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4. In city $\mathrm{N}$, only blondes and brunettes live. Blondes always lie, while brunettes always tell the truth. Every day, the residents dye their hair the opposite color. On one Monday in October, everyone born in the fall was asked: "Were you born this month?" - and 200 residents answered "yes," while no one ans... | 4. Answer. 0.
Solution. Note that on Monday and Friday of the same week, the residents have the same hair color: Monday - Tuesday - Wednesday - Thursday - Friday. Then, if the same people answer differently on Monday and Friday, it means that the month has changed, and the brunettes will answer "no" to the same questi... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. A tournament of dodgeball was held at school. In each game, two teams competed. 15 points were awarded for a win, 11 for a draw, and no points for a loss. Each team played against each other once. By the end of the tournament, the total number of points scored was 1151. How many teams were there? | 5. Answer: 12 teams.
Solution. Let there be $\mathrm{N}$ teams. Then the number of games was $\mathrm{N}(\mathrm{N}-1) / 2$. For each game, a total of 15 or 22 points are scored. Therefore, the number of games was no less than 53 $(1151 / 22)$ and no more than $76(1151 / 15)$.
Note that if there were no more than 10 ... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. A polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-8$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smallest... | Answer: (a) -24. (b) 6.
Solution. (a) Since the graph of $G(x)$ is symmetric with respect to the line $x=-8$, the points at which it takes the same value must be divided into pairs of symmetric points, except possibly one point that lies on the axis of symmetry. Therefore, the middle of the five given points should li... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.3.1. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-8$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -24 . (b) 6. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.3.4. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-6$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -18 . (b) 6. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.7.4. Let $N$ be the least common multiple of ten different natural numbers $a_{1}<a_{2}<a_{3}<\ldots<a_{10}$.
(a) (2 points) What is the smallest value that $N / a_{6}$ can take?
(b) (2 points) Identify all possible values of $a_{6}$ in the interval $[1 ; 2000]$, for which the value of $N / a_{1}$ can take... | Answer: (a) 5. (b) $504,1008,1512$. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the $3000-$th year, the World Hockey Championship will be held under new rules: 12 points will be awarded for a win, 5 points will be deducted for a loss, and no points will be awarded for a draw. If the Brazilian national team plays 38 matches in this championship, scores 60 points, and loses at least once, how ... | Solution. Let Brazil win in x matches and lose in y matches. We form the equation $12 x-5 y=60$. We see that $12 \mathrm{x} \vdots 12$ and $60 \vdots 12$. GCD(5, 12)=1, i.e., $y \vdots 12$. Possible: a) $y=12$. Then we get the equation $12 x-60=60$. Thus, $x=10$. This is possible. b) $y=24$. We get the equation: $12 x-... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Find all natural $k$ such that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first).
(V. Senderov) | Answer. $k=1$.
Solution. Let $n \geqslant 2$, and $2=p_{1}<p_{2}<\ldots<p_{k}$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation $(*)$ would be divisible by $q$, which is impossible. Therefore, $a>p_{k}$.
Without loss of generality, we can ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. On the board, a certain natural number $N$ was written nine times (one under the other). Petya appended a non-zero digit to the left or right of each of the 9 numbers; all the appended digits are different. What is the maximum number of prime numbers that could result among the 9 obtained numbers?
(I. Efremov) | Answer: 6.
Solution. Let $S$ be the sum of the digits of the number $N$. Then the sums of the digits of the obtained numbers will be $S+1, S+2, \ldots, S+9$. Three of these sums will be divisible by 3. By the divisibility rule for 3, the corresponding three numbers on the board will also be divisible by 3. Since these... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.4. All cells of a $100 \times 100$ square table are numbered in some order with numbers from 1 to 10000. Petya colors the cells according to the following rules. Initially, he colors $k$ cells at his discretion. Then, on each move, Petya can color one more uncolored cell with number $a$ if at least one of the two con... | Answer. $k=1$.
Solution. First, let's prove the following statement.
Lemma. For any two cells $A$ and $B$, there exists a cell $C$ such that by coloring it, one can then color both $A$ and $B$ (possibly $C$ coincides with $A$ or $B$).
Proof. We can assume that the number $a$ of cell $A$ is less than the number $b$ o... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.2 About numbers a and b, it is known that the system of equations
$$
\left\{\begin{array}{l}
y^{2}=x^{2}+a x+b \\
x^{2}=y^{2}+a y+b
\end{array}\right.
$$
has no solutions. Find a. | Solution: Since the system has no solutions, in particular, there are no solutions with $x=y$. When $x=y$, both equations of the system are equivalent to the equation $a x+b=0$. This linear equation has no roots only when its slope coefficient $a$ is zero.
## Criteria:
- Points are not deducted for the absence of an ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4 Another participant in the competition for meaningless activity marked the centers of 13 cells in a grid rectangle of size $(N-1) \times(N+1)$ such that the distance between any two marked points is greater than 2. What is the smallest value that $N$ can take? | Solution: We will show that it is impossible to mark cells in a $6 \times 8$ rectangle (and thus in any smaller size) in such a way. Indeed, let's divide the rectangle into $2 \times 2$ squares. In each of them, the pairwise distances between the centers of the cells do not exceed $\sqrt{2}$, so no more than one cell i... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.3. Consider natural numbers $a, b$, and $c$ such that the fraction
$$
k=\frac{a b+c^{2}}{a+b}
$$
is a natural number less than $a$ and $b$. What is the smallest number of natural divisors that the number $a+b$ can have?
(P. Kozlov) | Answer. Three divisors.
First solution. Since the number $a+b$ is greater than one, it has at least two distinct divisors. We will prove that there cannot be exactly two, i.e., that the number $a+b$ cannot be prime. Multiplying the equality from the condition by the denominator, we get $a b+c^{2}=k a+k b$ or, equivale... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. Natural numbers starting from 1 are written in a row. This results in a sequence of digits: 1234567891011121314... What digit is in the 2021st position? | Answer. 1.
Solution. Note that the sum of the digits of all single-digit and two-digit numbers is $1892021$. Therefore, the digit in the 2021st position belongs to the recording of some three-digit number.
