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9.4 Around a circle, 100 integers are written, the sum of which is 1. A chain is defined as several numbers (possibly one) standing consecutively. Find the number of chains, the sum of the numbers in which is positive.
# | # Answer: 4951
Solution. Let's divide all chains (except the chain consisting of all numbers) into pairs that complement each other. If the sum of the numbers in one chain of the pair is $s$, and in the second is $t$, then $s+t=1$. Since $s$ and $t$ are integers, exactly one of them is positive. Therefore, exactly hal... | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,798 |
9.5 The square of a natural number a, when divided by a natural number n, gives a remainder of 8. The cube of the number a, when divided by n, gives a remainder of 25. Find n. | Answer: $n=113$
Solution. Note that the number $x=a^{6}-8^{3}=(a^{2})^{3}-8^{3}=(a^{2}-8)(a^{4}+8a^{2}+64)$ is divisible by $n$. Also note that the number $y=a^{6}-25^{2}=(a^{3})^{2}-25^{2}=(a^{3}-25)(a^{3}+25)$ is divisible by $n$. Then the difference $x-y=(a^{6}-8^{3})-(a^{6}-25^{2})=625-512=113$ is divisible by $n$... | 113 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,799 |
1. The number 1232 is divisible by the sum of its digits $1+2+3+2=8$. What is the next number that is divisible by the sum of its digits? | Answer: 1233.
Solution: One can notice that the next number fits, as 1233 is divisible by the sum of its digits $1+2+3+3=9$. Indeed, $1233: 9=137$.
Comment: One could have avoided performing the division and instead used the divisibility rule for 9. | 1233 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,800 |
2. Vasya constructed a robot, but he messed up the program, so the robot makes random moves that somehow always lead it to the right point.
A move is a one-kilometer displacement either north (N), east (E), south (S), or west (W).
The robot reached its goal in 22 moves:
# EENENNWNENESEEEEESSSWN.
How many moves woul... | Answer: 10.
Solution: Let's count the number of each type of letter. $E$ appears 10 times, $W-2$ times, $N-6$ times, $S-$ 4 times. In total, the robot's goal is to move 8 km east and 2 km north. Therefore, without extra moves, the route would consist of $8+2=10$ moves. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,801 |
3. A farmer wants to start growing watermelons. He wants to sell at least 10,000 watermelons every year. Watermelons are grown from seeds (one seed grows into one watermelon). Each watermelon can produce 250 good seeds that can be planted the following year, but then this watermelon cannot be sold. What is the minimum ... | Answer: 10041.
Solution: Let the smallest number of seeds the farmer needs to buy be $10000+x$. Then he will be able to sell 10000 watermelons, and the remaining $x$ watermelons can be used for seeds. For the seeds to be sufficient for the next season, the inequality $250 x \geqslant 10000+x$ must hold. From this, $x ... | 10041 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,802 |
4. For the function $f(x)$, it is known that it is odd, i.e., $f(-x) = -f(x)$ for every real $x$. Additionally, it is known that for every $x$, $f(x+5) = f(x)$, and also $f(1 / 3) = 2022$, and $f(1 / 2) = 17$. What is the value of
$$
f(-7) + f(12) + f\left(\frac{16}{3}\right) + f\left(\frac{9}{2}\right) ?
$$ | Answer: 2005.
Solution. Let's find the values of $f(-7)+f(12)$, $f\left(\frac{16}{3}\right)$, and $f\left(\frac{9}{2}\right)$ separately.
- $f(-7)+f(12)=f(-7+5)+f(12-2 \cdot 5)=f(-2)+f(2)=0$;
- $f\left(\frac{16}{3}\right)=f\left(5+\frac{1}{3}\right)=f\left(\frac{1}{3}\right)=2022$.
- $f\left(\frac{9}{2}\right)=f\left... | 2005 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,803 |
5. Each of the diagonals of the inscribed quadrilateral $A B C D$ is the bisector of one of the angles from which it is drawn, and also divides the second angle in the ratio $1: 2$. What can the angle $\angle A$ of this quadrilateral be? Express the answer in degrees. List all possible options. If necessary, round the ... | Answer: $72, 108, \frac{720}{7}, \frac{540}{7}$.
Solution. Let in the inscribed quadrilateral $P Q R S$, the diagonals are angle bisectors of angles $P$ and $Q$, and also divide angles $R$ and $S$ in the ratio $1: 2$.
Recall that an inscribed angle is equal to half the arc it subtends. Since $P R$ and $Q S$ are bisec... | 72,108,\frac{720}{7},\frac{540}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,804 |
6. For the numbers $1000^{2}, 1001^{2}, 1002^{2}, \ldots$, the last three digits are discarded. How many of the first terms of the resulting sequence form an arithmetic progression? | Answer: 32.
Solution. $(1000+k)^{2}=1000000+2000 k+k^{2}$. As long as $k^{2}<1000$, after discarding the last three digits, the number $1000+2 k$ will remain, i.e., each term of the sequence will be 2 more than the previous one. When $k=31: k^{2}=961<1000$, and when $k=32: k^{2}=1024$. Therefore, the element obtained ... | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,805 |
7. Triangles $A B C$ and $A B D$ are inscribed in a semicircle with diameter $A B=5$. The perpendicular from $D$ to $A B$ intersects segment $A C$ at point $Q$, ray $B C$ at point $R$, and segment $A B$ at point $P$. It is known that $P R=27 / 10$, and $P Q=5 / 6$. Find the length of segment $D P$. If necessary, round ... | Answer: 1.5.
Solution: Since $AB$ is the diameter, then $\angle ADB = \angle ACD = 90^{\circ}$. Angles $\angle QAB$ and $\angle PRB$ are equal, as they are complemented by angle $CBA$ to $90^{\circ}$. Therefore, triangles $\triangle APQ$ and $\triangle RPB$ are similar by two angles. Thus, $\frac{AP}{PQ} = \frac{RP}{P... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,806 |
8. The sum of eight numbers is $\frac{4}{5}$. It turns out that the sum of any seven of these eight numbers is non-negative. What is the smallest value that one of these numbers can take? | Answer: $-4.8$.
