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$\underline{\text { Sabin A. }}$.
Ofenya bought a batch of pens at a wholesale market and offers customers either one pen for 5 rubles or three pens for 10 rubles. From each customer, Ofenya makes the same profit. What is the wholesale price of a pen? | According to the condition, the difference $10-5=5$ is the wholesale price of two pens.
## Answer
2 rubles 50 kopecks. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2 [Tournaments and tournament tables]
The Russian Chess Championship is held in a single round-robin format. How many games are played if 18 chess players participate?
# | The first method. Each participant must play 17 games, with two players in each game. Therefore, the total number of games is $18 \cdot 17: 2=153$.
The second method. In each game, one point is awarded. Suppose all games end in a draw. Then each participant will score $17: 2=8.5$ points. And the total number of points... | 153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$[$ Arithmetic of residues (other). ]
Find the remainder of the division of $2^{100}$ by 3. | $2^{100}=4^{50} \equiv 1^{50}=1(\bmod 3)$.
## Answer
1.
In a class of 30 people, can it be that 9 of them have 3 friends (in this class), 11 have 4 friends, and 10 have 5 friends?
## Solution
In the corresponding graph, there would be 30 vertices, 9 of which would have a degree of 3, 11 a degree of 4, and 10 a de... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left.\frac{[\text { Prime numbers and their properties }]}{[\underline{\text { Evenness and oddness }}]}\right]$
Find all such natural numbers $p$, that $p$ and $3 p^{2}+1$ are primes. | If $p$ is odd, then $3 p^{2}+1$ is even.
## Answer
$p=2$. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ]
Find the sum of the digits in the decimal representation of the number $4^{12} \cdot 5^{21}$.
# | Let's transform: $4^{12} \cdot 5^{21}=2^{3} \cdot 10^{21}=80 \ldots 0$.
## Answer
8. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5,6,7 |
Old cobbler Karl made boots and sent his son Hans to the market to sell them for 25 talers. At the market, two disabled men (one without a left leg, the other without a right leg) approached the boy and asked him to sell them each a boot. Hans agreed and sold each boot for 12.5 talers.
When the boy came home ... | Think about how much money Karl should have received, how much he actually received, and why.
## Solution
The disabled people paid 23 talers for the boots, but Karl received only 20, as Hans spent the remaining 3 talers on candies. Sitting in the pantry, Hans added the income (23 talers) to the expense (3 talers). Th... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Given a three-digit number ABB, the product of its digits is a two-digit number AC, the product of the digits of this number equals C (here, as in mathematical puzzles, the digits in the number notation are replaced by letters; identical letters correspond to identical digits, different letters correspond to different ... | Think about what $A$ is equal to.
## Solution
From the condition of the problem, it is clear that $A C=C$; then $A=1$ and $B B=10+C$, where $C$ is a digit. The last equation has a unique solution $B=4, C=6$. Therefore, the required number is 144.
## Answer
144. | 144 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7 chocolates are more expensive than 8 packs of cookies. What is more expensive - 8 chocolates or 9 packs of cookies?
# | Think about what is more expensive: $7 \cdot 8$ chocolate bars or $8 \cdot 8$ packs of cookies.
## Solution
7 chocolate bars are more expensive than 8 packs of cookies. Therefore, 56 chocolate bars cost more than 64 packs of cookies and even more than 63 packs of cookies. Hence,
8 chocolate bars cost more than 9 pac... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (miscellaneous).] $[$ Arithmetic. Mental arithmetic, etc. ]
In the room, there are 85 balloons — red and blue. It is known that: 1) at least one of the balloons is red; 2) in every arbitrarily chosen pair of balloons, at least one is blue. How many red balloons are in the room? | Think about whether there can be two red balls in the room.
## Solution
Since among any two balls, one is blue, there cannot be two red balls in the room. Therefore, there are 84 blue balloons and 1 red balloon in the room.
## Answer
1 ball. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
I walk from home to school in 30 minutes, while my brother takes 40 minutes. How many minutes will it take for me to catch up with my brother if he left the house 5 minutes before me? | If I had left 10 minutes later, I would have caught up with my brother in 30 minutes (by the school). Since 5 is half of 10, I will catch up with him twice as fast.
## Answer
In 15 minutes. | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2 [ Arithmetic. Mental calculation, etc. $\quad]$
One tea bag can brew two or three cups of tea. Milla and Tanya divided a box of tea bags equally. Milla brewed 57 cups of tea, and Tanya - 83 cups. How many tea bags could have been in the box? | Note: To brew 57 glasses, you need to have no less than $57: 3$ and no more than $57: 2$ tea bags.
## Solution
Mila brewed 57 glasses, which means she had no more than $57: 2=28.5$ tea bags. Tanya brewed 83 glasses, so she had no less than
$83: 3$ = 272/3 tea bags. Since the number of tea bags was the same (and an i... | 56 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Inequalities. Case Analysis ] [ Numerical Inequalities. Comparing Numbers. ]
Nine identical sparrows peck at fewer than 1001 grains, while ten such sparrows peck at more than 1100 grains. How many grains does each sparrow peck?
# | Notice, a sparrow can peck only an integer number of seeds.
## Solution
Since 10 sparrows peck more than 1100 seeds, then 9 sparrows will peck more than (1100 : 10) $9=990$ 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 ... | 111 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal number system ] [ Divisibility of numbers. General properties ]
In a 100-digit number 12345678901234...7890, all digits standing at odd positions were erased; in the resulting 50-digit number, all digits standing at odd positions were erased again, and so on. The erasing continued until there was nothing lef... | What characterizes the ordinal numbers of the digits that remain after the first deletion? And after the second?
## Solution
After the first deletion, only those digits will remain whose initial numbers are even, after the second, those whose initial numbers are divisible by 4, after the third - by 8, and so on. Befo... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
When dividing a certain number $m$ by 13 and 15, the same quotient was obtained, but the first division had a remainder of 8, while the second division had no remainder.
Find the number $m$. | How to determine how much the remainder of division by 13 is greater than the remainder of division by 15, if the quotient is known
## Solution
The number 13 is 2 less than 15. Therefore, with the same quotient \( n \), the remainder of division by 13 is \( 2n \) greater than the remainder of division by 15, i.e., \(... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In one American company, every employee is either a Democrat or a Republican. After one Republican decided to become a Democrat, the number of each became equal. Then, three more Republicans decided to become Democrats, and the number of Democrats became twice the number of Republicans. How many employees are there in ... | After the "conversion" of three Republicans, the number of Democrats at the firm became 6 more than the number of Republicans. According to the condition, this difference is equal to the number of Republicans. Therefore, the number of Republicans became 6, and the number of Democrats became 12.
