problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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75. In quadrilateral $A B C D$, it is given that $\widehat{D A B}=150^{\circ}$, $\widehat{D A C}+\widehat{A B D}=120^{\circ}, \widehat{D B C}-\widehat{A B \bar{D}}=60^{\circ}$. Find $\widehat{B D C}$. | 75. Let $\widehat{A B D}=\alpha, \widehat{B D C}=\varphi$. By the condition, $\widehat{D A C}=$ $=120^{\circ}-\alpha, \widehat{B A C}=30^{\circ}+\alpha, \widehat{A D B}=30^{\circ}-\alpha, \widehat{D B C}=60^{\circ}+\alpha$. Using the Law of Sines for triangles $A B C, B C D, A C D$, we get
$$
\begin{aligned}
& \frac{|... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
35*. What is the greatest number of rays in space that form pairwise obtuse angles? | 35. It is easy to see that it is possible to draw four rays, each pair of which forms obtuse angles: for example, it is sufficient to connect the center $O$ of a regular tetrahedron $ABCD$ with all its vertices (it is clear, for example, that $\angle AOB > \angle AO_1B = 90^\circ$, where $O_1$ is the center of the face... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
52. Given a triangle $T$.
a) Place a centrally symmetric polygon $m$ of the largest possible area inside $T$.
What is the area of $m$ if the area of $T$ is 1?
b) Enclose $T$ in a convex centrally symmetric polygon $M$ of the smallest possible area.
What is the area of $M$ if the area of $T$ is 1? | 52. a) If $O$ is the center of the sought centrally symmetric polygon $m$, then by symmetrically reflecting the given $\triangle ABC \equiv T$ together with $m$ (which will transform into itself), we can see that $m$ is also inscribed in $\triangle A_{1} B_{1} C_{1} \equiv T_{1}$, which is symmetric to $\triangle ABC$ ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
90. On a plane, there are 4 points $A_{1}, A_{2}, A_{3}, A_{4}$, the distance between any two of which is not less than 1. What is the maximum possible number of line segments $A_{i} A_{j}$ of length 1 connecting these points pairwise? | 90. The number $v_{2}(4)<6$, otherwise the number of diameters of a system of four points on a plane could equal the "complete" number of six segments connecting our points, and by the result of problem 87 a) this is not the case. However, the number of segments equal to 1, connecting the points of our system of four p... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. a) Points $A$ and $B$ move uniformly and with equal angular velocities along circles centered at $O_{1}$ and $O_{2}$, respectively (clockwise). Prove that vertex $C$ of the equilateral triangle $A B C$ also moves uniformly along some circle.
b) The distances from a fixed point $P$ on the plane to two vertices $A, B... | 6. a) Let $\overrightarrow{\mathrm{O}_{1} \mathrm{O}_{3}}$ be the vector obtained by rotating $\overrightarrow{O_{1} O_{2}}$ by $60^{\circ}$ (in the same direction as the rotation that transforms $\overrightarrow{A B}$ into $\overrightarrow{A C}$). Points $A^{\prime}$ and $B^{\prime}$ are the images of $A$ and $B$ unde... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Kolya and Petya are dividing $2 n+1$ nuts, $n \geqslant 2$, with each wanting to get as many as possible. There are three ways of dividing (each goes through three stages).
1st stage: Petya divides all the nuts into two parts, each containing no fewer than two nuts.
2nd stage: Kolya divides each part again into t... | 10. Answer: the most profitable way for Kolya is the first one; with the second and third, he will get one nut less with correct play. (In general, as we will see, the dispute in this division is over one nut.)
No matter what piles (of $a$ and $b$ nuts, $a<b$) form after Petya's first move, Kolya can split the larger ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
21. Let's take any 1962-digit number divisible by 9. The sum of its digits will be denoted by $a$, the sum of the digits of $a$ by $b$, and the sum of the digits of $b$ by $c$. What is $c$?[^0] | 21. Answer. $c=9$. The sum of the digits of any number gives the same remainder when divided by 9 as the number itself (PZ). On the other
Fig. 30 hand, $a \leqslant 1962.9<19999$, so $b \l... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
23. What is the maximum area that a triangle with sides \(a, b, c\) can have, given the following constraints:
\[
0 \leqslant a \leqslant 1 \leqslant b \leqslant 2 \leqslant c \leqslant 3 \text { ? }
\] | 23. Answer: 1. Among triangles with two sides $a, b$ that satisfy the conditions $0<a \leqslant 1,1 \leqslant b \leqslant 2$, the one with the largest area is the right triangle with legs $a=1, b=2$ (indeed, $s \leqslant a b / 2 \leqslant 1$, since the height dropped to side $b$ is no more than $a$). The third side of ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
43. For each of the numbers from 1 to 1000000000, the sum of its digits is calculated, and for each of the resulting billion numbers, the sum of its digits is calculated again, and so on, until a billion single-digit numbers are obtained. Which number will there be more of: 1 or 2? | 43. Answer: the number of ones will be one more than the number of twos.
Any number gives the same remainder when divided by 9 as the sum of its digits (DS). Therefore, in our problem, ones come from numbers that give a remainder of 1 when divided by 9, i.e., from the numbers 1, 10, 19, 28, ..., 999999991, 1000000000,... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
46. Solve the equation in integers
$$
\sqrt{\sqrt{x+\sqrt{x+\sqrt{x+\ldots+\sqrt{x}}}}}=y
$$ | 46. Answer: the only solution is $x=y=0$.
Let $\boldsymbol{x}$ and $\boldsymbol{y}$ be integers satisfying the condition. After a series of squaring, we are convinced that $\sqrt{x+\sqrt{x}}=\boldsymbol{m}$ and $\sqrt{x}=k$ are integers, and
$$
m^{2}=k(k+1)
$$
If $k>0$, then it must be that $k^{2}<m^{2}<(k+1)^{2}$, ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
53. What is the smallest number of non-overlapping tetrahedra into which a cube can be divided? | 53. It is easy to see that a cube can be divided into 5 tetrahedra. In Fig. 40, these are the tetrahedra $A A^{\prime} B^{\prime} D^{\prime}, A B^{\prime} B C, A C D D^{\prime}, B^{\prime} C^{\prime} D^{\prime} C$, and $A C D^{\prime} B^{\prime}$.
Let us now prove that it is impossible to divide the cube into fewer te... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
64*. Can 1965 points be placed in a square with a side of 1 so that any rectangle of area $1 / 200$ with sides parallel to the sides of the square contains at least one of these points? | 64. Answer: it is possible.
