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4. Sasha, Lёsha, and Kolya started a 100 m race at the same time. When Sasha finished, Lёsha was ten meters behind him, and when Lёsha finished, Kolya was ten meters behind him. How far apart were Sasha and Kolya when Sasha finished? (It is assumed that all the boys run at constant, but of course, not equal speeds.)
|
Answer: 19 m.
Solution: Kolya's speed is 0.9 of Lesha's speed. At the moment when Sasha finished, Lesha had run 90 m, and Kolya had run $0.9 \cdot 90=81$ m. Therefore, the distance between Sasha and Kolya was 19 m.
|
19
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The museum has 16 halls, arranged as shown in the diagram. In half of them, paintings are exhibited, and in the other half, sculptures. From any hall, you can go to any adjacent one (sharing a common wall). During any tour of the museum, the halls alternate: a hall with paintings - a hall with sculptures - a hall with paintings, and so on. The tour starts in hall A, where paintings are displayed, and ends in hall B.
a) Mark with crosses all the halls where paintings are exhibited. Solution. See the diagram.
b) A tourist wants to visit as many halls as possible (travel from hall A to hall B), but visit each hall no more than once. What is the maximum number of halls he can visit? Draw one of his longest routes and prove that he could not have visited more halls.
|
Answer: 15.
Solution: One of the possible routes is shown in the diagram. Let's prove that if a tourist wants to visit each hall no more than once, they will not be able to see more than 15 halls. Note that the route starts in a hall with paintings (A) and ends in a hall with paintings (B). Therefore, the number of halls with paintings that the tourist has passed is one more than the number of halls with sculptures. Since the number of halls with paintings that the tourist could pass is no more than 8, the number of halls with sculptures is no more than 7. Thus, the route cannot pass through more than
15 halls.
## 7th Grade. Recommendations for Checking.
Each problem is scored out of 7 points. Each score is an integer from 0 to 7. Below are some guidelines for checking. Naturally, the problem setters cannot foresee all cases. When evaluating a solution, it should be determined whether the provided solution is generally correct (although it may have flaws) - in which case the solution should be scored no less than 4 points. Or if it is incorrect (although it may have significant progress) - in which case the score should be no more than 3 points.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Sasha, Lёsha, and Kolya start a 100 m race at the same time. When Sasha finished, Lёsha was ten meters behind him, and when Lёsha finished, Kolya was ten meters behind him. How far apart were Sasha and Kolya when Sasha finished? (It is assumed that all the boys run at constant, but of course, not equal speeds.)
|
Answer: 19 m.
Solution: Kolya's speed is 0.9 of Lesha's speed. At the moment when Sasha finished, Lesha had run 90 m, and Kolya had run $0.9 \cdot 90=81$ m. Therefore, the distance between Sasha and Kolya was 19 m.
|
19
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Each of the 10 dwarfs either always tells the truth or always lies. It is known that each of them loves exactly one type of ice cream: butter, chocolate, or fruit. First, Snow White asked those who love butter ice cream to raise their hands, and everyone raised their hands, then those who love chocolate ice cream - and half of the dwarfs raised their hands, then those who love fruit ice cream - and only one dwarf raised his hand. How many of the dwarfs are truthful?
#
|
# Answer. 4.
Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of raised hands increases by 1. Since there were 6 "extra" hands raised, 6 gnomes lied, and 4 told the truth.
## 8th Grade. Grading Recommendations.
Each problem is scored out of 7 points. Each score is an integer from 0 to 7. Some guidelines for grading are provided below. Naturally, the creators cannot foresee all cases. When evaluating a solution, it should be determined whether the provided solution is generally correct (although it may have flaws) - in which case the solution should be scored at least 4 points. Or if it is incorrect (although it may have significant progress) - in which case the score should not exceed 3 points.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. We will call a number palindromic if it reads the same from left to right as it does from right to left. For example, the number 12321 is palindromic. How many five-digit palindromic numbers are there that are divisible by 5?
|
Answer: 100.
Solution: A number that is divisible by 5 must end in 5 or 0. A mirrored number cannot end in 0, as then it would have to start with 0. Therefore, the first and last digits are 5. The second and third digits can be anything from the combination 00 to the combination 99 - a total of 100 options. Since the fourth digit repeats the second, there will be a total of 100 different numbers.