## Answer
18 employees... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ $\left[\begin{array}{ll}\text { Irreducible fractions }\end{array}\right]$
What number should be subtracted from the numerator of the fraction ${ }^{537} / 463$ and added to the denominator to obtain $1 / 9$ after simplification? | Notice that $537+463=1000$.
## Solution
The sum of the numerator and the denominator will not change if the same number is subtracted from one and added to the other. Since this sum is 1000, the fraction before simplification should be $100 / 900$, and to obtain it, the number 437 must be subtracted from one and adde... | 437 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (miscellaneous).] [ Arithmetic. Mental arithmetic, etc. ]
Rabbits are sawing a log. They made 10 cuts. How many chunks did they get?
# | Into how many parts is a log divided by the first cut? How does the number of pieces change after each subsequent cut?
## Solution
The number of chunks is always one more than the number of cuts, since the first cut divides the log into two parts, and each subsequent cut adds one more chunk.
## Answer
11 chunks. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Recurrence relations ] $[\quad \underline{\text { C }}$ mean values $]$
Thirteen turkey chicks pecked at grains. The first turkey chick pecked 40 grains; the second - 60, each subsequent chick pecked the average number of grains pecked by all previous chicks. How many grains did the 10th turkey chick peck? | Recall the properties of the arithmetic mean.
## Solution
The third turkey chick pecked $(40+60): 2=50$ points. Each subsequent one also pecked 50 grains: if you add a number equal to the arithmetic mean of a group of numbers to that group, the arithmetic mean of the new group will be equal to the arithmetic mean of ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2 [Arithmetic. Mental arithmetic, etc.]
There are nuts in the 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? | Of course, we can set up a system of equations, but let's try to do without it.
## Solution
Let's combine both given conditions and obtain the following statement: "In the first and second boxes, there are 6 kg + 10 kg fewer nuts than in the first, second, and two-thirds of the third box." From this, it follows that ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Decimal numeral system]
A three-digit number. A three-digit number starts with the digit 4. If this digit is moved to the end of the number, the resulting number is 0.75 of the original number. Find the original number.
# | Let's write the equation $\underline{x} \underline{y}_{\mathbf{z}}=0.75 \cdot \underline{x} \underline{y}$, transform it to $\underline{x} \underline{y}_{\mathbf{2}} \cdot 10+4=0.75 \cdot\left(\underline{x} \underline{y}_{\mathbf{2}}+400\right)$, from which we get $\underline{x} \underline{y}_{\mathbf{z}}=32$.
## Answ... | 432 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a bookshelf, four volumes of Astrid Lindgren's collected works stand in order, with 200 pages in each volume. A worm living in this collection gnawed its way from the first page of the first volume to the last page of the fourth volume. How many pages did the worm gnaw through?
# | Try to recall how volumes from a collected works set stand on a bookshelf.
## Solution
Note that when the volumes are placed in order on the shelf, the first page of the 1st volume touches the last page of the 2nd volume, and the last page of the 4th volume touches the first page of the 3rd volume. Thus, the worm onl... | 400 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}\text { The ratio of volumes } \\ {[\text { Arithmetic. Mental calculation, etc. ]] }\end{array}\right]$
When Gulliver arrived in Lilliput, he found that everything there was exactly 12 times shorter than in his homeland. Can you tell how many Lilliputian matchboxes would fit into Gulliver's mat... | Notice that 12 Lilliputian matchboxes should fit into a Gulliver's matchbox in width, 12 - in length, and 12 - in height.
## Solution
In a Gulliver's matchbox, 12 Lilliputian matchboxes should fit in width, 12 - in length, and 12 - in height. In total, $12 \cdot 12 \cdot 12=1728$ matchboxes.
$$
\text { Send a commen... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.v.
In all entrances of the house, there is the same number of floors, and on each floor, there is the same number of apartments. The number of floors in the house is greater than the number of apartments per floor, the number of apartments per floor is greater than the number of entrances, and the number ... | $105=3 \cdot 5 \cdot 7$.
## Solution
Let $p$ denote the number of entrances in the building, $f$ the number of floors, and $k$ the number of apartments per floor. Then $p \cdot f \cdot k=105=3 \cdot 5 \cdot 7$, where the numbers $3,5,7-$ are prime. Considering that $1<p<k<f$, we get $f=7$.
## Answer
7 floors. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Andjensss A.
$N$ friends simultaneously learned $N$ pieces of news, with each person learning one piece of news. They started calling each other and exchanging news.
Each call lasts 1 hour. Any number of news items can be shared in one call.
What is the minimum number of hours required for everyone to learn all the ... | a) The news known to one of the friends will be known to no more than two (including the first) after 1 hour, no more than four after the second hour, ..., and no more than 32 after the 5th hour. Therefore, it will take no less than 6 hours.
We will show that 6 hours are sufficient. The conversations can proceed accor... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A number is guessed from 1 to 144. You are allowed to select one subset of the set of numbers from 1 to 144 and ask whether the guessed number belongs to it. For an answer of "yes," you have to pay 2 rubles, and for an answer of "no" - 1 ruble. What is the smallest amount of money needed to surely guess the number?
# | Let $a_{1}=2, a_{2}=3, a_{i}=a_{i-1}+a_{i-2}$ for $i \geq 2$. Then $a_{10}=144$. We will prove by induction that among no fewer than $a_{i}$ numbers, the guessed number cannot be found by paying less than $i+1$ rubles.
For $i=1$ and $i=2$, this is true.
Suppose there are no fewer than $a_{i}$ numbers. Then either the... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## Similar triangles (other). $\quad[$ [Ratio of areas of triangles with a common base or common height] $]$ Difficult [The area of a figure is equal to the sum of the areas of the figures into which it is divided Cauchy's inequality
The diagonals of a convex quadrilateral $ABCD$ intersect at point $E$. It is known th... | Prove that $B C \| A D$.
## Solution
According to problem 35162, $B C \| A D$.
