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Task 5. Replace in the equation
$$
\text { PIE = SLICE + SLICE + SLICE + ... + SLICE }
$$
identical letters with identical digits, and different letters with different digits, so that the equation is true, and the number of "slices of pie" is the largest possible. | Answer. The maximum number of "pieces" is seven, for example: ПИРОГ $=95207$, КУСОК $=13601$.
Solution. An example for seven "pieces" is given above. We will show that there cannot be more than seven "pieces". For this, it is convenient to rewrite the condition as a multiplication example: ПИРОГ $=$ КУСОК $\cdot n$, w... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. First-year students admitted to the university were distributed into study groups so that each group had the same number of students. Due to a reduction in the number of specialties, the number of groups decreased by 9, and all first-year students were redistributed into groups; as a result, the groups were again eq... | Let the new number of groups be $n$, then initially there were ( $n+9$ ) groups. In each group, there was an equal number of students, so $2376: n$ and $2376:(n+9)$. We factorize 2376 into prime factors ( $2376=2^{3} \cdot 3^{3} \cdot 11$ ) and list all divisors: $1,2,4,8,3,6,12,24,9,18,36,72,27,54$, $108,216,11,22,44,... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of the first thirty-three terms of an arithmetic progression, given that the sum of the seventh, eleventh, twentieth, and thirtieth terms of this progression is 28. | Answer. $S_{33}=231$.
Solution Let $a_{k}$ be the $k$-th term of the arithmetic progression, and $d$ be its common difference. Then, according to the condition, $a_{7}+a_{11}+a_{20}+a_{30}=28$, from which we have $a_{7}+\left(a_{7}+4 d\right)+\left(a_{7}+13 d\right)+\left(a_{7}+23 d\right)=28 \Leftrightarrow a_{7}+10 ... | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. An infinite geometric progression consists of natural numbers. It turned out that the product of the first four terms equals param1. Find the number of such progressions.
The infinite geometric progression consists of positive integers. It turned out that the product of the first four terms equals param1. Find the ... | Solution
If $b_{1}$ is the first term of the progression and $q$ is its common ratio, then the product of the first four terms of the progression is $b_{1}^{4} q^{6}$. Therefore, $b_{1}^{2} q^{3}=2^{100} \cdot 3^{150}$. Hence, $b_{1}=2^{a} 3^{b}, q=2^{c} 3^{d}$, and we obtain the system: $2 a+3 c=100, 2 b+3 d=150$. Th... | 442 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given a function $f: \sqcup \rightarrow \sqcup$ such that $f(1)=1$, and for any $x \in \sqcup, y \in \sqcup$ the equality $f(x)+f(y)+x y+1=f(x+y)$ holds. Find all integers $n$ for which the equality $f(n)=$ param 1 holds. In the answer, write down the sum of cubes of all such values of $n$.
Function $f: \sqcup \rig... | # Solution
Substituting $y=1$ into the given functional equation, we get: $f(x)+f(1)+x+1=f(x+1)$, which means $f(k)-f(k-1)=k+1$. Therefore, $f(n)-f(1)=(f(n)-f(n-1))+(f(n-1)-f(n-2))+\ldots$ $+f(2)-f(1)=n+1+n+\ldots+2$. Hence, $f(n)=\frac{n^{2}+3 n-2}{2}$. Reasoning similarly,
we get $f(n)=\frac{n^{2}+3 n-2}{2}$ for $n ... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. What is the smallest possible area of the pentagon $A B C D E$, given that the quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas param1 and... | # Solution
Let the areas of parallelograms ABPM and DCPN, triangle BCP, and quadrilateral MPNE be $S_{1}, S_{2}, S_{3}$, and $S_{4}$, respectively. According to the problem, $A M \sqcup B \mathcal{B}$ and $P N \sqcup C \mathcal{D}$. Therefore, $\mathrm{AE} \sqcup \mathcal{B} N \cup C D$. Similarly, $D E \sqcup C M \sq... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Let param1. What is the largest possible value of param2?
It is given that param1. Find the largest possible value of param2.
| param 1 | param 2 | Answer |
| :---: | :---: | :---: |
| $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=3$ | $6 \sin x$ | 4 |
| $\frac{25 \sin ^{2} x-37+40 \cos x}{... | # Solution
Notice that $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=\frac{6-(3 \sin x-2)^{2}}{2+(3 \cos x+\sqrt{5})^{2}}$. The obtained expression can equal 3 only if $3 \sin x-2=0$ and $3 \cos x+\sqrt{5}=0$. Therefore, the expression $6 \sin x$ can only take the value 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The clock hand points to 12. Jack writes a sequence consisting of param 1 symbols, each symbol being plus or minus. After that he gives this sequence to a robot. The robot reads it from right to left. If he sees a plus he turns the clock hand $120^{\circ}$ clockwise and if he sees a minus he turns it $120^{\circ}$ coun... | # Solution
Let $a_{n}$ be the number of sequences of length $n$ that result in the arrow pointing at 12 o'clock, and $b_{n}$ be the number of sequences of length $n$ that result in the arrow pointing at 4 or 8 o'clock. It is not difficult to understand that $a_{n+1}=2 b_{n}, b_{n+1}=a_{n}+b_{n}$. From this, we get tha... | 682 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. Let $S(k)$ denote the sum of all the digits in the decimal representation of a positive integer $k$. Let $n$ be the smallest positive integer satisfying the condition $S(n)+S(n+1)=$ param1. As the answer to the problem, write down a five-digit number such that its first two digits coincide with the first two digits... | # Solution
Let $R$ be the radius of the circumcircle of triangle KBM and $r$ be the radius of the incircle of triangle ABC. First, we will prove that the centers of the circles mentioned in the problem statement coincide. Indeed, the center $I$ of the incircle of triangle $ABC$ lies on the bisector of angle $BAC$, whi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
14. In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. Find the area of pentagon $A B C D E$ if it is known that quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas param1 and param2, respectivel... | # Solution
Let the areas of parallelograms ABP M and DCPN, triangle BCP, and quadrilateral
$A E \sqcup B N N C D$. Similarly, $D E \sqcup C M \sqcup\{B$. Then M PNE is also a parallelogram,... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
17. The difference of the squares of two different real numbers is param 1 times greater than the difference of these numbers, and the difference of the cubes of these numbers is param 2 times greater than the difference of these numbers. By how many times is the difference of the fourth powers of these numbers greater... | # Solution
From the equalities $a^{2}-b^{2}=k(a-b)$ and $a^{3}-b^{3}=m(a-b)$ (in the problem $k=37, m=1069$), it follows that $a+b=k$ and $a^{2}+a b+b^{2}=m$. Squaring the first equality and subtracting the second from it, we get: $a b=k^{2}-m$. Then the ratio $\left(a^{4}-b^{4}\right):\left(a^{2}-b^{2}\right)=a^{2}+b... | 769 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Out of two hundred ninth-grade students, 80% received excellent grades on the first exam, 70% on the second exam, and 59% on the third exam. What is the smallest number of participants who could have received excellent grades on all three exams?
