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5.1. Find the smallest natural solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}>2^{2017}$. Answer. 17. | Solution. The value $x=17$ is clearly a solution. If $x \leqslant 16$, then we have
$$
2^{x}+2^{x+1}+\ldots+2^{x+2000} \leqslant 2^{16}+2^{17}+\ldots+2^{2015}+2^{2016}<2^{2017}
$$
since this inequality reduces sequentially to the following: $2^{16}+2^{17}+\ldots+2^{2015}<$ $2^{2017}-2^{2016}=2^{2016}, 2^{16}+2^{17}+\... | 17 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,974 |
6.1. How many triangles with integer sides have a perimeter equal to 27? (Triangles that differ only in the order of the sides - for example, $7,10,10$ and $10,10,7$ - are considered the same triangle.) | Answer: 19.
Solution: Arrange the sides in ascending order: $a \leqslant b \leqslant c$. Then the smaller side $a$ does not exceed 9. If $a=1$ or $a=2$, there is one such triangle. If $a=3$ or $a=4$, there are two such triangles. If $a=5$ or $a=6$, there are three such triangles. If $a=7$, there are four such triangle... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,975 |
7.1. Solve the equation $\frac{n!}{2}=k!+l!$ in natural numbers, where $n!=1 \cdot 2 \cdot \ldots n$. If there are no solutions, write 0; if there is one solution, write $n$; if there are multiple solutions, write the sum of the values of $n$ for all solutions. Recall that a solution is a triplet $(n, k, l)$; if soluti... | Answer: 10 (all triples of solutions $(n, k, l):(3,1,2),(3,2,1),(4,3,3)$).
Solution. Note that if $k4$, then $n!>4 \cdot(n-1)!\geqslant 2 \cdot(k!+l!)$, so there are no such solutions. If $n=2$, we get $1=k!+l!$ - no solutions; if $n=3$, the equation $3=k!+l!$ has two solutions: $k=1, l=2$ and $k=2, l=1$; if $n=4$, th... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,976 |
2. First-grader Petya was laying out a contour of an equilateral triangle with the chips he had, so that each of its sides, including the vertices, contained the same number of chips. Then, with the same chips, he managed to lay out the contour of a square in the same way. How many chips does Petya have, if each side o... | Answer: 24.
Solution. Let the side of the triangle contain $x$ chips, and the side of the square - $y$ chips. The total number of chips, counted in two ways, is $3 x-3=4 y-4$ (we account for the corner chips being counted twice). From the problem statement, it follows that $y=x-2$. Therefore, we get the equation $3(x-... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,977 |
1. Given a parallelogram $A B C D$ and points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are chosen such that point $A$ is the midpoint of segment $D D_{1}$, point $B$ is the midpoint of $A A_{1}$, point $C$ is the midpoint of $B B_{1}$, and point $D$ is the midpoint of $C C_{1}$. Find the area of $A_{1} B_{1} C_{1} D_{1}$, gi... | # Answer: 5
## Solution:
Notice that triangles $\triangle A B C$ and $\triangle A B D_{1}$ are congruent, hence their areas are equal. Triangles $\triangle A B D_{1}$ and $\triangle A_{1} ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,978 |
3. Solve the system $\left\{\begin{array}{l}x^{2}=2 \sqrt{y^{2}+1} ; \\ y^{2}=2 \sqrt{z^{2}-1}-2 ; \\ z^{2}=4 \sqrt{x^{2}+2}-6 .\end{array}\right.$ | Answer: $(\sqrt{2}, 0, \sqrt{2}),(\sqrt{2}, 0,-\sqrt{2}),(-\sqrt{2}, 0, \sqrt{2}),(-\sqrt{2}, 0,-\sqrt{2})$.
Solution: Let $a=\sqrt{x^{2}+2}, b=\sqrt{y^{2}+1}, c=\sqrt{z^{2}-1}$. We get the system: $\left\{\begin{array}{l}a^{2}-2=2 b \\ b^{2}-1=2 c-2 \\ c^{2}+1=4 a-6\end{array}\right.$
Add all the equations and move ... | (\sqrt{2},0,\sqrt{2}),(\sqrt{2},0,-\sqrt{2}),(-\sqrt{2},0,\sqrt{2}),(-\sqrt{2},0,-\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,980 |
4. Kolya started playing $W o W$ at the moment when the hour and minute hands were opposite. He finished playing after an integer number of minutes, and at the end, the minute hand coincided with the hour hand. How long did he play (if it is known that he played for less than 12 hours) | Solution: The minute hand catches up with the hour hand at a speed of $\frac{11^{\circ}}{2}$ /min. For them to coincide, the difference in their angles of rotation should be $180+360 k$. This value is a multiple of 11 when $k=5,16, \ldots$. According to the problem, only $k=5$ fits, which gives us 6 o'clock. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,981 |
5. Find all pairs of natural numbers $(m, n)$ for which the equality
$$
n \cdot(n-1) \cdot(n-2) \cdot(n-3)=m \cdot(m-1)
$$
holds. | Answer: $(1,1),(2,1);(3,1)$.
Solution: Rewrite the equation as
$$
\left(n^{2}-3 n\right) \cdot\left(n^{2}-3 n+2\right)=m^{2}-m
$$
and let $N=n^{2}-3 n+1$, then, by completing the squares, we get $N^{2}-1=\left(m-\frac{1}{2}\right)^{2}-\frac{1}{4}$. Multiplying by 4, we obtain $(2 N)^{2}-(2 m-1)^{2}=3$. The number $2... | (1,1),(2,1);(3,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,982 |
6. Two circles of radii $R$ and $R^{\prime}$ touch each other externally at point $P$ and touch the line $l$ at points $A$ and $B$, respectively. Let $Q$ be the point of intersection of the line $B P$ with the first circle. Determine the distance from the line $l$ to the point $Q$.
^{4}+(y-z)^{4}+(z-x)^{4}\right)$ is the square of some integer. | Solution: Let $a=x-y, b=y-z, c=z-x$. Then $\sigma_{1}=a+b+c=0$. Let $ab+ac+bc=\sigma_{2}, abc=\sigma_{3}$ and express through them: $\frac{1}{2}\left(a^{4}+b^{4}+c^{4}\right)=\sigma_{2}^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 5,984 |
8. In how many different ways can integers $a, b, c \in [1,100]$ be chosen so that the points with coordinates $A(-1, a), B(0, b)$, and $C(1, c)$ form a right triangle? | Answer: 974
Solution: $A B^{2}=1+(b-a)^{2}, B C^{2}=1+(c-b)^{2}, A C^{2}=4+(c-a)^{2}$.
