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Problem 8-5. 8 8-5. A divisor of a natural number is called proper if it is different from 1 and the number itself. Find all natural numbers for which the difference between the sum of the two largest proper divisors and the sum of the two smallest proper divisors is a prime number. | Answer: $12(2,3,4,6)$.
Solution. One of two cases applies.
A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,144 |
Problem 8-6. There are several coins, each worth an integer number of tugriks. It is known that with these coins, you can make any other sum from 1 to 51 tugriks inclusive, except for the sum of 50 tugriks. Is it necessarily possible to make a sum of exactly 100 tugriks with these coins? | Answer: yes.
Solution. Let's take a set of coins $A$ with a total value of 51 tugriks. We will prove that $A$ consists of one coin worth 51 tugriks. Then it will remain to add it to the set worth 49 tugriks. Let $k$ tugriks be the value of the cheapest coin in the set $A$. If $k$ but not in the general case or minor g... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,145 |
11.1. On the coordinate plane, a set of points defined by the equation $x=y^{2}$ is drawn. A circle of radius 5 with center at the point $(11 ; 1)$ intersects this set at points $A, B, C$ and $D$. Prove that all points $A, B, C, D$ lie on one parabola, i.e., on a curve defined by the equation $y=a x^{2}+b x+c$, and fin... | Answer. $y=\frac{1}{2} x^{2}-\frac{21}{2} x+\frac{97}{2}$.
Solution. The coordinates of points $A, B, C, D$ are solutions to the system
$$
\left\{\begin{array}{l}
y^{2}=x \\
(x-11)^{2}+(y-1)^{2}=25
\end{array}\right.
$$
Expanding the brackets in the second equation and substituting $y^{2}$ from the first, we get
$$... | \frac{1}{2}x^{2}-\frac{21}{2}x+\frac{97}{2} | Geometry | proof | Yes | Yes | olympiads | false | 20,147 |
11.2. Through the vertices of a regular hexagon, 6 different parallel lines are drawn. Can it happen that all pairwise distances between these lines are integers?
# | # Answer. It can.
Solution. Let $A B C D E F$ be an arbitrary regular hexagon. Draw a line $m_{A}$ through vertex $A$ and point $G$ - the midpoint of side $B C$. Draw lines through the other vertices parallel to $m_{A}$. Denote these lines $m_{B}, m_{C} \ldots$ Drop perpendiculars $D S, C R, B P, C Q$ (see figure).
L... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,148 |
11.3. The sequence $a_{n}$ is constructed as follows: $a_{1}, a_{2}$ are arbitrary real numbers, and for $n \geq 3$, the number $a_{n}$ is the smallest of the numbers $\left|a_{i}-a_{j}\right|, 1 \leq$ $i<j \leq n-1$. For example, if $a_{1}=6, a_{2}=\frac{19}{2}$, then we get the sequence $6, \frac{19}{2}, \frac{7}{2},... | Solution. Let $S_{n}$ be the set of all numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1, \quad n \geq 3$. Then $a_{n}$ is the smallest number in the set $S_{n}$. Since $S_{n} \subset S_{n+1}$, then $a_{n} \geq a_{n+1}$, i.e., starting from $a_{3}$, the sequence is non-increasing. Therefore, $a_{9} \geq a_{10}=1$... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,149 |
11.4. A polyhedron is inscribed in a sphere of radius $R$, and its volume is numerically equal to the area of its surface.
a. Prove that $R>3$.
b. Can $R$ be greater than 1000? | # Solution.
Solution to part a:
Let $O$ be the center of the sphere, $S$ be the surface area of the polyhedron, and $V$ be its volume. Number the faces of the polyhedron from 1 to $n$. For each $i$, we can form a pyramid with vertex $O$ and base coinciding with the $i$-th face of the polyhedron. Denote the volume of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,150 |
11.5. Let $p>2$ be an integer not divisible by 3. Prove that there exist integers $a_{1}, a_{2}, \ldots, a_{k}$ such that
$$
-\frac{p}{2}<a_{1}<a_{2}<\cdots<a_{k}<\frac{p}{2}
$$
and the product
$$
\frac{p-a_{1}}{\left|a_{1}\right|} \cdot \frac{p-a_{2}}{\left|a_{2}\right|} \cdots \frac{p-a_{k}}{\left|a_{k}\right|}
$$... | Solution. FIRST METHOD. Let us take as numbers $a_{i}$ all numbers from the interval $\left(-\frac{p}{2}, \frac{p}{2}\right)$ such that $p \equiv a_{i}(\bmod 3)$. We will prove that they satisfy the condition of the problem. Note the following:
- All numbers $\left|a_{i}\right|$ belong to the interval $\left(0, \frac{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 20,151 |
11.6. On a grid board of size $2 \times n$ cells, some cells are painted black. A coloring is called correct if there are no two adjacent cells among the painted ones. (Adjacent are cells that share a common side.) A coloring in which no cell is painted is also considered correct.
Let $A_{n}$ be the number of correct ... | Answer: $\pm 1$.
Solution. Consider two types of grid boards: a regular board $F_{n}$ of size $2 \times n$, and a board $G_{n}$ of size $2 \times n$ with one corner cell removed. In addition to the quantities $A_{n}$ and $B_{n}$, introduce the quantities $C_{n}$ and $D_{n}$ - the number of correct colorings of the boa... | \1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,152 |
# 2. Task 2
In a dumpling shop, you can order dumplings in portions of 6, 9, and 20 pieces. Thus, not every number of dumplings can be ordered with these sets, for example, 1, 2, 3, 4, 5, 7, and 8 cannot be bought. What is the largest number of dumplings that cannot be ordered in the dumpling shop? | # 2. Task 2
In a dumpling shop, you can order dumplings in portions of 6, 9, and 20 pieces. Thus, not every number of dumplings can be ordered with these sets, for example, 1, 2, 3, 4, 5, 7, and 8 cannot be bought. What is the largest number of dumplings that cannot be ordered in the dumpling shop?
## Answer: 43 | 43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,154 |
2. Natural numbers $x$ and $y$ are such that the following equality holds:
$$
x^{2}-3 x=25 y^{2}-15 y
$$
How many times greater is the number $x$ than the number $y$? | Answer. 5
Solution. The equality $x^{2}-3 x=25 y^{2}-15 y$ is equivalent to the equality $(5 y-x)(5 y+x-3)=$ 0. Since $x$ and $y$ are natural numbers, $x, y \geq 1$. Therefore, the second bracket $5 y+x-3 \geq 3$, in particular, is non-zero. Therefore, the first bracket $5 y-x$ is zero, meaning $x$ is 5 times $y$.
Cr... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,164 |
3. One side of the rectangle is 5 times longer than the other. Show how to cut this rectangle into 5 pieces and assemble them into a square. The pieces can be flipped and rotated, but they cannot be placed on top of each other, and there should be no uncovered areas inside the square. | Answer. See the figure.
Criteria.
$(-)$
$(-$.
$(-/+)$ Correct idea of cutting, but the sizes of the parts are incorrect.
$(+/ 2)$
(+/-) The correct method of cutting is provided and the s... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 20,165 |
4. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha played 7 ma... | Answer: 4
Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.
Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and M... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,166 |
5. Nезнayka came up with an entertainment for himself. He writes on the board an expression
$$
((((0 \ldots) \ldots) \ldots .) \ldots) \ldots)
$$
where the number of parentheses he chooses at his discretion. Then, instead of each ellipsis, he inserts a plus or multiply sign and a natural number from 1 to 9, with each... | Answer. One of the possible answers: $(((((0+3)+2) \cdot 8)+4) \cdot 9)+7) \cdot 5)=2015$
Solution.
