problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
|
A city in the shape of a triangle is divided into 16 triangular blocks, and at the intersection of any two streets, there is a square (there are a total of 15 squares in the city). A tourist started touring the city from a certain square and ended the tour at a different square, visiting each square exactly once. Pr... | Paint the blocks in black and white in a checkerboard pattern. If a tourist has passed through two streets of one block, then he has made a turn of $120^{0}$.
## Solution
There are 15 areas in total, so the tourist has passed through 14 streets connecting pairs of adjacent areas. We will paint the blocks in black and... | 4 | Combinatorics | proof | Yes | Yes | olympiads | false |
[ Methods for solving problems with parameters ] [ Phase plane of coefficients ]
Plot on the phase plane $O p q$ the set of points $(p, q)$ for which the equation $x^{3}+p x+q=0$ has three distinct roots belonging to the interval $(-2,4)$. | See solutions of problems $\underline{61272}, \underline{61273}$.
## Answer
The set of points defined by the inequalities $4 p^{3}+27 q^{2}<0,-4 p-64<q<2 p+8$.[^0]
How many roots does the equation $8 x\left(1-2 x^{2}\right)\left(8 x^{4}-8 x^{2}+1\right)=1$ have on the interval $[0,1]$?
## Solution
Notice that $8 x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Franklin 5.
A convex $n$-gon $P$, where $n>3$, is cut into equal triangles by diagonals that do not intersect inside it.
What are the possible values of $n$, if the $n$-gon is cyclic? | Lemma. Let a convex $n$-gon be cut into equal triangles by diagonals that do not intersect inside it. Then, for each of the triangles in the partition, at least one side is a side (not a diagonal) of the $n$-gon.
Proof. Let a triangle in the partition have angles $\alpha \leq \beta \leq \gamma$ with vertices $A, B, C$... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
All integers from 1 to 100 are written in a string in an unknown order. With one question about any 50 numbers, you can find out the order of these 50 numbers relative to each other. What is the minimum number of questions needed to definitely find out the order of all 100 numbers?
# | To find the desired order $a_{1}, a_{2}, \ldots, a_{100}$ of numbers in a row, it is necessary that each pair $\left(a_{i}, a_{i+1}\right), i=1,2, \ldots, 99$, appears in at least one of the sets about which questions are asked; otherwise, for two sequences $a_{1}, \ldots, a_{i}, a_{i+1}, \ldots, a_{100}$ and $a_{1}, \... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Ionin Yu.i.
In each cell of an infinite sheet of graph paper, a number is written such that the sum of the numbers in any square, the sides of which lie along the grid lines, does not exceed one in absolute value.
a) Prove the existence of a number $c$ such that the sum of the numbers in any rectangle, the sides of w... | Suppose in a certain rectangle with sides $a$ and $b (a < b)$. We construct four squares, each of which has three sides along some three sides of this rectangle $a \times b$; then the lines on which the fourth sides of these squares lie form a new rectangle with sides $2b - a$ and $|2a - b|$ (see Fig. 1 and 2; the case... | 3 | Combinatorics | proof | Yes | Yes | olympiads | false |
Zaslavsky A.A.
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... | a) If two teams have scored the same number of points, then the difference between the number of draws they have is a multiple of 3.
The number of draws a team has 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 $... | 10 | Combinatorics | proof | Yes | Yes | olympiads | false |
Chebotarev A.S.
On a plane, there is a circle. What is the minimum number of lines that need to be drawn so that, by symmetrically reflecting the given circle relative to these lines (in any order a finite number of times), it can cover any given point on the plane? | 1) We will prove that three lines are sufficient. Let the horizontal line $a$ and line $b$, forming a $45^{\circ}$ angle with line $a$, contain the center $O$ of the circle, and let line $c$ be parallel to $a$ and be at a distance of $0.5R$ from it, where $R$ is the radius of the given circle (see figure).
The composi... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ostrovsky $M$.
A number from 1 to 144 is guessed. 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. You have to pay 2 rubles for a "yes" answer and 1 ruble for a "no" answer. What is the minimum 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 $i \geq 3$. Then $a 10=144$. We will prove by induction that among not less 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 not less than $a_{i}$ numbers. Then either the... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tamarkin D:
In the class, there are 16 students. Every month, the teacher divides the class into two groups.
What is the minimum number of months that must pass so that every two students end up in different groups at some point? | Example. The figure shows how to divide a class into two groups so that any two students are in different groups in at least one of the four months. Each student corresponds to a column in the table, and each month corresponds to a row. A zero in a cell of the table means that the student is in the first group, and a o... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokaeva I.
Let $F_{1}, F_{2}, F_{3}, \ldots$ be a sequence of convex quadrilaterals, where $F_{k+1}$ (for $k=1,2,3, \ldots$) is obtained by cutting $F_{k}$ along a diagonal, flipping one of the parts, and gluing it back along the cut line to the other part. What is the maximum number of different quadrilaterals that t... | Let $ABCD$ be the original quadrilateral $F_{1}$. We can assume that each time the half of the quadrilateral containing side $CD$ is flipped, while side $AB$ remains stationary. In this process, the sum of angle $A$ and the opposite angle does not change. Additionally, the set of side lengths does not change. However, ... | 6 | Geometry | math-word-problem | 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 $G=\overline{\boldsymbol{g g 9 g g}}$ 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 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
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 $n$ fights, 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 $n$ fights ( $u_{k}-$ Fibonacci numbers).
Base case $\left(n=1, u_{3}=2\right)... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A.K.
On the plane, there is an open non-intersecting broken line with 31 segments (adjacent segments do not lie on the same straight line). Through each segment, a line containing this segment was drawn. As a result, 31 lines were obtained, some of which may have coincided. What is the smallest number of different lin... | Evaluation. Except for the ends, the broken line has 30 vertices, and each is the intersection of two lines. If there are no more than eight lines, then there are no more than $7 \cdot 8: 2=28$ intersection points - a contradiction.
