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1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be co... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. If there are three even and three odd numbers, then each of these triples forms a cycle. Suppose there are four numbers $a, b, c, d$ of the same parity. Then all of them are pairwise connected, and we again get two th... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-f(x)=0$ has exactly three solutions. Find the ordinate of the vertex of the trinomial $f(x)$. | # Answer: 0.
Solution. Suppose the leading coefficient of the trinomial is positive. Note that $(f(x))^{3}-f(x)=f(x) \cdot(f(x)-1) \cdot(f(x)+1)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-1$, and fewer roots than the equation $f(x)=1$. It is also clear that no two equations have common roots. Ther... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers, the difference of whose square roots is less than 1? | Answer: 45.
Solution. We will show that $k=45$ works. Let's divide the numbers from 1 to 2016 into 44 groups:
\[
\begin{aligned}
& (1,2,3), \quad(4,5,6,7,8), \quad(9,10, \ldots, 15), \ldots, \quad\left(k^{2}, k^{2}+1, \ldots, k^{2}+2 k\right), \ldots, \\
& (1936,1937, \ldots, 2016)
\end{aligned}
\]
Since there are 4... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. What is the maximum number of different numbers from 1 to 1000 that can be chosen so that the difference between any two chosen numbers is not equal to any of the numbers 4, 5, 6. | Answer: 400.
Solution. Consider ten consecutive natural numbers. We will prove that no more than four of them are selected. If at least five numbers are selected, then three of them have the same parity, but then their pairwise differences cannot be only 2 and 8. Indeed, if $a<b<c$, then $b-a=2$ and $c-a=8$, but then ... | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the maximum number of chess kings that can be placed on a $12 \times 12$ board so that each king attacks exactly one of the others? | Answer: 56.
Solution. Note that two kings attack each other if and only if their cells share at least one vertex. For each pair of attacking kings, mark the vertices of the cells they occupy. In this case, no fewer than six vertices are marked for each such pair. Since different pairs of kings mark different vertices ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. From the numbers 1, 2, 3, ..., 2016, $k$ numbers are chosen. What is the smallest $k$ such that among the chosen numbers, there will definitely be two numbers whose difference is greater than 672 and less than 1344? | Answer: 674.
Solution. Let $n=672$. Then $2 n=1344$ and $3 n=2016$. Suppose it is possible to choose $674=n+2$ numbers such that no required pair of numbers can be found among them. Let $m$ be the smallest of the chosen numbers. Then the numbers $m+n+1, m+n+2$, $\ldots, m+2 n-1$ are not chosen. Remove them and the num... | 674 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given a board $2016 \times 2016$. For what smallest $k$ can the cells of the board be colored in $k$ colors such that
1) one of the diagonals is painted in the first color;
2) cells symmetric with respect to this diagonal are painted in the same color;
3) any two cells located in the same row on opposite sides of th... | Answer: 11.
Solution. Let the cells of the diagonal running from the top left corner to the bottom right corner be painted in the first color. Denote by $C_{i}$ the set of colors in which the cells of the $i$-th row, located to the left of the diagonal of a single color, are painted. We will prove that $C_{i} \neq C_{... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. At a cactus lovers' meeting, 80 cactus enthusiasts presented their collections, each consisting of cacti of different species. It turned out that no species of cactus is present in all collections at once, but any 15 people have cacti of the same species. What is the smallest total number of cactus species that can ... | Answer: 16.
Solution. We will show that 16 cacti were possible. Number the cacti from 1 to 16. Let the 1st cactus lover have all cacti except the first; the 2nd cactus lover have all except the second cactus; the 15th cactus lover have all except the fifteenth cactus; and the cactus lovers from the 16th to the 80th ha... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-4 f(x)=0$ has exactly three solutions. How many solutions does the equation $(f(x))^{2}=1$ have? | Answer: 2.
Solution. Suppose the leading coefficient of the polynomial is positive. Note that $(f(x))^{3}-4 f(x)=f(x) \cdot(f(x)-2) \cdot(f(x)+2)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-2$, and fewer roots than the equation $f(x)=2$. It is also clear that no two equations have common roots. The... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers $a$ and $b$ such that $|\sqrt[3]{a}-\sqrt[3]{b}|<1$? | Answer: 13.
Solution. We will show that $k=13$ works. Divide the numbers from 1 to 2016 into 12 groups:
\[
\begin{aligned}
& (1,2,3,4,5,6,7), \quad(8,9, \ldots, 26), \quad(27,28, \ldots, 63), \ldots \\
& \left(k^{3}, k^{3}+1, \ldots,(k+1)^{3}-1\right), \ldots,(1728,1729, \ldots, 2016)
\end{aligned}
\]
Since there ar... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The cells of a $9 \times 9$ board are painted in black and white in a checkerboard pattern. How many ways are there to place 9 rooks on the same-colored cells of the board so that they do not attack each other? (A rook attacks any cell that is in the same row or column as itself.) | Answer: $4!5!=2880$.
Solution. Let the upper left corner of the board be black for definiteness. Note that black cells come in two types: cells with both coordinates even, and cells with both coordinates odd. If a rook is on a black cell with both even coordinates, then all black cells it attacks also have even coordi... | 2880 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On an island, there live only knights, who always tell the truth, and liars, who always lie. One fine day, 30 islanders sat around a round table. Each of them can see everyone except themselves and their neighbors. Each person in turn said the phrase: "All I see are liars." How many liars were sitting at the table? | Answer: 28.
Solution. Not all of those sitting at the table are liars (otherwise they would all be telling the truth). Therefore, there is at least one knight sitting at the table. Everyone he sees is a liar. Let's determine who his neighbors are. Both of them cannot be liars (otherwise they would be telling the truth... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In the cells of a $2015 \times n$ table, non-negative numbers are arranged such that in each row and each column there is at least one positive number. It is known that if a cell contains a positive number, then the sum of all numbers in its column is equal to the sum of all numbers in its row. For which $n$ is this... | Answer: $n=2015$.
