problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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[ Combinations and Permutations ]
In how many ways can 4 paints be chosen from 7 different ones?
# | Reasoning analogously to problem $\underline{30687}$, we get that the number of these ways ( $C_{7}^{4}$ ) is $\frac{7 \cdot 6 \cdot 5 \cdot 4}{4!}$.
## Answer
35 ways. | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Pascal's Triangle and Newton's Binomial ]
How many times greater is the sum of the numbers in the 101st row of Pascal's Triangle compared to the sum of the numbers in the 100th row?
# | Let in the 100th row stand the numbers $c_{0}, c_{1}, \ldots, c_{100}$. Then in the 101st row stand the numbers $c_{0}, c_{0}+c_{1}, c_{1}+c_{2}, \ldots, c_{99}+$ $c_{100}, c_{100}$
their sum is $2 c_{0}+2 c_{1}+\ldots+2 c_{100}$.
## Otvet
Twice. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Graph Theory (Miscellaneous)]
Several Top Secret Objects are connected by an underground railway in such a way that each Object is directly connected to no more than three others, and from each Object, it is possible to reach any other by underground travel with no more than one transfer. What is the maximum number o... | Evaluation. From this Object, you can reach three Objects in one "move", and with a transfer - to another $2 \cdot 3=6$ Objects. Therefore, there are no more than 10 Objects.
An example with 10 Objects is shown in the figure.
 In Wonderland, there are three cities $A, B$, and $C$. There are 6 roads from city $A$ to city $B$, and 4 roads from city $B$ to city $C$.
How many ways are there to travel from $A$ to $C$?
b) In Wonderland, they built another city $D$ and several new roads - two from $A$ to $D$ and two from $... | б) $K 6 \cdot 4=24$ old methods were added with $2 \cdot 2=4$ new ones.
## Answer
a) 24; b) 28 ways. | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In an ordinary set of dominoes, there are 28 tiles. How many tiles would a set of dominoes contain if the values indicated on the tiles ranged not from 0 to 6, but from 0 to 12?
# | Try to understand why a standard set of dominoes has exactly 28 tiles.
## Solution
In such a set, there would be 13 "doubles." The number of other tiles is equal to the number of pairs from 13 numbers, which is $13 \cdot 12: 2$ = 78.
## Answer
91 tiles. | 91 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2 [Rectangles and Squares. Properties and Characteristics]
Auto: Dorricheno $C$.
Which are more numerous: rectangles with integer sides and a perimeter of 1996, or rectangles with integer sides and a perimeter of 1998?
(Rectangles $a \times b$ and $b \times a$ are considered the same.) | If the perimeter of a rectangle is 1996, then the sum of the lengths of its adjacent sides is 998.
## Solution
If the perimeter of a rectangle is 1996, then the sum of the lengths of its adjacent sides is 998. This means the length of the shorter side can take values from 1 to 499. If the perimeter of the rectangle i... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Semenova $M$.
On a globe, 17 parallels and 24 meridians are drawn. Into how many parts is the surface of the globe divided? | Solve the problem for two meridians and one parallel.
## Solution
Meridians divide the globe into 24 parts (slices), and parallels divide each slice into $17+1=18$ parts. In total, $18 \cdot 24=432$ parts.
## Answer
Into 432 parts.
## Problem | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
How many integers exist from 0 to 999999, in the decimal representation of which there are no two adjacent identical digits?
# | The number of $n$-digit numbers with such a property is $9^{n}$: the first place can be any of 9 digits (all except 0), the second place can also be any of 9 digits (all except the first one), and so on.
## Answer
$10+9^{2}+9^{3}+9^{4}+9^{5}+9^{6}=\frac{9^{7}-1}{8}=597871$ numbers. | 597871 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Classical combinatorics (other).]
Given a six-digit phone number. From how many seven-digit numbers can it be obtained by erasing one digit
# | How many ways are there to insert one more digit into a phone number?
## Solution
Let $A B C D E F$ be a six-digit phone number. Let's count how many seven-digit numbers can be obtained from it by adding one digit. Before the number, you can prepend any of the 10 digits, but the number obtained by prepending the digi... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}[\text { Combinatorics (other) })] \\ [\text { Extreme Principle (other) })]\end{array}\right.$
In how many ways can the numbers from 1 to 100 be permuted so 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 |
8,9,10 |
How many six-digit numbers exist where each subsequent digit is less than the previous one? | The first method. Each such number corresponds uniquely to the selection of six digits from the number 9876543210. The second method. To obtain a number with "decreasing digits," you need to strike out any four digits from the number 9876543210.
## Answer
$$
C_{10}^{6}=C_{10}^{4}=210 \text { numbers. }
$$
Problem | 210 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Pascal's Triangle and the Binomial Theorem ]
Why do the equalities $11^{2}=121$ and $11^{3}=1331$ resemble rows of Pascal's Triangle? What is $11^{4}$? | $11^{2}=(10+1)^{2}=100+2 \cdot 10+1, 11^{3}=(10+1)^{3}=1000+3 \cdot 100+3 \cdot 10+1$.
$11^{4}=C_{4}^{0} \cdot 10^{4}+C_{4}^{1} \cdot 10^{3}+C_{4}^{2} \cdot 10^{2}+C_{4}^{3} \cdot 10+C_{4}^{4}=14641$.
## Answer
$11^{4}=14641$. | 14641 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Rule of product]
There are four weights and a two-pan balance without a pointer. How many different weights can be accurately weighed with these weights if
a) the weights can only be placed on one pan of the balance;
b) the weights can be placed on both pans of the balance
# | б) We can distribute the weights into 3 piles in $3^{4}=81$ ways (see problem $\underline{60348}$). We will set aside the first pile, and place the other two on the scales (the case where both of these piles are empty should be discarded). Since it does not matter which pans these piles end up on, 80 should be divided ... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Classical combinatorics (other) $]$
In the cabinet of Anchuria, there are 100 ministers. Among them, there are both corrupt and honest ministers. It is known that among any ten ministers, at least one minister is a corrupt one. What is the smallest number of corrupt ministers that can be in the cabinet?