In the square $0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1$ and on the line $y=1 / 2$, uniformly place $c_{0}=200$ points: ( $\left.k / 201 ; 1 / 2\right), k=1,2, \ldots$ $\ldots, 200$. Then on each of the lines $y=1 / 4$ and $y=3 / 4$, place $c_{1}=100$ points ( $\left.k / 101 ; 1... | 1704 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
90. In a sequence of integers (positive), each term, starting from the third, is equal to the absolute difference of the two preceding ones.
What is the maximum number of terms such a sequence can have if each of its terms does not exceed 1967? | 90. Answer: 2952. We will prove that the length (number of terms) of a sequence satisfying the condition of the problem, where the largest term is the second and equals $n$, does not exceed $d_{n} = [3(n+1) / 2]$, and for the sequence $n-1, n, 1, \ldots, \ldots, 1,1$ the length is exactly $d_{n}$.
We will reason by in... | 2952 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
92. Three consecutive vertices of a rhombus lie on the sides $A B, B C, C D$ of a given square with side 1. Find the area of the figure filled by the fourth vertices of such rhombi. | 92. Answer: $S=1$.
Let $K, L, N$ be the vertices of the rhombus on the sides $A B, B C$, and $A D$ of the square (Fig. $52, a$). Note that the length $K B$ is equal to the distance from point $M$ to the line $A D$. Therefore, if we fix point $K$, the possible positions of point $M$ fill a certain segment $M_{1} M_{2}$... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
99. In a regular $n$-gon ( $n>5$ ), the difference between the largest and smallest diagonals is equal to the side. Find $n$. | 99. Answer: $n=9$.
Let $a_{n}$ be the side length, and $D_{n}$ and $d_{n}$ be the lengths of the largest and smallest diagonals of a regular $n$-gon. For $n=4$ and $n=5$, all diagonals are equal. For $n=6$ and $n=7$, $D_{n}-a_{n}=2 A K=D_{8}-d_{8}$. For $n=9$ (Fig. 55,6), similarly, we get $\angle A B K=30^{\circ}$, s... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
123. In a certain state, the airline system is arranged in such a way that any city is connected by air routes to no more than three other cities, and from any city to any other, one can travel with no more than one layover.
What is the maximum number of cities that can be in this state? | 123. Answer: 10 cities.
From any city $A$, one can reach no more than three cities, and from each of them no more than two (excluding $A$). Thus, the total number of cities is no more than $1+3+3 \cdot 2=$ $=10$.
The example in Fig. 60 shows that the required system of air routes in a state with ten cities exists.
$... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
126*. In the country's football championship, 20 teams are participating. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | 126. Answer: 90 games.
Let among any three teams, there be two that have already played against each other. Choose a team $A$ that has played the least number of games $-k$. Each of the $k$ teams that have already played with $A$, as well as team $A$ itself, has played no fewer than $k$ games. From the $(19-k)$ teams ... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
131. How many sides in a convex polygon can be equal in length to the longest diagonal? | 131. Answer: no more than two. An example of a polygon with two sides equal to the largest diagonal is shown in Fig. 61.
Suppose there are more than two such sides. We select two of them, $A B$ and $C D$, which do not share any vertices (this is possible, [^3] since a polygon with diagonals is not a triangle). Then at... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
161. Find the greatest integer $x$ such that the number
$$
4^{27}+4^{1000}+4^{x}
$$
is a perfect square. | 161. Answer: $x=1972$. Since $4^{27}+4^{1000}+4^{x}=$ $=2^{54}\left(1+2 \cdot 2^{1945}+2^{2 x-54}\right)$, the expression in parentheses will be a perfect square when $2 x-54=2 \cdot 1945$, i.e., when $x=1972$. If $x>1972$, then $2^{2(x-27)}<1+2 \cdot 2^{1945}+2^{2(x-27)}<\left(2^{x-27}+1\right)^{2}$, i.e., the given n... | 1972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
179. The Tennis Federation has assigned qualification numbers to all its tennis players: the strongest player gets the first number, the next strongest gets the second number, and so on. It is known that in matches where the qualification numbers differ by more than 2, the player with the lower number always wins. A to... | 179. Answer! The highest possible number of the winner is 20.
Since a tennis player with number $k$ can lose (not counting stronger ones) only to the $(k+1)$-th and $(k+2)$-th tennis players, the number of the strongest winner after each round cannot increase by more than 2. Thus, the number of the winner of the entir... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
190. Among the numbers of the form $36^{k}-5^{l}$, where $k$ and $l$ are natural numbers, find the smallest in absolute value. Prove that the found number is indeed the smallest. | 190. Answer: $11=36-5^{2}$.
The last digit of the number $36^{h}=6^{2 k}$ is 6, the last digit of the number $5^{t}$ is 5. Therefore, the number $\left|6^{2 k}-5^{t}\right|$ ends in either 1 (if $6^{2 k}>5^{t}$) or 9 (if $6^{2 k}<5^{t}$).
The equation $6^{2 k}-5^{t}=1$ is impossible, because then it would be $5^{l}=\... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
226. In a regular 1976-gon, the midpoints of all sides and the midpoints of all diagonals are marked. What is the greatest number of marked points that can lie on one circle? | 226. Answer: 1976. All marked points, except for the center $O$ of the 1976-gon, lie 1976 each on 987 circles with center $O$. Any other circle $\gamma$ intersects each of these 987 circles at two points; besides these intersection points, there can be only one more marked point on $\gamma$: $O$. Therefore, there are n... | 1976 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
238. Several black and white chips are arranged in a circle. Two players take turns performing the following operation: the first player removes all black chips that have a white neighbor (at least on one side), and the second player then removes all white chips that have a black neighbor. They continue doing this unti... | 238. Answers: a) yes, b) 8 moves (each player - per move).
Fig. 102 shows an example of arranging 41 chips; next to each chip is its rank-number, indicating how many moves before the end it will be removed. The construction of such an arrangement is convenient "from the end"; to the remaining last black chip of rank 0... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
252. Let $a_{n}$ denote the integer closest to $\sqrt{n}$. Find the sum
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{1980}}
$$ | 252. Answer: 88.
Each number $k=1,2,3, \ldots$ appears in the sequence $(a_{n})$ $2 k$ times, since the condition $a_{n}=k$ is equivalent to
$$
k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2}, \quad \text { or } \quad k^{2}-k<n \leqslant k^{2}+k
$$
Therefore, in the sum
$$
\begin{aligned}
&\left(\frac{1}{a_{1}}+\frac{1}{a_{2}... | 88 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
270. A kangaroo jumps in the corner $x \geqslant 0, y \geqslant 0$ of the coordinate plane $O x y$ as follows: from the point ( $x ; y$ ) the kangaroo can jump to the point ( $x-5$; $y+7)$ or to the point $(x+1 ; y-1)$, and jumping to points with a negative coordinate is not allowed. From which initial points $(x ; y)$... | 270. Answer: the set of points from which one cannot escape *to infinity* has an area of 15; this is the stepped figure $T$, shown in Fig. 112.
