problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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Each of the three cutlets needs to be fried on a pan for five minutes on each side. Only two cutlets fit on the pan. Can all three cutlets be fried in less than 20 minutes (neglecting the time for flipping and moving the cutlets)? | It is not necessary to immediately fry a cutlet on the other side once one side is fried.
## Solution
In the first 5 minutes, we fry the first two cutlets on one side. Then we flip the first cutlet, remove the second one, and place the third. We fry for another 5 minutes. Now the first cutlet is ready, and the second... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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, that is, they only know Latin. This means that $75-15=$ $60 \%$ speak both languages.
## Answer
$60 \%$.
## Problem | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$[$ Mathematical logic (other).]
About mushrooms. In the basket, there are 30 mushrooms. Among any 12 of them, there is at least one russula, and among any 20 mushrooms, there is at least one boletus. How many russulas and how many boletuses are in the basket?
# | Note: Among any 12 mushrooms, there is at least one russula.
## Solution
Since among any 12 mushrooms there is at least one russula, there cannot be more than 11 boletus mushrooms. And since among any 20 mushrooms there is at least one boletus, there cannot be more than 19 russulas. But since there are 30 mushrooms i... | 19 | 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 |
From the four inequalities $2x > 70$, $x < 25$, $x > 5$, and $x > 5$, two are true and two are false. Find the value of $x$, given that it is an integer. | Given that $x$ is an integer, the inequalities can be rewritten as: $x>35, x>6$ and $x>5$. If it is not true that $x>5, x>6$ and $x>35$ - one is true and two are false.
Since the third inequality implies the first two, and the second implies the first, there is only one option: the first inequality is true, and the ot... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2+ $[$ Prime numbers and their properties $]$
Four kids were discussing the answer to a problem.
Kolya said: "The number is 9."
Roman: "It's a prime number."
Katya: "It's an even number."
And Natasha said that the number is divisible by 15.
One boy and one girl answered correctly, while the other two were wrong. ... | If Kolya answered correctly, then both girls were wrong, since the number 9 is odd and does not divide by 15. This means that Roman gave the correct answer. But a prime number does not divide by 15, and the only even prime number is 2.
## Answer
2. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$2+$
12 candidates for mayor were talking about themselves. After some time, one said: "They lied to me once." Another said: "Now it's twice." "Now it's three times" - said the third, and so on up to the 12th, who said: "Now they have lied 12 times." At this point, the host interrupted the discussion. It turned out th... | ## Solution
Assume that the first candidate lied. This means that before his statement, the number of false statements was not equal to 1. After his statement, this number increased by 1, so it became not equal to 2. Therefore, the second candidate also lied. Continuing this reasoning, we get that all candidates lied,... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Symmetric Strategy ]
Two players play the following game. Each player, in turn, crosses out 9 numbers (of their choice) from the sequence $1,2, \ldots, 100,101$. After eleven such crossings, two numbers will remain. The first player is awarded as many points as the difference between these remaining numbers. Prove t... | The first move is to cross out numbers from the middle.
## Solution
The first move is to cross out 9 numbers from 47 to 55. The remaining numbers can be paired as follows: 1 - 56, 2 - 57, ..., 46 - 101. After each move by the second player, the first player can cross out numbers in such a way that in each pair, eithe... | 52 | Combinatorics | proof | Yes | Yes | olympiads | false |
Bogdanov I.I.
In some cells of a $10 \times 10$ table, several crosses and several zeros are placed. It is known that there is no line (row or column) completely filled with the same symbols (crosses or zeros). However, if any symbol is placed in any empty cell, this condition will be violated. What is the minimum num... | Let us assume that we have filled the table according to the condition, and cell $A$ is free. Since placing any symbol in it should result in a line of identical symbols, there must be a line containing it where all other cells are filled with crosses, and the same is true for noughts (these lines must, of course, be a... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (miscellaneous).]
On an island, there live liars and knights, a total of 2001 people. Knights always tell the truth, while liars lie. Each resident of the island stated: "Among the remaining residents of the island, more than half are liars." How many liars are there on the island?
# | Take one knight and one liar. From the statements made by them, lower and upper bounds on the number of liars can be derived.
## Solution
All the inhabitants of the island cannot be liars, otherwise each of them would be telling the truth. Let's take some knight. From his statement, it follows that the number of liar... | 1001 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Need to find out a five-digit phone number by asking questions that can only be answered with "yes" or "no." What is the minimum number of questions needed to guarantee this (assuming the answers to the questions are correct)?
# | We need to ask questions in such a way that each subsequent question approximately halves the number of remaining possible options.
## Solution
Initially, when we know nothing about the phone number, there are $10^{5}=100000$ possible options for the phone number. We ask the question: "Is the number greater than 5000... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
There are two hourglasses - one for 7 minutes and one for 11 minutes. An egg needs to boil for 15 minutes. How can you measure this time using the available hourglasses?
# | With the help of two different hourglasses, you can measure not only the time equal to their "sum," but also the time equal to their "difference."
## Solution
"Start" both hourglasses simultaneously. When the 7-minute hourglass runs out, flip it and let it run for 4 more minutes until the 11-minute hourglass runs out... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The tower clock strikes three times in 12 s. How long will it take to strike six times?
# | What is the interval between two consecutive strikes
## Solution
Since 3 strikes of the clock take 12 seconds, the interval between two consecutive strikes is 6 seconds. From the first strike to the second is 6 seconds, and from the second to the third is also 6 seconds. Six strikes, however, occur with 5 intervals. ... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a physics class, the teacher set up the following experiment. He placed 16 weights with masses of $1, 2, 3, \ldots, 16$ grams on a balance scale so that one of the pans was heavier. Fifteen students took turns leaving the classroom and each took one weight, and after each student left, the balance changed its positi... | Since at any given moment the masses on the scales differed by at least 1 gram, in order for the opposite scale to outweigh, it is necessary to remove a weight of no less than two grams. Therefore, no student could have taken a 1-gram weight when leaving the class.