Answer: 18. | Let $M_{i}$ be the number of students who received excellent grades only on the $i$-th exam; $M_{ij}$ be the number of students who received excellent grades only on exams $i$ and $j$; $M_{123}$ be the number of students who received excellent grades on all three exams. Then, according to the problem,
$$
\left\{\begin... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Piglet ran down a moving escalator and counted 66 steps. Then he ran up the same escalator at the same speed relative to the escalator and counted 198 steps. How many steps would he have counted if he had gone down a stationary escalator?
Answer. 99. | Let \( u \) be the speed of Piglet, \( v \) be the speed of the escalator (both measured in steps per unit time), and \( L \) be the length of the escalator (in steps). Then, the time Piglet spent descending the moving escalator is \( \frac{L}{u+v} \), and during this time, he counted \( \frac{L}{u+v} \cdot u \) steps.... | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The remainder of dividing a certain natural number $n$ by 22 is 7, and the remainder of dividing $n$ by 33 is 18. Find the remainder of dividing $n$ by 66. | Answer: 51.
Solution: According to the condition $n=22l+7, \quad l \in Z$ and $n=33m+18, m \in Z$. By equating these two expressions, we get $22l+7=33m+18, 2l=3m+1$. Since the left side of the equation is even, the right side must also be divisible by 2, so $m$ is an odd number, i.e., $m=2q+1, q \in Z$. Then $n=33(2q+... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Out of three hundred eleventh-grade students, excellent and good grades were received by $77 \%$ on the first exam, $71 \%$ on the second exam, and $61 \%$ on the third exam. What is the smallest number of participants who could have received excellent and good grades on all three exams?
Answer: 27. | Let $M_{i}$ be the number of students who received excellent grades only on the $i$-th exam; $M_{ij}$ be the number of students who received excellent grades only on exams $i$ and $j$; $M_{123}$ be the number of students who received excellent grades on all three exams. Then, according to the conditions,
\[
\left\{\be... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Winnie the Pooh ran down a moving escalator and counted 55 steps. Then he ran up the same escalator at the same speed relative to the escalator and counted 1155 steps. How many steps would he have counted if he had gone down a stationary escalator?
Answer: 105. | Let \( u \) be the speed of Winnie the Pooh, \( v \) be the speed of the escalator (both measured in steps per unit time), and \( L \) be the length of the escalator (in steps). Then, the time Winnie the Pooh spent descending the moving escalator is \( \frac{L}{u+v} \), and during this time, he counted \( \frac{L}{u+v}... | 105 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Let $S(k)$ denote the sum of all the digits in the decimal representation of a positive integer $k$. Let $n$ be the smallest positive integer satisfying the condition $S(n)+S(n+1)=$ param1. As the answer to the problem, write down a five-digit number such that its first two digits coincide with the first two digits... | # Solution
Let $R$ be the radius of the circumcircle of triangle KBM and $r$ be the radius of the incircle of triangle ABC. First, we will prove that the centers of the circles mentioned in the problem statement coincide. Indeed, the center $I$ of the incircle of triangle $ABC$ lies on the bisector of angle $BAC$, whi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given a strictly increasing function $f: \mathbb{N}_{0} \rightarrow \mathbb{N}_{0}$ (where $\mathbb{N}_{0}$ is the set of non-negative integers), which satisfies the relation $f(n+f(m))=f(n)+m+1$ for any $m, n \in \mathbb{N}_{0}$. Find all possible values that $f(2023)$ can take.
(T.A. Garmanova) | Solution. 1) Substituting $m=0, n=0$, we get $f(f(0))=f(0)+1$.
If $f(0)=0$, then we get $f(0)=f(0)+1$, which is impossible.
2) Let $f(0)=a$, then $a \in \mathbb{N}$. From the first point, we get that $f(a)=a+1$. If we substitute $m=0, n=a$, then we get that $f(2a)=f(a)+1=a+2$. Therefore, the values of the function at... | 2024 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. What is the smallest number of different integers that need to be taken so that among them one can choose both a geometric and an arithmetic progression of length $5$?
(M.A. Evdokimov) | Solution. Let's provide an example of six integers that satisfy the condition: $-8, -2, 1, 4, 10, 16$. The numbers $1, -2, 4, -8, 16$ form a geometric progression, while the numbers $-8, -2, 4, 10, 16$ form an arithmetic progression.
We will show that no five distinct integers satisfy the condition of the problem. Sup... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number $N>9$, which is not divisible by 7, but if any of its digits is replaced by 7, the resulting number is divisible by 7. | (M. A. Evdokimov) Solution. Let the smallest such number be of the form $\overline{a_{1} a_{2} \ldots a_{n}}$. From the condition, it follows that among its digits there are no 0 and 7. If the number contains digits 8 or 9, then they can be replaced by 1 or 2, respectively, to obtain a smaller number with the same prop... | 13264513 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the relay race Moscow - Petushki, two teams of 20 people each participated. Each team divided the distance into 20 not necessarily equal segments and distributed them among the participants so that each ran exactly one segment (the speed of each participant is constant, but the speeds of different participants ma... | Answer: 38 overtakes.
Solution. First, let's prove that no more than 38 overtakes occurred. Note that between the start and the first overtake, and between two consecutive overtakes, at least one of the teams must have changed the runner. There were 19 changes of runners in each team, meaning a total of 38 changes, so... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On an island, there live chameleons of five colors. When one chameleon bites another, the color of the bitten chameleon changes according to some rule, and the new color depends only on the color of the biter and the color of the bitten. It is known that 2023 red chameleons can agree on a sequence of bites, after wh... | # Answer. For $k=5$. Solution.
First, let's provide an example of rules under which the described recoloring would require at least 5 red chameleons. Let's number the colors so that red is the first color and blue is the last. Then, let the rules be as follows: if a chameleon of color $k1$ can reduce the number of its... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (B. Novikov) On each cell of an $8 \times 8$ board, a guard is placed. Each guard can look in one of four directions (along the lines of the board) and watch all the guards on the line of his sight. For what largest $k$ can the guards' gazes be directed so that each guard is watched by at least $k$ other guards? | Solution: Answer: 5. We will prove that $k \leq 5$. For this, assume that $k \geqslant 6$. Consider the guards standing at the corners of the board. Each of them is watched by at least 6 guards, and these guards must stand at the edge of the board. Moreover, if a guard sees one of the corner guards, they do not see the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the largest natural number $n$ with the following property: for any odd prime $p$ less than $n$, the difference $n-p$ is also a prime number.
(I. Akulich) | Answer: 10
Solution: Suppose that $n>10$. Notice that the numbers $n-3, n-5, n-7$ are all greater than three, and one of them is divisible by three, and consequently composite. Contradiction. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Among any five nodes of a regular grid paper, there will definitely be two nodes, the midpoint of the segment between which is also a node of the grid paper. What is the minimum number of nodes of a grid made of regular hexagons that need to be taken so that among them, there will definitely be two nodes, the midpoi... | # Answer. 9
Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.