If triangle $A B C$ is a right triangle with hypotenuse $A C$, then by the Pythagorean theorem $A C^{2}=A B^{2}+B C^{2}, 1+(b-a)^{2}+1+(b-c)^{2}=4+(a-c)^{2}$, which simplifies to $(b-a)(b-c)=1$. Since both factors are integers, we ... | 974 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,985 |
1. A father has three sons, born on the same day but in different years. The youngest son is 2 years old. In 12 years, the father's age will be equal to the sum of the ages of his three sons. Determine the current ages of the middle and oldest sons, given that the father is currently 33 years old. | Answer: 3 and 4. Solution. In 12 years, the father will be 45 years old, and the youngest son will be 14. The remaining sons will be a total of 31 years old. Since each is older than 14 years, this is only possible if they are 15 and 16 years old. Therefore, they are currently 3 and 4 years old. | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,986 |
2. A stack of A4 sheets was folded in half and folded in two (resulting in an A5 booklet). After that, the pages of the resulting booklet were renumbered: $1,2,3, \ldots$ It turned out that the sum of the numbers on one of the sheets was 74. How many sheets were in the stack? | Answer: 9. Solution. Note that the sum of the numbers on each leaf is the same. If the last page has the number $n$, then on the first leaf, the numbers will be $1,2, n-1$ and $n$. Therefore, $2n + 2 = 74$, from which we get that the number of pages $n=36$. Four pages are placed on each leaf, so there were 9 leaves. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,987 |
4. The Tale of the Greedy Bear Cubs. Two bear cubs found a large round piece of cheese. They wanted to divide it equally, but they couldn't—each was afraid the other would get more. Suddenly, out of nowhere, a fox approached them.
- Let me divide the cheese equally for you.
- That's great! - the bear cubs rejoiced. - ... | Answer: 680g. Solution. We will analyze from the end. At the last moment, the bear cubs had pieces of 20g each, which means before that, they had pieces of 40g and 20g. We will denote this as $(20,20) \leftarrow(40,20)$. Then the complete sequence of actions will look like this:
$(20,20) \leftarrow(40,20) \leftarrow(4... | 680g | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,989 |
5. In a $3 \times 3$ table, the following numbers are written:
| 10 | 20 | 40 |
| :---: | :---: | :---: |
| 32 | 61 | 91 |
| 100 | 1000 | 2000 |
It is allowed to swap any two numbers in one move. What is the minimum number of moves required to achieve that the sum of the numbers in each column is divisible by $3 ?$ | Answer: 2. Solution.
| 1 | 2 | 1 |
| :--- | :--- | :--- |
| 2 | 1 | 1 |
| 1 | 1 | 2 |
Let's write down the remainders of these numbers when divided by 3. It is not hard to notice that the sum of the numbers is divisible by 3 if and only if the sum of the corresponding remainders is divisible by 3. The remainders equa... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,990 |
6. Find the number of integers from 1 to 3400 that are divisible by 34 and have exactly 2 odd natural divisors. For example, the number 34 has divisors $1,2,17$ and 34, exactly two of which are odd. | Answer: 7. Solution. It is obvious that if a number is divisible by 34, then its divisors will always be 1 and 17. According to the condition, there should be no other odd divisors, so these must be numbers of the form $17 \cdot 2^{k}$, $k \geqslant 1$ (and only they). These numbers fall within the specified range for ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,991 |
7. It is known that the fraction $\frac{a}{b}$ is less than the fraction $\frac{c}{d}$ and $b>d>0$. Determine which is smaller: the arithmetic mean of these two fractions or the fraction $\frac{a+c}{b+d}$. | Answer: The fraction $\frac{a+c}{b+d}$ is smaller. Solution. $\frac{a}{b}<\frac{c}{d} \Leftrightarrow a d<b c \Leftrightarrow a d(b-d)<$ $b c(b-d) \Leftrightarrow a b d+b c d<b^{2} c+a d^{2} \Leftrightarrow 2 a b d+2 b c d<a b d+b c d+b^{2} c+a d^{2}$ $\Leftrightarrow 2(a+c) b d<(a d+b c)(b+d) \Leftrightarrow \frac{a+c... | \frac{+}{b+}<\frac{1}{2}(\frac{}{b}+\frac{}{}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,992 |
1. Nезнayka jumped from his house to Znayka's house. For the first three-quarters of the path, he jumped with jumps the length of two of his usual steps, and for the remaining quarter of the path, he jumped with jumps the length of three of his usual steps. It turned out that the number of jumps of two steps was 350 mo... | Answer variant l: 1200.
Answer variant 2: 1800.
Answer variant $3: 2040$.
Answer variant 4: 900.
Scoring evaluation: 15 баллов - correct solution and correct answer; 10 баллов - arithmetic error in otherwise correct solution | 1200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,993 |
8. What is the least number of odd numbers that can be chosen from the set of all odd numbers lying between 16 and 2016, so that no one of the chosen numbers is divisible by any other?
| Solution. We will select all odd numbers from 673 to 2015, their quantity is
$\frac{2015-671}{2}=672$. None of these numbers is divisible by another, because in division, the result should be an odd number, which is at least 3, but $673-3>2015$.
We will show that it is impossible to select more than 672 numbers satis... | 672 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,995 |
1.1. (4 points) Find the least common multiple of the numbers 28 and 72. | Answer: 504.
Solution. We have $28=2^{2} \cdot 7, 72=2^{3} \cdot 3^{2}$, therefore $\operatorname{LCM}(28,72)=2^{3} \cdot 3^{2} \cdot 7=504$. | 504 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,996 |
1.4. Find the greatest common divisor of the numbers 144 and 120. | Answer: 24.
Solution. We have $144=2^{4} \cdot 3^{2}, 120=2^{3} \cdot 3 \cdot 5$, so $\text{GCD}(144,120)=2^{3} \cdot 3=24$. | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,997 |
2.1. (16 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the proportion (in percent) of masters... | Answer: 76.
Solution. Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from which we find $19x = 6y$. Therefore, the proportion of masters is $\frac{y}{x+y} = \frac{19y}{19x + 19y} = \frac{19y}{25y} = 0.76$, i.e., $76\%$. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,998 |
3.1. (16 points) A square with a side of 36 cm was cut into three rectangles such that the areas of all three rectangles are equal and any two rectangles share a common boundary segment. What is the total length (in cm) of the cuts made? | Answer: 60.
Solution. One cut has a length of 36 cm and separates one of the rectangles (see fig.). The remaining 2 rectangles have a common side $x$, so they are both $18 \times x$ cm in size. Therefore,
 An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,000 |
5.1. (16 points) A squirrel jumps once a minute to a distance of 1 along the number line, starting from point 0, and in any direction. She jumped for 40 minutes, after which she fell asleep. How many points on the number line could the squirrel have fallen asleep in? | Answer: 41.
Solution. We will solve the problem in a general form. Let the squirrel have jumped for $N$ minutes. If $N$ is even $(N=2 n)$, then after $2 n$ steps, the squirrel can only reach a point with an even coordinate in the range from $-2 n$ to $2 n$, and it can reach any of these points (to a point with coordin... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,001 |
6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then ad... | Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 peop... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,002 |
7.1. (16 points) Vasya's parents allowed him to buy himself two toys. In the store, there are 7 different remote-controlled cars and 5 different construction sets. In how many ways can he choose his gift? | Answer: 66.