From any point outside $T$, one can reach the region $x \geq 5$ in several steps $(1; -1)$, and then make steps $(-5; 7) + 5(1; -1) = (0; 2)$
$\nabla$ Even more interesting s... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
281*. The final sequence $a_{1}, a_{2}, \ldots, a_{n}$ of numbers 0 and 1 must satisfy the following condition: for any integer $k$ from 0 to $n-1$, the sum
$$
a_{1} a_{k+1}+a_{2} a_{k+2}+\ldots+a_{n-k} a_{n}
$$
is an odd number.
a) Come up with such a sequence for $n=25$.
b) Prove that such a sequence exists for s... | 281. The value specified in the problem condition
$$
p_{k}=a_{1} a_{k+1}+a_{2} a_{k+2}+\ldots+a_{n-k} a_{n}
$$
is conveniently calculated as follows: the sequence $A_{n}=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ is signed under itself with a shift of $k$ digits; in this case, $p_{k}=p_{k}\left(A_{n}\right)$ is the nu... | 1201 | Combinatorics | proof | Yes | Yes | olympiads | false |
306. We will say that a number has the property $\mathrm{P}(k)$ if it can be factored into the product of $k$ consecutive natural numbers, all greater than 1.
a) Find $k$ such that some number $N$ simultaneously has the properties $\mathrm{P}(k)$ and $\mathrm{P}(k+2)$.
b) Prove that there are no numbers that simultane... | 306. a) $k=3 ; 720=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6=8 \cdot 9 \cdot 10$.
b) If $m(m+1)=n(n+1)(n+2)(n+3)$, then $m^{2}+m+1=\left(n^{2}+3 n+1\right)^{2}$, which is impossible. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
322. Find at least one natural number $n$ such that each of the numbers $n, n+1, n+2, \ldots, n+20$ has a common divisor with the number $30030=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ that is greater than one. | 322. Answer: for example, 9440. Let $m=2 \cdot 3 \cdot 5 \cdot 7 \cdot k$. By choosing $k$ such that $m-1$ is divisible by 11, and $m+1$ is divisible by 13, we get that the number $n=m-10$ satisfies the condition of the problem (P2). | 9440 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
325. a) Find the smallest possible value of the polynomial
$$
P(x, y)=4+x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2}
$$
b) ${ }^{*}$ Prove that this polynomial cannot be represented as a sum of squares of polynomials in the variables $x, y$. | 325. a) Answer: 3 at $x=y=1$.
From the inequality of the arithmetic mean, it follows that $1+x^{2} y^{4}+x^{4} y^{2} \geqslant 3 x^{2} y^{2}$.
b) Let $P(x, y)=g_{1}^{2}(x, y)+g_{2}^{2}(x, y)+\ldots+g_{n}^{2}(x, y)$, where $g_{i}(x, y), i=1,2, \ldots, n-$ are polynomials. Since $P(x, 0)=$ $=P(0, y)=4$, the polynomials... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
337. Natural numbers from 1 to 1982 are arranged in some order. A computer scans pairs of adjacent numbers (the first and second, the second and third, etc.) from left to right up to the last pair and swaps the numbers in the scanned pair if the larger number is to the left. Then it scans all pairs from right to left f... | 337. Answer: 100. Consider the number $a$-the largest of the first 99 numbers, and the number $b$-the smallest of the last 1882 numbers and ensure that $a<100<b$. | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
389. The sequence $x_{n}$ is defined recursively: $\quad x_{1}=1 ; x_{2}=1 ; x_{n+2}=x_{n+1}^{2}-\frac{1}{2} x_{n}, \quad$ if $n \geqslant 1$. Prove that the sequence $x_{n}$ has a limit, and find it.
390 *. In the white cells of a chessboard of size $1983 \times 1984$, numbers 1 or -1 are written such that for any bl... | 389. Answer: $\lim _{n \rightarrow \infty} x_{n}=0\left(\left|x_{n+7}\right| \leqslant \frac{1}{4}\left(\frac{5}{6}\right)^{n}\right.$ for $\left.n \geqslant 1\right)$. | 0 | Algebra | proof | Yes | Yes | olympiads | false |
432. Milk is poured into 30 glasses. A boy is trying to make the amount of milk equal in all glasses. For this, he takes any two glasses and pours milk from one to the other until the amount of milk in them is equal. Is it possible to pour milk into the glasses in such a way that the boy cannot achieve his goal, no mat... | 432. Answer: it is possible. You need to pour 100 g of milk into all the glasses except one, and into the remaining one - 200 g
256 | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
456. Every evening, Uncle Chernomor appoints 9 or 10 out of 33 bogatyrs for duty, at his discretion. What is the smallest number of days after which it can happen that each of the bogatyrs has been on duty the same number of times? | 456. Answer: 7 days. Let $k$ be the number of days when 9 bogatyrs were on duty, and $l$ be the number of days when 10 bogatyrs were on duty, with each of them being on duty $m$ times. Then $9 k + 10 l = 33 m$. For $m=1$ there are no solutions, but for $m=2$ we have $k=4$ and $l=3$. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.3. Replace the ellipsis with such a natural number $p$ so that the answer to the following question is unique: how many lines are drawn on the plane, if it is known that they intersect at ... different points? | 1.3. Answer. $p=2$.
We will prove that if there are two intersection points, there can only be three lines. Suppose there are only two intersection points $A$ and $B$, and let $l_{1}$ and $l_{2}$ be the lines intersecting at point $A$, and $l_{3}$ be a line not passing through $A$ (such a line exists, otherwise all li... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.9. What is the maximum number of strings connecting adjacent nodes in a volleyball net with square cells that can be cut so that the net does not fall apart into separate pieces? The size of the net is $10 \times 100$ cells.
## PARALLEL SEGMENTS | 1.9. Answer. 1000.
Suppose that so many strings have been torn that the net has not yet split into pieces, but no more strings can be torn. This means that there are no closed loops of strings left in the net. Prove that in this case, the number of unbroken strings is one less than the total number of nodes (including... | 1000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.1. The altitudes of triangle $ABC$ intersect at point $O$. It is known that $OC = AB$. Find the angle at vertex $C$. | ## 5.1. Answer. $45^{\circ}$.
