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2. Petya came up with a four-digit number, in which all the digits are different. It is known that the sum of the first three digits of this number is divisible by 9 and the sum of the last three digits of this number is divisible by 9. What values can the sum of all the digits of this number take? Find all possible va... | 2. Answer. The sum of the digits is 18.
According to the condition, the sum of the first three digits is divisible by 9, and the sum of the last three is also divisible by 9. Therefore, the difference between the first and the last digit is divisible by 9. Since all digits are different, the first digit is 9, and the ... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $ABC$, $AB = 3 \cdot BC$, point $M$ is the midpoint of side $AB$, and $BD$ is the angle bisector. Find the angle $MDB$.
---
Here is the translation of the given text, maintaining the original formatting and structure. | 4. Answer. $90^{\circ}$.
Let $X$ be the midpoint of segment $M B$, and draw segment $D X$. By the property of the angle bisector, we have
$$
A D: D C = A B: B C = 3: 1
$$
On the other hand,... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. For what least natural $n$ do there exist such natural $a$ and $b$ that
$$
\text { GCD }(a, b)=999 \text { and } \text{LCM}(a, b)=n!
$$
(here $n!=1 \cdot 2 \cdot \ldots \cdot n)$ ? | 2. For what smallest natural $n$ do there exist such natural $a$ and $b$ that
$$
\text { GCD }(a, b)=999 \text { and } \text{LCM}(a, b)=n!
$$
(Here $n!=1 \cdot 2 \cdot \ldots \cdot n) ?$
ANSWER. $n=37$.
SOLUTION. Since $\text{LCM}(a, b)$ is divisible by $\text{GCD}(a, b)=999=27 \cdot 37$, and the number 37 is prime... | 37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Vova has 19 math grades in his journal, all twos and threes, and the first four grades are twos. It turned out that among the quartets of consecutive grades, all 16 possible combinations of four twos and threes are present. What are Vova's last four grades? | 5. Vova has 19 math grades in his journal, all twos and threes, and the first four grades are twos. It turned out that among the quartets of consecutive grades, all 16 possible combinations of four twos and threes are present. What are Vova's last four grades?
ANSWER. 3222.
SOLUTION. Note that there are exactly 16 qu... | 3222 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The lateral sides $K L$ and $M N$ of trapezoid $K L M N$ are equal to 15 and 12, respectively, and the base $L M=3$. The bisector of angle $N K L$ passes through the midpoint of side $M N$. Find the area of the trapezoid. Answer: 80. | # Solution.
Let $Q$ be the midpoint of segment $M N$. Extend the bisector of angle $N K L$ to intersect line $L M$ at point $P$. $\angle N K Q = \angle Q K L$.
$\angle N K Q = \angle K P L$ ... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Each cell of a $5 \times 5$ square contains a number. The numbers in any row and any column form an arithmetic progression. The numbers in the corners of the square are $1, 25, 17, 81$. What number is in the central cell? Don't forget to justify your answer.