Example - in the figure.
. Hackers have prepared 100 viruses, numbered them, and at different times, in a random order, they launch each virus on the computer with the same number. If a ... | Let's prove several statements.
1) Every computer is infected by viruses at least twice.
Consider, for example, computer 1. If the first virus to infect it came from computer 100, then it will also be infected by virus 1. If the first virus to infect computer 1 is virus 1, then the virus that first infected computer ... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Pairing and grouping; bijections] Proof by contradiction
What is the smallest number of weights in a set that can be divided into 3, 4, and 5 equal-mass piles? | Answer: 9. First, let's prove that the set cannot contain fewer than nine weights. Suppose this is not the case, that is, there are no more than eight. Let $60 m$ be the total mass of all the weights in the set. First, note that the set cannot contain weights with a mass greater than 12 t (since the set can be divided ... | 9 | Number Theory | proof | Yes | Yes | olympiads | false |
|
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.
## An... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Rabbits are sawing a log again, but now both ends of the log are secured. Ten middle chunks have fallen, while the two at the ends remain secured. How many cuts did the rabbits make?
# | How many logs did the rabbits get?
## Solution
The rabbits got 12 logs - 10 fallen and 2 secured. Therefore, there were 11 cuts.
## Answer
11 cuts.
## Problem | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Cutting (other).]
What is the maximum number of pieces into which a round pancake can be divided using three straight cuts?
# | Recall problem 11.
## Solution
If from three lines each pair intersects inside the pancake, it will result in 7 pieces (see Fig. 11.4). If, however, any two of these lines are parallel or intersect outside the pancake, there will be fewer pieces.
## Answer
7 pieces. | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10 guests came to visit and each left a pair of galoshes in the hallway. All pairs of galoshes are of different sizes. The guests began to leave one by one, putting on any pair of galoshes that they could fit into (i.e., each guest could put on a pair of galoshes not smaller than their own). At some point, it was disco... | If there are more than five guests left, then the guest with the smallest size of galoshes among the remaining guests can put on the largest of the remaining galoshes.
## Solution
Let's number the guests and their pairs of galoshes from 1 to 10 in ascending order of the size of the galoshes. Suppose there are 6 guest... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left.\quad \begin{array}{ll}{\left[\begin{array}{l}\text { Principles of divisibility by 3 and 9 } \\ \text { [Examples and counterexamples. Constructions] }\end{array}\right]}\end{array}\right]$
Write the number 2013 several times in a row so that the resulting number is divisible by 9. | The number 201320132013 is divisible by 9, since the sum of its digits is equal to $(2+0+1+3) \cdot 3=18$.
## Answer
Given a sheet of graph paper. Each node of the grid is marked with some letter. What is the smallest number of different letters needed to mark these nodes so that on any segment (going along the sides... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Chuk and Gek were decorating the Christmas tree with their mother. To prevent them from fighting, their mother gave each of them the same number of branches and the same number of ornaments. Chuk tried to hang one ornament on each branch, but he was short of one branch. Gek tried to hang two ornaments on each branch, b... | Try to do as Chuk did - hang one toy on each branch.
## Solution
Let's try to do as Chuk did — hang one toy on each branch, then one toy will be left over. Now, let's take two toys — one that is left over, and another one from one of the branches. If we now hang these toys as the second ones on the branches that stil... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The scent from a blooming lily-of-the-valley bush spreads in a radius of 20 m around it. How many blooming lily-of-the-valley bushes need to be planted along a straight 400-meter alley so that every point along it smells of lily-of-the-valley
# | Notice that to meet the conditions of the problem, the distance between adjacent lilies of the valley should not exceed $40 \mathrm{M}$.
## Solution
Let's mentally move the leftmost 20 m of the alley to the right end. Then we will have segments of the alley that are 40 m long, and to the right of each segment, a lily... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Combinatorics (miscellaneous). ] [Principle of the extreme (miscellaneous) ]
How many ways are there to rearrange the numbers from 1 to 100 such that adjacent numbers differ by no more than 1?
# | Where can the number 1 be placed?
## Solution
Next to the number 1, only the number 2 can stand, so 1 must be at the edge. Suppose 1 is at the beginning. Then the next number is 2, the next is 3 (no other numbers can be next to 2), the next is 4, and so on. We get the arrangement $1, 2, \ldots, 99, 100$.
If 1 is at ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Case Analysis ] [ Proof by Contradiction ]
In the cells of a $3 \times 3$ table, numbers are arranged such that the sum of the numbers in each column and each row is zero. What is the smallest number of non-zero numbers that can be in this table, given that this number is odd?
# | ## Example.
| 0 | -1 | 1 |
| :---: | :---: | :---: |
| -1 | 2 | -1 |
| 1 | -1 | 0 |
Evaluation. We will prove that it is impossible to use fewer non-zero numbers.
If the table contains exactly one non-zero number, then the sum of the numbers in the row containing this number is not zero. Suppose the table contains e... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Pe
In the country of Distant, a province is called large if more than $7 \%$ of the country's population lives in it. It is known that for each large province, there are two provinces with a smaller population such that their combined population is greater than that of this large province. What is the smallest number ... | We will order the provinces by increasing population. The first and second provinces are not large, as for each of them, there will not be two provinces with a smaller population. In the third province, less than $14\%$ of the population lives, since in both provinces with a smaller population, the total is no more tha... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Chessboards and chess pieces ] [ Examples and counterexamples. Constructions ]
Author: S. Preskova
Under one of the cells of an $8 \times 8$ board, a treasure is buried. Under each of the other cells, there is a sign indicating the minimum number of steps required to reach the treasure from that cell (one step all... | Let's dig up the corner cell $U$. Suppose there is a sign there. All cells at the specified distance from $U$ form a diagonal perpendicular to the main diagonal drawn from $U$. Let's dig up the corner cell $W$ on the same side as $U$. If there is also a sign there, then another diagonal perpendicular to the first one i... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a company of 10 people, 14 pairwise arguments have occurred. Prove that it is still possible to form a group of three friends.