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Each of the 10 dwarfs either always tells the truth or always lies. It is known that each of them loves exactly one type of ice cream: butter, chocolate, or fruit. First, Snow White asked those who love butter ice cream to raise their hands, and everyone raised their hands, then those who love chocolate ice cream - and half of the dwarfs raised their hands, then those who love fruit ice cream - and only one dwarf raised their hand. How many of the dwarfs are truthful?
|
# Answer. 4.
Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of raised hands increases by 1. Since 6 "extra" hands were raised, 6 gnomes lied, and 4 told the truth.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4. 2011 Warehouses are connected by roads in such a way that from any warehouse you can drive to any other, possibly by driving along several roads. On the warehouses, there are $x_{1}, \ldots, x_{2011}$ kg of cement respectively. In one trip, you can transport an arbitrary amount of cement from any warehouse to another warehouse along the road connecting them. In the end, according to the plan, there should be $y_{1}, \ldots, y_{2011}$ kg of cement on the warehouses respectively, with
$$
x_{1}+x_{2}+\ldots+x_{2011}=y_{1}+y_{2}+\ldots+y_{2011}
$$
What is the minimum number of trips required to fulfill the plan for any values of the numbers $x_{i}$ and $y_{i}$ and any road network?
(P. Karasev)
|
Answer. In 2010 trips.
Solution. First, we show that the plan cannot always be completed in 2009 trips. Suppose (with any road scheme) that initially all the cement is located at one warehouse $S$, and it needs to be evenly distributed to all warehouses. Then, cement must be delivered to each warehouse, except $S$, in some trip; it is clear that these 2010 trips are distinct, so the total number of trips must be at least 2010.
We need to show that the plan can always be completed in 2010 trips. We will prove by induction on $n$ that with $n$ warehouses, it is always possible to manage with $n-1$ trips. The base case for $n=1$ is obvious.
Let $n>1$. Since from any warehouse you can reach any other, there exists a route that passes through all warehouses (possibly more than once). Consider any such route and the warehouse $A$ that appears on this route the latest for the first time. Then, if we remove warehouse $A$ and all roads leading from it, it is still possible to reach any warehouse from any other (using the previous roads of the route).
We can assume that $A$ is the warehouse with number $n$. If $y_{n} \leqslant x_{n}$, then we will transport $x_{n}-y_{n}$ kg of cement from $A$ to any connected warehouse, and then forget about it and all roads leading from it. By the induction hypothesis, for the remaining warehouses, the plan can be completed in $(n-1)-1$ trips. In total, through $(n-2)+1$ trips, the required distribution of cement will be achieved.
If $y_{n}>x_{n}$, then we have already shown that from the distribution where the $i$-th warehouse has $y_{i}$ kg, it is possible to achieve the distribution where the $i$-th warehouse has $x_{i}$ kg in $n-1$ trips. By performing all these transports in reverse order (and in the opposite direction), we will implement the required plan.
Comment. Only the answer - 0 points.
Only an example is provided, showing that the plan cannot always be completed in 2009 trips - 2 points.
Only proved that 2010 trips are always sufficient - 5 points.
|
2010
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Given natural numbers $M$ and $N$, both greater than ten, consisting of the same number of digits, and such that $M = 3N$. To obtain the number $M$, one of the digits of $N$ must be increased by 2, and each of the other digits must be increased by an odd digit. What digit could the number $N$ end with? Find all possible answers.
(N. Agakhanov)
|
Answer. The digit 6.
Solution. By the condition, $M=3 N$, so the number $A=M-N=2 N$ is even. However, by the condition, the number $A$ is composed of odd digits and the digit 2. Therefore, $A$ ends in 2. Thus, the number $N$, which is half of $A$, ends in either 1 or 6.
We will show that $N$ cannot end in 1. If $N$ ends in 1, then when it is doubled, there is no carry from the last digit to the second-to-last digit. This means that the second-to-last digit of the number $A=2 N$ would be even, but it must be odd. This is a contradiction.
Remark. Pairs of numbers $N$ and $M$ as described in the condition do exist, for example, $N=16, M=48$. Moreover, there are infinitely many such pairs. All suitable numbers $N$ can be described as follows: the first digit is 1 or 2, followed by several (possibly zero) digits, each of which is 5 or 6, and the last digit is 6.