Solution. Vasya can choose one car and one constructor in $5 \cdot 7=35$ ways, two different cars in $-\frac{7 \cdot 6}{2}=21$ ways (one car can be chosen in 7 ways, one of the remaining in 6 ways, but in this way each pair of cars is counted exactly twice), and two constructors in $-\frac{5 \cdot 4}{2}=10... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,003 |
4. The Tale of the Greedy Bear Cubs. Two bear cubs found a large round piece of cheese. They wanted to divide it equally, but they couldn't—each was afraid the other would get more. Suddenly, out of nowhere, a fox approached them.
- Let me divide the cheese equally for you.
- That's great! - the bear cubs rejoiced. - ... | Answer: 680g. Solution. We will analyze from the end. At the last moment, the bear cubs had pieces of 20g each, which means before that, they had pieces of 40g and 20g. We will denote this as $(20,20) \leftarrow(40,20)$. Then the complete sequence of actions will look like this:
$(20,20) \leftarrow(40,20) \leftarrow(4... | 680g | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,005 |
1.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five... | Answer: 7111765.
Solution. Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive... | 7111765 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,007 |
2.1. Find all values of $a$ for which the quadratic function $f(x)=a x^{2}-2 a x+1$ takes values, the modulus of which does not exceed 2, at all points of the interval $[0 ; 2]$. In your answer, specify the total length of the intervals to which the found values of $a$ belong. | Answer: 4 (the desired set of values for $a:[-1 ; 0) \cup(0 ; 3]$).
Solution. For the function $f(x)$ to be quadratic, it is necessary that $a \neq 0$. The vertex of the parabola, which is the graph of the given function, is located at the point $(1 ; 1-a)$. Since $f(0)=f(2)=1$, the condition of the problem is satisfi... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,008 |
2.2. Find all values of $a$ for which the quadratic function $f(x)=a x^{2}-4 a x+1$ takes values, the modulus of which does not exceed 3, at all points of the interval $[0 ; 4]$. In your answer, specify the total length of the intervals to which the found values of $a$ belong. | Answer: 1.5 (the set of values for $a$: $\left[-\frac{1}{2} ; 0\right) \cup(0 ; 1]$ ). | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,009 |
2.3. Find all values of $a$ for which the quadratic function $f(x)=a x^{2}+2 a x-1$ takes values, the modulus of which does not exceed 3, at all points of the interval $[-2 ; 0]$. In your answer, specify the total length of the intervals to which the found values of $a$ belong. | Answer: 6 (the set of desired values of $a$: $[-4 ; 0) \cup(0 ; 2]$). | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,010 |
2.4. Find all values of $a$ for which the quadratic function $f(x)=a x^{2}+4 a x-1$ takes values, the modulus of which does not exceed 4, at all points of the interval $[-4 ; 0]$. In your answer, specify the total length of the intervals to which the found values of $a$ belong. | Answer. 2 (the set of desired values of $a: \left[-\frac{5}{4} ; 0\right) \cup\left(0 ; \frac{3}{4}\right]$). | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,011 |
3.1. Find the largest integer solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}<1$. | Answer: -2001.
Solution. By factoring out $2^{x}$ and using the formula for the sum of the first terms of a geometric progression, we can transform the inequality to $2^{x} \cdot\left(2^{2001}-1\right)<1$. For $x \geqslant-2000$, this inequality is not satisfied. If, however, $x=-2001$, we get $1-2^{-2001}<1$, which i... | -2001 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,012 |
5.1. How many triangles with integer sides have a perimeter equal to 2017? (Triangles that differ only in the order of their sides, for example, 17, 1000, 1000 and 1000, 1000, 17, are counted as one triangle.) | Answer: 85008.
Solution. Let the sides of the triangle be $a, b, c$ ordered such that $a \leqslant b \leqslant c$. Then the smaller side $a$ does not exceed $2017 / 3=672 \frac{1}{3}$.
Consider the case of even $a$, i.e., $a=2 k$. Then for $k=1, \ldots, 252$, there will be $k$ triangles each, totaling $253 \cdot 252 ... | 85008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,013 |
6.1. Solve the equation $\frac{n!}{2}=k!+l!$ in natural numbers, where $n!=1 \cdot 2 \ldots n$. If there are no solutions, indicate 0; if there is one solution, indicate $n$; if there are multiple solutions, indicate the sum of the values of $n$ for all solutions. Recall that a solution is a triplet ( $n, k, l$); if so... | Answer: 10 (all triples of solutions $(n, k, l):(3,1,2),(3,2,1),(4,3,3)$).
Solution. Note that if $k4$, then $n!>4 \cdot(n-1)!\geqslant 2 \cdot(k!+l!)$, so there are no such solutions. If $n=2$, we get $1=k!+l!$ - no solutions; if $n=3$, the equation $3=k!+l!$ has two solutions: $k=1, l=2$ and $k=2, l=1$; if $n=4$, th... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,014 |
7.1. In an acute-angled triangle $A B C$, the median $B M$ and the altitude $C H$ were drawn. It turned out that $B M=C H=\sqrt{3}$, and $\angle M B C=\angle A C H$. Find the perimeter of triangle $A B C$. | Answer: 6.
Solution. We will prove that triangle $ABC$ is equilateral, with all sides equal to 2. Drop a perpendicular $MK$ from point $M$ to side $AB$. In triangle $BMK$, the leg $KM$ is half the hypotenuse $MB$, so $\angle KBM = 30^{\circ}$. Let $O$ be the intersection of $MB$ and $CH$, then $\angle MOC = 60^{\circ}... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,015 |
Problem 1. Three friends, weightlifters A, B, and C, came to a competition. They all competed in the same weight category, and one of them became the winner. If the weight lifted by weightlifter A is added to the weight lifted by weightlifter B, the total is 220 kg. If the weights lifted by weightlifters A and C are ad... | Answer: 135.
Solution. The sum of three weights is $(220+240+250) / 2=355$. Therefore, V lifted $355-220=135$, B lifted $355-240=115$, A lifted $355-250=105$. The winner is $\mathrm{B}=135$. | 135 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,016 |
Problem 2. How many solutions in integers does the equation
$$
\frac{1}{2022}=\frac{1}{x}+\frac{1}{y} ?
$$ | Answer: 53.
Solution. By eliminating the denominators, we obtain the equation
\[
(x-2022)(y-2022)=2022^{2}.
\]
Since \(2022^{2}=2^{2} \cdot 3^{2} \cdot 337^{2}\), the number \(x-2022\) can have the factors \(2^{0}, 2^{1}, 2^{2}\) - a total of three options. Similarly for the other factors. The number \(x-2022\) thus... | 53 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,017 |
Problem 3. Can four identical rectangles be placed on a plane so that no vertex is common to all rectangles, but any two rectangles have exactly one common vertex? (Rectangles can overlap each other.) | Answer: Yes.
