problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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Ya. I. V.
From the volcano station to the summit of Stromboli volcano, it takes 4 hours to walk along the road, and then - 4 hours along the path. At the summit, there are two craters. The first crater erupts for 1 hour, then remains silent for 17 hours, then erupts again for 1 hour, and so on. The second crater erupt... | The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the eruption of the first crater, this crater will not be dangerous.
Movement along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the eruption of... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ $\left.\frac{\text { Evenness and Oddness }}{\text { Case Enumeration }}\right]$
Find such a three-digit number $A^{2}$, which is a perfect square, that the product of its digits equals $A-1$.
# | Let $A^{2}=100 x+10 y+z$. Since $100 \leq A^{2} \leq 599$.
In the second case, we check the squares of the numbers 19 and 29; only 19 fits.
## Answer
361. | 361 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Binkov A, A:
In a football championship, 16 teams participated. Each team played against each of the others once, with 3 points awarded for a win, 1 point for a draw, and 0 points for a loss. We will call a team successful if it scored at least half of the maximum possible number of points. What is the maximum number ... | Each team played 15 games and therefore could have earned a maximum of $15 \cdot 3=45$ points. Thus, a team is successful if it has at least 23 points.
But one of the teams scored no more than the average possible number of points. Even if all matches were successful, the average is
$15 \cdot 1.5=22.5$
We will show ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
P.I. Find the largest natural number from which it is impossible to obtain a number divisible by 11 by erasing digits.
# | If in the decimal representation of a number there is a digit 0 or two identical digits, then by erasing the other digits, we can obtain a number that is a multiple of 11. Therefore, the desired number is at most nine digits long, and all its digits are distinct. The largest such number is 987654321. Let's prove that i... | 987654321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The angle formed by the rays $y=x$ and $y=2x$ for $x \geq 0$ cuts two arcs on the parabola $y=x^{2}+p x+q$. These arcs are projected onto the $Ox$ axis. Prove that the projection of the left arc is 1 unit shorter than the projection of the right arc. | The abscissas $x_{1}$ and $x_{2}$ of the points of intersection of the parabola and the line $y=x$ satisfy the equation $x^{2}+(p-1) x+q=0$. By Vieta's theorem, $x_{1}+x_{2}=1-p$. Similarly, we obtain that the abscissas $x_{3}$ and $x_{4}$ of the points of intersection of the parabola and the line $y=2 x$ are related b... | 1 | Geometry | proof | Yes | Yes | olympiads | false |
Kenodarov R.G.
Prove that if $a, b, c$ are positive numbers and $ab + bc + ca > a + b + c$, then $a + b + c > 3$.
# | According to the problem $30865(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \geq 3(a b+b c+c a)>3(a+b+c)$. Since $a+b$ $+c>0$, we get $a+b+c>3$ | 3 | Inequalities | proof | Yes | Yes | olympiads | false |
Bogdanov I.I.
The denominators of two irreducible fractions are 600 and 700. Find the smallest possible value of the denominator of their sum (in irreducible form).
# | Let our fractions be ${ }^{a} / 600$ and ${ }^{b} / 700$. Then $a$ is coprime with 6, and $b$ is coprime with 7. Therefore, the numerator of their sum ${ }^{7 a+6 b / 4200}$ is coprime with both $6=2 \cdot 3$ and 7. Since $4200=2^{3 \cdot 3} \cdot 7 \cdot 5^{2}$, this means that the denominator after simplification wil... | 168 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Khachaturyan A.V.
Petr was born in the 19th century, and his brother Pavel - in the 20th century. Once, the brothers met to celebrate their shared birthday. Petr said: "My age is equal to the sum of the digits of the year of my birth." - "Mine too," replied Pavel. How much younger is Pavel than Petr? | Let Peter and Pavel be born in the years $\overline{18 x y}$ and $\overline{19 u v}$, respectively. At the time of their meeting, Peter and Pavel were $1 + 8 + x + y$ and $1 + 9 + u + v$ years old, respectively. We can determine the year of their meeting in two ways. Since Peter's age at that time was equal to the sum ... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[formulas for abbreviated multiplication (other)] [Problems on maximum and minimum]
On the board, 101 numbers are written: $1^{2}, 2^{2}, \ldots, 101^{2}$. In one operation, it is allowed to erase any two numbers and write down the absolute value of their difference instead.
What is the smallest number that can resul... | From four consecutive squares (in three operations), the number 4 can be obtained: $(n+3)^{2}-(n+2)^{2}-((n+$ $\left.1)^{2}-n^{2}\right)=(2 n+5)-(2 n+1)=4$.
We can get 24 such fours from the numbers $6^{2}, 7^{2}, \ldots, 101^{2}$. 20 fours can be turned into zeros by pairwise subtraction. From the numbers $4,9,16,25$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Beroov S.L.
The numbers $a$ and $b$ are such that each of the two quadratic trinomials $x^{2} + a x + b$ and $x^{2} + b x + a$ has two distinct roots, and the product of these trinomials has exactly three distinct roots. Find all possible values of the sum of these three roots. | From the condition, it follows that the quadratic polynomials $x^{2}+a x+b$ and $x^{2}+b x+a$ have a common root $x_{0}$, as well as roots $x_{1}$ and $x_{2}$ different from it, respectively; in particular, $a \neq b$. The common root is also a root of the difference of these quadratics, that is, $(a-b)\left(x_{0}-1\ri... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Discrete distribution ] $[$ Equations in integers $]$
In a class of 25 children, two are randomly chosen for duty. The probability that both duty students will be boys is $3 / 25$.
How many girls are in the class
# | Let there be $n$ boys in the class, then the number of ways to choose an ordered pair of duty students from them is $n(n-1)$, the number of ways to choose an ordered pair from the entire class is $25 \cdot 24$. Therefore, the probability that two boys will be chosen is $n(n-1):(25 \cdot 24)=3 / 25$. Hence, $n(n-1)=72=9... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Gálnierin $\Gamma$.
Each face of a rectangular parallelepiped $3 \times 4 \times 5$ is divided into unit squares. Is it possible to inscribe a number in each square such that the sum of the numbers in each checkerboard ring of width 1, encircling the parallelepiped, equals 120? | For example, fill in 9 in all the squares of the faces $4 \times 5$, 8 in all the squares of the faces $3 \times 5$, and 5 in all the squares of the faces $3 \times 4$:
$2(4 \cdot 9+3 \cdot 8)=2(5 \cdot 9+3 \cdot 5)=2(5 \cdot 9+4 \cdot 5)=120$.