Comment. A correct answer and an example of a number $N$ ending in the digit 6 - 1 point.
It is established that the last digit of the number $M$ is 2 more than the last digit of the number $N$ - 1 point.
It is shown that the last digit of the number $N$ can only be 1 or 6 - 2 points.
Points for various advancements are cumulative.
Note that the problem does not require providing an example of such a number. It is sufficient to prove that no digit other than 6 can be the last digit.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Pentagon $A B C D E$ is inscribed in circle $\omega$. Diagonal $A C$ is the diameter of circle $\omega$. Find $\angle B E C$, if $\angle A D B=20^{\circ}$.
|
Answer: $70^{\circ}$.
Solution. Fig. 4. Since $\angle A D B=20^{\circ}$, the arc $A B$ is $40^{\circ}$. Since $A C$ is a diameter, the arc $A B C$ is $180^{\circ}$, so the arc $B C$ is $180^{\circ}-40^{\circ}=140^{\circ}$. The angle $B E C$ subtends the arc $B C$, which means it is equal to $140^{\circ} / 2=70^{\circ}$.
## Criteria
4 p. Correct solution.
0 p. Only the correct answer.
Fig. $4:$ to problem 2
|
70
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. On the edge $A A^{\prime}$ of the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge length 2, a point $K$ is marked. In space, a point $T$ is marked such that $T B=\sqrt{11}$ and $T C=\sqrt{15}$. Find the length of the height of the tetrahedron $T B C K$, dropped from vertex $C$.
|
Answer: 2.
Fig. 5: to problem 5
Solution. Notice that
$$
T B^{2}+B C^{2}=11+4=15=T C^{2}
$$
From this, by the converse of the Pythagorean theorem, it follows that angle $T B C$ is a right angle. Therefore, $T B \perp B C$, which means $T$ lies in the plane of the face $A A^{\prime} B^{\prime} B$. Thus, $B C$ is the height dropped from vertex $C$, and its length is 2.
Remark. There are two possible positions of point $T$, symmetric with respect to the plane $K B C$.
## Criteria
## 4 p. Correct solution.
0 p. Only the correct answer.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Inside the magician's hat, there live 100 rabbits: white, blue, and green. It is known that if 81 rabbits are randomly pulled out of the hat, there will definitely be three of different colors among them. What is the minimum number of rabbits that need to be taken out of the hat to ensure that there are definitely two of different colors?
#
|
# Answer: 61.
Solution. We will prove that if 61 rabbits are randomly pulled out of the hat, then among them there will be two of different colors. Suppose the opposite: let there be $a \geqslant 61$ rabbits of some color (for example, white). Let the second color by number of rabbits be blue. Then there are at least $\frac{100-a}{2}$ blue rabbits in the hat. Therefore, the total number of white and blue rabbits is at least
$$
a+\frac{100-a}{2}=\frac{100+a}{2} \geqslant \frac{161}{2}=80.5
$$
Since the number of rabbits is an integer, the total number of white and blue rabbits is at least 81, which contradicts the condition.
We will show that 60 rabbits may not be enough. Suppose there are 60 white rabbits and 20 blue and 20 green rabbits in the hat. Then it is possible that all the pulled-out rabbits are white. On the other hand, if 81 rabbits are pulled out, then among them there will definitely be rabbits of all three colors.
## Criteria
4 points. Correct solution.
3 points. Proven that 61 rabbits are enough.
1 point. Shown that 60 rabbits may not be enough.
0 points. Only the correct answer.
|
61
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In an $n \times n$ square, there are 1014 dominoes (each covering two adjacent cells). No two dominoes share any points (even corner points). For what smallest $n$ is this possible?
|
Answer. For $n=77$.
Solution. Attach four cells to the right and below each domino so that they form a $2 \times 3$ rectangle (if the domino is vertical) or a $3 \times 2$ rectangle (if it is horizontal). If the rectangles of two dominoes have at least one common cell, then the dominoes have a common point. Therefore, if 1014 dominoes, which do not have common points, fit into an $n \times n$ square, then all the constructed rectangles of 6 cells each must fit into a $(n+1) \times (n+1)$ square, obtained by adding a row below and a column to the right of the $n \times n$ square. Hence,
$$
(n+1)^{2} \geqslant 6 \cdot 1014=6084=78^{2}
$$
Thus, $n \geqslant 77$.