Solution. See the drawing.
Other examples are possible. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,018 |
Problem 4. For an infinite sequence of numbers $x_{1}, x_{2}, x_{3}, \ldots$, for all natural $n \geq 4$, the relation $x_{n}=x_{n-1} \cdot x_{n-3}$ holds. It is known that $x_{1}=1, x_{2}=1, x_{3}=-1$. Find $x_{2022}$. | # Answer: 1.
Solution. All members of the sequence have a modulus of 1. Mark the positive ones with a plus and the negative ones with a minus. Write out the beginning of the sequence:
$$
[++---+-]++---+-++\ldots
$$
It is clear that three consecutive members completely determine how the sequence will look further. Th... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,019 |
Problem 5. From the digits $a, b, c, d, e$, a five-digit number $\overline{a b c d e}$ is formed. For the two-digit numbers $\overline{a b}, \overline{b c}, \overline{c d}, \overline{d e}$, formed from the same digits, it is known that
$$
(\overline{a b}+\overline{b c})(\overline{b c}+\overline{c d})(\overline{c d}+\o... | Answer: 12345 or 21436.
Solution. Note that $157605=3 \cdot 5 \cdot 7 \cdot 19 \cdot 79$. To start, we need to understand which combinations of prime factors correspond to the sums of two-digit numbers. The sum of two-digit numbers cannot exceed $99+99=198$, and $79 \cdot 3$ is already greater than 200, so one of the ... | 12345 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,020 |
Problem 6. In a square room, there is a light bulb on each wall that can shine in one of the seven colors of the rainbow. There are no light bulbs in the room that shine the same color. In one move, a person can change the color of one of the light bulbs to a color that is not used by any light bulb in the room at the ... | Answer: 8 moves.
Solution. Each light bulb on the walls of the room must change color 6 times, and there are four walls, so the total number of color changes is at least 24. On the other hand, in one move, the color of three light bulbs changes, so the number of moves is at least 8. An example for 8 moves can be provi... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,021 |
Task 1. Find the smallest natural number that has the following property: the remainder of its division by 20 is one less than the remainder of its division by 21, and the remainder of its division by 22 is 2. | Answer: 838.
Solution. The desired number is $20k + a = 21l + a + 1 = 22m + 2$, where $0 \leqslant a \leqslant 19$ and $l, k, m \geqslant 0$. From the first equation and the congruence modulo 20, we get that $l + 1 \equiv 0 \pmod{20}$. Since we are looking for the smallest number, let's try $l = 19$, if this $l$ does ... | 838 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,022 |
Problem 2. A triangular pyramid, all edges of which are 6 cm long, stands on a flat table. The pyramid is rolled over its edges 6 times in such a way that one of its vertices remains stationary, and it is not rolled over the same edge twice in a row. Find the length of the trajectory along which the mobile vertex of th... | Answer: $\left(\pi-\arccos \frac{1}{3}\right) \cdot 12 \sqrt{3}$ cm.
Solution. Let the pyramid be $A B C D$. Suppose $A$ is the fixed vertex, and vertex $D$ does not touch the table surface initially. Let the first rolling be over the edge $A B$. During this rolling, vertex $D$ moves along an arc of a circle with the ... | (\pi-\arccos\frac{1}{3})\cdot12\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,023 |
Problem 3. Find the three last digits of the number $10^{2022}-9^{2022}$. | Answer: 119.
Solution. Since $A=10^{2022}-(10-1)^{2022}=10^{2022}-10^{2022}+2022 \cdot 10^{2021}-C_{2022}^{2} \cdot 10^{2022}+\ldots+$ $C_{2022}^{3} \cdot 10^{3}-C_{2022}^{2} \cdot 10^{2}+C_{2022}^{1} \cdot 10-1$, then $A(\bmod 1000) \equiv-C_{2022}^{2} \cdot 100+C_{2022}^{1} \cdot 10-1(\bmod 1000) \equiv$ $-\frac{202... | 119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,024 |
Problem 4. Around a circle, 2022 ones are written. Two players take turns: in one move, a player erases two adjacent numbers and writes their sum in place (once). The player who gets the number 4 wins. If at the end of the game only one number remains that is not 4, the game ends in a draw. Can any of the players ensur... | Answer: The second player will win.
Solution. The second player wins. Strategy: mirror the first player's moves symmetrically relative to the center of the circle until after the first player's move, either two adjacent twos or a three adjacent to a one appear. In these cases, the second player takes the initiative an... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,025 |
Problem 5. The number $a$ is such that the equation $t^{3}+4 t^{2}+4 t+a=0$ has three real roots $x<y<z$. Determine the numerical values that the expression
$$
A=x^{3}-4 y^{2}-4 z^{2}-4 y-4 z+32
$$
can take. | Answer: $\left(14 \frac{22}{27} ; 16\right)$.
Solution. Let $P(t)=t^{3}+4 t^{2}+4 t+a=(t-x)(t-y)(t-z)$. Then $x+y+z=-4, xy+$ $yz+xz=4, xyz=-a$ (this follows from Vieta's theorem or by directly comparing the left and right sides). Next, $A=x^{3}-4 y^{2}-4 z^{2}-4 y-4 z+32=\left(-4 x^{2}-4 x-a\right)-4 y^{2}-4 z^{2}-4 y... | (14\frac{22}{27};16) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,026 |
Problem 6. Find the coordinates of all points in the plane that are equidistant from all points of intersection of the parabolas given in the Cartesian coordinate system on the plane by the equations $y=2 x^{2}-1$ and $x=4 y^{2}-2$.
# | # Answer: $\left(\frac{1}{8} ; \frac{1}{4}\right)$.
Solution. By plotting the parabolas with the given equations, it is easy to notice that they have 4 common points. Therefore, there can be at most one point equidistant from them. Rewrite the equations of the parabolas as $x^{2}-\frac{1}{2} y-\frac{1}{2}=0$ and $y^{2... | (\frac{1}{8};\frac{1}{4}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,027 |
Problem 7. There is a certain number of identical plastic bags that can be nested inside each other. If all the other bags end up inside one of the bags, we will call this situation a "bag of bags." Calculate the number of ways to form a "bag of bags" from 10 bags.
## Explanation. Denote a bag with parentheses.
If th... | Solution. If $\Pi_{n}$ denotes the number of ways for $n$ packages, then:
$$
\begin{gathered}
\Pi_{1}=1, \Pi_{2}=1, \Pi_{3}=2, \Pi_{4}=4, \Pi_{5}=9, \Pi_{6}=20, \Pi_{7}=48, \Pi_{8}=115, \Pi_{9}=286 \\
\Pi_{10}=719
\end{gathered}
$$
The problem is solved by enumerating the cases. For example, if we take $\Pi_{5}$:
$\... | 719 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,028 |
1. Detective Podberezyakov is pursuing Maksim Detochkin (each driving their own car). At the beginning, both were driving on the highway at a speed of 60 km/h, with Podberezyakov lagging behind Detochkin by 2 km. Upon entering the city, each of them reduced their speed to 40 km/h, and upon exiting the city, finding the... | Answer: 1 km.
