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3. Anya and Danya together weigh 82 kg, Danya and Tanya - 74 kg, Tanya and Vanya - 75 kg, Vanya and Manya - 65 kg, Manya and Anya - 62 kg. Who is the heaviest and how much does he/she weigh? | Answer. Vanya weighs 43 kg.
Solution. Adding the weights given in the condition: $82+74+75+65+62=358$, we get the doubled weight of all the children. That is, all the children together weigh $358 / 2=179$.
Anya, Danya, Tanya, and Vanya together weigh $82+75=157$, so Manya weighs 179 $157=22$.
Similarly, we find that... | 43 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A student passed 31 exams over 5 years of study. Each subsequent year, he passed more exams than the previous year, and in the fifth year, he passed three times as many exams as in the first year. How many exams did he pass in the fourth year? | # Answer: 8.
Solution: Let $a, b, c, d, e$ be the number of exams taken in each year of study. According to the problem, $a+b+c+d+e=31, a<b<c<d<e$.
Replace the numbers $b, c, d, e$ in the equation with definitely not larger values:
$a+(a+1)+(a+2)+(a+3)+3a \leq 31$. We get $7a \leq 25$. Replacing these numbers with d... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A circle with center at point $O$ is circumscribed around quadrilateral $A B C D$. The diagonals of the quadrilateral are perpendicular. Find the length of side $B C$, if the distance from point $O$ to side $A D$ is 1. | Answer: $B C=2$
Solution. Let $O E \perp A D$, then $O E=1$. Draw a line through point $A$ perpendicular to $A D$, which intersects the circle at point $M$. Then $D M$ is a diameter and $D O=O M$. Since $\angle D B A=\angle D M A$ (as inscribed angles subtending the same arc), $\angle M D A=90^{\circ}-\angle D M A$ an... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.3
In a convex pentagon $P Q R S T$, angle $P R T$ is half the size of angle $Q R S$, and all sides are equal. Find angle $P R T$.
## Number of points 7 | Answer:
$30^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle P R Q+\angle T R S=\angle P R T(*)$.
Figure a
. How many minutes will it take him to trace all the lines of a checkered square $5 \times 5$? | Answer: 30 minutes.
Solution. Note that the grid rectangle $3 \times 7$ consists of four horizontal lines of length 7 and eight vertical lines of length 3. Then the total length of all lines is $4 \cdot 7 + 8 \cdot 3 = 52$. It turns out that Igor spends $26 : 52 = 1 / 2$ minutes on the side of one cell.
The rectangle... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other? | Answer: 8.
Solution: Note that it is impossible to arrange all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor, the card with the number 1. Therefore, both cards 5 and 7 must be at the ends, and the card with the number 1 must be adjacent ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. Karlson counts 200 buns baked by Fräulein Bock: «one, two, three, ..., one hundred and nine, one hundred and ten, two hundred». How many words will he say in total? (Each word is counted as many times as it was said.) | Answer. 443 words.
Solution. One word is required to pronounce 29 numbers: $1,2,3,4,5,6,7,8,9,10,11,12$, $13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,100,200$.
Among the first 99 numbers, the number of those pronounced in two words: $99-27=72$, thus, the number of words required for their pronunciation is $2 \cdot 7... | 443 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Ivan and Petr are running in the same direction on circular tracks with a common center, and initially, they are at the minimum distance from each other. Ivan completes one full circle every 20 seconds, while Petr completes one full circle every 28 seconds. After what least amount of time will they be at the maximum... | Answer: 35 seconds.
Solution. Ivan and Petr will be at the minimum distance from each other at the starting points after the LCM $(20,28)=140$ seconds. In this time, Ivan will complete 7 laps, and Petr will complete 5 laps relative to the starting point. Consider this movement in a reference frame where Petr is statio... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Rational numbers $a, b$ and $с$ are such that $(a+b+c)(a+b-c)=2 c^{2}$. Prove that $c=0$.
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Translation:
4. Rational numbers $a, b$ and $c$ are such that $(a+b+c)(a+b-c)=2 c^{2}$. Prove that $c=0$. | Solution. The initial equality is equivalent to the following $(a+b)^{2}-c^{2}=2 c^{2}$, or $(a+b)^{2}=3 c^{2}$. If $c \neq 0$, we get $((a+b) / c)^{2}=3 .|(a+b) / c|={ }^{-}$. On the left, we have a rational number, since the sum, quotient, and absolute value of rational numbers are rational, while on the right, we ha... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
7.1. Find the number of three-digit numbers for which the second digit is less than the third by 3. | 7.1. The first digit of the number can be chosen in 9 ways (it can be any digit from 1 to 9), the second digit in 7 ways (it can be any digit from 0 to 6), and the third digit is uniquely determined. We get $9 \cdot 7 \cdot 1=63$ three-digit numbers that satisfy the condition of the problem.