Let $D$ be the foot of the perpendicular dropped from point $A$. Prove that triangles $C O D$ and $A B D$ are congruent. Then show that $\angle O B D=45^{\circ}$, and from this, deduce that the required angle is also $45^{\circ}$.
Another solution can be obtained by almost verbatim repea... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.3. Find the angle $C$ of triangle $ABC$, if the distance from vertex $C$ to the orthocenter of the triangle is equal to the radius of the circumscribed circle. | 5.3. Answer. $60^{\circ}$.
1st solution. Let $O$ be the center of the circumcircle of triangle $ABC$, $M$ be the orthocenter, and $CC'$ be the diameter of the circumcircle (see figure a). Inscribed angles $ABC$ and $AC'C$ are equal, so the acute angles $MCB$ and $ACO$, which complement them to $90^{\circ}$, are also e... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.5. In triangle $A B C$ with an angle of $120^{\circ}$ at vertex $A$, the angle bisectors $A A_{1}, B B_{1}$ and $C C_{1}$ are drawn. Find the angle $C_{1} A_{1} B_{1}$. | 5.5. Answer. $90^{\circ}$.
Let the length of side $AB$ be denoted by $c$, $AC$ by $b$, and $BC$ by $a$. First, we will prove that $AA_1 = \frac{bc}{b+c}$.
Fig. 5.5.
Extend $AA_1$ and draw... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.7. In an isosceles triangle \(ABC\) with lateral sides \(AB = BC\), the angle \(ABC\) is \(80^\circ\). Inside the triangle, a point \(O\) is taken such that the angle \(OAC\) is \(10^\circ\) and the angle \(OCA\) is \(30^\circ\). Find the angle \(AOB\). | 5.7. Answer. $70^{\circ}$.
Let $K$ be the point of intersection of the altitude dropped from vertex $B$ and the bisector of angle $OAB$ (see figure). First, show that point $K$ lies on the extension of $OC$. Then prove that triangles
. A seventh penny, also lying on the table, rolls without slipping along the outer side of the chain, touching each of the six pennies in the chain in... | 6.3. Answer. 4 turns.
From figure $a$, it can be seen that during the time the moving coin, depicted with a dashed line, rolls along the arc $\alpha$ of the stationary coin with center $O$, it rotates by an angle of $2 \alpha$: in this figure, $M^{\prime} A^{\prime}$ is the new position of the radius $M A$, the radii ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. In a city, there are 10 streets parallel to each other and 10 others intersect them at right angles. What is the minimum number of turns a closed route can have, passing through all intersections? | 8.5. Answer. 20.
It is easy to provide an example with 20 turns. We will prove that fewer than 20 turns are not possible. Consider 10 streets of a certain direction. If the route passes through each of them, then there are already at least two turns of the route on each of them, and the proof is complete. If there is ... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Let $S(x)$ denote the sum of the digits of a natural number $x$. Solve the equations:
a) $x+S(x)+S(S(x))=1993$
b)* $x+S(x)+S(S(x))+S(S(S(x)))=1993$.
2*. It is known that the number $n$ is the sum of the squares of three natural numbers. Show that the number $n^{2}$ is also the sum of the squares of three natural ... | 1. a) According to the divisibility rule for 3 (see fact 6), the numbers $x$ and $S(x)$ give the same remainder when divided by 3. The same remainder will also be given by the number $S(S(x))$. Therefore, the sum
$$
x + S(x) + S(S(x))
$$
is divisible by 3 (since it is the sum of three numbers with the same remainder ... | 1963 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Petya has a total of 28 classmates. Any two of the 28 have a different number of friends in this class. How many friends does Petya have? | 4. Petya's classmates can have $0,1,2, \ldots, 28$ friends - a total of 29 options. However, if someone is friends with everyone, then everyone has at least one friend. Therefore, either there is someone who is friends with everyone, or there is someone who is not friends with anyone. In both cases, there are 28 option... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Each pair of numbers $x$ and $y$ is assigned a number $x * y$. Find $1993 * 1935$, given that for any three numbers $x, y$ and $z$ the identities $x * x=0$ and $x *(y * z)=(x * y)+z$ are satisfied. | 5. Let's take $y=z$ in the second identity. Then we get
$$
(x * y)+y=x *(y * y)=x * 0
$$
Thus, $x * y=x * 0 - y$. It remains to compute $x * 0$. For this, take $x=y=z$ in the second identity:
$$
x * 0=x *(x * x)=x * x + x=0 + x=x.
$$
Thus, $x * y=x * 0 - y=x - y$. Therefore, $1993 * 1935=1993 - 1935=58$.
Comment. ... | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A cooperative receives apple and grape juice in identical barrels and produces an apple-grape drink in identical cans. One barrel of apple juice is enough for exactly 6 cans of the drink, and one barrel of grape juice is enough for exactly 10 cans. When the recipe for the drink was changed, one barrel of apple juice... | 1. The first method. For one can of the drink, $\frac{1}{6}$ of a barrel of apple juice and $\frac{1}{10}$ of a barrel of grape juice are used, so the volume of the can is
$$
\frac{1}{6}+\frac{1}{10}=\frac{4}{15}
$$
of the barrel's volume.
After changing the recipe, for one can of the drink, $\frac{1}{5}$ of a barre... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A student did not notice the multiplication sign between two three-digit numbers and wrote a single six-digit number, which turned out to be seven times greater than their product. Find these numbers. | 2. First method. Let $x, y$ be the desired three-digit numbers. If we append three zeros to the number $x$, we get the number $1000 x$, and if we append $y$, we get $1000 x + y$ (see fact 11).
Thus, the student wrote the number $1000 x + y$. According to the problem, this number is seven times greater than $x \cdot y$... | 143 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5*. Find the largest natural number, not ending in zero, which, when one (not the first) digit is erased, decreases by an integer factor.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 5. Let $x$ be the deleted digit, $a$ be the part of the number to the left of $x$, and $c$ be the part of the number to the right of $x$. Then the number has the form $\overline{a x c}$, see fact 11. Suppose the digit $x$ is in the $(n+1)$-th place (counting from the right). Then
$$
\overline{a x c}=a \cdot 10^{n+1}+x... | 180625 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A rectangle of size $1 \times k$ for any natural number $k$ will be called a strip. For which natural numbers $n$ can a rectangle of size $1995 \times n$ be cut into pairwise distinct strips? | 3. Idea of the solution: take the maximum strip (equal to the maximum side of the rectangle). The remaining strips will be combined in pairs, giving the sum of the maximum strip. If we have filled the rectangle, the problem is solved; otherwise, reasoning with areas shows that the rectangle cannot be cut into different... | 3989 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Is it enough to make a closed rectangular box from all sides, enclosing no less than 1995 unit cubes, a) 962; b) 960; c) 958 square units of material? | 4. It is sufficient to take a box of size $11 \times 13 \times 14$. Its volume is 2002, which is sufficient; and the total area of its walls is $2 \cdot(11 \cdot 13+11 \cdot 14+13 \cdot 14)=958$.