Solution: At the moment Detochkin entered the city, Podberezyakov was 2 km behind him. Podberezyakov covered this distance in 2/60 hours, during which time Detochkin traveled $\frac{2}{60} \cdot 40$ km. That is, the distance was multiplied by 40/60. Similarly, when exiting the city and transitioning to t... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,030 |
3. And our cat gave birth to kittens yesterday! It is known that the two lightest kittens weigh a total of 80 g, the four heaviest weigh 200 g, and the total weight of all the kittens is 500 g. How many kittens did the cat give birth to? | Answer: 11.
Solution: The two lightest weigh 80 g, so the others weigh no less than 40 g each. Similarly, we find that all except the 4 heaviest weigh no more than 50 g. Consider the kittens that are not among the 2 lightest and the 4 heaviest. Their total weight is 500-200-80=220g. There must be 5 of them, because if... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,032 |
4. Find a two-digit number, the digits of which are different and the square of which is equal to the cube of the sum of its digits. | Answer: 27.
Solution: Let's write the condition as $\overline{A B}^{2}=(A+B)^{3}$. Note that the sum of the digits ( $A+B$ ) does not exceed 17, i.e., $\overline{A B}^{2} \leq 17^{3}$. Moreover, this number $\overline{A B}^{2}=n^{6}$, where $n-$
is some natural number that does not exceed 4. However, 1 and 2 do not wo... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,033 |
5. On a circle, 25 points are marked, painted either red or blue. Some of the points are connected by segments, with one end of each segment being blue and the other end red. It is known that there do not exist two red points that belong to the same number of segments. What is the maximum possible number of red points? | Answer: 13.
Solution: Let's take 13 red and 12 blue points. The first red point is not connected to any other, the second is connected to one blue, ..., the 13th is connected to 12 blue points. Obviously, there cannot be more red points, because if there are more than 13, the number of connection options is less than ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,034 |
6. Pete and Vasya are playing a game. On the board, there is a number: 11223334445555666677777. In one move, it is allowed to erase any number of identical digits. The player who erases the last digit wins. Pete goes first. Can he play in such a way as to guarantee a win? | Answer: Yes, he can.
Solution: On his first move, Petya can, for example, erase all the digits 7. The remaining digits can be written as:
113335555 | 666644422.
After this, Petya symmetrically (relative to the middle line) repeats Vasya's moves. | Yes,hecan | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,035 |
1. At $12: 00$ a raft departed from point A to point B, located downstream along the river. At $12: 45$ a motorboat departed from A in pursuit of the raft, caught up with the raft at 13:00, and continued its journey. Upon arriving at point B, the boat turned around and headed back, meeting the raft at 14:00. Determine ... | Answer: 15:00. Solution. The boat caught up with the raft in 15 minutes, which means its speed downstream is 4 times the speed of the river current. If we consider a coordinate system associated with the raft, it is not difficult to understand that the boat arrived at point B at 13:30, i.e., spent 45 minutes traveling ... | 15:00 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,036 |
2. An isosceles triangle with sides $A B=B C=3, A C=4$ has an inscribed circle that touches the sides of the triangle at points $K, L$, and $M$. Find the ratio of the areas $S(\triangle A B C): S(\triangle K L M)$. | Answer: $\frac{9}{2}$. Solution. Since the triangle is isosceles, the inscribed circle touches the base at its midpoint, so $A M=M C=$ $A K=C L=2, B K=B L=1$. We will find the areas of the triangles (in relation to the area of $\triangle A B C): S(\triangle A K M)=S(\triangle C L M)=\frac{1}{3} S(\triangle A B C)$, $S(... | \frac{9}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,037 |
3. Find the sum
$$
\begin{aligned}
& \frac{1}{(\sqrt[4]{1}+\sqrt[4]{2})(\sqrt{1}+\sqrt{2})}+\frac{1}{(\sqrt[4]{2}+\sqrt[4]{3})(\sqrt{2}+\sqrt{3})}+ \\
& +\ldots+\frac{1}{(\sqrt[4]{9999}+\sqrt[4]{10000})(\sqrt{9999}+\sqrt{10000})}
\end{aligned}
$$ | Answer: 9. Solution. If we multiply the numerator and denominator of the fraction $\frac{1}{(\sqrt{n}+\sqrt{n}+1)(\sqrt{n}+\sqrt{n+1})}$ by $\sqrt[4]{n}-\sqrt[4]{n+1}$, we get $\sqrt[4]{n+1}-\sqrt[4]{n}$. The specified
 are arranged such that each number is three times smaller than the number in the adjacent cell to the right and twice as large as the number in the adjacent cell below. Find the number in the central cell if it is known that the sum of all numbers in the ta... | Answer: $\frac{36}{341}$. Solution. Let the number in the lower left corner be denoted by $x$. We obtain the equation $x(1+2+4+8+16)(1+3+9+27+81)=11$, from which $x=\frac{1}{341}$. The number in the center is $36 x=\frac{36}{341}$. | \frac{36}{341} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,039 |
5. In trapezoid $A B C D$, the base $A D$ is four times larger than the base $B C$, and the angle $\angle B C D$ is twice the angle $\angle B A D$. Find the ratio $C D$ : $P Q$, where $P Q-$ is the midline of the trapezoid. | Answer: 6:5. Solution. Let $B C=a, A D=4 a$, then $P Q=(a+4 a) / 2=$ $2.5 a$. Draw the bisector $C L$ of angle $\angle B C D$ (see figure). It will divide the trapezoid into a parallelogram and an isosceles triangle $\triangle C L D$ with sides $C D=L D=3 a$. Therefore, $C D: P Q=3 a: 2.5 a=6: 5$.