Criteria.
$(-)$ Incorrect answer.
$(-$.
$(-/+)$
$(+/ 2)$
$(+/-)$
(+.) It is clear from the solution that the author understood the correct answer, but there is a typo in the answer itself.
$(+)$ Correct answer. | (((((0+3)+2)\cdot8)+4)\cdot9)+7)\cdot5)=2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,167 |
6. Given a triangle $ABC$, in which $AB = BC$ and $\angle ABC = 90^{\circ}$. In it, the altitude $BH$ is drawn. On side $CA$, a point $P$ is chosen such that $AP = AB$, and on side $CB$, a point $Q$ is chosen such that $BQ = BH$. Prove that the lines $PQ$ and $AB$ are parallel. | Solution. We will prove that $B P$ is the bisector of angle $H B C$. Indeed,
$$
\begin{aligned}
\angle H B P & =180^{\circ}-\angle B H P-\angle H P B \quad \text { (by the theorem on the sum of angles in a triangle) } \\
& =90^{\circ}-\angle A P B \quad(\text { since } B H \text { is the altitude }) \\
& =90^{\circ}-\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,168 |
# 1. Task 1.1
Timur and Alexander are counting the trees growing around their house. Both are moving in the same direction but start counting from different trees. What is the total number of trees growing around the house if the tree that Timur called the 12th, Alexander counted as $33-\mathrm{m}$, and the tree that... | # 1. Task 1.1
Timur and Alexander are counting the trees growing around their house. Both are moving in the same direction but start counting from different trees. What is the total number of trees growing around the house if the tree that Timur called the 12th, Alexander counted as $33-\mathrm{m}$, and the tree that... | 118 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,169 |
# 3. Problem 3.1
Given four numbers $a, b, c, d$. Each pair of them was added, resulting in 6 sums: $a+b, b+c, c+d, a+c, b+d, a+d$. The four smallest of these sums are $7, 15, 43, 47$. Find the sum of the original four numbers $a+b+c+d$. If the answer is not an integer, write it as a decimal. | # 3. Problem 3.1
Given four numbers $a, b, c, d$. Each pair of them was added, resulting in 6 sums: $a+b, b+c, c+d, a+c, b+d, a+d$. The four smallest of these sums are $7, 15, 43, 47$. Find the sum of the original four numbers $a+b+c+d$. If the answer is not an integer, write it as a decimal. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,171 |
# 8. Problem 8.1
Pentagon $A B C D E$ is circumscribed around a circle.
Angles $\angle A B C, \angle B A E, \angle C D E$ are each $104^{\circ}$. Find $\angle A D B$. Write the answer in degrees (the number only, without specifying the unit of measurement).
# | # 8. Problem 8.1
Pentagon $A B C D E$ is circumscribed around a circle.
Angles $\angle A B C, \angle B A E, \angle C D E$ are each $104^{\circ}$. Find $\angle A D B$. Write the answer in degrees (the number only, without specifying the unit of measurement).
## Answer: 38 | 38 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,176 |
8.2. Prove that the number $10^{\left(10^{\left(10^{2013}\right)}\right)}+10^{\left(10^{2013}\right)}+10^{2013}-1$ is not prime. | Answer: divisible by 11.
CRITERIA
b - complete solution with a minor gap in justification.
d - divisibility by 11 is claimed without justification.
SOLUTION. $10^{\left(10^{\left(10^{2013}\right)}\right)}+10^{\left(10^{2013}\right)}+10^{2013}-1=100^{a}-1+10^{2013}\left(10^{b}+1\right)$, where $a=$ $10^{\left(10^{20... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 20,179 |
8.3. Is it possible to cut a circle into parts in such a way that a) the center of the circle is on the boundary of each part and b) from some of the parts obtained as a result of the cutting, a regular hexagon inscribed in this circle can be formed? If it is possible, describe the cutting and indicate how to form the ... | Answer: Inscribed in a circle with center $O$ and radius $r$ is a regular hexagon $A B C D E F$. Cut the circle along segments $A O, C O, E O$ and arcs of circles with centers $A, C$, and $E$, radius $r$, and endpoints at points $B, D$, and $F$ (on the left picture). Also, cut the regular hexagon along arcs of circles ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,180 |
8.4. Find all integer solutions $(x, y)$ of the equation $2 x^{2}-y^{2}=2^{x+y}$ and prove that there are no others. | Answer: $(1,0),(1,-1),(-1,1),(-3,4)$.
Criteria
b - complete solution with minor gaps in justification.
c - all answers obtained.
cd - complete solution with minor gaps in justification and loss of an answer.
d - at least one answer obtained in the original problem (excluding cases of erroneous transformation of th... | (1,0),(1,-1),(-1,1),(-3,4) | Number Theory | proof | Yes | Yes | olympiads | false | 20,181 |
8.5. How many different (i.e., not equal to each other) acute triangles with integer side lengths and a perimeter of 24 exist? List the lengths of the three sides of all these triangles and prove that there are no others.
ANSWER: 6.
CRITERIA
b - correct criterion for acuteness and either a correct list of acute tria... | Solution. Note that a triangle with sides $a \geqslant b \geqslant c$ is obtuse if and only if $a^{2}>b^{2}+c^{2}$. Indeed, if $A B C$ is a triangle with sides $a \geqslant b \geqslant c$ opposite vertices $A, B$, and $C$ respectively, and $A B^{\prime} C$ is a triangle with a right angle $\angle A$ and $A B^{\prime}=A... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,182 |
8.6. The upper half-plane is divided into square cells (see figure). A domino tile occupies two adjacent cells. Is it possible to fill some of the cells with non-overlapping domino tiles so that in each row and each column there is an odd number of filled cells? If it is possible, describe the configuration; if not, pr... | Solution. Let's arrange, for example, four dominoes as the two lower pairs of touching dominoes in the figure. Then, for each natural number $n$, add a pair of dominoes obtained by shifting the left pair $2n$ cells to the left and $4n$ cells up, as well as a pair of dominoes obtained by shifting the right pair $2n$ cel... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,183 |
Problem 9.1. (15 points) Artem, Borya, Vadim, and Grisha returned from the forest where they had been collecting mushrooms. If Artem had collected 2 times less and Borya 2 times more, then together they would have as many as Vadim and Grisha together. And if Vadim had collected 2 times less and Grisha 2 times more, the... | Solution. Let Artem, Borya, Vadim, Grisha have collected $a, b, v, g$ mushrooms respectively. From the conditions, it follows that $\frac{a}{2}+2 b=v+g$ and $a+b=\frac{v}{2}+2 g$. If the first equation, multiplied by 2, is subtracted from the second, we get $3 b=\frac{3}{2} v$, from which $v=2 b$. Then from the second ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,184 |
Problem 9.2. (15 points) Does there exist a 1000-digit natural number, consisting of non-zero digits, that is divisible by the sum of its digits? | Answer: Yes.
Solution. Consider an arbitrary number $A$ that ends with 31253125. Note that it is divisible by $3125=5^{5}$, since the number $x=31253125=3125 \cdot 10001$ is clearly divisible by 3125, and the remaining $A-x$ ends with eight zeros, i.e., it is also divisible by $5^{5}$.