# | The total number of ways to choose a company of three people is $C_{10}^{3}=120$. Each quarrel destroys no more than eight such companies, so the number of destroyed companies is no more than $8 \cdot 14=112$. Therefore, at least 8 friendly companies remain. | 8 | Combinatorics | proof | Yes | Yes | olympiads | false |
Laiko 0.
Around a round table, there is a company of thirty people. Each of them is either a fool or smart. Everyone sitting is asked: Is your right neighbor smart or a fool? In response, a smart person tells the truth, while a fool can say either the truth or a lie. It is known that the number of fools does not excee... | When $F=8$. If $F=0$, then one can point to any person sitting at the table. Let now $F \neq 0$. We divide all those sitting at the table into non-empty groups of consecutive smart and consecutive foolish people; the number of these groups is denoted by $2k$ ($k$ groups of smart people and $k$ groups of foolish people)... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Fomin d:
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 200 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 200 can be balanced by some number of weights from the set, and in a unique way (the b... | The correct set should correspond to the factorization of the number 201 (see the solution of problem $\underline{98056}$), and it only factors into two factors: $201=3 \cdot 67$.
## Answer
a) Two weights of 67 g and 66 weights of 1 g or 66 weights of 3 g and two of 1 g.
b) 3 sets. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Fomin D:
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 500 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 500 can be balanced by some number of weights from the set, and in a unique way (the b... | Let the largest weight of a weight in some correct set be $M$ (grams). This means that any smaller weight can be balanced by smaller weights. Let the weight of all smaller weights be $m$. Clearly, $m \geq$ $M-1$. But if $m \geq M$, then we have two ways to balance the weight $M+r$, where $r$ is the remainder of the div... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Rubanov I.S. }}$
Microcalculator MK-97 can perform only three operations on numbers stored in memory:
1) check if two selected numbers are equal,
2) add selected numbers,
3) find the roots of the equation $x^{2}+a x+b=0$ for selected numbers $a$ and $b$, or display a message if there are no roots.... | By adding $x$ to itself, we get $2x$. We compare $x$ and $2x$. If they are equal, then $x=0$. Otherwise, we find the roots of the equation $y^{2}+2xy+x=0$. The discriminant of this equation is $4(x^{2}-x)$, so the roots are equal if and only if $x=1$. | 1 | Logic and Puzzles | 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 smaller 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).
, let's take row $A$, in which the leftmost cell is marked, row $B$,... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given $n$ sticks. From any three, an obtuse triangle can be formed. What is the largest possible value of $n$? | From three sticks of lengths $a \leq b \leq c$, a triangle can be formed if $a+b>c$. According to the cosine theorem, this triangle is obtuse if and only if $a^{2}+b^{2}<c^{2}$. For the given problem, we need to check if $a_{4}^{2}+a_{3}^{2} \geqslant 2 a_{3}^{2}>2 a_{2}^{2}+2 a_{1}^{2}$. On the other hand, $a_{5}<a_{1... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kondakov G.V.
Divide the segment $[-1,1]$ into black and white segments so that the integrals of any a) linear function;
b) quadratic trinomial over the white and black segments are equal.
# | Since the integral of a polynomial over an interval is equal to the increment of its antiderivative, our task is a special case of problem $\underline{98268}$.
## Answer
For example, a) intervals $[-1,-1 / 2],[1 / 2,1]$ - black, $[-1 / 2,1 / 2]-$ white.
b) intervals $[-1,-3 / 4],[-1 / 4,0],[1 / 4,3 / 4]$ - white, $[... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Frankin B.R. }}$
Given a polynomial $P(x)$ with real coefficients. An infinite sequence of distinct natural numbers $a_{1}, a_{2}, a_{3}, \ldots$ is such that
$P\left(a_{1}\right)=0, P\left(a_{2}\right)=a_{1}, P\left(a_{3}\right)=a_{2}$, and so on. What degree can $P(x)$ have? | The constant clearly does not satisfy the condition.
For example, the polynomial $P(x)=x-1$ works.
Note that the leading coefficient of the polynomial $P$ is positive (otherwise, $P(x)x$ for each $x \in N$. Suppose the sequence exists. Starting from some index, the terms less than $N$ will end, that is, there will be... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shirov V.
On a circle, $2 N$ points are marked ($N$ is a natural number). It is known that through any point inside the circle, no more than two chords with endpoints at the marked points pass. We will call a matching a set of $N$ chords with endpoints at the marked points such that each marked point is the endpoint o... | Let the marked points be $A_{1}, A_{2}, \ldots, A_{2 N}$ in the order of a clockwise traversal of the circle. We will prove by induction on $N$ that the number of even matchings is one more than the number of odd matchings. For $N=1$, the statement is obvious: there is only one matching, and it is even.
Inductive step... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Ribamko A.V.
In the lower left corner of a $n \times n$ chessboard, there is a knight. It is known that the minimum number of moves it takes for the knight to reach the upper right corner is equal to the minimum number of moves it takes to reach the lower right corner. Find $n$.
# | Let $n$ be even. In this case, the left lower field has the same color as the right upper one, while the right lower field has a different color. After each move, the knight lands on a field of the opposite color. Therefore, any path from the left lower corner to the right upper one consists of an even number of moves,... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berov s.L.