Answer: 63 | 63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. A plot of $80 \times 50$ meters is allocated for gardens and is fenced on the outside. How should 5 straight fences of the same length be installed inside the plot to divide it into 5 rectangular plots of equal area? | 7.3. One of the possible solutions is shown in the figure. The fences have a length of 40 m, their ends are marked with bold dots (the horizontal segment is composed of two fences).
 | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.4. The monkey becomes happy when it eats three different fruits. What is the maximum number of monkeys that can be made happy with 20 pears, 30 bananas, 40 peaches, and 50 tangerines? | 7.4. Let's set the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed each monkey no more than one tangerine, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. We will show how to satisfy 45 monkeys:
5 monkeys eat a pear, a banana, ... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.5. Zya decided to buy a crumblik. In the store, they also sold kryambliks. Zya bought a kryamblik and received coupons worth $50\%$ of the cost of the purchased kryamblik. With these coupons, he was able to pay $20\%$ of the cost of the crumblik. After paying the remaining amount, he bought the crumblik as well. By w... | 7.5. From the condition of the problem, it follows that $50 \%$ of the cost of a kryamblik is equal to $20 \%$ of the cost of a krumblik. This means that the kryamblik constitutes $40 \%$ of the cost of the krumblik. Zya paid the full cost of the kryamblik and the remaining $80 \%$ of the cost of the krumblik. Thus, he... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. What is the sum of the digits of the number $A=100^{40}-100^{30}+100^{20}-100^{10}+1$? | # Answer: 361.
Solution. The number is the sum of three numbers: a number composed of 20 nines followed by 60 zeros, a number composed of 20 nines followed by 20 zeros, and finally the number 1. All nines and the one fall on the zeros of the other addends, so there is no carry-over, and the answer is $180+180+1=361$.
... | 361 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Given three quadratic trinomials $f(x)=a x^{2}+b x+c, g(x)=b x^{2}+c x+a, h(x)=c x^{2}+$ $a x+b$, where $a, b, c$ are distinct non-zero real numbers. From them, three equations were formed: $f(x)=g(x), f(x)=h(x), g(x)=h(x)$. Find the product of all roots of these three equations, given that each of them has two d... | Answer. 1.
Solution. Since it is known that all equations have roots, we can use Vieta's theorem. Then the product of all roots will be equal to $\frac{c-a}{a-b} \cdot \frac{a-b}{b-c} \cdot \frac{b-c}{c-a}=1$.
Comment. Correct answer without justification - 0 points. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. Toshi is traveling from point A to point B via point C. From A to C, Toshi travels at an average speed of 75 km/h, and from C to B, Toshi travels at an average speed of 145 km/h. The entire journey from A to B took Toshi 4 hours and 48 minutes. The next day, Toshi travels back at an average speed of 100 km/h. The ... | Answer: 290 km.
Solution. Let x km be the distance between B and C, and y km be the distance between A and C. From the system of equations $x / 145 + y / 75 = 24 / 5$ and $(x + y) / 100 = 2 + y / 70$, we find: $x = 290$ and $y = 210$.
. Let the white square be in the bottom-lef... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.1. Mother gives pocket money to her children: 1 ruble to Anya, 2 rubles to Borya, 3 rubles to Vitya, then 4 rubles to Anya, 5 rubles to Borya and so on until she gives 202 rubles to Anya, and 203 rubles to Borya. By how many rubles will Anya receive more than Vitya? | # Answer: 68
Solution. Note that Anya will receive money one more time than Vitya. If we remove the first ruble, then each subsequent time Anya receives one more ruble than Vitya. Thus, it remains to determine how many such times there were. Apart from 1 ruble and 2 rubles, all the "moves" by mom break down into tripl... | 68 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. In a hotel, rooms are arranged in a row in order from 1 to 10000. Masha and Alina checked into the hotel in two different rooms. The sum of the room numbers they are staying in is 2022, and the sum of the room numbers of all the rooms between them is 3033. In which room is Masha staying, if her room has a lower nu... | Answer: 1009
Solution. If there is exactly one room between Masha's and Alina's rooms, then it is room number 3033, but then Masha's and Alina's rooms are 3032 and 3034, and their sum is not 2022.