Comments. $1^{\circ}$. How can one guess such a solution? It is known that among all parallelepipeds with a given volume, t... | 958 | Geometry | MCQ | Yes | Yes | olympiads | false |
6. A line cuts off triangle $A K N$ from a regular hexagon $A B C D E F$ such that $A K+A N=A B$. Find the sum of the angles under which segment $K N$ is seen from the vertices of the hexagon ( $\angle K A N+\angle K B N+\angle K C N+\angle K D N+\angle K E N+$ $+\angle K F N$).
## 9 t h g r a d e | 6. Let's assume that $N$ lies on $A B$, and $K$ lies on $A F$ (Fig. 60). Note that $F K = A N$. We choose point $P$ on $B C$, point $R$ on $C D$, point $S$ on $D E$, and point $T$ on $E F$ such that the equalities $F K = A N = B P = C R = D S = E T$ hold. Then $\angle K B N = \angle T A K$, $\angle K C N = \angle S A T... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In a $10 \times 10$ grid, the centers of all unit squares are marked (a total of 100 points). What is the minimum number of lines, not parallel to the sides of the square, needed to cross out all the marked points? | 2. Let's draw all lines parallel to one of the diagonals of the square and containing more than one of the marked points - there are 17 such lines. The un-
Fig. 77 erased will be the two c... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given an equilateral triangle $A B C$. Side $B C$ is divided into three equal parts by points $K$ and $L$, and point $M$ divides side $A C$ in the ratio $1: 2$, starting from vertex $A$. Prove that the sum of angles $A K M$ and $A L M$ is $30^{\circ}$. | 4. Without loss of generality, we can assume that point $K$ is closer to point $B$ than point $L$ (Fig. 72). Then triangle $M K C$ is equilateral (since $M C = K C, \angle M C K = 60^{\circ}$). Therefore, $A B \| M K$ (since $\left.\angle M K C = \angle A B C = 60^{\circ}\right)$. This means that angles $A K M$ and $B ... | 30 | Geometry | proof | Yes | Yes | olympiads | false |
6. Ali-Baba and the bandit are dividing a treasure consisting of 100 gold coins, arranged in 10 piles of 10 coins each. Ali-Baba chooses 4 piles, places a cup next to each, and sets aside several coins from each pile (at least one, but not the entire pile). The bandit must then rearrange the cups, changing their initia... | 6. We will show that Ali-Baba can achieve no more than 4 coins in 7 piles, while the robber can ensure that there are no piles with fewer than 4 coins. Therefore, Ali-Baba will take $100 - 7 \cdot 4 = 72$ coins.
First, we will prove that the robber can act in such a way that there are no piles with fewer than 4 coins.... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. From the volcano station to the summit of Stromboli volcano, it takes 4 hours to walk along the road, and then 4 hours along the path. At the summit, there are two craters. The first crater erupts for 1 hour, then remains silent for 17 hours, then erupts again for 1 hour, and so on. The second crater erupts for 1 ho... | 2. The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the eruption of the first crater, it will not be dangerous.
Movement along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the eruption of the s... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5*. In rhombus $A B C D$ the measure of angle $B$ is $40^{\circ}, E-$ is the midpoint of $B C, F$ - is the foot of the perpendicular dropped from $A$ to $D E$. Find the measure of angle $D F C$. | 5. Let lines $D E$ and $A B$ intersect at point $G$ (Fig. 83). Then triangles $D E C$ and $B E G$ are congruent by the second criterion. Therefore, $B G=C D=B A$. Hence, points $A, G$, and $C$ lie on a circle with center at point $B$, and $A G$ is the diameter. Since $\angle A F G=90^{\circ}$, point $F$ lies on the sam... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A road 1 km long is fully illuminated by street lamps, each of which illuminates a section of the road 1 m long. What is the maximum number of street lamps that can be on the road, given that after turning off any street lamp, the road will no longer be fully illuminated? | 3. Let's number the street lamps with natural numbers in the order of their placement along the road. If the segments illuminated by the $n$-th and $(n+2)$-th street lamps intersect (at least at one point), then the $(n+1)$-th street lamp can be turned off. Therefore, segments with different odd numbers do not intersec... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On the board in the laboratory, two numbers are written. Every day, senior researcher Petya erases both numbers from the board and writes down their arithmetic mean and harmonic mean instead. On the morning of the first day, the numbers 1 and 2 were written on the board. Find the product of the numbers written on th... | 1. The product of the numbers on the board does not change. Indeed,
$$
\frac{a+b}{2} \cdot \frac{2}{1 / a+1 / b}=\frac{a+b}{2} \cdot \frac{2 a b}{a+b}=a b
$$
Therefore, on the 1999th day, the product will be the same as it was on the first day. See also fact 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3 *. Find all pairs of natural numbers $x, y$ such that $x^{3}+y$ and $y^{3}+x$ are divisible by $x^{2}+y^{2}$. | 3. Let's first prove that $x$ and $y$ are coprime. Assume the opposite. Then $x$ and $y$ are divisible by some prime number $p$. Let $p$ enter the prime factorizations of $x$ and $y$ with powers $a \geqslant 1$ and $b \geqslant 1$ respectively (see fact 10). Without loss of generality, we can assume that $a \geqslant b... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Pete's bank account contains 500 dollars. The bank allows only two types of transactions: withdrawing 300 or adding 198 dollars. What is the maximum amount Pete can withdraw from his account if he has no other money? | 4. Since 300 and 198 are divisible by 6, Petya will only be able to withdraw an amount that is a multiple of 6 dollars (see fact 5). The maximum number that is a multiple of 6 and does not exceed 500 is 498.
Let's show how to withdraw 498 dollars. We will perform the following operations: $500-300=200, 200+198=398, 39... | 498 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4*. Find all such positive integers $k$ for which the number
$$
\underbrace{1 \ldots 1 \overbrace{2 \ldots 2}^{k}}_{2000}-\underbrace{2 \ldots 2}_{1001}
$$
is a perfect square. | 4. Let $n=1000$. Consider two cases.