Solution. We will prove by induction on $m+n$ that for an $m \times n$ table, the specified arrangement is possible only when $m=n$. For $m+n=2$, the statement is obvious. Consider an $m \times n$ table. Let the sum of the numbers in the first row be $a$. Consider all rows and columns whose sum of nu... | 2015 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. How many ways are there to color the cells of a $10 \times 10$ board in blue and green so that in each $2 \times 2$ square there are two blue and two green cells? | Answer: $2^{11}-2=2046$.
Solution. Note that if the colors of three cells in a $2 \times 2$ square are known, the color of the remaining cell is uniquely determined. Also, the colors of the two remaining cells are uniquely determined if there are two adjacent cells of the same color in a $2 \times 2$ square.
Consider... | 2046 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On an island, there live only knights, who always tell the truth, and liars, who always lie, and there are at least two knights and at least two liars. One fine day, each islander, in turn, pointed to each of the others and said one of two phrases: "You are a knight!" or "You are a liar!" The phrase "You are a liar!... | Answer: 526.
Solution. Let $r$ and $\ell$ denote the number of knights and liars, respectively. Note that a knight will say to another knight and a liar will say to another liar: "You are a knight!", while a knight will say to a liar and a liar will say to a knight: "You are a liar!" Therefore, the number of liar-knig... | 526 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In an $8 \times 8$ table, natural numbers are arranged. The numbers in cells symmetric with respect to both diagonals of the table are equal. It is known that the sum of all numbers in the table is 1000, and the sum of the numbers on the diagonals is 200. For what smallest number $M$ can we assert that the sum of th... | Answer: $M=288$.
Solution. Consider the upper half of the table. Let the numbers be arranged as shown in the figure. Due to symmetry, the marked numbers completely determine the placement of the remaining numbers in the table.
| $a_{1}$ | $b_{1}$ | $b_{2}$ | $b_{3}$ | $c_{3}$ | $c_{2}$ | $c_{1}$ | $d_{1}$ |
| :--- | ... | 288 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Real numbers $a, b, c$ and $d$ satisfy the condition $a^{6}+b^{6}+c^{6}+d^{6}=64$. Find the maximum value of the expression $a^{7}+b^{7}+c^{7}+d^{7}$. | Answer: 128
Solution. By the condition $a^{6} \leqslant a^{6}+b^{6}+c^{6}+d^{6}=64$, therefore $a \leqslant 2$. Similarly, we get that $b \leqslant 2, c \leqslant 2$ and $d \leqslant 2$. Consequently,
$$
a^{7}+b^{7}+c^{7}+d^{7}=a \cdot a^{6}+b \cdot b^{6}+c \cdot c^{6}+d \cdot d^{6} \leqslant 2\left(a^{6}+b^{6}+c^{6}... | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Find all natural numbers $n$ such that the number $2^{n}+n^{2}+25$ is a cube of a prime number. | Answer: $n=6$
Solution. Let $2^{n}+n^{2}+25=p^{3}$ for some prime number $p$. Since $p>3$, $p$ is an odd prime number. Then $n$ is an even number, and $2^{n}$ gives a remainder of 1 when divided by three. If $n$ is not divisible by three, then $n^{2}$ gives a remainder of 1 when divided by three, and then $2^{n}+n^{2}... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find all values of $a$ for which the quadratic trinomials $x^{2}+2 x+a$ and $x^{2}+a x+2=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial. | Answer: $a=-4$
Solution. If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
Therefore, we need to find such numbers $a$ for which $2^{2}-2 a=a^{2}-2 \cdot 2$. Thus, we need to solve the... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find all values of $a$ for which the quadratic trinomials $x^{2}-6 x+4 a$ and $x^{2}+a x+6=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial. | Answer: $a=-12$
If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
Therefore, we need to find such numbers $a$ for which $6^{2}-8 a=a^{2}-2 \cdot 6$. Thus, we need to solve the equation... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On an island, there live liars and knights, a total of 2021 people. Knights always tell the truth, and liars always lie. Every resident of the island knows whether each person is a knight or a liar. One fine day, all the residents of the island lined up. After that, each resident of the island stated: "The number of... | Answer: 1010
Solution. The rightmost islander cannot be telling the truth, since there is no one to the left of him, and in particular, there are no liars. Therefore, he is a liar. The leftmost islander cannot be lying, since there is at least one liar to the left of him, and there is no one to the right of him, and i... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. (15 points) If all the trees on one hectare of forest are cut down, then 100 cubic meters of boards can be produced. Assuming that all the trees in the forest are the same, are evenly distributed, and that 0.4 m$^{3}$ of boards can be obtained from each tree, determine the area in square meters on which one tree gro... | Answer: 40.
Solution. Let's find out how many trees grow on one hectare of forest: $\frac{100 \mathrm{m}^{3}}{0.4 \mathrm{M}^{3}}=250$. Let's recall that 1 ha is $100 \mathrm{~m} \times 100 \mathrm{~m}=10000 \mathrm{~m}^{2}$. Thus, one tree grows on $\frac{10000 \mathrm{M}^{2}}{250}=40 \mathrm{M}^{2}$. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) Witch Gingema enchanted the wall clock so that the minute hand moves in the correct direction for five minutes, then three minutes in the opposite direction, then again five minutes in the correct direction, and so on. How many minutes will the hand show after 2022 minutes from the moment it pointed exac... | Answer: 28.
Solution. In 8 minutes of magical time, the hand will move 2 minutes in the clockwise direction. Therefore, in 2022 minutes, it will complete 252 full eight-minute cycles and have 6 minutes left. Since $252 \cdot 2=60 \cdot 8+24$, the hand will travel 8 full circles, plus 24 minutes, and then 5 minutes in ... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. (50 points) Two multiplication examples of two natural numbers were written on the board, and in one of the examples, the numbers were the same. The troublemaker Petya erased both the original numbers and almost all the digits of the multiplication results, leaving only the last three digits: ... 689 and ... 759. Ca... | Answer: It could be 689.