Proof. Introduce the origin at one of the grid nodes and denote by $\vec{a}$ and $\vec{b}$ the radius vectors to the two nearest nodes (see picture). Th... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given a polynomial of degree 2022 with integer coefficients and leading coefficient 1. What is the maximum number of roots it can have on the interval $(0,1)$?
(A. Kanel-Belov) | Answer: 2021
Solution. If the interval $(0,1)$ contains all 2022 roots of the polynomial, then by Vieta's theorem, the constant term of the polynomial must be equal to their product, and thus will also lie in the interval $(0,1)$ and will not be an integer. We will prove that the polynomial $P(x)=x^{2022}+(1-$ $4042 x... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. M. Evdokimov
}
A natural number is written on the board. If the last digit (in the units place) is erased, the remaining non-zero number will be divisible by 20, and if the first digit is erased, the remaining number will be divisible by 21. What is the smallest number that can be written on the board if its second... | Answer: 1609.
Solution.
The second to last digit of the number is 0, since the number without the last digit is divisible by 20. Therefore, the number is at least four digits. Note that the number remaining after erasing the last digit cannot be 100 according to the condition. Also, this number cannot be 120 or 140, ... | 1609 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. proposed by A. Shen
There is an infinite one-way strip of cells, numbered with natural numbers, and a bag with ten stones. Initially, there are no stones in the cells of the strip. The following actions are allowed:
- moving a stone from the bag to the first cell of the strip or back;
- if there is a stone in the ... | Answer: $\partial a$.
Solution. Note that for each action there is an inverse to it. Therefore, if we get from situation $A$ to situation $B$ by following the rules, we can also get from situation $B$ to situation $A$ by following the rules.
We will show by induction that if there is a reserve of $n$ stones, then, ac... | 1000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. A polyhedron with vertices at the midpoints of the edges of a certain cube is called a cuboctahedron. When the cuboctahedron is intersected by a plane, a regular polygon is obtained. What is the maximum number of sides this polygon can have?
(M. A. Evdokimov) | Solution.
Let the edge of the original cube, from which the cuboctahedron is obtained, be 1. Consider the sections of the cuboctahedron by a plane parallel to the base of the cube at a distan... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (I. Akulich) Find the largest natural $n$ with the following property: for any odd prime $p$ less than $n$, the difference $n-p$ is also a prime number. | Solution: Answer: 10. Indeed, $10=3+7=5+5$. We will prove that numbers greater than 10 do not work. Let $n$ be a number greater than 10. Note that the numbers $3, 5, 7$ give different remainders when divided by 3. Then the numbers $n-3, n-5, n-7$ give different remainders when divided by 3, meaning one of them is divis... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Sasha writes down the numbers 1, 2, 3, 4, 5 in some order, places the arithmetic operation signs "++", "-", "×" and parentheses, and looks at the result of the obtained expression. For example, he can get the number 8 using the expression $(4-3) \times (2+5)+1$. Can he get the number 123?
Forming numbers from ... | Answer: 2 yes
Solution. For example, $3 \times(2 \times 4 \times 5+1)=123$. | 123 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In Alik's collection, there are two types of items: badges and bracelets. There are more badges than bracelets. Alik noticed that if he increases the number of bracelets by a certain (not necessarily integer) factor, without changing the number of badges, then his collection will have 100 items. And if, conversely, ... | Solution. Let Alik have $x$ badges and $y$ bracelets, and the increase occurs by a factor of $n$. Then we get the system
$$
x+n y=100, \quad n x+y=101
$$
By adding and subtracting the equations and eliminating $n$, we obtain
$$
\frac{201}{x+y}-\frac{1}{x-y}=2
$$
We transform this equation to the form
$$
(201-2 u)(... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The Sultan gathered 300 court wise men and proposed a test. There are 25 different colors of hats, known in advance to the wise men. The Sultan informed them that one of these hats would be placed on each of the wise men, and if the number of hats of each color is written down, all the numbers will be different. Eac... | (A. V. Grialko) Solution. Since $0+1+2+\ldots+24=300$, the quantities of caps of different colors take all values from 0 to 24.
Next, each sage counts the number of caps of each color. For two colors, the quantities of caps coincide, and the sage understands that he is wearing a cap of one of these two colors. It only... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Provide an example of a nine-digit natural number that is divisible by 2 if the second (from the left) digit is erased, by 3 if the third digit is erased in the original number, ..., and by 9 if the ninth digit is erased in the original number. | 1. Answer. For example, 900900000.
Note. In fact, there are 28573 numbers that satisfy the conditions of the problem, the smallest of which is 100006020, and the largest is 999993240. | 900900000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Tanya was sequentially writing down numbers of the form $n^{7}-1$ for natural numbers $n=2,3, \ldots$ and noticed that for $n=8$ the resulting number is divisible by 337. For what smallest $n>1$ will she get a number divisible by $2022?$ | (T. A. Garmanova) Solution. Let the natural number $n$ be such that $n^{7}-1$ is divisible by $2022=2 \cdot 3 \cdot 337$. Then $n^{7}-1$ is divisible by 2 and 3, so $n$ is an odd number, having a remainder of 1 when divided by 3. In addition, $n^{7}-1$ is divisible by 337. Note that if two numbers are congruent modulo ... | 79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. (A. Zaslavsky) In a certain state, there are 32 cities, each of which is connected by a road with one-way traffic. The Minister of Communications, a secret villain, decided to organize traffic in such a way that after leaving any city, it would be impossible to return to it. To achieve this, he can change th... | Solution. We will prove the general formula. Let there be $2^{n}$ cities in the state. Then the prime minister can achieve the desired result in no more than $2^{n-2}\left(2^{n}-n-1\right)$ days.
Lemma. Suppose there are $2k$ cities in the state, each two of which are connected by a one-way road. Choose half of them w... | 208 | Combinatorics | proof | Yes | Yes | olympiads | false |
4. First solution. Let $a_{1}, a_{2}, \ldots, a_{15}$ be the days of the month that were sunny. Then Andrey Stepanovich will not drink a single drop from the 1st to the $a_{1}$-th day, he will drink one drop per day from the $a_{1}$-th day to the $a_{2}$-th day (including the $a_{1}$-th day but not the $a_{2}$-th day),... | The second solution. First, let's consider the situation where the first fifteen days of April were cloudy, and the last fifteen were sunny. It is easy to check that both characters in the problem will drink $1+2+3+\ldots+15=120$ drops of valerian - indeed, equally.
Let's swap some sunny day $s$ with some cloudy day $... | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. At the vertices of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ (Fig. 62), the following masses are placed: $2, 6, 2, 3$. Calculate the moment of inertia of the resulting system of material points relative to the point $S$. (The side of each square on Fig. 62 is 1 unit.) (O. 117, P. 177.)