$1^{\circ} . k>n$. Then
$$
\underbrace{1 \ldots 12 \ldots 2}_{2 n}-\underbrace{2 \ldots 2}_{n+1}=\underbrace{1 \ldots 1}_{2 n-k} \overbrace{2 \ldots 2}^{k-(n+1)} \underbrace{0 \ldots 0}_{n+1} .
$$
This number ends with $n+1=1001$ zeros. But if a number is a square of a natural nu... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the elections to the 100-seat parliament, 12 parties participated. Parties that received strictly more than $5 \%$ of the voters' votes enter the parliament. Among the parties that entered the parliament, seats are distributed proportionally to the number of votes they received (i.e., if one party received $x$ ti... | 2. Idea of the solution: The Party of Mathematics Enthusiasts (PME) will receive the maximum number of seats in parliament if the total number of votes cast for non-qualifying (i.e., receiving no more than $5 \%$ of the votes) parties is maximized.
If 10 parties receive exactly $5 \%$ of the votes each, and two, inclu... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. What is the maximum number of knights that can be placed on a $5 \times 5$ board such that each one attacks exactly two others? (Provide an example and explain why it is not possible to place more knights.)
## 9 t h g r a d e | 6. Fig. 116 shows the arrangement of 16 knights that satisfies the problem's condition. We will show that it is impossible to place more knights. Let's color the cells of the board in black and white, as shown in Fig. 116. Note that the number of knights on black cells is equal to the number of knights on white cells. ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Six words are given: ZANOZA, ZIPUNY, KAZINO, KEFAL', OTMEL', SHELEST. In one step, you can replace any letter in any of these words with any other letter (for example, in one step, you can get the word ZKNOZA from ZANOZA). How many steps are needed to make all the words the same (nonsense words are allowed)? Provide... | 3. Let's write the words in a column:
|  | | | |
| :---: | :---: | :---: | :---: |
After all the letter replacements in each column, the letters should become the same. The number of rep... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In a convex quadrilateral $A B C D$, points $E$ and $F$ are the midpoints of sides $B C$ and $C D$ respectively. Segments $A E, A F$, and $E F$ divide the quadrilateral into 4 triangles, the areas of which are consecutive natural numbers. What is the maximum possible value of the area of triangle $A B D$? | 3. Let the areas of the triangles be $n, n+1, n+2$, $n+3$. Then the area of the quadrilateral $A B C D$ is $4 n+6$. It is easy to see that the area of triangle $B C D$ is four times the area of triangle $E C F$, so this area is at least $4 n$. Therefore,
$$
S_{A B D}=S_{A B C D}-S_{B C D} \leqslant(4 n+6)-4 n=6 .
$$
... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the city of Udoyev, mayoral elections proceed as follows. If in a given round of voting no candidate receives more than half of the votes, then a subsequent round is held with the participation of all candidates except the one who received the fewest votes. (No two candidates ever receive the same number of votes... | 5. a) Ostap could not take the last, 2002nd place in the first round, as otherwise he would have been immediately eliminated from the list of candidates. Therefore, $k \leqslant 2001$.
Suppose all candidates in the first round received almost the same number of votes, Ostap took the second-to-last place, and in each s... | 2001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. There are 4 people in the family. If Masha's scholarship is doubled, the total income of the entire family will increase by $5 \%$, if instead, Mom's salary is doubled - then by $15 \%$, if Dad's salary is doubled - then by $25 \%$. By what percentage will the family's total income increase if Grandpa's pension is d... | 1. The first method. If Masha's scholarship is doubled, the family income will increase by the amount of this scholarship. Therefore, Masha's scholarship constitutes $5 \%$ of the total income. Similarly, Mom's salary is $15 \%$, and Dad's is $25 \%$. The remaining share $100 \% - 5 \% - 15 \% - 25 \% = 55 \%$ is attri... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In the country, there are 15 cities, some of which are connected by air routes belonging to three airlines. It is known that even if any one of the airlines ceases operations, it will still be possible to travel from any city to any other (possibly with layovers), using the flights of the remaining two airlines. Wha... | 5. We will prove that fewer than 21 airlines are not sufficient. First, note that if 15 cities are connected by airlines in such a way that one can travel from any city to any other, then there are no fewer than 14 airlines. Indeed, starting from an arbitrary city, we will try to visit all the others, and visiting each... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. For a convex polyhedron, the internal dihedral angle at each edge is acute. How many faces can the polyhedron have? | 5. For each of the faces, consider the vector of the external normal, i.e., a vector perpendicular to this face and directed outward from the polyhedron.
$1^{\circ}$. Let's prove that the angle between any two external normals is obtuse or straight. Suppose this is not the case, and there exist two faces $\Gamma_{1}$ ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. On the shore of a round island called Somewhere, there are 20 villages, each inhabited by 20 wrestlers. A tournament was held where each wrestler faced all wrestlers from all other villages. Village A is considered stronger than Village B if at least $k$ matches between wrestlers from these villages end with a victo... | 6. Let's provide an example showing that the described situation is possible when $k \leqslant 290$. We will order all the wrestlers by strength and renumber them in ascending order of strength (the first being the weakest). We will call the 210 weakest wrestlers novices, and the 190 strongest - masters. In particular,... | 290 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. An extrasensory perception (ESP) practitioner has a deck of 36 cards face down in front of him (four suits, nine cards of each suit). He names the suit of the top card, after which the card is revealed to him. Then he names the suit of the next card, and so on. The task of the ESP practitioner is to guess the suit a... | 6. a) Note that guessing 18 cards is not difficult. Indeed, the first two backs can "encode" the suit of the second card (by associating each suit with one of the four possible arrangements of the two backs), the next two backs can encode the suit of the fourth card, and so on.