First solution. Statement: at least one of the two last digits of the square of a natural number is necessarily even. Then the second number is eliminated. Example for the first number: $133 \cdot 133=17689$.
Proof of the statement: the statement is true for even numbers, let's consider the s... | 689 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. How many ways are there to color the cells of a $10 \times 10$ board in blue and green such that in each $2 \times 2$ square there are two blue and two green cells? | Answer: $2^{11}-2=2046$.
Solution. Note that if the colors of three cells in a $2 \times 2$ square are known, the color of the remaining cell is uniquely determined. Also, the colors of the two remaining cells are uniquely determined if there are two adjacent cells of the same color in a $2 \times 2$ square.
Consider... | 2046 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second digit does not have to be different. For example, 101 and 222 have this form, while 220 and 123 do not. Similarly, we define the form of the number $\overline{\overline{b a b c}}$. How many nu... | Answer: 180.
Solution. Numbers divisible by $5$ are those ending in 0 or 5, so we have two options for $c$. For $a$, we have 9 options, as the number cannot start with zero, and the value of $b$ can be anything.
Thus, we get that the total number of such numbers is $2 \cdot 9 \cdot 10=180$. | 180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) Nезнayka came up with a password for his email consisting of five characters. Deciding to check the reliability of this password, he calculated all possible combinations that can be formed from these five characters. In the end, he got 20 different combinations. Is a password with this number of combinat... | For example, error.
Solution. The maximum number of different combinations that can be formed from 5 symbols equals $5!(n!=1 \cdot 2 \cdots n$ - the factorial of number $n$, i.e., the product of all natural numbers from 1 to $n$ inclusive). This number is obtained when all 5 symbols are distinct.
Suppose among the gi... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Masha has 1000 beads in 50 different colors, 20 beads of each color. What is the smallest $n$ such that for any way of making a necklace from all the beads, one can choose $n$ consecutive beads, among which there are beads of 25 different colors? | Answer: $n=462$.
First solution. Let's call a segment of the necklace of length $m$ a set of $m$ consecutive beads. If the beads in the necklace are arranged in groups of 20 of the same color, then a segment of length 461 cannot contain more than 24 different colors. Therefore, $n \geqslant 462$. Consider a segment of... | 462 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the minimum number of chips that can be placed in the cells of a $99 \times 99$ table so that in each $4 \times 4$ square there are at least eight chips? | Answer: 4801.
Solution. Add a row and a column with number 100 to the table. Place a chip in each of their cells. Divide the expanded table into 625 squares $4 \times 4$. In each square, there can be no more than eight empty cells, so there are no more than 5000 empty cells in the entire table. Therefore, the total nu... | 4801 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Boy Tolya loves to swim. When he comes to one grandmother's cottage, he swims in the Volkhov and floats downstream from one beach to another in 18 minutes. On the way back, it takes him exactly 1 hour. When he came to another grandmother's cottage, he swam the same distance downstream in the Luga River in 20 minutes... | Answer: 45 minutes.
Solution. Let the distance between the beaches be $x$ km. Then in Volkhov, Tolya swims downstream at a speed of $x / 18$ km/min, and upstream at a speed of $x / 60$ km/min. Therefore, Tolya's own speed is $\frac{1}{2}(x / 60 + x / 18) = 13 x / 360$ km/min. In Luga, Tolya swims downstream at a speed... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The cells of a $20 \times 20$ table are painted in $n$ colors, and there are cells of each color. In each row and each column of the table, no more than six different colors are used. What is the largest $n$ for which this is possible? | Answer: 101.
Solution. Suppose there is a coloring with 102 colors. Then there are two rows that together use at least 12 colors (otherwise, the total number of colors does not exceed $11+5 \cdot 18=101$). Let's assume for definiteness that these are the first and second rows. By the condition, no more than six differ... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) We will call a date diverse if in its representation in the form DD/MM/YY (day-month-year) all digits from 0 to 5 are present. How many diverse dates are there in the year 2013? | Answer: 2.
Solution. Note that in any date of 2013 in the specified format, the digits 1 and 3 are used, so for the day and month of a diverse date, the digits left are 0, 2, 4, and 5.
Let $Д_{1}$ and $Д_{2}$ be the first and second digits in the day's notation, and $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ be the first ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (10 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different six-letter words can be formed from the letters of the word СКАЛКА? And how many seven-letter words from the letters of the word ТЕФТЕЛЬ? In your answer, indicate the quotient of dividing the larger of the foun... | # Answer: 7.
Solution. The word СКАЛКА has six letters, but the letter А and the letter K each appear twice. Therefore, the number of different words will be $\frac{6!}{2!\cdot 2!}$. In
the word ТЕФТЕЛЬ, there are seven letters, and the letters $\mathrm{T}$ and $\mathrm{E}$ each appear twice, so the number of differen... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) Several positive numbers are written on the board. The sum of the five largest of them is 0.29 of the sum of all the numbers, and the sum of the five smallest is 0.26 of the sum of all the numbers. How many numbers are written on the board? | Answer: 18.
Solution. Let the total number of numbers be $k+10$. From the condition, it is clear that $k>0$. We can assume that the sum of all numbers is 1 (otherwise, we divide each number by this sum, and the condition of the problem remains). We order the numbers in ascending order:
$$
a_{1} \leqslant a_{2} \leqsl... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Students' written work is evaluated on a two-point grading system, i.e., a work will either be credited if it is done well, or not credited if it is done poorly. The works are first checked by a neural network, which gives an incorrect answer in $10 \%$ of cases, and then all works deemed uncredited are ... | Answer: 66.
Solution. We will represent the students' works in the diagram below.
Neural network errors can:
a) classify all 10% of good works as bad;
b) classify part of the good works as bad and part of the bad works as good;
c) classify all 10% of bad works as good.
 Let a sequence of non-negative integers be given
$$
k, k+1, k+2, \ldots, k+n
$$
Find the smallest $k$, for which the sum of all numbers in the sequence is equal to 100.
