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6. Several cells on a $14 \times 14$ board are marked. It is known that no two of the marked cells are in the same row and the same column, and also that a knight can, starting from any marked cell, reach any other marked cell via marked cells. What is the maximum possible number of marked cells? | Answer: 14.
Sketch of the solution.
Estimate. In each row, there is no more than one marked cell (field), and therefore no more than 14 cells are marked.
Example in the image.
 | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. Ten football teams each played one game against each of the others. As a result, each team ended up with exactly $x$ points. What is the greatest possible value of $x$? (Win - 3 points, draw - 1 point, loss - 0 points.) | Answer: 13.
Evaluation. Let's prove that $x$ cannot be greater than 13. Indeed, in each match, either 3 points are awarded (if one of the teams wins) or 2 points (if there is a draw). In total, $\frac{10 \cdot 9}{2}=45$ matches were played, meaning no more than 135 points were awarded, that is, the total points scored... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. From points $A$ and $B$, two cars set off towards each other simultaneously with constant speeds. One hour before the first car arrived at $B$ and four hours before the second car arrived at $A$, they met. Find the ratio of the speeds of the cars. | 3. Answer: the speed of the first car is twice as high.
Let the speed of one car be $k$ times the speed of the other. Since they started at the same time, one car will travel a distance $k$ times greater. After this, the segments will switch, and the slower car, whose speed is $k$ times less, will have to cover a dist... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The numbers $p$ and $b$ are roots of the quadratic equation $x^{2}+2020 a x+c=0, a \neq 0$. Find the sum of the roots of the quadratic equations $a x^{2}+b x+d=0$ and $a x^{2}+p x+q=0$, if each of them has 2 distinct real roots. | Solution. Since $p$ and $b$ are the roots of the quadratic equation $x^{2}+2020 a x+c=0$, by Vieta's theorem, $p+b=-2020 a$. Let $x_{1}$ and $x_{2}$ be the roots of the equation $a x^{2}+b x+d=0$, and $x_{3}$ and $x_{4}$ be the roots of the equation $a x^{2}+p x+q=0$. Then, by Vieta's theorem, $x_{1}+x_{2}=-\frac{b}{a}... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On his birthday, Piglet baked a large cake weighing 10 kg and invited 100 guests. Among them was Winnie-the-Pooh, who is fond of sweets. The birthday boy announced the rule for dividing the cake: the first guest cuts a piece of the cake that is $1 \%$, the second guest cuts a piece of the cake that is $2 \%$ of the ... | Solution. The first guests in the queue receive increasingly larger pieces of the pie because the remaining part of the pie is large at the initial stages of division. However, since the remaining part of the pie decreases, there will come a point when guests start receiving smaller pieces of the pie. At which guest wi... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. On the website of the football club "Rostov," a poll is being conducted to determine which of the $m$ football players the website visitors consider the best at the end of the season. Each visitor votes once for one player. The website displays the rating of each player, which is the percentage of votes cast for the... | Solution. Let $a$ be the greatest loss of a percentage point due to rounding when determining a footballer's rating. Then, according to rounding rules, $aa m \geqslant 5$, which means $0.5 m>5$ or $m>10$.
We will show that a solution exists when $m=11$. For example, let $m=11$ and 73 visitors voted, with 33 of them vo... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 10.1
For all real $x$ and $y$, the equality $f\left(x^{2}+y\right)=f(x)+f\left(y^{2}\right)$ holds. Find $f(-1)$.
## Number of points 7 | Answer 0.
Solution
Substituting $x=0, y=0$, we get $f(0)=f(0)+f(0)$, that is, $f(0)=0$.
Substituting $x=0, y=-1$, we get $f(-1)=f(0)+f(1)$, that is, $f(-1)=f(1)$.
Substituting $x=-1, y=-1$, we get $f(0)=f(-1)+f(1)$. Therefore, $2 f(-1)=0$.
# | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A positive number $x$ was increased by $69 \%$. By what percentage did the number $\sqrt{\frac{x}{5}}$ increase? | # Solution.
The increased number will be $1.69 x$, then from the proportion
$\sqrt{\frac{x}{5}}-100 \%$, we determine $y=\frac{1.3 \sqrt{\frac{x}{5}} \cdot 100}{\sqrt{\frac{x}{5}}}=130$ (\%), so the difference will be
equal to $30 \%$.
Answer: by $30 \%$.
Instructions. Only the answer - 0 points; the answer obtain... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In each cell of a $10 \times 10$ board, there is a grasshopper. At the whistle, each grasshopper jumps over one cell diagonally (not to the adjacent diagonal cell, but to the next one). As a result, some cells may end up with more than one grasshopper, while some cells will be unoccupied. Prove that in this case, th... | # Solution.