When only two cards remain in the deck, ... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. A square sheet of checkered paper $8 \times 8$ was folded several times along the grid lines so that a $1 \times 1$ square was obtained. It was then cut along a segment connecting the midpoints of two opposite sides of the square. Into how many pieces could the square have split as a result? | 2. Let the cut be vertical (the case of a horizontal cut is analogous). Draw vertical segments in all $1 \times 1$ squares, connecting the midpoints of opposite sides. Notice that when folding along the grid lines, these
. Therefore,
$$
x^{3}=10^{4} a+100 b+c44$;
2) $x=45(x-1=44), 45^{3}=91125, a=9, b=11, c... | 91125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
35. Given a line $l$, two points $A$ and $B$ on the same side of it, and a segment $a$. Find a segment $X Y$ on the line $l$ of length $a$ such that the length of the broken line $A X Y B$ is the smallest (Fig. 42). | 35. Since the length of the segment $X Y$ is equal to $a$, it is necessary that the sum $A X + B Y$ be minimal. Suppose that the segment $X Y$ is laid out. The sliding symmetry with the axis $l$ "of the parallel translation magnitude $a$ translates point $B$ to $B'$, and point $Y$ to $X$ (Fig. 144); therefore, $B Y = B... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
42. Given three lines $l_{1}, l_{2}$ and $l_{3}$ and three points $A, B$ and $C$-one point on each of the lines. Draw a line $m$ intersecting lines $l_{1}, l_{2}$ and $l_{3}$ at points $X, Y$ and $Z$ such that $A X=B Y=C Z$. | 42. Let the lines $l_{1}, l_{2}$ and $l_{3}$ not all be parallel to each other, for example, $l_{3}$ is not parallel to either $l_{1}$ or $l_{2}$. Suppose the problem is solved (Fig. 155). By Theorem 1, there exists
 | Solution 4. Let the factors have the form $2^{a_{1}} 5^{b_{1}}, 2^{a_{2}} 5^{b_{2}}$ and $2^{a_{3}} 5^{b_{3}}$. Then $a_{1}+a_{2}+a_{3}=6$ and $b_{1}+b_{2}+b_{3}=6$. Here, the numbers $a_{i}$ and $b_{i}$ can be zero. If $a_{1}=k$, then for the decomposition $a_{2}+a_{3}=6-k$ we get $7-k$ options. Therefore, for the dec... | 139 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-4. How many natural ${ }^{2}$ numbers less than a thousand are there that are not divisible by 5 or 7? | Solution 4. Answer: 686 numbers. First, let's strike out from the set of numbers $1,2, \ldots, 999$ the numbers that are multiples of 5; their quantity is $\left[\frac{999}{5}\right]=199$. Then, from the same set of numbers $1,2, \ldots, 999$, let's strike out the numbers that are multiples of 7; their quantity is $\le... | 686 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3-5. The factorial of a number $n$ is defined as the product of all integers from 1 to $n$ inclusive. Find all three-digit numbers that are equal to the sum of the factorials of their digits. | Solve 5. Answer: 145. Let $N=100x+10y+z$ be the desired number, for which $N=x!+y!+z!$. The number $7!=5040$ is four-digit, so no digit of the number $N$ exceeds 6. Therefore, the number $N$ is less than 700. But then no digit of the number $N$ exceeds 5, since $6!=720$. The inequality $3 \cdot 4!=72<100$ shows that at... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. Find a four-digit number that is a perfect square, and such that the first two digits are the same as each other and the last two digits are also the same. | Solution 1. Let $a$ be the first and second digits, $b$ be the third and fourth. Then the given number is equal to $11(b+100a)$, so $b+100a=11x^2$ for some natural number $x$. Moreover, $100 \leqslant b+100a \leqslant 908$, which means $3 \leqslant x \leqslant 9$. By calculating the squares of the numbers $33, 44, \ldo... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3-4. How many pairs of integers $x, y$, lying between 1 and 1000, are there such that $x^{2}+y^{2}$ is divisible by 7. | Solve 4. Answer: $142^{2}=20164$. The number $x^{2}+y^{2}$ is divisible by 7 if and only if both numbers $x$ and $y$ are divisible by 7. Indeed, the square of an integer when divided by 7 gives remainders of 0, 2, and 4. The number of integers between 1 and 1000 that are divisible by 7 is 142. Therefore, the desired nu... | 20164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3-5. How many positive integers $x$, less than 10000, are there for which $2^{x}-x^{2}$ is divisible by 7? | Solve 5. Answer: 2857. The remainders of the division by 7 of the numbers $2^{x}$ and $x^{2}$ repeat with periods of 3 and 7, respectively, so the remainders of the division by 7 of the numbers $2^{x} - x^{2}$ repeat with a period of 21. Among the numbers $x$ from 1 to 21, the numbers that give equal remainders from th... | 2857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3-4. Construct triangle $ABC$ given points $M$ and $N$ - the feet of the altitudes $AM$ and $BN$ - and the line on which side $AB$ lies. | Sol 4. Points $M$ and $N$ lie on a circle with diameter $AB$. The center $O$ of this circle is the intersection of line $AB$ and the perpendicular bisector of segment $NM$, so we can construct it. Points $A$ and $B$ are the points of intersection of line $AB$ and the circle centered at $O$ passing through point $M$. Th... | 63 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3-2. Find a three-digit number, every integer power of which ends in the same three digits as the original number (in the same order). | Let $N$ be the desired number. Then $N^{2}-N=N(N-1)$ is divisible by 1000. The numbers $N$ and $N-1$ are coprime, so one of them is divisible by 8, and the other by 125. Let's first assume $N=125 k$. Then $k \leqslant 8$. Among the numbers $125 k-1, k=1, \ldots, 8$, only the number 624 is divisible by 8. Now let $N-1=1... | 376 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. Given 6 digits: $0,1,2,3,4,5$. Find the sum of all four-digit even numbers that can be written using these digits (the same digit can be repeated in a number). | Solution 1. Answer: 1769580.
We will separately calculate the sum of thousands, hundreds, tens, and units for the considered numbers. The first digit can be any of the five digits $1,2,3,4,5$. The number of all numbers with a fixed first digit is $6 \cdot 6 \cdot 3=108$, since the second and third places can be any of... | 1769580 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-2. Two identical polygons were cut out of cardboard, aligned, and pierced with a pin at some point. When one of the polygons is rotated around this "axis" by $25^{\circ} 30^{\prime}$, it aligns again with the second polygon. What is the smallest possible number of sides of such polygons? | Solution 2. Answer: 240. First, note that $\frac{1}{360} \cdot 25 \frac{1}{2}=\frac{17}{240}$, and the numbers 17 and 240 are coprime. Consider a ray emanating from the "axis" to the vertex of the first polygon. Rotations of this ray around the "axis" by angles $k \cdot 25^{\circ} 30^{\prime}$, where $k=1,2, \ldots, 24... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3-3. A circle with a radius equal to the height of a certain equilateral triangle rolls along the side of this triangle. Prove that the arc intercepted by the sides of the triangle on the circle is always $60^{\circ}$. | Solution 3. Let the angular measure of the arc cut off by the sides of the given equilateral triangle $ABC$ be denoted by $\alpha$. We will assume that the circle is tangent to the side $BC$. Consider the arc cut off by the extensions of the sides of the triangle $ABC$ on the circle, and denote its angular measure by $... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
3-ча 3. Find a four-digit number that, when divided by 131, gives a remainder of 112, and when divided by 132, gives a remainder of 98. | Solve 3. Answer: 1946. Let $N$ be the desired number. By the condition, $N=131 k+112=132 l+98$, where $k$ and $l$ are natural numbers. Moreover, $N<10000$, so $l=\frac{N-98}{132}<\frac{10000-98}{132} \leqslant 75$. Further, $131 k+112=$ $132 l+98$, so $131(k-l)=l-14$. Therefore, if $k \neq l$, then $|l-14| \geqslant 13... | 1946 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3-5. The city's bus network is organized as follows: 1) from any stop to any other stop, you can get without transferring;
2) for any pair of routes, there is, and only one, stop where you can transfer from one of these routes to the other;
3) on each route, there are exactly three stops.