| 1 | | | | 25 |
| :--- | :--- | :--- | :--- | :--- |
... | 2. Answer: 31. Solution. In any arithmetic progression $a_{5}=a_{1}+4 d, a_{3}=a_{1}+2 d \Rightarrow$ $a_{3}=\left(a_{1}+a_{5}\right) / 2$. Therefore, in the third cell of the first column stands the number 41, in the third cell of the last column stands the number 21, and in the middle cell of the second row (i.e., in... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest natural number that has exactly 12 different natural divisors, the largest prime divisor of which is the number 101, and the last digit is zero. | Solution. Let $n$ be the desired number. According to the condition, $n: 101, n: 2, n: 5$. Consider the number $m=2 \cdot 5 \cdot 101$, note that it has exactly 8 different natural divisors $(1$, $2,5,101,2 \cdot 5, \ldots, 2 \cdot 5 \cdot 101$ ), so $n>m$. Since $n$ is the smallest natural number satisfying the condit... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. How many four-digit numbers exist where the digit in the thousands place is greater than the digit in the hundreds place? | Solution. The digit in the thousands place can take one of 9 possible values: $1,2,3, \ldots, 9$ (we cannot take 0, since the number is four-digit). For each of these options, we can specify the corresponding number of options for the hundreds digit: $1,2,3$, ..., 9. That is, in total $1+2+\ldots+9=45$ options. The oth... | 4500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. What is the minimum number of chips that need to be taken so that in any arrangement of them on the cells of a chessboard, there would necessarily be 4 chips standing in a row horizontally? | Solution. Consider one row. To ensure that there are 4 chips in it, there must be no less than 7 chips (6 chips can be placed in two groups of 3 chips). If there are a total of $6 \cdot 8+1=49$ chips, then by the pigeonhole principle, there will definitely be a row with no fewer than 7 chips. On the other hand, it is i... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. We consider all possible $100-$-digit numbers, where each digit is equal to 1 or 2. For each, the remainder when divided by 1024 is calculated. How many different remainders are there among these? | 5. Solution: Subtract the number $11 \ldots 1$ (one hundred ones) from all the numbers. The numbers under consideration have turned into numbers consisting of zeros and ones from 0 to $11 \ldots 1$. We will show that if two of these numbers have differences in the last 10 digits, then they have different remainders whe... | 1024 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. In a supermarket, there are two types of fruit sets for sale. The first type of set consists of 3 apples and 15 oranges and costs 360 rubles. The second type of set consists of 20 apples and 5 oranges and costs 500 rubles. Fruits are sold only in sets, and sets cannot be divided. Seryozha came to the supermarket a... | Solution: Let Sergey buy $x$ sets of the first type and $y$ sets of the second type. Then he bought $3x + 20y$ apples and $15x + 5y$ oranges. From the problem's condition, it follows that these numbers are equal, that is, $15y = 12x$ or $5y = 4x$. The smaller $y$ is, the smaller $x$ is, and the smaller the amount paid.... | 3800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. On a line, several points were marked. Then, between each pair of neighboring points, one more point was marked, and this operation was repeated once more. As a result, 101 points were obtained. How many points were marked initially? | 9.1. Answer: 26. Solution. Let there initially be $k$ points marked. Then $k-1$ more points were added to them (one between the first and second, second and third, $\ldots, k-1$-th and $k$-th marked points), and then another $(k+(k-1))-1=$ $2 k-2$ points. In total, the number of points became $4 k-3$. Solving the equat... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. The average height of 11 football players on the team is 182 cm. During the match, the referee sent off one player, and the average height of the remaining players became 181 cm. What is the height of the player who was sent off? | Answer: 192 cm. Solution. Let $S$ be the sum that results from adding the heights of the 10 remaining football players. Then $\frac{S}{10}=181$ and $\frac{S+x}{11}=182$, where $x$ is the height of the removed football player. From this, $S=1810$ and $x=182 \cdot 11-S=2002-1810=192$. | 192 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. Petya wrote down all natural numbers from 1 to $n$ in a row on the board and counted the total number of digits written. It turned out to be 777. What is $n$? | Answer: 295. Solution: Since a total of 777 digits have been written, $n$ must be a three-digit number: indeed, in the case of a two-digit $n$, no more than $9+2 \cdot 90=189$ digits would have been written, and in the case of a four-digit (or more) number, more than $9+2 \cdot 90+3 \cdot 900=2889$ digits would have be... | 295 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. In a bag for a bingo game, there are 90 barrels with numbers from 1 to 90. What is the minimum number of barrels that need to be drawn at random from the bag to guarantee getting a barrel with a number divisible by 3 or 5 (or both 3 and 5)? | Answer: 49. Solution. We will call numbers that are divisible by 3 or 5 desirable. The number of numbers from 1 to 90 that are divisible by 3 is 30 ( =90: 3 ), and those divisible by 5 is 18 ( =90:5). If we add $30+18=48$, then all numbers that are divisible only by 3 and only by 5 will be counted once, and numbers tha... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. How many six-digit natural numbers exist, each of which has adjacent digits with different parity | Answer: 28125. Solution. If the first (most significant) digit is even, then it can be chosen in four ways $(2,4,6,8)$, and all subsequent ones can be chosen in five ways (possible candidates for the second, fourth, and sixth digits are $1,3,5,7,9$, and for the third and fifth - $0,2,4,6,8$). In the end, by the rule of... | 28125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. Points $A, B, C, D$ are located at the vertices of rectangle $ABCD$, its sides and diagonals $AC$ and $BD$ are roads. The first car traveled the route $B \rightarrow$ $C \rightarrow A \rightarrow D$ in one hour, and the second car traveled the route $D \rightarrow B \rightarrow C \rightarrow A$ in one hour. After ... | Answer: In 40 minutes.