Let $x$ be some three-digit number, then the sum of the digits in the sequence from 1 to $x$ is $n=189+3(x-99)$.... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The bathtub fills with cold water in 6 minutes 40 seconds, with hot water - in 8 minutes. In addition, if the plug is removed from a full bathtub, the water will drain in 13 minutes 20 seconds. How long will it take to fill the bathtub completely, provided that both taps are open, but the bathtub is not plugged? | Solution: First, we will convert the time in seconds to minutes: 6 minutes 40 seconds will be replaced by $6+2 / 3$, or $20 / 3$, and 13 minutes 20 seconds will be replaced by $13+1 / 3$, or $40 / 3$. Then, in one minute, the cold water will fill $3 / 20$ of the bathtub, the hot water will fill $1 / 8$ of the bathtub, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Six people are standing in a circle, each of whom is either a knight - who always tells the truth, or a liar - who always lies. Each of them said one of two phrases: "There is a liar next to me" or "There is a liar opposite me." What is the minimum number of liars among them? Provide an example and prove that there ... | Answer: 2.
Solution. Let's number all the people standing clockwise (this way, people with numbers 1 and 4, 2 and 5, 3 and 6 will stand opposite each other).
Zero liars is obviously impossible (then there would be only knights and no one could say any of the phrases).
If there is one liar, let's say his number is 1,... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2. In Solar City, 6 dwarfs eat donuts daily, 8 dwarfs eat donuts every other day, and the rest do not eat donuts at all. Yesterday, 11 dwarfs ate donuts. How many dwarfs will eat donuts today? | Answer: 9.
Solution: Of the 11 dwarfs who ate donuts yesterday, 6 dwarfs eat them daily, so the remaining $11-6=5$ eat them every other day. Therefore, these five will not eat donuts today, while the other $8-5=3$ from those who eat every other day will. So today, these three will eat donuts, as well as the six who al... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. A round table was sat by 10 people - liars and knights. Liars always lie, while knights always tell the truth. Each of them was given a coin. Then each of those sitting passed their coin to one of their two neighbors. After that, 5 people said: “I have one coin,” while the other 5 said: “I have no coins.” What is ... | # Answer: 7.
Solution. After passing the coins, each person sitting at the table can have 0, 1, or 2 coins. The total number of coins will be 10. Note that if a person lies, they will state a number of coins that differs from the actual number by 1 or 2. Since the total number of coins based on the answers differs fro... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.3 In triangle $A B C$, the median $A M$ is drawn (point $M$ lies on side $\mathrm{BC}$). It is known that angle $C A M$ is $30^{\circ}$, and side $A C$ is 2. Find the distance from point $B$ to the line $A C$.
Omвem: 1. | Solution: See fig.
Triangle СKM is equal to triangle ВHM (these are right triangles, the hypotenuses СM and ВM of which are equal, and the angles are the same).
Therefore, $\mathrm{BH}=$ СK. But in triangle СKA, the leg СK lies opposite the angle $30^{\circ}$ and is equal to half the hypotenuse АC: СK=1.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-5. In a tournament, 6 teams $P, Q, R, S, T$ and $U$ participate, and each team must play against every other team exactly once. Each day, they are divided into 3 pairs, and all three matches are played simultaneously. The "Sports" channel has chosen which match it will broadcast each day:
$$
\begin{array}{c|c|c|c|c}... | Answer. Only in the 1st.
Solution. Let's look at team $P$: on the 1st, 3rd, and 5th days, it will play against teams $Q, T$, and $R$. Therefore, in the remaining two days, it must play against teams $S$ and $U$. Since on the 2nd day $S$ plays against $R$, $P$ has no choice but to play against $U$ on the 2nd day, and a... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11-1. Twins Paolo and Sevilla are celebrating their birthday at a cafe with friends. If the final bill amount is divided equally among everyone, then each person should pay 12 euros. But if the bill is divided equally among everyone except Paolo and Sevilla, then each person should pay 16 euros. How many friends came t... | Answer: 6.
Solution. Let $n$ be the number of friends who arrived. Then we get the equation $12(n+2)=16n$, from which $n=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Variant 1.
It is known that
$$
\left(x^{2}-x+3\right)\left(y^{2}-6 y+41\right)\left(2 z^{2}-z+1\right)=77
$$
Find $\frac{x y}{z}$. | # Answer: 6.
Solution.
$$
\begin{aligned}
& x^{2}-x+3=(x-0.5)^{2}+2.75 \geq 2.75 \\
& y^{2}-6 y+41=(y-3)^{2}+32 \geq 32 \\
& 2 z^{2}-z+1=2(z-0.25)^{2}+0.875 \geq 0.875
\end{aligned}
$$
Therefore, $\left(x^{2}-x+3\right)\left(y^{2}-6 y+41\right)\left(2 z^{2}-z+1\right) \geq 77$ and if at least one of the three inequa... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Variant 1.
Given a parallelogram $A B C D$. Let $B P$ and $C Q$ be the perpendiculars dropped from vertices $B$ and $C$ to diagonals $A C$ and $B D$ respectively (point $P$ lies on segment $A C$, and point $Q$ lies on segment $B D$). Find the ratio $\frac{10 B D}{A C}$, if $\frac{A P}{A C}=\frac{4}{9}$ and $\frac{D... | Answer: 6.
Solution: Let $O$ be the point of intersection of the diagonals. Note that points $B, C, Q, P$ lie on the same circle (segment $B C$ is seen from points $P$ and $Q$ at a right angle). Therefore, triangles $B O P$ and $C O Q$ are similar. Let $A C=2 a, B D=2 b$. Then $P O=a-\frac{8 a}{9}=\frac{a}{9}, Q O=$ $... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. Draw a row of 11 circles, each of which is either red, blue, or green. Moreover, among any three consecutive circles, there should be at least one red, among any four consecutive circles, there should be at least one blue, and there should be more than half green. How many red circles did you get? | Answer: 3 red circles
Hint. The circles are arranged only as follows: ZZKSKZKSKZZ.
Solution. (1) Three non-overlapping triplets of circles can be identified, each containing at least one red circle. Therefore, there are no fewer than three red circles. (2) Two non-overlapping quartets of circles can be identified, ea... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. Represent the number 32 as the product of three integer factors, the sum of which is 3. What is the smallest of the factors? | Answer: -4.