Let $B C = x$. From the similarity of triangles $B E C$ and $D E A$, it follows that $S_{B E C} = \frac{B E}{E D} \cdot S_{D C E} = \frac{x}{3}, S_{D E A} = \frac{D E}{B E} \cdot S_{A B E} = \frac{3}{x}$.
By the condition $S_{A B C D} \l... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Chessboards and chess pieces $]$ [ Examples and counterexamples. Constructions ] [ Evenness and oddness
Authors: Pechkovsky A.N., Itenberg I.
Given an infinite grid paper with a cell side equal to one. The distance between two cells is defined as the length of the shortest path of a rook from one cell to another (t... | An example of coloring a square grid in four colors, such that any two cells at a distance of 6 are colored differently, is shown in the figure on the left.
Another example. Introduce a coor... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Vasilev N.B.
What is the maximum number of parts into which the coordinate plane $xy$ can be divided by the graphs of 100 quadratic trinomials of the form
$y=a_{n} x^{2}+b_{n} x+c_{n}(n=1,2, \ldots, 100) ?$ | We will prove by induction that $n$ parabolas of the specified type can divide the plane into no more than $n^{2}+1$ parts.
Base case. One parabola divides the plane into $2=1^{2}+1$ parts.
Inductive step. The $n$-th parabola intersects each of the others at no more than two points. These intersection points, of whic... | 10001 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
At the ends of the segment, two ones are written. In the middle between them, their sum - the number 2 - is written. Then, in the middle between each pair of adjacent written numbers, their sum is written again, and so on, 1973 times. How many times will the number 1973 be written?
# | Let's write down the lines of numbers that form as a result of each step. We will get some table. Let's try to find out how many times the number $n$ will appear in the $n$-th line of this table (it is clear that in each subsequent line - with a number greater than $n$ - the number $n$ will appear exactly as many times... | 1972 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Beshkariev V.P.
The sum of the tangents of angles measuring $1^{\circ}, 5^{\circ}, 9^{\circ}, 13^{\circ}, \ldots, 173^{\circ}, 177^{\circ}$ is 45. Prove this. | Author: Berkolayko S.T.
Let's use the known identity $\operatorname{tg} \alpha+\operatorname{tg}\left(\alpha+60^{\circ}\right)+\operatorname{tg}\left(\alpha-60^{\circ}\right)=3 \operatorname{tg} 3 \alpha$ (which can be verified by expressing both sides in terms of $\operatorname{tg} \alpha$). From this, it follows tha... | 45 | Algebra | proof | Yes | Yes | olympiads | false |
Given a $15 \times 15$ board. Some pairs of centers of adjacent cells by side are connected by segments such that a closed non-self-intersecting broken line is formed, which is symmetric with respect to one of the diagonals of the board. Prove that the length of the broken line is no more than 200. | Clearly, the broken line intersects the diagonal. Let $A$ be one of the vertices of the broken line lying on the diagonal.
We will move along the broken line until we first reach again a vertex $B$ lying on the diagonal. By symmetry, if we move along the broken line from $A$ in the other direction, $B$ will also be th... | 200 | Geometry | proof | Yes | Yes | olympiads | false |
Tokarev S.i.
A set of five-digit numbers $\left\{N_{1}, N_{k}\right\}$ is such that any five-digit number, all digits of which are in non-decreasing order, coincides in at least one digit with at least one of the numbers $N_{1}, N_{k}$. Find the smallest possible value of $k$. | A set with the specified properties cannot consist of a single number. Indeed, for each $N=\overline{a b c d e}$, there is a number $G=\overline{g g g 9 g}$ that differs from $N$ in all digits, where $g$ is a non-zero digit different from $a, b, c, d, e$. We will show that the numbers $N_{1}=13579$ and $N_{2}=12468$ fo... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Murashkin M.V.
In a racing tournament, there are 12 stages and $n$ participants. After each stage, all participants receive points $a_{k}$ depending on the place $k$ they occupy (the numbers $a_{k}$ are natural, and $a_{1}>a_{2}>\ldots>a_{n}$). For what smallest $n$ can the tournament organizer choose the numbers $a_{... | Evaluation. Suppose there are no more than 12 participants.
Let one of the participants (call him A) win all 11 stages, and each of the remaining participants at least once took the last place.
Then participant $A$ after 12 stages will score no less than $11 a_{1} + a_{n}$ points, while each of the remaining particip... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Inequalities problems. Case analysis ]
A circular road is divided into kilometer segments by poles, and it is known that the number of poles is even. One of the poles is painted yellow, another is painted blue, and the rest are painted white. We will call the distance between poles the length of the shortest arc con... | Let there be $-2n$ poles on a circular road. We calculate the sum of distances from the blue pole to all others: $2(1 + 2 + \ldots + (n-1)) + n = n(n-1) + n = n^2$. Therefore, $n^2 > 2008$. Since the distance from the blue pole to the yellow one does not exceed $n$, then $n^2 - n \leq 2008$, which means $n(n-1) \leq 20... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bverov S.L.
On some cells of a $10 \times 10$ board, $k$ rooks were placed, and then all cells that are attacked by at least one rook (a rook attacks the cell it stands on as well) were marked. For what largest $k$ can it happen that after removing any rook from the board, at least one marked cell will no longer be un... | Evaluation. Let's consider the placement of $k$ rooks satisfying the condition. There are two possible cases.
1) In each column, there is at least one rook. Then the entire board is under attack, and a rook can be removed from any column that has at least two rooks. Therefore, in this case, there is exactly one rook i... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Blinkov A.d:
a) In a football tournament in a round-robin format, 75 teams participated. A team received 3 points for a win, 1 point for a draw, and 0 points for a loss. It is known that every two teams scored a different number of points. Find the smallest possible difference in points between the teams that took fir... | b) The minimum gap between the first and last place cannot be less than $n-1$. We will prove that for $n>3$ it is possible to construct a tournament scheme to achieve a gap of $n-1$ (if $n=2$ or 3, the minimum gap is obviously 3 points).
By induction, we will construct a table for $n$ teams, where the results are all ... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Kanel-Belov A.Y. }}$
For each non-negative integer $i$, define the number $M(i)$ as follows: write the number $i$ in binary form; if the number of ones in this representation is even, then $M(i)=0$, and if it is odd - then 1 (the first terms of this sequence: $0,1,1,0,1,0,0,1, \ldots$).
a) Conside... | a) All binary numbers ending in 01 satisfy the condition $M(i)=M(i+1)$. Among four consecutive numbers, exactly one ends in 01, so there are 250 such numbers. In addition, numbers ending in 0111 (there are no fewer than $248: 4=62$) and ending in 011111 (there are no fewer than $60: 4=15$) also fit. In total, $250+62+1... | 325 | Combinatorics | proof | Yes | Yes | olympiads | false |
Tokarev S.i.