We will paint the cells of the board black and white, as shown in the figure. As a result, 60 cells will be painted black, and 40 cells will be painted white. Notice that from a black cell, the grasshopper can only jump to a white cell, and from a white cell, it can only jump to a black cell. Consequently,... | 20 | Combinatorics | proof | Yes | Yes | olympiads | false |
9.2. Seryozha chose two different natural numbers $a$ and $b$. He wrote down four numbers in his notebook: $a, a+2, b$ and $b+2$. Then he wrote on the board all six pairwise products of the numbers from the notebook. What is the maximum number of perfect squares that can be among the numbers on the board?
(S. Berlov)
... | # Answer. Two.
Solution. Note that no two squares of natural numbers differ by 1, because $x^{2}-y^{2}=(x-y)(x+y)$, where the second bracket is greater than one. Therefore, the numbers $a(a+2)=(a+1)^{2}-1$ and $b(b+2)=(b+1)^{2}-1$ are not squares. Moreover, the numbers $ab$ and $a(b+2)$ cannot both be squares, otherwi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.3. Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation of the num... | # Answer. $k=6$.
Solution. By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).
We will use the following known fact: t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Thirty beads (blue and green) were laid out in a circle. For 26 beads, the neighboring one was blue, and for 20 beads, the neighboring one was green. How many blue beads were there? | Answer: 18 blue beads.
Solution. Let's find the number of beads that have both a blue and a green bead next to them: $26+20-30=$ 16. The number of beads that have only blue beads next to them is $26-16=10$. The number of blue beads is $\frac{10 \cdot 2+16}{2}=18$. Here is an example of such an arrangement. Let's denot... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 1. Clone 1
The figure is divided into 7 equal squares and several rectangles. The perimeter of rectangle A is 112 cm. What is the perimeter of rectangle B? Express your answer in centimeters.
![](https://cdn.mathpix.com/cropped/2024_05_06_1f50971f08007f421b41g-01.jpg?height=485&width=483&top_left_y=1008&top_left_x=... | # Answer: 168
## Solution
1st method. Rectangles A and B are composed of identical squares: Rectangle A consists of three, and Rectangle B consists of four. Let's call the side of such a square a "stick" and count the perimeter in sticks. The perimeter of Rectangle B is four sticks more than the perimeter of Rectangl... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3. Clone 1
A rope was divided into 19 equal parts and arranged in a snake-like pattern. After that, a cut was made along the dotted line. The rope split into 20 pieces: the longest of them is 8 meters, and the shortest is 2 meters. What was the length of the rope before it was cut? Express your answer in meters.
—settled, one person per room. Each of these 25 people said: “At least one of the neighboring rooms to mine is occupied by a liar.” What i... | Answer: 18 knights.
Solution: Note that each knight must have at least one neighbor who is a liar. We will show that there must be at least 7 liars (thus showing that there are no more than 18 knights).
First, consider dividing the rooms into 6 groups (2 rooms, marked in gray, do not belong to any group). In each gro... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.4 The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ satisfies the relations $\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}-1} \cdot \mathrm{a}_{\mathrm{n}-3}$ for $\mathrm{n}=4,5,6, \ldots$ Find $\mathrm{a}_{2019}$ if it is known that $\mathrm{a}_{1}=1,... | Solution. It is clear that all members of this sequence are equal to $\pm 1$. We find:
$$
\begin{aligned}
& a_{n}=\left(a_{n-1}\right) \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.3. For breakfast, Karlson ate $40 \%$ of the cake, and Little One ate 150 g. For lunch, Fräulein Bock ate $30 \%$ of the remainder and another 120 g, and Matilda licked the remaining 90 g of crumbs from the cake. What was the initial mass of the cake? | Answer: 750 g.
Solution. First method (solving "from the end").
1) $90+120=210$ (g) of the cake remained after Fröken Bok ate $30\%$ of the remainder.
Since Fröken Bok ate $30\%$ of the remainder, 210 g is $70\%$ of the remainder.
2) $210 \div 0.7 = 300$ (g) of the cake was left before Fröken Bok started her lunch.
... | 750 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In a correspondence mathematics olympiad, out of 500 participants, exactly 30 did not like the problem conditions, exactly 40 did not like the organization of the event, and finally, exactly 50 did not like the method of determining the winners of the olympiad. We will call an olympiad participant "significantly dis... | Solution. In total, there were $120=30+40+50$ "dissatisfactions" expressed, so there cannot be more than 60 "significantly dissatisfied" individuals, because otherwise they would have more than 120 "dissatisfactions" in total, which contradicts the condition. It remains to show that there could have been 60 of them. Fo... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The sides of the quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ have the following lengths: $A B=9, B C=2$, $C D=14, D A=5$. Find the length of the diagonal $\boldsymbol{A} \boldsymbol{C}$, if it is known that it is an integer. | Solution. Apply the triangle inequality to $\triangle ABC$ and to $\triangle ACD$: for the first, we will have that $AB + BC > AC$, that is, $AC < AB + BC = 11$; for the second, we will have that $AC + CD > AD$, that is, $AC > CD - DA = 9$. Therefore, $9 < AC < 11$, from which $AC = 10$.