Consider room B, whose number is 1 more than Masha's room. Also consider room V, whose number is 1 less than Alina's room... | 1009 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. In a batch, there are 90000 boxes of vaccine weighing 3300 grams each, and 5000 small boxes weighing 200 grams each. What is the minimum number of temperature-controlled containers needed if, according to the new rules, no more than 100 kilograms can be placed in each? | Answer: 3000
Solution. Note that no more than 30 boxes of vaccines can be placed in one container. Thus, less than $90000 / 30=3000$ containers would not be enough. Also, note that if a container holds 30 boxes, there is still room for 5 small boxes, so all these small boxes will fit into the free spaces. | 3000 | Other | math-word-problem | Yes | Yes | olympiads | false |
9.5. What is the largest number of different natural numbers that can be chosen so that the sum of any three of them is a prime number? | Answer: 4 numbers. Example: $1,3,7,9$. Indeed, the numbers $1+3+7=11, 1+3+9=13$, $1+7+9=17, 3+7+9=19$ are prime. Evaluation. Note that among five natural numbers, it is always possible to choose three whose sum is a composite number. Consider the remainders of these five numbers when divided by 3. If there are three id... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. Ten guides were leading ten tour groups through the halls of the Hermitage. The groups did not necessarily have equal numbers, but on average, there were 9 people in each group. When one of the groups finished their tour, the average number of tourists per group decreased to 8 people. How many people were in the g... | Answer: 18.
Solution: The total number of excursionists initially was $9 \cdot 10=90$. After one of the groups finished the excursion, $8 \cdot 9=72$ excursionists remained. Therefore, the group that finished the excursion had $90-72=18$ excursionists.
Comment: A correct answer without justification - 0 points. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. In the school gym, there is one arm wrestling table. The physical education teacher organized a school tournament. He calls any two participants who have not yet met each other for a match. There are no draws. If a participant loses twice, they are eliminated from the tournament. After 29 matches were held, all pa... | Answer: 16.
Solution: Each participant is eliminated after exactly two losses. In the situation where two "finalists" remain, the total number of losses is 29. If $n$ people have been eliminated from the tournament, then they have collectively suffered $2 n$ losses, while the "finalists" could have $0(0+0), 1(0+1)$ or... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.4. Four pirates divided a treasure of 100 coins. It is known that among them, there are exactly two liars (who always lie) and exactly two knights (who always tell the truth).
They said:
First pirate: “We divided the coins equally.”
Second pirate: “Everyone has a different number of coins, but each got at least 15... | Answer: 40 coins
Solution. Note that the first and fourth pirates could not have told the truth simultaneously.
If the first pirate is a knight, then everyone received 25 coins (such a situation is possible). In this case, the maximum number is 25.
If the fourth pirate is a knight, then each received no more than 35... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. One hundred non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among these 100 written numbers? | Answer: 50.
Solution: Note that two consecutive numbers cannot both be positive (i.e., natural numbers). Suppose the opposite. Then their product is positive, and the number before them (counterclockwise) is also a natural number. Since it is greater than the product of these two natural numbers, it is greater than ea... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Sergey wrote down the numbers from 500 to 1499 in a row in some order. Under each number, except the leftmost one, he wrote the GCD of that number and its left neighbor, obtaining a second row of 999 numbers. Then he applied the same rule to get a third row of 998 numbers, from it a fourth row of 997 numbers, and so... | Answer: 501 (in the second variant 10001).
Let's prove that all numbers in the 501st row are already equal to 1. Indeed, if there is a number $d \neq 1$ in it, then in the 500th row there are 2 numbers divisible by $d$, in the 449th row - 3 such numbers, ..., in the 1st row there are 501 such numbers. But no number $d... | 501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a regular 20-gon, four consecutive vertices $A, B, C$ and $D$ are marked. Inside it, a point $E$ is chosen such that $A E=D E$ and $\angle B E C=2 \angle C E D$. Find the angle $A E B$. | Answer: $39^{\circ}$ (in the $2^{nd}$ variant: $36^{\circ}$).
Note that $ABCD$ is an isosceles trapezoid with angles $\angle ABC = \angle DBC = 180^{\circ} \cdot 18 / 20 = 162^{\circ}$. Point E lies on the perpendicular bisector of the base $AC$, and therefore, triangle $BEC$ is isosceles. Draw the height $EH$ in it, ... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The exam consists of $N \geqslant 3000$ questions. Each of the 31 students has learned exactly 3000 of them, and every question is known by at least 29 students. Before the exam, the teacher openly laid out all the question cards in a circle. He asked the students to point to one of the questions and explained that ... | Answer: $N=3100$.
Each student does not know $N-3000$ questions and thus can mentally mark exactly that many cards which do not suit them as the initial one. Together, they can indicate no more than $31(N-3000)$ different cards. If $31(N-3000)<N$, then the students can indicate a card that suits everyone. This inequal... | 3100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Ten different natural numbers are such that the product of any 5 of them is even, and the sum of all 10 numbers is odd. What is their smallest possible sum | Answer: 65.
Solution. The product of 5 odd numbers is odd $\Rightarrow$ among the given ten numbers, there are no more than 4 odd numbers. 4 odd numbers are also impossible, since then the sum of all numbers would be even, while according to the condition, it is odd. Therefore, there are a maximum of 3 odd numbers. If... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. Solve the inequality: $\sqrt{(x-2)^{2}\left(x-x^{2}\right)}<\sqrt{4 x-1-\left(x^{2}-3 x\right)^{2}}$. | 11.1. Answer: $x=2$.