Now, let's make the sum of the ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,185 |
Problem 9.4. (15 points) Point $O$ is the center of the circumcircle of an acute triangle $ABC$. On the sides $AB$ and $BC$, there are points $N$ and $M$ respectively such that $\angle BAC = \angle NOA$ and $\angle BCA = \angle MOC$. Point $K$ is the center of the circumcircle of triangle $MBN$. Prove that $AK = CK$.
... | Solution. Let $\angle B A C=\alpha, \angle A B C=\beta, \angle B C A=\gamma$; we have $\alpha+\beta+\gamma=180^{\circ}$. Since $O$ is the center of the circumcircle of triangle $A B C$, then $O A=O B=O C$, $\angle A O B=2 \gamma, \angle A O C=2 \beta, \angle O B A=\angle O A B=90^{\circ}-\gamma, \angle O A C=\angle O C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,187 |
Problem 9.5. (20 points) Real numbers $a, b, c, d$ are such that $a+b=\frac{9}{c-d}$ and $c+d=$ $=\frac{25}{a-b}$. What is the smallest value that the quantity $a^{2}+b^{2}+c^{2}+d^{2}$ can take? | # Answer: 34.
Solution. If the given equalities are multiplied by the denominators of the corresponding fractions and added, we get $2(a c-b d)=34$. We will prove that $a^{2}+b^{2}+c^{2}+d^{2} \geqslant 2(a c-b d)$. This follows from $a^{2}+c^{2} \geqslant 2 a c$ (equivalent to $(a-c)^{2} \geqslant 0$) and $b^{2}+d^{2... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,188 |
Problem 9.6. (20 points) There were $n$ identical-looking coins weighing $x_{1}, x_{2}, \ldots, x_{n}$ grams (the weights of the coins are pairwise distinct positive real numbers), and also weightless stickers with numbers $x_{1}, x_{2}, \ldots, x_{n}$. At night, a lab assistant weighed the coins and labeled them with ... | Answer: 2 yes
Solution. Consider the weights 10, 20, 25, 30, 35, 201, 203, 207 (weights will be measured in grams hereafter). We will perform two checks:
$$
\begin{aligned}
10+20+30 & =25+35 \\
10+25+201 & <30+207
\end{aligned}
$$
First, consider the first weighing. We will prove that if some three coins balance som... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,189 |
1. Kolya came up with an entertainment for himself: he rearranges the digits in the number 2015, then places a multiplication sign between any two digits. In this process, none of the resulting two factors should start with zero. Then he calculates the value of this expression. For example: $150 \cdot 2=300$, or $10 \c... | Answer: 1050
Solution. Note three facts (their justification is obvious). First, in the case of the maximum product, the digits in each of the factors are in descending order. That is, 0 should be at the end of one of the factors. Second, it does not matter at the end of which of the two factors to place 0, the produc... | 1050 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,190 |
3. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha - 7 matches... | Answer: 4
Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.
Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and M... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,192 |
4. The numbers $x$ and $y$ are such that $x+y=xy=17$. Find the value of the expression:
$$
\left(x^{2}-17 x\right)\left(y+\frac{17}{y}\right)
$$ | Answer: -289
Solution. $x-17=-y, \frac{17}{y}=x$. The desired expression is:
$$
x(x-17)(y+x)=-x y \cdot 17=-289 .
$$
Criteria.
$(-)$
$(-$.
(-/+) Serious arithmetic error affecting the solution process, or several arithmetic errors.
$(+/ 2)$
(+/-) One arithmetic error in an otherwise correct solution (for exampl... | -289 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,193 |
5. Given a triangle $A B C, \angle B=90^{\circ}$. On the sides $A C, B C$ points $E$ and $D$ are chosen respectively, such that $A E=E C, \angle A D B=\angle E D C$. Find the ratio $C D: B D$. | Answer. $2: 1$
Solution. Construct triangle $A^{\prime} B C$, symmetric to the given one with respect to side $B C$. Points $A^{\prime}, D, E$ lie on the same line, since $\angle A^{\prime} D B=\angle E D C$. Therefore, $D$ is the point of intersection of the medians $A^{\prime} E$ and $C B$ of triangle $A A^{\prime} ... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,194 |
# 4. Problem $4.1 *$
In the class, 6 students received a grade of 5, 7 received a grade of 4, and 1 received a grade of 3. The teacher told them to form pairs with different grades, where the student with the better grade would explain to the student with the worse grade where they made a mistake.
In how many ways co... | # 4. Problem $4.1 *$
In the class, 6 students received a grade of 5, 7 received a grade of 4, and 1 received a grade of 3. The teacher told them to form pairs with different grades, where the student with the better grade would explain to the student with the worse grade where they made a mistake.
In how many ways co... | 5040 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,199 |
1. Kolya came up with an entertainment for himself: he rearranges the digits in the number 2015, then places a multiplication sign between any two digits and calculates the value of the resulting expression. For example: $150 \cdot 2=300$, or $10 \cdot 25=250$. What is the largest number he can get as a result of such ... | Answer: 1050
Solution. Note three facts (their justification is obvious). First, in the case of the maximum product, the digits in each of the factors are in descending order. That is, 0 should be at the end of one of the factors. Second, it does not matter at the end of which of the two factors to place 0, the produc... | 1050 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,206 |
2. The numbers $x$ and $y$ are such that $x+y=xy=17$. Find the value of the expression:
$$
\left(x^{2}-17 x\right)\left(y+\frac{17}{y}\right)
$$ | Answer: -289
Solution. $x-17=-y, \frac{17}{y}=x$. The desired expression is:
$$
x(x-17)(y+x)=-x y \cdot 17=-289
$$
Criteria.
$(-)$
$(-$.
(-/+) Serious arithmetic error affecting the solution process, or several arithmetic errors.
$(+/ 2)$
(+/-) One arithmetic error in an otherwise correct solution.
(+.) Omissi... | -289 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,207 |
3. Given a triangle $A B C, \angle B=90^{\circ}$. On the sides $A C, B C$ points $E$ and $D$ are chosen respectively, such that $A E=E C, \angle A D B=\angle E D C$. Find the ratio $C D: B D$. | Answer: $2: 1$
Solution. Construct triangle $A^{\prime} B C$, symmetric to the given one with respect to side $B C$. Points $A^{\prime}, D, E$ lie on the same line, since $\angle A^{\prime} D B = \angle E D C$. Therefore, $D$ is the point of intersection of the medians $A^{\prime} E$ and $C B$ of triangle $A A^{\prime... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,208 |
4. In the country of Lumumba, there are four national currencies: bananas (B), coconuts (K), raccoons $(\mathcal{E})$, and dollars (\$). The exchange rates for these currencies (the same in all exchange offices in the country) are given below:
$$
\mathrm{B} \underset{\frac{1}{2}}{\stackrel{2}{\rightleftarrows}} \mathr... | Solution. Consider the cycle $\mathrm{K} \rightarrow \mathcal{E} \rightarrow$ B $\rightarrow$ K. This cycle allows increasing the amount of currency. Indeed, if initially there were 11 coconuts, they can be exchanged for 1 raccoon, which in turn can be exchanged for 6 bananas, which can be exchanged for 12 coconuts. Th... | 200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,209 |
6. Find all natural numbers $n$ such that the number $n^{2}+77 n$ is a perfect square of a natural number. | Answer: 4, 99, 175, 1444.