Answer: $AC = 10$. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Unlucky Emelya was given several metal balls, from which he broke 3 of the largest ones (their mass was $35 \%$ of the total mass of all the balls), then lost 3 of the smallest ones, and brought the remaining balls (their mass was $8 / 13$ of the unbroken ones) home. How many balls were given to Emelya | Answer: 10 balls.
## Solution:
Let the total mass of all the balls be M. Of these, Emelya broke balls with a total mass of $\frac{35}{100} M = \frac{7}{20} M$, did not break balls with a total mass of $\frac{13}{20} M$, brought home balls with a total mass of $\frac{8}{13} \cdot \frac{13}{20} M = \frac{8}{20} M$, and... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. On the road between cities A and B, there are some poles in certain places, and two numbers are written on each pole: how many kilometers from the pole to city A, and how many kilometers from the pole to city B. A tourist walking from one city to the other saw a pole where one of the numbers was three times the othe... | Answer: 7.
Solution: Without loss of generality, we can assume that the tourist saw the first post when he was closer to city A. Let the distance from the first post to A be $x$ kilometers, then the distance from it to B is $3x$. If the numbers on the second post are $y$ and $3y$, then $x + 3x = y + 3y$, since the sum... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. A four-digit number $n$ was doubled and 1000 was added to the result. As a result, a four-digit number was obtained, written with the same digits as $n$, but in reverse order. Find all possible values of $n$. | Answer: 2996.
Solution. Let $a, b, c, d$ be the number of thousands, hundreds, tens, and units in the number $n$ respectively. Write the condition in the form:
| $a$ | $b$ | $c$ | $d$ |
| ---: | ---: | ---: | ---: |
| $a$ | $b$ | $c$ | $d$ |
| $+\quad 1$ | 0 | 0 | 0 |
| $d$ | $c$ | $b$ | $a$ |
Since $a$ is the last ... | 2996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. In the basket, there are oranges and bananas. If you add as many oranges as there are currently bananas (in pieces), then the percentage of oranges
will be twice as much as it would be if you added as many bananas as there are currently oranges. What is the current percentage of oranges in the basket? | Answer: 50.
Solution. Let $a$ be the number of oranges, and $b$ be the number of bananas in the basket. If we add as many oranges as there are currently bananas, then there will be $a+b$ oranges out of $a+2b$ fruits, and the proportion of oranges will be $(a+b)/(a+2b)$. If we add as many bananas as there are currently... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. Find the value of the expression $x-\sqrt{2022 x}+2023$,
if $x-\sqrt{\frac{2022}{x}}=2023$. | Answer: 2024.
Solution. Transform the condition $x-\sqrt{\frac{2022}{x}}=2023$ given that $x>0$. We get: $x^{2}-\sqrt{2022 x}=2023 x ; x^{2}-2022 x-x-\sqrt{2022 x}=0$;
$$
(x-\sqrt{2022 x})(x+\sqrt{2022 x})-(x+\sqrt{2022 x});(x-\sqrt{2022 x}-1)(x+
$$
$\sqrt{2022 x})=0$.
The second factor is positive for all $x>0$, w... | 2024 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. At competitions, an athlete's performance is evaluated by 7 judges, each of whom gives a score (an integer from 0 to 10). To obtain the final score, the best and worst scores from the judges are discarded, and the arithmetic mean is calculated. If the average score were calculated based on all seven scores, the at... | Answer: 5.