# | There cannot be more than nine honest ministers: ten honest ministers would contradict the condition.
On the other hand, the presence of exactly nine honest ministers does not contradict the condition.
## Answer
91. | 91 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ [ product rule ]
In the USA, the date is typically written as the month number, followed by the day number, and then the year. In Europe, however, the day comes first, followed by the month and the year. How many days in a year cannot be read unambiguously without knowing which format it is written in
# | Obviously, these are the days whose number can be the number of the month, that is, takes a value from 1 to 12. Such days are $12 \cdot 12=144$. But the days, for which the number coincides with the month number, are understood unambiguously. Therefore, the number of days sought is $144-12=132$. | 132 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
There were seven boxes. In some of them, seven more boxes (not nested inside each other) were placed, and so on. In the end, there were 10 non-empty boxes.
How many boxes are there in total? | At each operation, one empty box is filled. Since there are now 10 non-empty boxes, 10 operations have been performed. With each operation, seven boxes were added. Therefore, in the end, there are $7+10 \cdot 7=77$ boxes. | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$3-$
The city plan has a layout represented by a $5 \times 10$ rectangular grid. One-way traffic is in effect on the streets: travel is only allowed to the right and upwards. How many different routes are there from the bottom-left corner to the top-right corner? | Each route can be associated with a string of 15 letters "P" and "V" ("P" means moving one cell to the right, "V" - one cell up).
## Solution
Each route can be associated with a string of 15 letters "P" and "V" (see hint), with exactly 5 "P" letters and 10 "V" letters. Conversely, each such string of letters can uniq... | 3003 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4+ Classical combinatorics (other).
A natural number $b$ is called lucky if for any natural number $a$, such that $a^{5}$ is divisible by $b^{2}$, the number $a^{2}$ is divisible by $b$.
Find the number of lucky natural numbers less than 2010. | Lemma. A number $b$ is lucky if and only if each prime number enters the prime factorization of $b$ with one of the following exponents: $0,1,2,3,4,6,8$.
Proof. Let's call a non-negative integer $k$ happy if there does not exist an integer $m$ such that $2 m < k < 2 m + 2$. If $k > 4$, then choose the largest number $... | 1961 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Degree of a vertex Descent method Authors: Karpov D.V., Karasev R. In the country $N 1998$ there are 1998 cities, and from each city, there are non-stop flights to three other cities (all flights are two-way). It is known that from any city, with several transfers, one can fly to any other city. The Ministry of Secu... | Consider a graph where the vertices are cities and the edges are air routes. This will result in a connected graph where the degrees of the vertices are 3.
Suppose that in this graph there are two intersecting (by vertex) cycles (see figure). Consider the vertex $O$, where they branch out. This vertex, obviously, has ... | 200 | Combinatorics | proof | Yes | Yes | olympiads | false |
Announce $A$.
$N$ friends simultaneously learned $N$ pieces of news, with each person learning one piece of news. They started calling each other and exchanging news.
Each call lasts 1 hour. Any number of news items can be shared in one call.
What is the minimum number of hours required for everyone to learn all the... | a) 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 5th hour. Therefore, it will take no less than 6 hours.
We will show that 6 hours are sufficient. The conversations can procee... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
Polina has a deck of 36 cards (4 suits with 9 cards each). She chooses half of the cards she wants and gives them to Vasilisa, and keeps the other half for herself. Each turn, the players take turns opening one card of their choice (the opponent sees the suit and rank of the opened card), starting with ... | If Polina takes all the hearts, all the aces, all the kings, and all the queens, then Vasilisa will not be able to score points on the ace, king, and queen of hearts, i.e., she will score no more than 15 points.
Now let's prove that for any choice of Polina, Vasilisa can earn at least 15 points. Lay out the cards in t... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
How many pairs of integers $x, y$, between 1 and 1000, exist such that $x^{2}+y^{2}$ is divisible by 7. | The number $x^{2}+y^{2}$ is divisible by 7 if and only if both numbers $x$ and $y$ are multiples of 7 (see problem 108745). The number of integers between 1 and 1000 that are multiples of 7 is 142 (since $1000=142 \cdot 7+6$). Therefore, the desired number is $142^{2}$.
## Answer
$142^{2}=20164$ pairs. | 20164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ [Decimal numeral system ] [ Problems with constraints ]
Find the number of five-digit numbers in decimal notation that contain at least one digit 8. | Find the number of all five-digit numbers and five-digit numbers that do not contain the digit 8.
## Solution
There are a total of 90000 five-digit numbers (see the solution to problem 60336). Let's find the number of five-digit numbers that do not contain the digit 8. The first digit of such a number cannot be 0 or ... | 37512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Gapochnik A.I.
How many integers from 1 to 1997 have a sum of digits that is divisible by 5? | Consider 10 consecutive numbers, starting from a number ending in zero and ending with a number ending in nine. The sums of the digits of these numbers also form 10 consecutive numbers, so among them, exactly two numbers are divisible by 5. The numbers from 0 to 1999 are divided into 200 such tens, therefore, among the... | 399 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Combinations and Permutations ]
[ $\underline{\text { Directed Graphs }}$ ]
In the discussion, 15 deputies participated. Each of them, in their speech, criticized exactly $k$ of the remaining 14 deputies.
For what smallest $k$ can we assert that there will be two deputies who criticized each other? | Consider a directed graph where the vertices correspond to deputies, and an edge leading from $A$ to $B$ means that deputy $A$ has criticized deputy $B$.