## Answer
A 1-gram weight remained on the scales. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
On each of 2013 cards, a number is written, and all these 2013 numbers are distinct. The cards are face down. In one move, it is allowed to point to ten cards, and in response, one of the numbers written on them will be reported (it is unknown which one).
For what largest $t$ is it guaranteed to be poss... | 1) Let's first show that it is impossible to guess $1987=2013-26$ cards. We will number the cards $A_{1}, \ldots, A_{2013}$, place the numbers from 1 to 2013 on them in the same order, and indicate how to answer so that none of the numbers on the cards $A_{1}, \ldots, A_{27}$ can be determined.
For each $i=1, \ldots, ... | 1986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
In a row from left to right, there are 31 wallets, each containing 100 coins. From one of the wallets, some coins were moved: one coin to each of the wallets to the right of it. With one question, you can find out the total number of coins in any set of wallets. What is the minimum number of questions n... | It is enough to get an answer to the question "How many coins are there in total in the wallets with odd numbers?" Indeed, if the answer to it is " $1600+n$ " ( $n>0$ ), then the coins were moved from the wallet with an even number, to the right of which there were exactly $n$ wallets with odd numbers - that is, from t... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shen A.H.
There is an infinite one-way strip of cells, numbered with natural numbers, and a bag with ten stones. Initially, there are no stones in the cells of the strip. The following actions can be performed:
- move a stone from the bag to the first cell of the strip or vice versa;
- if there is a stone in the cell... | Note that for every action, there is an inverse action. Therefore, if we move from situation \$A\$, acting according to the rules, to situation \$ \mathrm{~B} \$$, then from situation \$ \mathrm{~B} \$$ we can return to situation \$ \mathrm{~A} \$$ by acting according to the rules.
We will show by induction that if th... | 1023 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Fundamental Theorem of Arithmetic. Prime Factorization]
In the country of Anchuria, where President Miraflores is in power, the time for new presidential elections is approaching. The country has exactly 20 million voters, of which only one percent support Miraflores (the regular army of Anchuria). Miraflores, of co... | We will divide the voters into groups of 5 people. In 66666 such groups, we can place 3 military personnel. As a result, we get 66666 military electors out of 4 million.
We will divide these 4 million electors into groups of 4 people, placing 3 military personnel in 22222 of them. As a result, we get 22222 military el... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Evaluation + example
Find the smallest $n$ such that any convex 100-gon can be obtained as the intersection of $n$ triangles. Prove that for smaller $n$ this cannot be done with any convex 100-gon.
# | Answer: $n=50$. First, note that 50 triangles are sufficient. Indeed, let $\Delta_{k}$ be the triangle whose sides lie on the rays $A_{\mathrm{k}} A_{\mathrm{k}-1}$ and $A_{\mathrm{k}} A_{\mathrm{k}+1}$ and which contains the convex polygon $A_{1} \ldots A_{100}$. Then this polygon is the intersection of the triangles ... | 50 | Geometry | proof | Yes | Yes | olympiads | false |
$\underline{\text { Kanel-Belov A.Y. }}$
On a sphere, five points are marked, no three of which lie on a great circle (a great circle is a circle formed by the intersection of the sphere and a plane passing through its center). Two great circles, not passing through the marked points, are called equivalent if one can ... | a) Let's move on to dual objects: each circle corresponds to a pair of opposite points on the sphere such that the diameter connecting them is perpendicular to this circle; conversely, each point corresponds to a great circle. Then the problem reduces to a dual one: points are considered equivalent if one can be transl... | 290 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Karasov $P$.
On a line, 100 sets $A_{1}, A_{2}, . ., A_{100}$ are chosen, each of which is the union of 100 pairwise non-intersecting segments. Prove that the intersection of the sets $A_{1}, A_{2}, \ldots, A_{100}$ is the union of no more than 9901 pairwise non-intersecting segments (a point is also considered a segm... | Let sets $A$ and $B$ on a line be unions of $m$ and $n$ segments, respectively. Then $A \cap B$ is a union of no more than $m+n-1$ segments. It is clear that $A \cap B$ is also a union of segments. Let the number of these segments be $k$. The endpoints of the segments in $A \cap B$ are endpoints of segments in $A$ or $... | 9901 | Combinatorics | proof | Yes | Yes | olympiads | false |
A spider web has the form of a grid of $100 \times 100$ nodes (in other words, it is a grid of 99×99 cells). In one of its corners sits a spider, and in some 100 nodes, flies are stuck to the web. In one move, the spider can move to any adjacent node. Can the spider guarantee to eat all the flies, spending no more than... | b) Let's divide the grid into 10 horizontal strips, each with 1000 nodes ($9 \times 99$ cells). Suppose the spider is sitting in the bottom-left corner. Initially, it moves from left to right along the bottom edge of the lower strip until it sees a fly stuck in the same strip above it. Then it crosses the strip vertica... | 1980 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
In a line, all integers from 1 to 100 are written 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 determine 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 $a_{1}, \... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7 wolves eat 7 sheep in 7 days. How many days will it take for 9 wolves to eat 9 sheep? | The number of wolves has increased by the same factor as the number of sheep, so the time of consumption will not change.
## Answer
In 7 days. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
I.v.
In the multiplication example written on the board, the hooligan Petya changed two digits. It became 4$\cdot$5$\cdot$4$\cdot$5$\cdot$4 = 2247. Restore the original example.