## Answer
It is possible. | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A magpie cooked porridge and fed her chicks. The third chick received as much porridge as the first two combined. The fourth one received as much as the second and third. The fifth one received as much as the third and fourth. The sixth one received as much as the fourth and fifth. The seventh one did not get any - the... | Replacing the first two chicks with the third, and the sixth with the fifth and fourth, we see that the total amount of porridge is twice as much as what the third, fourth, and fifth chicks received. But this is four times more than what the fifth chick received, so the magpie cooked $10 \cdot 4=40$ g of porridge.
## ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ [trigonometric inequalities ]
Solve the inequality $\sin \frac{x}{x^{2}+1}+\frac{x^{2}+1}{x} \cos \frac{x}{x^{2}+1}>0$ | 1) Note that $x \neq 0$ and $t=\frac{x}{x^{2}+1}=\frac{1}{x+\frac{1}{x}}$. Since $|x+1 / x| \geq 2$, then $|t| \leq 1 / 20$, so $00$ and $\cos t>0$. Moreover, $1 / t>0$, which means $\sin t+1 / t \cos t>0$. Therefore, for all $x>0$, the original inequality holds.
3) If $x>0$, then $-1 / 2 \leq t0.1 / t [ [various and G... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[ Ordinary fractions ] [ Inclusion-exclusion principle ]
How many irreducible fractions with numerator 2015 are there, which are less than $1 / 2015$ and greater than $1 / 2016$? | Let $a>0$ be the denominator of the desired fraction, then $\frac{1 / 2016 < 2015 / a < 1 / 2015}{2015^{2} < a < 2015 \cdot 2016} 2015^{2} < a$ $<2015^{2}+2015$. Therefore, the desired values of $a$ are numbers of the form $2015^{2}+n$, where $n$ is a natural number, $1 \leq n \leq$ 2014 and $\text{GCD}(2015; n)=1$.
S... | 1440 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Khachaturyan A.V. }}$
The robot invented a cipher for writing words: he replaced some letters of the alphabet with single or double-digit numbers, using only the digits 1, 2, and 3 (different letters he replaced with different numbers). First, he wrote himself in code: РОБОТ = 3112131233. After enc... | Let's consider the word ROBOT $=3112131233$. It contains 5 letters and 10 digits, so all codes are two-digit and can be determined easily. Let's write down all twelve possible codes and the letters we definitely know:
$$
\begin{array}{llll}
1= & 11= & 21= & 31=\mathrm{P} \\
2= & 12=\mathrm{O} & 22= & 32= \\
3= & 13=\m... | 2232331122323323132 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ $\quad$ Arithmetic progression $\quad]$ [Integer and fractional parts. Archimedes' principle ]
For the numbers $1000^{2}, 1001^{2}, 1002^{2}, \ldots$, the last two digits are discarded. How many of the first terms of the resulting sequence form an arithmetic progression? | The general term of the original sequence is: $a_{n}=\left(10^{3}+n\right)^{2}=10^{6}+2 n \cdot 10^{3}+n^{2}\left(a_{0}-\right.$ the first term $)$.
Let the general term of the resulting sequence be denoted by $b_{n}$, then
$b_{n}=\left[\frac{a_{n}}{100}\right]=\left[\frac{10^{6}+2 \cdot n \cdot 10^{3}+n^{2}}{100}\ri... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovelev A.B.
Artemon gave Malvina a bouquet of little red flowers and black roses. Each black rose has 4 pistils and 4 stamens, and two leaves on the stem. Each little red flower has 8 pistils and 10 stamens, and three leaves on the stem. The number of leaves in the bouquet is 108 less than the number of pistils. H... | Note that for each flower, the number of stamens is twice the difference between the number of pistils and the number of leaves.
Therefore, the same ratio is true for the entire bouquet.
## Answer
216 stamens. | 216 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Processes and operations ] $]$ [equations in integers]
Shagreen leather fulfills wishes, but after each wish, its area decreases: either by 1 dm² in the ordinary case, or by half - if the wish was cherished. Ten wishes reduced the area of the leather threefold, the next few - sevenfold, and after a few more wishes, ... | Let the original area of the skin be $S$ dm² ($S>0$). After the area was reduced threefold and then sevenfold, it became $S / 21$ dm². This number must be an integer; otherwise, it would be impossible to get 0 by dividing by 2 and subtracting 1.
Consider the first 10 wishes. Among them, there cannot be more than one u... | 42 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Average values $\quad]$ [ Area of a circle, sector, and segment ]
At a familiar factory, metal disks with a diameter of 1 m are cut out. It is known that a disk with a diameter of exactly 1 m weighs exactly 100 kg. During manufacturing, there is a measurement error, and therefore the standard deviation of the radius... | Given $\mathrm{E} R=0.5 \mathrm{~m}, \mathrm{D} R=10^{-4} \mathrm{~M}^{2}$. Let's find the expected value of the area of one disk: $\mathrm{E} S=\mathrm{E}\left(\pi R^{2}\right)=\pi \mathrm{E} R^{2}=\pi\left(\mathrm{D} R+\mathrm{E}^{2} R\right)=\pi\left(10^{-4}+0.25\right)=0.2501 \pi$.
Therefore, the expected value of... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Sedrakyan $H$.
Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be? | We can assume that the vertex of the first parabola is the point $(0,0)$. Let the vertex of the second be $(a, b)$. Then the equations of the parabolas are: $y=p x^{2}$ and $y=q(x-a)^{2}+b$, with $b=p a^{2}$ and $0=q a^{2}+b$. From this, $(p+q) a^{2}=0$. If $a=0$, then $b=$ 0, but the vertices are different. Therefore,... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A kangaroo jumps along a straight line. Pushing off with its left foot, it jumps 3 meters, with its right foot - 5 meters, and with both feet - 7 meters.
How can it cover exactly 200 meters in 30 jumps? | If all 30 jumps are 7 m, then the kangaroo will cover 10 m more than needed. Replacing a 7 m jump with a 5 m jump reduces this distance by 2 m, and replacing it with a 3 m jump reduces it by 4 m. The number 10 can be represented as the sum of twos and fours in three ways: $2+2+2+2+2$, $2+2+2+4$, and $2+4+4+4$. Hence, t... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ $[$ graphs and GMT on the coordinate plane $]$ Cauchy's Inequality $\quad]$
The graph of the linear function $y=k x+k+1$, where $k>0$, intersects the coordinate axes at points $A$ and $B$.