+6 \cdot 2^{2}+2\left(3^{2}+4^{2}\right)+3 \cdot 8^{2}=316
\end{aligned}
$$ | 316 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. There are two solutions of wine spirit in water ${ }^{1}$ ): The first solution weighs 4002 and contains $30 \%$ spirit, the second solution weighs 6002 and contains $80 \%$ spirit.
From these two solutions, one solution is made. What percentage of spirit will it contain? (O. 117, U. 126, P. 215.$)$ | 2. a) Let's represent the solutions using material points. For this, we will take a segment $AB$ of one unit length (Fig. 172). Let point $A$ represent pure water
[Boda] (cimupm.)
Fig. 172.... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. What is the smallest number of triangular pyramids (tetrahedrons) into which a cube can be divided? | 14. It is easy to see that the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ can be divided into five tetrahedra: if we cut off the tetrahedra $B A C B_{1}$ and $D A C D_{1}$, as well as the tetrahedra $A_{1} B_{1} D_{1} A$ and $C_{1} B_{1} D_{1} C$, we will have one more (fifth) tetrahedron $A C B_{1} D_{1}$ left (Fig. 51; t... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15. On the plane, there are a (generally non-convex) quadrilateral and a pentagon, and no vertex of one lies on the side of the other. What is the maximum possible number of intersection points of their sides? | 15. Note that no line can intersect the sides of a polygon in an odd number of points, since otherwise, moving along this line in a certain direction - after entering the polygon for the last time, we would not be able to leave its boundaries. Therefore, each side of the quadrilateral can have no more than four interse... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16. Every two adjacent sides of a flat (self-intersecting!) 14-sided polygon are mutually perpendicular; no two sides of it lie on the same line. What is the maximum possible number of self-intersection points of the sides of such a polygon? | 16. Let's agree to consider that all sides of the 14-sided polygon are either "horizontal" or "vertical." Clearly, in this case, exactly 7 sides are horizontal (and 7 are vertical): after all, from each vertex, one horizontal (and one vertical) side extends. Summing over all 14 vertices, we count 14 horizontal sides; b... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18. a) Into how many parts can a plane be divided by two closed curves, one of which is a circle and the other is the boundary of a square?
b) $* *$ Into how many parts can space be divided by two closed surfaces, one of which is a sphere and the other is the surface of a cube? | 18. a) The boundary of a square and a circle can divide the plane into parts of the following types:
1) the part located outside the square and outside the circle; such a part is always one;
2) the part located inside the square and inside the circle; if the square does not intersect the circle, then there are no such ... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
49. Let $h_{a}=\beta_{b}=m_{c}$, where $h_{a}=A P, \beta_{b}=B L$ and $m_{c}=C F-$ are the altitude, bisector, and median drawn from three different vertices of some triangle $A B C$. What value can the ratio of the greatest side of triangle $A B C$ to its smallest side have? | 49. First of all, note that if for triangle \(ABC\) the inequality \(m_{a}b = CA\) holds. This immediately follows from the known formula \(^{2}\) \(m_{a} = \frac{1}{2} \sqrt{2 b^{2} + 2 c^{2} - a^{2}}\), but it can also be proven without calculations. Translate triangle \(ABC\) parallel to vector \(\overline{ED}\) to ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In space, there are 4 points not lying in the same plane. How many planes can be drawn equidistant from these points? | 1. Since the four points do not lie in the same plane, the plane equidistant from these points cannot be located on the same side of all of them. Therefore, there are only two possible cases: 1) three points lie on one side of the considered plane, and the fourth point lies on the other side, and 2) two points lie on e... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. How many spheres exist that touch all the faces of the given triangular pyramid $T$?
| 3. The task is to determine how many points (centers of the sought spheres) are equidistant from the four faces of the pyramid.
The geometric locus of points equidistant from the faces of a given dihedral angle is a plane passing through the edge of the dihedral angle and bisecting this angle — the bisector plane of t... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In how many different ways can the faces of a cube be painted using six given colors (each face must be painted entirely with one color), if only those colorings are considered different that cannot be made to coincide by rotating the cube? | 4. Suppose the faces of a cube are painted green, blue, red, yellow, white, and black. Let's place the cube so that the green face is at the bottom. In this case, the top face can be painted one of the five remaining colors. It is clear that no two colorings, in which the top (i.e., opposite the green) face is painted ... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In how many different ways can 30 workers be divided into 3 teams of 10 people each? | 5. 10 workers, forming the first brigade, can be chosen from thirty in $C_{30}^{10}=\frac{30 \cdot 29 \cdot \ldots \cdot 21}{10!}$ ways. After this, from the remaining 20 workers, 10 workers for the second brigade can be chosen in $C_{20}^{10}=\frac{20 \cdot 19 \cdots 11}{10!}$ ways. By combining each way of forming th... | 2775498395670 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. The commission consists of 11 people. The materials that the commission works on are stored in a safe. How many locks should the safe have, and how many keys should each member of the commission be provided with, so that access to the safe is possible when a majority of the commission members gather, but not possibl... | 7. According to the problem statement, for any five members of the committee, there should be a lock for which the key is absent from all these committee members (but is present with each of the six absent members, since the presence of any of the nine absent members already makes the meeting of the committee possible)... | 252 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. The numbers from 1 to 1000 are written in a circle. Starting from the first, every 15th number is crossed out (i.e., the numbers $1, 16, 31$, etc.), and during subsequent rounds, already crossed-out numbers are also taken into account. The crossing out continues until it turns out that all the numbers to be crossed ... | 8. After the first cycle, all numbers that give a remainder of 1 when divided by 15 will be crossed out; the last such number will be 991. The first number to be crossed out during the second cycle will be $991+15-1000=6$; subsequently, during the second cycle, all numbers that give a remainder of 6 when divided by 15 ... | 800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. All numbers from 1 to 10000000000 are written in a row. Which numbers will be more - those in which the digit 1 appears, or those in which 1 does not appear? | 9. First solution. Let's calculate the number of numbers in our sequence that do not contain the digit 1. Add the number 0 (zero) at the very beginning of this sequence and omit the last number 10000000000; we will get a sequence of $10^{10}$ numbers, containing one more number, in the notation of which the digit 1 doe... | 6513215600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. All integers from 1 to 222222222 are written in sequence. How many times does the digit 0 appear in the recording of these numbers?
## 2. Factorization of numbers into products and their decomposition into sums
In solving some of the subsequent problems, the following notations prove useful.