# | In the original example, at least one of the multipliers is even.
## Solution
In the resulting example, three multipliers are even, which means that at least one was even in the original example. Therefore, the product was an even number, meaning the last digit of the product was changed. Thus, no more than one digit... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order? | Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Algorithm. Number the chips in order from 0 to 99. Color the cells in four colors: $a b c d a b c d . . . d$. ... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Classical combinatorics (miscellaneous).]
$[$ Inclusion-exclusion principle $]$
From the sequence of natural numbers, all numbers that are squares or cubes of integers have been erased. Which of the remaining numbers is in the hundredth position? | Consider the first hundred natural numbers. Among these numbers, there are ten squares (from 1 to $10^{2}=100$) and four cubes (from 1 to $4^{3}=64$). Note that two of these numbers, namely, 1 and 64, are both squares and cubes. Thus, from the first hundred, we have crossed out 12 numbers. Among the next twelve numbers... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Kalinin }}$ D.:
In the room, there are 20 chairs of two colors: blue and red. On each chair sits either a knight or a liar. Knights always tell the truth, and liars always lie. Each of the seated individuals claimed that they were sitting on a blue chair. Then they somehow rearranged themselves, af... | Initially, all knights sit on blue chairs, and all liars on red ones. Therefore, the number of knights who moved to red chairs is equal to the number of liars who moved to blue chairs. Both groups claimed they were sitting on red chairs. In total, 10 people said they were sitting on red chairs. Therefore, the number of... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Rebus $]$
Authors: Kumygein A.K., Raskina I.V.
The magician taught Kastanka to bark as many times as he showed her secretly from the audience. When Kastanka correctly answered how much two times two is in this way, he hid a delicious cake in a suitcase with a combination lock and said:
- The eight-digit code for t... | a) Note that KE and KS replace different numbers, but swapping them does not change the product UCHU. Therefore, for each solution to the puzzle, there is a paired solution where the digits corresponding to E and S are swapped. Thus, it is impossible to uniquely reconstruct the solution the magician intended.
b) After... | 2021 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The kindergarten received cards for reading lessons: some have "MA" written on them, and others have "NYA". Each child took three cards and started forming words from them. It turned out that 20 children can form the word "MAMA", 30 children can form the word "NYANYA", and 40 children can form the word "MANYA". How man... | Notice that each child has three cards, and there are two different inscriptions on them.
## Solution
Since each child has three cards, and there are only two inscriptions in total, it is necessary that two inscriptions must match, meaning each can form either the word MAMA (20 such children) or the word NYANYA (30 s... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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 conditions?
(The kin... | 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 |
The family approached the bridge at night. Dad can cross it in 1 minute, Mom in 2, the child in 5, and Grandma in 10 minutes. They have one flashlight. The bridge can only hold two people at a time. How can they cross the bridge in 17 minutes? (If two people cross together, they move at the slower of their speeds. Movi... | Here is the sought-after algorithm:
dad and mom - 2 min dad back with the flashlight -1 min, baby and grandma - 10 min, mom back with the flashlight -2 min, dad and mom - 2 min.
In total 17 minutes | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.V.
In Mexico, ecologists have succeeded in passing a law according to which each car must not be driven at least one day a week (the owner reports to the police the car's number and the "day off" for the car). In a certain family, all adults wish to drive daily (each for their own business!). How many car... | b) If no more than one car "rests" each day, then there are no more than 7 cars in total.
## Solution
a) Five cars are not enough, because on the day when one of the cars is "resting," someone will have no car to ride in. Six cars, obviously, are enough.
b) If no more than one car "rests" each day, then the total nu... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3 $[\quad$ Evaluation + example $\quad]$
There are 100 boxes, numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the ... | Since the order of the answers to his questions is unknown to the audience, he must make a mistake-free choice, knowing only the number of "no" answers. If $N$ notes are sent, the number of "no" answers heard can take any integer value from 0 to $N$, meaning there are $N+1$ possible outcomes. This number must determine... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Around a round table, 30 people are sitting - knights and liars (knights always tell the truth, while liars always lie). It is known that each of them has exactly one friend at the same table, and a knight's friend is a liar, while a liar's friend is a knight (friendship is always mutual). When asked "Is your friend si... | All those sitting at the table are paired as friends, which means there are an equal number of knights and liars. Consider any pair of friends. If they are sitting next to each other, the knight will answer "Yes" to the given question, and the liar will answer "No." If they are not sitting next to each other, their ans... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Mathematical logic (other) ] [ Divisibility of numbers. General properties ]
On a meadow, ladybugs gathered. If a ladybug has six spots on its back, it always tells the truth, and if it has four spots, it always lies, and there were no other ladybugs on the meadow. The first ladybug said: "Each of us has the same nu... | If the first ladybug tells the truth, then the second and third should also tell the truth, as they should have the same number of spots on their backs as the first. But the second and third ladybugs contradict each other, so at least one of them is lying. Therefore, the first ladybug is also lying.
Suppose each of th... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Raskin M.A.
On an island, there live chameleons of five colors. When one chameleon bites another, the color of the bitten chameleon changes according to some rule, and the new color depends only on the color of the biter and the color of the bitten. It is known that $\$ 2023 \$$ red chameleons can agree on a sequence ... | Let's start with an example of rules under which at least 5 red chameleons would be necessary for the described recoloring. Let's number the colors so that red is the first color and blue is the last. Then, let the rules be as follows: a chameleon of color $k 1$ can reduce the number of its color by a bite from a chame... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ] Joke problems
Calculate the product
$$
\left(100-1^{2}\right)\left(100-2^{2}\right)\left(100-3^{2}\right) \ldots\left(100-25^{2}\right)
$$
# | Pay attention to the factors hidden behind the ellipsis.
## Solution
The product is 0, since among the factors there will be (100 - $10^{2}$).