^{2}+3 z^{2}$, we get $4-4 z+3 z^{2}$. This is a quadratic function, which reaches a minimum of $\frac{8}{3}$ at $z=\frac{2}{3}$. | \frac{8}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,041 |
7. Given the polynomial $P(x)=x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$, it is known that $P(2014)=1, P(2015)=2, P(2016)=3, P(2017)=4$, $P(2018)=5$. Find $P(2013)$. | Answer: $-5!=-120$. Solution. Consider the polynomial $P(x)-x+2013$, the numbers $2014, \ldots 2018$ are its roots. Therefore (since the leading coefficient is 1) it can be represented as $P(x)-x+2013=$ $(x-2014)(x-2015) \cdot \ldots \cdot(x-2018)$. Substituting $x=2013$, we get $P(2013)-2013+2013=-120$. | -120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,042 |
8. An equilateral triangle $\triangle A_{1} B_{1} C_{1}$ is inscribed in a circle of unit radius. A circle is inscribed in the triangle $\triangle A_{1} B_{1} C_{1}$, and the points of tangency are denoted as $A_{2}, B_{2}$, and $C_{2}$. A circle is inscribed in the triangle $\triangle A_{2} B_{2} C_{2}$, and the point... | Answer: $\left(\frac{4 \pi}{3}-\sqrt{3}\right) \cdot\left(1-4^{-100}\right)$. Solution. Note that at each step, the size of the circle and the triangle is halved. Therefore, the areas of the specified parts form a geometric progression with a common ratio of $\frac{1}{4}$. The area of the first one can be obtained by s... | (\frac{4\pi}{3}-\sqrt{3})\cdot(1-4^{-100}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,043 |
9. Solve the equation $f(x)=0$, given that the function $f(x)$ is even and for any $x, y$ the equality $f(x+y)=f(x)+f(y)+$ $xy+2014$ holds. | Answer: $\pm 2 \sqrt{1007}$. Solution. Substituting $x=y=0$, we get that $f(0)=$ -2014. Substitute $y=-x$ and use the evenness of $f(x)$, then $-2014=$ $f(0)=2 f(x)-x^{2}+2014$, from which $f(x)=\frac{x^{2}}{2}-2014$. Setting it to zero and solving the obtained equation, we find $x= \pm 2 \sqrt{1007}$. | \2\sqrt{1007} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,044 |
1. Solve the cryptarithm (the same letters correspond to the same digits, different letters correspond to different digits)
$$
\mathrm{MSU}+\mathrm{MSU}+\mathrm{MSU}+\mathrm{MSU}+\mathrm{OLYMP}+\mathrm{OLYMP}=\mathrm{MOSCOW}
$$ | Answer: $C=5, L=7, M=1, O=9, P=2, S=4, U=3, W=6, Y=0,143+143+$ $143+143+97012+97012=194596$
## Solution: | 143+143+143+143+97012+97012=194596 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,045 |
3. A flea jumps along the number line, and the length of each jump cannot be less than $n$. It starts its movement from the origin and wants to visit all integer points belonging to the segment $[0,2013]$ (and only them!) exactly once. For what greatest value of $n$ will it be able to do this? | Answer: $n=1006$.
Solution: For $n=1006$, a path can be constructed as follows:
$$
0 \rightarrow 1007 \rightarrow 1 \rightarrow 1008 \rightarrow \ldots \rightarrow 1005 \rightarrow 2012 \rightarrow 1006 \rightarrow 2013
$$
We will prove that $n$ cannot be greater than 1006. Indeed, suppose $n \geqslant 1007$. Then t... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,047 |
4. Solve the system $\left\{\begin{array}{l}x^{2}-2 y+1=0 ; \\ y^{2}-4 z+7=0 \\ z^{2}+2 x-2=0 .\end{array}\right.$ | Answer: $x=-1, y=1, z=2$.
Solution: Adding all three equations, we get $x^{2}-4 x+y^{2}-2 y+z^{2}-4 z+6=0$. By completing the square, we obtain $(x+1)^{2}+(y-1)^{2}+(z-2)^{2}=0$, from which $x=-1$, $y=1, z=2$. Verification shows that the specified values are a solution to the system. | -1,1,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,048 |
5. Find the number of 9-digit numbers in which each digit from 1 to 9 appears exactly once, the digits 1, 2, 3, 4, 5 are arranged in ascending order, and the digit 6 appears before the digit 1 (for example, 916238457). | Answer: 504
Solution: Note that after arranging the digits $7,8,9$, the remaining digits are uniquely determined. Therefore, the number of such numbers is $7 \cdot 8 \cdot 9=504$. | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,049 |
6. Given a parallelogram $A B C D$ and points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are chosen such that point $A$ is the midpoint of segment $D D_{1}$, point $B$ is the midpoint of $A A_{1}$, point $C$ is the midpoint of $B B_{1}$, and point $D$ is the midpoint of $C C_{1}$. a) Prove that $A_{1} B_{1} C_{1} D_{1}$ is als... | # Answer: b) 5
## Solution:
a) From the properties of a parallelogram, it follows that $A B = C D, \angle B A D_{1} = 180^{\circ} - \angle B A D = 180^{\circ} - \angle B C D = \angle D C ... | 5 | Geometry | proof | Yes | Yes | olympiads | false | 6,050 |
1.1. From the digits 1, 3, and 5, different three-digit numbers are formed, each of which has all distinct digits. Find the sum of all such three-digit numbers. | Answer: 1998 (all possible numbers: $135,153,315,351,513,531$ ). | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,052 |
1.2. From the digits 1, 2, and 5, different three-digit numbers are formed, each of which has all distinct digits. Find the sum of all such three-digit numbers. | Answer: 1776 (all possible numbers: 125, 152, 215, 251, 512, 521). | 1776 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,053 |
3.1. The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $577 a, \frac{2020 b}{7}, \frac{c}{7}$ is an arithmetic progression. Find the common ratio of the geometric progression. | Answer: 4039.
Solution. Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{2020 a q}{7}=577 a+\frac{a q^{2}}{7} \Leftrightarrow q^{2}-4040 q+4039=0$, from which $q=1$ or $q=4039$. A decreasing geometric progression can only occur wh... | 4039 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,056 |
4.1. Find the area of the figure defined on the coordinate plane by the system
$$
\left\{\begin{array}{l}
2 \pi\left(x^{2}+y^{2}\right) \leqslant 15 \\
x^{4}-y^{4} \leqslant x y-x^{3} y^{3}
\end{array}\right.
$$ | Answer: 3.75.
Solution. The first set is the interior of a circle with radius $\sqrt{\frac{15}{2 \pi}}$. The second inequality is equivalent to
$$
x^{3}\left(x+y^{3}\right) \leqslant y\left(x+y^{3}\right) \quad \Leftrightarrow \quad\left(x^{3}-y\right)\left(x+y^{3}\right) \leqslant 0
$$
By plotting the curves $y=x^{... | 3.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,057 |
5.1. What is the maximum volume of the pyramid $S A B C$, where $A B=5, A C=8$ and $\sin \angle B A C = \frac{4}{5}$, and all lateral edges $S A, S B, S C$ form the same angle with the base plane, not exceeding $60^{\circ} ?$ | Answer. $10 \sqrt{\frac{137}{3}} \approx 67.58$.