Solution. Let the numbers be denoted by $a_{1}, a_{2}, \ldots, a_{8}$. According to the condition, $a_{1}+a_{2}+\ldots+a_{8}=\frac{4}{5}$. Note that the sum of all numbers except one is non-negative, so each number does not exceed $\frac{4}{5}$ (and conversely, if each number does not exceed $\frac{4}{... | -4.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,807 |
5. Each of the diagonals of the inscribed quadrilateral $A B C D$ is the bisector of one of the angles from which it is drawn, and also divides the second angle in the ratio $1: 3$. What can the angle $\angle A$ of this quadrilateral be? Express the answer in degrees. List all possible options. If necessary, round the ... | Answer: $108,72,60,120$. | 108,72,60,120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,808 |
5. Each of the diagonals of the inscribed quadrilateral $A B C D$ is the bisector of one of the angles from which it is drawn, and also divides the second angle in the ratio $2: 3$. What can the angle $\angle A$ of this quadrilateral be? Express the answer in degrees. List all possible options. If necessary, round the ... | Answer. $80,100, \frac{1080}{11}, \frac{900}{11}$. | 80,100,\frac{1080}{11},\frac{900}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,809 |
5. Each of the diagonals of the inscribed quadrilateral $A B C D$ is the bisector of one of the angles from which it is drawn, and also divides the second angle in the ratio $1: 5$. What can the angle $\angle A$ of this quadrilateral be? Express the answer in degrees. List all possible options. If necessary, round the ... | Answer. $45, 135, \frac{225}{2}, \frac{135}{2}$. | 45,135,\frac{225}{2},\frac{135}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,810 |
1. Which of the numbers is greater: $77^{7}$ or $7^{77} ?$ | Answer. The second number is greater.
Solution. $7^{10}>7^{2}>11$, therefore $7^{11}=7 \cdot 7^{10}>7 \cdot 11=77$. From this, it follows that $7^{77}=\left(7^{11}\right)^{7}>77^{7}$.
Grading criteria.
Correct answer and proof - 7 points.
Correct reasoning and incorrect answer at the end - 6 points.
Only answer - ... | 7^{77}>77^7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,811 |
3. While tidying up the children's room before the guests arrived, mom found 9 socks. Among any four socks, at least two belong to the same owner. And among any five socks, no more than three belong to the same owner. How many children scattered the socks, and how many socks does each child own?
Answer. There are thre... | Solution. No child owned more than three socks, as otherwise the condition "among any five socks, no more than three belonged to one owner" would not be met. There are a total of 9 socks, so there are no fewer than three children. On the other hand, among any four socks, there are two socks belonging to one child, so t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,813 |
4. Given a cube. $A, B$ and $C$ are the midpoints of its edges (see figure). What is the angle $ABC$?
Answer. $120^{\circ}$. | Solution. 1st method. Draw diagonals $D E \| B C$ and $E F \| A B$ and let $K$ be the point on the extension of diagonal $D E$ beyond point $E$ (see figure). Then $\angle A B C = \angle F E K$. But triangle $D E F$ is equilateral, so
(a+b)}=\frac{a-b}{(a+c)(b+c)}$. From this: $(b+c)(b-c)=(a-b)(a+b)$ or $b^{2}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,815 |
6. How many natural numbers $n$ exist for which $4^{n}-15$ is a square of an integer? | Answer. Two.
Solution. Let $4^{n}-15=x^{2}$, where $x$ is an integer. It is obvious that $x \neq 0$. If $x$ is negative, then $(-x)^{2}$ is also equal to $4^{n}-15$; therefore, we will assume that $4^{n}-15=x^{2}$, where $x$ is a natural number. From the equation $2^{2 n}-15=x^{2}$, we get: $2^{2 n}-x^{2}=15$, and usi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,816 |
# Task № 3.1
## Condition:
Grandma is embroidering her grandchildren's names on their towels. She embroidered the name "ANNA" in 20 minutes, and the name "LINA" in 16 minutes. She spends the same amount of time on the same letters, and possibly different times on different letters. How long will it take her to embroi... | Express the answer in minutes.
Answer: 12
Exact match of the answer - 1 point
Solution.
Since the grandmother spends the same amount of time on identical letters, she will spend $20: 2=10$ minutes on the syllable "NA". Then she will spend
$16-10=6$ minutes on the syllable "LI". Therefore, she will spend $6+6=12$ m... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,824 |
# Task № 3.3
## Condition:
Petya is teaching his cat to read and for this, he makes him cards with words, burning them. He burned the word "MAMA" in 20 minutes, and the word "MISKA" in 35 minutes. He spends the same amount of time on identical letters, and possibly different times on different letters. How long will ... | Express the answer in minutes.
Answer: 50
Exact match of the answer - 1 point
Solution by analogy with problem № 3.1.
## Condition:
Grandma is embroidering her grandchildren's names on their towels. She embroidered the name "LANA" in 15 minutes, and the name "ALLA" in 12 minutes. She spends the same amount of time... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,825 |
# Task № 4.1
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a five-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars above me than knights below me!"
How many liars can live in this building? | # Answer: 3
## Exact match of the answer -1 point
## Solution.
Notice that the person living on the 5th floor is definitely lying, as there is no one living above them, including liars. Therefore, there are 0 liars above them, and 0 or more knights below. Now consider the resident on the 1st floor. They definitely t... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,826 |
# Task № 4.3
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a six-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars above me than knights below me!" How many liars can live in this building? | Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task No. 4.1.
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a five-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars b... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,828 |
# Task № 5.1
## Condition:
In a correct equation, identical digits were replaced with the same letters, and different digits with different letters. The result is
$$
P+\mathrm{P}+\mathrm{A}+3+Д+\mathrm{H}+\mathrm{U}+\mathrm{K}=\mathrm{U} \mathrm{U}
$$
What can U be equal to? | # Answer:
$\circ 1$
$\circ 2$
$\checkmark 3$
$\circ 4$
० 5
० 6
○ 7
○ 8
$\circ 9$
$\circ 0$
Exact match of the answer - 1 point
## Solution.
Notice that the expression involves 9 different letters, meaning all digits except one are used. On the left, we have the sum of eight different single-digit numbers. ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,829 |
# Task No. 6.1
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
Now it is easy to count the resulting pieces. For convenience, they are highl... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,830 |
# Task No. 7.1
## Condition:
On the Misty Planet, santiks, kubriks, and tugriks are in circulation. One santik can be exchanged for 1 kubrik or 1 tugrik, 1 kubrik can be exchanged for 3 santiks, and 1 tugrik can be exchanged for 4 santiks. No other exchanges are allowed. Jolly U, initially having 1 santik, made 20 ex... | # Answer: 6
## Exact match of the answer -1 point
## Solution.
To increase the number of santiks, we need to exchange them for kubriks or tugriks. In essence, we need to perform a double exchange, that is, first convert a santik into a tugrik or kubrik at a 1:1 ratio, and then increase the number of coins. Since the... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,832 |
# Task No. 7.2
Condition:
In the city of Abracodabra, funtics, tubrics, and santics are in circulation. One funtic can be exchanged for 1 tubric or 1 santic, 1 tubric for 5 funtics, and 1 santic for 2 funtics. No other exchanges are allowed. Lunatic, initially having 1 funtic, made 24 exchanges and now has 40 funtics... | Answer: 9
Exact match of the answer -1 point
Solution by analogy with task No. 7.1.