## Solution
If each deputy has criticized 8 others, then the number of edges in the graph is $15 \cdot 8 = 120$, which is greater than the number of pairs $C_{15}^{... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Sorting in ascending (descending) order ] [ Classical combinatorics (other). $\quad]$
a) A traveler stopped at an inn, and the owner agreed to accept rings from a golden chain the traveler wore on his wrist as payment for his stay. However, he set a condition that the payment should be daily: each day the owner shou... | a) It is enough to cut two rings so that pieces of three and six rings are separated. On the third day, the traveler gives the piece of three rings and receives two rings as change, and on the sixth day, the piece of six rings and receives five rings as change.
b) Arrange the resulting pieces of the chain (not countin... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Divisibility of numbers. General properties ] $[$ Classical combinatorics (other) $]$
How many natural numbers less than a thousand are not divisible by either 5 or 7? | First, we will strike out from the set of numbers $1,2, \ldots, 999$ the numbers that are multiples of 5; their quantity is $[999 / 5]=199$. Then, from the same set of numbers $1,2, \ldots, 999$, we will strike out the numbers that are multiples of 7; their quantity is $[999 / 7]=142$. In this process, numbers that are... | 686 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Layouts and partitions ] [ Product rule $\quad]$
How many four-digit numbers (from 0001 to 9999) exist such that the sum of the first two digits equals the sum of the last two digits
# | Let the sum of the first two digits be $n$, and the sum of the last two digits also be $n$. The value of $n$ ranges from 1 to 18. If the number of two-digit numbers whose digits sum to $n$ is $a_{n}$, then the desired number is $a_{1}^{2}+a_{2}^{2}+\ldots+a_{18}^{2}$. A two-digit number whose digits sum to $n$ consists... | 669 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Fomin S.B.
Among ten-digit numbers, which are there more of: those that can be represented as the product of two five-digit numbers, or those that cannot be represented this way
# | The number of ten-digit numbers that can be represented as the product of two five-digit numbers is no more than the number of all pairs of five-digit numbers (including pairs of identical numbers). (In fact, there are significantly fewer: firstly, not all products of two five-digit numbers will be ten-digit numbers, a... | 110 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ $[\quad$ Rule of product $\quad]$
a) Leshа climbs a ladder with 10 steps. In one jump, he can either go up one step or two steps. In how many ways can Leshа climb the ladder?
b) When descending the same ladder, Leshа skips some steps (he can even skip all 10). In how many ways can he descend the ladder?
# | a) Let $a_{n}$ denote the number of ways to climb a staircase of $n$ steps, adhering to the conditions of the problem.
Obviously, $a_{1}=1, a_{2}=2$.
Suppose Petya jumps onto a staircase of $n>2$ steps. If the first jump is to two steps, then he has $n-2$ steps left to jump, and the number of ways to complete the asc... | 512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shaovalov A.v.
Let's call a ticket with a number from 000000 to 999999 excellent if the difference between some two adjacent digits of its number is 5.
Find the number of excellent tickets. | Let's calculate the number of non-distinct tickets. As the first digit, we can choose any of the 10 digits. Digits that differ by 5 form pairs: 0-5, 1-6, 2-7, 3-8, 4-9, so when the first digit is chosen, for the second digit in a non-distinct ticket, we can choose any of the 9 digits (excluding the one that pairs with ... | 409510 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Counting in two ways ] $\left[\begin{array}{ll}\text { Trees }\end{array}\right]$
A travel agency ran a promotion: "Buy a trip to Egypt, bring four friends who also buy a trip, and get the cost of your trip back." During the promotion, 13 buyers came on their own, and the rest were brought by friends. Some of them b... | Each of the $x$ "lucky ones" brought 4 friends. Therefore, the number of "brought" customers is $4x$, and 13 more came on their own, so the total number of tourists was $13 + 4x$. On the other hand, $x$ people brought new customers, while 100 did not, meaning the total number of tourists was $x + 100$. Thus, $13 + 4x =... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Calculate the coefficient of $x^{100}$ in the polynomial $\left(1+x+x^{2}+\ldots+x^{100}\right)^{3}$ after combining all like terms.
# | The coefficient of $x^{100}$ is equal to the number of solutions of the equation $p+q+r=100$ in non-negative integers.
## Solution
Multiplying the polynomial $1+x+x^{2}+\ldots+x^{100}$ by itself twice, we get the sum of monomials of the form $x^{p} x^{q} x^{r}$, where $p$, $q, r$ independently take values from 0 to 1... | 5151 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A 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 could form the word "MAMA", 30 children could form the word "NYANYA", and 40 children could form the word "MANYA". Ho... | 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 child can form either the word MAMA (20 such children) or the word NYANYA... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the old manor, the house is surrounded in a circle by tall trees - pines, spruces, and birches. There are a total of 96 trees. These trees have a strange property: of the two trees growing one apart from any coniferous tree - one is coniferous, and the other is deciduous, and of the two trees growing three apart fro... | Let's mentally remove half of the trees - those planted every other one. Then 48 trees will remain, and the condition will be as follows: of the two trees growing next to a coniferous one, one is coniferous and the other is a birch, and of the two trees growing one apart from the coniferous one, one is coniferous and t... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tompsongo 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 the 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 con... | 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 |
Find such a three-digit number that, when doubled, we get the number expressing the number of digits required to write all consecutive integers from one to this sought three-digit number (inclusive).
# | Let $x$ be the desired number. There are 9 single-digit numbers, 90 two-digit numbers, and the number of three-digit numbers up to $x$ inclusive is $x-99$. The number of digits required to write all consecutive numbers from 1 to $x$ is $9 + 2 \cdot 90 + 3 \cdot (x - 99) = 3x - 108$. We obtain the equation $2x = 3x - 10... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Prozvolov V.V.