Example: $32=(-4) \cdot(-1) \cdot 8$.
Solution. The given factorization is unique. This can be proven.
If all three factors are positive, then the largest of them is not less than 4, and the sum is greater than 3, which contradicts the condition. Therefore, two of the factors are negative, and the third ... | -4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.7. Petya told Misha that in his class exactly two thirds of all the girls are blondes, exactly one seventh of the boys are blonds, and in total, a third of the class has light hair. Misha said: "You once told me that there are no more than 40 people in your class. 0 ! I know how many girls are in your class!" How man... | Answer: 12 girls
Solution. Let there be $x$ girls and $y$ boys in the class. From the problem statement, we have the following relationship:
$\frac{2}{3} x+\frac{1}{7} y=\frac{1}{3}(x+y)$,
which, after transformation, becomes $7 x=4 y$.
From the condition and the derived relationship, it follows that the number $x$... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. Represent the number 36 as the product of three integer factors, the sum of which is 4. What is the smallest of the factors? | Answer: -4.
Example: $36=(-4) \cdot(-1) \cdot 9$.
Solution. The given factorization is unique. This can be proven.
If all three factors are positive, then the largest of them is not less than 4 (since $3^{3}<36$), and the sum is greater than 4, which contradicts the condition. Therefore, two of the factors are negat... | -4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. Given a parallelogram $A B C D, \angle D=100^{\circ}, B C=12$. On side $A D$ there is a point $L$ such that $\angle A B L=50^{\circ}, L D=4$. Find the length of $C D$. | Answer: 8.
Solution. By the property of a parallelogram, $\angle A B C=\angle D=100^{\circ}, A D=B C=12$ and $C D=A B$. Therefore, $\angle C B L=\angle A B C-\angle A B L=100^{\circ}-50^{\circ}=50^{\circ}$ and $A L=A D-L D=12-4=8$. Since $\angle A L B=\angle C B L$ (as alternate interior angles when $A D$ and $B C$ ar... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. Four boys and three girls went to the forest to pick mushrooms. Each found several mushrooms, in total they collected 70. No two girls collected the same amount, and any three boys together brought no fewer than 43 mushrooms. The number of mushrooms collected by any two children differed by no more than 5 times. M... | Answer: 5 mushrooms.
Solution. Any three boys collected at least 43 mushrooms together, so there is a boy who collected no less than 15 mushrooms (since $14 \cdot 3 < 43$). Therefore, this boy and the other three collected no less than $15 + 43 = 58$ pieces.
If there is a boy who collected no less than 15 pieces, the... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7.5 (7 points)
On a plane, 6 lines are drawn and several points are marked. It turned out that on each line exactly 3 points are marked. What is the minimum number of points that could have been marked? | # Solution:
The vertices of the triangle, the midpoints of its sides, and the point of intersection of the medians - 7 points lying in threes on 6 lines (3 sides and 3 medians).
P.S. It doesn't have to be medians specifically.
Proof of the estimate: If we have a point through which at least 4 lines pass, then we wil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Misha has a $7 \times 7$ square of paper, all cells of which are white. Misha wants to color $N$ cells black. What is the smallest $N$ for which Misha can color the cells so that after coloring, no completely white rectangle with at least ten cells can be cut out from the square? | Answer: 4.
Solution. Divide the $7 \times 7$ square into 5 rectangles: four $3 \times 4$ rectangles (each corner of such a rectangle coincides with one of the corners of the $7 \times 7$ square) and a $1 \times 1$ square. If only three cells are colored, there will be a white rectangle consisting of 12 cells. Example ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the side $A D$ of the square $A B C D$, point $K$ is marked, and on the extension of ray $A B$ beyond point $B$ - point $L$. It is known that $\angle L K C=45^{\circ}, A K=1, K D=2$. Find $L B$. | Answer: $L B=2$.
Fig. 1: to the solution of problem 5
Solution. Note that $\angle L A C=45^{\circ}=\angle L K C$, which implies that quadrilateral $L A K C$ is cyclic. Then $\angle K C L=90^... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. An excursion group of 6 tourists is visiting attractions. At each attraction, three people take photos, while the others photograph them. After what minimum number of attractions will each tourist have photos of all other participants in the excursion?
Answer: after 4 attractions. | Solution. Evaluation. A total of $6 \cdot 5=30$ photographs need to be taken (considering only photographs between two people $A$ and $B$, that is, if person $A$ photographs 3 other participants $B, C, D$ in one photograph - this counts as 3 photographings $A \rightarrow B, A \rightarrow C, A \rightarrow D$).
At one l... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of the numbers $1-2+3-4+5-6+\ldots+2013-2014$ and $1+2-3+4-5+6-\ldots-2013+2014$. | 1. Answer: 2.
Notice that for each term of the first sum, except for 1, there is an opposite term in the second sum. The sum of opposite numbers is 0. Therefore, the total sum is $1+1=2$.
Grading criteria:
Correct answer with proper justification: 7 points.
Incorrect answer with the right idea in the justification:... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Misha painted all integers in several colors such that numbers whose difference is a prime number are painted in different colors. What is the smallest number of colors that Misha could have used? Justify your answer. | 5. Answer: 4 colors
Evaluation. Consider the numbers $1,3,6,8$. The difference between any two of them is a prime number, which means that all of them must be of different colors, and at least four colors are needed.
Example. Paint numbers of the form $4 \mathrm{k}$ in the first color, numbers of the form $4 \mathrm{... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. How to measure 8 liters of water when you are near a river and have two buckets with a capacity of 10 liters and 6 liters? (8 liters of water should end up in one bucket).
| Solution. Let's write the sequence of filling the buckets in the form of a table:
| | Bucket with a capacity of 10 liters | Bucket with a capacity of 6 liters | Comment |
| :--- | :--- | :--- | :--- |
| Initially | 0 liters | 0 liters | |
| Step 1 | 10 liters | 0 liters | Filled the first bucket from the river |
| S... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Snow White entered a room where there were 30 chairs around a round table. Some of the chairs were occupied by dwarfs. It turned out that Snow White could not sit down without having someone next to her. What is the minimum number of dwarfs that could have been at the table? (Explain how the dwarfs should have been ... | Answer: 10.