## Answer
0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Excellent student Polycarp and poor student Kolya were forming the smallest 5-digit number consisting of different even digits. Polycarp formed his number correctly, while Kolya made a mistake. However, it turned out that the difference between Kolya's number and the correct answer was less than 100. What numbers did P... | Think about what the first two digits of Polycarp's number and the last two digits of Kolya's number should be.
## Solution
If we proceed as in the previous problem, Polycarp should have formed the number 02468, but the first digit cannot be zero, so Polycarp formed the number 20468. Let's try to find Kolya's number.... | 20486 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ [equations in integers $]$ ]
In the room, there are three-legged stools and four-legged chairs. When people sat on all these seats, there were 39 legs in the room.
How many stools are in the room? | According to the condition, there are five- and six-legged creatures in the room, with a total of 39 legs. The number of legs of the five-legged creatures ends in 0 or 5. But it cannot end in 0: then the number of legs of the six-legged creatures would end in 9. In such a case, the number of five-legged creatures can b... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2In one move, you are allowed to either double the number or erase its last digit. Is it possible to get from the number 458 to the number 14 in several moves?
# | Try to multiply the original number by 2 until the first digit of the result becomes 7.
## Solution
You can first double the number, then strike out the last digit, or you can do the opposite - strike out the last digit first, then double the number. This has almost no effect on the value of the first digit of the re... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
I bought a lottery ticket, the sum of the digits of its five-digit number turned out to be equal to the age of my neighbor. Determine the number of this ticket, given that my neighbor solved this problem without difficulty.
# | Notice, there cannot be two unequal digits in the ticket.
## Solution
The ticket number is 99999. If there were at least two unequal digits in the ticket, they could be swapped, and the neighbor would not be able to solve the problem for sure. If all digits are equal but less than 9, there is always a possibility to ... | 99999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Folklore
In a herd consisting of horses, two-humped and one-humped camels, there are a total of 200 humps.
How many animals are in the herd if the number of horses is equal to the number of two-humped camels? | Let each two-humped camel "share" a hump with a horse. Then each animal will have one hump, so the number of humps will be equal to the number of animals in the herd.
## Oтвет
200. | 200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
|Rabbits sawed several logs. They made 10 cuts and got 16 chunks. How many logs did they saw
# | Recall problem 89914 - the problem about how the rabbits made 10 cuts to split one log.
## Solution
From each log, you get one more chunk than the number of cuts. Since there are 6 more chunks, it means there were 6 logs. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A person has 10 friends and over several days invites some of them to visit so that the company never repeats (on some days he may not invite anyone). How many days can he do this?
# | The first method. For each of the 10 friends, there are two possibilities: to invite them or not. Therefore, the total number of ways to choose a group of friends is $2^{10}$.
The second method. The problem is equivalent to finding the number of subsets of a set with 10 elements. As is known, there are $2^{10}$ of the... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the school canteen queue, Vika, Sonya, Borya, Denis, and Alla are standing. Vika is in front of Sonya but after Alla; Borya and Alla are not standing next to each other; Denis is not standing next to Alla, Vika, or Borya. In what order are the kids standing?
# | Recall problems $71,144,195$.
## Solution
For convenience, let's repeat the conditions: 1) Vika stands in front of Sonya but after Alла; 2) Borya and Alла do not stand next to each other; 3) Denis does not stand next to Alла, Vika, or Borya. From condition 1, it follows that the girls stand in the following order: Al... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Arithmetic. Mental calculation, etc. } \\ {[\underline{\text { Rebus }}]}\end{array}\right.$
Authors: Galnierein G.A., Grieorenko D.:
2002 is a palindrome year, meaning it reads the same backward as forward. The previous palindrome year was 11 years earlier (1991). What is the maximum n... | Let the current year be a palindrome of the form $\overline{a b b a}$. When will the next such year occur? Consider two cases:
a) $b=9$ (year of the form $\overline{a 99 a}, a<9)$. Then, 11 years later, another palindromic year will occur: $\overline{(a+1) 00(a+1)}$. For example, the years 3993 and 4004.
b) $b<9$. In... | 109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems ] [Algorithm theory]
Pouring milk. From an 8-liter bucket filled with milk, you need to measure out 4 liters using empty 3-liter and 5-liter buckets. | Let's record the solution in a table.
| Eight-liter bucket | Five-liter bucket | Three-liter bucket |
| :---: | :---: | :---: |
| 8 | | |
| 3 | 5 | |
| | 5 | |
| 3 | 2 | 3 |
| :---: | :---: | :---: |
| 6 | 2 | |
| 6 | | 2 |
| 1 | 5 | 2 |
| 4 | 4 | 3 | | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Three frogs on a swamp jumped in turn. Each landed exactly in the middle of the segment between the other two. The length of the second frog's jump is 60 cm. Find the length of the third frog's jump. | Regardless of how the frogs were sitting initially, after the first jump they will be on one straight line, with the first (A) in the middle
A B
Now the second frog (B) jumps. It flies a dis... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Berdnikov A.
Sasha writes a sequence of natural numbers on the board. The first number $N>1$ is written in advance. New natural numbers he gets by subtracting from the last written number or adding to it any of its divisors greater than 1. Can Sasha write the number 2011 on the board at some point for any natural $N>1... | First method. By adding $N$ each time, we get 2011N. By subtracting 2011 each time, we get 2011.
Second method. If $N$ is odd, add $N$ and get an even number. By adding or subtracting twos from it, we get 4022. Subtracting 2011, we get 2011.