It remains to provide an example where this is possible for $n=77$. For this, place the dominoes horizontally in the first row, starting from the first cell, with a gap of one cell; a total of 26 dominoes will fit. Do the same with all other odd-numbered rows. The total number of dominoes will then be $26 \cdot 39=1014$; they obviously do not have common points.
|
77
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (7 points) Percival's castle had a square shape. One day, Percival decided to expand his domain and added a square extension to the castle. As a result, the perimeter of the castle increased by $10 \%$. By what percentage did the area of the castle increase?
|
Answer: $4 \%$.
Solution. Let the width of the castle be $a$, and the width of the extension be $b$. Then the original perimeter is $4 a$, and the final perimeter is $4 a+2 b$. Therefore:
$$
1.1 \cdot 4 a=4 a+2 b \Leftrightarrow b=0.2 a
$$
From this, the area of the castle becomes $a^{2}+(0.2 a)^{2}=1.04 a^{2}$, which means the area has increased by $4 \%$.
Criteria. Any correct solution: 7 points.
The side of the extension is correctly found, but the further solution is missing or incorrect: 4 points.
Only the correct answer is provided: 0 points.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (7 points) It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)$ take?
## Answer: 0.
|
Solution. Note that the equality $a^{2}+b=b^{2}+c$ can be written as: $a^{2}-b^{2}=c-b$. Similarly, we have $b^{2}-c^{2}=a-c, c^{2}-a^{2}=b-a$. Substituting these equalities into the desired expressions, we get that
$$
a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0
$$
Criteria. Any correct solution: 7 points.
Only the correct answer is provided: 0 points.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (7 points) Lёsha did not hesitate to calculate the sum
$$
9+99+999+\ldots+\underbrace{9 \ldots 9}_{2017}
$$
and wrote it on the board. How many times is the digit 1 written in the final result?
|
Answer: 2013.
Solution. Transform the expression:
$$
\begin{aligned}
9+99+999+\ldots+\underbrace{9 \ldots 9}_{2017} & =(10-1)+(100-1)+\ldots+\left(10^{2017}-1\right)= \\
& =\underbrace{1 \ldots 10}_{2017}-2017=\underbrace{1 \ldots 1}_{2013} 09093 .
\end{aligned}
$$
Criteria. Any correct solution: 7 points.
It is shown that the original sum is equal to $\underbrace{1 \ldots 10}_{2017}-2017$, but further solution is missing or contains an arithmetic error: 5 points. Only the correct answer is provided: 1 point.
|
2013
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (7 points) Several sages lined up in a column. All of them wore either black or white caps. It turned out that among any 10 consecutive sages, there were an equal number of sages with white and black caps, while among any 12 consecutive sages - not an equal number. What is the maximum number of sages that could be
Answer: 15 sages.
|
Solution. We will prove that there cannot be more than 15 sages. Suppose the opposite, that there are at least 16 sages. Sequentially number all the sages. Consider nine consecutive sages. If we add one of the two neighboring sages to them, then among them there will be an equal number of sages with white and black hats, so on any sages, between whom there are 9 sages, hats of the same color are worn.
Without loss of generality, let the first sage wear a black hat. Then the eleventh sage also wears a black hat. If the twelfth sage wears a white hat, then among the first twelve sages there will be an equal number of white and black hats. Therefore, the twelfth sage wears a black hat, from which it follows that the second sage also wears a black hat. Similarly, considering the sages from the second to the eleventh, we get that the third and thirteenth sages wear black hats. Considering the sages from the third to the twelfth, we get that the fourth and fourteenth sages wear black hats. Similarly, the fifth and fifteenth, and the sixth and sixteenth sages wear black hats. But then among the first ten sages, the first six wear black hats, so there will be more black hats. Contradiction.
15 sages can be: let the first 5 and the last 5 sages wear black hats, and the remaining 5 wear white hats. It is easy to see that the condition of the problem will be satisfied.
Criteria. Any correct solution: 7 points.
Proved that there cannot be more than 15 sages, but no example is given of how to put the hats on 15 sages: 6 points.
Proved that on two sages, between whom stand 9 sages, hats of the same color are worn, but further reasoning is missing or incorrect: 2 points.