Solution: Suppose there are no fewer than six dancers. Let $A, a, S_{A}$ be the best score, the worst score, and the sum of all non-discarded scores of the winner, respectively, and $B, b, S_{B}$ be the same for the last athlete. Instead of averages, the dancers can be ranked by the sum of all scores or the... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3 In a pond with carp, 30 pikes were released, which began to gradually eat each other. A pike is considered full if it has eaten three other pikes (hungry or full), and each full pike ate exactly one carp for dessert (hungry pikes did not eat carp). What is the maximum number of carp that could have been eaten? Just... | Solution. The maximum number of pikes eaten is 29, so no more than $29: 3=9.6666 \ldots$ pikes can be satiated, which means no more than 9 pikes (and, accordingly, no more than 9 roaches can be eaten). 9 roaches can be eaten as follows (the method is not unique). Choose 9 pikes and denote them as $A_{1}, A_{2}, \ldots,... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.4 On the table, there are 2020 boxes, some of which contain candies, while the others are empty. On the first box, it is written: "All boxes are empty." On the second: "At least 2019 boxes are empty." On the third: "At least 2018 boxes are empty," and so on, up to the 2020th, which says: "At least one box is empty." ... | Solution. Note that if the inscription "At least $A$ boxes are empty" is true for some $A$, then all subsequent inscriptions are also true. This means that the boxes with candies are all boxes starting from some number $N$. Then the number of boxes without candies is exactly $N-1$. The inscription on the $N$-th box rea... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5 A businessman was driving to a business meeting. He calculated that if he traveled at a speed of 90 km/h, he would arrive an hour earlier, and if he traveled at 60 km/h, he would be an hour late. What is the minimum speed he should travel to arrive on time? Justify your answer. | Solution. Method 1. Suppose that three businessmen, each in their own car, were heading to a meeting. The first (Speedster) at a speed of 90 km/h, the second (Turtle) at a speed of 60 km/h, and the third - Punctual, whose speed we need to find. Consider the moment when Speedster arrives at the meeting place. The meetin... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.6 On a horizontal line, points $A$ and $B$ are marked, the distance between which is 4. Above the line, two semicircles with a radius of 2 are constructed, centered at points
A and B. Additionally, one circle, also with a radius of 2, is constructed, for which the point of intersection of these semicircles is the lo... | Solution. Method 1. Let the midpoint of segment $A B$ be $E$, the center of the circle be $O$, and the second points of intersection of the circle and the semicircles be $C$ and $D$ (see figure). Since the radii of the semicircles and the radius of the circle are equal, segments $A D, E O$, and $B C$ will be equal to e... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find all roots of the equation $(x-a)(x-b)=(x-c)(x-d)$, given that $a+d=b+c=2016$ and $a \neq c$ (the numbers themselves are not given). | Solution. Expanding the brackets $\mathrm{x}^{2}-(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}=\mathrm{x}^{2}-(\mathrm{c}+\mathrm{d}) \mathrm{x}+\mathrm{cd}$, i.e., $\quad(c+d-a-b) x=c d-a b$. By the condition $c - a = d - b$ - let's denote this by $r$. Then $a=c-r, d=b+r$ and $c d-a b=c(b+r)-(c-r) b=(b+c) r=2016 r$. ... | 1008 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
At the entrance to the amusement park, they sell children's and adult tickets. One children's ticket costs 600 rubles. Alexander bought 2 children's and 3 adult tickets, while Anna bought 3 children's and 2 adult tickets. It is known that Alexander paid 200 rubles more than Anna. How much did Alexander p... | Answer: 3600
## Solution.
Let $A$ be the cost of a children's ticket, and $B$ be the cost of an adult ticket. We calculate the difference $3B + 2A - 2B - 3A = B - A = 200$ rubles. This means the difference between the cost of an adult ticket and a children's ticket is 200 rubles. Then Alexander paid for the tickets $... | 3600 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Variant 1.
On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, 2022 natives gathered around a round table, and each of them made the following statement:
"I am sitting next to a knight and a liar!"
It is known that three knights made a mistake (i.e., accid... | Answer: 1349
## Solution.
Let's divide all the people sitting at the table into blocks consisting of natives of the same type sitting in a row. Then, blocks of liars can only consist of 1 person; otherwise, a liar sitting at the edge of this block would tell the truth. If a block of knights consists of two people, th... | 1349 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Variant 1.
On the coordinate plane, the graphs of three reduced quadratic trinomials are drawn, intersecting the y-axis at points $-15, -6, -27$ respectively. For each trinomial, the coefficient of $x$ is a natural number, and the larger root is a prime number. Find the sum of all roots of these trinomials. | Answer: -9.
## Solution.
The parabola intersects the $O y$ axis at a point whose ordinate is equal to the free term. Therefore, our trinomials have the form $x^{2}+a x-15, x^{2}+b x-6, x^{2}+c x-27$. Let's denote their larger roots by $p, q, r$ respectively. By Vieta's theorem, the second root of the first trinomial ... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1.
In the figure, an example is given of how 3 rays divide the plane into 3 parts. Into what maximum number of parts can 11 rays divide the plane?
 | Answer: 56.
## Solution.
If the $(n+1)$-th ray is drawn so that it intersects all $n$ previous rays, then $n$ intersection points on it divide it into $n+1$ segments. The segment closest to the vertex does not add new parts to the partition, while each of the other $n$ segments ( $n-1$ segments and 1 ray) divides som... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
In the "Triangle" cinema, the seats are arranged in a triangular shape: in the first row, there is one seat with number 1, in the second row - seats with numbers 2 and 3, in the third row - 4, 5, 6, and so on (the figure shows an example of such a triangular hall with 45 seats). The best seat in the cine... | Answer: 1035
Solution.
1st method.
Note that the number of rows in the cinema cannot be even, otherwise there would be no best seat. Let the total number of rows in the cinema be $2 n+1$, then the best seat is in the $n+1$ row. If we remove this row, the triangle can be divided into 4 parts, and the number of seats ... | 1035 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Variant 1.