Seryozha chose two different natural numbers $a$ and $b$. He wrote down four numbers in his notebook: $a, a+2, b$ and $b+2$. Then he wrote on the board all six pairwise products of the numbers from the notebook. What is the maximum number of perfect squares that can be among the numbers on the board? | Note that no two squares of natural numbers differ by 1: $x^{2}-y^{2}=(x-y)(x+y)$, and the second bracket is greater than one. Therefore, the numbers
$a(a+2)=(a+1)^{2}-1$ and $b(b+2)=(b+1)^{2}-1$ are not squares.
The numbers $a b$ and $a(b+2)$ cannot both be squares; otherwise, their product $a^2 b(b+2)$ would also b... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Rubanov I.S.
For four different integers, all their pairwise sums and pairwise products were calculated.
The obtained sums and products were written on the board. What is the smallest number of different numbers that could end up on the board?
# | If we take the numbers $-1,0,1,2$, then, as is easy to check, each of the numbers written on the board will be equal to $-2,-1,0$, 1,2 or 3 - a total of 6 different values.
We will show that fewer than six different numbers on the board could not have occurred. Let the numbers taken be $ac+d$. If $ac+d$.
The second m... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Combinatorics (miscellaneous).]
$[$ Estimation + example $]$
In a pond, 30 pikes were released, which gradually eat each other. A pike is considered full if it has eaten at least three pikes (full or hungry). What is the maximum number of pikes that can become full? | The number of pikes eaten is not less than three times the number of satiated ones.
## Solution
Let $s$ be the number of satiated pikes. Then they together have eaten no less than $3 s$ pikes. Since each pike can only be eaten once, and at least one pike remains at the end, $3 s<30$. Therefore, $s \leq 9$.
We will p... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
For what value of the parameter $m$ is the sum of the squares of the roots of the equation $x^{2}-(m+1) x+m-1=0$ the smallest?
# | The sum of the squares of the roots $\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(m+1)^{2}-2(m-1)=m^{2}+3$ is minimal when $m=0$. Note that in this case, the equation
$x^{2}-x-1=0$ has roots.
## Answer
When $m=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
Cheburashka has a set of 36 stones with masses of 1 g, 2 g, ..., 36 g, and Shapoklyak has super glue, one drop of which can glue two stones together (thus, three stones can be glued with two drops, and so on). Shapoklyak wants to glue the stones in such a way that Cheburashka cannot select one or several ... | Example. By gluing stones with masses 1 and 18, 2 and 17, ..., 9 and 10, respectively, Cheburashka will get a set where each stone weighs from 19 to 36 grams, so one stone will be too little, and two will be too many.
Another way. By gluing all the stones with odd masses in pairs, Cheburashka will get a set where all ... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5,6 |
| :---: | :---: | :---: |
| | Classical combinatorics (miscellaneous). | |
| | Examples and counterexamples. Constructions | |
| | Estimation + example | |
Each of thirty sixth-graders has one pen, one pencil, and one ruler. After their participation in the Olympiad, it turned out that 26 students lost the... | From the condition, it follows that four sixth-graders have a pen, seven have a ruler, and nine have a pencil. Thus, at least one item can be owned by no more than $4+7+9=20$ people. Therefore, no fewer than $30-20=10$ people have lost all three items.
All three items will be lost by exactly 10 people if each of the o... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Does there exist such an $N$ and $N-1$ infinite arithmetic progressions with differences $2,3,4, \ldots, N$, such that every natural number belongs to at least one of these progressions? | Let $N=12$. Each natural number can be written in the form $12 k+r$, where $r$ is one of the numbers $0,1, \ldots, 11$. All numbers for which $r$ is even belong to the progression $2,4,6, \ldots$; all numbers for which $r$ is a multiple of $3$ belong to the progression $3,6,9, \ldots$ The remaining numbers have $r=1,5,... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Tolony A.K.
In a certain kingdom, there were 32 knights. Some of them were vassals of others (a vassal could have only one suzerain, and a suzerain was always richer than his vassal). A knight who had at least four vassals bore the title of baron. What is the maximum number of barons that could be under these conditio... | Evaluation. 8 barons should have 32 vassals, and the richest knight cannot be anyone's vassal. Example. Let 24 knights be vassals of six barons, and all these barons be vassals of the richest Baron. In total, 7 barons.
## Answer
7 barons. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berlov S.L.
A natural number $n$ is called good if every its natural divisor, increased by 1, is a divisor of the number $n+1$.
Find all good natural numbers. | Clearly, $n=1$ satisfies the condition. Also, all odd prime numbers satisfy it: the divisors of such a number $p$, increased by 1, are 2 and
$p+1$; both of them divide $p+1$.
On the other hand, any number $n$ that satisfies the condition has a divisor 1; hence, $n+1$ is divisible by 1 +1, which means $n$ is odd.
Sup... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On the plane, $n$ lines are drawn such that every two intersect, but no four pass through the same point. There are a total of 16 intersection points, and through 6 of them, three lines pass. Find $n$.
# | "Move" the given construction in such a way that any two lines still intersect, but no three lines pass through the same point. Then, if some three lines intersected at a certain point $O$, now instead of one point $O$, there will be three points of pairwise intersection of these lines. Therefore, as a result of "movin... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
In a $10 \times 10$ square, all cells of the left upper $5 \times 5$ square are painted black, and the rest of the cells are white. Into what maximum number of polygons can this square be cut (along the cell boundaries) so that in each polygon, the number of black cells is three times less than the number... | In each polygon of the partition, there must be cells of both colors. This means that there must be a black cell adjacent to a white one. However, there are only 9 such cells. See the example of cutting into 9 polygons in the figure.
![](https://cdn.mathpix.com/cropped/2024_05_06_c07b2c4be0d3dcbcc4e5g-27.jpg?height=42... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Frankin B.R. }}$ 5.