Given fifty different natural numbers, twenty-five of which do not exceed 50, and the rest are greater than 50 but do not exceed 100. Moreover, no two of them differ exactly by 50. Find the sum of these numbers. | Subtract 50 from each number that is greater than 50. We will get 50 different numbers, that is, numbers from 1 to 50. Their sum is $1+2+\ldots+50=25 \cdot 51$, and the sum of the original numbers is $-25 \cdot 51+25 \cdot 50=25 \cdot 101$.
## Answer
2525. | 2525 | Number Theory | 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 place? | Let's 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 n... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Given a regular nonagon.
In how many ways can three of its vertices be chosen so that they form the vertices of an isosceles triangle? | For every two vertices of the nonagon, there exists exactly one vertex that is equidistant from them, so each of the resulting isosceles but not equilateral triangles is uniquely determined by its base. The number of ways to choose two points out of nine is $C_{9}^{2}=36$. However, with this method of counting, each of... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tsar Gvidon had 5 sons. Among his descendants, 100 each had exactly 3 sons, and the rest died childless.
How many descendants did Tsar Gvidon have
# | When counting the descendants, don't forget about the sons of Gvidon.
## Solution
Every descendant of King Gvidon is either a son of one of his descendants or a son of Gvidon himself. According to the condition, all the descendants of Gvidon had a total of 300 sons. And Gvidon himself had 5 sons, so the total number ... | 305 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$3+$
$$
\text { [ Recurrence Relations ] }
$$
a) The "Tower of Hanoi" puzzle consists of eight disks arranged in decreasing size on one of three pegs. The task is to move the entire tower to another peg, moving only one disk at a time and never placing a larger disk on a smaller one. Prove that the puzzle has a solut... | a) Let the minimum number of steps to move a tower of $n$ disks be $K_{n}$. Express $K_{n+1}$ in terms of $K_{n}$. Consider a tower of $n+1$ disks. We can divide the entire procedure into three stages.
Stage 1 - from the beginning to the first move of the bottom disk. By the end of this stage, the tower of $n$ disks s... | 4373 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
|
| | $[$ Pigeonhole Principle (other) $)]$ | |
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 |
Zamyatin V.
Vladimir wants to make a set of cubes of the same size and write one digit on each face of each cube so that he can use these cubes to form any 30-digit number. What is the smallest number of cubes he will need? (The digits 6 and 9 do not turn into each other when flipped.)
# | No less than 30 units, twos, ..., nines, and no less than 29 zeros. In total, no less than 50 cubes. It is not difficult to arrange no less than 30 instances of each digit on 50 cubes so that the digits on each cube do not repeat.
## Answer
50 cubes. | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$3+$ [ Examples and counterexamples. Constructions ]
In a square table of size $100 \times 100$, some cells are shaded. Each shaded cell is the only shaded cell either in its column or in its row. What is the maximum number of cells that can be shaded
# | Example. We will color all the cells of one row and all the cells of one column, except for their common cell. In this case, the condition of the problem is satisfied and exactly 198 cells are colored.
Evaluation. For each colored cell, we will highlight the line (row or column) in which it is the only colored one. In... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\begin{aligned} & \text { [Systems of points and segments. Examples and counterexamples] } \\ & {[\quad \underline{\text { Classical combinatorics (other) }}]}\end{aligned}$
On a plane, 10 equal segments were drawn, and all their points of intersection were marked. It turned out that each point of intersection divide... | On each segment, there are no more than two points. On the other hand, each intersection point belongs to at least two segments. Therefore, there are no more than $10 \cdot 2: 10=10$ points. An example with 10 points is shown in the figure.
.
Evaluation. Suppose there is a string with an arbitrary arrangement of crosses and noughts. We will make allowed moves, moving the noughts to the right or left edge so that there are n... | 200 | Combinatorics | 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 are 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 |
3+
Out of 101 Dalmatians, 29 have a spot only on the left ear, 17 have a spot only on the right ear, and 22 Dalmatians have no spots on their ears.
How many Dalmatians have a spot on the right ear? # | Subtract from 101 the number of Dalmatians that have a spot only on their left ear, and the number of those that have no spots at all:
$101-22-29=50$.
## Answer
50 Dalmatians.
## Problem | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tomonigo A.K.
How many different ways are there to break down the number 2004 into natural addends that are approximately equal? There can be one or several addends. Numbers are considered approximately equal if their difference is no more than 1. Ways that differ only in the order of the addends are considered the sa... | Each method is uniquely determined by the number of summands. Indeed, let 2004 be split into $n$ summands, $r$ of which are equal to $q+1$, and the rest are equal to $q(0 \leq r<n)$. Then $2004=q n+r$. Thus, the numbers $q$ and $r$ are the quotient and remainder of the division of 2004 by $n$, they are uniquely determi... | 2004 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Ribalko A.V.
In the National Basketball Association, there are 30 teams, each of which plays 82 games with other teams in the regular season. Can the Association's management divide the teams (not necessarily equally) into Eastern and Western Conferences and create a schedule such that the games between teams from di... | Let $x$ and $y$ be the total number of matches played within the Eastern and Western conferences, respectively, and $z$ be the number of matches between teams from different conferences. We need to prove that the equation $z = 1/2(x + y + z)$ is impossible.
Each of the $k$ teams in the Eastern conference plays 82 game... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
At a chamber music festival, six musicians gathered. At each concert, some of the musicians perform, while the others listen from the audience. What is the minimum number of concerts required for each of the six musicians to have listened to (from the audience) all the others? | A total of $6 \cdot 5=30$ auditions are needed. No more than 9 auditions can take place in one concert. Therefore, there must be at least four concerts.
Example: Musicians with numbers (4,5,6), (2,3,6), (1,3,5), and (1,2,4) perform in the concerts.