## Answer
For any. | 2011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Three mad painters started painting the floor each in their own color. One managed to paint 75% of the floor red, another 70% green, and the third 65% blue. What part of the floor is definitely painted with all three colors? | Evaluation. 25% of the floor is not painted red, 30% of the floor is not painted green, and 35% of the floor is not painted blue.
25+30+35=90. From this, it follows that at least 10% of the floor is painted with all three colors.
An example where exactly 10% is painted with all three colors is clear from the evaluati... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Andjans A.
In a table with $m$ rows and $n$ columns, a horizontal move is defined as a permutation of the elements in the table such that each element remains in the same row it was in before the permutation; similarly, a vertical move is defined ("row" in the previous definition is replaced with "column"). Determine ... | The case $m=1$ or $n=1$ is trivial.
Let $m \neq 1$ and $n \neq 1$. We will show that two moves are insufficient. Suppose in the initial table, the top-left
square is \begin{tabular}{|l|l|}
\... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Author: Tolingo A.K.
Geologists took 80 cans of different known weights on an expedition (a list is available). After some time, the labels on the cans became unreadable, and only the supply manager knows what is in each can. He can prove this to everyone (i.e., justify what is in each can) without opening the cans an... | a) For simplicity, let's mentally add one empty can (with a mass of 0). The first weighing is organized by the caretaker as follows: 27 of the heaviest cans are placed on one pan, and 27 of the lightest cans on the other. This results in the greatest possible difference. The other members of the expedition must acknowl... | 98 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
Given $M$.
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)... | 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 insufficient to weigh 4 g, $p_{2} \leq 4=2^{2}$. Continuing... | 146 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zkov G.
A bank serves a million customers, the list of whom is known to Ostap Bender. Each has a six-digit PIN code, and different customers have different codes. In one move, Ostap Bender can choose any customer he has not yet chosen and peek at the digits of the code at any $N$ positions (he can choose different pos... | It is not difficult to do when $N=3$. Since any combination of the first three digits occurs exactly 1000 times, by looking at these digits for everyone except Koreiko, Bender will know them for Koreiko as well. Then it is sufficient to look at the last three digits of Koreiko's code.
We will prove that when $N<3$, it... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
There are 2004 boxes on the table, each containing one ball. It is known that some of the balls are white, and their quantity is even. You are allowed to point to any two boxes and ask if there is at least one white ball in them. What is the minimum number of questions needed to guarantee identifying a box containing a... | We will number the boxes (and thus the balls in them) from 1 to 2004 and denote a question by a pair of box numbers. We will call non-white balls black.
We will show that a white ball can be found in 2003 questions.
We will ask the questions $(1,2),(1,3),(1,2004)$. If all answers are positive, then the first ball is ... | 2003 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Two sages are playing the following game. The numbers $0,1,2, \ldots, 1024$ are written down. The first sage crosses out 512 numbers (of his choice), the second crosses out 256 of the remaining, then the first crosses out 128 numbers, and so on. On the tenth step, the second sage crosses out one number; two numbers rem... | Answer: with correct play, the difference between the remaining numbers is 32 (as they say, the "cost of the game" is 32). The following considerations lead to this answer: the first player aims to maximize the difference between the remaining numbers and should always try to "thin out" the remaining set of numbers as ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogosiov I.I.
On the reverse sides of 2005 cards, different numbers are written (one on each). In one question, it is allowed to point to any three cards and find out the set of numbers written on them. What is the minimum number of questions needed to find out what numbers are written on each card?
# | Let there be $N$ questions asked. It is clear that each card participates in at least one question; otherwise, we would not be able to determine the number on it.
Suppose there are $k$ cards that participate in exactly one question.
Then, two such cards cannot appear in the same question. Indeed, if two such cards we... | 1003 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the city of Udoyev, mayoral elections proceed as follows. If in a given round of voting no candidate receives more than half of the votes, a subsequent round is held with all candidates exce... | a) Ostap could not take the last, 2002nd place in the first round, as otherwise he would have been immediately eliminated from the candidates. Therefore, $k \leq 2001$.
Suppose all candidates in the first round received almost the same number of votes, Ostap took the second-to-last place, and in each subsequent round,... | 2001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a box, there are cards numbered with natural numbers from 1 to 2006. On the card numbered 2006 lies the card numbered 2005, and so on down to 1. In one move, it is allowed to take one top card (from any box) and place it either at the bottom of an empty box, or on a card with a number one greater. How many empty box... | The answer follows from the general fact: let the number of cards be $n$, where $2^{\mathrm{k}-1} \leq n < 2^{\mathrm{k}}$. If we assume that all cards can be placed in $k$ boxes, then card $a$ still needs to be moved so that all cards end up in one box. For this, at some point, it will be necessary to occupy the "top"... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
Ali-Baba and the bandit are dividing a treasure consisting of 100 gold coins arranged in 10 piles of 10 coins each. Ali-Baba chooses 4 piles, places a cup next to each, and sets aside several coins from each pile (not less than one, but not the entire pile). The bandit must then rearrange the cups, ch... | Let's show that Ali-Baba can achieve that in 7 piles there are no more than 4 coins, while the robber can achieve that there are no piles containing fewer than 4 coins. Therefore, Ali-Baba will take $100 - 7 \cdot 4 = 72$ coins.
First, let's prove that the robber can act in such a way that there are no piles containin... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
\left[\begin{array}{l}{\left[\begin{array}{l}\text { Order relation } \\ \text { [ }\end{array}\right]} \\ {[\text { Problems with inequalities. Case analysis }]} \\ {[\text { Examples and counterexamples. Constructions }]} \\ {[\text { Estimation + example }}\end{array}\right]
Author: Tokoreve C.i.
In a class of 30 ... | Students who perform better than most of their friends are called good. Let $n$ be the number of good students, and $k$ be the number of friends each student has. We will prove that $n \leq 25$. For this, we will consider two cases.