The symbol $[x]$ (re... | 10. Obviously, among the first 222222222 integers, there will be 22222222 numbers ending in zero (numbers 10, $20,30, \ldots, 222222220$). Further, among them, there will be 2222222 numbers ending in the digits 00 (numbers $100,200,300, \ldots, 222222200$); 2222222 numbers ending in the digits 01 (numbers 101, 201, 301... | 175308642 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. a) How many integers less than 1000 are not divisible by 5 or 7?
b) How many of these numbers are not divisible by 3, 5, or 7? | 11. a) There are a total of 999 numbers less than 1000 (numbers $1,2,3, \ldots, 999$). Now, let's strike out those that are divisible by 5; there are $\left[\frac{999}{5}\right]=199$ such numbers (numbers $5,10,15,20, \ldots, 995=199.5$). Next, let's strike out all numbers divisible by 7; there are $\left[\frac{999}{7}... | 457 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12*. How many integers are there that are less than the number 56700000 and coprime with it? | 12. The problem is very close to the previous one. The factorization of the number 56700000 is $56700000=2^{5} \cdot 3^{4} \cdot 5^{5} \cdot 7$; thus, the problem reduces to determining how many numbers less than 56700000 are not divisible by 2, 3, 5, or 7.
Let's list all numbers from 1 to 56700000. We will strike out... | 12960000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. How many positive integers $x$, less than 10000, are there for which the difference $2^{x}-x^{2}$ is not divisible by 7? | 13. $2^{0}=1$ gives a remainder of $1$ when divided by 7, $2^{1}=2$ gives a remainder of $2$, $2^{2}=4$ gives a remainder of $4$, $2^{3}$ gives a remainder of $1$ again, $2^{4}$ gives a remainder of $2$ again, $2^{5}$ gives a remainder of $4$ again, $2^{6}$ gives a remainder of $1$ for the third time, and so on. Thus, ... | 7142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. How many different pairs of integers $x, y$, lying between 1 and 1000, are there such that $x^{2}+y^{2}$ is divisible by 49? | 14. If $x^{2}+y^{2}$ is divisible by 49, then $x^{2}+y^{2}$ is also divisible by 7. But $x^{2}$, when divided by 7, can only give remainders of $0, 1, 4$, or 2 (see the solution to problem 13). The remainder from dividing $x^{2}+y^{2}$ by 7 is equal to the sum of the remainders from dividing the numbers $x^{2}$ and $y^... | 10153 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
15. In how many ways can a million be factored into three factors? Factorizations that differ only in the order of the factors are considered the same. | 15. Since a million is equal to $2^{6} \cdot 5^{6}$, any of its divisors has the form $2^{\alpha} \cdot 5^{\beta}$, and the factorization of a million into 3 factors has the form
$$
1000000=\left(2^{\alpha_{1}} \cdot 5^{\beta_{2}}\right)\left(2^{\alpha_{2}} \cdot 5^{\beta_{2}}\right)\left(2^{\alpha_{3}} \cdot 5^{\beta... | 139 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
17. How many different pairs of integers $A, B$ exist for which the least common multiple is 59400000? | 17. The prime factorization of the number 59400000 is
$$
59400000=2^{6} \cdot 3^{3} \cdot 5^{5} \cdot 11
$$
Therefore, if the least common multiple of two numbers $A$ and $B$ is 59400000, then it must be
$$
A=2^{x_{1}} \cdot 3^{\beta_{1}} \cdot 5 \gamma_{1} \cdot 11^{\delta_{1}}, \quad B=2^{\alpha_{2}} \cdot 3^{\bet... | 1502 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
18. Find the coefficients of $x^{17}$ and $x^{18}$ after expanding the brackets and combining like terms in the expression
$$
\left(1+x^{5}+x^{7}\right)^{20}
$$ | 18. Let's repeat 20 times the trinomial $1+x^{3}+x^{7}$ and multiply it according to the usual rules. Each term of the resulting sum will represent a product of 20 factors, each equal to either 1, or $x^{5}$, or $x^{7}$. It is essential that in these products of twenty factors, any of the three expressions can stand in... | 3420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
19. In how many ways can 20 kopecks be exchanged for coins worth 5 kopecks, 2 kopecks, and 1 kopeck? | 19. Let's list all the ways to change 20 kopecks. In such a change, we can use either four 5-kopeck coins, or three such coins, or two, or one, or none.
There is only one way to change 20 kopecks using four 5-kopeck coins (since four 5-kopeck coins already make 20 kopecks).
If we use three 5-kopeck coins, we need to ... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
22**. In how many ways can you make up a ruble using coins of 1, 2, 5, 10, 20, and 50 kopecks? | 22. Since from coins of 10, 20, and 50 kopecks, only a whole number of tens of kopecks can be formed, then from 1, 2, and 5 kopeck coins, a whole number of tens of kopecks must also be formed. Therefore, the following cases are possible:
1) The ruble is composed of coins worth 10, 20, and 50 kopecks,
2) Coins worth 10,... | 4562 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
24. How many integer solutions does the inequality
$$
|x|+|y|<100 ?
$$
have? Here, for $x \neq y$, the solutions $x, y$ and $y, x$ should be considered different. | 24. We need to find the number of pairs of integers \(x, y\) such that \(|x| + |y|\) is 0, 1, 2, 3, ..., or 99. Let's count the number of pairs \(x, y\) for which \(|x| + |y| = k\). Here, \(|x|\) can take \(k + 1\) different values, namely \(0, 1, 2, \ldots, k-1, k\); in this case, \(|y|\) will be \(k, k-1, k-2, \ldots... | 19801 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
32. a) What is the maximum number of rooks that can be placed on a chessboard of $n^{2}$ squares so that no two rooks threaten each other? In how many different ways can this be done?
b) What is the minimum number of rooks that can be placed on a chessboard of $n^{2}$ squares so that these rooks threaten all the squar... | 32. a) A chessboard consisting of $n^{2}$ squares (see Fig. 39, where the case $n=8$ is shown) contains $n$ rows and $n$ columns. To ensure that no two rooks placed on this board threaten each other, it is necessary that no two rooks stand on the same row or the same column. It is clear, therefore, that the total numbe... | 33514312 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
47. a) How many different squares, in terms of size or position, consisting of whole cells, can be drawn on a chessboard of 64 cells?
b) The same question for a chessboard of $n^{2}$ cells. | 47. a) This problem is close to the previous one. The number of all possible squares consisting of $k^{2}$ cells that can be chosen on a chessboard of 64 cells is $(9-k)^{2}$ (see the solution to problem 46 a)). From this, it follows that the number of all squares is $8^{2}+7^{2}+6^{2}+\ldots+1^{2}=$
$$
=64+49+36+25+1... | 204 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
69. What is more likely: to win 3 games out of 4 or 5 games out of 8 against an equally matched opponent? | 69. The total number of possible outcomes of a sequence of four matches we get by combining a win or loss in the first match with a win or loss in the second, third, and fourth matches. This number is equal to $2^{4}=16$. All these outcomes are equally probable, as the opponents are of equal strength, and for each sepa... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
103*. The bus network of the city, consisting of several (more than two) routes, is organized in such a way that:
$1^{\circ}$ each route has at least three stops,
$2^{\circ}$ from any stop to any other stop, one can travel without transferring, and
$3^{\circ}$ for each pair of routes, there is one (and only one) sto... | 103. a) Let \( n \) be the number of stops on one of the bus routes in the city. We need to prove that in this case, each route will have exactly \( n \) stops and that exactly 2 routes will pass through each stop.