A billiard table has the shape of a rectangle $2 \times 1$, with pockets located at the corners and at the midpoints of the longer sides. What is the smallest number of balls that need to be placed inside the rectangle so that each pocket lies on a line with some two balls? | Let there be only three balls. A line passing through a pair of balls inside a rectangle intersects the boundary of the rectangle exactly at two points. We have six pockets, so we need at least three lines. Three balls will give three lines only if these lines form a triangle. However, there are only seven lines passin... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the board, two-digit numbers are written. Each number is composite, but any two numbers are coprime.
What is the maximum number of numbers that can be written? | Evaluation. Since any two written numbers are coprime, each of the prime numbers 2, 3, 5, and 7 can appear in the factorization of no more than one of them. If there are five or more numbers on the board, then all prime factors in the factorization of one of them must be at least 11. But this is a composite number, so ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the expression $10: 9: 8: 7: 6: 5: 4: 3: 2: 1$, parentheses were placed such that the value of the expression is an integer.
What is the smallest number that could have been obtained?
# | In order for the value of the expression to be an integer, after placing the parentheses and writing the resulting expression as a common fraction, the number 7 must end up in the numerator. Therefore, the value of this expression is no less than 7. This can be achieved, for example, as follows:
$10: 9:(8: 7:(6:(5: 4:... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zaslavsky A.A.
In a volleyball tournament, each team met every other team once. Each match consisted of several sets - until one of the teams won three sets. If a match ended with a score of $3: 0$ or $3: 1$, the winning team received 3 points, and the losing team received 0. If the set score was
$3: 2$, the winner r... | Evaluation. If there are no more than three teams, then the "Simpletons" won all the matches, which means they have the most points. Contradiction.
If there are four or five teams, then each team will play three or four matches. This means that the "Cunning" won no more than one match and scored a maximum of $5=3+1+1$... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
In each cell of a $1000 \times 1000$ square, a number is inscribed such that in any rectangle of area $s$ that does not extend beyond the square and whose sides lie along the cell boundaries, the sum of the numbers is the same. For which $s$ will the numbers in all cells necessarily be the same? | It is clear that when $s=1$, the numbers in all cells are the same.
Let $s>1$ and $p$ be a prime divisor of $s$. In cells where the sum of the coordinates is divisible by $p$, we write ones, and in the other cells, we write zeros. In any rectangle $T$ of area $s$, one of the sides is divisible by $p$, so it is divided... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Andjans 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 news?... | a) A piece of 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 fifth hour. Therefore, it will take no less than 6 hours.
We will show that 6 hours are sufficient. The conversations can proc... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow Metro from "Taganskaya" station to "Kievskaya" station, and in the evening - back (see diagram).
Upon e... | Let $p$ be the probability that the Scientist boards a train going clockwise. Then the expected travel time from "Taganskaya" to "Kievskaya" is $11 p + 17(1-p) = 17 - 6p$.
On the return journey from "Kievskaya" to "Taganskaya," the expected travel time is $17 p + 11(1-p) = 11 + 6p$.
According to the condition, $(11 +... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the class, there are fewer than 30 people. The probability that a randomly chosen girl is an excellent student is $3 / 13$, and the probability that a randomly chosen boy is an excellent student is $4 / 11$. How many excellent students are there in the class? | The probability that a randomly chosen girl is an excellent student is the ratio of the number of excellent girl students to the total number of girls in the class. Therefore, the number of girls is divisible by 13, meaning it is either 13 or 26.
Similarly, the number of boys is either 11 or 22. Considering that there... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The probability that a purchased light bulb will work is 0.95.
How many light bulbs need to be bought so that with a probability of 0.99, there will be at least five working ones among them?
# | Let's take 6 light bulbs. The probability that at least 5 of them will be working is the sum of the probabilities that exactly 5 of them will be working and that all 6 will be working, which is $6 \cdot 0.95^{5} \cdot 0.05 + 0.95^{6} = 0.9672$.
Let's take 7 light bulbs. The desired probability is $21 \cdot 0.95^{5} \c... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
According to the conditions of the chess match, the winner is the one who outperforms the opponent by two wins. Draws do not count. The probabilities of winning for the opponents are equal. The number of decisive games in such a match is a random variable. Find its mathematical expectation. | Let $X$ be the number of decisive games. At the beginning of the match, the difference in the number of wins between the two participants is zero. Let's list the possible cases of two decisive games, denoting a win by the first participant as 1 and a win by the second participant as 2: 11, 12, 21, 22. Two of the four c... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
At a familiar factory, they cut out metal disks with a diameter of 1 m. 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 is 10 mm. Engineer Sidorov believes that a stack of 100 disks will ... | 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:
$\mathrm{ES}=\mathrm{E}\left(\pi R^{2}\right)=\pi \mathrm{E} R^{2}=\pi\left(\mathrm{D} R+\mathrm{E}^{2} R\right)=\pi\left(10^{-4}+0.25\right)=0.2501 \pi$.
Therefore, the expected value of... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the Valley of Five Lakes, there are five identical lakes, some of which are connected by streams (on the diagram, dashed lines indicate possible "routes" of the streams). Small carp are born only in lake $S$. While growing up, a carp crosses from one lake to another exactly four times via some stream (the carp choos... | The transition from one lake to another will be called a route of length $n$ if it passes through $n$ streams. Let's prove several statements.
1. There is no route of length 2 from $S$ to any of the lakes $A, C$, and $D$.
Proof. Suppose it is possible to sail from lake $S$ to lake $A$ through one intermediate lake (s... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Every day, the dog Patrick gnaws one slipper from the available supply in the house. With a probability of 0.5, Patrick wants to gnaw a left slipper and with a probability of 0.5 - a right slipper. If the desired slipper is not available, Patrick gets upset. How many pairs of identical slippers need to be bought so tha... | Let in a week Patrick wants to eat $S$ left and $7-S$ right slippers. We need to find such $k$ that the inequality $\mathrm{P}(S \leq k \cap 7-S \leq k) \geq 0.8$ holds. Rewrite the event in parentheses: $P(7-k \leq S \leq k)$. It is clear that $7-k \leq k$, that is, $k \geq 4$.