1) $k \geq 8$. Then the five worst students in the class are not good.
2) $k \leq 7$. ... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Head A.C.
Dima calculated the factorials of all natural numbers from 80 to 99, found the numbers that are the reciprocals of these factorials, and printed the resulting decimal fractions on 20 infinite tapes (for example, on the last tape, the number $\frac{1}{99!}=0, \underbrace{00 \ldots 00}_{155 \text { zeros }} 10... | $N=155$. Let on ribbons, on which numbers $1 / k!$ and $1 / l!$ ( $k < l$ ) are written, there are pieces of 155 characters that are indistinguishable. Since $1 / 100! < \frac{1}{10^{156}}$, therefore $N<156$.
Thus, a piece of 156 characters is always sufficient to determine from which strip it was cut. On the other h... | 155 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Combinations and Permutations ] $[$ Graph Theory (other) ]
In the parliament, there are 30 deputies. Any two of them are either friends or enemies, and each one is friends with exactly six others. Any three deputies form a committee. Find the total number of committees in which all three members are pairwise friends... | In how many committees does a deputy serve? In how many committees do the remaining two members - the deputy's friend and enemy - serve?
## Solution
Let's denote the deputies as points. We will connect the points with a red segment if the corresponding deputies are friends, and with a blue segment if they are enemies... | 1990 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7,8
In how many ways can the word "строка" be read, moving right or down?:
C T P O K A
T P O K A
P O K A
O K A
K A
A | If moving from the first letter "C", each time two options are provided - to go right or down.
It is clear that movement can only start from the first letter "C". With each move, we descend to the next diagonal, and we are faced with a choice - to go right or down. With each of these two options, we again descend to t... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
\begin{aligned} & \text { [Combinations and Permutations]} \\ & {[\text { Rule of Product }]}\end{aligned}
Nina has 7 different chocolate candies, and Kolya has 9 different caramels. In how many ways can they exchange five candies with each other? | See the solution to problem 30690.
## Answer
$C_{7}^{5} C_{9}^{5}=C_{7}^{2} C_{9}^{4}=2646$ ways.
## Problem | 2646 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can 12 white and 12 black checkers be placed on the black squares of a chessboard?
# | First, we need to choose 12 fields for the white checkers, and then from the remaining 20 fields, choose 12 fields for the black checkers.
## Answer
$$
C_{32}^{12} \cdot C_{20}^{12}=\frac{32!}{12!\cdot 12!\cdot 8!} \text { ways. }
$$
 $]
In a bag, there are 70 balls that differ only in color: 20 red, 20 blue, 20 yellow, and the rest are black and white.
What is the smallest number of balls that need to be drawn from the bag, without seeing them, to ensure that among them there are at least 10 balls of the same ... | By drawing 37 balls, we risk getting 9 red, blue, and yellow balls each, and there will be no ten balls of one color. If we draw 38 balls, however, the total number of red, blue, and yellow balls among them will be no less than 28, and the number of balls of one of these colors will be no less than ten (since $28>3 \cd... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let's call a natural number "remarkable" if it is the smallest among all natural numbers with the same digit sum as it has.
How many three-digit remarkable numbers exist?
# | "Remarkable" numbers with the sum of digits from 1 to 18 are either single-digit or two-digit. Therefore, remarkable three-digit numbers have a sum of digits from 19 to 27. Each such sum corresponds to one remarkable number. Consequently, there are nine three-digit "remarkable" numbers.
## Answer
9 numbers. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a chess club, 2 girls and 7 boys are attending. To participate in a competition, a team of four people needs to be formed, which must include at least one girl. In how many ways can this be done? | The team includes either one girl or two. Let's consider both cases. If the team has two girls, then two boys can be added to them in $\mathrm{C}_{7}^{2}$ ways. If the team includes only one girl (who can be chosen in two ways), then the team can be supplemented with three boys in $\mathrm{C}_{7}^{3}$ different ways. T... | 91 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
What is the maximum number of white and black chips that can be placed on a chessboard so that in each row and each column there are exactly twice as many white chips as black ones? | Evaluation. The number of chips on each vertical is a multiple of 3, meaning there are no more than 6, and on the entire board - no more than 48. Example: 32 white chips are placed on white squares, and 16 black chips are placed along the main "black" diagonal and along two parallel diagonals of "length" 4.
## Answer
... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Folklore
The vertices and midpoints of the sides of a regular decagon are marked (that is, a total of 20 points are marked).
How many triangles exist with vertices at the marked points? | A triangle is uniquely determined by its three vertices. From twenty points, three can be chosen in $C_{20}^{3}=$ 1140 ways. Considering that three points lying on one side of a decagon do not form a triangle, we get: $1140-10=1130$ triangles.
## Answer
1130 triangles. | 1130 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the warehouse, there are 200 boots of sizes 41, 42, and 43, and among these 600 boots, 300 are left and 300 are right.
Prove that from them, it is possible to form at least 100 complete pairs of shoes.
# | In each size of some boots, there are fewer right or left ones. Let's list these types of boots by size. Some type, for example, left, will repeat at least twice, for example, in sizes 41 and 42. But since the total number of left boots in these sizes is not less than 100 (left boots of size 43 are no more than 200), w... | 100 | Combinatorics | proof | Yes | Yes | olympiads | false |
$\begin{aligned} & \text { [Combinations and Permutations]} \\ & {[\text { Problems with Constraints }]}\end{aligned}$
There are 12 books on a shelf. In how many ways can five books be chosen from them such that no two of them are next to each other? | By putting the remaining books in correspondence with white balls, and the removed ones with black balls, we obtain a particular case of problem $\underline{60403}$.