The left side of the original inequality is defined at $x=2$ and. At the point $x=2$, the inequality is true. We will prove that there are no solutions on the interval $[0 ; 1]$. For this, we will square both sides and bring the inequality to the form $x^{2}(1-x)<-1$. The last inequality is not sa... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. All horizontal and vertical distances between adjacent points are equal to 1. What is the area of the triangle with vertices at the black points? | Answer. 1.
Solution. The area of the triangle can be found, for example, by subtracting from half the area of the square the area of the square and the areas of two right triangles. We get $S=10-4-2-3=1$. The figure can be divided in other ways.
Comment. An answer without justification - 0 points. A partition that al... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. There were 1009 gnomes with 2017 cards, numbered from 1 to 2017. Ori had one card, and each of the other gnomes had two. All the gnomes knew only the numbers on their own cards. Each gnome, except Ori, said: "I am sure that I cannot give Ori any of my cards so that the sum of the numbers on his two cards would be 20... | Answer: 1009
Solution: If one of the gnomes has a card with the number 1, then he must be sure that Ori does not have a card with the number 2017. This can only be certain if the card with the number 2017 is also in the hands of this gnome. Similarly, the gnome with the card that has the number 2 has another card with... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Vanya wrote down a four-digit number, subtracted a two-digit number from it, multiplied the result by a two-digit number, divided by the sum of two single-digit numbers, added a single-digit number, and then divided the result by the sum of three single-digit numbers. To write all the numbers, he used only one digit... | Answer: 2017; any digit.
Solution. Let the digit be $a$. We get $\left(\frac{(\overline{a a a a}-\overline{a \bar{a}} \cdot \cdot \overline{a a}}{a+a}+a\right):(a+a+a)$. Then
$$
\overline{a a a a}-\overline{a a}=\overline{a a 00}=a \cdot 1100 \Rightarrow \frac{\bar{a}}{a a}=a \cdot 11
$$
The numerator of the fractio... | 2017 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The giants were prepared 813 burgers, among which are cheeseburgers, hamburgers, fishburgers, and chickenburgers. If three of them start eating cheeseburgers, then in that time two giants will eat all the hamburgers. If five take on eating hamburgers, then in that time six giants will eat all the fishburgers. If sev... | Answer: 252 fishburgers, 36 chickenburgers, 210 hamburgers, and 315 cheeseburgers.
Solution: Let $a, b, c$, and $d$ be the quantities of cheeseburgers, hamburgers, fishburgers, and chickenburgers, respectively. According to the problem, $a + b + c + d = 813$. The statement that while three people eat cheeseburgers, tw... | 252 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle AB... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $ABCD$, it is known that $\angle BAC = \angle CAD = 60^{\circ}$, $AB + AD = AC$. It is also known that $\angle ACD = 23^{\circ}$. How many degrees does the angle $ABC$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
. 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 th... | Answer: 9 knights.
Solution. Estimation. Note that none of the knights could have said the phrase "My number is greater than 10," otherwise the number they thought of would indeed be greater than 10. But then he could not have said any of the phrases "My number is less than 1," "My number is less than 2," ..., "My num... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.3. The cells of a $2 \times 2019$ table must be filled with numbers (exactly one number in each cell) according to the following rules. In the top row, there should be 2019 real numbers, none of which are equal, and in the bottom row, there should be the same 2019 numbers, but in a different order. In each of the 20... | Answer: 2016.
Solution. Estimation. We will prove that in the first row of the table, where numbers are arranged according to the rules, there are no fewer than three rational numbers (and, accordingly, no more than 2016 irrational numbers). Each number appearing in the table is written in exactly two cells, one of wh... | 2016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Petya was exchanging stickers. He trades one sticker for 5 others. At first, he had 1 sticker. How many stickers will he have after 50 exchanges? | Answer: 201.
Solution: After each exchange, the number of Petya's stickers increases by 4 (one sticker disappears and 5 new ones appear). After 50 exchanges, the number of stickers will increase by 50*4=200. Initially, Petya had one sticker, so after 50 exchanges, he will have $1+200=201$. | 201 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. One day Uncle Fyodor weighed Sharik and Matroskin. It turned out that Sharik is 6 kg heavier than Matroskin, and Matroskin is three times lighter than Sharik. How much did Matroskin weigh? | Answer: 3 kg.
Solution: Since Matroskin is three times lighter than Sharik, Matroskin is lighter than Sharik by two of his own weights. According to the condition, this is equal to 6 kg, i.e., Matroskin weighs $6: 2=3$ kg. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Mom bought a box of lump sugar (sugar in cubes). The children first ate the top layer - 77 cubes, then the side layer - 55 cubes, and finally, the front layer. How many sugar cubes are left in the box? | Answer: 300 or 0.