# | # Answer: 9.
Solution. The given sequence of numbers is an arithmetic progression with $n+1$ terms. Its sum is $\frac{(n+1)(2 k+n)}{2}$. Therefore, the condition can be rewritten as $(n+1)(2 k+n)=200$, where $n$ is a non-negative integer. It is clear that $k$ decreases as $n$ increases. Thus, we need to find the large... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (30 points) Point $M$ is the midpoint of the hypotenuse $A C$ of the right triangle $A B C$. Points $P$ and $Q$ on lines $A B$ and $B C$ respectively are such that $A P = P M$ and $C Q = Q M$. Find the measure of angle $\angle P Q M$, if $\angle B A C = 17^{\circ}$.
 Let's call the tail of a natural number any number that is obtained from it by discarding one or several of its first digits. For example, 234, 34, and 4 are tails of the number 1234. Does there exist a six-digit number without zeros in its decimal representation that is divisible by each of its tails? | Answer: Yes (721875 fits).
Solution. Suppose the required number exists. Let's write it as $A=$ $=\overline{a_{5} a_{4} \ldots a_{0}}$. Then $A$ is divisible by its five-digit tail, that is, by $\overline{a_{4} \ldots a_{0}}$. Therefore, $\overline{a_{4} \ldots a_{0}}$ divides the difference between $A$ and its tail, ... | 721875 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (15 points) In a bookstore, there is a rule for "summing" discounts: "if different types of discounts apply to an item, they are applied sequentially one after another." For example, if two discounts A% and B% apply to an item, the first discount is applied to the original price, and the second discount is applied t... | Answer: $50 \%$.
Solution. Note that the sequence of applying discounts is irrelevant, since applying a discount of A\% is equivalent to multiplying the cost of the item by ( $1-\frac{A}{100}$ ). Let $R$ be the magnitude of the "random" discount. Then, as a result of "summing" the three discounts, we get
$$
230 \cdot... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different four-letter words can be formed from the letters of the word КАША? And from the letters of the word ХЛЕБ? In your answer, indicate the sum of the found numbers. | Answer: 36.
Solution. In the word ХЛЕБ, all letters are different. Therefore, by rearranging the letters, we get $4 \cdot 3 \cdot 2 \cdot 1=24$ words. From the word КАША, we can form 12 words. Indeed, for the letters K and Ш, there are $4 \cdot 3=12$ positions, and we write the letters А in the remaining places. Thus,... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (35 points) Journalists have found out that
a) in the lineage of Tsar Pafnuty, all descendants are male: the tsar himself had 2 sons, 60 of his descendants also had 2 sons each, and 20 had 1 son each;
b) in the lineage of Tsar Zinovy, all descendants are female: the tsar himself had 4 daughters, 35 of his descenda... | Answer: At Zinovy.
Solution. We need to find the total number of children in the lineage of Pafnuty, including the children of the king himself. It is equal to $60 \cdot 2+20 \cdot 1+2=142$. The total number of children in the lineage of King Zinovy is $4+35 \cdot 3+35 \cdot 1=144$. | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (50 points) A $5 \times 5$ square of cells was cut into several pieces of different areas, each consisting of an integer number of cells. What is the maximum number of pieces that could result from such a cutting? | Answer: 6.
Solution. We will show that there cannot be more than 6 parts. Indeed, the total area of 7 parts cannot be less than $1+2+3+4+5+6+7=28$, which exceeds the area of the square.
Now we will show how a $5 \times 5$ square can be cut into 6 parts of different sizes:
 From the numbers 1 to 200, one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. What is the maximum number of numbers that can be in a group with this property? | Answer: 40.
Solution. Suppose that the group selected a number $A$, which gives a remainder $i \neq 0$ when divided by 5. If there is another number $B$ in the group, then it must give a remainder $5-i$ when divided by 5, so that $A+B$ is divisible by 5. We will show that there cannot be any other numbers in this grou... | 40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (50 points) Each of the 12 knights sitting around a round table has thought of a number, and all the numbers are different. Each knight claims that the number they thought of is greater than the numbers thought of by their neighbors to the right and left. What is the maximum number of these claims that can be true? | Answer: 6.
Solution. Let's renumber the knights in order with numbers from 1 to 12. In the pairs $(1,2),(3,4)$, $\ldots,(11,12)$, at least one of the knights is lying (specifically, the one who guessed the smaller number). Therefore, there can be no more than 6 true statements. Now let's provide an example where exact... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. On a $5 \times 7$ grid, 9 cells are marked. We will call a pair of cells with a common side interesting if exactly one cell in the pair is marked. What is the maximum number of interesting pairs that can be? | Answer: 35.
Solution. Let's call two cells adjacent if they share a side. The number of interesting pairs containing a given marked cell is no more than 4, and for a boundary cell, it is no more than 3. Then the total number of interesting pairs does not exceed $9 \cdot 4 = 36$. At the same time, if there are two adja... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a+b}{c}\right)^{4}+\left(\frac{b+c}{d}\right)^{4}+\left(\frac{c+d}{a}\right)^{4}+\left(\frac{d+a}{b}\right)^{4}
$$ | Answer: 64.
Solution. By Cauchy's inequality for means,
$$
A \geqslant 4 \cdot \frac{(a+b)(b+c)(c+d)(d+a)}{a b c d}=64 \cdot \frac{a+b}{2 \sqrt{a b}} \cdot \frac{b+c}{2 \sqrt{b c}} \cdot \frac{c+d}{2 \sqrt{c d}} \cdot \frac{d+a}{2 \sqrt{d a}} \geqslant 64
$$
Equality is achieved when $a=b=c=d=1$. | 64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Quadrilateral $A B C D$ is inscribed in a circle. A tangent line $\ell$ is drawn to this circle at point $C$. Circle $\omega$ passes through points $A$ and $B$ and is tangent to line $\ell$ at point $P$. Line $P B$ intersects segment $C D$ at point $Q$. Find the ratio $\frac{B C}{C Q}$, given that $B$ is the point o... | Answer: 1.