## Answer
In 4 concerts. | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
Consider an $8 \times 8$ board, the cells of which are not yet colored. In how many ways can the board be colored in black and white so that there are 31 black cells and no two black cells share a side? (Two colorings are considered different if there is a cell that is white in one of these colorings an... | Let's divide the board into 16 $2 \times 2$ squares. It is clear that one of the squares is special, containing only one black cell, while the other fifteen squares each contain two black cells, located in opposite corners. The coloring of a non-special square uniquely extends to an adjacent (by side) non-special squar... | 68 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Folkoro
In a certain state, the airline system is arranged in such a way that each city is connected by air routes
to no more than three other cities, and from any city, you can reach any other city with no more than one
transfer. What is the maximum number of cities that can be in this state? | Evaluation. From a fixed city $A$, one can directly reach no more than three cities, and with one transfer - no more than $3 \cdot 2=6$ additional cities. Thus, the total number of cities can be no more than ten. For an example of a network of 10 cities, see the figure below.
# | Evaluation. In each match, either 3 or 2 points are awarded. A total of $10 \cdot 9: 2=45$ matches were played, so no more than 135 points were awarded. Thus, $10 x \leq 135$, which means $x \leq 13.5$. Since $x$ is an integer, $x \leq$ 13.
Example. Arrange the teams in a circle and sequentially divide them into 5 pai... | 13 | Other | math-word-problem | Yes | Yes | olympiads | false |
3 [Pairings and Groupings; Bijections $]
Three people are playing table tennis, with the player who loses a game giving up their place to the player who did not participate in it. In the end, it turned out that the first player played 10 games, the second - 21. How many games did the third player play?
# | The first player plays the rarest every second game.
## Solution
According to the condition, the second player played 21 games, so there were at least 21 games in total. Out of every two consecutive games, the first player must participate in at least one, which means there were no more than $2 \cdot 10+1=21$ games. ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Ten numbers are written around a circle, the sum of which is 100. It is known that the sum of any three consecutive numbers is not less than 29.
Indicate the smallest number $A$ such that in any such set of numbers, each number does not exceed $A$. | Each number from the given set of numbers is complemented to 100 by three "neighbors" of threes.
## Solution
Let $X$ be the largest of the numbers. The remaining numbers will be divided into three "neighbor" triples. The sum of the numbers in each such triple is at least 29, therefore,
$X \leq 100-3 \cdot 29=13$.
A... | 13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Around a round table, 10 boys and 15 girls sat down. It turned out that there were exactly 5 pairs of boys sitting next to each other.
How many pairs of girls sitting next to each other are there?
# | How many groups of boys sitting next to each other are there at the table?
## Solution
Groups of boys sitting next to each other alternate with groups of girls sitting next to each other. Let the number of groups of boys sitting next to each other be $k$. Then the number of groups of girls sitting next to each other ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $M$ - be a finite set of numbers. It is known that among any three of its elements, there will be two whose sum belongs to $M$.
What is the maximum number of elements that can be in $M$? | Consider either the four largest or the four smallest numbers.
## Solution
Example of a set of 7 elements: $\{-3,-2,-1,0,1,2,3\}$.
We will prove that the set $M=\left\{a_{1}, a_{1}, \ldots, a_{n}\right\}$ of $n>7$ numbers does not have the required property. We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8,9 |
| :---: | :---: | :---: |
Ten points are marked on a circle. How many non-closed, non-self-intersecting nine-segment broken lines with vertices at these points exist? | The first point can be chosen in ten ways. Each of the following eight points can be chosen in two ways: it must be adjacent to one of the previously chosen points (otherwise, a self-intersecting broken line would result). Since the beginning and end are distinguishable in this counting, but not in the broken line, the... | 1280 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Pascal's Triangle and Newton's Binomial ]
How many rational terms are contained in the expansion of
a) $(\sqrt{2}+\sqrt[4]{3})^{100}$
b) $(\sqrt{2}+\sqrt[8]{3})^{300}$? | a) For the number $(\sqrt{2})^{100-k}(\sqrt[4]{3})^{k}$ to be rational, $k$ must be a multiple of 4. There are 26 such integers: $0, 4, 8$, ..., 96, 100.
b) For the number $(\sqrt{2})^{300-k}(\sqrt[3]{3})^{k}$ to be rational, $k$ must be a multiple of 3, and $300-k$ must be even. Therefore, $k$ must be a multiple of 6... | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How to connect 50 cities with the smallest number of air routes so that you can travel from any city to any other, making no more than two transfers?
# | Let's highlight one city and connect it by air routes to each of the other 49 cities. For this, 49 air routes will be required.
A smaller number of air routes cannot suffice. Indeed, according to problem 31098 g), in a connected graph with 50 vertices, there are no fewer than 49 edges. | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$[$ Classical combinatorics (miscellaneous).]
In a dark room, on a shelf, there are four pairs of socks of two different sizes and two different colors lying in a random order. What is the smallest number of socks that need to be moved from the shelf to a suitcase, without leaving the room, to ensure that the suitcase... | First, note that seven socks are sufficient. Indeed, only one sock is not taken, which means all pairs of socks except one are taken. Therefore, in the suitcase, there are two pairs of different colors and sizes, specifically the pair that differs from the one not taken only by color, and the pair that differs from the... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Tasks with constraints $]$
A New Year's garland hanging along the school corridor consists of red and blue bulbs. Next to each red bulb, there is definitely a blue one. What is the maximum number of red bulbs that can be in this garland if there are 50 bulbs in total?
# | Let's calculate the minimum number of blue bulbs that can be in the garland. We can assume that the first bulb is red. Since there must be a blue bulb next to each red bulb, three red bulbs cannot go in a row. Therefore, among any three consecutive bulbs, at least one bulb must be blue. Then, among the first 48 bulbs, ... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the number of zeros with which the number $11^{100}-1$ ends.