The probability on the left side of the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On board an airliner, there are $2 n$ passengers, and the airline has loaded $n$ portions of chicken and $n$ portions of fish for them. It is known that a passenger prefers chicken with a probability of 0.5 and fish with a probability of 0.5. We will call a passenger dissatisfied if they are left with what they do not ... | 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". $\xi=0$... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In Anchuria, a checkers championship is taking place in several rounds. The days and cities for the rounds are determined by a draw. According to the championship rules, no two rounds can take place in the same city, and no two rounds can take place on the same day. Among the fans, a lottery is organized: the main priz... | In an $8 \times 8$ table, you need to select $k$ cells such that no more than one cell is selected in any row or column. The number of tours $k$ should be chosen so that the number $N k$ of possible selection options is maximized.
$$
N_{k}=C_{8}^{k} A_{8}^{k}=\frac{8!\cdot 8!}{(8-k)!(8-k)!k!}\left(C_{8}^{k}-\text { th... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Ilya Muromets meets the three-headed Zmei Gorynych. And the battle begins. Every minute Ilya cuts off one of Zmei's heads. With a probability of $1 / 4$, two new heads grow in place of the severed one, with a probability of $1 / 3$ - only one new head, and with a probability of $5 / 12$ - no heads at all. The Zmei is c... | Strikes by Ilya Muromets, in which the number of heads changes, are called successful.
Let's find the probability that at some point there will be a last successful strike. This means that starting from this point, there will be no more successful strikes, that is, all strikes will be unsuccessful. The probability of ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A deck of playing cards is laid out on the table (for example, in a row). On top of each card, a card from another deck is placed. Some cards may have matched. Find:
a) the expected value of the number of matches;
b) the variance of the number of matches.
# | a) Let's number the pairs from 1 to $N$ (we don't know how many there are) in the order they lie on the table. Let
the indicator $I_{k}$ be 1 if the two cards in the $k$-th pair are the same, and 0 if the cards in the $k$-th pair are different.
Obviously, $\mathrm{P}\left(I_{k}=1\right)=1 / N$. Therefore, $\mathrm{E}... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
The game takes place on a $9 \times 9$ grid of squared paper. Two players take turns. The player who starts the game places crosses in free cells, while his partner places noughts. When all cells are filled, the number of rows and columns $K$ in which there are more crosses than noughts, and the numbe... | One possible strategy for the first player: the first move is to the center of the board; thereafter, for each move the second player makes to any cell, the first player responds by moving to the cell symmetric to the center. This is possible because the second player always breaks the symmetry.
In the final position,... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Two players take turns coloring the sides of an $n$-gon. The first player can color a side that borders with zero or two colored sides, the second player - a side that borders with one colored side. The player who cannot make a move loses. For which $n$ can the second player win, regardless of how the ... | For $n=3$, it is obvious that the first player wins, while for $n=4$, the second player wins. Let's show how the first player wins for $n>4$. After the first move of the second player, two adjacent sides are painted. The first player can paint a side "one apart" from them, creating an unpainted "hole" of one side. This... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.B.
The weight of each weight in the set is a non-integer number of grams. They can balance any integer weight from 1 g to 40 g (weights are placed on one pan of the scales, the weight to be measured - on the other). What is the smallest number of weights in such a set? | Example 1. Let's take weights of $1,1,3,5,11,21,43$ g. The first two can measure any integer weight up to 2 g. Therefore, the first three can measure up to 5 g, the first four up to 10 g, the first five up to 21 g, the first six up to 42 g, and all seven up to 85 g. If we reduce the weight of each weight by half, all t... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Fomin D:
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 200 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 200 can be balanced by some number of weights from the set, and in a unique way (the b... | The correct set should correspond to the factorization of the number 201 (see the solution to problem $\underline{98056}$), and it only factors into two factors: $201=3 \cdot 67$.
## Answer
a) Two weights of 67 grams and 66 weights of 1 gram or 66 weights of 3 grams and two of 1 gram.
b) 3 sets.
Author: Fomin D:
G... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Fomin }}$ D:
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 500 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 500 can be balanced by some number of weights from the set, an... | Let the largest weight of a weight in some correct set be $M$ (grams). This means that any smaller weight can be balanced by smaller weights. Let the weight of all smaller weights be $m$. Clearly, $m \geq M-1$. But if $m \geq M$, then we have two ways to balance the weight $M+r$, where $r$ is the remainder of the divis... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Twelve chairs are arranged in a row. Sometimes a person sits on one of the free chairs. At this point, exactly one of his neighbors (if they were present) stands up and leaves. What is the maximum number of people that can be sitting at the same time, if initially all the chairs were empty?
# | Evaluation. It is impossible for all chairs to be occupied simultaneously, because at the moment when a person sits on the last unoccupied chair, one of his neighbors will stand up. Therefore, the number of people sitting simultaneously cannot exceed 11.
Example. Let's show how to seat 11 people. Number the chairs fro... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Evdokiyov M.A.