Can natural numbers from 1 to 81 be written in the cells of a $9 \times 9$ table so that the sum of the numbers in each $3 \times 3$ square is the same
# | Table $T$, depicted in the left figure, contains each of the numbers $1,2, \ldots, 9$ nine times and has the property that the sum of the numbers in each $3 \times 3$ square is 36.
| 1 | 4 | 7 | 1 | 4 | 7 | 1 | 4 | 7 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2 | 5 | 8 | 2 | 5 | 8 | 2 | 5 | 8... | 369 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Muamiev D:
What is the maximum number of colors in which all cells of a $10 \times 10$ board can be painted so that in each row and each column there are cells of no more than five different colors? | An example of coloring in 41 colors is shown in the figure (unmarked cells are colored in the 41st color).
| 1 | 2 | 3 | 4 | | | | | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 5 | 6 | 7 | 8 | | | | | |
| | | 9 | 10 | 11 | 12 | | | | |
| | | | 13 | 14 ... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Auto: GToooanov A.C.
The sum of the digits in the decimal representation of a natural number $n$ is 100, and the sum of the digits of the number $44 n$ is 800. What is the sum of the digits of the number $3 n$? | Note that $44 n$ is the sum of 4 instances of the number $n$ and 4 instances of the number $10 n$.
If we add these numbers digit by digit, then in each digit place, there will be the sum of the quadrupled digit from the same place in the number $n$ and the quadrupled digit from the next place. If no carries occur in t... | 300 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many roots does the equation $\sin x = x / 100$ have?
# | The problem is easily solved graphically. Since the graphs of the sine function and the function $y = x / 100$ are symmetric with respect to the origin, it is sufficient to consider the right part of the graphs. The maximum value of the sine function is 1. Therefore, the points of intersection of the graphs will be wit... | 63 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
The figure shows a track scheme for karting. The start and finish are at point $A$, and the karting driver can make as many laps as they want, returning to the starting point.
The young dr... | Let $M_{n}$ denote the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is either $A B$, $B A$, or a loop $B B$). Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
One can reach point $B$ in... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In Anchuria, a unified state exam is taking place. The probability of guessing the correct answer to each question on the exam is 0.25. In 2011, to obtain a certificate, one needed to answer... | If a graduate guesses the answers, the Unified State Exam (EGE) can be considered as a Bernoulli scheme with the probability of success $p=$ 0.25 and the probability of failure $q=0.75$. In 2011, to pass the exam, one needed to answer at least three questions correctly. It is more convenient to find the probability of ... | 2012 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
A natural number $a>1$ was appended to itself to form the number $b$, which is divisible by $a^{2}$. Find all possible values of the number $b / a^{2}$. | If the number \(a\) is \(n\)-digit, then \(\frac{b}{a^{2}}=\frac{\left(10^{n}+1\right) a}{a^{2}}=\frac{10^{n}+1}{a}\). Clearly, \(a \neq 10^{n-1}\), so \(1<\frac{10^{n}+1}{a}<10\). The number \(10^{n}+1\) (and hence the quotient) is not divisible by 2, 3 (the sum of the digits is 2), or 5, so
## Answer
$$
\begin{alig... | 140 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The equation with integer coefficients $x^{4}+a x^{3}+b x^{2}+c x+d=0$ has four positive roots, counting multiplicities.
Find the smallest possible value of the coefficient $b$ under these conditions. | Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of the equation (some of them may coincide). By Vieta's theorem, $b=x_{1} x_{2}+x_{1} x_{3}$ $+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}$
$d=x_{1} x_{2} x_{3} x_{4}$, hence $b$ and $d$ are positive. Notice that
$$
\frac{b}{\sqrt{d}}=\left(\sqrt{\frac{x_{1} x_{2}}{x_... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evvoconiov M.A.
An abstract artist took a wooden cube 5×5×5, divided each face into unit squares, and painted each of them in one of three colors - black, white, or red - so that there are no adjacent squares of the same color. What is the minimum number of black squares that could result? (Squares that share a side a... | Evaluation. Three squares at the vertex of a cube form a cycle of adjacent squares of length 3. Around it, another cycle of length 9 is formed from adjacent cells. And around that, a cycle of length 15. Taking three such cycles around two opposite vertices of the cube, and two small cycles around the remaining vertices... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ $\quad$ Number of divisors and their sum of a number [Fundamental Theorem of Arithmetic. Factorization into prime factors ]
a) Find the number $k$, which is divisible by 2 and 9 and has exactly 14 divisors (including 1 and $k$ ).
b) Prove that if 14 is replaced by 15, the problem will have multiple solutions, and i... | Let $M=p_{1}^{\alpha_{1}} \ldots p_{r}^{\alpha_{r}}$ be the prime factorization of the number $M$; for definiteness, we will assume that $\alpha_{1} \geq \alpha_{2} \geq \ldots \geq \alpha_{r} \geq 1$. According to the formula from problem $\underline{60537}$ a) the total number of divisors of $M$ is $\left(\alpha_{1}+... | 1458 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$[\underline{\text { equations in integers }}]$
Solve the equation $\underbrace{\sqrt{n+\sqrt{n+\ldots \sqrt{n}}}}_{1964 \text { times }}=m$ in integers
# | Let's solve the more general equation $A_{y}(x)=z$ in integers, where $A_{y}(x)=\underbrace{\sqrt{x+\sqrt{x+\ldots \sqrt{x}}}}_{y \text { times }}$. Clearly, $x+A_{y-1}(x)=z^{2}$, or $A_{y-1}(x)=z^{2}-x$.