Solution. The angle between the tangent $BD$ and the chord $AB$ of the circle $\omega$ is equal to the inscribed angle that subtends $AB$, so $\angle APB = \angle ABD = \angle AC... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. On a $7 \times 7$ checkerboard, 14 cells are marked. We will call a pair of cells with a common side interesting if at least one cell in the pair is marked. What is the maximum number of interesting pairs that can be? | # Answer: 55.
Solution. Let's call two cells adjacent if they share a side. The number of interesting pairs containing a given marked cell is no more than 4, and for a boundary cell, no more than 3. Then the total number of interesting pairs does not exceed $14 \cdot 4 = 56$. At the same time, if there are two adjacen... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a^{2}+b^{2}}{c d}\right)^{4}+\left(\frac{b^{2}+c^{2}}{a d}\right)^{4}+\left(\frac{c^{2}+d^{2}}{a b}\right)^{4}+\left(\frac{d^{2}+a^{2}}{b c}\right)^{4}
$$ | Answer: 64.
Solution. By the Cauchy inequalities for means,
$$
A \geqslant 4 \cdot \frac{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+d^{2}\right)\left(d^{2}+a^{2}\right)}{c d \cdot a d \cdot a b \cdot b c}=64 \cdot \frac{a^{2}+b^{2}}{2 a b} \cdot \frac{b^{2}+c^{2}}{2 b c} \cdot \frac{c^{2}+d^{2}}{2 c ... | 64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. A circle $\omega$ is circumscribed around triangle $A B C$. Tangents to the circle, drawn at points $A$ and $B$, intersect at point $K$. Point $M$ is the midpoint of side $A C$. A line passing through point $K$ parallel to $A C$ intersects side $B C$ at point $L$. Find the angle $A M L$. | Answer: $90^{\circ}$.
Solution. Let $\alpha=\angle A C B$. The angle between the tangent $A K$ and the chord $A B$ of the circle $\omega$ is equal to the inscribed angle that subtends $A B$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. On the table, there are three spheres and a cone (with its base $k$ on the table), touching each other externally. The radii of the spheres are 20, 40, and 40, and the radius of the base of the cone is 21. Find the height of the cone. | Answer: 28.
Solution. Let $O$ be the center of the base of the cone, $h$ be its height, and $2 \alpha$ be the angle of inclination of the cone's generators to the table. Consider the sectio... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Each cell of a $5 \times 6$ table is colored in one of three colors: blue, red, or yellow. In each row of the table, the number of red cells is not less than the number of blue cells and not less than the number of yellow cells, and in each column of the table, the number of blue cells is not less than the number of... | Answer: 6.
| $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{C}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BC$ intersects circle $\omega_{1}$ agai... | Answer: $180^{\circ}$.
Solution 1. Since $\angle B C O=\angle P C O_{1}$, the isosceles triangles $B O C$ and $P O_{1} C$ are similar, from which $\angle B O C=\angle P O_{1} C$. Similarly, it can be verified that $\angle A O D=\angle Q O_{2} D$. Then the segments $O_{1} P$ and $O_{2} Q$ are parallel to the line $A B$... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Each cell of a $5 \times 5$ table is colored in one of three colors: blue, red, or yellow. In each row of the table, the number of yellow cells is not less than the number of red cells and not less than the number of blue cells, and in each column of the table, the number of red cells is not less than the number of ... | Answer: 5.
| $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| :---: | :---: | :---: | :---: | :---: |
| $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\m... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BD$ intersects circle $\omega_{2}$ agai... | Answer: $180^{\circ}$.
Solution 1. Since $\angle A C O=\angle Q C O_{1}$, the isosceles triangles $A O C$ and $Q O_{1} C$ are similar, from which $\angle A O C=\angle Q O_{1} C$. Similarly, ... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the cells of a $5 \times 5$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table. | Answer: 48.
Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 5. Since all these sums are distinct, the minimal possible set of their values is $\{5,6, \ldots, 13,14\}$. By adding the sums of the rows and columns of the table, we get twice the... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a+\frac{1}{b}\right)^{3}+\left(b+\frac{1}{c}\right)^{3}+\left(c+\frac{1}{d}\right)^{3}+\left(d+\frac{1}{a}\right)^{3}
$$ | Answer: 32.
Solution 1. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(2 \sqrt{\frac{a}{b}}\right)^{3}+\left(2 \sqrt{\frac{b}{c}}\right)^{3}+\left(2 \sqrt{\frac{c}{d}}\right)^{3}+\left(2 \sqrt{\frac{d}{a}}\right)^{3} \geqslant 32\left(\fr... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given an acute-angled triangle $A B C$. A circle with diameter $B C$ intersects sides $A B$ and $A C$ at points $D$ and $E$ respectively. Tangents drawn to the circle at points $D$ and $E$ intersect at point $K$. Find the angle between the lines $A K$ and $B C$. | Answer: $90^{\circ}$.
Solution 1. Let $\alpha=\angle D B E, \beta=\angle B C D, \gamma=\angle C B E, O$ - the intersection point of segments $C D$ and $B E$. The angle between the tangent $D... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the cells of a $4 \times 6$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table. | Answer: 43.
Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 4. Since all these sums are different, the minimal possible set of their values is $\{4,5, \ldots, 12,13\}$. By adding the sums of the rows and columns of the table, we get twice th... | 43 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a^{2}+\frac{1}{b c}\right)^{3}+\left(b^{2}+\frac{1}{c d}\right)^{3}+\left(c^{2}+\frac{1}{d a}\right)^{3}+\left(d^{2}+\frac{1}{a b}\right)^{3}
$$ | Answer: 32.