Let's denote the route with \( n \) stops by \( a \), and the stops on this route by \( A_{1}, A_{2}, \l... | 8 | Combinatorics | proof | Yes | Yes | olympiads | false |
1.2. Given a cube with an edge of 1. Find the angle and distance between the skew diagonals of two of its adjacent faces. | 1.2. Consider the diagonals $A B_{1}$ and $B D$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Since $B_{1} D_{1} \| B D$, the angle between the diagonals $A B_{1}$ and $B D$ is equal to the angle $A B_{1} D_{1}$. But the triangle $A B_{1} D_{1}$ is equilateral, so $\angle A B_{1} D_{1}=60^{\circ}$.
It is easy to chec... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.10. At the base of a regular triangular prism lies a triangle $ABC$ with side $a$. Points $A_{1}, B_{1}$, and $C_{1}$ are taken on the lateral edges, the distances from which to the plane of the base are $a / 2$, $a$, and $3 a / 2$. Find the angle between the planes $ABC$ and $A_{1} B_{1} C_{1}$.
## § 3. Lines formi... | 1.10. Let $O$ be the point of intersection of the lines $A B$ and $A_{1} B_{1}$, $M$ be the point of intersection of the lines $A C$ and $A_{1} C_{1}$. First, we will prove that $M O \perp O A$. For this, we will take points $B_{2}$ and $C_{2}$ on the segments $B B_{1}$ and $C C_{1}$ such that $B B_{3}=C C_{2}=A A_{1}$... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.28. Through the vertex $A$ of a right circular cone, a section of maximum area is drawn. Its area is twice the area of the section passing through the axis of the cone. Find the angle at the vertex of the axial section of the cone. | 2.28. Consider an arbitrary section passing through the vertex $A$. This section is a triangle $A B C$, and its sides $A B$ and $A C$ are the generators of the cone, i.e., they have a constant length. Therefore, the area of the section is proportional to the sine of the angle $B A C$. The angle $B A C$ varies from $0^{... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.11. The centers of three spheres, with radii of 3, 4, and 6, are located at the vertices of an equilateral triangle with a side length of 11. How many planes exist that are tangent to all three spheres simultaneously?
## § 3. Two intersecting circles lie on the same sphere | 4.11. Consider a plane tangent to all three given spheres, and draw a plane through the center of the sphere of radius 3 parallel to it. The resulting plane is tangent to the spheres of radii $4 \pm 3$ and $6 \pm 3$, concentric with the spheres of radii 4 and 6. When the signs of the number 3 are the same, the tangency... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.19. Chord $A B$ of a sphere with radius 1 has a length of 1 and is positioned at an angle of $60^{\circ}$ to the diameter $C D$ of this sphere. It is known that $A C=\sqrt{2}$ and $A C<B C$. Find the length of the segment $B D$. | 4.19. Let $O$ be the center of the sphere. Take a point $E$ such that $\overrightarrow{C E}=\overrightarrow{A B}$. Since $\angle O C E=60^{\circ}$ and $C E=1=O C$, then $O E=1$. Point $O$ is equidistant from all vertices of the parallelogram $A B E C$, so $A B E C$ is a rectangle and the projection $O_{1}$ of point $O$... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.21. In the tetrahedron \(ABCD\), the plane angles at vertex \(A\) are right angles, and \(AB = AC + AD\). Prove that the sum of the plane angles at vertex \(B\) is \(90^\circ\). | 6.21. Let's take points $P$ and $R$ on rays $A C$ and $A D$ such that $A P = A R = A B$, and consider the square $A P Q R$. It is clear that $\triangle A B C = \triangle R Q D$ and $\triangle A B D = \triangle P Q C$, and therefore, $\triangle B C D = \triangle Q D C$. Thus, the sum of the planar angles at vertex $B$ i... | 90 | Geometry | proof | Yes | Yes | olympiads | false |
6.35. a) Prove that the sum of the cosines of the dihedral angles of a regular tetrahedron is 2.
b) The sum of the plane angles of a trihedral angle is \(180^{\circ}\). Find the sum of the cosines of its dihedral angles.
## § 5. Orthocentric Tetrahedron
Definition. A tetrahedron is called orthocentric if all its alt... | 6.35. a) Let $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$ and $\mathbf{e}_{4}$ be unit vectors, non-perpendicular to the faces and directed outward. Since the areas of all faces are equal, then $\mathbf{e}_{1}+\mathbf{e}_{2}+\mathbf{e}_{3}+\mathbf{e}_{4}=0$ (see problem 7.19). Therefore, $0=\left|e_{1}+\mathbf{e}_{... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
7.37. What is the maximum number of planes of symmetry that a spatial figure consisting of three pairwise non-parallel lines can have?
Symmetry with respect to a line $l$ is a transformation of space that maps a point $X$ to a point $X^{\prime}$ such that the line $l$ passes through the midpoint of the segment $X X^{\... | 7.37. Let $P$ be a plane of symmetry of a figure consisting of three pairwise non-parallel lines. There are only two possible cases: 1) each given line is symmetric relative to $P$; 2) one line is symmetric relative to $P$, and the other two lines are symmetric to each other.
In the first case, either one line is perp... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.26. a) How many pairwise non-equal spatial quadrilaterals exist with one and the same set of side vectors?
b) Prove that the volumes of all tetrahedra defined by these spatial quadrilaterals are equal. | 8.26. a) We fix one of the vectors of the sides. It can be followed by any of the three remaining vectors, and then by any of the two remaining ones. Therefore, there are exactly 6 different quadrilaterals.
b) Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and $\mathbf{d}$ be the given vectors of the sides. Consider the pa... | 6 | Geometry | proof | Yes | Yes | olympiads | false |
9.9. a) Prove that it is possible to choose 8 vertices of a dodecahedron such that they are the vertices of a cube. How many ways can this be done?
b) Prove that it is possible to choose 4 vertices of a dodecahedron such that they are the vertices of a regular tetrahedron. | 9.9. a) From the solution to problem 9.2, it is clear that there exists a cube whose vertices are located at the vertices of the dodecahedron. Moreover, one edge of the cube is located on each face of the dodecahedron. It is also clear that choosing any of the five diagonals of a certain face of the dodecahedron as an ... | 5 | Geometry | proof | Yes | Yes | olympiads | false |
11.18. The height of a regular quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ is half the length of the side of the base. Find the maximum value of the angle $A_{1} M C_{1}$, where $M$ is a point on the edge $A B$. | 11.18. Let $A A_{1}=1, A M=x$. Introduce a coordinate system, the axes of which are parallel to the edges of the prism. The vectors $\overrightarrow{M A}_{\text {}}$ and $\overrightarrow{M C}_{1}$ have coordinates $(0,1,-x)$ and ( $2,1,2-x$ ); their scalar product is equal to $1-2 x+x^{2}=(1-x)^{2} \geqslant 0$. Theref... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15.22. What is the smallest number of tetrahedra into which a cube can be cut? | 15.22. If a tetrahedron \(A^{\prime} B C^{\prime} D\) is cut out from the cube \(A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}\), the remaining part of the cube splits into 4 tetrahedra, i.e., the cube can be cut into 5 tetrahedra.