Solution. FIRST METHOD. Let $n(n+77)=m^{2}$, where $m, n$ are natural numbers. Consider 4 cases:
- $\gcd(n, 77)=1$. In this case, the numbers $n$ and $n+77$ are coprime, and thus each of them is a perfect square: $n=p^{2}, n+77=q^{2}$. Hence, $q^{2}-p^{2}=77$, i.e., either $q-p=1, q+p=77$, o... | 4,99,175,1444 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,211 |
# 2. Problem 2
Angles $A, B, C$ of triangle $A B C$ are $44^{\circ}, 66^{\circ}$, and $70^{\circ}$ respectively. The bisector of angle $A B C$ and the perpendicular bisector of side $A C$ intersect at point $D$. How many degrees is angle $A D C$? (Write the answer without the degree symbol.)
# | # 2. Task 2
Angles $A, B, C$ of triangle $A B C$ are $44^{\circ}, 66^{\circ}$, and $70^{\circ}$ respectively.
The bisector of angle $A B C$ and the perpendicular bisector of side $A C$ intersect at point $D$. How many degrees does angle $A D C$ measure? (Write the answer without the degree symbol.)
## Answer: 114 | 114 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,213 |
# 9. Problem 9
Vasya throws three dice (cubes with numbers from 1 to 6 on the faces) and adds the numbers that come up. Additionally, if all three numbers are different, he can throw all three dice again and add the numbers that come up to the already accumulated sum, and continue doing so until at least two of the th... | # 9. Problem 9
Vasya throws three dice (cubes with numbers from 1 to 6 on the faces) and adds the numbers that come up. Additionally, if all three numbers are different, he can throw all three dice again and add the numbers that come up to the already accumulated sum, and continue doing so until at least two of the th... | 23.625 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,220 |
7.1. In the expression
$$
(* *+*)(* *+*)=* * * *
$$
insert digits instead of asterisks so that a correct equality is obtained and no more than 4 different digits are used. (A number cannot start with zero). | Solution. For example, $(90+9)(10+1)=1089$. Or: $(99+1)(10+9)=1900$. Criteria.
(+.) Correct left side.
$(+)$ Correct answer. | 1089 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,222 |
7.3. There are 4 watermelons, any two of which have different weights. Show how to find the two heaviest watermelons in 4 weighings using balance scales without weights. | Solution. First, let's show how to find the lightest watermelon out of 3 with 2 weighings. Let the weights of the 3 watermelons be $a, b, c$. Weigh watermelons $a$ and $b$. If $a > b$, then weigh $b$ and $c$. If $b > c$, then $c$ is the lightest, and if $b < c$, then $b$ is the lightest. If on the first weighing $a < b... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 20,224 |
7.4. An art gallery has the shape of a 9-sided polygon (not necessarily convex). It turned out that with any placement of two point light sources inside the gallery, some point of the gallery will remain unlit. Draw how such a gallery could look. Explain why two light sources are not enough to illuminate it. (The walls... | Solution. A possible gallery is shown in the figure. Indeed, note that to illuminate each of the three corners in each of the shaded areas, there must be a light source in each area. However, since the areas do not overlap, each of the three areas must have its own source. Therefore, there must be at least three.
Answer: 1. | Solution. We will provide two solutions to the problem.
First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:
$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,228 |
Problem 11-2. Mr. $A$ stood at the point with coordinates $(0,0)$ for an hour. During the same hour, moving uniformly and in a straight line, Mr. $B$ traveled from the point $(22,0)$ to the point $(2,20)$. During this same hour, Miss $C$, also moving uniformly and in a straight line, traveled from the point $(30,4)$ to... | Answer: 53.
Solution. Informally speaking, the problem in this task is not to find a path of calculations leading to the answer; but to find a path to the answer that passes through not too many intermediate calculations. Let's show how to do this.
As is known, the area of a triangle formed by vectors $\overrightarro... | 53 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,229 |
Problem 11-3. From $n$ regular hexagons with side 1, a polygon on the plane was made by gluing hexagons along their sides. Any two hexagons either have exactly one common side or have no common points at all. There are no holes inside the polygon. Moreover, each hexagon has at least one side lying on the boundary of th... | Answer: $2 n+6$ for $n \geq 2$, $6$ for $n=1$.
Solution. It is easy to understand that if a point is an endpoint for at least one side of a hexagon, it is an endpoint for two or three sides: more than three is impossible because all angles are $120^{\circ}$.
Let's draw a bit. Imagine that in each point that is an end... | 2n+6forn\geq2,6forn=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,230 |
Problem 11-4. Through the vertices of triangle $A B C$, three parallel lines $a, b, c$ are drawn respectively, not parallel to the sides of the triangle. Let $A_{0}, B_{0}, C_{0}$ be the midpoints of sides $B C, C A, A B$. Let $A_{1}, B_{1}, C_{1}$ be the points of intersection of the pairs of lines $a$ and $B_{0} C_{0... | Solution. We will provide a trigonometric solution. It is sufficient to prove that
$$
\frac{B_{0} A_{1}}{C_{0} A_{1}} \cdot \frac{C_{0} B_{1}}{A_{0} B_{1}} \cdot \frac{A_{0} C_{1}}{B_{0} C_{1}}=1
$$
Then, by Ceva's theorem, the lines $A_{0} A_{1}, B_{0} B_{1}, C_{0} C_{1}$ intersect at one point.
Let us assume, with... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,231 |
Problem 11-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not in... | Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,232 |
8.1. In the expression
$$
(* *+*)(* *+*)=* * * *
$$
insert digits instead of asterisks so that a correct equality is obtained and no more than 4 different digits are used. (A number cannot start with zero). | Solution. For example, $(90+9)(10+1)=1089$. Or: $(99+1)(10+9)=1900$.
Criteria.
(+.) Correct left side.
$(+)$ Correct answer. | 1089 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,234 |
8.2. It is known that none of the numbers $a, b, c$ are integers. Can it happen that each of the numbers $a b, b c, c a, a b c$ is an integer? | Solution. Yes, it can. For example, let's choose three different prime numbers $p_{1}$, $p_{2}$, $p_{3}$ and consider the numbers
$$
a=\frac{p_{1} p_{2}}{p_{3}}, b=\frac{p_{1} p_{3}}{p_{2}}, c=\frac{p_{2} p_{3}}{p_{1}}
$$
each of which is obviously not an integer. This triplet satisfies the conditions of the problem.... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,235 |
8.4. Two players play a game as follows: a chip is placed at point $A$ on the diagram shown below. They take turns moving the chip from the point where it stands to one of the points named on the diagram, connected to it by a line segment. No segment can be traversed more than once. The player who cannot make a move lo... | Solution. Let's call the degree of a vertex the number of segments coming out of it. Note that the game must end at vertex $A$. Indeed, the degree of vertex $A$ was 4 at the beginning, and it became 0 at the end, so it changed by an even number, which immediately implies that we must end precisely at it. We will prove ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,237 |
8.5. Along the shore of a round lake, apple trees grow. Petya and Vasya start walking from point $A$ on the shore in opposite directions along the shore and count all the apple trees they encounter, as well as all the apples growing on the trees. Meeting at some point $B$, they compared their results. It turned out tha... | Solution. Let there be $n$ apple trees growing along the shore in total. On the way from $A$ to $B$, Vasya counted twice as few apple trees as Petya, and together they counted all the apple trees growing on the shore. Therefore, Vasya counted $\frac{n}{3}$, and Petya - $\frac{2 n}{3}$ apple trees. Similarly, on the way... | 3 | Other | math-word-problem | Yes | Yes | olympiads | false | 20,238 |
8.6. The altitudes of an acute-angled scalene triangle $ABC$ intersect at point $H$. $I$ is the incenter of triangle $ABC$, $O$ is the circumcenter of triangle $BHC$. It is known that point $I$ lies on the segment $OA$. Find the angle $BAC$. | Answer: $60^{\circ}$
Solution. Note that point $O$ lies on the perpendicular bisector of segment $B C$. On the other hand, point $O$ lies on the bisector of angle $A$. However, these two lines intersect at a point lying on the circumcircle of triangle $A B C$, and dividing the arc $B C$ in half. Therefore, point $O$ l... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,239 |
# 1. Task 1
On the coordinate plane, points $A=(-3,-4), C=(1.5,-2)$ are given. Find the minimum length of the broken line $A B C$, where point $B$ lies on the horizontal coordinate line $O x$. If the answer is a fractional number, it should be written as a decimal using a dot, for example, $« 0.15 »$. | # 1. Task 1
On the coordinate plane, points $A=(-3,-4), C=(1.5,-2)$ are given. Find the minimum length of the broken line $A B C$, where point $B$ lies on the horizontal coordinate line $O x$. If the answer is a fractional number, it should be written as a decimal using a dot, for example, $« 0.15 »$.