Solution: The diagonals of a rectangle have the same length and are longer than any of its sides. In one hour, together the two cars would travel three times the length of side $B C$ and three times the diagonal, since one car travels two sides equal to $B C$ and one diagonal in an hour, while t... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A straight road passes not far from Mount Fuji. The driver of a car noticed the mountain 60 km to the north, and an hour later - 45 km to the west. What is the shortest distance the car passed from Mount Fuji? | Answer: 36 km.
Solution: The mountain and the observation points are at the vertices of a right-angled triangle with legs of 60 and 45. The hypotenuse in this case is 75. By equating the areas of the triangle calculated in different ways, we get that the product of the legs is equal to the product of the hypotenuse an... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. This year, the son and daughter are so many years old that the product of their ages is 7 times less than the father's age. And in three years, the product of their ages will already be equal to the father's age. Find the father's age. | Answer: 21 years
Solution. Let $n$ be the son's age, $m$ be the daughter's age, then the father's age is $7 m n$. In three years, the son's age will be $n+3$, the daughter's age will be $m+3$, and the father's age will be $7 \mathrm{mn}+3$, and according to the condition, the equation $(n+3)(m+3)=7 m n+3$ will be sati... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A right triangle $A B C$ (cathetus $B C$ is greater than cathetus $A C$) is inscribed in a circle. A point $D$ is chosen on side $B C$ such that $B D=A C$, and point $M$ is the midpoint of the arc $A C B$. Find the angle $C D M$. | Answer: $45^{\circ}$.
Solution. Note that $A B$ is the diameter of the circumscribed circle. Connect point $M$ with points $A, B, C$, and $D$. Since arcs $A M$ and $B M$ are equal, the chords $A M$ and $B M$ that subtend them are also equal. Segments $B D$ and $A C$ are equal by the condition. Finally, angles $M B C$ ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. When a certain number is divided by 13 and 15 with a remainder, the same incomplete quotients are obtained. Find the largest such number. | Answer: 90. Suppose the number a satisfies the condition of the problem. Then the equalities $a=13 d+b=15 d+c$ hold, where $d$ is the quotient from the condition of the problem. From this, $2 d=b-c \leq 12$, i.e., $d \leq 6$, and the remainder $b$ from dividing the number by 13 is no more than 12. For $d=6$ and $b=12$,... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. Oleg drew an empty $50 \times 50$ table and wrote a non-zero number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers written... | Answer: 1250 works.
Solution. First, we show that there are at least 1250 irrational numbers in the table. Suppose along the left side of the table, there are $x$ irrational and $50-x$ rational numbers. Then along the top side, there are $50-x$ irrational and $x$ rational numbers. Since the product of a non-zero ratio... | 1250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Does there exist a four-digit natural number with distinct non-zero digits that has the following property: if this number is added to the same number written in reverse order, the result is divisible by $101 ?$ | Answer. It exists.
Solution. For example, the number 1234 works. Indeed, $1234+4321=5555=101 \cdot 55$.
