $K$ is the number of detachments. But if there is only one detachment, then, since $240=33 \cdot 7+9$, Chernomor will receive only 9 c... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Blinkov A. A:
The teams held a football tournament in a round-robin format (each team played one match against every other team, with 3 points for a win, 1 point for a draw, and 0 points for a loss). It turned out that the sole winner scored less than $50 \%$ of the maximum possible points for one participant. What is... | Let's prove that there could not have been fewer than six teams. If, for example, there were five teams in the tournament, then they played $5 \cdot 4: 2=10$ matches and scored a total of at least 20 points. Therefore, the sole winner scored more than $20: 5=4$ points. However, according to the conditions, he scored no... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berruv S.l.
On the board, 100 pairwise distinct natural numbers $a_{1}, a_{2}, \ldots, a_{100}$ were written. Then, under each number $a_{i}$, a number $b_{i}$ was written, obtained by adding to $a_{i}$ the greatest common divisor of the remaining 99 original numbers. What is the smallest number of pairwise distinct n... | If $a_{100}=1$ and $a_{i}=2 i$ for $i=1,2, \ldots, 99$, then $b_{1}=b_{100}=3$, so among the numbers $b_{i}$ there will be no more than 99 different ones.
We will prove that among the numbers $b_{i}$ there will always be 99 different ones. We can assume that $a_{1}a_{j} \geq a_{i}+d_{i}=b_{i}$, from which $b_{i} \neq ... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Murashkin M.V.
On each cell of a $10 \times 10$ board, there is a chip. It is allowed to choose a diagonal with an even number of chips and remove any chip from it.
What is the maximum number of chips that can be removed from the board using such operations? | Let's call a (non-)even diagonal the one on which there is a (non-)even number of chips (at the moment). After removing a chip, an even diagonal becomes odd, and an odd one becomes even. Therefore, the number of odd diagonals does not decrease. At the beginning, there are 20 odd diagonals on the board, so there will be... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.910 |
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99 children are standing in a circle, and each initially has a ball. Every minute, each child with a ball throws their ball to one of the two neighbors; if two balls land with the same child, one of these balls is lost irretrievably. What is the minimum ti... | Number the children and balls clockwise from 1 to 99.
Example. Suppose children 1 and 2 toss the first ball to each other. The other balls with odd numbers are always thrown counterclockwise until they reach the second child, who discards them (this happens after an odd number of minutes, and at that moment he also re... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Boogonov I.I.
The Tsar summoned two sages. He gave the first 100 blank cards and ordered him to write a positive number on each (the numbers do not have to be different), without showing them to the second. Then the first can tell the second several different numbers, each of which is either written on some card or eq... | Let's prove that it is sufficient to lose 101 hairs for each sage. Let the first write the numbers $1, 2, 4, \ldots, 2^{99}$ on the cards, and inform the second of these numbers and their sum. When the second hears the number 1, he will understand that there is a card not exceeding 1. Hearing the next number $2^{k}$, h... | 101 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Font der Flas:
In the vertices of a cube, the numbers $1^{2}, 2^{2}, \ldots, 8^{2}$ (one number per vertex) are placed. For each edge, the product of the numbers at its ends is calculated. Find the maximum possible sum of all these products. | We will color the vertices of a cube in two colors such that the ends of each edge are of different colors. Let the numbers $a_{1}, a_{2}, a_{3}, a_{4}$ be placed in the vertices of one color, and the numbers $b_{1}, b_{2}, b_{3}, b_{4}$ in the vertices of the other color, with numbers having the same indices placed in... | 9420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are $1, 2, \ldots, 11$ kg. He also has a bag that will tear if more than 11 kg is placed in it.
Archimedes has learned the weights of all the ingots and wants ... | Let Archimedes first put ingots weighing 1, 2, 3, and 5 kg into the bag, and then put ingots weighing 1, 4, and 6 kg. In both cases, the bag does not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if Archimedes used ingots weighing \( w_{1}, \ldots, w_{6} \) kg instead of ing... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Ten dogs and cats were fed 56 biscuits. Each cat received 5 biscuits, and each dog - 6. How many
were dogs and how many were cats? | How many biscuits would be needed if all the animals were dogs?
## Solution
If only dogs were being fed, 106=60 biscuits would be needed. The extra 4 biscuits are required because a dog eats one more biscuit than a cat. This means there were 4 cats, and consequently, 6 dogs.
## Answer
6 dogs and 4 cats. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
There were 35 yellow and white dandelions growing on the lawn. After eight white dandelions flew away, and two yellow ones turned white, the number of yellow dandelions became twice the number of white ones. How many white and how many yellow dandelions were there on the lawn at first?
# | How many dandelions are left on the lawn after eight white ones have flown away?
## Solution
In the end, there are 27 dandelions left on the lawn - 18 yellow and 9 white. This means that at the beginning, there were $18+2=$ 20 yellow and $9+8-2=15$ white dandelions on the lawn.
## Answer
20 yellow and 15 white dand... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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