## Condition:
On the planet Mon Calamari, dataries, flans, and pegats are in circulation. One datary can be exchanged for 1 flan or 1 pegat, 1 flan - for 2 dataries, and 1 pegat - for 4 dataries. No other exchanges are allowed. Mera... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,833 |
# Task No. 8.1
## Condition:
Postman Pechkin thinks that his watch is 10 minutes slow, but in reality, it is 20 minutes fast. He agreed with Uncle Fyodor to meet at the post office at $12:00$, and then called and said that he wouldn't make it on time and would be 17 minutes late. At what time will Pechkin actually ar... | Write the answer in the format HH:MM.
Answer: $11:47$
Exact match of the answer -1 point
## Solution.
Since Pechkin's clock is 20 minutes fast, but he thinks it is 10 minutes slow, he adds half an hour to any actual time. Thus, at 11:30, Pechkin thinks it is already 12:00.
Accordingly, thinking that he is 17 minut... | 11:47 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,834 |
# Task No. 8.2
## Condition:
Dr. Watson thinks that his watch is 5 minutes slow, but in reality, it is 13 minutes fast. He agreed to meet Sherlock Holmes at 16:00, and then called and said that he wouldn't make it on time and would be 20 minutes late. At what time will Watson actually arrive for the meeting? | Write the answer in the format HH:MM.
Answer: 16:02
Exact match of the answer - 1 point
Solution by analogy with problem № 8.1.
# | 16:02 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,835 |
# Task № 8.3
## Condition:
Donkey Eeyore thinks that his clock is 20 minutes slow, but in reality, it is 15 minutes fast. He agreed to meet Piglet on the meadow at 10:00, but then realized he had more things to do and would leave 30 minutes later than planned. What time will Eeyore actually arrive at the meeting? Wri... | Answer: $9: 55$
Exact match of the answer - 1 point
Solution by analogy with problem № 8.1.
## Condition:
Hemul thinks that his watch is fast by 10 minutes, but in reality, it is slow by 15 minutes. He agreed with Snusmumrik to meet by the river at 14:00, and then decided to come 5 minutes earlier. At what time wil... | 9:55 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,836 |
1. $\left(7\right.$ points) Calculate $\frac{(2009 \cdot 2029+100) \cdot(1999 \cdot 2039+400)}{2019^{4}}$. | # Solution.
$2009 \cdot 2029+100=(2019-10) \cdot(2019+10)+100=2019^{2}-10^{2}+100=2019^{2}$.
$1999 \cdot 2039+400=(2019-20) \cdot(2019+20)+400=2019^{2}-20^{2}+400=2019^{2}$.
Then $\frac{2019^{2} \cdot 2019^{2}}{2019^{4}}=1$
Answer. 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,837 |
2. (7 points) Find all natural solutions to the equation $2 n-\frac{1}{n^{5}}=3-\frac{2}{n}$.
# | # Solution.
1 method. $2 n-\frac{1}{n^{5}}=3-\frac{2}{n}, 2 n-3=\frac{1}{n^{5}}-\frac{2}{n}, 2 n-3=\frac{1-2 n^{4}}{n^{5}}$.
For $n=1$ the equality is true, for $n>1 \quad 2 n-3>0, \frac{1-2 n^{4}}{n^{5}}<0$.
Answer. $n=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,838 |
3. (7 points) A traveler was riding in a bus and saw a two-digit number on a kilometer post. He fell asleep and woke up an hour later to see a three-digit number on the kilometer post, where the first digit was the same as the second digit an hour ago, the second digit was zero, and the third digit was the same as the ... | Solution. Let the first number the traveler saw be $\overline{x y}=10 x+y$. After an hour, the number became $\overline{y o x}=100 y+x$. After 2 hours, it became $\overline{y z x}=100 x+10 z+x$. Since the bus's speed is constant, $\overline{y o x}-\overline{x y}=\frac{1}{2} \cdot(\overline{y z x}-\overline{y o x})$, th... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,839 |
5. (7 points) In an acute-angled triangle $A B C$, the altitude $A H$, the greatest of the altitudes, is equal to the median $B M$. Prove that the angle $\angle A B C$ is less than $60^{\circ}$. | # Solution.
Drop perpendiculars from point $M$ to sides $B C$ and $A B$. $M K=\frac{1}{2} A H$, by the condition, $A H=B M$.
Therefore, $M K=\frac{1}{2} B M, \angle C B M=30^{\circ}$.
Simil... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,841 |
1. Can the first 8 natural numbers be arranged in a circle so that each number is divisible by the difference of its neighbors? | Answer: Yes.
Solution: For example, the numbers can be arranged as follows: $1,5,2,7,3,8,4,6$ (numbers 6 and $1-$ are adjacent).
Comment: Answer only - 0 points.
Correct example, even without verification - 7 points.
Note: The required arrangement is not unique. | Yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,842 |
2. For natural numbers $a$ and $b$, it is known that $5a-1$ is divisible by $b$, $a-10$ is divisible by $b$, but $3a+5$ is not divisible by $b$. What values can the number $b$ take? | Answer: 49.
Solution. The number $(5 a-1)-5 \cdot(a-10)=49$ is divisible by $b$, so either $b=1$, or $b=7$, or $b=49$. If the number $(3 a+5)-3 \cdot(a-10)=35$ were divisible by $b$, then $3 a+5$ would be divisible by $b$, which is not the case. Therefore, the options $b=1$ and $b=7$ are impossible. The numbers $a=10$... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,843 |
3. In triangle $A B C$, the angle bisectors $A A_{1}, B B_{1}$, and $C C_{1}$ are drawn. It is known that in triangle $A_{1} B_{1} C_{1}$, these lines are also angle bisectors. Is it true that triangle $A B C$ is equilateral? | Answer: Yes.
First solution. Consider triangles $A B_{1} A_{1}$ and $A C_{1} A_{1}$. Since by the condition $\angle B_{1} A A_{1}=\angle A_{1} A C_{1}$ and $\angle B_{1} A_{1} A=\angle A A_{1} C_{1}$, these triangles are equal by side and two angles. Therefore, $A B_{1}=A C_{1}$, so triangle $A B_{1} C_{1}$ is isoscel... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,844 |
4. In a train, there are several identical carriages. Let's call a carriage crowded if at least half of the seats are occupied. Prove that the percentage of people traveling in crowded carriages is not less than the percentage of crowded carriages. | First solution. Let the total number of wagons be $n$, the first $k$ of which are crowded, and the number of people in the $i$-th wagon be $a_{i}$. Then $100 \frac{a_{1}+\ldots+a_{k}}{a_{1}+\ldots+a_{n}}$ is the percentage of people traveling in crowded wagons, and $100 \frac{k}{n}$ is the percentage of crowded wagons.... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 13,845 |
5. There is an infinite grid plane, none of the points of which are painted blue. For one ruble, you can choose a cell and paint all its sides blue, even if some side was already painted. What is the minimum amount of money you need to pay to get a grid square $1001 \times 1001$, all grid lines inside and on the bounda... | Answer: 503000 rubles.