What is the smallest possible area of triangle $A B O$ ( $(O$ - the origin)? | The abscissa of the point of intersection of the graph with the $O X$ axis is $-\left(1+\frac{1}{k}\right)$. The ordinate of the point of intersection with the $O Y$ axis is $k+1$. Therefore,
$S_{A B O}=1 / 2 O A \cdot O B=1 / 2(k+1)(1+1 / k)=1 / 2(2+k+1 / k)$. The minimum value of the expression $k+1 / k$ is achieved... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
In the Country of Fools, coins of 1, 2, 3, ..., 19, 20 soldo (no others) are in circulation. Buratino had one coin. He bought an ice cream and received one coin as change. He bought the same ice cream again and received change in three different coins. Buratino wanted to buy a third such ice cream, but ... | Paying with three different coins amounts to no less than $1+2+3=6$ soldos. Since this amount was not enough to buy the ice cream, it must cost no less than 7 soldos.
The ice cream cannot cost more than 7 soldos; otherwise, two ice creams would cost no less than $8+8=$ 16 soldos. However, Buratino had only one coin, w... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Runner $\underline{\text { A.V. }}$.
Find the smallest natural number divisible by 80, in which two of its different digits can be swapped so that the resulting number is also divisible by 80.
# | Note that both before and after the rearrangement of digits, the number is divisible by 10 and therefore must end in 0. We will show that there are no three-digit numbers with the property described in the condition. Indeed, if \(100a + 10b = 80k\) and \(100b + 10a = 80l\), where \(a \geq 1\). In this case, \(900(a-1) ... | 1520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A team of several workers can complete a task in 7 full days. The same team, without two people, can complete the same task in several full days, and the same team, without six people, can also complete the same task in several full days. How many workers are in the team? (The productivity of the workers is the same.)
... | Let there be $n$ workers in the team. The daily productivity of one worker is called the norm.
Over 7 days, the reduced teams will fall short by 14 and 42 norms, respectively. This means that $n-2$ is a divisor of 14, and $n-6$ is a divisor of 42. Since $n>6$, then $n-2=7$ or 14, that is, $n=9$ or 15. Since 42 is divi... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Several natural numbers are written in a row with a sum of 2019. No number and no sum of several consecutive numbers equals 40. What is the maximum number of numbers that could have been written? | Lemma. The sum of any forty consecutive numbers is not less than 80.
Proof. Let the numbers $a_{1}, \ldots, a_{40}$ be written consecutively. Among the numbers $b_{0}=0, b_{1}=a_{1}, b_{2}$
b... | 1019 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (other).]
Kolya and Katya study in the same class. There are twice as many boys as girls in this class. Kolya has 7 more classmates than classmates of the opposite gender. How many female classmates does Katya have? | Since Kolya has 7 more classmates who are boys than classmates who are girls, there are 8 more boys than girls in this class. In addition, there are twice as many of them as there are girls. Therefore, there are 16 boys and 8 girls.
## Answer
7 girl classmates. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
A dog and a cat simultaneously grabbed a sausage loaf with their teeth from different sides. If the dog bites off its piece and runs away, the cat will get 300 g more than the dog. If the cat bites off its piece and runs away, the dog will get 500 g more than the cat. How much sausage will be left if b... | If two (identical) dogs grab a sausage from both sides, there will be a piece of 300 g between them. If two cats grab a sausage from both sides, there will be a piece of 500 g between them (see figure).
$.
## Answer
7.
Send a comment | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Residue arithmetic (other).]
How many zeros does the number $9^{999}+1$ end with?
# | $9^{999}+1=(9+1)\left(9^{998}-9^{997}+\ldots-9+1\right), 9^{998}-9^{997}+\ldots-9+1 \equiv(-1)^{998}-(-1)^{997}+\ldots-(-1)+1=999(\bmod 10)$. Therefore, $9^{999}+1$ is divisible by 10, but not by 100.
## Answer
By one. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ [Decimal numeral system ]
The numbers $2^{2000}$ and $5^{2000}$ are written in sequence. How many digits are written in total?
# | If the number $2^{2000}$ contains m digits, then $10^{\mathrm{m}-1}<2^{2000}<10^{\mathrm{m}}$.
## Solution
Let the number $2^{2000}$ contain $\mathrm{m}$ digits, and the number $5^{2000}$ contain $\mathrm{n}$ digits. Then the following inequalities hold: $10^{\mathrm{m}-1}<2^{2000}<10^{\mathrm{m}}, 10^{\mathrm{n}-1}<... | 2001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Grazing Problems.]
The grass grows uniformly and quickly across the entire meadow. It is known that 70 cows would eat it all in 24 days, and 30 cows would take 60 days.
How many cows would eat it all in 96 days?
# | Let a cow eat 1 portion of grass per day. In $60-24=36$ days, the meadow grew $30 \cdot 60-70 \cdot 24=120$ portions. This means, in addition to the 1800 portions eaten by 30 cows over 60 days, another 120 portions will grow over the additional $96-60=36$ days. In total, 1920 portions. Over 96 days, these will be eaten... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shapavaooov A.B.
a) On each of the fields of the top and bottom horizontal rows of an $8 \times 8$ chessboard, there is a token: white ones at the bottom, black ones at the top. In one move, it is allowed to move any token to an adjacent free cell vertically or horizontally. What is the minimum number of moves require... | a) Evaluation. To reach the opposite side of the board, a piece needs to make seven vertical moves. However, at least one of the two pieces standing on the same vertical must make a horizontal move (otherwise, they would not be able to pass each other). Therefore, together these pieces will make no fewer than 15 moves.... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Chekanov Yu.V. }}$ Yu.
A rectangle of size $1 \times k$ for any natural $k$ will be called a strip. For which natural $n$ can a rectangle of size $1995 \times n$ be cut into pairwise distinct strips? | Idea of the solution: we take the maximum strip (equal to the maximum side of the rectangle). The remaining strips will be combined in pairs, giving the sum of the maximum strip. If we have filled the rectangle, the problem is solved; otherwise, reasoning with areas shows that the rectangle cannot be cut into different... | 3989 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Font der Flasch D:
A square board is divided by a grid of horizontal and vertical lines into $n^{2}$ cells with a side length of 1. For what largest $n$ can $n$ cells be marked so that every rectangle of area at least $n$ with sides along the grid lines contains at least one marked cell? | Evaluation. It is clear that if $n$ cells are marked in such a way that the condition of the problem is satisfied, then in each row and each column there is exactly one marked cell. Assuming that $n \geq 3$ (it is obvious that $n=2$ is not the largest), let's take row $A$, in which the leftmost cell is marked, row $B$,... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.I.
Find the smallest natural number that can be represented as the sum of 2002 natural addends with the same sum of digits and as the sum of 2003 natural addends with the same sum of digits.