Solution. The maximum volume is achieved at the angle $\alpha=60^{\circ}$, then the height of the pyramid is $R \operatorname{tg} \alpha=R \sqrt{3}$, where $R$ is the radius of the circumscribed circle around the triangle. Under the given conditions, there are two possi... | 10\sqrt{\frac{137}{3}}\approx67.58 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,058 |
5.2. What is the maximum volume of the pyramid $S A B C$, where $A B=5, A C=8$ and $\sin \angle B A C = \frac{3}{5}$, and all lateral edges $S A, S B, S C$ form the same angle with the base plane, not exceeding $60^{\circ} ?$ | Answer. $10 \sqrt{51} \approx 71.41$. | 10\sqrt{51} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,059 |
5.3. What is the maximum volume of the pyramid $S A B C$, where $A B=3, A C=5$ and $\sin \angle B A C = \frac{4}{5}$, and all lateral edges $S A, S B, S C$ form the same angle with the base plane, not exceeding $60^{\circ}$? | Answer. $\frac{5 \sqrt{39}}{2} \approx 15.61$. | \frac{5\sqrt{39}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,060 |
5.4. What is the maximum volume of the pyramid $S A B C$, where $A B=3, A C=5$ and $\sin \angle B A C = \frac{3}{5}$, and all lateral edges $S A, S B, S C$ form the same angle with the base plane, not exceeding $60^{\circ} ?$ | Answer. $\frac{5 \sqrt{174}}{4} \approx 16.49$. | \frac{5\sqrt{174}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,061 |
6.1. On 19 cards, the numbers $15,16,17, \ldots, 33$ are written respectively (one number per card). Members of the math club Vasya, Petya, and Misha decided to divide all these cards among themselves so that each of them gets at least one card and no one ends up with a pair of cards where the difference between the nu... | Answer: 4596.
Solution. According to the condition, each participant will have either only even-numbered cards or only odd-numbered cards. We choose a participant (3 ways), and give them all 9 cards with even numbers. The remaining 10 cards with odd numbers are distributed between the two others, which can be done in ... | 4596 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,062 |
7.1. The numbers $p$ and $q$ are chosen such that the parabola $y=p x-x^{2}$ intersects the hyperbola $x y=q$ at three distinct points $A, B$, and $C$, and the sum of the squares of the sides of triangle $A B C$ is 324, while the centroid of the triangle is 2 units away from the origin. Find the product $p q$. | Answer: 42.
Solution. The system $y=p x-x^{2}, x y=q$ reduces to the cubic equation $x^{3}-p x^{2}+q=0$, which has, according to the problem, three distinct roots $x_{1}, x_{2}, x_{3}$, since the abscissas of any two different points on the curve $y=p x-x^{2}$ are distinct. By Vieta's theorem,
$$
x_{1}+x_{2}+x_{3}=p,... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,063 |
8.1. For all triples $(x, y, z)$ satisfying the system
$$
\left\{\begin{array}{l}
2 \sin x=\operatorname{tg} y \\
2 \cos y=\operatorname{ctg} z \\
\sin z=\operatorname{tg} x
\end{array}\right.
$$
find the smallest value of the expression $\cos x-\sin z$. | Answer: $-\frac{5 \sqrt{3}}{6} \approx-1.44$.
Solution. Squaring the first equation and adding 1 to both sides, taking into account the second equation, we have
$$
4 \sin ^{2} x+1=\operatorname{tg}^{2} y+1=\frac{1}{\cos ^{2} y}=4 \operatorname{tg}^{2} z
$$
Performing the same procedure with the third equation, we fi... | -\frac{5\sqrt{3}}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,064 |
8.2. For all triples $(x, y, z)$ satisfying the system
$$
\left\{\begin{array}{l}
\sqrt{3} \sin x=\operatorname{tg} y \\
2 \sin y=\operatorname{ctg} z \\
\sin z=2 \operatorname{tg} x
\end{array}\right.
$$
find the smallest value of the expression $\cos x-\cos z$. | Answer. $-\frac{7 \sqrt{2}}{6} \approx-1.65$. | -\frac{7\sqrt{2}}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,065 |
8.3. For all triples $(x, y, z)$ satisfying the system
$$
\left\{\begin{array}{l}
2 \cos x=\operatorname{ctg} y \\
2 \sin y=\operatorname{tg} z \\
\cos z=\operatorname{ctg} x
\end{array}\right.
$$
find the smallest value of the expression $\sin x+\cos z$. | Answer. $-\frac{5 \sqrt{3}}{6} \approx-1.44$. | -\frac{5\sqrt{3}}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,066 |
8.4. For all triples $(x, y, z)$ satisfying the system
$$
\left\{\begin{array}{l}
\sqrt{3} \cos x=\operatorname{ctg} y \\
2 \cos y=\operatorname{tg} z \\
\cos z=2 \operatorname{ctg} x
\end{array}\right.
$$
find the smallest value of the expression $\sin x+\sin z$. | Answer. $-\frac{7 \sqrt{2}}{6} \approx-1.65$. | -\frac{7\sqrt{2}}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,067 |
9.1. How many natural numbers $n \in [20182019 ; 20192018]$ are there for which the number $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}\right]$ is even? (Here, $[x]$ denotes the greatest integer not exceeding $x$.) | Answer: 4999.
Solution. Consider the sequences
$$
x_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}, \quad y_{n}=\left(\frac{1-\sqrt{5}}{2}\right)^{n}, \quad a_{n}=x_{n}+y_{n}
$$
Since $-1<\frac{1-\sqrt{5}}{2}<0$, we have $y_{n} \in(0 ; 1)$ for even $n$, and $y_{n} \in(-1 ; 0)$ for odd $n$. Then $\left[x_{n}\right]=\left[... | 4999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,068 |
10.1. Find the smallest solution of the inequality
$$
\frac{-\log _{2}(120-2 x \sqrt{32-2 x})^{2}+\left|\log _{2} \frac{120-2 x \sqrt{32-2 x}}{\left(x^{2}-2 x+8\right)^{3}}\right|}{5 \log _{7}(71-2 x \sqrt{32-2 x})-2 \log _{2}(120-2 x \sqrt{32-2 x})} \geqslant 0
$$ | Answer: $-13-\sqrt{57} \approx-20.55$.
Solution. Let $f(x)=-\log _{2}(x+120)$. Then, taking into account the condition $71-2 x \sqrt{32-2 x}>0$, the original inequality can be rewritten as
$$
\frac{2 f(-2 x \sqrt{32-2 x})+\left|3 f\left(x^{2}-2 x-112\right)-f(-2 x \sqrt{32-2 x})\right|}{\left(2 \log _{2} 7\right) \cd... | -13-\sqrt{57}\approx-20.55 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,069 |
10.2. Find the smallest solution of the inequality
$$
\frac{-\log _{2}(105+2 x \sqrt{x+19})^{3}+\left|\log _{2} \frac{105+2 x \sqrt{x+19}}{\left(x^{2}+x+3\right)^{4}}\right|}{9 \log _{5}(76+2 x \sqrt{x+19})-4 \log _{2}(105+2 x \sqrt{x+19})} \geqslant 0
$$ | Answer: $\frac{-21+\sqrt{33}}{2} \approx-7.63$. | \frac{-21+\sqrt{33}}{2}\approx-7.63 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,070 |
10.3. Find the greatest solution of the inequality
$$
\frac{-\log _{3}(100+2 x \sqrt{2 x+25})^{3}+\left|\log _{3} \frac{100+2 x \sqrt{2 x+25}}{\left(x^{2}+2 x+4\right)^{4}}\right|}{3 \log _{6}(50+2 x \sqrt{2 x+25})-2 \log _{3}(100+2 x \sqrt{2 x+25})} \geqslant 0
$$ | Answer. $12+4 \sqrt{3} \approx 18.93$. | 12+4\sqrt{3}\approx18.93 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,071 |
1. The sum of positive integers is 11. In the first part of this equation, identical numbers are hidden behind cards with the same letters, and different numbers - behind cards with different letters.