A pirate has five bags of coins, each containing 30 coins. He knows that one bag contains gold coins, another contains silver coins, a third contains bronze coins, and each of the two remaining bags contains an equal number of gold, silver, and bronze coins. You can simultaneously take any number of coi... | Example. Let's take one coin from each bag. Among these five coins, there are coins of all three types, so there is only one coin of a certain type. If it is, for example, a gold coin, then it was taken from the bag with gold coins. Indeed, for each coin from the "mixed" bag, there is a matching one from the correspond... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Folkior
There is a set of tickets numbered from 1 to 30 (numbers can repeat). Each student drew one ticket. The teacher can perform the following operation: read out a list of several (possibly one) numbers and ask their owners to raise their hands. How many times does he need to perform this operation to find out the... | We will encode the tickets with binary numbers from 00001 to 11110. On the $k$-th stage, the teacher includes in the list all numbers whose $k$-th bit is one (for example, on the 3rd stage, the numbers are $4,5,6,7,12,13,14,15,20,21,22,23$, $28,29,30$). After the fifth stage, the teacher learns the binary code of the n... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the room, there are 12 people; some of them are honest, meaning they always tell the truth, while the rest always lie. "There is not a single honest person here," said the first. "There is no more than one honest person here," said the second. The third said that there are no more than two honest people, the fourth ... | Note that if someone among those present lied, then all the previous ones lied as well. Such people are in the room; otherwise, the first one told the truth, and according to his words, there are no honest people in the room. For the same reason, there must be honest people in the room.
Let there be $x$ liars in the r... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Symmetric Strategy ]
There are two piles of stones: one has 30, the other has 20. On a turn, you are allowed to take any number of stones, but only from one pile. The player who cannot make a move loses.
# | The first one wins. With the first move, he equalizes the number of stones in the piles, after which he plays as in problem 10. | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Combinatorics (miscellaneous) $]$ $[$ Estimation + example ]
In a pond, 30 pikes were released, which gradually eat each other. A pike is considered full if it has eaten at least three pikes (full or hungry). What is the maximum number of pikes that can become full? | The number of pikes eaten is not less than three times the number of satiated ones.
## Solution
Let $s$ be the number of satiated pikes. Then they together have eaten no less than $3 s$ pikes. Since each pike can only be eaten once, and at least one pike remains at the end, $3 s<30$. Therefore, $s \leq 9$.
We will p... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On Uncle Fyodor's birthday, Postman Pechkin wants to find out how old he is. Sharik says that Uncle Fyodor is more than 11 years old, and cat Matroskin claims that he is more than 10 years old. How old is Uncle Fyodor, given that exactly one of them is wrong? Justify your answer. | Note that if Sharik did not make a mistake, then Matroskin did not make a mistake either, which contradicts the condition. Therefore, Sharik must have lied, while Matroskin told the truth. Thus, Uncle Fyodor is more than 10 years old, but not more than 11. Therefore, Uncle Fyodor is 11 years old.
## Problem | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
There are five chain links, each with 3 rings. What is the minimum number of rings that need to be unlinked and relinked to connect these links into one chain?
# | For connecting two links, one ring is required.
## Solution
Example: we unlock 3 rings from one link. The remaining 4 links are connected using the three unlocked rings.
Evaluation: if fewer than 3 rings are unlocked, at least 5 separate links will remain, which would require at least 4 unlocked rings to connect - a... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
|
There are 85 balloons in the room - 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 |
How can you use a balance scale without weights to divide 24 kg of nails into two parts - 9 and 15 kg?
# | Try to weigh out 12 kg first, then -6 kg, and then -3 kg.
## Solution
Weigh out 12 kg of nails and set them aside. From the remaining 12 kg, weigh out 6 kg and set them aside in a different place. From the remaining 6 kg, weigh out 3 kg and combine them with the 6 kg that were set aside. We get the desired 9 kg of na... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Five first-graders stood in a line and held 37 flags. All those to the right of Tanya had 14 flags, to the right of Yasha - 32, to the right of Vera - 20, and to the right of Maksim - 8. How many flags does Dasha have?
# | ## Solution
Obviously, the more flags to the right of a first-grader, the "further left" their place in the line. Someone is standing to the right of Maksim (otherwise there would be no flags to his right). But everyone except Dasha is definitely standing to the left of Maksim. Therefore, Dasha is standing to the righ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\begin{aligned} & {[\text { Mathematical Logic (other) }]} \\ & {[\text { Arithmetic. Mental calculation, etc. }]}\end{aligned}$
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 |
[Mathematical logic (miscellaneous).]
In the wallet, there are 2 coins with a total value of 15 kopecks. One of them is not a five-kopeck coin. What are these coins?
# | Two coins can only be 5 kopecks and 10 kopecks. If one of them is not a five-kopeck coin, then it is a ten-kopeck coin, and the other one is a five-kopeck coin. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[
There are pan scales without weights and 3 visually identical coins, one of which is counterfeit: it is lighter than the genuine ones (the genuine coins weigh the same). How many weighings are needed to determine the counterfeit coin?
# | When searching for a counterfeit coin among three coins, try placing one coin on each pan of the balance.
## Solution
We will need only 1 weighing. Place one coin on each pan of the balance. If one of the pans is lighter, the counterfeit coin is on it. If the balance is even, the counterfeit coin is the one that was ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (other)]
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. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the garden of Anya and Vitya, there were 2006 rose bushes. Vitya watered half of all the bushes, and Anya watered half of all the bushes. It turned out that exactly three bushes, the most beautiful ones, were watered by both Anya and Vitya. How many rose bushes remained unwatered? | Vitya watered 1003 bushes, of which 1000 he watered alone, and three - together with Anya. Similarly, Anya watered 1003 bushes, of which 1000 she watered alone, and three - with Vitya. Therefore, together they watered $1000+1000+3=2003$ bushes. Thus, 2006 - 2003 = 3 rose bushes were left unwatered.
## Answer
3 bushes... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Golenischeva-Kumuzova T.I.
Yura has a calculator that allows multiplying a number by 3, adding 3 to a number, or (if the number is divisible by 3) dividing the number by 3. How can Yura use this calculator to get from the number 1 to the number 11? | Comment. Note that on Yura's calculator, any number can be increased by $1: (x \cdot 3+3): 3=x+1$. Therefore, in principle, from one, any natural number can be obtained on it.