Solution. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(\frac{2 a}{\sqrt{b c}}\right)^{3}+\left(\frac{2 b}{\sqrt{c d}}\right)^{3}+\left(\frac{2 c}{\sqrt{d a}}\right)^{3}+\left(\frac{2 d}{\sqrt{a b}}\right)^{3} \geqslant 32\le... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Point $M$ is the midpoint of side $AB$ of triangle $ABC$. A circle $\omega_{1}$ is drawn through points $A$ and $M$, tangent to line $AC$, and a circle $\omega_{2}$ is drawn through points $B$ and $M$, tangent to line $BC$. Circles $\omega_{1}$ and $\omega_{2}$ intersect again at point $D$. Point $E$ lies inside tri... | Answer: $180^{\circ}$
Solution 1. The angle between the tangent $A C$ and the chord $A M$ of circle $\omega_{1}$ is equal to the inscribed angle in $\omega_{1}$ that subtends $A M$, i.e., $... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. What is the minimum number of cells that need to be marked in a $7 \times 7$ table so that every vertical or horizontal strip $1 \times 4$ contains at least one marked cell? | Answer: 12.
Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given numbers $x, y, z \in\left[0, \frac{\pi}{2}\right]$. Find the minimum value of the expression
$$
A=\cos (x-y)+\cos (y-z)+\cos (z-x)
$$ | Answer: 1.
Solution. We can assume that $x \leqslant y \leqslant z$, since the expression $A$ does not change under pairwise permutations of the variables. Notice that
$$
\cos (x-y)+\cos (z-x)=2 \cos \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The first cosine in the right-hand side is positive a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. What is the minimum number of cells that need to be marked in a $9 \times 9$ table so that each vertical or horizontal strip $1 \times 5$ contains at least one marked cell? | Answer: 16.
Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given numbers $x, y, z \in [0, \pi]$. Find the minimum value of the expression
$$
A=\cos (x-y)+\cos (y-z)+\cos (z-x)
$$ | Answer: -1.
Solution. We can assume that $x \leqslant y \leqslant z$, since the expression $A$ does not change under pairwise permutations of the variables. Notice that
$$
\cos (x-y)+\cos (z-x)=2 \cos \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The first cosine in the right-hand side is non-negat... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A circle $\omega$ with center at point $O$ is circumscribed around triangle $ABC$. Circle $\omega_{1}$ touches the line $AB$ at point $A$ and passes through point $C$, while circle $\omega_{2}$ touches the line $AC$ at point $A$ and passes through point $B$. A line through point $A$ intersects circle $\omega_{1}$ ag... | Answer: $90^{\circ}$.
Solution 1. Let $\alpha=\angle B A C$. The angle between the tangent $A B$ and the chord $A C$ of circle $\omega_{1}$ is equal to the inscribed angle that subtends $A ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given nine-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It turned out that the product $mn$ consists of an odd number of digits and reads the same from left to right and from right to left. Find the largest number $m$ for which this is possible. | Answer: 220000001.
Solution. Let $m=\overline{a_{8} \ldots a_{0}}, n=\overline{a_{0} \ldots a_{8}}$. Since the number $m n$ contains an odd number of digits, it is a seventeen-digit number. Write $m n=\overline{b_{16} \ldots b_{0}}$. We will show by induction that
$$
b_{k}=a_{0} a_{8-k}+a_{1} a_{9-k}+\ldots+a_{k-1} a... | 220000001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a $3 \times 3$ table, 9 numbers are arranged such that the six products of these numbers in the rows and columns of the table are all different. What is the maximum number of these numbers that can equal one? | Answer: 5.
Solution. Let's call the index of the table the total number of its rows and columns consisting of ones. According to the condition, the index does not exceed 1. Let $n$ be an element of the table different from 1. Then in one row or one column with $n$ there is another number not equal to 1 (otherwise the ... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Point $O$ is the center of the circumcircle of triangle $ABC$. On the circumcircle of triangle $BOC$, outside triangle $ABC$, point $X$ is chosen. On the rays $XB$ and $XC$ beyond points $B$ and $C$, points $Y$ and $Z$ are chosen respectively such that $XY = XZ$. The circumcircle of triangle $ABY$ intersects side $A... | Answer: $180^{\circ}$.
Solution. Note that $\angle B Y T=\angle B A T$ as inscribed angles subtending the same arc. Since quadrilateral $B O C X$ is cyclic, we get
$$
180^{\circ}-\angle B X C=\angle B O C=2 \angle B A C=2 \angle B Y T
$$
On the other hand, triangle $Y X Z$ is isosceles, so $180^{\circ}-\angle Y X Z=... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In a $3 \times 4$ table, 12 numbers are arranged such that all seven sums of these numbers in the rows and columns of the table are distinct. What is the maximum number of numbers in this table that can be zero? | Answer: 8.
Solution. Let's call the index of the table the total number of its zero rows and columns. According to the condition, the index does not exceed 1. Let $n$ be a non-zero element of the table. Then in the same row or in the same column with $n$ there is another non-zero number (otherwise the sums in the row ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given numbers $x, y, z \in [0, \pi]$. Find the maximum value of the expression
$$
A=\sin (x-y)+\sin (y-z)+\sin (z-x)
$$ | Answer: 2.
Solution 1. We can assume that $x \leqslant y$ and $x \leqslant z$, since the expression $A$ does not change under cyclic permutation of the variables. Notice that
$$
\sin (x-y)+\sin (z-x)=2 \sin \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The argument of the sine on the right-hand sid... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Point $O$ is the center of the circumcircle of triangle $A B C$. Points $Q$ and $R$ are chosen on sides $A B$ and $B C$ respectively. Line $Q R$ intersects the circumcircle of triangle $A B R$ again at point $P$ and intersects the circumcircle of triangle $B C Q$ again at point $S$. Lines $A P$ and $C S$ intersect a... | Answer: $90^{\circ}$.
Solution. Note that $\angle C S Q=\angle C B Q$ and $\angle A P R=\angle A B R$ as inscribed angles subtending the same arc. Therefore, $\angle K S P=\angle K P S$, whi... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. (40 points) For the quadratic function $p(x)=a x^{2}+b x+c$, for some integers $n$, the equality $p(n)=p\left(n^{2}\right)$ holds. Provide an example of the function $p(x)$ for which the number of such numbers $n$ is the largest. What is this largest number of numbers $n$? | Answer: The maximum number of numbers $n$ is 4. An example of a function: $p(x)=x^{2}-6 x+1$.