Consider the equation: $\quad \mathbf{C}+\mathbf{y}+\mathbf{M}+\mathbf{M}+\mathbf{A}=11$.
Can you tell which number i... | Answer: 1. Solution. If $M=1$, then the sum $C+y+A=9$, which is only possible with the set of numbers 2, 3, 4 (in any order). If $M=2$, then the sum $C+y+A=7$, which is impossible, as this sum is not less than $1+3+4=8$. If $M>2$, this is even more impossible. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,073 |
# 5. Place the signs $\times$ and : instead of the squares in the expression
$$
1 \square 3 \square 3^{2} \square 3^{4} \square 3^{8} \square 3^{16} \square 3^{32} \square 3^{64}=3^{99}
$$
such that the value of the expression becomes $3^{99}$. | Answer: $1 \times 3: 3^{2}: 3^{4}: 3^{8} \times 3^{16} \times 3^{32} \times 3^{64}=3^{99}$. Solution. The problem reduces to choosing between the signs (plus) and (minus) in the identity $0 \pm 1 \pm 2 \pm 4 \pm 8 \pm 16 \pm 32 \pm 64=99$. This can be done simply by trial and error. However, we can also denote the sum ... | 1\times3:3^{2}:3^{4}:3^{8}\times3^{16}\times3^{32}\times3^{64}=3^{99} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,074 |
6. Pete was given a new electric jigsaw on his birthday, with a feature to count the length of the cuts made. To try out the gift, Pete took a square piece of plywood with a side of 50 cm and cut it into squares with sides of 10 cm and 20 cm. How many squares in total were obtained, if the electric jigsaw shows a total... | Answer: 16. Solution. The perimeter of the figures must be taken into account, in addition to the perimeter of the original square, from which we get that the total perimeter of the resulting squares is $280 \cdot 2 + 200 = 760$. Now, we can denote the number of squares through $x$ and $y$ respectively and solve the sy... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,075 |
3. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{224}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$ ANSWER:60. | Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$. | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,078 |
4. An isosceles triangle can be cut into two isosceles triangles (not necessarily equal). Find the values that the smallest angle of such a triangle can take. In your answer, specify the smallest of these values in degrees, multiplied by 6006.
ANSWER: 154440. | Solution: Consider triangle $ABC$ with angles $\angle A=\angle C=\alpha, \angle B=180^{\circ}-2 \alpha$. For the line to divide the triangle into two, it must pass through one of the vertices.
Consider the case when it passes through vertex A and divides the triangle into two: $ADB$ and $ADC$ (see figure).
^{2}+$ $(y-1)^{2}=25$. This is possible when one of the terms equals 25 and the other equals 0 (4 cases), or when one equals 16 and the other equals 9 (8 cases). | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,080 |
7. Find all natural numbers N such that the remainder of dividing 2017 by $N$ is 17. In your answer, specify the number of such $N$.
ANSWER 13. | Solution: As N, all positive divisors of the number 2000 greater than 17 will work. The divisors of 2000 have the form $2^{a} \cdot 5^{b}$, where $a=0,1,2,3,4$ and $b=0,1,2,3$. There will be 20 of them in total, but we need to exclude $1,2,4,5,8,10$ and 16. | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,081 |
8. On a circle, 2017 different points $A_{1}, \ldots, A_{2017}$ are marked, and all possible chords connecting these points pairwise are drawn. A line is drawn through the point $A_{1}$, not passing through any of the points $A_{2}, \ldots A_{2017}$. Find the maximum possible number of chords that can have at least one... | Solution: Let $k$ points be located on one side of the given line, then $2016-k$ points are on the other side. Thus, $k(2016-k)$ chords intersect the line and another 2016 pass through point $A_{1}$. Note that $k(2016-k)=-(k-1008)^{2}+1008^{2}$. The maximum of this expression is achieved when $k=1008$. | 1018080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,082 |
5.1. $[5-6.4$ (а. - 10 points, б. - 20 points)] On a grid paper, a figure is drawn (see the picture). It is required to cut it into several parts and assemble a square from them (parts can be rotated, but not flipped). Is it possible to do this under the condition that a) there are no more than four parts; b) there are... | Answer: a) yes; b) yes.
Solution. Possible cutting options for parts a) and b) are shown in Fig. 5, a) and b).
Fig. 5, a)
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,084 |
6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then ad... | Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 peop... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,085 |
2. Which of the numbers is greater: $\underbrace{\sqrt{17 \sqrt{13 \sqrt{17 \sqrt{13 \sqrt{17 \ldots}}}}}}_{2018 \text { root signs }}$ or $17 \sqrt[3]{\frac{13}{17}}$? | Answer: the second.
Solution. Let $A$ be the first number, $B$ be the second. Then
Since
$$
\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\ldots+\frac{1}{2^{2017}}<\frac{1}{2}+\frac{1}{8}+\frac{1}{... | the | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,087 |
3. In triangle $ABC$, whose area is 20, the median $CD$ is drawn. Find the radius of the circumcircle of triangle $ABC$, given that $AC=\sqrt{41}$, and the center of the circle inscribed in triangle $ACD$ lies on the circumcircle of triangle $BCD$.
---
The translation maintains the original text's line breaks and for... | Answer: $\frac{41}{10}$ or $\frac{41}{8}$.
Solution. Let $Q$ be the center of the circle inscribed in triangle $ACD$. Then the segments $AQ$ and $CQ$ are the bisectors of angles $BAC$ and $ACD$ respectively, by the properties of inscribed angles $\angle DBQ = \angle DCQ$. Therefore, triangles $ABQ$ and $ACQ$ are equal... | \frac{41}{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,088 |
4. Photographs are archived in the order of their numbering in identical albums, with exactly 4 photographs per page. In this case, the 81st photograph in sequence landed on the 5th page of one of the albums, and the 171st photograph landed on the 3rd page of another. How many photographs can each album hold?
# | # Answer: 32.
Solution. Let $x, y$ be the album numbers in which the 81st and 171st photos are placed, respectively, and $n>4$ be the number of pages in the album. Then $4 n(x-1)+16<81 \leqslant 4 n(x-1)+20, 4 n(y-1)+8<171 \leqslant 4 n(y-1)+12$, i.e., $61 \leqslant 4 n(x-1)<65, 159 \leqslant 4 n(y-1)<163$. Therefore,... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,089 |
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