Solution. A box has three dimensions: height, width, and depth. To find out how many cubes are in the top layer, you need to multiply the width by the depth, and for the side layer, multiply the height by the depth. After the top layer is eaten, the height decreases by 1, while the depth remains the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Cut a square with a side of 4 into rectangles, the sum of the perimeters of which is 25. | For example, two rectangles $2 \times 0.5$ and one rectangle $3.5 \times 4-$ cm. The total perimeter is $2 * 2 * (2 + 0.5) + 2 * (3.5 + 4) = 25$.
 | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Someone wrote down two numbers $5^{2020}$ and $2^{2020}$ in a row. How many digits will the resulting number contain? | Solution. Let the number $2^{2020}$ contain $m$ digits, and the number $5^{2020}$ contain $n$ digits. Then the following inequalities hold: $10^{m-1}<2^{2020}<10^{m}, 10^{n-1}<5^{2020}<10^{n}$ (the inequalities are strict because the power of two or five is not equal to the power of ten). Multiplying these inequalities... | 2021 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The older brother noticed that in 10 years, the younger brother will be as old as he is now, and his age will be twice the current age of the younger brother. How old is the younger brother now? | Answer: 20 years.
Solution. Let the current age of the younger brother be $x$ years, and the older brother be $y$ years. In 10 years, the younger brother's age will be $y$ years, and the older brother's age will be $2x$ years. Since the age of each has changed by 10 years, we have the equations: $y+10=2x, x+10=y$. By ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A rectangle $10 \times 20$ is divided into unit squares. How many triangles are formed after drawing one diagonal?
 | Answer: 220.
Solution. The figure shows one of the obtainable triangles. All such triangles are right-angled, and the vertex of the right angle can be any lattice node,
except those lying on the diagonal. There are a total of $21 \times 11$ nodes, and 11 of them are on the
diagonal, so the number of triangles is $2... | 220 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.1. (7 points)
Prove that $\sqrt{\frac{11 \ldots 1}{2 n \text { digits }}-\underbrace{22 \ldots .2}_{n \text { digits }}}=\underbrace{33 \ldots 3}_{n \text { digits }}$. | # Solution:
$$
\begin{aligned}
& \sqrt{\underbrace{11 \ldots 1}_{2 n \text { digits }}-\underbrace{22 \ldots 2}_{n \text { digits }}}=\sqrt{\frac{1}{9} \cdot \underbrace{99 \ldots 9}_{2 n \text { digits }}-\frac{2}{9} \cdot \underbrace{99 \ldots 9}_{n \text { digits }}}=\frac{1}{3} \sqrt{\left(10^{2 n}-1\right)-2\lef... | 42 | Number Theory | proof | Yes | Yes | olympiads | false |
# 10.5. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars al... | Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The f... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Emperor Pea has a worm in one of his rejuvenating apples. He has 13 apples in total, and they are arranged in a circle in a special box for rejuvenating apples. To find the worm, Emperor Pea decided to use a balance scale. He knows that all the apples weigh the same, except for the one with the worm, which is heavie... | # Solution.
Let's number the apples clockwise. The neighboring apples will be numbered 1 and 2, 2 and 3, ..., 12 and 13, 13 and 1.
Weigh apples 1, 2, 3, 4, 5, 6 against 7, 8, 9, 10, 11, 12.
If they are equal, then the worm is in apple 13.
Suppose 1, 2, ..., 6 are heavier.
Then after the apples are put back and the... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3. In 60 chandeliers (each chandelier has 4 lampshades), lampshades need to be replaced. Each electrician spends 5 minutes replacing one lampshade. A total of 48 electricians will be working. Two lampshades in a chandelier cannot be replaced simultaneously. What is the minimum time required to replace all the lampsha... | Answer: 25 minutes
Solution. Let's show how to proceed. First, 48 electricians replace one lampshade in 48 chandeliers, which takes 5 minutes, and 48 chandeliers have one lampshade replaced, while 12 have none replaced. Then, 12 electricians replace lampshades in the chandeliers that haven't been replaced yet, while t... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.4. Find the perimeter of a rectangle if the sum of the lengths of two of its sides is 10 dm, and the sum of the lengths of three of its sides is 14 dm. | Answer: 18 dm, 19 dm or 20 dm
Solution. If the sum of two adjacent sides is 10 dm, then the perimeter is 20 dm, which does not contradict the condition on the sum of three sides. If the sum of opposite sides is 10 dm, then each of these sides is 5 dm. In this case, the adjacent side to them is 4 dm or 4.5 dm. | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-1-1. Select the numbers that serve as counterexamples to the given statement: “If the sum of the digits of a natural number is divisible by 27, then the number itself is divisible by $27$.”
a) 81 ; b) 999 ; c) 9918 ; d) 18 . | Answer: only 9918.