How many bus routes are ther... | Solve 5. Answer: 7. We will prove that if the given conditions are satisfied, then the number of stops $n$ and the number of routes $N$ are related by the formula $N=n(n-1)+1$. Let $a$ be one of the routes, and $B$ be a stop that route $a$ does not pass through. Each route passing through $B$ intersects route $a$. Ther... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. In a chess tournament, students from grades IX and X participated. There were 10 times more students from grade X than from grade IX, and they scored 4.5 times more points in total than all the students from grade IX. How many points did the students from grade IX score? Find all solutions. | Solution 1. Let $x$ be the number of ninth graders in the tournament. Then there were a total of $11x$ participants, and they scored $\frac{11x(11x-1)}{2}$ points. According to the problem, the ratio of the number of points scored by the ninth graders to the number of points scored by the tenth graders is $1:4.5$. Ther... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-4. In the city, there are 57 bus routes. It is known that:
1) from any stop to any other stop, one can travel without transferring;
2) for any pair of routes, there is one, and only one, stop where one can transfer from one of these routes to the other;
3) each route has no fewer than three stops.
How many stops do... | Solution 4. Answer: 8. We will prove that if the given conditions are satisfied, then the number of stops $n$ and the number of routes $N$ are related by the formula $N=n(n-1)+1$. First, we will show that if a route has $n$ stops, then any other route also has $n$ stops, and furthermore, each stop is served by $n$ rout... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. Determine the coefficients that will stand by $x^{17}$ and $x^{18}$ after expanding the brackets and combining like terms in the expression
$$
\left(1+x^{5}+x^{7}\right)^{20}
$$ | Solution 1. The number 18 cannot be represented as the sum of the numbers 5 and 7, so the coefficient of $x^{18}$ will be zero.
The number 17 can be represented as the sum of the numbers 5 and 7 as follows: $17=7+5+5$; this representation is unique up to the order of the terms. In one of the 20 expressions $1+x^{5}+x^... | 3420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-rd 5. From point \(A\) to other points, one can get there in two ways:
1. Exit immediately and walk.
2. Call a car and, after waiting for a certain amount of time, ride in it.
In each case, the mode of transportation that requires the least time is used. It turns out that
\begin{tabular}{|c|c|}
\hline if the fina... | Solution 5. Answer: 25 min.
Let \(v\) be the pedestrian's speed, \(V\) be the car's speed, and \(T\) be the waiting time for the car. (We will measure speed in km/h and time in hours.) If walking, the time required for 1.2 km and 3 km would be \(1 / v\), \(2 / v\), and \(3 / v\) hours, respectively, and if traveling b... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. In a convex 13-sided polygon, all diagonals are drawn. They divide it into polygons. Let's take among them a polygon with the largest number of sides. What is the maximum number of sides it can have | Solve 1. Answer: 13. From each vertex of the original 13-gon, no more than two diagonals emerge that are sides of the considered polygon. Each diagonal corresponds to two vertices, so the number of sides of the considered polygon does not exceed 13. An example of a regular 13-gon shows that the number of sides of the p... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3-4. On a circle, there are 20 points. These 20 points are connected by 10 (non-intersecting - ed. note) chords. In how many ways can this be done? | Solution 4. Answer: 16796. Let \(a_{n}\) be the number of ways to connect \(2 n\) points on a circle with \(n\) non-intersecting chords. It is clear that \(a_{1}=1\) and \(a_{2}=2\). We will show that
\[
a_{n}=a_{n-1}+a_{n-2} a_{1}+a_{n-3} a_{2}+\cdots+a_{1} a_{n-2}+a_{n-1}.
\]
Fix one of the \(2 n\) points. The chor... | 16796 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. In a convex 1950-gon, all diagonals are drawn. They divide it into polygons. Among them, we take the polygon with the largest number of sides. What is the maximum number of sides it can have? | Solution 1. Answer: 1949. The same reasoning as in the solution to problem 1 for grades 7-8 shows that the resulting polygon has no more than 1950 sides, and if the number of its sides is 1950, then exactly two diagonals emanate from each vertex of the original polygon, bounding the resulting polygon. Suppose two diago... | 1949 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3-5. There is a piece of chain consisting of 60 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, \(\ldots, 60\) g can be formed (an unbuckled link also weighs 1 g) | Solution 5. Answer: 3 links. Let's determine the greatest \(n\) for which it is sufficient to break \(k\) links of an \(n\)-link chain so that all weights from 1 to \(n\) can be formed from the resulting parts. If \(k\) links are broken, then any number of links from 1 to \(k\) can be formed from them. But \(k+1\) link... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3-4. There is a piece of chain consisting of 150 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, \(\ldots, 150\) g can be formed (an unbuckled link also weighs 1 g)? | Solution 4. Answer: 4 links. According to the solution of problem 5 for grades \(7-8\) for a chain consisting of \(n\) links, where \(64 \leqslant n \leqslant 159\), it is sufficient to unfasten 4 links. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3-rd 5. When dividing the polynomial \(x^{1951}-1\) by \(x^{4}+x^{3}+2 x^{2}+x+1\), a quotient and a remainder are obtained. Find the coefficient of \(x^{14}\) in the quotient. | Solve 5. Answer: -1. The equalities \(x^{4}+x^{3}+2 x^{2}+x+1=\left(x^{2}+1\right)\left(x^{2}+x+1\right)\) and \(x^{12}-1=(x-1)\left(x^{2}+x+\right.\)
1) \(\left(x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)\) show that
\[
\begin{aligned}
x^{4}+x^{3}+2 x^{2}+x+1 & =\frac{x^{12}-1}{(x-1)\left(x^{3}+1\ri... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-5. Determine the maximum value of the ratio of a three-digit number to the number equal to the sum of the digits of this number. | Solve 5. Answer: 100. The sum of the digits of a three-digit number \(100a + 10b + c\) is \(a + b + c\). It is clear that \(\frac{100a + 10b + c}{a + b + c} \leqslant 100\). Moreover, for the number 100, the given ratio is 100.
\section*{8th Grade} | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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