# | Use the binomial expansion of the expression $(10+1)^{100}$.
## Solution
$S=(10+1)^{100}-1=\ldots+(100 \cdot 99 \cdot 98: 6) \cdot 10^{3}+(100 \cdot 99: 2) \cdot 10^{2}+100 \cdot 10+1=\ldots+(33 \cdot 49) \cdot 10^{5}+495000+1000=A$ +496000, where all terms in place of the ellipsis are divisible by $10^{4}$, meaning ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the country of Drevland, there are 101 cities, and some of them are connected by roads. At that, any two cities are connected by exactly one path.
How many roads are there in this country?
# | See problems $\underline{30784}$ and $\underline{31098}$ b).
## Answer
100 roads. | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Planar graphs. Euler's formula ]
In the country of Lake, there are seven lakes connected by ten non-intersecting channels, such that one can sail from any lake to any other. How many islands are there in this country?
# | Consider a planar graph, where the vertices correspond to lakes, and the edges - to channels. Let it divide the plane into $F$ pieces. By Euler's formula (problem $\underline{30759}$) $7-10+F=2$, so $F=5$. One of these pieces is the mainland, and the rest are islands.
## Answer
4 islands. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[$ Inclusion-Exclusion Principle $]$
Petya plans to spend all 90 days of his vacation in the village and during this time, he will go swimming every second day (i.e., every other day), go to the store for groceries every third day, and solve math problems every fifth day. (On the first day, Petya did the first, secon... | Let's number the days from 0 to 89. Petya swims on days with even numbers, goes to the store on days with numbers divisible by 3, and solves problems on days with numbers divisible by 5.
Thus, we need to answer two questions:
1) How many even numbers from 0 to 89 are not divisible by either 3 or 5?
2) How many odd nu... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
In the spring round of the 2000 Cities Tournament, senior high school students from country $N$ were given six problems. Each problem was solved by exactly 1000 students, but no two students solved all six problems together. What is the minimum possible number of senior high school students from country... | Evaluation. Since each problem was solved by someone, no student could have solved five problems. If someone solved exactly four problems, then no student could have solved both the fifth and sixth problems together. Thus, there are 1000 students who solved the fifth problem, and another 1000 who solved the sixth. In t... | 2000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Frankin B.R.
In a tournament with 20 athletes, 10 referees were involved. Each athlete played against each other once, and each match was officiated by exactly one referee. After each game, both participants took a photo with the referee. A year after the tournament, a stack of all these photos was found. It turned ou... | Let's call a person suspicious if we cannot determine whether they are an athlete or an referee. Note that each referee was only photographed with athletes, that is, no more than 20 people. On the other hand, each athlete was photographed with all other athletes and at least one referee, that is, no fewer than 20 peopl... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A.K.
In the plane, there is an open, non-self-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 differe... | 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.
 In the store "Everything for Tea," there are 5 different cups and 3 different saucers. How many ways can you buy a cup?
b) The store also has 4 teaspoons. How many ways can you buy a set consisting of a cup, a saucer, and a teaspoon?
c) The store still sells 5 cups, 3 saucers, and 4 teaspoons. How many ways can yo... | a) Let's choose a cup. To it, we can choose any of the three saucers. Therefore, there are 3 different sets, so_
b) Any of the 15 sets from part a) can be complemented with a spoon in 4 ways. Therefore, the total number of possible sets is_
c) There are three different cases: the first - a cup with a saucer is purc... | 47 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2 [ Problems with constraints ]
From a class of 28 students, 4 people are appointed to duty in the cafeteria.
a) In how many ways can this be done?
b) How many ways are there to form a team of duty students that includes Kolya Vasin, a student from this class? | б) If Kolya is already chosen, then we need to choose 3 people from the remaining 27.
## Answer
a) $C_{28}^{4}=20475$ ways; b) $C_{27}^{3}=2925$ ways. | 2925 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can a committee of three people be formed, choosing its members from four married couples;
# | Choose first the family, a representative of which will not enter the commission, and then in each of the remaining families -
## Answer
$4 \cdot 2^{3}=32$ ways.
## Problem | 32 | Combinatorics | 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 able... | 1) Let's first show that it's 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, 9... | 1986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Zakorko P.
Karabas-Barabas has a large plot of land in the shape of a convex $\$ 12 \$$-sided polygon, with lanterns at each vertex. Karabas-Barabas needs to place a finite number of lanterns inside the plot, divide it into triangular plots with vertices at the lanterns, and distribute these plots to the actors of the... | It is not difficult to provide an example in which there are $\$ 24 \$$ such plots:
Let's try to prove that it is not possible to have more than $\$ 24 \$$ plots. First, consider a more gen... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Lifshits $\underline{\text { K. }}$.
Hydras consist of heads and necks (each neck connects exactly two heads). With one sword strike, one can cut all the necks coming out of some head $A$ of the hydra. But in this case, from head $A$, one neck will instantly grow to all heads with which $A$ was not connected. Heracles... | Let's move on to a graph where heads are vertices, necks are edges, and a strike on the necks coming out of head $A$ is called inverting vertex $A$.
If there is a vertex $X$ of degree no more than 10, it is sufficient to invert its neighbors, and it will separate. If there is a vertex connected to all vertices except ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kazitsyna T.v.
Borya and Misha are traveling by train and counting the poles outside the window: "one, two, ...". Borya cannot pronounce the letter "R", so he skips numbers that contain the letter "R" and immediately says the next number without the letter "R". Misha cannot pronounce the letter "SH", so he skips numbe... | Borya pronounces numbers that do not contain the digits 3 and 4 - among the first hundred numbers, there are $(10-2)^{2}=64$ such numbers (there are 8 options for both the tens and units digits), so in fact, there were 64 poles.
Misha skips numbers that contain the digit 6 in their notation. Therefore, after counting ... | 81 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A point exits from the origin on a line and makes $a$ steps one unit to the right, $b$ steps one unit to the left in some order, with $a > b$. The range of the point's movement is defined as the difference between the greatest and the smallest coordinates of the point during the entire movement.
a) Find the greatest p... | b) At the end of its path, the point inevitably has the coordinate $a - b$ (regardless of the order of steps to the right and left). Therefore, the span cannot be less than $a - b$.