Solution option 1. The statements about 81 and 18 are definitely not counterexamples, as for them, the premise "if the sum of the digits of a natural number is divisible by 27" is not satisfied.
The statement about 999 is not a counterexample, as it does not contradict our statement.
Finally, the ... | 9918 | Number Theory | MCQ | Yes | Yes | olympiads | false |
10-2-1. Petya writes down a sequence of numbers: if the current number is equal to $x$, then the next one is $\frac{1}{1-x}$. The first number in the sequence is 2. What is the five hundredth number? | Answer: -1.
Solution variant 1. Let's list the first few terms of the obtained sequence:
$$
2,-1,1 / 2,2, \ldots
$$
We see that the sequence has entered a cycle with a period of 3. Since the number 500 when divided by 3 gives a remainder of 2, the 500th term will be the same as the 2nd. That is, the five hundredth n... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10-3-1. Non-negative integers $a, b, c, d$ are such that
$$
a b+b c+c d+d a=707
$$
What is the smallest value that the sum $a+b+c+d$ can take? | Answer: 108.
Solution variant 1. The given equality can be rewritten as
\[
(a+c)(b+d)=7 \cdot 101
\]
where the numbers 7 and 101 are prime. Therefore, either one of the expressions in parentheses is 1 and the other is 707, or one of the expressions in parentheses is 7 and the other is 101. In the first case, \(a+b+c... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10-4-1. Anya and Borya are playing rock-paper-scissors. In this game, each player chooses one of the figures: rock, scissors, or paper. Rock beats scissors, scissors beat paper, and paper beats rock. If the players choose the same figure, the game ends in a tie.
Anya and Borya played 25 rounds. Anya chose rock 12 time... | Answer: 16.
Solution version 1. Note that, since there were no draws, when Anya chose rock, Borya must have chosen scissors or paper. Anya chose rock 12 times. Borya chose scissors or paper a total of $9+3=12$ times. Therefore, all these cases must have occurred in the rounds where Anya chose rock.
This means that in... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-6-1. In the figure, there are two circles with centers $A$ and $B$. Additionally, points $A, B$, and $C$ lie on the same line, and points $D, B$, and $E$ also lie on the same line. Find the degree measure of the angle marked with a “?”.
. Angles $ABE$ and $DBC$ are equal as vertical angles. Note that triangle $DBC$ is isosceles wi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-7-1. At a large round table, 90 people are feasting, facing the center of the table: 40 barons, 30 counts, and 20 marquises. On a signal, exactly those who have both neighbors—left and right—with the same title should stand up. What is the maximum number of people who can stand up?
For example, for a count to stand... | Answer: 86.
Solution Variant 1. We ask half of the revelers, standing every other person, to step forward. Then all the people will be divided into two circles: an inner and an outer one. Note that a person in the inner circle has both neighbors in the outer circle, and they are also neighbors there. Similarly for a p... | 86 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-7-3. At a large round table, 55 people are feasting, facing the center of the table: 25 barons, 20 counts, and 10 marquises. On a signal, exactly those who have both neighbors—left and right—with the same title should stand up. What is the maximum number of people who can stand up?
For example, for a count to stand... | Answer: 52.
Solution variant 3. We will place people in a new circle every other person: $1,3,5,7$, $\ldots, 55,2,4,6, \ldots, 54$, back to 1. It is sufficient to find the maximum number of neighbors with the same title in the new circle.
It is clear that there are neighbors of different titles in the new circle. It ... | 52 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-8-1. There is a magical grid sheet of size $2000 \times 70$, initially all cells are gray. The painter stands on a certain cell and paints it red. Every second, the painter takes two steps: one cell to the left and one cell down, and paints the cell red where he ends up after the two steps. If the painter is in the ... | Answer: 14000.
Solution version 1. We need to understand how many cells will be painted by the time the painter returns to the initial cell. Note that after every 2000 moves, the painter returns to the starting column, and after every 140 moves, he returns to the starting row. Therefore, he will return to the initial ... | 14000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.3. Inside parallelogram $A B C D$, a point $E$ is chosen such that $A E=D E$ and $\angle A B E=90^{\circ}$. Point $M$ is the midpoint of segment $B C$. Find the angle $D M E$. | Answer: $90^{\circ}$.
First solution. Let $N$ be the midpoint of segment $A D$. Since triangle $A E D$ is isosceles, its median $E N$ is also an altitude, that is, $E N \perp A D$. Therefore, $N E \perp B C$ (see Fig. 2).