On side $A B$ of parallelogram $A B C D$, a point $F$ is chosen, and on the extension of side $B C$ beyond vertex $B$, a point $H$ is chosen such that $A B / B F = B C / B H = 5$. Point $G$ is chosen so that $B F G H$ is a parallelogram. $G D$ intersects $A C$ at point $X$. Find $A X$, if $A C = 100$. | Answer: 40.
Solution.
In parallelogram $ABCD$, draw diagonal $BD$, and in parallelogram $BFGH$, draw diagonal $GB$. Let $BD$ intersect $AC$ at point $O$. We will prove that $AC \| GB$. Tri... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) In rectangle $A B C D$, side $A B$ is equal to 6, and side $B C$ is equal to 11. Bisectors of the angles from vertices $B$ and $C$ intersect side $A D$ at points $X$ and $Y$ respectively. Find the length of segment $X Y$.
 The knightly tournament lasts exactly 7 days. By the end of the fourth day, Sir Lancelot had not yet faced one quarter of the total number of participants. And by this time, Sir Tristan had fought exactly one seventh of the knights that Sir Lancelot had faced. What is the minimum number of knights that co... | Answer: 20.
Solution. Let Lancelot not have fought with $x$ knights. Then the total number of knights is $4 x$, and Lancelot fought with $3 x-1$ knights (the total number minus $x$ and Lancelot himself). Then Tristan fought with $\frac{3 x-1}{7}$ knights. To find the smallest possible number of knights, we need to fin... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (7 points) In triangle $A B C$, the median $A M$ is drawn. Find the angle $A M C$, if angles $B A C$ and $B C A$ are equal to $45^{\circ}$ and $30^{\circ}$ respectively.
Answer: $135^{\circ}$. | Solution. Let $B H$ be the height of triangle $A B C$. According to the problem, angle $B A C$ is $45^{\circ}$, so $B H = A H$. In triangle $C B H$, the leg $B H$ lies opposite the angle $30^{\circ}$, so $B C = 2 B H$. The median $H M$ of the right triangle $B H C$ is equal to half the hypotenuse $B C$.
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA + ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
![](https://cdn.mathpix.com/cropped/2024_05_06_4984b677816e04f7f1c9g-35.jpg?height=305&width=40... | Answer: 972.
Solution. Let's mark the center of the circle $I$, as well as the points of tangency $P, Q, K$ with the sides $B C$, $A D, A B$ respectively. Note that $I P \perp B C, I Q \perp A D$, i.e., points $P, I, Q$ lie on the same line, and $P Q$ is the height of the given trapezoid, equal to the diameter of its ... | 972 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.6. Thirty girls - 13 in red dresses and 17 in blue dresses - were dancing in a circle around a Christmas tree. Later, each of them was asked if their right neighbor was in a blue dress. It turned out that those who answered correctly were only the girls standing between girls in dresses of the same color. How many gi... | Answer: 17.
Solution: Consider any girl. The colors of the dresses of her left and right neighbors could have been: blue-blue, blue-red, red-blue, red-red. The girl answered "yes" in exactly the first two cases; therefore, she said "yes" exactly when her left neighbor was wearing a blue dress.
Thus, since exactly 17 ... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.8. In the cells of an $8 \times 8$ board, the numbers 1 and -1 are placed (one number per cell). Consider all possible placements of the figure $\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). We will call such a placement unsuccessful if the sum of the numb... | Answer: 36.
Solution: We will show that in each "cross" of five cells on the board, there will be at least one unsuccessful placement. Suppose the opposite; let the numbers in the outer cells of the cross be \(a, b, c, d\), and the number in the central cell be \(e\); denote by \(S\) (M. Antipov) the sum of all these ... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.4. A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar? | Answer: In the sixth jar, there are -16 beetles.
Solution. Let there be $x$ beetles in the first jar, then in the second jar there are no fewer than $x+1$ beetles, in the third jar no fewer than $x+2$ beetles, and so on. Thus, in the tenth jar there are no fewer than $x+9$ beetles. Therefore, the total number of beetl... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1 In the glass, there was a solution in which water made up $99 \%$. The glass with the solution was weighed, and the weight turned out to be 500 gr. After that, some of the water evaporated, so that in the end, the proportion of water was 98\%. What will be the weight of the glass with the resulting solution, if the... | Answer: 400 g.
Indication: Initially, the water was $0.99 \cdot(500-300)=198$ (g), and the substance was $200-198=2$ (g). After the water evaporated, 2 g of the substance make up $100 \%-98 \%=2 \%$ of the solution, so the entire solution weighs 100 g, and together with the glass, it weighs 400 g. | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Usain runs one lap around the school stadium at a constant speed, while photographers Arina and Marina are positioned around the track. For the first 4 seconds after the start, Usain was closer to Arina, then for 21 seconds he was closer to Marina, and then until the finish, he was closer to Arina again. How long do... | # Solution:
It is not hard to see that regardless of Arina and Marina's positions, the entire circle of the school stadium is divided into two equal parts - half of the circle is closer to Arina and the other half is closer to Marina (this is half of the shorter arc between Arina and Marina and half of the longer arc ... | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.5. Thirty girls -13 in red dresses and 17 in blue dresses - were dancing in a circle around a Christmas tree. Later, each of them was asked if their right neighbor was in a blue dress. It turned out that those who answered correctly were only the girls standing between girls in dresses of the same color. How many gi... | # Answer: 17.