Solution: If there were three consecutive empty chairs at the table in some place, Snow White could sit down in such a way that no one would sit next to her. Therefore, in any set of three consecutive chairs, at least one must be occupied by a dwarf. Since there are 30 chairs in total, there cannot be fewe... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. How to measure 2 liters of water when you are near a river and have two buckets with a capacity of 10 liters and 6 liters? (2 liters of water should end up in one bucket).
| Solution. Let's write the sequence of filling the buckets in the form of a table:
| | Bucket with a capacity of 10 liters | Bucket with a capacity of 6 liters | Comment |
| :---: | :---: | :---: | :---: |
| Initially | 0 liters | 0 liters | |
| Step 1 | 10 liters | 0 liters | The first bucket is filled from the rive... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Each of the 10 dwarfs either always tells the truth or always lies. It is known that each of them loves exactly one type of ice cream: butter, chocolate, or fruit. First, Snow White asked those who love butter ice cream to raise their hands, and everyone raised their hands, then those who love chocolate ice cream - ... | # Answer. 4.
Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of rai... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Each of the 10 dwarfs either always tells the truth or always lies. It is known that each of them loves exactly one type of ice cream: butter, chocolate, or fruit. First, Snow White asked those who love butter ice cream to raise their hands, and everyone raised their hands, then those who love chocolate ice cream - ... | # Answer. 4.
Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of rai... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.1. Given natural numbers $M$ and $N$, both greater than ten, consisting of the same number of digits, and such that $M = 3N$. To obtain the number $M$, one of the digits of $N$ must be increased by 2, and each of the other digits must be increased by an odd digit. What digit could the number $N$ end with? Find all p... | Answer. The digit 6.
Solution. By the condition, $M=3 N$, so the number $A=M-N=2 N$ is even. However, by the condition, the number $A$ is composed of odd digits and the digit 2. Therefore, $A$ ends in 2. Thus, the number $N$, which is half of $A$, ends in either 1 or 6.
We will show that $N$ cannot end in 1. If $N$ e... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the edge $A A^{\prime}$ of the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge length 2, a point $K$ is marked. In space, a point $T$ is marked such that $T B=\sqrt{11}$ and $T C=\sqrt{15}$. Find the length of the height of the tetrahedron $T B C K$, dropped from vertex $C$. | Answer: 2.
Fig. 5: to problem 5
Solution. Notice that
$$
T B^{2}+B C^{2}=11+4=15=T C^{2}
$$
From this, by the converse of the Pythagorean theorem, it follows that angle $T B C$ is a right... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) Percival's castle had a square shape. One day, Percival decided to expand his domain and added a square extension to the castle. As a result, the perimeter of the castle increased by $10 \%$. By what percentage did the area of the castle increase? | Answer: $4 \%$.
Solution. Let the width of the castle be $a$, and the width of the extension be $b$. Then the original perimeter is $4 a$, and the final perimeter is $4 a+2 b$. Therefore:
$$
1.1 \cdot 4 a=4 a+2 b \Leftrightarrow b=0.2 a
$$
From this, the area of the castle becomes $a^{2}+(0.2 a)^{2}=1.04 a^{2}$, whi... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)$ take?
## Answer: 0. | Solution. Note that the equality $a^{2}+b=b^{2}+c$ can be written as: $a^{2}-b^{2}=c-b$. Similarly, we have $b^{2}-c^{2}=a-c, c^{2}-a^{2}=b-a$. Substituting these equalities into the desired expressions, we get that
$$
a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In a football tournament where each team played against each other once, teams A, B, C, D, and E participated. For a win, a team received 3 points, for a draw 1 point, and for a loss 0 points. In the end, it turned out that teams A, B, C, D, and E each had 7 points. What is the maximum number of points that team $\m... | Answer: 7 points.
Solution: In a match where one of the teams won, the teams together score 3 points, in a match that ended in a draw - 2 points. Since 7 is not divisible by 3, the team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament.... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 2. Solve the equation
$$
\sqrt{3 x-2-x^{2}}+\sqrt{x^{2}-4 x+3}=\sqrt{2}(1-\sqrt{x})
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{c}
3 x-2-x^{2} \geq 0 \\
x^{2}-4 x+3 \geq 0
\end{array}\right.
$$
we obtain that the domain of the function on the left side of the equation is $\{1\}$. The domain of the function on the right side of the equation is the numerical ray $[0 ;+\infty)$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.3. Each of the thirteen dwarfs is either a knight, who always tells the truth, or a liar, who always lies. One day, all the dwarfs in turn made the statement: “Among the statements made previously, there are exactly two more false ones than true ones.” How many knights could there have been among the dwarfs? | Answer: 6
Solution. The first two statements are obviously false, as there were fewer than two statements made before each of them. The third statement is true, as there were 2 false statements and zero true statements made before it. The fourth statement is false, as it adds one true statement to the two false ones, ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. At each vertex of a cube lives a number, not necessarily positive. All eight numbers are distinct. If a number is equal to the sum of the three numbers living in the adjacent vertices, then it is happy. What is the maximum number of happy numbers that can live at the vertices of the cube? | Answer: 8.
Solution. See, for example, Fig. 7.5a. It is easy to verify that each vertex of the cube contains a lucky number.
There are other examples as well. Let's understand how they are structured (this was not required of the olympiad participants). Let's denote the numbers at the vertices of the cube (see Fig. 7... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.1. In the numerical example АБВ $+9=$ ГДЕ, the letters А, Б, В, Г, Д, and Е represent six different digits. What digit is represented by the letter Д? | Answer: 0.
Solution: In the addition, the second digit of the first addend АБВ has changed (Д instead of Б). This could only happen if 1 was carried over from the units place to the tens place during the addition. However, the first digit also changed (Г instead of А). This means that 1 was also carried over from the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.5. In the room, there are 10 people - liars and knights (liars always lie, and knights always tell the truth). The first said: "In this room, there is at least 1 liar." The second said: "In this room, there are at least 2 liars." The third said: "In this room, there are at least 3 liars." And so on,
up to the tenth,... | Answer: 5.