Since $A D \| B C$ and $B M = M C = A N = N D = A D / 2$, quadrilaterals $A B M N$ and $B M D N$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. A confectionery factory produces $N$ types of candies. For New Year, the factory gave each of 1000 school students a gift containing candies of several types (the compositions of the gifts could be different). Each student noticed that for any 11 types of candies, they received a candy of at least one of these typ... | Answer. $N=5501$.
Solution. Let $A_{1}, A_{2}, \ldots, A_{N}$ be the sets of students who did not receive candies of the 1st, 2nd, ..., $N$-th types, respectively. According to the problem, all these sets are distinct; moreover, each student is contained in no more than ten of them. Therefore, the total number of elem... | 5501 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On a coordinate plane with the origin at point $O$, a parabola $y=x^{2}$ is drawn. Points $A, B$ are marked on the parabola such that $\angle A O B$ is a right angle. Find the smallest possible value of the area of triangle $A O B$. | 5. Answer: 1. Solution. Let $A\left(a, a^{2}\right), B\left(b, b^{2}\right)$ be arbitrary points on the parabola. Then, by the Pythagorean theorem, $a^{2}+a^{4}+b^{2}+b^{4}=(a-b)^{2}+\left(a^{2}-b^{2}\right)^{2} \Leftrightarrow a b=-1$. From this, we find the expression for the area of the triangle: $2 S=\sqrt{a^{2}+a^... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. The numbers $1,2, \ldots, 2016$ are written on a board. It is allowed to erase any two numbers and replace them with their arithmetic mean. How should one proceed to ensure that the number 1000 remains on the board? | 6. Solution. We will first act according to the following scheme: $1,2,3, \ldots, n-3, n-2, n-1, n \rightarrow 1,2,3, \ldots, n-3, n-1, n-1 \rightarrow 1,2,3, \ldots, n-3, n-1$. Acting in this way, we will arrive at the set $1,2,3, \ldots, 1000,1002$. Then we will act according to the same scheme from the other end:
\... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Karlson counts 200 buns baked by Fräulein Bock: «one, two, three, ..., one hundred and nine, one hundred and ten, ..., one hundred and ninety-eight, one hundred and ninety-nine, two hundred». How many words will he say in total? (Each word is counted as many times as it was said.) | Answer: 443 words.
Solution. One word will be required to pronounce 29 numbers: $1,2,3,4,5,6,7$, $8,9,10,11,12,13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,100,200$.
Among the first 99 numbers, the number of those pronounced in two words: $99-27=72$, thus, the number of words required for their pronunciation is $2 \c... | 443 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. In the notebook, all irreducible fractions with the numerator 15 are written down, which are greater than $\frac{1}{16}$ and less than $\frac{1}{15}$. How many such fractions are written down 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 difficult t... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. Lёsha colors cells inside a $6 \times 6$ square drawn on graph paper. Then he marks the nodes (intersections of the grid lines) to which the same number of colored and uncolored squares are adjacent. What is the maximum number of nodes that can be marked? | Answer: 45.
## Solution.
Estimation. Each grid node belongs to one, two, or four squares.
The corner vertices of the original square are adjacent to only one small square, so Lёsha will not be able to mark them. Therefore, the maximum number of marked nodes does not exceed $7 \cdot 7-4=45$.
Example.
, the third is in the 49th position, and so on. The ninth nine (from the number 89) is in the $9+2 \cdot 80=$ 169th position (numbers from 1 to 9 occ... | 174 | Number Theory | proof | Yes | Yes | olympiads | false |
3.1. From a square grid with a side of 40, a rectangle $36 \times 37$ was cut out, adjacent to one of the corners of the square. Grisha wants to color a five-cell cross in the remaining piece. In how many ways can he do this?