Thus, if the number $A_{y}(x)=z$ is an integer, then the number $A_{y-1}(x)=\left(z^{2}-x\right)$ is also an integ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all natural numbers $x$, satisfying the conditions: the product of the digits of the number $x$ is equal to $44x - 86868$, and the sum of the digits is a cube of a natural number. | Let $\overline{a_{n}^{\ldots}-\ldots a_{0}^{-}}$ be the decimal representation of the number $a$. Then $a \geq a_{n} \cdot 10^{n} \geq a_{n} a_{n-1} \cdot 9^{n} \geq a_{n} a_{n-1} \ldots a_{0}$. From this it follows that $x \geq$ $44 x$ - 86868, from which $x \leq 2020$. On the other hand, the product of the digits of ... | 1989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Khachaturyan A.V. }}$
13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." That ch... | It is clear that there were both boys and girls at the table. A group of boys sitting next to each other is followed by a group of girls, then boys again, then girls, and so on (a group can consist of one person). Groups of boys and girls alternate, so their number is even. The incorrect statements were made at the tra... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Some text was encrypted by mapping each letter to some (possibly the same) letter in such a way that the text can be uniquely decrypted. Prove that there exists a number $\mathrm{N}$ such that after $\mathrm{N}$-fold application of the encryption, the original text will definitely be obtained. Find the smallest such va... | If the letter $a_{1}$ is encrypted into the letter $a_{2}$, the letter $a_{2}$ is encrypted into the letter $a_{3}, \ldots$, the letter $a_{k}$ is encrypted into the letter $a_{1}$, then after $k-$ times encryption we will get the original text.
## Solution
According to the condition, the encryption allows for a uniq... | 144403552893600 | Number Theory | proof | Yes | Yes | olympiads | false |
Shapovalov A.V.
The price of a standard lunch at the "Buratino" tavern depends only on the day of the week. Anya dined for 10 consecutive days starting from July 10 and paid 70 soldo. Vanya also paid 70 soldo for 12 meals starting from July 12. Tanya paid 100 soldo for 20 meals starting from July 20. How much will San... | Let's create a calendar from July 10 to August 16, conditionally considering August 10 as the first day of the week. Each row of the calendar corresponds to the same day of the week (see Fig. a). Let the cost of lunch on the first day of the week be $a_{1}$ soldo, on the second day $-a_{2}$ soldo, ..., on the seventh d... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
16 cards with integers from 1 to 16 are laid face down in a \(4 \times 4\) table so that cards with consecutive numbers are adjacent (touching by a side). What is the minimum number of cards that need to be flipped simultaneously to definitely determine the location of all numbers (regardless of how the... | Evaluation. Number the cells as shown in Figure 1.
| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 2 | 1 | 4 | 3 |
| 5 | 6 | 7 | 8 |
| 6 | 5 | 8 | 7 |
Fig. 1
Fig. 2
$-th list contained those who were in the $k$-th list and those they had defeated. It turned out that for eac... | Consider a directed graph where the vertices are chess players, and arrows point from the winner to the loser. The condition means that for each chess player, there is another player who can be reached by exactly 11 arrows (this, in particular, means that from each chess player, one can reach any other player). Conside... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the largest natural number in which each non-edge digit is less than the arithmetic mean of the adjacent digits.
# | The differences between adjacent digits of a number decrease when viewing the decimal representation of the number from left to right.
## Solution
Let \(a_{1} a_{2} \ldots a_{k}\) be the decimal representation of a number, each non-edge digit of which is less than the arithmetic mean of the adjacent digits. Then
\(a... | 96433469 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Gooovanov A.S.
In an infinite increasing sequence of natural numbers, each number is divisible by at least one of the numbers 1005 and 1006, but none are divisible by 97. Additionally, any two consecutive numbers differ by no more than \( k \). What is the smallest \( k \) for which this is possible? | Let's denote our sequence as $\left(a_{n}\right)$. It is clear that $a_{1}D$ (while $a_{n} \neq D$ by the condition). However, the largest numbers less than $D$ and divisible by 1005 and 1006 are $D-1005$ and $D-1006$, respectively; therefore, $a_{n} \leq D - 1005$. Similarly, $a_{n+1} \geq D + 1005$; hence, $a_{n+1} -... | 2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bervov S.l.
In some cells of a $10 \times 10$ board, $k$ rooks were placed, and then all cells that are attacked by at least one rook were marked (it is assumed that a rook attacks the cell it stands on). For what largest $k$ can it happen that after removing any rook from the board, at least one marked cell will no l... | Let's consider the placement of $k$ rooks satisfying the condition. There are two possible cases.
1. Suppose there is at least one rook in each column. Then the entire board is under attack, and a rook can be removed from any column that has at least two rooks. Therefore, in this case, there is exactly one rook in eac... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tasks with inequalities. Case analysis Examples and counterexamples. Constructions
Author: Shapovosov A.B.
The plan of the shah's palace is a $6 \times 6$ square, divided into rooms of size $1 \times 1$. There is a door in the middle of each wall between the rooms. The shah told his architect: "Knock down some walls... | Consider an arbitrary path from the bottom-left corner of the palace to the top-right corner. Since it is necessary to "climb" 5 horizontal levels and "shift to the right" 5 vertical levels, one would have to pass through at least 10 doors, visiting at least 11 rooms (including the starting and ending rooms).
11 rooms... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $A$ be the sum of the digits of the number $4444^{4444}$, and $B$ be the sum of the digits of the number $A$. Find the sum of the digits of the number $B$.
# | The sum of the digits is comparable to $4444^{4444} \equiv (-2)^{4444} = (2^6)^{740} \cdot 2^4 \equiv 1^{740} \cdot 7 \pmod{9}$. Moreover, $4444^{4444} < 10^{4 \cdot 5000}$, meaning this number has no more than 20000 digits. Therefore, $A \leq 180000, B \leq 45$, and the sum of the digits of the number $B$ is no more t... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many natural numbers $x$, less than 10000, are there for which $2^{x}-x^{2}$ is divisible by 7? | The remainders of the division by 7 of the numbers $2^{x}$ and $x^{2}$ repeat with periods of 3 and 7, respectively, so the remainders of the division by 7 of the numbers $2^{x}$ and $x^{2}$ repeat with a period of 21. Among the numbers $x$ from 1 to 21, equal remainders from the division by 7 of the numbers $2^{x}$ an... | 2857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8,9,10 |
| | $[$ Rating + example $]$ | |
Auto:: Shapovesov A.B.
There is a board $1 \times 1000$, initially empty, and a pile of $n$ chips. Two players take turns. The first player, on their turn, "places" no more than 17 chips on any free cell of the board (they can take all 17 from the pile, or part of them - fr... | a) Let's outline the strategy of the first player. Initially, he builds 12 series of 8 chips each, such that adjacent series are separated by one space, sequentially restoring a removed series and placing another one. Then, after restoring the configuration following the second player's move, he inserts two chips into ... | 98 | Other | proof | Yes | Yes | olympiads | false |
Razin M.