Solution: Let there be $k$ liars in the room. Then the first $k$ people told the truth (and thus were knights), while the remaining $(10-k)$ lied (and were liars). Therefore, $k=10-k$, from which $k=5$.
Comment: The answer is obtained by considering an example -3 points. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. On the island, there live knights who always tell the truth and liars who always lie. In the island's football team, there are 11 people. Player number 1 said: "In our team, there are as many knights as there are liars." Player number 2 said: "In our team, the number of knights and the number of liars differ by one,... | 4. Answer: There are either no knights at all, or there is only one and he plays under number 10. The two answers cannot both be true, as they contradict each other. This means there can be no more than one true answer. If there is no true answer, then the team consists entirely of liars, in which case indeed none of t... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2 Out of 24 matches, a figure in the form of a $3 \times 3$ square is laid out (see figure), the side length of each small square is equal to the length of a match. What is the smallest number of matches that can be removed so that there are no whole $1 \times 1$ squares left, formed from matches.
![](https://cdn.ma... | Answer: 5 matches.
Solution: Estimation: It is impossible to manage with four matches, as by removing a match, we "ruin" no more than two squares (each match is a side of one or two adjacent squares), but initially, we have 9 small squares. Example for 5 matches:
$ and $g(x)$ each have two roots, and the equalities $f(1)=g(2)$ and $g(1)=f(2)$ hold. Find the sum of all four roots of these trinomials. | Answer: 6.
First solution. Let $f(x)=x^{2}+a x+b, g(x)=x^{2}+$ $+c x+d$. Then the conditions of the problem can be written as
$$
1+a+b=4+2 c+d \quad \text { and } \quad 4+2 a+b=1+c+d
$$
Subtracting the second equation from the first, we get $-3-a=3+c$, which means $a+c=-6$. By Vieta's theorem, $-a$ is the sum of the... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Each of the 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some integer. Then the first said: “My number is greater than 1”, the second said: “My number is greater than 2”, \ldots, the tenth said: “My number is greater than 10”. After that, all ten, sp... | Answer: 8 knights.
Solution. We will prove that none of the knights could have said either of the phrases "My number is greater than 9" or "My number is greater than 10." Indeed, if this were possible, the integer thought of by the knight would be at least 10. But then he could not have said any of the phrases "My num... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. For what least natural $n$ do there exist integers $a_{1}, a_{2}, \ldots, a_{n}$ such that the quadratic trinomial
$$
x^{2}-2\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} x+\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$
has at least one integer root?
(P. Kozlov) | Answer. For $n=6$.
Solution. For $n=6$, we can set $a_{1}=a_{2}=a_{3}=a_{4}=1$ and $a_{5}=a_{6}=-1$; then the quadratic trinomial from the condition becomes $x^{2}-8 x+7$ and has two integer roots: 1 and 7. It remains to show that this is the smallest possible value of $n$.
Suppose the numbers $a_{1}, a_{2}, \ldots, ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. In a notebook, a triangular grid is drawn (see figure). Tanya placed integers at the nodes of the grid. We will call two numbers close if they are in adjacent nodes of the grid. It is known that
- the sum of all ten numbers is 43;
- the sum of any three numbers such that any two of them are close is 11.
... | Answer: 10.
Solution. Let's denote the numbers by variables as shown in the figure.
Then
\[
\begin{gathered}
a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}=c_{1}+c_{2}+c_{3}=11 \\
\left(a_{1}+a_{2}+a... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. The teacher wrote a fraction on the board, where the numerator and the denominator are natural numbers. Misha added 30 to the numerator of the given fraction and wrote the resulting fraction in his notebook, while Lesha subtracted 6 from the denominator of the fraction written on the board and also wrote t... | Answer: 5.
Solution. Let $\frac{a}{b}$ be the original fraction. Then Misha wrote down the fraction $\frac{a+30}{b}$ in his notebook, and Lёsha wrote down $-\frac{a}{b-6}$.
Let's write the equation
$$
\frac{a+30}{b}=\frac{a}{b-6}
$$
Transforming it, we get
$$
\begin{gathered}
(a+30)(b-6)=a b \\
a b+30 b-6 a-180=a ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Pantelej and Gerasim received 20 grades each in November, and Pantelej received as many fives as Gerasim received fours, as many fours as Gerasim received threes, as many threes as Gerasim received twos, and as many twos as Gerasim received fives. At the same time, their average grade for November is the same. How m... | Solution. Let's add one to each of Gerasim's grades. His total score will increase by 20. On the other hand, it will become greater than Pantelej's total score by four times the number of Pantelej's twos.
Answer: 5 twos. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. In triangle $\mathrm{ABC}$, $\mathrm{AC}=1$, $\mathrm{AB}=2$, $\mathrm{O}$ is the point of intersection of the angle bisectors. A segment passing through point O and parallel to side $\mathrm{BC}$ intersects sides $\mathrm{AC}$ and $\mathrm{AB}$ at points K and M, respectively. Find the perimeter of triangle $\mat... | 8.3. $\angle \mathrm{KCO}=\angle \mathrm{BCO}=\angle \mathrm{KOC}$ (alternate interior angles). Therefore, OK = KC, and similarly BM = OM. Then
$$
A K+A M+K M=A K+K C+A M+B M=3 .
$$
Answer: 3. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.2. We will call a natural number interesting if the sum of its digits is a prime number. What is the maximum number of interesting numbers that can be among five consecutive natural numbers?
(
#
| # Answer: 4.
Solution. Among five consecutive natural numbers, there can be 4 interesting numbers. For example, the numbers 199, 200, 201, 202, 203 (with digit sums 19, 2, 3, 4, and 5) will work.
Now, let's prove that all 5 numbers cannot be interesting. Among our five numbers, there are three that lie within the sam... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. At the New Year's celebration, schoolchildren organized an exchange game: if they were given five tangerines, they would exchange them for three crackers and a candy, and if they were given two crackers, they would exchange them for three tangerines and a candy. Father Frost played this game with them several times ... | Solution. Ded Moroz conducted 50 exchanges, as he was given 50 candies. At the end of the game, he had no crackers left, meaning he exchanged all of them back. For every two exchanges of tangerines for crackers (ten given - six received), he conducted three exchanges of crackers for tangerines (six given - nine receive... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10-1. Piglet has balloons of five colors. He managed to arrange them in a row in such a way that for any two different colors in the row, there will always be two adjacent balloons of these colors. What is the minimum number of balloons Piglet could have? | Answer: 11 balls.