# | Let for a natural number $n$ the following representations hold: $n=a_{1}+\ldots+a_{2002}=b_{1}+\ldots+b_{2003}$.
We will use the fact that each of the numbers $a_{1}, a_{2002}$ gives the same remainder when divided by 9 as the sum of its digits; let this remainder be $r(0 \leq r \leq 8)$, and the corresponding remain... | 10010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Malinnikova E.
At the enterprise, 50,000 people are employed. For each of them, the sum of the number of their immediate supervisors and their immediate subordinates is 7. On Monday, each employee of the enterprise issues an order and gives a copy of this order to each of their immediate subordinates (if any). Then, e... | If in an enterprise there are $k$ supreme chiefs, then every worker should eventually see at least one of the $k$ orders issued by these chiefs on Monday. On Monday, no more than $7 k$ workers saw the orders, on Tuesday - no more than $7 k \cdot 6$, on Wednesday - no more than $7 k \cdot 36$ workers. All those who saw ... | 97 | Combinatorics | proof | Yes | Yes | olympiads | false |
Zamyatin B.
For what least $n$ can a square $n \times n$ be cut into squares $40 \times 40$ and $49 \times 49$ so that squares of both types are present? | Note that for $n=2000=40 \cdot 49+40$ the required cutting exists (see the left figure).
Assume that there exists a square $n \times n$, where $n>2000$. This leads to a contradiction.
## Ans... | 2000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.B.
55 boxers participated in a tournament with a "loser leaves" system. The fights proceeded sequentially. It is known that in each match, the number of previous victories of the participants differed by no more than 1. What is the maximum number of fights the tournament winner could have conducted? | We will prove by induction that
a) if the winner has conducted no less than $n$ battles, then the number of participants is no less than $u_{n+2}$;
b) there exists a tournament with $u_{n+2}$ participants, the winner of which has conducted $n$ battles ( $u_{k}-$ Fibonacci numbers).
Base case $\left(n=1, u_{3}=2\righ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Store A. }}$.
A straight stick 2 meters long was sawn into $N$ sticks, the length of each of which is expressed in whole centimeters. For what smallest $N$ can it be guaranteed that, using all the resulting sticks, one can, without breaking them, form the contour of some rectangle? | Let $N \leq 101$. We will cut a stick into $N-1$ sticks of length 1 cm and one stick of length $201-N$ cm. From the resulting set, it is impossible to form a rectangle, since each side of the rectangle must be less than half the perimeter, and thus the stick of length $201-N \geq 100$ cm cannot be part of any side.
Th... | 102 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Highimiy and.
In Anchuria, a day can be either clear, when the sun shines all day, or rainy, when it rains all day. And if today is not like yesterday, the Anchurians say that the weather has changed today. Once, Anchurian scientists established that January 1 is always clear, and each subsequent day in January will b... | Let's correspond to each January of the year an ordered set $\left(a_{1}, a_{2}, \ldots, a_{31}\right)$ of zeros and ones as follows: if the $k$-th of January is a clear day, then $a_{k}=1$, otherwise $a_{k}=0$. By condition, January 1st is always clear, so $a_{1}=1$. We will correspond to the weather set of the curren... | 2047 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Monotonicity, boundedness
Cauchy's inequality
$F(x)$ is an increasing function defined on the interval $[0,1]$. It is known that the range of its values belongs to the interval $[0,1]$. Prove that for any natural number $n$, the graph of the function can be covered by $N$ rectangles, the sides of which are parallel to... | Let's "plug" the "holes" in the graph with vertical segments (by connecting the left and right limits at the point of discontinuity). If we now swap the coordinate axes, we get the graph of a non-decreasing continuous function. Therefore, we can consider the original function to be continuous. For the same reasons, we ... | 290 | Calculus | proof | Yes | Yes | olympiads | false |
[ Properties of polynomial coefficients]
Find the sum of all coefficients of the polynomial $\left(x^{2}-3 x+1\right)^{100}$ after expanding the brackets and combining like terms.
# | The sum of the coefficients of a polynomial is equal to the value of the polynomial at $x=1$.
## Solution
This sum is equal to the value of the polynomial at $x=1$, that is, $(1-3 \cdot 1+1)^{100}=1$.
## Answer
1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ [ Linear inequalities and systems of inequalities] ]
The number x is natural. Among the statements
1) $2 x>70$,
2) $x>100$,
3) $3 x>25$,
4) $x \geq 10.5$,
5) $x>5$ three are true and two are false. What is the value of $x$? | Write down all inequalities with respect to $x$.
## Solution
Let's write the first and third inequalities as $x>35, x>25 / 3$. We will mark the numbers $5, 25 / 3, 10, 35$, 100 on the number line; these points divide the line into six intervals (two of which are infinite). It remains to consider all intervals and fin... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Environmentalists have protested against the large volume of logging. The chairman of the forestry company calmed them down as follows: "In the forest, $99\%$ are pines. Only pines will be cut, and after the logging, the percentage of pines will remain almost unchanged - pines will be 98\%". What part of the forest is ... | The number of "non-pines" remains constant, while the proportion of "non-pines" increased from one percent to two...
## Solution
We will call trees that are not pines oaks. Before the logging, oaks made up $1 \%$ of all trees, and after logging, they will make up $2 \%$. This means that the total number of trees will... | 53 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (other) $)]$
The scale's needle is off. When one bunch of bananas was placed on the scale, it showed 1.5 kg. When a larger bunch of bananas was placed on the scale, it showed 2.5 kg. When both bunches of bananas were weighed together, the scale showed 3.5 kg. What is the actual weight of the bunches of... | The pointer shifts twice for the total weight of $1.5+2.5=4$ kg, but only once for the weight of 3.5 kg. Therefore, the pointer shift is $4-3.5=0.5$ kg. Consequently, the correct weight of the bundles is half a kilogram less than what the scales show.
## Answer
1 and 2 kg. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ [motion problems ]
A cyclist left point $A$ for point $B$. At the same time, a pedestrian left point $B$ for point $A$ to meet the cyclist. After they met, the cyclist turned back, while the pedestrian continued his journey. It is known that the cyclist returned to point $A$ 30 minutes earlier than the pedestrian, a... | Before the meeting, the pedestrian had walked a distance five times less than the cyclist had ridden, that is, $1 / 6$ of the entire path. By the time the cyclist returned to point $A$, the pedestrian had walked the same distance again, meaning he still had $2 / 3$ of the path left. For this, he spent 30 minutes. Conse... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ [motion problems ]
A swimmer is swimming upstream against the current of the Neva River. Near the Palace Bridge, he lost an empty flask. After swimming another 20 minutes against the current, he noticed the loss and turned back to catch up with the flask; he caught up with it near the Lieutenant Schmidt Bridge. What... | Relative to the flask, the swimmer moves at a constant speed in both directions, so he will catch up with the flask in 20 minutes. Therefore, the current "travels" 2 km in 40 minutes.