. Due to the cut-out, $36 \times 37-1$ ways are lost (the center of the cross cannot be in the cut-out rec... | 113 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4.1. Every evening, starting from September 1st, little Anton ate one pastry. After eating another pastry, he noticed that during this entire time, he had eaten 10 delicious pastries (the rest seemed tasteless to him). But among any seven consecutive pastries he ate, no fewer than three turned out to be delicious. What... | Answer: 26
Solution. We will show that among 27 pastries, there will be no fewer than 11 delicious pastries. Let's number the pastries from 1 to 27. Note that among the first seven pastries, at least three are delicious, among the next seven as well, and among the pastries numbered from 15 to 21, there are at least th... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.1. Find the smallest natural number $n$ such that the natural number $n^{2}+14 n+13$ is divisible by 68. | Answer: 21
Solution. Note that $n^{2}+14 n+13=(n+1)(n+13)$, and $68=4 \cdot 17$. Therefore, one of the numbers $n+1$ and $n+13$ must be divisible by 17. Moreover, $n$ must be odd; otherwise, $n+1$ and $n+13$ would both be odd. The smallest such natural number $n$ is $2 \cdot 17-13=21$. Note that it works because $21^{... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. On September 2nd, Robin Bobin ate 12 chickens, and starting from September 3rd, he ate every day as many as he had already eaten on average in September. On September 15th, Robin Bobin ate 32 chickens. How many chickens did he eat on September 1st? | Answer: 52
Solution. Let at the beginning of the $k$-th day he on average ate $x$ chickens. This means that by this point, he had eaten $(k-1) x$ chickens. Then on the $k$-th day, he also ate $x$ chickens, which means that over $k$ days, he ate $k x$ chickens. This means that after the $k$-th day, the average number o... | 52 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. In quadrilateral $A B C D$, $A B=B C=C D$. Let $E$ be the intersection point of $A B$ and $C D$ (with $B$ between $A$ and $E$). It turns out that $A C=C E$ and $\angle D B C=15^{\circ}$. Find $\angle A E D$. | Answer: $50^{\circ}$
Solution. We will solve the general problem. Let $\angle D B C=\alpha$, and $\angle B E C=\beta$. From the isosceles property of $\triangle D C B$, we get $\angle B D C=\alpha$. From the isosceles property of $\triangle A C E$, we get $\angle B E C=\angle B A C$, and from the isosceles property of... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.1. There are 15 rectangular sheets of paper. In each move, one of the sheets is chosen and divided by a straight cut, not passing through the vertices, into two sheets. After 60 moves, it turned out that all the sheets are either triangles or hexagons. How many hexagons? | Answer: 25
Solution. Let after 60 cuts, $x$ hexagons and $75-x$ triangles are formed (there were 15 sheets, and with each cut, the number of sheets increases by 1). Now let's count the number of sides. Initially, there were $15 \cdot 4=60$ sides, and with each cut, the number of sides increases by 4 (the cut divides t... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. There are 2001 coins on the table. Two players play the following game: they take turns, on each turn the first player can take any odd number of coins from 1 to 99, the second player can take any even number of coins from 2 to 100, and so on. The player who cannot make a move loses. Who wins with correct play? | 5. The first one wins. Winning strategy: the first player should take 81 coins from the table on the first move. On each subsequent move, if the second player takes x coins, then the first player should take (101 - x) coins. He can always do this because if x is an even number from 2 to 100, then (101 - x) is an odd nu... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. Given a convex quadrilateral $ABCD$, $X$ is the midpoint of diagonal $AC$. It turns out that $CD \parallel BX$. Find $AD$, if it is known that $BX=3, BC=7, CD=6$.
 | Answer: 14.
Solution. Double the median $B X$ of triangle $A B C$, to get point $M$. Quadrilateral $A B C M$ is a parallelogram (Fig. 1).
Notice that $B C D M$ is also a parallelogram, since segments $B M$ and $C D$ are equal in length (both 6) and parallel. This means that point $M$ lies on segment $A D$, since $A M... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation
$$
y=\frac{1}{5} x^{2}+a x+b
$$
passes through points $B$ and $C$. Additionally, the vertex of this parabola (po... | Answer: 20.
Solution. Note that the parabola is symmetric with respect to the vertical axis passing through its vertex, point $E$. Since points $B$ and $C$ are on the same horizontal line, they are symmetric with respect to this axis. This means that this axis passes through the midpoint of $B C$, and therefore, throu... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is such that $\angle ADC = 2 \angle CAD = 82^{\circ}$. Inside the trapezoid, a point $T$ is chosen such that $CT = CD, AT = TD$. Find $\angle TCD$. Give your answer in degrees.
.
. From this similarity and the cyclic nature of the pentagon
. It is known that $AB = AN$, $BC = MC$. The circumcircles of triangles $ABM$ and $CBN$ intersect at points $B$ and $K$. How many degrees does the angle $AKC$ measure if $\angle ABC = 68^\circ$?
# | # Answer: 110
## Solution
We will call the equilateral triangles that make up the original hexagon unit triangles. Consider each of the highlighted triangles inside the hexagon separately. The smallest one coincides with the unit triangle and has an area of 10. The medium triangle is inscribed in a hexagon made up of... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. In a class, there are 18 children. The parents decided to gift the children from this class a cake. To do this, they first asked each child the area of the piece they wanted to receive. After that, they ordered a square cake, the area of which is exactly equal to the sum of the 18 named numbers. However, upon seei... | Answer. $k=12$.
Solution. We always assume that the area of the cake is 1. We will show that for some children's requests, the parents will not be able to cut out more than 12 required pieces. Choose a number $1 / 15 > x > 1 / 16$. Suppose that 15 main children ordered a piece of cake with an area of $x$ each (and the... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
, adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. In the chat of students from one of the schools, a poll was being held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the poll.
Then, 80 more people participated in the poll, voting only for October 22. After that, th... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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