There is a set of 20 weights with which any integer weight from 1 to 1997 g can be measured (weights are placed on one pan of the scales, the weight to be measured - on the other). What is the minimum possible weight of the heaviest weight in such a set, if:
a) the weights in the set are all integers,
b) th... | Let's order the weights of the weights (in grams) in ascending order: $p_{0} \leq p_{1} \leq \ldots \leq p_{19}$. Clearly, $p_{0} \leq 1$ (otherwise, it's impossible to weigh a load of 1 g). Similarly, $p_{1} \leq 2$ (to weigh 2 g). Since these two weights are not enough to weigh 4 g, $p_{2} \leq 4=2^{2}$. Continuing t... | 146 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Knop K.A.
The re-attestation of the Council of Sages happens as follows: the king lines them up in a single column and puts a cap on each one, either white, blue, or red. All sages can see the colors of the caps of all the sages in front of them, but they cannot see the color of their own cap or those of the sages beh... | All 100 sages cannot be guaranteed to survive since no one among them can see the color of the last sage's hat.
Let's show how 99 sages can be saved. We will assign the numbers 0, 1, and 2 to the white, blue, and red colors, respectively. In the first minute, the sage standing at the back should shout out the color co... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.I.
The path from platform $A$ to platform $B$ was traveled by the electric train in $X$ minutes ($0<X<60$). Find $X$, given that the angle between the hour and minute hands was $X$ degrees both at the moment of departure from $A$ and at the moment of arrival at $B$. | Since more than an hour passes between two consecutive overtakes of the hour hand by the minute hand, no more than one overtake occurred during the specified time of the electric train's movement. Let $O$ be the center of the clock face, $T_{A}$ and $T_{B}$ be the points where the tip of the hour hand was located when ... | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Let $\Phi$.
On the board, $N \geq 9$ different non-negative numbers, each less than one, are written. It turns out that for any eight different numbers on the board, there is a ninth, different from them, such that the sum of these nine numbers is an integer. For which $N$ is this possible? | Clearly, when $N=9$, the required is possible - it is sufficient to write 9 different positive numbers with a unit sum on the board. We will show that when $N>9$, the required is impossible.
Assume the contrary; let $S$ be the sum of all numbers on the board. Choose any numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.I.
Doughnut was snacking at a roadside cafe when a bus drove past him. Three doughnuts after the bus, a motorcycle passed Doughnut, and another three doughnuts later, a car passed. Past Syrup, who was snacking at another cafe on the same road, they drove by in a different order: first the bus, three doughnut... | Let at the moment when the bus passed by Donut, the motorcycle had $x$ km left to the first cafe. This means that in the time of three donuts, the car travels $x$ km. The car moves twice as fast as the motorcycle and in the same time travels $2x$ km. It travels the same distance in the next time of three donuts. Theref... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Solve the equation $2 \sin \pi x / 2 - 2 \cos \pi x = x^{5} + 10 x - 54$.
# | Transferring all terms to one side, we get the equation $x^{5}+10 x-54-2 \sin \pi x / 2+2 \cos \pi x=0$. Consider the function
$f(x)=x^{5}+10 x-54-2 \sin \pi x / 2+2 \cos \pi x$. Note that $f^{\prime}(x)=5 x^{4}+10-\pi \cos \pi x / 2-2 \sin \pi x>5 x^{4}+10-3 \pi>0$. Therefore, $f(x)$ is an increasing function, so the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
From Zlatoust to Miass, a "GAZ", a "MAZ", and a "KamAZ" set off simultaneously. The "KamAZ", upon reaching Miass, immediately turned back and met the "MAZ" 18 km from Miass, and the "GAZ" - 25 km from Miass. The "MAZ", upon reaching Miass, also immediately turned back and met the "GAZ" 8 km from Miass. What is the dist... | Let the distance between the cities be $x$ km, and the speeds of the trucks: "GAZ" - $g$ km/h, "MAZ" - $m$ km/h, "KAMAZ" - $k$ km/h. For each pair of vehicles, we equate their travel time until they meet. We get
$\frac{x+18}{k}=\frac{x-18}{m}, \frac{x+25}{k}=\frac{x-25}{g}$ and $\frac{x+8}{m}=\frac{x-8}{g}$. From this... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Numerical tables and their properties ]
Each cell of a $7 \times 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells, and not less than the number of green cells, while in each column, the number of... | 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones.
In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones.
Thus, in the table, ther... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.g.
What is the maximum number of knights that can be placed on a chessboard so that each one attacks no more than seven of the others
# | Example of placing 60 knights: all cells are occupied except for the cells of the central $2 \times 2$ square.
We will show that with a placement satisfying the condition, no fewer than four cells will remain free (i.e., it is impossible to place more than 60 knights). Consider the central $4 \times 4$ square. If all ... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$Among the actors of Karabas Barabas theater, a chess tournament was held. Each participant played exactly one game with each of the others. One solido was given for a win, half a solido for a draw, and nothing for a loss. It turned out that among any three participants, there would be a ... | Example. Let's denote the participants with letters A, B, V, G, D. Suppose A won against B, B won against V, V won against G, G won against D, D won against A, and all other matches ended in a draw. The condition of the problem is satisfied.
Evaluation. From the condition, it follows that for this tournament, two stat... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Eevokimov $M$.
In each cell of a $5 \times 5$ board, there is either a cross or a zero, and no three crosses stand in a row either horizontally, vertically, or diagonally. What is the maximum number of crosses that can be on the board? | Example of placing 16 crosses according to the condition, see the figure on the left.
| $x$ | $x$ | 0 | $x$ | $x$ |
| :---: | :---: | :---: | :---: | :---: |
| $x$ | $x$ | 0 | $x$ | $x$ |
| 0 | 0 | 0 | 0 | 0 |
| $x$ | $x$ | 0 | $x$ | $x$ |
| $x$ | $x$ | 0 | $x$ | $x$ |
. Find the maximum po... | Example. Let's highlight the "border" of width 5 in a $200 \times 200$ square. This border consists of four corner squares $5 \times 5$ and four rectangles $5 \times 190$. Place 3800 chips in these four rectangles: red chips in the left and top rectangles, and blue chips in the right and bottom rectangles. It is easy t... | 3800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Number Theory. Divisibility (other).] Decimal representation of a number $\quad]$
Auto: E.Eogokimo M.A.