Solution. Consider any girl. The colors of the dresses of her left and right neighbors could have been: blue-blue, blue-red, red-blue, red-red. The girl answered "yes" in exactly the first two cases; therefore, she said "yes" exactly when her left neighbor was wearing a blue dress.
Thus, since exactly 1... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.8. On a circle of length 2013, 2013 points are marked, dividing it into equal arcs. A chip is placed at each marked point. We define the distance between two points as the length of the shorter arc between them. For what largest $n$ can the chips be rearranged so that there is again one chip at each marked point, an... | Answer. $n=670$.
Solution. Let's number the points and the chips placed on them in a clockwise direction with consecutive non-negative integers from 0 to 2012. Consider an arbitrary permutation and the chips with numbers 0, 671, and 1342, initially located at the vertices of an equilateral triangle. The pairwise dista... | 670 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. A paper rectangle $3 \times 7$ was cut into squares $1 \times 1$. Each square, except those that stood at the corners of the rectangle, was cut along both diagonals. How many small triangles were obtained? | Answer: 68.
Solution. Note that the total number of squares cut along the diagonals is $3 \cdot 7-4=17$. Each of them is cut into 4 small triangles. Therefore, there will be $4 \cdot 17$ - 68 small triangles. | 68 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Alla thought of a three-digit number, in which there is no digit 0, and all digits are different. Bella wrote down the number in which the same digits are in reverse order. Galia subtracted the smaller number from the larger one. What digit stands in the tens place of the resulting difference?
# | # Answer: 9
Solution. Since we are subtracting a smaller number from a larger one, the digit in the hundreds place of the first number is greater than that of the second. Then, in the units place, conversely, the digit of the first number is smaller than that of the second. Therefore, when subtracting, we will have to... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Masha has 4 pieces of red plasticine, 3 pieces of blue plasticine, and 5 pieces of yellow plasticine. First, she divided each non-red piece of plasticine in half, and then she divided each non-yellow piece of plasticine in half. How many pieces of plasticine did Masha get | Answer: 30 pieces of plasticine.
Solution. After Masha's first action, the number of blue and yellow pieces of plasticine doubles. They become 6 and 10, respectively. After Masha's second action, the number of red and blue pieces of plasticine doubles. They become 8 and 12, respectively. Then the total number of piece... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A square area was paved with square tiles (all tiles are the same). A total of 20 tiles adjoin the four sides of the area. How many tiles were used in total?
 | Answer: 36 tiles.
Solution. Let's call the tiles adjacent to the top side of the platform "top tiles". Similarly, we define "right", "left", and "bottom" tiles. Notice that if we add the number of top tiles, the number of bottom tiles, the number of right tiles, and the number of left tiles, we will count the corner t... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Merlin decided to weigh King Arthur on enchanted scales that always err by the same weight in the same direction. When Merlin weighed Arthur, they showed a weight of 19 stones. Then Merlin weighed the royal horse and got a weight of 101 stones. Finally, Merlin weighed Arthur on the horse, and the scales show... | Answer: 13 stones.
Solution. Note that if we add 19 stones and 101 stones, we get the combined weight of Arthur and the horse, to which the scale error has been added (or subtracted) twice. Meanwhile, 114 stones is the combined weight of Arthur and the horse, to which the scale error has been added only once. Therefor... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Peter has 5 rabbit cages (the cages are in a row). It is known that there is at least one rabbit in each cage. We will call two rabbits neighbors if they sit either in the same cage or in adjacent ones. It turned out that each rabbit has either 3 or 7 neighbors. How many rabbits are sitting in the central ca... | Answer: 4 rabbits.
Solution. Let's number the cells from 1 to 5 from left to right.
Notice that the neighbors of the rabbit in the first cell are all the rabbits living in the first two cells. The neighbors of the rabbit in the second cell are all the rabbits living in the first three cells. The third cell cannot be ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. In the queue for the school cafeteria, 16 schoolchildren are standing in such a way that boys and girls alternate. (The first is a boy, followed by a girl, then a boy again, and so on.) Any boy who is followed by a girl in the queue can swap places with her. After some time, it turned out that all the girls ... | Answer: 36 exchanges.
Solution. Note that in the end, each boy will make one exchange with each girl who is in line after him. That is, the first boy will make 8 exchanges, the second - 7, the third - 6, and so on. Then the total number of exchanges is
$$
8+7+6+5+4+3+2+1=36.
$$
It is not hard to understand that if e... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. Seryozha placed numbers from 1 to 8 in the circles so that each number, except one, was used exactly once. It turned out that the sums of the numbers on each of the five lines are equal. Which number did Seryozha not use?