## Answer
3 km $/$ h. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Motion task ]
Two hunters set out at the same time towards each other from two villages, the distance between which is 18 km. The first walked at a speed of 5 km/h, and the second at 4 km/h. The first hunter took a dog with him, which ran at a speed of 8 km/h. The dog immediately ran towards the second hunter, met h... | It will be two hours before the hunters meet, during which time the dog ran 16 km.
## Answer
16 km.
Prove that if $a$ and $b$ are integers and $b \neq 0$, then there exists a unique pair of numbers $q$ and $r$ such that $a = bq + r, 0 \leq r < |b|$. | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
To assemble a car, Leonid needed to buy several screws and dowels. When he approached the cash register, it turned out that the store was running a promotional offer, offering customers either a $15\%$ discount on the entire purchase or a $50\%$ discount on dowels. It turned out that the cost of the purchase with the d... | According to the condition, $15\%$ of the total purchase cost constitutes $50\%$ of the cost of the dowels. Therefore, the cost of the dowels is $30\%$ of the total purchase cost. Consequently, the (planned) costs of the screws and dowels are in the ratio of $7:3$.
## Answer
3 rubles.
Does there exist such an $x$ th... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
The sum of the subtrahend, minuend, and difference is 2016. Find the minuend.
# | The sum of the subtrahend and the difference is equal to the minuend.
## Otвет
1008. | 1008 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the competition, 50 shooters participated. The first scored 60 points; the second - 80; the third - the arithmetic mean of the points of the first two; the fourth - the arithmetic mean of the points of the first three. Each subsequent shooter scored the arithmetic mean of the points of all previous shooters. How man... | If a number equal to the arithmetic mean of a group of numbers is added to the group, then the arithmetic mean of the new group will be equal to the arithmetic mean of the initial group.
## Solution
The third player scored $(60+80): 2=70$ points. Each subsequent player also scored 70 points.
## Answer
70 points; 70... | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Average values $]$
Professor Tester conducts a series of tests, based on which he assigns the test subject an average score. After finishing his answers, John realized that if he had received 97 points for the last test, his average score would have been 90; and if he had received only 73 points for the last test, h... | A change of 24 points in the score changes the average by 3 points. This means there are $24 : 3 = 8$ tests.
## Answer
8 tests.
## Problem | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Motion Problems ]
A "Moskvich" car was sent from the post office to the airfield to transport the mail. The plane with the mail landed earlier than the scheduled time, and the mail was sent to the post office on a passing truck. After 30 minutes of driving, the truck met the "Moskvich" on the road, which took the ma... | How many minutes before the scheduled landing time did the "Moskvich" car meet the truck on the road?
## Solution
Since the "Moskvich" spent 20 minutes less on the entire trip (there and back), it spent 10 minutes less on the one-way trip. Therefore, the "Moskvich" met the truck 10 minutes before the scheduled landin... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The average age of eleven football players is 22 years. During the match, one of the players got injured and left the field. The average age of the remaining players on the field became 21 years. How old is the football player who got injured?
# | What is the total sum of the ages of 11 players on a team?
## Solution
First method. The total sum of the ages of 11 players is $11 \cdot 22 = 242$. After one player left, this sum became $10 \cdot 21 = 210$. By calculating the difference, we find that the player who left was 32 years old.
Second method. The sum of ... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Anthroov A.
What is the maximum number of white and black pawns that can be placed on a $9 \times 9$ checkerboard (a pawn, regardless of its color, can be placed on any cell of the board) so that no pawn attacks any other pawn (including those of the same color)? A white pawn attacks the two diagonally adjacent cells ... | An example with 56 pawns is shown in the figure.
Evaluation. Suppose in a rectangle of three rows and two columns, there are at least five pawns. Then on three cells of one color, there are... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
A natural number \$N\$ is divisible by 2020. In its decimal representation, all digits are distinct, and if any two of them are swapped, the resulting number is not divisible by 2020. For how many digits in the decimal representation of \$N\$ is this possible? | If the seventh digit from the right in a number is $\$ 2 \$$, and the third from the right is $\$ \mathrm{~b} \$$, then by swapping them, we change the number by \$b|cdot10^6-alcdot10^6+a|cdot10^2-blcdot10^2=(b-a)\} \backslash \operatorname { c d o t } ( 1 0 \wedge 4 - 1 ) \backslash \operatorname { c d o t } 1 0 0 \$ ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Does there exist a natural number that can be represented as the product of two palindromes in more than
100 ways? (A palindrome is a natural number that reads the same from left to right as from right to left.)
# | Consider a palindrome consisting of \$n\$ ones \$1_n = 1...1\$.
Method 1. If \$n\$ is divisible by \$k\$, then \$1_n\$ is divisible by the palindrome \$1_k\$, and the quotient is also a palindrome, consisting of ones separated by groups of \$k-1\$ zeros. It remains to choose a number \$n\$ that has more than 100 prope... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Neustroeva E.
In the relay race Moscow-Petushki, two teams of $\$ 20 \$$ people each participated. Each team divided the distance into $\$ 20 \$$ not necessarily equal segments and distributed them among the participants so that each ran exactly one segment (the speed of each participant is constant, but the speeds of... | First, let's prove that no more than $38$ overtakes occurred. Note that between the start and the first overtake, and between two consecutive overtakes, at least one of the teams must have changed the runner. There were $19$ runner changes in each team, which means a total of $38$, and thus, there were no more than $38... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kimoo A.V. The King walked around the chessboard, visiting each square exactly once and returning to the starting square on the last move. (The King moves according to the usual rules: in one move, he can move horizontally, vertically, or diagonally to any adjacent square.) When his path was drawn, sequentially connect... | The King made 64 moves. Since the length of each move is either 1 (a straight move) or $\sqrt{2}$ (a diagonal move), the total length of the path is certainly no less than 64. A path of length 64 is shown in the figure.
 rooks; b) queens that can be placed on an $8 \times 8$ chessboard so that each of these pieces is attacked by no more than one of the others? | a) Let $k$ rooks be placed while satisfying the condition. On each field where a rook stands, we write the number 0. In each of the $n$ columns, we perform the following operation: if there are two numbers in the column, we add 1 to both; if there is one number, we add 2 to it (we do nothing in an empty column). Then w... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
All integers are written in a row, starting from one. Determine which digit stands at the 206788th position.