Solution. Consider the balls of color $a$. Their neighbors must be balls of all 4 other colors. But one ball can have no more than two neighbors, so there must be at least 2 balls of color $a$. This is true for each of the 5 colors, so there must be at least 10 balls in total.
Notice that some ball ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-5. From 80 identical Lego parts, several figures were assembled, with the number of parts used in all figures being different. For the manufacture of the three smallest figures, 14 parts were used, and in the three largest, 43 were used in total. How many figures were assembled? How many parts are in the largest fig... | Answer: 8 figurines, 16 parts.
Solution. Let the number of parts in the figurines be denoted by $a_{1}43$, so $a_{n-2} \leq 13$.
Remove the three largest and three smallest figurines. In the remaining figurines, there will be $80-14$ - 43 $=23$ parts, and each will have between 7 and 12 parts. One figurine is clearly... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3.1 Two ants are running along a circle towards each other at constant speeds. While one of them runs 9 laps, the other runs 6. At the point where the ants meet, a red dot appears. How many red dots will there be on the circle?
, settled into 3 rooms of a hotel. When everyone gathered in their rooms, Basil, who was staying in the first room, said: "There are more liars than knights in this room right now. Although no - there are more ... | Answer: 9 knights.
Solution: Since Vasily's statements contradict each other, Vasily is a liar. Therefore, both of Vasily's statements (about each room) are false, and in each room (when he was there) the number of liars and knights was equal. This means that in each room, without Vasily, there was one more knight tha... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. It is known that the numbers EGGPLANT and FROG are divisible by 3. What is the remainder when the number CLAN is divided by 3? (Letters represent digits, the same letters represent the same digits, different letters represent different digits).
Answer: 0 | Solution. By the divisibility rule for 3, the sums B+A+K+L+A+Z+A+N and Z+A+B+A are divisible by 3, and therefore the difference of these sums $K+N+A+H$ is also divisible by 3, which by the rule means that the number KLAN is divisible by 3, hence the remainder is 0.
Criteria. Only the answer - 0 points | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. Find the value of the expression $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}$, if $a+b+c=0$. | Answer: 3.
Solution. $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+(a+b+c)(3 a b+3 b c+3 c a)-3 a b c$. Using the condition $a+b+c=0$, we get: $a^{3}+b^{3}+c^{3}=3 a b c$. Then $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}=\frac{a^{3}+b^{3}+c^{3}}{a b c}=\frac{3 a b c}{a b c}=3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. At present, the exchange rates of the US dollar and the euro are as follows: $D=6$ yuan and $E=7$ yuan. The People's Bank of China determines the yuan exchange rate independently of market conditions and adheres to a policy of approximate equality of currencies. One bank employee proposed the following scheme for ch... | # Solution.
Note that if $D$ and $E$ have different parity, their sum is odd, and if they have the same parity, their sum is even. The second number in the pair is always odd. Initially, we have a pair $О Н$ (Odd, Odd).
After one year
After two years $\quad \mathrm{H}+\mathrm{H}=$ О, Н. And so on.
That is, after an... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression
$$
a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right) ?
$$
take? | Answer: 0.
Solution. From the condition, it follows that $a^{2}-b^{2}=c-b$, $b^{2}-c^{2}=a-c$, and $c^{2}-a^{2}=b-a$. Therefore, $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0$.
## Grading Criteria:
+ a complete and justified solution is provided
the correct an... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.6. Waiting for customers, a watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). In the process, the seller wrote down on a piece of paper the weights he used. What is the... | Answer: 6 numbers.
Solution. Let's check that with weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, one can weigh any of the given watermelons. Indeed, $2=1+1 ; 4=3+1 ; 6=5+1 ; 8=7+1 ; 11=10+1$; $12=9+3 ; 13=10+3 ; 14=9+5 ; 15=10+5 ; 16=9+7 ; 17=10+7 ; 18=9+9 ; 19=10+9 ;$ $20=10+10$. Thus, 6 different numbers could... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The working day at the enterprise lasted 8 hours. During this time, the labor productivity was as planned for the first six hours, and then it decreased by $25 \%$. The director (in agreement with the labor collective) extended the shift by one hour. As a result, it turned out that again the first six hours were wor... | 2. Answer: by 8 percent. Solution. Let's take 1 as the planned labor productivity (the volume of work performed per hour). Then before the shift extension, workers completed $6+1.5=7.5$ units of work per shift. And after the extension, $8+2.1=8.1$ units. Thus, the overall productivity per shift became $8.1: 7.5 \times ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In pentagon $A B C D E \quad A B=A E=C D=1, \quad B C+D E=1$, $\angle A B C=\angle A E D=90^{\circ}$. Find the area of the pentagon. | 6. Answer: 1. Solution. Extend the segment $C B$ beyond point $B$ to segment $B F=D E$, and then draw segments $C A, C E$ and $A F$. It is easy to see that triangles $C D E$ and $A B F$ are equal. Therefore, $C E=A F$ and the areas of the pentagon and quadrilateral $A E C F$ coincide. But triangles $A E C$ and $A F C$ ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) In a building, on all floors in all entrances there is an equal number of apartments (more than one). Also, in all entrances there are an equal number of floors. The number of floors is greater than the number of apartments per floor, but less than the number of entrances. How many floors are there in the... | Answer: 11.
Solution. Let the number of apartments per floor be K, the number of floors be F, and the number of entrances be E. According to the condition, $1<K<F<E$. The number 715 can be factored into numbers greater than one in only one way: $715=5 \cdot 11 \cdot 13$. Therefore, $K=5, F=11, E=13$.
Criterion. Corre... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. Anya left the house, and after some time Vanya left from the same place and soon caught up with Anya. If Vanya had walked twice as fast, he would have caught up with Anya three times faster. How many times faster would Vanya have caught up with Anya (compared to the actual time) if, in addition, Anya walked twice... | Answer: 7.