We will prove that the cube cannot be cut into fewer than 5 tetrahedra. The face ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15.26. Into how many parts do the planes of the faces divide the space a) of a cube; b) of a tetrahedron? | 15.26. The planes of the faces of both polyhedra intersect only along the lines containing their edges. Therefore, the space divided into parts, into which the space is split, has a common point with the polyhedron. Moreover, to each vertex, each edge, and each face, one can correspond exactly one adjacent part to it, ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15.32. A plane intersects the lower base of a cylinder along its diameter and has only one common point with the upper base. Prove that the area of the cut-off part of the lateral surface of the cylinder is equal to the area of its axial section. | 15.32. Let 0 be the center of the lower base of the cylinder; $AB$ - the diameter along which the plane intersects the base; $\alpha$ - the angle between the base and the intersecting plane; $r$ - the radius of the cylinder. Consider an arbitrary generatrix $XY$ of the cylinder, having a common point $Z$ with the inter... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
15.34. In space, 4 points are given, not lying in the same plane. How many different parallelepipeds exist for which these points serve as vertices? | 15.34. Consider a parallelepiped for which the given points are vertices, and mark its edges connecting these points. Let \( n \) be the maximum number of marked edges of this parallelepiped emanating from one vertex; the number \( n \) can vary from 0 to 3. A simple enumeration shows that only the variants depicted in... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16.16. Given three pairwise tangent spheres $\Sigma_{1}$, $\Sigma_{2}$, $\Sigma_{3}$ and a set of spheres $S_{1}, S_{2}, \ldots, S_{n}$, such that each sphere $S_{i}$ is tangent to all three spheres $\Sigma_{1}$, $\Sigma_{2}$, $\Sigma_{3}$, as well as to the spheres $S_{i-1}$ and $S_{i+1}$ (it is meant that $S_{0}=S_{n... | 16.16. Consider the inversion with the center at the point of tangency of spheres $\Sigma_{1}$ and $\Sigma_{2}$. In this case, they transform into a pair of parallel planes, and the images of all other spheres touch these planes, and therefore their radii are equal. Thus, in a section by a plane equidistant from these ... | 6 | Geometry | proof | Yes | Yes | olympiads | false |
61. Determine the remainder of the division by 7 of the number
$$
3^{100}
$$
The 10-arithmetic and 7-arithmetic we have discussed are special cases of residue arithmetic modulo $m$, or $m$-arithmetic. Let $m$ be any positive integer. The elements of $m$-arithmetic are the numbers $0,1,2, \ldots, m-1$. Addition and mu... | 61. Let's compute $3^{100}$ in 7-arithmetic:
$$
3^{2}=2,3^{8}=6,3^{4}=4,3^{5}=5,3^{6}=1
$$
Therefore,
$$
3^{100}=3^{6 \cdot 16+4}=\left(3^{6}\right)^{16} \cdot 3^{4}=3^{4} \Rightarrow 4
$$
$14^{*}$ | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
112. Make sure that the last digits of the numbers in the Fibonacci sequence $^{\circ}$ repeat periodically. What is the length of the period?
Ensure that the last digits of the numbers in the Fibonacci sequence $^{\circ}$ repeat periodically. What is the length of the period? | 112. The last digits of the Fibonacci sequence $\varnothing^{0}$ themselves form a Fibonacci sequence in the sense of 10-arithmetic (sequence $\Phi_{10}^{0}$).
Let's write out the terms of this sequence:
$0,1,1,2,3,5,8,3,1,4,5,9,4,3,7$, $0,7,7,4,1,5,6,1,7,8,5,3,8,1,9$, $0,9,9,8,7,5,2,7,9,6,5,1,6,7,3$, $0,3,3,6,9,5,4,... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
117. Let three elements $x, y, z$ of the circular sequence $\breve{\Phi}_{m}$ be arranged such that the distance between $x$ and $y$ is equal to the distance between $y$ and $z$ (Fig. 54). Prove that if $x=y=0$, then $z=0$. | 117. The number $y=0$ is at an equal distance from $x$ and $z$. Therefore, based on problem 116, either $x+z=0$ or $x-z=0$. Since $x=0$, in both cases $z=0$, | 0 | Number Theory | proof | Yes | Yes | olympiads | false |
120. Prove that if a circular sequence without repetitions $\Phi_{m}$ contains zero and consists of an odd number of elements, then the number of its elements is three. | 120. If the circular sequence $\breve{\Phi}_{m}$ contains zero and consists of an odd number of terms, then there will be a pair of adjacent elements $x, y$ located at the same distance from zero (Fig. 143). According to problem 116, either