Solution: We will call a segment in the solution any segment connecting grid nodes at a distance equal to the side of a cell. Let $s$ be our $1001 \times 1001$ square. Any cell either has two common segments with the perimeter of $s$ (in which case it falls into a corner of $s$), or has no more ... | 503000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,846 |
1. Can the number 26 be represented as a sum of natural addends, the sum of the reciprocals of which equals 1?
# | # Answer: Possible.
Instructions. Example: $26=6+6+6+4+4.3 \cdot 1 / 6+2 \cdot 1 / 4=1 / 2+1 / 2=1$.
Criteria. If there is no correct example, then 0 points.
If there is a correct example and the calculation of the sum of reciprocals:
-7 points
One point is deducted for the absence of the calculation of the sum of... | 6+6+6+4+4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,847 |
2. The numbers from 1 to 8 are arranged in a circle. A number is called large if it is greater than its neighbors, and small if it is less than its neighbors. Each number in the arrangement is either large or small. What is the greatest possible sum of the small numbers? | Answer: 13.
Instructions. Adjacent numbers cannot be of the same type, so large and small numbers alternate, and there are four of each. 8 is large. 7 is also large, since a small number must be less than two numbers, and seven is less than only one. 1 and 2 are small. 6 and 5 cannot both be small, as they are less th... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,848 |
3. Find all pairs of numbers (a, b) for which the equality $(\mathrm{a}+\mathrm{b}-1)^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}-1$ holds. | Answer: all pairs of numbers $(1, t),(t, 1)$, where $t$ is any number.
By moving all terms to one side and using the difference of squares formula and factoring out the common factor, we get $2(a-1)(b-1)=0$. If the product is zero, then one of the factors is zero. Hence the answer.
Criteria. It is obtained that only o... | (1,),(,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,849 |
6. Do there exist natural numbers a and b such that a $<$ b and $\mathrm{b}^{2}+4 \mathrm{a}$ is a square of a natural number? | Answer: there are no such natural numbers.
Let such numbers exist. Note that $b^{2}<b^{2}+4 a<b^{2}+4 b+4=(b+2)^{2}$. Between the squares $\mathrm{b}^{2}$ and $(\mathrm{b}+2)^{2}$, there is only one square $(b+1)$, so $b^{2}+4 a=(b+1)^{2}$, from which $4 a=2 b+1$, but the equality of an even and an odd number is impos... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,850 |
11.1. On the plate, there are various pancakes with three fillings: 2 with meat, 3 with cottage cheese, and 5 with strawberries. Svetlana ate them all sequentially, choosing each subsequent pancake at random. Find the probability that the first and last pancakes eaten were with the same filling. | Answer: $\frac{14}{45}$.
Solution: Two pancakes with the same filling can be either with meat, cottage cheese, or strawberries. Let's calculate the probabilities of each of these events and add them up. Arrange the pancakes in the order they are eaten. The probability that the first pancake is with meat is $\frac{2}{1... | \frac{14}{45} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,851 |
11.2. Does there exist a number of the form $10000 \ldots 00001$ that can be represented as the sum $a!+b!+c!$, where $a, b, c$ are natural numbers? (The factorial of a natural number $n$ is defined as the product $n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$). | Answer. Does not exist.
Solution. Each number $n!$ for $n \geq 2$ is even. Therefore, among the numbers $a, b, c$, exactly one equals 1 (all three cannot equal 1, otherwise the sum of their factorials would be 3). Let's assume, for example, $c=1$. Then $a! + b! = 100 \ldots 00$. If both numbers $a$ and $b$ are not les... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,852 |
11.3. We consider all possible quadratic trinomials of the form $a x^{2}+b x+c$, where all coefficients $a, b, c$ are natural numbers not exceeding 100. Which trinomials are more numerous among them: those having at least one real root or those having none? | Answer. More than three terms without roots.
Solution. Let's find the number of pairs of natural numbers $(a, b)$, not exceeding 100, in which $b \geq 2a$. For $a=1$, there are 99 such pairs: $(1,2), (1,3), \ldots, (1,100)$; for $a=2$, there are 97: $(2,4), (2,5), \ldots, (2,100)$, and so on; for $a=50$, there is one ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,853 |
11.4. Real numbers $x, y, z$ are such that $x+y+z=2$ and $xy+yz+zx=1$. Find the greatest possible value of the quantity $x-y$. | Answer: $\frac{2 \sqrt{3}}{3}$.
Solution: Eliminate the variable $z$:
$$
1=x y+z(x+y)=x y+(2-x-y)(x+y)=2 x+2 y-x^{2}-y^{2}-x y
$$
from which
$$
x^{2}+y^{2}+x y+1=2 x+2 y
$$
Let $a=x+y$ and $b=x-y$, express everything in terms of $a$ and $b$:
$$
a^{2}-\frac{a^{2}-b^{2}}{4}+1=2 a \Leftrightarrow 3 a^{2}+b^{2}+4=8 a... | \frac{2\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,854 |
11.5. Given an acute-angled triangle $K L M$. Circles centered at $K$ and $M$ pass through point $L$, intersect again at point $P$, and intersect the circumcircle $\omega$ of triangle $K L M$ at points $S$ and $T$. Segment $L P$ intersects circle $\omega$ at point $O$. Prove that $O$ is the center of the circumcircle o... | Solution. First, we will prove that $O T=O P$. For this, we do not need the circle centered at point $K$. Consider the drawing without it. Let the central angle $L M T$ in the circle centered at $M$ be $2 \alpha$. Then the inscribed angle $T P L$ is $\alpha$. In the circle $\omega$, the angles $L M T$ and $L O T$ are i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,855 |
9.1 In a $7 \times 7$ grid, several cells were marked, with at least one cell marked in each row and each column. It turned out that in any row of the grid, an even number of cells were marked, and in any column, the number of marked cells is divisible by 3. What is the smallest and largest number of cells in the grid ... | Solution: Let's start with a lower bound estimate. By summing the number of cells in each row, we get that the total number of cells in the table is even. Summing by columns, we also get that the total number of cells is divisible by 3. Therefore, it is divisible by 6. Since the sum by columns is at least \(3 \cdot 7 =... | 2442 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,856 |
9.2 In Pokémon hunting, 11 adults and $n$ children participated. Together, they caught $n^{2}+3 n-2$ Pokémon, with all adults catching the same number, and all children catching the same number, but each child catching 6 fewer than an adult. How many children participated in the game? | Solution: Let each child catch $m$ pokemons. Then $n m+11(m+6)=n^{2}+3 n-2$. From this, $(n+11) m=n^{2}+3 n-68$. Therefore, the right side is divisible by $n+11$. We have $n^{2}+3 n-68=$ $n(n+11)-8(n+11)+20$, so 20 is divisible by $n+11$. The only divisor of 20 greater than 10 is 20 itself, so $n+11=20, n=9$.