Solution: Let Ane's speed be $v$, and Vanya's speed be $V$. The distance Ane has walked is proportional to $v$, while Vanya catches up with her at a speed of $u=V-v$. When Vanya doubles his speed, this difference triples, i.e., $u+V=2V-v=3u$. From this, we get $V=2u=2v$. If Vanya's speed doubles and Ane's s... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. Solve the equation $\sqrt{2 x-1}+\sqrt[3]{x}=\sqrt[4]{17-x}$. | Answer: 1.
Solution. It is easy to verify that $x=1$ is a root of the equation.
Since the left side of the equation is an increasing function (as the sum of two increasing functions), and the right side is a decreasing function, there are no other roots.
Comment. The correct answer without proof of its uniqueness - ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. In a chess tournament, each of the 10 players played one game against each other, and Petya came in last place (scored fewer points than any other participant). Then two players were disqualified, and all points earned in matches against them were annulled, and these two players were removed from the table. It tur... | Answer: 4 points.
Solution. In a tournament with 10 players, played in a single round, $\frac{10 \cdot 9}{2}=45$ points are distributed. Therefore, there will be a player who scores no more than $45: 10=4.5$ points. This means that the player who finished in absolute last place scored no more than 4 points. Similarly,... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. In the notebook, all irreducible fractions with the numerator 15 are written, which are greater than $\frac{1}{16}$ and less than $\frac{1}{15} \cdot$ How many such fractions are written in the notebook? | Answer: 8 fractions.
Solution. We look for all suitable irreducible fractions of the form $\frac{n}{15}$. Since $\frac{1}{16} < \frac{15}{n}$, then $\frac{15}{225} > \frac{15}{n}$, and $n > 225$. Therefore, $225 < n < 240$. The fraction $\frac{n}{15}$ is irreducible, meaning $n$ is not divisible by 3 or 5. It is not d... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.5. In an isosceles triangle \(ABC\), the angle \(A\) at the base is \(75^\circ\). The bisector of angle \(A\) intersects side \(BC\) at point \(K\). Find the distance from point \(K\) to the base \(AC\), if \(BK=10\). | Answer: 5.
Solution. In triangle $ABC$, $\angle B = (180^0 - 75^0 - 75^0) = 30^0$. Drop perpendiculars from point $K$: $KH$ to side $AB$, and $KN$ to side $AC$. In the right triangle $HBK$, the hypotenuse $BK$ is 10. Angle $B$ is 30 degrees, and the side opposite to it is half the hypotenuse, so $HK = 5$. Triangles $A... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. Given a triangle $A B C$. A line parallel to $A C$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively, and the median $A M$ at point $Q$. It is known that $P Q=3$, and $Q T=5$. Find the length of $A C$. | Answer: $AC=11$.
Solution. First method. Draw a line through point $Q$ parallel to $BC$ (where $N$ and $L$ are the points of intersection of this line with sides $AB$ and $AC$ respectively, see Fig. 9.3a). Since $AM$ is the median of triangle $ABC$, then $LQ=NQ$. Moreover, $PT \| AC$, so $PQ$ is the midline of triangl... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.2. There are nuts in three boxes. In the first box, there are six nuts fewer than in the other two boxes combined, and in the second box, there are ten fewer nuts than in the other two boxes combined. How many nuts are in the third box? Justify your answer. | Solution: Let there be $x$ nuts in the first box, $y$ and $z$ in the second and third boxes, respectively. Then the condition of the problem is defined by the equations $x+6=y+z$ and $x+z=y+10$. From the first equation, $x-y=z-6$, and from the second, $x-y=10-z$. Therefore, $z-6=10-z$, from which $z=8$.
Answer: 8 nuts... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For what least natural $\mathrm{k}$ does the quadratic trinomial
$$
y=k x^{2}-p x+q \text { with natural coefficients } p \text{ and } q \text{ have two }
$$
distinct positive roots less than 1? | Solution. First, let's estimate the coefficient $\mathrm{k}_{\text{from below. Let }} 0k^{2} x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right) \geq 1 \Rightarrow k>4 \text{. Thus, we have obtained the }$
$$
$$
lower bound. To show the accuracy of the estimate, it is sufficient to construct an example with $k=5: y=5 x... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Among the numbers from 1 to $10^{23}$, which are more numerous - those with a two-digit sum of digits or those with a three-digit sum? Answer: those with a three-digit sum. | Solution. The number $10^{23}$ has a digit sum of 1, so we will not consider it. We will write numbers as 23-digit numbers, adding leading zeros. We will call digits "complementary" if their sum is 9. To each number with a two-digit digit sum, we will correspond a number by replacing each digit with its "complementary"... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.2. Petya has 25 coins, each with a denomination of $1, 2, 5$, or 10 rubles. Among these coins, 19 are not two-ruble coins, 20 are not ten-ruble coins, and 16 are not one-ruble coins. How many five-ruble coins does Petya have? | Answer: 5.
Solution. Since among Petya's coins, 19 are not two-ruble coins, Petya has a total of 6 two-ruble coins. Similarly, it follows that Petya has 5 ten-ruble coins and 9 one-ruble coins.
Thus, Petya has $25-6-5-9=5$ five-ruble coins. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. By September 1st, Vlad bought himself several ballpoint and gel pens. He noticed that if all the pens he bought were gel pens, he would have paid 4 times more than he actually did. And if all the pens were ballpoint pens, the purchase would have cost him 2 times less than the actual price. How many times m... | Answer: 8.
Solution. If all the pens were gel pens, their price would be 4 times the actual price, which in turn is 2 times more than if all the pens were ballpoint pens. Therefore, gel pens cost $4 \cdot 2=8$ times more than ballpoint pens. Consequently, one gel pen is 8 times more expensive than one ballpoint pen. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.7. Jerry the mouse decided to give Tom the cat a square cake $8 \times 8$ for his birthday. In three pieces marked with the letter "P," he put fish, in two pieces marked with the letter "K," he put sausage, and in one piece, he added both, but did not mark it (all other pieces are without filling). Jerry also... | # Answer: 5.