Solution: Since $0=0^{2}$ and $1=1^{2}$, for any function $p(x)$ there are at least two such numbers $n$.
Write the equality $p(n)=p\left(n^{2}\right)$ and transform it:
$$
a n^{2}+b n+c=a n^{4}+b n^{2}+c \Leftrightarrow a ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (20 points) On the board, all two-digit numbers divisible by 5, where the number of tens is greater than the number of units, were written down. There turned out to be $A$ such numbers. Then, all two-digit numbers divisible by 5, where the number of tens is less than the number of units, were written down. There tur... | Answer: 413.
Solution: We will write two-digit numbers in the form $\overline{x y}$, where $x$ is the number of tens, and $y$ is the number of units.
Let's calculate what $A$ is. We need two-digit numbers divisible by 5, i.e., numbers where $y$ is either 0 or 5. Note that if $y=0$, then $x$ can take any value from 1 ... | 413 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (20 points) Find the number of different four-digit numbers that can be obtained by rearranging the digits of the number 2021 (including this number itself). | Answer: 9.
Solution: The number of options can be found by enumerating the permutations of the digits: 2021, 2012, 2201, 2210, 2102, 2120, 1022, 1202, 1220.
The number of options can also be calculated using combinatorial methods. The position of zero can be chosen in three ways, as it should not be the first. Then, ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) The pensioners of one of the planets in the Alpha Centauri system enjoy spending their free time solving cybrow solitaire puzzles: they choose natural numbers from a certain interval $[A, B]$ in such a way that the sum of any two of the chosen numbers is not divisible by a given number $N$. Last week, th... | Answer: 356.
Solution: For $k=0,1, \ldots, 10$, let $I_{k}$ be the set of all numbers in $[A, B)$ that give a remainder of $k$ when divided by 11. Since $A$ and $B$ are multiples of 11, all sets $I_{k}$ contain an equal number of elements. Therefore, all numbers in $[A, B)$ that are not multiples of 11 can be paired a... | 356 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) Before the geometry lesson, the teacher wrote on the board the values of all the angles (in degrees) of a certain convex polygon. However, the duty students erased one of the written numbers. When the lesson began, it turned out that the sum of the remaining numbers was 1703. What number did the duty stu... | Answer: 97.
Solution: Let the polygon have $n$ vertices. Since the $n$-gon is convex, each of its angles is less than $180^{\circ}$, and the sum of all angles is $(n-2) \cdot 180^{\circ}$. Therefore, the sum of all angles of the polygon minus one lies in the interval from $180(n-3)$ to $180(n-2)$. Then
$$
180(n-3)<17... | 97 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. All three-digit numbers from 100 to 999 are written in a row without spaces. Kostya underlined \( k \) consecutive digits in this sequence, and Andrey underlined other \( k \) consecutive digits in this sequence. It turned out that the \( k \)-digit numbers underlined by the boys are equal. For what largest \( k \) ... | Answer: $k=5$.
Solution. An example of a five-digit number that satisfies the condition of the problem is 22923. Indeed, Kostya could underline the fragment «22923» at the junction of the numbers 229 and 230, and Andrey - at the junction of the numbers 922 and 923.
Now we will show that it is not possible to choose a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. All four-digit numbers from 1000 to 9999 are written in a row without spaces. Kostya underlined \( k \) consecutive digits in this sequence, and Andrey underlined other \( k \) consecutive digits in this sequence. It turned out that the \( k \)-digit numbers underlined by the boys are equal. For what largest \( k \)... | Answer: $k=7$.
Solution. An example of a seven-digit number that satisfies the condition of the problem is 2229223. Indeed, Kostya could underline the fragment «2229223» at the junction of the numbers 2229 and 2230, and Andrey - at the junction of the numbers 9222 and 9223.
Now we will show that it is impossible to c... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the maze diagram in Fig. 1, each segment (link) is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, d3, a5.
## Solution.
Fig. 2: Parts of the maze
 is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 8): part $\mathcal{K}$ is room $K$, the long dead-end corridor that exits from it to the left, and the dead-en... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In the maze diagram in Fig. 13, each segment (link) is a corridor, and each circle is a small room. Some rooms have beacons that hum, each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the shortest ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
## Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 14): part $\mathcal{K}$ is room $K$ and the dead-end corridors that lead from it to the right and down, par... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In each cell of a $100 \times 100$ table, a natural number is written. In each row, there are at least 10 different numbers, and in any four consecutive rows, there are no more than 15 different numbers. What is the maximum number of different numbers that can be in the table? | Answer: 175
Solution. In one line, there are no less than 10 different numbers, so in the next three lines together, there appear no more than 5 new numbers. Therefore, the first four lines contain no more than 15 different numbers, and each of the following three lines adds no more than 5 new numbers, making the tota... | 175 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find all primes $p$ for which the numbers $p+1$ and $p^{2}+1$ are double the squares of natural numbers.
# | # Answer: $p=7$
First solution. Let $p+1=2 x^{2}$ and $p^{2}+1=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Therefore, either $y-x$ or $y+x$ is divisible by $p$. From the inequality $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. No... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In each cell of a $75 \times 75$ table, a natural number is written. In each row, there are at least 15 different numbers, and in any three consecutive rows, there are no more than 25 different numbers. What is the maximum number of different numbers that can be in the table? | # Answer: 385
Solution. In one line, there are no less than 15 different numbers, so in the next two lines together, there appear no more than 10 new numbers. Therefore, the first three lines contain no more than 25 different numbers, and each of the following two lines adds no more than 10 new numbers, making the tot... | 385 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find all primes $p$ for which the numbers $p+7$ and $p^{2}+7$ are double the squares of natural numbers. | Answer: $p=11$
First solution. Let $p+7=2 x^{2}$ and $p^{2}+7=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Since $p-$ is odd, $p \geqslant 3$ and $2 p^{2}>p^{2}+7$. Therefore, $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. Note tha... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In an acute-angled triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are dropped. A point $T$ is chosen on the plane such that the lines $T A$ and $T B$ are tangents to the circumcircle of triangle $A B C$, and point $O$ is the center of this circle. The perpendicular dropped from point $T$ to the li... | Answer: $90^{\circ}$
Solution. Let $\angle B A C=\alpha$ and $\angle A C B=\gamma$. Triangles $A O T$ and $B O T$ are congruent by three sides, and the inscribed angle $\angle A C B$ subten... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. What is the minimum number of cells that need to be marked in a $15 \times 15$ table so that each vertical or horizontal strip $1 \times 10$ contains at least one marked cell. | Answer: 20
Solution. Let's cut the $15 \times 15$ table without the central $5 \times 5$ square into 20 rectangles of $1 \times 10$ (see the left figure). Therefore, we will need to mark at least 20 cells. An example with 20 cells: all cells of two parallel diagonals of length 10 are marked (see the right figure).