## Answer: 7.5 | 7.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,240 |
9.1. It is known that none of the numbers $a, b, c$ are integers. Can it happen that each of the numbers $a b, b c, c a, a b c$ is an integer? | Answer. It can.
Solution. For example, let's choose three different prime numbers $p_{1}, p_{2}, p_{3}$ and consider the numbers
$$
a=\frac{p_{1} p_{2}}{p_{3}}, b=\frac{p_{1} p_{3}}{p_{2}}, c=\frac{p_{2} p_{3}}{p_{1}}
$$
each of which is obviously not an integer. This triplet satisfies the conditions of the problem.... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,250 |
9.2. Point $B$ is the midpoint of segment $A C$. Square $A B D E$ and equilateral triangle $B C F$ are located in the same half-plane relative to line $A C$. Find (in degrees) the measure of the acute angle between lines $C D$ and $A F$. | Answer: $75^{\circ}$.
Solution. Note that points $A D F C$ lie on the same circle with center at point $B$. The arc $F C$ is $60^{\circ}$. Therefore, the angle $F A C$ is $30^{\circ}$ (by the inscribed angle theorem). Triangle $B D C$ is isosceles with a right angle at vertex $B$. Therefore, the angle $D C A$ is $\pi ... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,251 |
9.3. Two players play a game as follows: a chip is placed at point $A$ on the diagram shown below. They take turns moving the chip from the point where it stands to one of the points named on the diagram, connected to it by a line segment. No segment can be traversed more than once. The player who cannot make a move lo... | Answer. The second player wins.
Solution. Note that the game must end at vertex $A$. Indeed, the degree of vertex $A$ was 4 at the beginning, and 0 at the end, so it changed by an even number, which immediately implies that we must end precisely at this vertex. We will prove that by the end of the game, all edges have... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,252 |
9.4. The lines containing the altitudes of the non-isosceles triangle \(ABC\) intersect at point \(H\). \(I\) is the incenter of triangle \(ABC\), \(O\) is the circumcenter of triangle \(BHC\). It is known that point \(I\) lies on the segment \(OA\). Find the angle \(BAC\). | Answer: $60^{\circ}$
Solution. Note that point $O$ lies on the bisector (perpendicular through the midpoint) of segment $B C$. On the other hand, point $O$ lies on the bisector of angle $A$. However, these two lines intersect at a point lying on the circumcircle of triangle $B A C$, and dividing the arc $B C$ in half.... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,253 |
9.5. The cells of a chessboard are painted in 3 colors - white, gray, and black - in such a way that adjacent cells, sharing a side, must differ in color, but a sharp change in color (i.e., a white cell adjacent to a black cell) is prohibited. Find the number of such colorings of the chessboard (colorings that coincide... | Solution. To avoid confusion with the natural coloring of the chessboard, let's choose three colors: red, blue, and gray. Suppose the field a1 is painted gray. Then, according to our coloring rules, all black fields of the board (recall that a1 is a black field) automatically become gray, while the color of the remaini... | 2^{33} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,254 |
9.6. The sequence $a_{n}$ is constructed as follows: $a_{1}, a_{2}$ are arbitrary real numbers, and for $n \geq 3$, the number $a_{n}$ is the smallest of the numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1$. For example, if $a_{1}=6, a_{2}=\frac{19}{2}$, then we get the sequence $6, \frac{19}{2}, \frac{7}{2}, \f... | Solution. Let $S_{n}$ be the set of all numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1, n \geq 3$. Then $a_{n}$ is the smallest number in the set $S_{n}$. Since $S_{n-1} \subset S_{n}$, then $a_{n-1} \geq a_{n}$, i.e., starting from $a_{3}$, the sequence is non-increasing. Therefore, $a_{9} \geq a_{10}=1$. Furt... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,255 |
Problem 7.1. (15 points) Find the smallest ten-digit natural number, all digits of which are different, such that when all even digits are erased, 97531 remains, and when all odd digits are erased, 02468 remains. | Answer: 9024675318.
Solution. Let $A=9024675318$. Consider an arbitrary number $N=\overline{a_{1} a_{2} \ldots a_{10}}$ that satisfies the conditions of the problem. Clearly, $a_{1}=9$, since $a_{1} \neq 0$. Let $7=a_{k}$ for some $k$. We will consider several cases.
- If $k>6$, then $N=9024687531>A$.
- If $kA$.
- If... | 9024675318 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,256 |
Problem 7.2. (15 points) Given an isosceles triangle $A B C(A B=B C)$. On the sides $A B, B C, A C$ points $K, L, M$ are marked respectively such that $\angle A K M=90^{\circ}, \angle B L K=90^{\circ}$ and $K M=K L$. What is the measure of angle $C M L ?$ | Answer: $90^{\circ}$.
Solution. From the condition, it follows that $\angle M K L + \angle L K B = 90^{\circ} = \angle L K B + \angle L B K$, from which $\angle M K L = \angle L B K$ (Fig. 1). Triangles $A B C$ and $M K L$ are isosceles with equal angles at the vertex, so the base angles are also equal: $\angle K L M ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,257 |
Problem 7.3. (15 points) Several boxes are stored in a warehouse. It is known that there are no more than 60 boxes, and each of them contains either 59 apples or 60 oranges. After a box with a certain number of oranges was brought to the warehouse, the number of fruits in the warehouse became equal. What is the smalles... | Answer: 30.
Solution. Let the number of boxes with oranges be denoted by $n$, the number of boxes with apples by $m$, and the number of oranges in the brought box by $x$. According to the problem, $59 m = 60 n + x$. By moving $59 n$ to the left side, we get $59(m - n) = n + x$. From this, it follows that $n + x$ is a ... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,258 |
Problem 7.4. (15 points) On an island, there live knights who always tell the truth, and liars who always lie. One day, 100 residents of this island lined up, and each of them said one of the following phrases:
- "To the left of me, there are as many liars as there are knights."
- "To the left of me, there are 1 more ... | Answer: 50.
Solution. Suppose there are no fewer than 51 knights in the row. Consider the rightmost of them. To his left, there are at least 50 knights and no more than 49 liars, so he could not have said any of the listed phrases, leading to a contradiction. Therefore, the total number of knights cannot exceed 50.
N... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,259 |
Problem 7.5. (20 points) Andrey wrote down 6 consecutive four-digit numbers in a row in ascending order on the board. Then he wrote one of their prime divisors under each of these numbers, and all the prime divisors turned out to be different. After that, Andrey erased the original 6 numbers and invited Boris into the ... | Answer: always.
Solution. Let the smallest of the numbers initially written by Andrei be denoted by $x$. Then the next numbers are $x+1, x+2, x+3, x+4, x+5$ respectively. Let their prime divisors be $p_{0}, p_{1}, \ldots, p_{5}$ respectively.
Suppose Boris cannot uniquely reconstruct Andrei's numbers. This means that... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,260 |
4. To protect valuable equipment from bad weather, a tent in the shape of a pyramid is required, with a rectangle as its base, and one of the lateral edges should be perpendicular to the base. Find the maximum possible volume of the tent, given that no edge of the pyramid should be longer than 2 meters.
$8 /(9 \sqrt{3... | Solution. (1) Let the base be a rectangle with sides $x$ and $y$, and height $z$. Then the conditions of the problem are equivalent to finding $x, y, z$ such that $\frac{x y z}{3} \Rightarrow \max$ while simultaneously satisfying the conditions $x>0, y>0, z>0, x^{2}+y^{2}+z^{2} \leqslant 4$. (2) Let the tent of maximum... | \frac{8}{9\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,262 |
7. On side $AB$ of triangle $ABC$, a point $D$ is taken. In triangle $ADC$, the angle bisectors $AP$ and $CQ$ are drawn. On side $AC$ of triangle $ADC$, a point $R$ is taken such that $PR \perp CQ$. It is known that the angle bisector of angle $D$ of triangle $BCD$ is perpendicular to segment $PB$, $AB=18$, $AP=12$. Fi... | Solution. Let $A B=n, A P=m, A D=c, A C=b, D P=p, C P=q, B D=D P=p, C P=C R=q$.
Using the property of the angle bisector and the formula for the length of the angle bisector, we get $\left\{\begin{array}{l}\frac{c}{b}=\frac{p}{q}, \\ c+p=A B=n, \\ m^{2}=b c-p q,\end{array}\right.$
$\left\{\begin{array}{l}c=\frac{b p}... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,264 |
8. Find all minimum points and maximum points of the function $f(x)=\left(x^{2}-3 x-9\right) \cdot \sqrt{x^{2}-6 x+9}$. $x_{\min } \in\{0 ; 4\}, x_{\max } \in\{3\}$. | Solution. (1) $\left(x^{2}-3 x-9\right) \cdot \sqrt{x^{2}-6 x+9}=\left(x^{2}-3 x-9\right) \cdot \sqrt{(x-3)^{2}}=\left(x^{2}-3 x-9\right) \cdot |x-3|$ $=\left\{\begin{array}{c}-\left(x^{2}-3 x-9\right)(x-3) \text { for } x \leqslant 3, \\ \left(x^{2}-3 x-9\right)(x-3) \text { for } x \geqslant 3\end{array}=\left\{\begi... | x_{\}\in{0;4},x_{\max}\in{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 20,265 |
10. All edges of the regular quadrilateral pyramid $S A B C D$ are equal to $b$. The height of the regular quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ is equal to the height of the specified pyramid. The base $A B C D$ is common to both. (1) A plane $\alpha$ is drawn through points $A$ and $C_{1}$ parallel to... | # Solution.
(1) $B D=b \sqrt{2}, D Q=Q S, B P=P S, P Q \| B D$. Triangles $S B D$ and $S P Q$ are similar with a similarity ratio of $2: 1: P Q=B D / 2=b / \sqrt{2}$.
(2) $A C_{1}=b \sqrt{5 ... | b^{2}\frac{\sqrt{5}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,266 |
1. Three different positive numbers are three consecutive terms of an arithmetic progression. Can these same three numbers be three (not necessarily consecutive) terms of a geometric progression? | Answer. They can.
Solution. Consider the geometric progression $1, q, q^{2}, q^{3}$. Choose $q$ such that the numbers $1, q, q^{3}$ form an arithmetic progression. For this, the equation $1+q^{3}=2 q \Leftrightarrow\left(q^{2}+q-1\right)(q-1)=0$ must be satisfied. One of the roots of this equation is $q=\frac{\sqrt{5}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,267 |
2. A circle is circumscribed around triangle $A B C$ with angle $\angle B=60^{\circ}$. Tangents to the circle, drawn at points $A$ and $C$, intersect at point $B_{1}$. On the rays $A B$ and $C B$, points $A_{0}$ and $C_{0}$ are marked respectively such that $A A_{0}=A C=C C_{0}$. Prove that points $A_{0}, C_{0}, B_{1}$... | Solution. By the theorem on the angle between a tangent and a chord, we have $\angle A C B_{1}=\angle C A B_{1}=$ $\angle A B C=60^{\circ}$, i.e., triangle $A B_{1} C$ is equilateral. Then $A A_{0}=A C=A B_{1}$, i.e., triangle $A_{0} A B_{1}$ is isosceles. If we denote $\angle B A C=\alpha, \angle B C A=\beta$, then we... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,268 |
3. The function $f(x)$, defined for all real $x$, is even. Moreover, for any real $x$, the equality
$$
f(x)+f(10-x)=4
$$
a) Provide an example of such a function that is not a constant.
b) Prove that any such function is periodic. | Solution.
a) For example, $f(x)=2+\cos \left(\frac{\pi x}{10}\right)$. The evenness is obvious, let's check the second condition:
$f(x)+f(10-x)=4+\cos \left(\frac{\pi x}{10}\right)+\cos \left(\frac{\pi(10-x)}{10}\right)=4+\cos \left(\frac{\pi x}{10}\right)+\cos \left(\pi-\frac{\pi x}{10}\right)=0$,
since $\cos (\pi-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,269 |
4. Petya wants to test the knowledge of his brother Kolya, the winner of the "Higher Test" mathematics Olympiad. For this, Petya thought of three natural numbers $a, b, c$, and calculated $x = \text{GCD}(a, b)$, $y = \text{GCD}(b, c)$, $z = \text{GCD}(c, a)$. Then he wrote three rows of five numbers each on the board:
... | Answer. $x=8, y=14, z=18$.
Solution. We will use the following statement:
Lemma. If two of the numbers $x, y, z$ are divisible by some natural number $m$, then the third is also divisible by $m$.
Proof. Suppose, for example, $x$ and $y$ are divisible by $m$.
$$
\left\{\begin{array} { l }
{ x : m \Rightarrow a : m ... | 8,14,18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,270 |
5. Two corridors, each 1 m high and 1 m wide, run perpendicular to each other on the first and second floors of a building. The separating floor has been removed, creating a 1 m x 1 m hole in the floor of one and the ceiling of the other. What is the maximum length of the beam that can be passed from one corridor to th... | Answer: $\frac{1}{2} \cdot(2+\sqrt[3]{4})^{\frac{3}{2}}$.