## Answer
$\mathrm{C}_{8}^{5}=56$ ways. | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the pond, 30 pikes were released, which gradually eat each other. A pike is considered full if it has eaten no less than 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 pikes.
## Solution
Let $s$ be the number of satiated pikes. Then they together have eaten no fewer than $3s$ pikes. Since each pike can only be eaten once, and at least one pike remains at the end, $3s < 30$. Therefore, $s \leq 9$.
We will... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Each of the thirty sixth-graders has one pen, one pencil, and one ruler. After their participation in the Olympiad, it turned out that 26 students lost a pen, 23 - a ruler, and 21 - a pencil. Find the smallest possible number of sixth-graders who lost all three items. | 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 |
Lobanov M. S.
Around a round rotating table, on which there are 8 white and 7 black cups, 15 gnomes are sitting. They have put on 8 white and 7 black caps. Each gnome takes a cup whose color matches the color of their cap and places it in front of them, after which the table is rotated randomly. What is the maximum nu... | Let's consider an arbitrary arrangement of cups and write down their colors in a row. Below this row, we will also write down all its different cyclic shifts — a total of 14. We will count how many color matches there will be in the same position in the original arrangement and in the arrangements obtained by shifts. F... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## In the city, there are 57 bus routes. It is known that:
1) from any stop, you can get to any other stop without transferring;
2) for each pair of routes, there is one, and only one, stop where you can transfer from one of these routes to the other
3) each route has no fewer than three stops.
How many stops does ea... | Let there be $n$ stops on some route $a$. Take a stop $B$ that route $a$ does not pass through. From $B$, there is a route to each of the $n$ stops on route $a$, and there is exactly one such route, since two different routes cannot have two common stops. Each route passing through $B$ intersects route $a$. Therefore, ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
In the Prostokvashino primary school, there are only 20 children. Any two of them have a common grandfather.
Prove that one of the grandfathers has at least 14 grandchildren in this school.
# | Consider any schoolchild $A$ and his two grandfathers $X, Y$. If not all schoolchildren are grandchildren of one of them (in which case there is nothing to prove), then there exists a schoolchild $B$ who is not a grandson of $X$ (then he must be a grandson of $Y$), and a schoolchild $C$ who is not a grandson of $Y$ (th... | 14 | Combinatorics | proof | Yes | Yes | olympiads | false |
Rubbanov I.S.
A square is divided by lines into 25 smaller square cells. In some cells, one of the diagonals is drawn such that no two diagonals share a common point (even a common endpoint). What is the maximum possible number of drawn diagonals?
# | Example with 16 diagonals, see the figure.
Estimate. Suppose it was possible to draw 17 diagonals. We will present two ways to arrive at a contradiction.
First method. Each diagonal has tw... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berr S.L.
100 people came to a party. Then those who had no acquaintances among the attendees left. Then those who had exactly one acquaintance among the remaining people also left. Then, similarly, those who had exactly 2, 3, 4, ..., 99 acquaintances among the remaining people at the time of their departure, also lef... | It is not difficult to check that if all the guests, except for two people $A$ and $B$, were acquainted with each other, then at the end, everyone except $A$ and $B$ should remain, that is, 98 people.
Let's prove that 99 people could not remain. It is clear that person $A$, who initially had the fewest acquaintances (... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
There are 1955 points. What is the maximum number of triples that can be chosen from them so that any two triples have exactly one point in common?
# | Let's highlight one of the triples $\{A, B, C\}$. Each of the other triples contains exactly one of the points $A, B, C$. Let's consider two cases.
1) All other triples contain point $A$. Then there are no more than the number of non-intersecting pairs formed from 1954 points, which is no more than 977. This number of... | 977 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{ll}[\text { Classical combinatorics (other) })]\end{array}\right]$
On a $11 \times 11$ grid, 22 cells are marked such that exactly two cells are marked in each row and each column. Two arrangements of marked cells are equivalent if, by swapping any number of columns among themselves and any number... | Consider a bipartite graph, where the vertices are the rows and columns of the given board, and a row and a column are connected by an edge if and only if the cell at their intersection is marked. Note that permutations of the rows and columns of the original table correspond to renumberings of the vertices in each par... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Karashev $P$.
In each cell of a table consisting of 10 columns and $n$ rows, a digit is written. It is known that for any row $A$ and any two columns, there exists a row that differs from $A$ exactly in these two columns. Prove that $n \geq 512$. | Let $R_{0}$ be the first row of the table. Consider any set of an even number of columns and number them from left to right: $C_{1}, \ldots, C_{2 m}$. Then there is a row $R_{1}$ in the table that differs from $R_{0}$ exactly in columns $C_{1}$ and $C_{2}$; further, there is a row $R_{2}$ that differs from $R_{1}$ exac... | 512 | Combinatorics | proof | Yes | Yes | olympiads | false |
On a circle, there are 20 points. These 20 points are connected by 10 chords, none of which share endpoints and do not intersect.
How many ways can this be done
# | Let $a_{n}$ be the number of ways to connect $2 n$ points on a circle with $n$ non-intersecting chords. It is clear that $a_{1}=1$ and $a_{2}=2$. We will show that
$a_{n}=a_{n-1}+a_{n-2} a_{1}+a_{n-3} a_{2}+\ldots+a_{1} a_{n-2}+a_{n-1}$
Fix one of the $2 n$ points. The chord emanating from it divides the circle into ... | 16796 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.ya.
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) Consider the finite sequence $... | 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 |
A sheet of a notebook was colored in 23 colors by cells. A pair of colors is called good if there exist two adjacent cells colored with these colors. What is the minimum number of good pairs? | Evaluation. Let's draw a graph with 23 vertices (colors) and connect the vertices corresponding to good pairs with edges.