## Crite... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,857 |
9.3 On the board, an acute-angled $\triangle A B C$ was drawn, the feet of the altitudes $A_{1}$ and $B_{1}$ were marked, as well as the midpoint $C_{1}$ of side $A B$. Then the entire drawing was erased, except for points $A_{1}, B_{1}$ and $C_{1$. Can the original $\triangle A B C$ be restored? | Solution 1: Consider an arbitrary acute-angled $\triangle ABC$ with angle $\angle C=60^{\circ}$. Since angles $\angle A A_{1} B$ and $\angle B B_{1} A$ are right, the quadrilateral $A B_{1} A_{1} B$ is inscribed and $\angle A_{1} B_{1} C=\angle A B C$. Then triangles $A B C$ and $A_{1} B_{1} C$ are similar with a coeff... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,858 |
9.4 Will there be three such irrational numbers that their sum and product are integers? | Solution 1: Let's take the numbers $\sqrt[3]{2}, \sqrt[3]{2}, -2 \sqrt[3]{2}$. Their sum is 0, and their product is -4.
Solution 2: Consider the polynomial $f(x)=x^{3}-30 x^{2}+31 x-1$. Since $f(0) < 0, f(2) > 0$, the equation $f(x)=0$ has three real roots. The numbers $\pm 1$ are not roots, so all roots of the equati... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,859 |
9.5 What is the largest number of non-overlapping groups into which all integers from 1 to 25 can be divided so that the sum of the numbers in each group is a perfect square? | Solution 1: A group consisting of a single number can only be formed by 5 squares. The remaining 20 numbers must be divided into groups of at least two. Therefore, there will be no more than 15 groups in total. Let's check that exactly 15 groups are not possible. Indeed, in such a case, the numbers 1, 4, 9, 16, 25 form... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,860 |
Problem 5. (I. Bogdanov) On an island, there live 2016 people, each of whom is either a knight or a liar. Knights always tell the truth, while liars always lie. Some of the residents are acquainted (if $\boldsymbol{A}$ is acquainted with $\boldsymbol{B}$, then $\boldsymbol{B}$ is acquainted with $\boldsymbol{A}$). Ever... | # Solution.
Lemma. Suppose in some company, every person in the company has recorded how many acquaintances they have in the company. Then the number of recorded odd numbers is even.
Proof. Consider the number of ordered pairs ( $A, B$ ) of acquaintances in this company. It is even, because with the pair ( $A, B$ ) t... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 13,862 |
5.1. The numbers $415, 43, 7, 8, 74, 3$ are written on cards (see figure). Arrange the cards in a row so that the resulting ten-digit number is the smallest possible. | Answer: $\square 3415 \square 43 \square 74 \square 7 \square 8$
The correct answer can simply be recorded as the number 3415437478.
+ correct answer provided
- incorrect answer or no answer provided | 3415437478 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,863 |
5.3. One side of a rectangle was increased by 3 times, and the other was reduced by 2 times, resulting in a square. What is the side of the square if the area of the rectangle is $54 \mathrm{m}^{2} ?$ | Answer: 9 m.
Let's reduce the side of the given rectangle by half (see Fig. 5.3a). Then the area of the resulting rectangle will be 27 m $^{2}$ (see Fig. 5.3b). Next, we will increase the other side by three times, that is, "add" two more rectangles (see Fig. 5.3c). The area of the resulting figure will become $27 \cd... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,865 |
5.4. The number 61 is written on the board. Every minute, the number is erased from the board and replaced with the product of its digits, increased by 13. That is, after one minute, the number on the board will be $19 (6 \cdot 1 + 13 = 19)$. What number will be on the board after an hour? | Answer: 16.
Let's consider the numbers that will be written on the board over the first few minutes:
| After one minute | $6 \cdot 1+13=\mathbf{1 9}$ |
| :---: | :---: |
| After two minutes | $1 \cdot 9+13=\mathbf{2 2}$ |
| After three minutes | $2 \cdot 2+13=\mathbf{1 7}$ |
| After four minutes | $1 \cdot 7+13=\math... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,866 |
# 1. CONDITION
Calculate $\operatorname{arctg} 1+\operatorname{arctg} 2+\operatorname{arctg} 3$. | Solution. Since $\operatorname{arctg} 1=\pi / 4$, it is sufficient to compute $\operatorname{arctg} 2 +$ $\operatorname{arctg} 3$. From the formula for the tangent of the sum of angles, it follows that if $\alpha=\operatorname{arctg} 2, \beta=$ $\operatorname{arctg} 3$, then $\operatorname{tg}(\alpha+\beta)=(2+3) /(1-2... | \frac{3\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,867 |
# 2. CONDITION
Do there exist two consecutive natural numbers, greater than 1000 000, such that the sums of their digits are perfect squares? | Solution. The desired numbers are, for example, $n=999999999$ and $\mathrm{n}+1=1000000000$, which have digit sums S(n) $=81, \mathrm{~S}(\mathrm{n}+1)=1$.
Answer: they exist. | theyexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,868 |
# 3. CONDITION
Prove that in the product $P=1! \times 2! \times \ldots \times 100!$, one of the factors can be erased so that the product of the remaining factors is a perfect square.
Proof. From the equality $(2k)! = (2k-1)! \times 2k$, it follows that the given product can be rewritten as $(1!)^2 \times 2 \times (3... | Answer: you can cross out 50 !.
# | 50! | Number Theory | proof | Yes | Yes | olympiads | false | 13,869 |
# 6. CONDITION
At the factory, there are exactly 217 women, among whom 17 are brunettes, and the remaining 200 are blondes. Before New Year's, all of them dyed their hair, and each of these women wrote on "VKontakte" the surnames of exactly 200 women from the factory, whom they believed to be definitely blondes. Each ... | Solution. According to the problem, the correct list of all 200 blondes will be on "Vkontakte" exactly for 17 female workers at the plant: brunettes will write exactly this list, and a blonde will never write it, as otherwise she would have to include herself in it. Therefore, if a certain list appears not 17 times, bu... | 13 | Combinatorics | proof | Yes | Yes | olympiads | false | 13,870 |
7.1 On an island, there live knights who always tell the truth, and liars who always lie. A traveler met islanders A and B. Islander A said: "At least one of us (A and B) is a liar." Can we determine whether A and B are knights or liars? | Solution: If A were a liar, then his statement would be a lie, but it turned out to be true. Therefore, A is a knight, and his statement is true. Consequently, 5 is a liar. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,871 |
7.2 Find all solutions to the numerical puzzle
$$
\mathrm{AX}+\mathrm{YX}=\mathrm{YPA}
$$
(different letters correspond to different digits, and the same letters correspond to the same digits). | Solution. It is clear that $\mathrm{Y}=1$ and A is an even digit. A cannot be less than 8, otherwise the result will not be a three-digit number. Therefore, $\mathrm{A}=8$.