Solution. Let's call a piece with fish and sausage a coveted piece.
According to the problem, in any $6 \times 6$ square, there are at least 2 pieces with fish. In any such square, at least one known piece with fish is already included; consider the squares that contain only 1 known piece of fish (all of... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.8. A natural number $n$ is called good if 2020 gives a remainder of 22 when divided by $n$. How many good numbers exist? | Answer: 10.
Solution. In the subsequent solution, the expression of the form $a^{k}$ - the number $a$ raised to the power of $k$ - is the number $a$ multiplied by itself $k$ times. We will also consider $a^{0}=1$. For example, $3^{2}=3 \cdot 3=9$, and $2^{0}=1$.
Since 2020 gives a remainder of 22 when divided by $n$,... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. Four children were walking along an alley and decided to count the number of firs planted along it.
- Anya said: "There are a total of 15 firs along the alley."
- Borya said: "The number of firs is divisible by 11."
- Vера said: "There are definitely fewer than 25 firs."
- Gena said: "I am sure that their... | Answer: 11.
Solution. Let $N-$ be the number of elms along the alley.
Suppose Genya told the truth, and $N$ is divisible by 22. But then $N$ is also divisible by 11, i.e., Borya also told the truth. But according to the problem, one of the boys was wrong. Therefore, Genya must have been wrong, but Borya was right. So... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. In a class, there are 20 students. Thinking about which girls to send a Valentine's card to on February 14, each boy made a list of all the girls in the class he finds attractive (possibly an empty list). It is known that there do not exist three boys whose lists have the same number of girls. What is the ... | Answer: 6.
Solution. Let the number of girls in the class be $d$. According to the condition, there are no three boys whose lists match in the number of girls, so there can be a maximum of 2 empty lists, 2 lists with one girl, 2 lists with two girls, ..., 2 lists with $d$ girls. This means that the number of lists, an... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. How many pairs of natural numbers $a$ and $b$ exist such that $a \geqslant b$ and
$$
\frac{1}{a}+\frac{1}{b}=\frac{1}{6} ?
$$ | Answer: 5.
Solution. From the condition, it follows that $\frac{1}{6}=\frac{1}{a}+\frac{1}{b} \leqslant \frac{2}{b}$, from which $b \leqslant 12$. Also, $\frac{1}{6}=\frac{1}{a}+\frac{1}{b}>\frac{1}{b}$, so $b>6$. Therefore, $b$ can take values from 7 to 12 inclusive.
Using $\frac{1}{a}=\frac{1}{6}-\frac{1}{b}=\frac{... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. Yura has $n$ cards, on which numbers from 1 to $n$ are written. After losing one of them, the sum of the numbers on the remaining cards turned out to be 101. What number is written on the lost card? | Answer: 4.
Solution. Suppose that $n \leqslant 13$. Then $1+2+\ldots+n=\frac{n(n+1)}{2} \leqslant 91101$, a contradiction.
Therefore, $n=14$, and the missing number is $1+2+\ldots+14-101=105-101=4$. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Around a round table, $2 n(n>5)$ people are sitting - knights and liars. Liars give a false answer to any question, while knights give a true answer. Each of them knows who is a knight and who is a liar. Each of them answered two questions: "Who is your left neighbor?", "Who is your right neighbor?". A wise man, who... | 5. Two adjacent knights (R), like two adjacent liars (L), give answers "R" and "R", while adjacent R and L give answers "L" and "L". If the number of L is less than \( n-1 \), then in a group of consecutive R, the outermost R can be replaced by L, and we get the same set of answers with a different number of R. Also, L... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5.1. In the cells of a $12 \times 12$ table, natural numbers are arranged such that the following condition is met: for any number in a non-corner cell, there is an adjacent cell (by side) that contains a smaller number. What is the smallest number of different numbers that can be in the table?
(Non-corner c... | Answer: 11.
Solution. First, let's provide an example with the arrangement of 11 different numbers.
| 1 | 2 | 3 | 4 | 5 | 6 | 6 | 5 | 4 | 3 | 2 | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 3 | 4 | 5 | 6 | 7 | 7 | 6 | 5 | 4 | 3 | 2 |
| 3 | 4 | 5 | 6 | 7 ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1.
It is known that the equations $x^{2}+(2 a-5) x+a^{2}+1=0$ and $x^{3}+(2 a-5) x^{2}+\left(a^{2}+1\right) x+a^{2}-4=0$ have common roots. Find the sum of these roots. | Answer: 9.
Solution. The left side of the second equation is obtained from the left side of the first equation by multiplying by $x$ and adding the expression $a^{2}-4$. Therefore, $a^{2}-4=0$, from which $a=2$ or $a=-2$. If $a=2$, then the first equation has no roots. Therefore, $a=-2$, and then we get that the first... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.7. On the board, 99 numbers were written, none of which are equal. In a notebook, $\frac{99 \cdot 98}{2}$ numbers were written—all differences between two numbers from the board (each time subtracting the smaller number from the larger one). It turned out that the number 1 was written exactly 85 times in the notebook... | Answer. $d=7$.
Solution. We will prove that $d \geqslant 7$. All numbers on the board can be divided into chains of numbers of the form $a, a+1, a+2, \ldots, a+t$ such that numbers from different chains do not differ by exactly 1. Such a partition is not difficult to construct by connecting any two numbers that differ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.9. Find the largest number $m$ such that for any positive numbers $a, b$, and $c$, the sum of which is 1, the inequality
$$
\sqrt{\frac{a b}{c+a b}}+\sqrt{\frac{b c}{a+b c}}+\sqrt{\frac{c a}{b+c a}} \geqslant m
$$
holds.
(l. Emelyanov) | Answer. $m=1$.
First solution. First, we prove that $m=1$ satisfies the requirements of the problem. Notice that $ab + c = ab + c(a + b + c) = (c + a)(c + b)$. Therefore,
\[
\begin{aligned}
& \sqrt{\frac{ab}{c + ab}} + \sqrt{\frac{bc}{a + bc}} + \sqrt{\frac{ca}{b + ca}} = \\
& = \sqrt{\frac{ab}{(c + a)(c + b)}} + \sq... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
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