# | Answer: the digit 7. There are exactly 9 single-digit numbers, 99 - 9 = 90 two-digit numbers, 999 - 99 - 9 = 900 three-digit numbers, 9000 four-digit numbers, and so on. Single-digit numbers will occupy the first 9 positions in the written sequence, two-digit numbers 90 * 2 = 180 positions, three-digit numbers 900 * 3 ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10,11
The game board has the shape of a rhombus with an angle of $60^{\circ}$. Each side of the rhombus is divided into nine parts. Lines are drawn through the division points, parallel to the sides and the short diagonal of the rhombus, dividing the board into triangular cells. If a chip is placed on a certain cell, ... | Let's replace the board with an equivalent $9 \times 9$ square board, where diagonals of the same direction are drawn in all cells (see Fig. 1).
Six chips are sufficient to cover all cells (see Fig. 1).
$. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Identical Transformations ]
Knowing that $x^{2}+x+1=0$, determine $x^{14}+1 / x^{14}$.
# | Given $x < 0$. Dividing the given equation term by term by $x$, we get $x+1 / x=-1$. Multiplying the given equation term by term by $x: x^{3}+x^{2}+x=0$, from which $x^{3}=-\left(x^{2}+x\right)=1$ (due to our equation).
$$
\begin{aligned}
x^{14}+1 / x^{14} & =\left(x^{3}\right)^{4} x^{2}+1 /\left(x^{3}\right)^{4} x^{2... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Дубановв И.с.
What is the maximum finite number of roots that the equation
$$
\left|x-a_{1}\right|+. .+|x-a 50|=\left|x-b_{1}\right|+. .+|x-b 50|
$$
can have, where $a_{1}, a_{2}, \ldots, a_{50}, b_{1}, b_{2}, \ldots, b_{50}$ are distinct numbers?
# | Let $f(x)=\left|x-a_{1}\right|+\ldots+|x-a 50|-\left|x-b_{1}\right|-. .-|x-b 50|$ and rewrite the original equation in the form $f(x)=0$.
Let $c_{1}<c_{2}<. .<c 100$ be all the numbers from the set $\left\{a_{1}, . ., a 50, b_{1}, \ldots, b 50\right\}$, ordered in ascending order. On each of the 101 intervals $\left[-... | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In some 16 cells of an $8 \times 8$ board, rooks were placed. What is the minimum number of pairs of rooks that could end up attacking each other? | Evaluation. Note that if there are $a$ rooks on a row, then there are $a-1$ pairs of rooks attacking each other in that row. Therefore, the number of pairs of rooks attacking each other horizontally is no less than the number of rooks minus the number of rows, which is no less than 8. Similarly, the number of pairs of ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Konagin S.V.
Find $x_{1000}$, if $x_{1}=4, x_{2}=6$, and for any natural $n \geq 3$, $x_{n}$ is the smallest composite number greater than $2 x_{n-1}-x_{n-2}$. | We will prove by induction that $x_{n}=1 / 2 n(n+3)$.
Base. For $n=3,4$, the formula is correct: $2 x_{2}-x_{1}=8$, so $x_{3}=9$; $2 x_{3}-x_{2}=12$, so $x_{4}=14$.
Induction step. $2 x_{n}-x_{n-1}=2 \cdot 1 / 2 n(n+3)-1 / 2(n-1)(n+2)=1 / 2(n+1)(n+4)-1$. By the condition, $x_{n+1}$ is the first composite number great... | 501500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\stackrel{\text { Senderov B.A. }}{ }$.
Find the smallest natural number that cannot be represented in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are natural numbers. | $1=\frac{4-2}{4-2}, 2=\frac{8-4}{4-2}, 3=\frac{8-2}{4-2}, 4=\frac{16-8}{4-2}, 5=\frac{32-2}{8-2}$
$6=\frac{16-4}{4-2}, \quad 7=\frac{16-2}{4-2}, 8=\frac{32-16}{4-2}, 9=\frac{128-2}{16-2}, \quad 10=\frac{64-4}{8-2}$.
Assume that $11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$. Without loss of generality, let $a>b, c>d$. Denote $... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A.
A natural number is written on the board. If the number $x$ is written on the board, then you can add the number $2 x+1$ or ${ }^{x} / x+2$. At some point, it was found that the number 2008 is on the board. Prove that it was there from the beginning. | We can assume that no "extra" numbers were written on the board, meaning all numbers "participated" in obtaining the number 2008. Note that all the numbers written are positive. Suppose at some point, a rational number is written on the board, in its irreducible form as
$x = \frac{p}{q}$. Then we can write the number ... | 2008 | Number Theory | proof | Yes | Yes | olympiads | false |
8,9 [ Investigation of a quadratic trinomial $]$ Authors: Kosukhin O.N., Bednov B.B. Three athletes started simultaneously from point $A$ and ran in a straight line to point $B$, each at their own constant speed. Upon reaching point $B$, each of them instantly turned around and ran back to the finish line at point $A$... | Let's assign numbers to the athletes in descending order of their starting speeds. We will draw graphs of their movements, plotting time on the x-axis and the distance to point $A$ on the y-axis. Let $O$ be the origin, $S$ the point on the y-axis corresponding to point $B \quad (O S=60 \mathrm{m}), K, L, M$ - points on... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{ll}{\left[\begin{array}{l}\text { Averages }\end{array}\right]} \\ {[\text { Pascal's Triangle and the Binomial Theorem }]}\end{array}\right]$
[ Products and Factorials $\quad]$
On board an airliner, there are $2 n$ passengers, and the airline has loaded $n$ portions of chicken and $n$ portions o... | a) The number of dissatisfied passengers can be any from 0 to $n$. In the case of $n=1$, everything is obvious: there is either no dissatisfied passenger or one, and both cases are equally likely. We will further assume that $n>1$.
Let us introduce the random variable $\xi$ "Number of dissatisfied passengers". $\xi=0$... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Akulich I.F. }}$
There are 40 identical gas cylinders, the pressure values in which are unknown to us and may be different. It is allowed to connect any cylinders to each other in a quantity not exceeding a given natural number $k$, and then disconnect them; in this case, the gas pressure in the co... | When $k=5$, the following method of equalizing pressures is suitable. By dividing the cylinders into 8 groups of 5 cylinders each, we equalize the pressures in the cylinders of each of these groups. Then we form 5 new groups, each consisting of 8 cylinders that previously belonged to different groups. We achieve equal ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a single-round football tournament, $n>4$ teams played. For a win, 3 points were awarded, for a draw 1, and for a loss 0. It turned out that all teams scored the same number of points.
a) Prove that there will be four teams with the same number of wins, the same number of draws, and the same number of losses.
b) F... | a) If two teams have scored the same number of points, then the difference in the number of draws between them is a multiple of 3.