Such a span can be achieved, for example, as follows: the point moves right-left exactly $b$ times, and then $a - b$ times to the right.
... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y. A cube with a side of 20 is divided into 8000 unit cubes, and a number is written in each cube. It is known that in each column of 20 cubes, parallel to the edge of the cube, the sum of the numbers is 1 (columns in all three directions are considered). In some cube, the number 10 is written. Through th... | Through the given cube K, one horizontal layer G and two vertical layers pass. The sum of all numbers in 361 vertical columns, not included in the last two layers, is 361. From this sum, we need to subtract the sum $S$ of the numbers lying in the cubes at the intersection of these columns with G (there are 361 such cub... | 333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A cube with a side of 10 is divided into 1000 smaller cubes with an edge of 1. In each small cube, a number is written, and the sum of the numbers in each column of 10 cubes (in any of the three directions) is 0. In one of the cubes (denoted as $A$), the number 1 is written. Three layers, parallel to the faces of the c... | Through the given cube $A$ pass one horizontal layer $G$ and two vertical layers. The sum of all numbers in 81 vertical columns not included in the last two layers is 0. From this sum, we need to subtract the sum $S$ of the numbers lying in the cubes at the intersection of these columns with $G$ (there are 81 such cube... | -1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Marachev A.
A cube $3 \times 3 \times 3$ is made from $1 \times 1 \times 1$ cubes. What is the maximum number of cubes that can be removed from it so that the remaining figure has the following two properties:
- from the side of each face of the original cube, the figure looks like a $3 \times 3$ square (looking perp... | Example. The layers from bottom to top are shown in the figure. The 13 remaining cubes are marked in black. Each layer has cubes in all columns and rows. When the layers are overlaid, a black $3 \times 3$ square is formed. Therefore, the first condition is met. The bottom and middle layers are connected and glued by th... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.i.
A chess piece can move 8 or 9 squares horizontally or vertically. It is forbidden to move to the same square twice.
What is the maximum number of squares this piece can traverse on a $15 \times 15$ board? (The traversal can start from any square.)
# | The central cell cannot be reached from any other cell. Cells in the central vertical (horizontal) line can only be reached from cells of this vertical line. Conversely, starting from such a cell, the figure will not leave the central vertical, meaning it will cover no more than 14 cells.
The remaining $225-29=196$ ce... | 196 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a $9 \times 9$ square table, 9 cells are marked at the intersections of the second, fifth, and eighth rows with the second, fifth, and eighth columns. How many ways are there to get from the bottom-left cell to the top-right cell, moving only through unmarked cells upwards or to the right? | Let $a_{i, j}$ be the number of paths from the bottom-left cell to the cell ( $i, j$ ), not passing through the marked cells. Note that if the cell ( $i, j$ ) is not marked, then $a_{i, j}=a_{i-1, j}+a_{i, j-1}$ (and if it is marked, then $a_{i, j}=0$ ). Now it is not difficult to calculate these numbers for each cell ... | 678 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Skeletons of Polyhedral Figures ]
A toy factory produces wireframe cubes, with small colorful beads located at their vertices. According to the GOST standard, each cube must use beads of all eight colors (white and the seven colors of the rainbow). How many different models of cubes can the factory produce?
# | If a cube is fixed, then 8 different balls can be placed at its vertices in 8! ways. But the cube can be rotated: each of its six faces can be made the bottom one and placed on it in four ways. Therefore, each cube corresponds to $6 \cdot 4=24$ "colorings", and the total number of models is $8! / 24=8 \cdot 7 \cdot 6 \... | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In Anchuria, a unified state exam is taking place. The probability of guessing the correct answer to each question on the exam is 0.25. In 2011, to obtain a certificate, one needed to answer three questions correctly out of 20. In 2012, the School Administration of Anchuria decided that three questions were too few. No... | If a graduate guesses the answers, the Unified State Exam (EGE) can be considered a Bernoulli scheme with a success probability of $p=0.25$ and a failure probability of $q=0.75$. In 2011, to pass the exam, one needed to answer at least three questions correctly. It is more convenient to find the probability of the oppo... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Coloring Vertex Degree
20 phones are connected by wires in such a way that each wire connects two phones, each pair of phones is connected by no more than one wire, and no more than two wires extend from each phone. It is necessary to color the wires (each wire entirely in one color) so that the wires extending from e... | We will represent phones as points and the wires connecting them as segments connecting these points. Since no more than two segments come out of each point, the resulting graph breaks down into open paths and simple cycles. Note that an open path can be painted with two colors, and a simple cycle with three. For this,... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can the numbers from 1 to 100 be arranged in a $2 \times 50$ rectangle so that any two numbers differing by 1 always end up in cells sharing a side?
# | Note that there is a one-to-one correspondence between the ways of arranging numbers and the rook's "tours" of the $2 \times 50$ board: if a method is given, we traverse the cells in the order of the numbers; if a tour is given, we number the cells in the order of the tour. This correspondence allows us to reason both ... | 4904 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
In a certain country, there are 100 cities (consider the cities as points on a plane). In a directory, for each pair of cities, there is a record of the distance between them (a total of 4950 records).
a) One record has been erased. Can it always be uniquely restored from the remaining ones?
b) Suppose... | a) Let 98 points lie on a straight line $l$, and two points $A$ and $B$ lie outside it. If the distance between $A$ and $B$ is unknown, it cannot be restored: by replacing point $B$ with $B^{\prime}$, symmetric to $B$ relative to $l$, the other distances will not change.
b) By induction, we will show that for $n \geq 4... | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Suppose we have 1,000,000 bus tickets numbered from 000000 to 999999. We will call a ticket lucky if the sum of the first three digits of its number equals the sum of the last three digits. Let \( N \) be the number of lucky tickets. Prove the following equalities:
a) \(\left(1+x+\ldots+x^{9}\right)^{3}\left(1+x^{-1}+... | a) By expanding the product $\left(1+\ldots+x^{9}\right)\left(1+\ldots+x^{9}\right)\left(1+\ldots+x^{9}\right)\left(1+\ldots+x^{-9}\right)\left(1+\ldots+x^{-9}\right)\left(1+\ldots+x^{-9}\right)$, we obtain a sum of expressions of the form $x^{k} x^{\prime} x^{l^{m}} x^{-n}{ }^{-p}{ }^{-p} x^{-q}$ (where $x^{k}$ is tak... | 55252 | Combinatorics | proof | Yes | Yes | olympiads | false |
9,10 | |
| $[\underline{\text { Problems with Constraints }}]$ | | |
Authors: Bugayenko V.O., Tokarev S.i.