$+\frac{$| $8 \mathrm{X}$ |
| :--- |
| $1 \mathrm{X}$ |}{$+1 \mathrm{P} 8$}
We see that when adding, there must be a carry of 1 from the units pl... | 89+19=108 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,872 |
7.317 students took a test. Each of them scored a whole number of points, and everyone had a different score. Each student scored less than the sum of any two. Could it have happened that Petya scored 15 points? | Solution. Let $\mathrm{a}_{1}a_{17}$ (each scored less than the sum of any two). From this, we conclude that $\mathrm{a}_{1}+\mathrm{a}_{2}>\mathrm{a}_{2}+15$, i.e., $\mathrm{a}_{1}>15$.
Thus, each, including Petya, scored more than 15 points. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,873 |
7.4 Vanya wrote the numbers $1,2,3, \ldots, 13$ in his notebook. He multiplied five of them by 3, and the rest by 7, then added all the products. Could the result have been 433? | Solution. Let's try to figure out the situation. Let A be the sum of the five numbers that need to be multiplied by 3, and B be the sum of the other eight numbers (which need to be multiplied by 7). Consider the sum after multiplication: 3A + 7B. We can write it as (taking into account that the sum of all thirteen numb... | 433 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,874 |
7.5 In triangle ABC, the median BM is drawn. It is known that $\angle \mathrm{BAC}=30^{\circ}, \angle \mathrm{BMC}=45^{\circ}$. Find angle BAC. | Solution. See fig.
C
$\triangle \mathrm{AMB}: \angle \mathrm{AMB}=135^{\circ} \Rightarrow \angle \mathrm{ABM}=15^{\circ}$
Let $\mathrm{AC}=2 \mathrm{~b}$. Then $\mathrm{AM}=\mathrm{MC}=\mat... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,875 |
# Problem №3
The base and the lateral side of an isosceles triangle are 34 and 49, respectively.
a) Prove that the midline of the triangle, parallel to the base, intersects the inscribed circle of the triangle.
b) Find the length of the segment of this midline that is contained within the circle. | # Answer: 8.
## Solution
a) Let $\mathrm{O}$ be the center of the inscribed circle in triangle $ABC$ with sides $AB = AC = 49$, $BC = 34$, and $AH$ be the height of the triangle. Points $M$ and $N$ are the midpoints of sides $AB$ and $AC$, respectively, and $K$ is the intersection point of $AH$ and $MN$. Since $MN$ i... | 8 | Geometry | proof | Yes | Yes | olympiads | false | 13,876 |
# Problem 4
Let $f(x)=x^{2}+3 x+2$. Calculate the expression
$\left(1-\frac{2}{f(1)}\right) \cdot\left(1-\frac{2}{f(2)}\right) \cdot\left(1-\frac{2}{f(3)}\right) \cdots\left(1-\frac{2}{f(2019)}\right)$. | Answer: $\frac{337}{1010}$
## Solution.
$$
\left(1-\frac{2}{f(n)}\right)=\frac{f(n)-2}{f(n)}=\frac{n^{2}+3 n}{n^{2}+3 n+2}=\frac{n(n+3)}{(n+1)(n+2)}
$$
Substituting this fraction for $\mat... | \frac{337}{1010} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,877 |
# Problem 5
Find all values of \(a\) for which the equation
$$
(\operatorname{tg} x+6)^{2}-\left(a^{2}+2 a+8\right)(\operatorname{tg} x+6)+a^{2}(2 a+8)=0
$$
has exactly two solutions on the interval \(\left[0 ; \frac{3 \pi}{2}\right]\). | Answer: $(-\sqrt{6} ;-2)(-2 ;-1) ; 4$.
## Solution.
Let $t=\operatorname{tg} x+6$, then the equation can be written as $t^{2}+\left(a^{2}+2 a+8\right) t+a^{2}(2 a+8)=0$, from which $t=2 a+8$ or $t=a^{2}$. Therefore, the solutions of the original equation are the solutions of the equations $\operatorname{tg} x=2 a+2$ ... | (-\sqrt{6};-2)(-2;-1);4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,878 |
1. An army of mice attacked a grain warehouse. To fight the mice, one cat was released into the warehouse on the first day, a second cat on the next day, and each subsequent day, one more cat was released. A mouse eats 100 grams of grain per day, and a cat eats 10 mice per day. After several days, there were no mice le... | Solution. Let the entire process last $n$ days, then by its end, there were $n$ cats on the warehouse, with the first cat eating $10 n$ mice, the second cat eating 10( $n-1)$ mice, $\ldots, n$-th cat - 10 mice, meaning there were a total of $10(1+2+\ldots+n)$ mice. Let's calculate how much grain the mice ate.
Ten mice... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,879 |
2. Find all pairs of real numbers $x, y$, satisfying the equation
$$
\frac{2 x}{1+x^{2}}=\frac{1+y^{2}}{2 y}
$$ | Solution. Transform the given equation to the form: $4 x y=\left(1+y^{2}\right)\left(1+x^{2}\right)$, which is equivalent to $1-2 x y+x^{2} y^{2}+x^{2}-2 x y+y^{2}=0$ or $(1-x y)^{2}+(x-y)^2=0$. From this, it follows that $x=y$ and $x y=1$, that is, the solutions to the given equation are two pairs: $x=y=1$ and $x=y=-1... | 1-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,880 |
3. In a convex quadrilateral $A B C D$, the midpoints of consecutive sides are marked: $M, N, K, L$. Find the area of quadrilateral $M N K L$, if $|A C|=|B D|=2 a,|M K|+|N L|=2 b$.
| Solution. Since $M N, N K, K L$ and $L M$ are the midlines of triangles $A B C, B C D, C D A$ and $D A B$ respectively, then $M N K L$ is a rhombus with side length $a$. Let $O$ be the point of intersection of its diagonals and consider triangle $M O N$. Since the diagonals of a rhombus are perpendicular to each other,... | b^{2}-^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,881 |
4. In a $25 \times 25$ table, integers from 1 to 25 are arranged, and in each row, the entire set of numbers appears. The table is symmetric with respect to the diagonal running from the top left to the bottom right. Prove that on this diagonal, all numbers from 1 to 25 also appear. | Solution. Suppose that not all numbers are found on the diagonal. Then the number that is missing from the diagonal can only be located above or below the diagonal. However, the table is symmetric relative to the diagonal, which means this number appears in the table an even number of times. But this is impossible, as ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,882 |
5. Does there exist an infinite increasing sequence of integers in which each element with an odd index (starting from the third) is the arithmetic mean, and each element with an even index is the geometric mean of its neighboring elements? | Solution. Yes, it exists. Here is one of the possible examples:
$$
1,1 \cdot 2,2^{2}, 2 \cdot 3,3^{2}, \ldots, n^{2}, n(n+1),(n+1)^{2}, \ldots
$$
Evaluation criteria. Only the answer "yes" - 0 points. Any reasoning that does not lead to the correct answer - 0 points. Several correct initial terms of the sequence with... | 1,1\cdot2,2^{2},2\cdot3,3^{2},\ldots,n^{2},n(n+1),(n+1)^{2},\ldots | Algebra | proof | Yes | Yes | olympiads | false | 13,883 |
3. Prove that the number $5^{120}+2^{150}$ is divisible by 13. | Solution. Since
$$
5^{120}+2^{150}=\left(5^{2}\right)^{60}+\left(2^{6}\right)^{25}=(13 \cdot 2-1)^{60}+(13 \cdot 5-1)^{25},
$$
the remainder of this number when divided by 13 is the same as that of the number $(-1)^{60}+(-1)^{25}$, which is 0.