Consider two arbitrary vertices $A$ and $B$. Take some cell $K$, colored in color $A$, and some cell $L$, colored in color $B$. Construct a "chain" of cells connecting $K$ with $L$. Each pair of ad... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Combinations and Permutations } \\ {\left[\begin{array}{l}\text { Partitions into Pairs and Groups; Bijections }]\end{array}\right]}\end{array}\right]$
There are 20 people - 10 boys and 10 girls. How many ways are there to form a company in which there would be an equal number of boys an... | For each such company, associate a set of 10 people, which includes all the girls who joined the company and all the boys who did not join it.
## Solution
Let there be some company consisting of $k$ boys and $k$ girls. We will associate it with a set of 10 people, which includes $k$ girls who joined the company and $... | 184756 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The international commission consists of nine people. The commission's materials are stored in a safe. How many locks should the safe have, how many keys for them need to be made, and how should they be distributed among the commission members so that access to the safe is possible only when at least six members of the... | According to the condition, any five people cannot open the safe. This means they do not have the key to a certain lock. At the same time, any other member of the commission must have this key. Therefore, we need to install $C_{9}^{5}=C_{9}^{4}$ locks. Four keys to each lock are given to a certain quartet of commission... | 126 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kalinin D.A.
Find the maximum number of colors in which the edges of a cube can be painted (each edge in one color) so that for each pair of colors there are two adjacent edges painted in these colors. Adjacent are considered edges that have a common vertex.
# | There are several ways to color the edges of a cube in six colors while satisfying the condition of the problem. Here is one of them:
We will show that more than six colors are not possible... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
In the language of the AU tribe, there are two letters - "a" and "u". Some sequences of these letters are words, and each word contains no fewer than one and no more than 13 letters. It is known that if you write down any two words in a row, the resulting sequence of letters will not be a word. Find the ... | If all sequences, the number of letters in which is not less than 7 and not more than 13, are words, then, obviously, the condition of the problem is satisfied; in this case, the number of such words is $2^{7}+\ldots+2^{13}=2^{14}-2^{7}$. It remains to show that this number is the largest possible.
The first method. T... | 16256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$4-$
[Examples and counterexamples. Constructions]
One hundred cards were laid out in a row in ascending order: $00,01,02,03, \ldots, 99$. Then the cards were rearranged so that each subsequent card was obtained from the previous one by increasing or decreasing exactly one of the digits by 1 (for example, after 29 ca... | Evaluation. After the rearrangement, the sums of the digits of adjacent numbers differ by one. Therefore, either the parity of the sum of the digits of each ticket number matches the parity of its position, or, conversely, these parities are different for all ticket numbers. Before the rearrangement, the parities match... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
Petya has a total of 28 classmates. Each of the 28 has a different number of friends in this class. How many friends does Petya have? | Petya's classmates can have $0,1,2, \ldots, 28$ friends - a total of 29 options. But if someone is friends with everyone, then everyone has at least one friend. Therefore, either someone is friends with everyone, or someone is not friends with anyone. In both cases, there are 28 options left: $1,2, \ldots, 28$ or $0,1,... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A.Y. of evnin
2000 people registered on a new website. Each of them invited 1000 people to be their friends. Two people are considered friends if and only if each of them invited the other to be a friend. What is the minimum number of pairs of friends that could have formed? | Evaluation. A total of 2,000,000 invitations were sent, while the number of pairs on the site is $1000 \cdot 1999 = 1999000$. There are 1000 more invitations than pairs, so within at least 1000 pairs, two invitations were sent. Therefore, at least 1000 pairs were formed.
Example: Place everyone at the vertices of a re... | 1000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Processes and Operations ] [ Invariants and Semi-invariants ]
On a bookshelf, 30 volumes of an encyclopedia are arranged in some order. In one operation, it is allowed to swap any two adjacent volumes. What is the minimum number of operations required to guarantee that all volumes can be arranged in the correct orde... | Let's call a disorder a pair of volumes where the volume with the larger number is to the left of the volume with the smaller number. In one operation, the number of disorders changes by no more than 1.
## Solution
Suppose we have some arrangement of volumes on a shelf. Consider all possible pairs of volumes (there a... | 435 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a football team (11 people), a captain and his deputy need to be selected. How many ways can this be done
# | The captain can be any of the 11 football players. After choosing the captain, the role of his deputy can be claimed by any of the 10 remaining players. Thus, there are $11 \cdot 10=110$ different variants.
## Answer
110 ways.
At the dance floor, $N$ young men and $N$ young women have gathered. In how many ways can ... | 110 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## [ Combinations and Permutations $]$
[Tournaments and tournament tables] The Russian Chess Championship is played 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 |
[ Graph Traversal ]
There is a group of islands connected by bridges in such a way that from any island, you can reach any other island. A tourist visited all the islands, crossing each bridge exactly once. He visited the Triplet Island three times. How many bridges lead to the Triplet Island if the tourist
a) did no... | a) If the Triple tourist entered the island 3 times and exited 3 times, that means he used 6 bridges.
b) In this case, the tourist entered the island twice, but exited three times.
## Answer
a) 6 bridges; b) 5 bridges; c) 4 bridges. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Combinations and Permutations ]
a) From a class of 30 students, two students need to be selected to participate in a mathematics olympiad. In how many ways can this be done?
b) In how many ways can a team of three students be selected from the same class? | a) The first student can be chosen in 30 ways, the second, independently of the choice of the first student, in 29 ways. In this case, each pair is counted twice. Therefore, the total number of ways is $30 \cdot 29: 2$.
b) Similarly, we get $30 \cdot 29 \cdot 28$ variants for the sequential selection of three students... | 4060 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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