## Solution.
Let $A$ and $B$ be the ends of the beam, with $B$ being the lower end, which first enters the lower corridor (we assume the beam is being transferred from the upper corridor to the lower one). For brevity, let $d=\frac{1}{2} \cdot(2+\sqrt[3]{4})^{... | \frac{1}{2}\cdot(2+\sqrt[3]{4})^{\frac{3}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,271 |
6. An $n \times n$ table is filled with natural numbers from 1 to 2016 such that no two identical numbers appear in the same row or column. Coincidence of numbers in different rows and columns is allowed. Let $f(n)$ be the number of such arrangements. For example, $f(1)=2016, f(2017)=0$.
a) Which is greater, $f(2015)$... | Answer. a) $f(2015)>f(2016)$, b) $f(1009)>f(1008)$.
Solution. Let $S_{n}$ denote the set of all required arrangements for an $n \times n$ table. Then $f(n)$, by definition, is the number of elements in the set $S_{n}$.
We introduce an operation $g$ over a table, which consists of removing the last (rightmost) column ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,272 |
1. Real numbers $x, y, z$ are chosen such that the following equalities hold: $x y + y z + z x = 4$, $x y z = 6$. Prove that for any such choice, the value of the expression
$$
\left(x y - \frac{3}{2}(x + y)\right)\left(y z - \frac{3}{2}(y + z)\right)\left(z x - \frac{3}{2}(z + x)\right)
$$
is always the same number,... | Answer: $\frac{81}{4}$.
Solution: From the condition, it follows that $x, y, z$ are non-zero numbers. From the given equalities, we get $x y=\frac{6}{z}$ and $x+y=\frac{4-x y}{z}=\frac{4}{z}-\frac{x y}{z}$. Substituting this into the first parenthesis, we obtain
$$
x y-\frac{3}{2}(x+y)=\frac{6}{z}-\frac{3}{2}\left(\f... | \frac{81}{4} | Algebra | proof | Yes | Yes | olympiads | false | 20,273 |
2. In the country of Limpopo, there are four national currencies: bananas (B), coconuts (K), raccoons $(\mathcal{E})$, and dollars (\$). The exchange rates for these currencies (the same in all exchange points in the country) are given below:
---
The number on the arrow indicates how many units, specified at the end ... | # Answer. It can.
Solution. Note that the problem statement does not explicitly state whether it is required to obtain no less than $200 \$$, or exactly $200 \$$$. The answer is positive for both interpretations. Although it is clear that a solution for the case "exactly $200 \$$" also solves the second case, we will ... | 200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,274 |
3. Given three points $A, B, C$, forming a triangle with angles $30^{\circ}, 45^{\circ}, 105^{\circ}$. Two of these points are chosen, and the perpendicular bisector of the segment connecting them is drawn, after which the third point is reflected across this perpendicular bisector. This results in a fourth point $D$. ... | Answer: 12 points.
## Solution. FIRST WAY. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,275 |
4. Provide an example of a function $f(x)$ that satisfies all three of the following conditions:
- the domain of the function $f(x)$ is the set of all real numbers $\mathbb{R}$,
- for any $b \in \mathbb{R}$, the equation $f(x)=b$ has exactly one solution,
- for any $a>0$ and any $b \in \mathbb{R}$, the equation $f(x)=... | Solution. Let's check the conditions.
- It follows directly from the definition that the function is defined for all $x \in \mathbb{R}$.
- If $b \neq 0$, then for $x \neq 0$ the equation $f(x)=b$ has a unique solution $x=\frac{1}{b}$, and $x=0$ is not a solution. That is, we have exactly one solution.
If $b=0$, then ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,276 |
5. Let $T_{k}$ denote the product of the first $k$ odd prime numbers: $T_{1}=3, T_{2}=$ $3 \cdot 5, T_{6}=3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$ and so on. For each natural number $k$, find the number of natural numbers $n$ such that the number $n^{2}+T_{k} n$ is a perfect square of a natural number. Solve the p... | Answer. $\frac{3^{k}-1}{2}$.
Solution. FIRST METHOD. Let's write the condition of the problem as an equation
$$
n^{2}+T_{k} n=m^{2} \text { for some } m \in \mathbb{N} \text {, }
$$
Let $d=\operatorname{GCD}\left(n, T_{k}\right), \quad \nu=\frac{n}{d}, r=\frac{T_{k}}{d}$. Then the numbers $d, \nu, r$ satisfy the sys... | \frac{3^{k}-1}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,277 |
6. In space, there are 270 spheres of equal radii, any two of which intersect. Prove that among them, one can select 10 spheres such that there exists a point belonging to all the selected spheres. | Solution. FIRST METHOD. Let the radii of the spheres be 1 (obviously, the specific value of the radii does not affect the solution). Choose one of them, let its center be \( O \). Then the centers of all other spheres are at a distance of no more than 2 from \( O \), and therefore the distance from \( O \) to any point... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 20,278 |
# 2. Task 2*
The number 2017 has 7 ones and 4 zeros in its binary representation. When will the next year come, in which the number of the year in binary representation will have no more ones than zeros? (Enter the year.) Points for the task: 8.
# | # 2. Task 2*
The number 2017 has 7 ones and 4 zeros in its binary representation. When will the next year come, in which the number of the year in binary representation will have no more ones than zeros? (Enter the year.) Points for the task: 8.
## Answer: 2048 | 2048 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,280 |
# 7. Problem 7*
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Find ... | # 7. Problem $7 *$
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fi... | 806 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,285 |
# 3. Problem 3.10
Given four numbers $a, b, c, d$. Each pair of them was added together to get 6 sums: $a+b, b+c, c+d, a+c, b+d, a+d$. The four smallest of these sums are $7, 15, 43, 47$. Find the sum of the original four numbers $a+b+c+d$. If the answer is not an integer, write it as a decimal. | # 3. Problem 3.10
Given four numbers $a, b, c, d$. Each pair of them was added together to get 6 sums: $a+b, b+c, c+d, a+c, b+d, a+d$. The four smallest of these sums are $7, 15, 43, 47$. Find the sum of the original four numbers $a+b+c+d$. If the answer is not an integer, write it as a decimal. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,291 |
# 8. Problem 8.10
Pentagon $A B C D E$ is circumscribed around a circle.
Angles $\angle A B C, \angle B A E, \angle C D E$ are each $104^{\circ}$. Find $\angle A D B$. Write the answer in degrees (the number only, without specifying the unit of measurement).
# | # 8. Problem 8.10
Pentagon $A B C D E$ is circumscribed around a circle.
Angles $\angle A B C, \angle B A E, \angle C D E$ are each $104^{\circ}$. Find $\angle A D B$. Write the answer in degrees (the number only, without the unit of measurement).
## Answer: 38 | 38 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,296 |
# 9. Problem 9.10
In a convex 10-gon \(A_{1} A_{2} \ldots A_{10}\), all sides and all diagonals connecting vertices every other one (i.e., \(A_{1} A_{3}, A_{2} A_{4}\), etc.) are drawn, except for the side \(A_{1} A_{10}\) and the diagonals \(A_{1} A_{9}, A_{2} A_{10}\).
We call a path leading from \(A_{1}\) to \(A_{... | # 9. Problem 9.10
In a convex 10-gon $A_{1} A_{2} \ldots A_{10}$, all sides and all diagonals connecting vertices every other one (i.e., $A_{1} A_{3}, A_{2} A_{4}$, etc.) are drawn, except for the side $A_{1} A_{10}$ and the diagonals $A_{1} A_{9}, A_{2} A_{10}$.
A path from $A_{1}$ to $A_{10}$ is defined as a non-se... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,297 |
# 1. Problem 1*
Let $S$ be the sum of the digits of the number $11^{2017}$. Find the remainder when $S$ is divided by 9. Points for the problem: 8.