A sequence of natural numbers is given, forming an arithmetic progression (with a non-zero difference), where the decimal representation of each number does not contain the digit 9.
a) Prove that the number ... | We can assume that the difference $d$ of the progression is positive and not a multiple of 10. Let $d$ be a $k$-digit number.
a) The numbers from A890...0 to A9...9 ($k+2$ nines) all contain nines. There are more than $d$ such numbers. The distance between two consecutive groups of such numbers is $89 \cdot 10^{k} \le... | 72 | Number Theory | proof | Yes | Yes | olympiads | false |
Grinchiin A.
There are 20 beads of ten colors, with two beads of each color. They are somehow distributed into 10 boxes. It is known that it is possible to choose one bead from each box such that all colors are represented. Prove that the number of ways to make such a selection is a non-zero power of two. | Clearly, there are no empty boxes. If a box contains only one bead, it must be chosen. Remove this box with the bead, as well as the second bead of the same color (from another box). There will be 18 beads left, distributed in 9 boxes. The number of ways to choose one bead from each box remains the same as in the origi... | 1024 | Combinatorics | proof | Yes | Yes | olympiads | false |
Berpow S.l.
At the alumni meeting, 45 people attended. It turned out that any two of them who had the same number of acquaintances among those present were not acquainted with each other. What is the maximum number of pairs of acquaintances that could have been among those who attended the meeting? | Since $45=1+2+3+\ldots+9$, we can divide 45 people into groups of $1,2, \ldots, 9$ people. Let people in the same group not know each other, while people in different groups know each other. Then each person in the $k$-th group has $45-k$ acquaintances.
Under this condition, the problem is solved, and the total number... | 870 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A magician with an assistant is going to perform the following trick. A spectator writes a sequence of $N$ digits on a board. The magician's assistant covers two adjacent digits with a black circle. Then the magician enters. His task is to guess both covered digits (and the order in which they are located). For what sm... | Suppose that for some value of $N$ the trick is possible. Then for each variant of the sequence with two closed digits (let their number be $k_{1}$), the magician can restore the original; hence, to each sequence with two closed digits, the magician can uniquely correspond the restored sequence of $N$ digits (let their... | 101 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8,9
[ Equilateral (regular) triangle ]
A line is drawn through vertex $C$ of the square $A B C D$, intersecting the diagonal $B D$ at point $K$, and the perpendicular bisector of side $A B$ at point $M$ ( $M$ is between $C$ and $K$ ). Find $\angle D C K$, if $\angle A K B=\angle A M B$. | Prove that triangle $A M B$ is equilateral.
## Solution
Since $\angle A K B=\angle A M B$, points $M, K, A$ and $B$ lie on the same circle. Therefore, $\angle M A B=\angle M K B=\angle A K B$ $=\angle A M B$, so $M B=A B$. Since $M B=M A$, triangle $A M B$ is equilateral. Thus, $\angle B K M=\angle A M B=60^{\circ}, ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Pythagorean Theorem (direct and inverse).
[ Law of Sines
[ Isosceles, inscribed, and circumscribed trapezoids ]
[ $\underline{\text { Chords and secants (other). }}$ ]
The lengths of two parallel chords of a circle are 40 and 48, and the distance between them is 22. Find the radius of the circle. | Draw the diagonal of the resulting trapezoid and find the radius of the circumscribed circle of one of the resulting triangles.
## Solution
Let $AB=48$ and $CD=40$ be the parallel chords. Find the desired radius $R$.
First method. Drop a perpendicular $CM$ to $AB$. Then $AM=44, BM=4, AC^2=AM^2+CM^2=44^2+22^2=$ $5 \c... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Two tangents drawn from one point
[ $\left.\quad \begin{array}{l}\text { Similar auxiliary triangles }\end{array}\right]$
[ Properties and characteristics of isosceles triangles. $]$
[ Trapezoids (other).
In trapezoid $A B C D$ with bases $A D$ and $B C$, the lateral side $A B$ is equal to 2. The bisector of angle $... | Use the similarity of triangles $B M H$ and $B A E$.
## Solution
Since $B E \| A D$, then $\angle A E B=\angle E A D=\angle B A E$. Therefore, triangle $A B E$ is isosceles.
Consequently, $B E=A B=2$ and $M H \| A E$.
Let $B M=B H=x$. Suppose $K$ is the point of tangency of the given circle with side $A E$. Then $A... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A circle with its center at the intersection of the diagonals $K M$ and $L N$ of an isosceles trapezoid $K L M N$ touches the smaller base $L M$ and the lateral side $M N$. Find the perimeter of trapezoid $K L M N$, given that its height is 36 and the radius of the circle is 11. | Prove that the lateral side of the trapezoid is equal to the larger base.
## Solution
Let $O$ be the point of intersection of the diagonals of the given trapezoid, $A$ be the point of tangency of the circle with the base $L M, A B$, and $L P$ be the heights of the trapezoid. Then
$A M: K B = A O: O B = 11: 25$.
Let... | 129 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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