The number of draws for a team is between 0 and $n-1$. Therefore, the number of groups, each of which consists of teams with the same number of wins, draws, and losses, does not exceed $k=... | 10 | Combinatorics | proof | Yes | Yes | olympiads | false |
Prove that among any ten consecutive natural numbers, there is a number that is coprime with the rest.
# | The greatest common divisor of two out of ten consecutive numbers does not exceed 9. Let's show that among 10 consecutive numbers, there will be one that is not divisible by the numbers 2, 3, 5, 7. It will satisfy the condition of the problem. Indeed, among these numbers, five are divisible by 2. Among the remaining nu... | 444 | Number Theory | proof | Yes | Yes | olympiads | false |
Find the smallest even natural number $a$ such that $a+1$ is divisible by $3, a+2$ by $5, a+3$ by $7, a+4$ by $11, a+5$ by 13. | First, let's find the numbers that satisfy the first two conditions: $a$ is even and $a+1$ is divisible by 3.
The smallest such number is -2. Therefore, the numbers that satisfy these conditions are of the form $6k+2$ (GCD $(2,3)=6)$.
Adding the third condition: we are looking for numbers of the form $6k+4$ that are ... | 788 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the remainder of the division of the polynomial $P(x)=x^{6 n}+x^{5 n}+x^{4 n}+x^{3 n}+x^{2 n}+x^{n}+1$ by $Q(x)=x^{6}+x^{5}+x^{4}+x^{3}+x^{2}$ $+x+1$, given that $n$ is a multiple of 7. | The difference $x^{k n}-1$ is divisible by $x^{7}-1$, and therefore also by $Q(x)$. Therefore, $P(x)-7=\left(x^{6 n}-1\right)+\left(x^{5 n}-1\right)+\ldots+\left(x^{n}-1\right)$ is divisible by $Q(x)$.
## Answer
7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Authors: Bogdanov I.i., Sukho K.
Given the polynomial $P(x)=a_{2 n} x^{2 n}+a_{2 n-1} x^{2 n-1}+\ldots+a_{1} x+a_{0}$, where each coefficient $a_{i}$ belongs to the interval $[100,101]$.
For what minimal natural $n$ can such a polynomial have a real root? | Example. Let's call a polynomial that satisfies the condition of the problem beautiful. The polynomial $P(x)=100\left(x^{200}+x^{198}+\right.$ $\left.\ldots+x^{2}+1\right)+101\left(x^{199}+x^{197}+\ldots+x\right)$ is beautiful and has a root -1. Therefore, for $n=100$ the required is possible.
Estimate. We will prove ... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Klepitsyn V.A. In the city of Flat, there is not a single tower. To develop tourism, the residents of the city are going to build several towers with a total height of 30 floors. Inspector Vysotnikov, climbing each tower, counts the number of shorter towers, and then adds up the resulting values. After that, the inspe... | Let some tower have a height $h>2$. We will remove the top floor from it, turning it into a new tower of height 1. How will this change the inspector's sum? From the shortened tower, towers of height $h-1$ are no longer visible. However, since $h>2$, from all these towers (including the shortened one), the new tower is... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Kosukhin O.N. }}$
Sasha found that there were exactly $n$ working number buttons left on the calculator. It turned out that any natural number from 1 to 99999999 can either be entered using only the working buttons, or obtained as the sum of two natural numbers, each of which can be entered using o... | Let's show that the conditions of the problem are met if the buttons with digits 0, 1, 3, $4, 5$ remain functional. Indeed, any digit from 0 to 9 can be represented as the sum of some two "working" digits. Let the number from 1 to 99999999 that we want to obtain consist of digits $a_{1}, a_{2}, \ldots, a_{8}$ (some of ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Golovnov A.S.
Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation o... | By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).
Then $a=\frac{m}{10^{30}-1}, \quad b=\frac{n}{10^{30}-1}$. We also ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Vasiniev N. 5.
Can natural numbers be placed in a $9 \times 9$ table such that the following conditions are simultaneously satisfied:
1) the products of the numbers in each row are the same for all rows;
2) the products of the numbers in each column are the same for all columns;
3) no two numbers are equal;
4) all nu... | As in problem 98107, we will construct a $9 \times 9$ Graeco-Latin square (see figure). In it:
1) each cell contains one Latin and one Greek letter;
2) in each row and each column, all letters are different;
3) in any two different cells, the pairs of letters are different.
| | $b$ | | $d \delta$ | | | $\eta$ | ... | 1121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Zvonkin }}$ D:
It is required to make a set of weights, each of which weighs an integer number of grams, with which it is possible to weigh any integer weight from 1 to 55 grams inclusive, even in the case where some weights are lost (the weights are placed on one pan of the scales, the weight to b... | a) The desired set of weights: $f_{1}=1, f_{2}=1, f_{3}=2, f_{4}=3, f_{5}=5, f_{6}=8, f_{7}=13, f_{8}=21, f_{9}=34, f_{10}=55$ (terms of the Fibonacci sequence, for which $\left.f_{n+2}=f_{n+1}+f_{n}\right)$.
By induction on $n$, we will prove that even after losing one weight from the set of weights $f_{1}, f_{2}, \l... | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Evenness and Oddness ] [ Monotonicity, Boundedness ]
Integers $a_{1}, a_{2}, \ldots, a_{n}$ are such that the equality
$$
a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}+\cdots+\frac{1}{a_{n}+\frac{1}{x}}}}=x
$$
is satisfied for all integer values of $x$ that belong to the domain of the fraction on the left side.
a) Prove t... | a) Let $n$ be odd. Consider the interval ( $b,+\infty$ ), where the number $b$ is greater than all the points of discontinuity of the function on the left side. The function $a_{n}+1 / x$ decreases on this interval, the function $a_{n-1}+\frac{1}{a_{n}+\frac{1}{x}}$ increases, ..., the function on the left side of the ... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
Shapovalov A.V.
In how many ways can the numbers from 1 to 100 be arranged in a $2 \times 50$ rectangle so that any two numbers differing by 1 always end up in cells that share a side? | Note that there is a one-to-one correspondence between the ways of arranging numbers and the rook's "tours" of the $2 \times 50$ board: if a method is given, we traverse the cells in the order of the numbers; if a tour is given, we number the cells in the order of the tour. This correspondence allows us to reason both ... | 4904 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Anton ran down a moving escalator and counted 30 steps. Then he decided to run up the same escalator at the same speed relative to the escalator and counted 150 steps. How many steps did he count when descending with a police officer on the stationary escalator?