Evaluation criteria: Finding the remainder of one of the numbers $5^{120}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,884 |
5. Curious Dima has 2022 identical buttons, with one side of each button being black and the other side white. Dima arranged the buttons in a circle such that 2021 buttons are white side up, and the last button is black side up. Dima came up with a game. In one move, you can simultaneously flip two buttons in only two ... | Solution. Let's number the buttons clockwise in order: 1, 2, 3, ..., 2022 (starting, for example, with a black button). We will divide all the buttons into two sets: $S_{\text {o }}$ - buttons with odd numbers, $\mathrm{S}_{\text {e }}$ - buttons with even numbers. In $\mathrm{S}_{\text {o }}$ and in $\mathrm{S}_{\text... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,885 |
Task No. 1.1
## Condition:
A worker is laying tiles so that tiles of the same color do not touch each other. He has tiles of three colors: blue, green, and orange. On the diagram, the colors of two tiles are indicated. What color tile will the worker place in the spot marked with a «?»?
We will sequentially color the tiles. Tile 1 can only be green because it borders orange and blue. Tile 2 borders green and blue, so it is orange. Then tile 3 is blue, tile 4 is... | notfound | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 13,886 |
# Task № 3.4
## Condition:
Cupcakes are sold in boxes of 2, eclairs - in boxes of 6, and gingerbread - in boxes of 15. You can only buy whole boxes, you cannot open them. Alice bought an equal number of cupcakes, eclairs, and gingerbread. What is the smallest number of boxes she could have taken | Answer: 22
Exact match of the answer - 1 point
Solution by analogy with task №3.1
# | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,887 |
# Task № 4.1
## Condition:
Fifth-graders Olya, Fyodor, Katya, and Dima participated in a school olympiad. Out of 8 problems offered at the olympiad, each of the students solved more than four. It is known that:
- Olya solved more problems than Katya,
- Fyodor and Dima solved the same number of problems,
- Dima solve... | # Solution.
If the boys solved 5 problems each, then Olya solved 6 problems, and Katya solved 27-5-5-6=11 problems, which cannot be.
If the boys solved 6 problems each, then Olya solved 7 problems, and Katya solved 27-6-6-7=8 problems, which cannot be, because Katya solved fewer problems than Olya.
If the boys solve... | Olya-8,Fyodor-7,Katya-5,Dima-7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,888 |
# Task № 4.2
## Condition:
Fifth-graders Olya, Fyodor, Katya, and Dima participated in a school olympiad. Out of 8 problems offered at the olympiad, each of the students solved more than four. It is known that:
- Olya solved more problems than Katya,
- Fyodor and Dima solved the same number of problems,
- Dima solve... | Solution by analogy with task №4.1
# | Olya-7,Fyodor-7,Katya-6,Dima-7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,889 |
# Task № 4.3
## Condition:
Fifth-graders Olya, Fyodor, Katya, and Dima participated in a school olympiad. Out of 8 problems offered at the olympiad, each of the students solved more than four. It is known that:
- Olya solved more than Katya,
- Fyodor and Dima solved the same number of problems,
- Dima solved one mor... | Solution by analogy with task №4.1
# | Olya-8,Fyodor-6,Katya-5,Dima-6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,890 |
# Task No. 5.1
## Condition:
A Dog, a Cat, and a Mouse are running around a circular lake. They all started simultaneously in the same direction from the same point and finished at the same time, each running at a constant speed. The Dog ran 12 laps, the Cat ran 6 laps, and the Mouse ran 4 laps. How many total overta... | # Answer: 13
## Exact match of the answer -1 point
## Solution.
At the moment when the faster runner catches up with the slower one, he is ahead by one lap. The Dog has lapped the Cat by 6 laps, meaning it has caught up with her 6 times, with the last catch-up at the finish line not counting as a lap, so the Dog has... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,892 |
# Task No. 5.3
## Condition:
A Dog, a Cat, and a Mouse are running around a circular lake. They all started in the same direction from the same point and finished at the same time, each running at a constant speed.
The Dog ran 12 laps, the Cat ran 7 laps, and the Mouse ran 3 laps. How many total overtakes were made ... | # Answer: 15
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,894 |
# Task № 5.4
## Condition:
A Dog, a Cat, and a Mouse are running around a circular lake. They started simultaneously in the same direction from the same point and finished simultaneously, all running at constant speeds.
The Dog ran 12 laps, the Cat ran 5 laps, and the Mouse ran 2 laps. How many total overtakes were ... | Answer: 17
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,895 |
# Task № 6.1
## Condition:
Anya, Tanya, and Vanya had identical cardboard squares with a side of 16 cm. Each of them cut off two rectangles from their square, as shown in the figure, and all 6 rectangles are the same. The perimeter of Anya's figure is 88 cm, and the perimeter of Vanya's figure is 82 cm.
=a x^{2}+b x, \quad g(x)=c x^{2}+d x, \quad f_{1}(x)=a x+b$, $g_{1}(x)=c x+d$, intersect at one point with a negative abscissa. Prove that if $a c \neq 0$, then $b c=a d$. | Solution. Let's find the abscissa $x_{0}$ of the intersection point of $f(x)$ and $f_{1}(x)$: $a x^{2} + b x = a x + b \Leftrightarrow (a x + b)(x - 1) = 0$. Therefore, either $x_{0} = 1$ or $x_{0} = -\frac{b}{a} (a c \neq 0)$. Since the abscissa of the intersection point is negative by condition, $x_{0} \neq 1$. Thus,... | Algebra | proof | Yes | Yes | olympiads | false | 13,899 | |
10.2. 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 one player was disqualified, and all points earned in matches against him were annulled, and this player was removed from the table. Could Petya hav... | Answer: Could not.
Solution. In a tournament with 10 players, played in 1 round, $\frac{10 \cdot 9}{2}=45$ points are distributed. Therefore, there will be a player who has scored no more than $45: 10=4.5$ points. This means that Pete, who took the absolute last place, scored no more than 4 points. Similarly, in a tou... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,900 |
10.3. Each of the 13 boys thought of an integer. It turned out that the sum of the numbers they thought of was 125. After that, each changed their number: either divided it by 3, or multiplied it by 5. Could the sum of the 13 resulting numbers be $175?$ | Answer: Could not.