![](https://cdn.mathpix.com/cropped/2024_05_06_03b105aa15b6031ee02fg-3.jpg?height=326&... | Answer: 6.
Solution. Note that each number is contained in exactly two lines. Therefore, the doubled sum of all used numbers is five times greater than the sum of the numbers on one line. Thus, the sum of all numbers is divisible by 5.
Note that $1+2+3+4+5+6+7+8=36$. Therefore, for the sum of the used numbers to be d... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.4. How to cut a $5 \times 5$ square with straight lines so that the resulting pieces can be used to form 50 equal squares? It is not allowed to leave unused pieces or overlap them. | Solution. First, cut the $5 \times 5$ square into 25 squares of $1 \times 1$, then cut each of the resulting squares along the diagonals into 4 triangles, from which, by attaching the longer sides of two triangles to each other, 2 squares can be formed:
![](https://cdn.mathpix.com/cropped/2024_05_06_a954f424d22e8250b8... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1. Each of 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some integer. Then the first said: “My number is greater than 1”, the second said: “My number is greater than $2 ”, \ldots$, the tenth said: “My number is greater than 10”. After that, all ten, sp... | Answer: 8 knights.
Solution. We will prove that none of the knights could have said either of the phrases "My number is greater than 9" or "My number is greater than 10." Indeed, if this were possible, the integer thought of by the knight would be at least 10. But then he could not have said any of the phrases "My num... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.3. Let's call the distance between two cells on a checkerboard the minimum number of moves a chess king can make to get from one to the other. Find the maximum number of cells that can be marked on a $100 \times 100$ board such that there are no two marked cells with a distance of 15 between them.
(I. Bogdanov) | Answer. $55^{2}=3025$ cells.
Solution. Divide the board into 9 squares $30 \times 30$, 6 rectangles $10 \times 30$, and one square $10 \times 10$ (see Fig. 5). In each $30 \times 30$ square, the cells are divided into $15^{2}$ groups of four such that the distance between any two cells in the same group is 15 (each gr... | 3025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $1-(2-(3-(\ldots 2010-(2011-(2012-x)) \ldots)))=1006$. | # Answer. $x=2012$
Solution. Opening the brackets, we get $1-2+3-4+\ldots+2011-2012+x=$ 1006; $-1006+x=1006 ; x=2012$. | 2012 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A road 28 kilometers long was divided into three unequal parts. The distance between the midpoints of the extreme parts is 16 km. Find the length of the middle part. | Answer: 4 km.
Solution. The distance between the midpoints of the outermost sections consists of half of the outer sections and the entire middle section, i.e., twice this number equals the length of the road plus the length of the middle section. Thus, the length of the middle section $=16 * 2-28=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In five 15-liter buckets, there are 1, 2, 3, 4, and 5 liters of water respectively. It is allowed to triple the amount of water in any container by pouring water from one other container (if there is not enough water to triple the amount, then it is not allowed to pour from this bucket). What is the maximum amount o... | # Answer
The answer depends on the interpretation of the condition -- whether it is allowed to pour NOT all the contents of the bucket (essentially -- whether it is possible to measure OUT ONE LITER of water)
A) If it is not allowed, then the answer is 9 liters
B) If it is allowed, then the answer is 12 liters
## S... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.1. While walking in the park, Seryozha and Misha stumbled upon a meadow surrounded by lindens. Seryozha walked around the meadow, counting the trees. Misha did the same but started from a different tree (although he went in the same direction). The tree that was the $20-\mathrm{th}$ for Seryozha was the $7-\mathrm{t... | # Answer: 100.
Solution. First method. Let there be $n$ trees growing around the glade. We will calculate in two ways the number of intervals between the two trees mentioned in the problem's condition. During Sergei's walk: $20-7=13$. During Misha's walk: $7+(n-94)=n-87$. Therefore, $n-87=13$, which means $n=100$.
Se... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. In triangle $A B C$, angle $C$ is $75^{\circ}$, and angle $B$ is $60^{\circ}$. The vertex $M$ of the isosceles right triangle $B C M$ with hypotenuse $B C$ is located inside triangle $A B C$. Find angle $M A C$. | Answer: $30^{\circ}$.
First method. From the problem statement, it follows that $\angle B A C=45^{\circ}$. Draw a circle with center $M$ and radius $M B=M C$ (see Fig. 11.2). Since $\angle B M C=90^{\circ}$, the larger arc $B C$ of this circle is the locus of points from which the chord $B C$ is seen at an angle of $4... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.6. On a circle, 20 points are marked. How many triples of chords with endpoints at these points exist such that each chord intersects each other (possibly at the endpoints)? | Answer: 156180.