## Answer
For example, $((1 \cdot 3 \cdot 3 \cdot 3)+3+3): 3=11$ or $(1 \cdot 3+3): 3+3+3+3=11$. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Zsspasii A.A.
The number of the current Olympiad (70) is formed by the last digits of the year of its holding, written in reverse order.
How many more times will such a situation occur in this millennium?
# | Let in some year the described coincidence occurred.
If the number of the Olympics is two-digit, then the sum of this number and the number formed by the last two digits of the year is divisible by 11 (the sum of two numbers consisting of digits $a$ and $b$ is $11(a+b)$). Since each year this sum increases by 2, the e... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Rubanov I.S.
Microcalculator MK-97 can perform only three operations on numbers stored in memory:
1) check if two selected numbers are equal,
2) add two selected numbers,
3) find the roots of the equation $x^{2}+a x+b=0$ for selected numbers $a$ and $b$, and if there are no roots, display a message about it.
The res... | By adding $x$ to itself, we get $2x$. We compare $x$ and $2x$. If they are equal, then $x=0$. Otherwise, we find the roots of the equation $y^{2}+2xy+x=0$. The discriminant of this equation is $4(x^{2}-x)$, so the roots are equal if and only if $x=1$.
Send a comment | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Anzhans A.
There are 100 silver coins, ordered by weight, and 101 gold coins, also ordered by weight. It is known that all coins have different weights. We have a two-pan balance that allows us to determine which of any two coins is heavier. How can we find the coin that ranks 101st in weight among all the coins with ... | Let's prove that if from $n$ silver and $n$ gold coins, the $n$-th by weight coin can be found in $k$ weighings, then the $2n$-th by weight coin can be found in $k+1$ weighings from $2n$ silver and $2n$ gold coins.
Indeed, suppose the $n$-th silver coin is heavier than the $n$-th gold coin. Then the $n$ first silver c... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Authors: Bogdanov I.Y., Knop K.A.
King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are 1, 2, ..., 11 kg. He also has a bag that will tear if more than 11 kg is placed in it. Archimedes has learned the weights of all the ingots and wants to prove ... | Let Archimedes first put ingots weighing 1, 2, 3, and 5 kg into the bag, and then put ingots weighing 1, 4, and 6 kg. In both cases, the bag will not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if Archimedes used ingots weighing \( w_1, \ldots, w_6 \) kg instead of ingots ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10,11
There is a piece of chain consisting of 150 links, each weighing 1 g. What is the smallest number of links that need to be broken so that from the resulting parts, all weights of 1 g, 2 g, 3 g, ..., 150 g can be formed (a broken link also weighs 1 g)? | Answer: 4 links. According to the solution of problem 5 for grades $7-8$, for a chain consisting of $n$ links, where $64 \leq n \leq 159$, it is sufficient to unfasten 4 links. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
Among 2000 indistinguishable balls, half are aluminum with a mass of 10 g, and the rest are duralumin with a mass of 9.9 g. It is required to separate the balls into two piles such that the masses of the piles are different, but the number of balls in them is the same. What is the smallest number of weigh... | Let's compare the mass of 667 balls with the mass of another 667 balls. If the masses of these two piles are not equal, the required condition is met.
Suppose the specified masses are equal. Then the mass of 666 balls that did not participate in the weighing is not equal to the mass of any 666 balls lying on one of th... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The Cantor Set. A segment of the number line from 0 to 1 is painted green. Then its middle part, the interval $(1 / 3 ; 2 / 3)$, is repainted red. Next, the middle part of each of the remaining green segments is also repainted red, and the same operation is performed on the remaining green segments, and so on to infini... | Numbers from 0 to 1 can conveniently be considered as infinite ternary fractions of the digits 0, 1, and 2. The numbers mentioned in point v) are those numbers whose ternary representation contains no 1.
| | Game Theory (Other) |
| :---: | :---: |
| | Similarity Criteria |
| | [ Auxiliary Similar Triangles |
| Prob... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical Logic (Miscellaneous).] $[$ Evenness and Oddness $]$
Author: Khaitumuren A.v.
13 children sat around 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... | 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 just one person). Groups of boys and girls alternate, so their number is even. Incorrect statements were made at the tr... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$:$ Folkiore
It is known that among 63 coins there are 7 counterfeit ones. All counterfeit coins weigh the same, all genuine coins also weigh the same, and a counterfeit coin is lighter than a genuine one. How can you determine 7 genuine coins in three weighings using a balance scale without weights? | 1) Let's set aside one coin and put 31 coins on each pan of the scales. If the pans balance, then we have set aside the counterfeit coin, and there are 3 counterfeit coins on each pan. If one of the pans is heavier, then there are no more than three counterfeit coins on it. Thus, after the first weighing, we will be ab... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Mathematical logic (other).]
Several natives met (each one is either a liar or a knight), and each one declared to all the others: “You are all liars.” How many knights were among them?
# | Think, for example, could there have been 3 knights?
## Solution
Two (or more) knights cannot be, because then the statement made by the knight would not be true. It is also impossible to have no knights at all - then the statements of all liars would be true. The only possibility left is - 1 knight, the rest are lia... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
At the end of the quarter, Vovochka wrote down his current grades for singing in a row and placed a multiplication sign between some of them. The product of the resulting numbers turned out to be 2007. What grade does Vovochka get for singing in the quarter? ("Kol" is not given by the singing teacher.)
# | $2007=3 \cdot 3 \cdot 223=9 \cdot 223=3 \cdot 669$. Since there is no grade 9, only the first option fits. Since Vovochka has more threes than twos, and the last grade, no matter how you rearrange the factors, is a three, we can hope that he will get a three for the quarter.
## Answer
Three. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.