![]... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. What is the minimum number of cells that need to be marked in a $20 \times 20$ table so that each vertical or horizontal strip of $1 \times 12$ contains at least one marked cell. | Answer: 32
Solution. Let's cut the $20 \times 20$ table without the central $4 \times 4$ square into 32 rectangles of $1 \times 12$ (see the left figure). Therefore, we will need to mark at least 32 cells. An example with 32 cells: all cells of three parallel diagonals of lengths 4, 16, and 12 are marked (see the righ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the leg $AC$ of the right triangle $ABC$ with hypotenuse $AB$, a point $P$ is marked. Point $D$ is the foot of the perpendicular dropped from vertex $A$ to the line $BP$, and point $E$ is the foot of the perpendicular dropped from point $P$ to the side $AB$. On the plane, a point $T$ is chosen such that the lines... | Answer: $90^{\circ}$
Solution. Let $\angle A B P=\varphi$ and $\angle C B P=\psi$. Triangles $A O T$ and $P O T$ are equal by three sides, and the inscribed angle $\angle A B P$ subtends ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The sum of positive numbers $a, b, c$ and $d$ is not less than 8. Find the minimum value of the expression
$\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}$. | Answer: 1
First solution. By the inequality of means for four numbers, we have
$$
\begin{aligned}
\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{a+b}{16}+\frac{a+c}{16} & +\frac{a+d}{16} \geqslant \\
& \geqslant 4 \sqrt[4]{\frac{a^{4}}{(a+b)(a+c)(a+d)} \cdot \frac{a+b}{16} \cdot \frac{a+c}{16} \cdot \frac{a+d}{16}}=\frac{a}{2}
... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and triangle $A P D$ is acute-angled. Points $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively. A perpendicular is drawn from point $E$ to line $A C$, and a perpendicular is drawn from point $F$ to line $B D$, these perp... | Answer: $90^{\circ}$
First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $T$ be the point of inte... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and the angle $A P B$ is obtuse. Points $E$ and $F$ are the midpoints of sides $A D$ and $B C$ respectively. A perpendicular is drawn from point $E$ to the line $A C$, and a perpendicular is drawn from point $F$ to the line $B D$, these p... | Answer: $90^{\circ}$
First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $B P$, respectively, and let $T$ be the point of inte... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. (30 points) Point $M$ is the midpoint of the hypotenuse $A C$ of the right triangle $A B C$. Points $P$ and $Q$ on lines $A B$ and $B C$ respectively are such that $A P = P M$ and $C Q = Q M$. Find the measure of angle $\angle P Q M$, if $\angle B A C = 17^{\circ}$.
 At the Journalism Faculty of the University of Enchanted Commonwealth, 4 chickens are applying. The faculty has 2 places in the daytime program and 3 places in the evening program. Assuming all 4 chickens will be admitted to the faculty, determine the number of outcomes in which exactly two chickens will... | Answer: 6.
Solution. The faculty has a total of 5 places, to which 4 applicants are applying. Since exactly two chickens will be admitted to the evening department, the other two will be admitted to the daytime department. The number of ways to choose 2 out of 4 applicants to be admitted to the daytime department is $... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (30 points) On the sides $B C$ and $A C$ of the isosceles triangle $A B C (A B = A C)$, points $D$ and $E$ were found respectively such that $A E = A D, \angle E D C = 18^{\circ}$. Find the measure of the angle $\angle B A D$. | Answer: $36^{\circ}$.
Solution. Denote the angles as indicated in the figure. The angle $A D C$ is an exterior angle for triangle $A D B$; hence, $\beta+18^{\circ}=\alpha+x$. Angles $A B C$ ... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. (40 points) For real numbers $a, b$ and $c$ it is known that $a b + b c + c a = 3$. What values can the expression $\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}$ take? | Answer: 6.
Solution. Consider the first term of the desired expression. Using the condition that $a b+b c+c a=3$. Then
$$
b^{2}+3=b^{2}+a b+b c+c a=(b+a)(b+c)
$$
Therefore,
$$
\frac{a\left(b^{2}+3\right)}{a+b}=\frac{a(b+a)(b+c)}{a+b}=a(b+c) \text {. }
$$
Similarly for the second and third terms, we get
$$
\frac{b... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $2 f(2 x-3)-f(3 x+1)$ also has exactly one root. Find the root of the trinomial $f(x)$. | Answer: -11.
Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=2 f(2 x-3)-f(3 x+1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$... | -11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In the cells of an $11 \times 11$ square, zeros and ones are arranged in such a way that in any figure of four cells in the shape of $\square$, the sum of the numbers is odd. (The figure can be rotated and flipped). What is the minimum number of ones that can be in such an arrangement? | Answer: 25
Solution. Place 25 figures in the square without any common cells (see the left figure). Each of them contains at least one unit, so the total number of units is no less than 25. A suitable arrangement of 25 units: in all cells with even coordinates (see the right figure).
=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $f(3 x+2)-2 f(2 x-1)$ also has exactly one root. Find the root of the trinomial $f(x)$. | Answer: -7.
Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=f(3 x+2)-2 f(2 x-1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$ ... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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