Solution. The ends of the sought chords can be 3, 4, 5, or 6 points. Let's consider these cases.
1) The ends of the chords are 3 points (see Fig. 11.6a). They can be chosen in $C_{20}^{3}$ ways. Each triplet of points can be connected by chords in a unique way.
According to the condition, in each such pair, there is no more than one athlete in a red costume. Therefore, amo... | 50 | Combinatorics | proof | Yes | Yes | olympiads | false |
6. (7 points) If the number $A$ is written on the board, you can add any of its divisors, except 1 and $A$ itself. Can you get 1234321 from $A=4$? Answer: Yes. | Solution. Adding a divisor $n$ to a number means adding $n$ to a number of the form $k n$. The result will be a number of the form $(k+1)n$. Note that the number 1234321 is divisible by 11. Then to the number $A=4=2 \cdot 2$, we will add 2 until we get the number $2 \cdot 11$: $2 \cdot 2 \rightarrow 2 \cdot 3 \rightarr... | 1234321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Point $\mathbf{E}$ is the midpoint of side AB of parallelogram ABCD. On segment DE, there is a point F such that $\mathrm{AD}=\mathbf{B F}$. Find the measure of angle CFD. | Solution: Extend $\mathrm{DE}$ to intersect line $\mathrm{BC}$ at point $\mathrm{K}$ (see the figure). Since $\mathrm{BK} \| \mathrm{AD}$, we have $\angle \mathrm{KBE} = \angle \mathrm{DAE}$. Additionally, $\angle \mathrm{KEB} = \angle \mathrm{DEA}$ and $\mathrm{AE} = \mathrm{BE}$, so triangles $\mathrm{BKE}$ and $\mat... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter Г on the board, at least one colored cell is found? | Solution: Consider a $2 \times 3$ rectangle. In it, obviously, a minimum number of cells need to be colored. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown in the figure.
... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Two painters are painting a 15-meter corridor. Each of them moves from the beginning of the corridor to its end and starts painting at some point until the paint runs out. The first painter has red paint, which is enough to paint 9 meters of the corridor; the second has yellow paint, which is enough for 10 m... | Answer: 5 meters.
Solution. The first painter starts painting 2 meters from the beginning of the corridor and finishes at $2+9=11$ meters from the beginning of the corridor.
The second painter finishes painting 1 meter from the end of the corridor, which is 14 meters from the beginning of the corridor, and starts at ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Lёsha bought a chocolate bar in the shape of a heart (see the picture on the right). Each whole small square of the bar weighs 6 g. How much does the entire bar weigh?
Answer: 240... | Solution. The tile consists of 32 whole squares and 16 triangles. Each triangle is half a square, meaning it weighs $6: 2=3$ g. Therefore, the weight of the chocolate tile is calculated as follows:
$$
32 \cdot 6+16 \cdot 3=240_{\Gamma} \text {. }
$$
## Criteria
4 p. The correct answer is obtained. | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. As is known, balance scales come to equilibrium when the weight on both pans is the same. On one pan, there are 9 identical diamonds, and on the other, 4 identical emeralds. If one more such emerald is added to the diamonds, the scales will be balanced. How many diamonds will balance one emerald? The answer ... | Answer: 3 diamonds.
Solution. From the condition of the problem, it follows that 9 diamonds and 1 emerald weigh as much as 4 emeralds. Thus, if we remove one emerald from each side of the scales, the equality will be preserved, that is, 9 diamonds weigh as much as 3 emeralds. This means that 3 diamonds weigh as much a... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Once, a team of Knights and a team of Liars met in the park and decided to ride the circular carousel, which can accommodate 40 people (the "Chain" carousel, where everyone sits one behind the other). When they sat down, each person saw two others, one in front of them and one behind them, and said: "At least one of... | Solution. Let's consider the initial seating arrangement. From the statement of each Liar, it follows that none of those sitting in front of or behind him is a Liar, i.e., each Liar is surrounded by two Knights. Knights and Liars cannot alternate, as in front of or behind each Knight, there must be at least one Knight.... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
Find the smallest natural number whose sum of digits is 47. | Answer: 299999.
Solution. To find the smallest number, you need to get by with as few digits as possible. The largest digit is 9, so you can't do with fewer than 6 digits ( $5 \cdot 9<47$ ). We cannot put less than 2 in the first place, and by taking 2, the other five digits are uniquely determined: they are nines. | 299999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Variant 1.
In the apartment, there are four square rooms, which are marked as room №1, №2, №3, №4, and a corridor (№5). The perimeter of room №1 is 16 m, and the perimeter of room №2 is 24 m. What is the perimeter of the corridor (№5)? Give your answer in meters.
, and the side of room No.2 is $24: 4=6$ meters. Then, the side of room No.3 is $6+4=10$ meters, so the side of room No.4 is $10+4=14$ meters. Therefore, the longer side of the corridor is $14+4=18$ meter... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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