Remark. A number $\overline{a b c d}$ with distinct non-zero digits satisfies the condition if and only if $a+d=b+c$.
Comment. Any correct example with verification that it fits - 7 points.
Any correct example wi... | 1234 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest number that contains only the digits 2 and 3 in equal quantities, and is divisible by 2 and 3. | Answer: 223332.
Sketch of the solution. If a number is divisible by 3, then the sum of its digits is divisible by 3, which means the number of twos is a multiple of three, and thus, there must be at least three twos. Therefore, the desired number is a six-digit number. Since it is divisible by 2, it ends in 2. The num... | 223332 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. A rectangle with sides $6 \mathrm{~cm}$ and $3 \mathrm{~cm}$ was cut into three rectangles of equal perimeter. What can the perimeter of these rectangles be? Find all possible answers. | Answer: 14 cm, 10 cm, 10.5 cm.
Sketch of the solution.
Let's cut a rectangle with sides $a$ and $b$. Cases 1 and 2. Cut the rectangle into three equal rectangles with two cuts parallel to side $a$.
, 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
3. (7 points) 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
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\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 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 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 |
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:
 | 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 |
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 |
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 |
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 |
4. Variant 1.
The numbers $96, 28, 6, 20$ were written on the board. One of them was multiplied, another was divided, a third was increased, and a fourth was decreased by the same number. As a result, all the numbers became equal to one number. Which one? | Answer: 24.
Solution: Addition and multiplication increase numbers, while subtraction and division decrease them, so the two larger numbers have decreased, and the smaller ones have increased. Now let's consider the two middle numbers 20 and 28. It is clear that a number was subtracted from 28, and the same number was... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1.
Petya wrote the numbers from 1 to 10 on cards and laid them out along the edge of a $3 \times 4$ rectangle. At first, one of the cards - with the number 7 - was opened (see the figure). When the other cards were opened, it turned out that the sum of the numbers in the top and bottom horizontal rows is th... | Answer: 23.
Solution. Let $B$ be the last number in the second horizontal row. Since the sum of all numbers from 1 to 10 is 55, we have $2A + B + 7 = 55$ or $B = 48 - 2A$. It follows that $B$ is an even number. To make $A$ the largest, $B$ needs to be the smallest. The smallest even number among the given numbers is 2... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 7. Variant 1.
100 natural numbers are written in a circle. It is known that among any three consecutive numbers, there is an even number. What is the smallest number of even numbers that can be among the written numbers? | Answer: 34.
Solution. Consider any 3 consecutive numbers. Among them, there is an even number. Fix it, and divide the remaining 99 into 33 groups of 3 consecutive numbers. In each such group, there is at least one even number. Thus, the total number of even numbers is at least $1+33=34$. Such a situation is possible. ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.3. Grandma has two balls of wool: a large one and a small one. From the large one, she can knit either a sweater and three socks, or five identical caps. $A$ from the small one - either half a sweater, or two caps. (In both cases, all the wool will be used.) What is the maximum number of socks that Grandma can knit u... | # Solution:
Method 1. Half a sweater requires as much wool as 2 hats, so a sweater requires as much wool as 4 hats. Then 4 hats and 3 socks require as much wool as 5 hats. Therefore, one hat is equivalent to three socks. In total, 5 + 2 = 7 hats or 21 socks can be knitted.
Method 2. Let the amount of wool required fo... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.5. Ten different natural numbers are such that the product of any five of them is even, and the sum of all ten is odd. What is the smallest possible sum of all these numbers? Justify your answer. | Solution: There cannot be more than four odd numbers (five odd numbers in a product give an odd number), and, moreover, their quantity must be odd. Therefore, there are either 3 or 1 odd numbers, and 7 or 9 even numbers, respectively. The sets with the smallest sum are 1, 3, 5, 2, 4, 6, 8, 10, 12, 14 and 1, 2, 4, 6, 8,... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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