Solution: Suppose the sum could have become 175. Multiply each of the resulting numbers by 3 (then the sum will be 525). This is equivalent to some of the numbers the boys thought of remaining unchanged, while others were multiplied by 15. If a number $x$ is multiplied by 15, the sum changes by $14x... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,901 |
10.4. The numbers $x, y, z$ are such that $2 x>y^{2}+z^{2}, 2 y>x^{2}+z^{2}, 2 z>y^{2}+x^{2}$. Prove that $x y z<1$. | First solution. From the condition, it follows that the numbers $x, y, z$ are positive. Note that $y^{2}+z^{2} \geq 2 y z$ (this inequality is equivalent to $(y-z)^{2} \geq 0$). Therefore, $x>y z$. Similarly, $y>x z, z>y x$. Multiplying the inequalities (both sides of the inequalities are positive), we get $x y z>(x y ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,902 |
10.5. Quadrilateral $A B C D$ is inscribed in circle $\Omega$ with center $O$, and $B D$ is the diameter of the circle. The rays $A B$ and $D C$ intersect at point $S$. Circle $\omega$, passing through points $A, O, C$, intersects segment $C D$ at point $M (M \neq C)$. Prove that $M$ is the midpoint of segment $D S$. | Solution. From the condition, it follows that the quadrilateral $A O M C$ is inscribed (see Fig. 3), so $\alpha=\angle O A C=180^{\circ}-\angle O M C=\angle O M D$ (adjacent angles). Triangle $O A C$ is isosceles ($O A=O C$ as radii). Therefore, $\angle O C A=\alpha$. Similarly, $\angle O C D=\angle O D C=\beta$. Then ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,903 |
1. Dima wrote a sequence of 0s and 1s in his notebook. Then he noticed that a 1 follows a 0 sixteen times, a 0 follows a 1 fifteen times, and a 0 follows 01 eight times. How many times does a 0 follow 11? | # Answer. 7.
Solution. The combination 01 appears 16 times in the tetrad, while the combination 10 appears 15 times. Therefore, the string starts with 0, meaning that before each combination 10 there is either 0 or 1. According to the condition, eight times it is 0, so the combination 110 appears $15-8=7$ times.
Crit... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,904 |
2. Masha drew a rhombus in her notebook and wrote four numbers at the vertices, the sum of which is 2021. Then she wrote the product of the numbers at the ends of each side of the rhombus and calculated the sum of the numbers written on the sides. Katya increased all the numbers written by Masha at the vertices by 1, a... | Answer: 4046.
Solution. Let Masha write the numbers $a, b, c, d$ at the vertices of the rhombus, then $a+b+c+d=2021$, and she obtained the sum of the products
$$
a b+b c+c d+d a=(a+c)(b+d)
$$
Since Katya increased each number by 1, her sum is
$$
(a+c+2)(b+d+2)=(a+c)(b+d)+2(a+b+c+d)+4
$$
greater than Masha's by $2(... | 4046 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,905 |
4. Sasha wrote a natural number $M$ on the board and calculated the sum of its digits, which turned out to be 100, and then noticed that if the number $M$ is multiplied by 5, the sum of the digits becomes 50. He then told this to Masha. Masha immediately declared that $M$ is even. Is Masha wrong? | Answer. No mistake.
Solution. Let $s(A)$ denote the sum of the digits of the number $A$. From the consideration of the column addition of two numbers $A$ and $B$, it follows that $s(A+B) \leqslant s(A)+s(B)$, and equality is achieved if and only if there are no carries across digits during the addition. Thus, from the... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,907 |
5. Petya and Vasya found two bags of caramels and one huge chocolate bar. They decided to divide what they found by playing the following game: on the first move, Petya takes several caramels from one of the bags (he chooses the bag from which he takes the caramels at his discretion) and moves the same number of carame... | Answer. a) Vasya; b) Petya.
Solution. a) Symmetric strategy. Let Vasya take as many candies on each move as Petya, but from the other bag. Obviously, in this case, Vasya can always make a counter move. Each time after Vasya's move, the difference in the number of candies in the bags will be the same as at the beginnin... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,908 |
1. Integers $\mathrm{a}, \mathrm{b}$, c and $\mathrm{d}$ satisfy the equation $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=\mathrm{d}^{2}$. Prove that the number abc is divisible by 4. (6 points) | Solution. The square of an even number is divisible by 4, while the square of an odd number gives a remainder of 1 when divided by 4. If the numbers a, b, c are all odd, then $d^{2}$ would have to give a remainder of 3 when divided by 4, which is impossible. If among the numbers $a, b, c$ there are two odd and one even... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,909 |
2. Prove that in any company there will be two people who have the same number of acquaintances in this company (if A is acquainted with B, then B is also acquainted with A). (6 points) | Solution. Let there be $\mathrm{k}$ people in the company. Then each person can have from zero to $(\mathrm{k}-$ 1) acquaintances. Suppose the opposite: the number of acquaintances is different for everyone. Then there will be a person with no acquaintances, a person with one acquaintance, and so on, finally, there wil... | \frac{2}{7}=\frac{1}{4}+\frac{1}{28} | Combinatorics | proof | Yes | Yes | olympiads | false | 13,910 |
5. On the bases $AB$ and $CD$ of trapezoid $ABCD$, points $K$ and $L$ are taken. Let $E$ be the point of intersection of segments $AL$ and $DK$, and $F$ be the point of intersection of $BL$ and $CK$. Prove that the sum of the areas of triangles $\triangle ADE$ and $\triangle BCF$ is equal to the area of quadrilateral E... | Solution.
We have $S_{\triangle A D K}=S_{\triangle A L K}$, since they share the same base $A K$ and have equal heights, which coincide with the distance between the parallel lines $A B$ and... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,911 |
8.2. Biathletes Arina and Lisa start on the same distance. Their starting numbers are two-digit numbers with the following peculiarity: if the sum of the digits of the starting number is added to the square of the difference of the digits of the number, the result is this number. Find the starting numbers of Arina and ... | # Solution.
Let 10a+b be the desired number. According to the condition $\mathrm{a}+\mathrm{b}+(\mathrm{a}-\mathrm{z})^{2}=10 \mathrm{a}+\mathrm{b}$ or $(a-в)^{2}=9 a$, from which it follows that $a-$ is a perfect square.
By checking the three cases ( $a=1, \mathrm{a}=4$ and $\mathrm{a}=9$ ), we get two solutions: 14... | 1490 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,912 |
8.3. Given an arbitrary trapezoid. Compare the difference of its lateral sides and the difference of the bases. Which is greater? Justify your answer. (7 points)
# | # Solution.
Let in trapezoid ABCD the larger base be AD, and the larger lateral side be AB. Draw segment BK $\| \mathrm{CD}$. Then, by the triangle inequality, $\mathrm{AB}<\mathrm{BK}+\mathrm{AK}$. Considering that $\mathrm{AK}=\mathrm{AD}-\mathrm{BC}$, and $\mathrm{BK}=\mathrm{CD}$, we get $\mathrm{AB}-\mathrm{CD}<\... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,913 |
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