We will call a nine-digit number beautiful if all its digits are different.
Prove that there exist at least a) 1000; b) 2018 beautiful numbers, each of which is divisible by 37. | a) Any nine-digit number $M$ is equal to the sum $10^6 A + 10^3 B + C = 999(1001 A + B) + (A + B + C)$, where $A$, $B$, and $C$ are numbers formed by the first three, the next three, and the last three digits of the number $M$, respectively. If we divide the digits from 1 to 9 into three triplets with a sum of 15 in ea... | 20736 | Number Theory | proof | Yes | Yes | olympiads | false |
Some cells of a $100 \times 100$ board are painted black. In all rows and columns that contain black cells, the number of black cells is odd. In each row that contains black cells, we place a ... | Remove from the board all rows and columns that do not contain black cells. Note that after this, the condition of the problem will still be satisfied. Now all black cells lie within some rectangle $n \times m$, in each row and each column of which there is at least one black cell.
By the condition, there is exactly o... | 1515 | Combinatorics | proof | Yes | Yes | olympiads | false |
[Problems on percentages and ratios $]$ Dirichlet's Principle (etc.) $\quad]$
## Author: : Sergei I.n
A council of 2000 deputies decided to approve the state budget, which contains 200 expenditure items. Each deputy prepared their own budget proposal, in which they indicated the maximum allowable amount of expenditur... | If $k \leq 1990$, it may happen that the first 10 deputies propose to allocate nothing for the first item of expenditure and allocate $S / 199$ for the rest. The next 10 deputies will allocate nothing for the second item and allocate $S / 199$ for the rest, and so on. As a result, an expenditure amount of $S / 199$ wil... | 1991 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Classical inequalities ]
Specify any positive integer $n$ for which
a) $1.001^{n}>10$;
b) $0.999^{n}<0.1$. | a) According to Bernoulli's inequality (see problem $\underline{30899})\left(1+\frac{1}{1000}\right)^{10000}>1+10000 \cdot \frac{1}{1000}>10$.
b) $\left(1-\frac{1}{1000}\right)\left(1+\frac{1}{1000}\right)<1$, therefore, $\left(1-\frac{1}{1000}\right)^{10000}<\left(1+\frac{1}{1000}\right)^{-10000}<1 / 10$.
## Answer
... | 10000 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Folkpor
Five identical balls are moving in one direction in a straight line at some distance from each other, and five other identical balls are moving towards them. The speeds of all the balls are the same. When any two balls collide, they fly apart in opposite directions with the same speed they had before the colli... | After the collision, the balls scatter at the same speed, so the situation will not change if we allow the balls to pass through each other upon collision, maintaining their speed. Then each ball rolling "to the right" will meet each of the balls rolling "to the left" once, meaning there will be 25 encounters.
## Answ... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ $\quad$ Rebus $\quad]$ [ Arithmetic progression ] Avor: Akumich i.f. To a natural number $A$, three digits were appended on the right. The resulting number turned out to be equal to the sum of all natural numbers from 1 to $A$. Find $A$.
# | Let the appended digits form the number $B, 0 \leq B \leq 999$. Then the resulting number is, on one hand, $1000 A+B$, and on the other hand, $-1+2+\ldots+A=\frac{1}{2} A(A+1)$. The equation $1000 A+B=\frac{1}{2} A(A+1)$ transforms into $A(A-1999)=2 B$, from which $0 \leq A(A-1999) \leq 1998$. Since the left inequality... | 1999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Combinations and Permutations ]
[ Decimal Number System ]
How many six-digit numbers exist where each subsequent digit is less than the previous one? | The first method. Each such number corresponds uniquely to the selection of six digits from the number 9876543210.
The second method. To obtain a number with "decreasing digits," you need to strike out any four digits from the number 9876543210.
## Answer
$C_{10}^{6}=C_{10}^{4}=210$ numbers. | 210 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ $[$ Pascal's Triangle and the Binomial Theorem ] [ Decimal Number System $]$
Why do the equalities $11^{2}=121$ and $11^{3}=1331$ resemble rows of Pascal's Triangle? What is $11^{4}$ equal to?
# | $11^{2}=(10+1)^{2}=100+2 \cdot 10+1,11^{3}=(10+1)^{3}=1000+3 \cdot 100+3 \cdot 10+1$.
$11^{4}=C_{4}^{0} \cdot 10^{4}+C_{4}^{1} \cdot 10^{3}+C_{4}^{2} \cdot 10^{2}+C_{4}^{3} \cdot 10+C_{4}^{4}=14641$
## Answer
$11^{4}=14641$. | 14641 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Evenness and Oddness }\end{array}\right]$
Several consecutive natural numbers are written on the board. Exactly 52\% of them are even. How many even numbers are written on the board? | Since the recorded natural numbers are consecutive, even and odd numbers alternate. According to the condition, there are more even numbers, which means the recorded sequence starts and ends with even numbers.
There is one more even number, meaning one number constitutes (52 - 48)% of their total quantity.
Therefore,... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
86088 Topics
Cockroach Valentin announced that he can run at a speed of 50 m/min. No one believed him, and rightly so: Valentin had everything mixed up and thought there were 60 cm in a meter and 100 seconds in a minute. What is the actual speed (in "normal" m/min) at which Cockroach Valentin runs? | Valentin runs $50 \cdot 60=3000$ cm in 100 s, which means his speed is 30 cm/s, or 18 m/min.
## Answer
18 m/min. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ] [Arithmetic. Mental calculation, etc.]
Given the number: $123456789101112 \ldots$. What digit is in the 2000th place? | Let's find out from which point the decimal representation of the given number will start to include three-digit numbers: 2000 - $9 \cdot 1$ $90 \cdot 2=1811$.
$1811: 3$ = 603 (remainder 2), which means that the 2000th position is the second digit of the 604th three-digit number. This number is 703, so the required di... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal number system ] $[$ Arithmetic. Mental calculation, etc. ]
A part of a book fell out. The first of the missing pages has the number 387, and the number of the last page consists of the same digits but in a different order. How many sheets fell out of the book? | Notice that when a part of a book falls out, the first of the fallen pages has an odd number, and the last one has an even number.