 A. P. Kiselev, Algebra, Part 2, 23rd edition, Uchpedgiz, M., 1946, p. 78.
18 Law. Zzya. E. Dynkin and V. Uspenskii
moment, the chip is located at point $a$ (Fig. 156). After each unit of time, it moves to... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In triangle $ABC$, the median $AD$ is drawn. $\widehat{D A C} + \widehat{A B C} = 90^{\circ}$. Find $\widehat{B A C}$, given that $|A B| = |A C|$. | 1. Take a point $A_{1}$ on the line $B A$ such that $\left|A_{1} B\right|=\left|A_{1} C\right|$. The points $A_{1}, A, D$ and $C$ lie on the same circle $\left(\widehat{D A_{1} C}=90^{\circ}-\widehat{A B C}=\widehat{D A C}\right)$. Therefore, $\widehat{A_{1} A C}=\widehat{A_{1} D C}=90^{\circ}$, which means $\widehat{B... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Three circles with radii 1, 2, and 3 touch each other externally. Find the radius of the circle passing through the points of tangency of these circles. | 2. Note that the circle passing through the points of tangency is inscribed in a triangle with vertices at the centers of the circles. By equating the expressions for the area of the triangle obtained using Heron's formula and as the product of the semiperimeter and the radius of the inscribed circle, we find \( r=1 \)... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. In a trapezoid, the diagonals are equal to 3 and 5, and the segment connecting the midpoints of the bases is equal to 2. Find the area of the trapezoid. | 7. Let (Fig. 2) $ABCD$ be the given trapezoid ($BCAD$). Draw a line through $C$ parallel to $BD$, and denote by $K$ the point of intersection of this line with the line $AD$. The area of $\triangle ACK$ is equal to the area of the trapezoid, the sides $AC$ and $CK$ are equal to the diagonals of the trapezoid, and the m... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. A circle of radius 1 is inscribed in triangle $A B C$, where $\cos \widehat{A B C}=0.8$. This circle touches the midline of triangle $A B C$, parallel to side $A C$. Find the length of side $A C$. | 9. It follows from the condition that the height to side $AC$ is equal to two diameters of the inscribed circle, i.e., 4. If $M, N$ and $K$ are the points of tangency with $AB, BC$ and $CA$, then
$$
|BM|=|BN|=r \operatorname{ctg} \frac{\widehat{ABC}}{2}=1 \cdot \sqrt{\frac{1+0.8}{1-0.8}}=\sqrt{9}=3
$$
If $|MA|=|AK|=x... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
58. Determine the measure of angle $\hat{A}$ of triangle $ABC$, if it is known that the bisector of this angle is perpendicular to the line passing through the point of intersection of the altitudes and the center of the circumscribed circle of this triangle. | 58. Let $O$ be the center of the circumscribed circle, $H$ the intersection point of the altitudes of $\triangle ABC$. Since the line $OH$ is perpendicular to the bisector of angle $A$, it intersects sides $AB$ and $AC$ at points $K$ and $M$ such that $|AK|=|AM|$. Thus, $|AO|=|OB|$ and $\widehat{AOB}=2 \hat{C}$ (assumi... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
61. In trapezoid $A B C D$, the lateral side $A B$ is perpendicular to $A D$ and $B C$, and $|A B|=\sqrt{|A D| \cdot|B C|}$. Let $E$ be the point of intersection of the non-parallel sides of the trapezoid, $O$ be the point of intersection of the diagonals, and $M$ be the midpoint of $A B$. Find $\widehat{E O M}$. | 61. Let $|A D|=a,|B C|=b$. Drop a perpendicular $O K$ from $O$ to $A B$. Now we can find $|B K|=\sqrt{\overline{a b}} \frac{b}{b+a},|B E|=\sqrt{\overline{a b}} \frac{b}{a-b}$, $|M K|=\frac{\sqrt{a b}}{2}-\sqrt{\overline{a b}} \frac{b}{b+a}=\sqrt{\overline{a b}} \frac{a-b}{2(a+b)},|E K|=|B E|+|B K|=$ $=\sqrt{a \bar{b}} ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
63. Two circles touch each other internally at point $A$. From the center of the larger circle, a radius $O B$ is drawn, touching the smaller circle at point $C$. Find $\widehat{B A C}$. | 63. If $O_{1}$ is the center of the smaller circle, and $\widehat{B O A}=\varphi$, then $\widehat{B A O}=90^{\circ}-\frac{\varphi}{2}$, $\widehat{C O_{1} A}=90^{\circ}+\varphi$, $\widehat{C A O_{1}}=45^{\circ}-\frac{\varphi}{2}$. Therefore,
$$
\begin{gathered}
\widehat{B A C}=\widehat{B A O}-\widehat{C A O_{1}}=45^{\c... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
64. Inside the square $A B C D$, a point $M$ is taken such that $\widehat{M A B}=60^{\circ}, \widehat{M C D}=15^{\circ}$. Find $\widehat{M B C}$. | 64. Construct an equilateral triangle $A B K$ on $A B$ inside the square. Then $\widehat{K A B}=60^{\circ}, \widehat{K C D}=15^{\circ}$, i.e., $K$ coincides with $M$. Answer: $30^{\circ}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
65. In triangle $ABC$ with angle $\widehat{ABC}=60^{\circ}$, the bisector of angle $A$ intersects $BC$ at point $M$. On side $AC$, a point $K$ is taken such that $\widehat{AM} K=30^{\circ}$. Find $\widehat{OKC}$, where $O$ is the center of the circumcircle of triangle $AMC$. | 65. If $\widehat{B A C}=2 \alpha$, it is easy to find that $\widehat{K M C}=\widehat{M K C}=$ $=30^{\circ}+\alpha$, i.e., $|M C|=|K C|$. Extend $M K$ until it intersects the circle at point $N$; $\triangle K M C$ is similar to $\triangle K A N$, so $|A N|=|K N|=R$ - the radius of the circle (since $\widehat{A M N}=30^{... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
66. Given a triangle $A B C$, where $|A B|=|A C|$, $\widehat{B A C}=80^{\circ}$. Inside the triangle, a point $M$ is taken such that $\widehat{M} \overrightarrow{B C}=30^{\circ}, \widehat{M C B}=10^{\circ}$. Find $\widehat{A M C}$. | 66. Let's draw (Fig. 12) the angle bisector of $\angle A$ and extend $B M$ until it intersects with the bisector at point N. Since $|B N|=|N C|$, then $\widehat{B N C}=120^{\circ} ;$ therefore, the angles $\widehat{B N A}, \widehat{C N A}$ are also $120^{\circ}$ each, $\widehat{N C A}=\widehat{N C M}=20^{\circ}, \quad$... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
67. In triangle $ABC$, given are $\widehat{ABC}=100^{\circ}, \widehat{ACB}=$ $=65^{\circ}$. On $AB$, a point $M$ is taken such that $\widehat{MCB}=55^{\circ}$, and on $AC$ - a point $N$ such that $\widehat{NBC}=80^{\circ}$. Find $\widehat{NMC}$. | 67. Describe a circle around $\triangle M C B$ (Fig. 13) and extend $B N$ until it intersects with the circle
Fig. 12. at point $M_{1} ;\left|C M_{1}\right|=|C M|$, since the angles subtend... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
68. In triangle $ABC$, it is given that $|AB|=|BC|, \widehat{ABC}=$ $=20^{\circ}$; on $AB$ a point $M$ is taken such that $\widehat{MCA}=60^{\circ}$; on side $CB$ - a point $N$ such that $\widehat{NAC}=50^{\circ}$. Find $\widehat{NMA}$. | 68. Let's take a point $K$ on $BC$ (Fig. 14) such that $\widehat{K A C}=60^{\circ}$, $M K \| A C$. Let $L$ be the intersection point of $A K$ and $M C$; $\triangle A L C$ is equilateral, $\triangle A N C$ is isosceles (calculate the angles), so $\triangle L N C$ is also isosceles, $\widehat{L C N}=20^{\circ}$. Now let'... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
69. In triangle $ABC$, given are $\widehat{ABC}=70^{\circ}, \widehat{ACB}=$ $=50^{\circ}$. On $AB$, a point $M$ is taken such that $\widehat{MCB}=40^{\circ}$, and on $AC$-point $N$ such that $\widehat{NBC}=50^{\circ}$. Find $\widehat{NMC}$. | 69. Let's take point $K$ (Fig. 15) such that $\widehat{K B C}=\widehat{K C B}=30^{\circ}$, and denote by $L$ the intersection point of the lines $M C$ and $B K$.
Fig. 15. Since $\triangle ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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