# | Anton ran up 5 times longer than he ran down (because he counted 5 times more steps). At the same time, 5 times more steps came towards him than "ran away" when descending. If Anton runs down 5 times, he will count 150 steps, and the same number of steps will "run away" from him as came towards him during the ascent. T... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Galperin G.A.
For each pair of numbers $x$ and $y$, a certain number $x^{*} y$ is assigned. Find 1993*1935, given that for any three numbers $x, y, z$ the identities $x^{*} x=0$ and $x^{*}\left(y^{*} z\right)=\left(x^{*} y\right)+z$ hold. | $x=x^{*} x+x=x^{*}\left(x^{*} x\right)=x^{*} 0=x^{*}\left(y^{*} y\right)=x^{*} y+y$. Therefore, $x^{*} y=x-y$. Hence $1993 * 1935=1993-1935=58$.
## Answer
58. | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Golovanov A.S. $^{2}$
Given three reduced quadratic trinomials: $P_{1}(x), P_{2}(x)$, and $P_{3}(x)$. Prove that the equation $\left|P_{1}(x)\right|+\left|P_{2}(x)\right|=$ $\left|P_{3}(x)\right|$ has no more than eight roots. | Each root of the given equation is a root of one of the quadratic trinomials $\pm P_{1} \pm P_{2} \pm P_{3}$ with some set of signs. There are 8 such sets, and all of them yield genuine quadratic trinomials, since the coefficient of $x^{2}$ is odd. However, two opposite sets of signs correspond to quadratic equations t... | 8 | Algebra | proof | Yes | Yes | olympiads | false |
ерешин D.A.
Once, Rabbit was in a hurry to meet with Donkey Eeyore, but Winnie-the-Pooh and Piglet unexpectedly came to him. Being well-mannered, Rabbit offered his guests some refreshments. Pooh tied a napkin around Piglet's mouth and single-handedly ate 10 pots of honey and 22 cans of condensed milk, with each pot o... | Clearly, Pooh and Piglet must finish eating at the same time; otherwise, one of them can help the other, thereby reducing the total time spent eating. Pooh alone spent 42 minutes eating. By eating a pot of honey, Piglet reduces Pooh's time by 2 minutes but spends 5 minutes doing so himself. By eating two cans of milk, ... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalovo $A . B$. The numbers from 1 to 37 were written in a row such that the sum of any initial segment of numbers is divisible by the next number in the sequence.
What number is in the third position if the first position is occupied by the number 37, and the second by 1? # | Let the last place in the line be occupied by the number $x$. The sum of all numbers in the line, except for $x$, is divisible by $x$; therefore, the sum of all numbers in the line, which is
$1+2+\ldots+37=37 \cdot 19$, is also divisible by $x$. Hence, $x=19$, since 37 is already placed in the first position. The thir... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Several peasants have 128 sheep. If any one of them has no less than half of all the sheep, the others conspire and expropriate him: each takes as many sheep as they already have. If two of them have 64 sheep each, they expropriate one of them. Seven expropriations have occurred. Prove that all the she... | After the first expropriation, the number of sheep that everyone except the expropriated has is even, and the total number of sheep is also even, which means the remainder of the expropriated is also even. Similarly, after the second expropriation, the number of sheep each has is divisible by 4, after the seventh - by ... | 128 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
Chkhhanov N.X.
In the natural number $A$, the digits were rearranged to obtain the number $B$. It is known that $A-B=\underbrace{1 \ldots 1}_{n}$. Find the smallest possible value of $n$.
# | Numbers obtained from each other by rearranging digits have the same remainder when divided by 9, which means their difference is divisible by 9. Therefore, the sum of the digits of the difference, equal to $n$, is also divisible by 9, hence $n \geq 9$.
The value $n=9$ can be obtained, for example, as follows: $901234... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
Tanya took a list of one hundred numbers $1,2,3, \ldots, 100$ and crossed out several of them. It turned out that no matter which two numbers from the remaining ones Tanya took as $\$ a \$$ and $\$ b \$$, the equation $\$ x \wedge 2+a x+b=0 \$$ has at least one real root. What is the maximum number of n... | The equation $\$ x^{\wedge} 2+\mathrm{ax}+\mathrm{b}=0 \$$ has real roots if and only if its discriminant $\$ \mathrm{D}=\mathrm{a} \wedge 2-4 \mathrm{~b} \$$ is non-negative. This means that any two of the remaining numbers must satisfy the inequality \$ $\mathrm{2} \wedge 2$ \geqslant $4 b \$$.
And here is the most ... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
We took three numbers $x, y, z$. We calculated the absolute values of the pairwise differences $x_{1}=|x-y|, y_{1}=|y-z|, z_{1}=|z-x|$. In the same way, from the numbers $x_{1}, y_{1}, z_{1}$, we constructed the numbers $x_{2}, y_{2}, z_{2}$, and so on. It turned out that for some $n$, $x_{\mathrm{n}}=x, y_{\mathrm{n}}... | Answer: $y=z=0$. The numbers $x_{n}, y_{\mathrm{n}}, z_{\mathrm{n}}$ are non-negative, so the numbers $x, y, z$ are also non-negative. If all the numbers $x, y, z$ were positive, then the largest of the numbers $x_{1}, y_{1}, z_{1}$ would be strictly less than the largest of the numbers $x, y$, $z$, and then the larges... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
From point A, 100 planes (the flagship and 99 additional planes) take off simultaneously. With a full tank of fuel, a plane can fly 1000 km. In flight, planes can transfer fuel to each other. A plane that has transferred fuel to others makes a gliding landing. How should the flight be organized so that the flagship fli... | Let's describe the optimal fuel exchange procedure. Initially, 100 aircraft take off with full tanks. As soon as possible, one of the aircraft distributes its fuel to the others, after which 99 aircraft have full tanks, and the freed-up aircraft lands. By this point, the aircraft will have flown \(1000 \cdot 1 / 100\) ... | 5187 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Anjans A.
$a_{1}, a_{2}, a_{3}, \ldots$ - an increasing sequence of natural numbers. It is known that $a_{a_{k}}=3 k$ for any $k$.
Find
a) $a_{100}$
b) $a_{1983}$. | a) Immediately, we notice that the sequence $a_{k}$ is strictly increasing. Indeed, the assumption $a_{k}=a_{k+1}=n$ immediately leads to a contradiction: $a_{n}=3 k=3(k+1)$.
Moreover, $a_{1}>1$ (otherwise, $a_{a_{1}}=a_{1}=1 \neq 3$). From this, it follows that $a_{k}>k$ for all $k$. On the other hand, $a_{1}<a_{a_{1... | 181 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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