## Solution
When a part of a book falls out, the first of the fallen pages has an odd number, and the last one has an even number (each page is numbered on both sides, and two numbers are... | 176 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Union, intersection, and difference of sets ] Problems on percentages and ratios $\quad]$
In the gymnasium, all students know at least one of the ancient languages - Greek or Latin, some know both languages. $85 \%$ of all students know Greek and $75 \%$ know Latin. What fraction of the students know both languages? | $100-85=15 \%$ of all the guys do not know Greek, which means they only know Latin. This means that $75-15=$ $60 \%$ speak both languages.
## Answer
$60 \%$.
35 ravens flew into the glade. Unexpectedly, the ravens took off and split into two flocks: one flock perched on the branches of an old birch tree, and the oth... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The bathtub fills with cold water in 6 minutes 40 seconds, and with hot water in 8 minutes. In addition, if the plug is removed from a full bathtub, the water will drain in 13 minutes 20 seconds. How long will it take to fill the bathtub completely, given that both taps are open, but the bathtub is not plugged?
# | Let's replace the time in seconds with time in minutes: 6 minutes 40 seconds will be replaced by $62 / 3$, or $20 / 3$, and 13 minutes 20 seconds by $40 / 3$. In one minute, the cold water will fill $3 / 20$ of the bathtub, the hot water will fill $-1 / 8$ of the bathtub, and $3 / 40$ of the bathtub will drain. Therefo... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Three cowboys walked into a saloon. One bought 4 sandwiches, a cup of coffee, and 10 donuts for a total of $1.69. The second bought 3 sandwiches, a cup of coffee, and 7 donuts for $1.26. How much did the third cowboy pay for a sandwich, a cup of coffee, and a donut? | How much do 8 sandwiches, 2 cups of coffee, and 20 donuts cost? And how much do 9 sandwiches, 3 cups of coffee, and 21 donuts cost?
## Solution
From the condition, it is clear that 8 sandwiches, 2 cups of coffee, and 20 donuts cost $2 \cdot 1.69 = 3.38$, and 9 sandwiches, 3 cups of coffee, and 21 donuts $-3 \cdot 1.2... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the books of Novgorod scribes of the $X V$ century, such measures of liquid are mentioned: barrel, nozzle, and bucket. From these same books, it became known that a barrel and 20 buckets of kvass are equal to three barrels of kvass, and 19 barrels, a nozzle, and 15.5 buckets are equal to twenty barrels and eight buc... | From the first condition, it follows that the barrel's capacity is 10 buckets, and from the second condition, that 7.5 buckets and one nozzle can fit into the barrel. Therefore, one nozzle holds 2.5 buckets, or a quarter of the barrel, meaning that 4 nozzles can fit into the barrel.
## Answer
They can. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find numbers equal to twice the sum of their digits.
# | $10 x+y=2(x+y)$. Or $8 x=y$. Since $y$ is a digit, the only answer is 18.
## Answer
18. | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Word Problems (miscellaneous) }) \\ {[\text { Common Fractions }}\end{array}\right]$
A flock of geese was flying, and a single goose was flying towards them and said: "Greetings, a hundred geese!" The leader of the flock replied to him: "No, we are not a hundred geese! You see, if there ... | Try to solve this problem using "parts".
## Solution
Let "so many" be 4 parts. Then "so many plus so many, plus half so many, plus a quarter so many" will be $4+4+2+1=11$ parts. This number of parts is equal to 99 geese. Therefore, one part is 9 geese, and there were 36 geese in the flock.
## Answer
36 geese. | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kovaldzhi A.k.
A hundred people answered the question: "Will the new president be better than the previous one?" Of them, $a$ people think it will be better, $b$ - that it will be the same, and $c$ - that it will be worse. Sociologists constructed two "optimism" indicators of the respondents: $m=a+b / 2$ and $n=a-c$. ... | We know that $a+b+c=100$ and $2a+b=80$. By subtracting the first equation from the second, we have: $a-c=-20$.
## Otвет
$n=-20$. | -20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7 wolves eat 7 sheep in 7 days. How many days will it take for 9 wolves to eat 9 sheep?
# | The number of wolves has increased by the same factor as the number of sheep, so the time of consumption will not change.
## Answer
In 7 days. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental arithmetic, etc.] $[$ Work problems $]$
Three diggers dug three holes in two hours. How many holes will six diggers dig in five hours?
# | First, calculate how many pits 6 diggers will dig in 2 hours.
## Solution
Six diggers will dig $2 \cdot 3=6$ pits in 2 hours. And in 5 hours, they will dig $2.5 \cdot 6=15$ pits.
## Answer
15 pits. | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left[\right.$ Arithmetic. Mental calculation, etc.] $\left[\begin{array}{l}\text { Work problems. }\end{array}\right]$
Three diggers dug three holes in three hours. How many holes will six diggers dig in five hours? | First, calculate how many pits 6 diggers will dig in 3 hours.
## Solution
Six diggers in 3 hours will dig $2 \cdot 3=6$ pits. And in 5 hours, they will dig $5 / 3 \cdot 6=10$ pits.
## Answer
10 pits. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A natural number was sequentially multiplied by each of its digits. The result was 1995. Find the original number.
# | Decompose the number 1995 into prime factors.
## Solution
$1995=3 \cdot 5 \cdot 7 \cdot 19$. We need to split this product into two groups: some factors will form the original number, and the other part will be its digits. It is clear that 19 will be part of the number (digits "19": no!). The remaining is a simple en... | 57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Several gnomes, loading their luggage onto a pony, set off on a long journey. They were spotted by trolls, who counted 36 legs and 15 heads in the caravan. How many gnomes and how many ponies were there?
# | There were 15 creatures in the caravan. If all of them were gnomes, they would have $15 \cdot 2=30$ legs; but in reality, there are 6 more legs, which means there were $6:(4-2)=3$ ponies and $15-3=12$ gnomes.
## Answer
12 gnomes and 3 ponies. | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Oleg collected a bag of coins. Sasha counted them, and it turned out that if all the coins were divided into five equal piles, there would be two extra coins. And if divided into four equal piles, there would be one extra coin. At the same time, the coins can be divided into three equal piles. What is the smallest numb... | If you add 3 coins, the resulting number of coins will be divisible by 3, 4, and 5, that is, by 60.
## Answer
57 coins.
Problem | 57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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