## Answer: 2
# | # 1. Problem 1*
Let $S$ be the sum of the digits of the number $11^{2017}$. Find the remainder when $S$ is divided by 9. Points for the problem: 8.
## Answer: 2
# | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,299 |
# 2. Task 2*
The number 2017 has 7 ones and 4 zeros in its binary representation. When will the next year come, in which the number of the year in binary representation will have no more ones than zeros? Enter the year. Points for the task: 8.
## Answer: 2048 | # 2. Task 2*
The number 2017 has 7 ones and 4 zeros in its binary representation. When will the next year come, in which the number of the year in binary representation will have no more ones than zeros? Enter the year. Points for the task: 8.
## Answer: 2048 | 2048 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,300 |
# 7. Problem $7 *$
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fi... | # 7. Problem $7 *$
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fi... | 806 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,305 |
3. Each move of a chess knight is a move of one square horizontally and two squares vertically, or vice versa - one square vertically and two squares horizontally. (In the diagram on the right, the knight marked with the letter $\mathrm{K}$ can move to any of the shaded squares in one move.)
A chess knight is placed i... | Answer: 2
Solution. Let's number the cells as shown in Figure 1. The knight can move from any cell to any cell with the same number, and cannot move to other cells.
| $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,309 |
4. The function $f(x)$, defined for all real $x$, is even. Moreover, for any real $x$, the equality
$$
f(x) + f(10 - x) = 4
$$
a) Provide an example of such a function that is not a constant.
b) Prove that any such function is periodic. | Solution.
a) For example, $f(x)=2+\cos \left(\frac{\pi x}{10}\right)$. The evenness is obvious, let's check the second condition:
$f(x)+f(10-x)=4+\cos \left(\frac{\pi x}{10}\right)+\cos \left(\frac{\pi(10-x)}{10}\right)=4+\cos \left(\frac{\pi x}{10}\right)+\cos \left(\pi-\frac{\pi x}{10}\right)=0$,
since $\cos (\pi-... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,310 |
5. Petya wants to test the knowledge of his brother Kolya, the winner of the "Higher Test" mathematics Olympiad. For this, Petya thought of three natural numbers $a, b, c$, and calculated $x = \text{GCD}(a, b)$, $y = \text{GCD}(b, c)$, $z = \text{GCD}(c, a)$. Then he wrote three rows of five numbers each on the board:
... | Answer. $x=8, y=14, z=18$.
Solution. We will use the following statement: If two of the numbers $x, y, z$ are divisible by some natural number $m$, then the third is also divisible by $m$.
Proof. Let's say, for example, $x$ and $y$ are divisible by $m$.
$$
\left\{\begin{array} { l }
{ x : m \Rightarrow a : m , b : ... | 8,14,18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,311 |
6. A table of size $n \times n$ is filled with natural numbers from 1 to 10 such that no two identical numbers appear in the same row or column. Coincidence of numbers in different rows and columns is allowed. Let $f(n)$ be the number of such arrangements. For example, $f(1)=10, f(11)=0$.
a) Which is greater, $f(9)$ o... | Answer. a) $f(9)>f(10)$, b) $f(6)>f(5)$.
Solution. Let $S_{n}$ denote the set of all required arrangements for an $n \times n$ table. Then $f(n)$, by definition, is the number of elements in the set $S_{n}$.
We introduce an operation $g$ on a table, which consists of removing the last (rightmost) column and the last ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,312 |
# Problem 1. (Stern A.)
Vasya added the same natural number, which is less than both the numerator and the denominator, to the numerator and the denominator of a proper fraction. As a result, the fraction increased by more than 50%. Vasya claims that if he subtracts this number from the numerator and the denominator o... | Answer: No. Solution:
$\left\{\begin{array}{l}\frac{a+n}{b+n}>\frac{3}{2} \frac{a}{b} \\ \frac{a-n}{b-n}>\frac{1}{2} \frac{a}{b}\end{array} \Rightarrow\left\{\begin{array}{c}2 a b+2 n b>3 a b+3 a n \\ 2 a b-2 n b>a b-a n\end{array} \Rightarrow 4 a b>4 a b+2 a n\right.\right.$, which is impossible
Criteria: Correct so... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,313 |
# Problem 2. (Folklore)
In a box, there are balls of two colors: blue and red (both colors are present). It is known that there are more blue balls, and two balls of the same color can be drawn with the same probability as two balls of different colors. What can the difference between the number of blue and red balls ... | Answer: Any natural number greater than 1. Solution:
Let blue be $m$, red be $n(m>n)$.
The probability of drawing two balls of the same color $\left(\frac{n}{n+m}\right)\left(\frac{n-1}{n+m-1}\right)+\left(\frac{m}{n+m}\right)\left(\frac{m-1}{n+m-1}\right)$
The probability of drawing two balls of different colors $\... | Anynaturalgreaterthan1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,314 |
# Problem 3. (Stern A.)
In a row, 2020 natural numbers are arranged such that among any six consecutive numbers, the first number is divisible by the last, and among any nine consecutive numbers, the last number is divisible by the first. Prove that the sum of the first hundred numbers is divisible by the sum of the l... | # Solution:
Instead of "a is divisible by $b$", it is sufficient that $a \geq b$, that is, $a_{n} \geq a_{n+5}, a_{n} \leq a_{n+8}$
Indeed, $a_{1} \geq a_{1+5 * 5=26} \geq a_{26-3 * 8=2}, \quad a_{2} \geq a_{17} \geq a_{1} \Rightarrow a_{1}=a_{2}$
That is, for any part of the sequence of 26 elements $a_{1}=a_{2}=a_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 20,315 |
# Problem 4. (Kuyanov 9.)
Find all quadruples of natural numbers $a, b, c, d$ for which the following equations are satisfied:
\[
\left\{
\begin{array}{l}
a+b=c d \\
c+d=a b
\end{array}
\right.
\] | Answer: $(2,2,2,2),(1,2,3,5),(2,1,3,5),(1,2,5,3),(2,1,5,3),(3,5,1,2),(5,3,1,2),(3,5,2,1),(5,3,2,1)$
Solution:
$\left\{\begin{array}{l}0=c d-a-b \\ 0=a b-c-d\end{array} \Rightarrow(a-1)(b-1)+(c-1)(d-1)=(c d-a-b)+(a b-c-d)+2=2\right.$
Each of $(a-1)(b-1)$ and $(c-1)(d-1)$ is a non-negative integer. If this is:
1 and ... | (2,2,2,2),(1,2,3,5),(2,1,3,5),(1,2,5,3),(2,1,5,3),(3,5,1,2),(5,3,1,2),(3,5,2,1),(5,3,2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,316 |
# Problem 5. (based on materials from the Ural Tournaments)
In triangle $ABC$, angle $B$ is $80^{\circ}$. On side $BC$, point $D$ is marked such that $AB = AD = CD$. On side $AB$, point $F$ is marked such that $AF = BD$. On segment $AC$, point $E$ is marked such that $AB = AE$. Find angle $AEF$.
 $n=404$? b) $n=406... | Answer: 202 colors in both parts. Solution:
Example for part a) with 202 colors:
Let's call a 198-block a sequence of 198 consecutive houses, where any pair of houses 99 apart is painted in a unique color (i.e., among all 404 houses, there are no other houses of that color).
Let's call a 2-block a pair of houses pai... | 202 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,318 |
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