An example of arranging 15 sages that satisfies the condition is given, but it is not proved that more sages cannot be placed: 1 point.
Only the correct answer is given: 0 points.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Variant 1.
Petya has four cards with digits $1,2,3,4$. Each digit appears exactly once. How many natural numbers greater than 2222 can Petya make from these cards?
|
Answer: 16.
Solution: Let's find out how many different numbers can be formed from these cards: the first digit can be chosen in 4 ways, the second can be appended in 3 ways, the third in 2 ways, and the last one is uniquely determined. That is, a total of 24 different numbers can be obtained (it is also possible to verify this explicitly by listing all suitable numbers). Let's find out how many of the listed numbers will not suit us. These are all numbers starting with 1, there are 6 of them. And 2 more numbers: 2134, 2143, because if the number starts with 2, the second digit can only be 1. Then the suitable numbers are $24-6-2=16$.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Variant 1.
Nine integers from 1 to 5 are written on the board. It is known that seven of them are not less than 2, six are greater than 2, three are not less than 4, and one is not less than 5. Find the sum of all the numbers.
|
Answer: 26.
Solution. A number not less than 5 is 5. There is exactly one number 5. Three numbers are not less than 4, so exactly two numbers are equal to 4. Six numbers are greater than 2, meaning all of them are not less than 3. Therefore, exactly three numbers are equal to 3. Seven numbers are not less than 2, so one number is equal to 2. In total, there are 9 numbers, hence two numbers are equal to 1. The sum of all numbers is $1+1+2+3+3+3+4+4+5=26$.
|
26
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. Option 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya said that from April 1 to April 20, the café was open for 17 days, and from April 10 to April 30, it was open for 18 days. It is known that he made a mistake once. What was the date of the last Tuesday in April?
|
Answer: 29
Solution: Since there are exactly 21 days from April 10 to April 30, each day of the week occurred exactly 3 times during this period. Therefore, this statement cannot be false. This means the first statement is false, and there were only 2 Mondays from April 1 to April 20 (there could not have been four, as at least 22 days would be needed for that). This could only happen if April 1 was a Tuesday. Therefore, the last Tuesday in April was the 29th.
|
29
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Variant 1.
A rectangle was cut into three rectangles, two of which have dimensions 9 m $\times$ 12 m and 10 m $\times$ 15 m. What is the maximum area the original rectangle could have had? Express your answer in square meters.
|
Answer: 330
Solution. Since the sizes of the two rectangles are fixed, in order for the original rectangle to have the maximum area, the third rectangle must have the largest area. Since the two given rectangles do not have the same sides, the largest area will be obtained by attaching the smaller side of one rectangle to the larger side of the other. This results in a rectangle of size $12 \times(9+15)$ or a rectangle of size $15 \times(10+12)$. The area of the first is 288, and the second is 330, so the answer is 330.
Variant 2.
A rectangle was cut into three rectangles, two of which have dimensions 8 m $\times 12$ m and 10 m $\times 14$ m. What is the maximum area the original rectangle could have had? Express your answer in square meters.
Answer: 308
|
330
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Variant 1.
In the addition example, where the numbers were written on cards, two cards were swapped, resulting in the incorrect expression: $37541+43839=80280$. Find the error and write down the correct sum.
|
Answer: 80380
Solution. Let's start checking the example from right to left. There are no errors in the units and tens place, but an error appears in the hundreds place. This means that one of the digits in this place $-2, 8$ or $5-$ is transposed.
Let's consider the following cases:
|
80380
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 6. Option 1.
Nезнайка named four numbers, and Ponchik wrote down all their pairwise sums on six cards. Then he lost one card, and the numbers left on the remaining cards were $270, 360, 390, 500, 620$. What number did Ponchik write on the lost card?
|
Answer: 530.
Solution. Let the original numbers be $a \leq b \leq c \leq d$. Suppose the card with the maximum sum is lost. Then this sum is $c+d$. Therefore, $a+b=270$ and $a+b+c+d>270+620=890$. On the other hand, the sum of all numbers on the cards is $3a+3b+2c+2d=270+360+390+500+620=2140$. We get that $2140>1780+2c+2d$, hence $c+d<180$. This is a contradiction. Similarly, we can prove that the card with the smallest numbers was not lost. Therefore, the sum of all numbers on the cards is $270+620=890$. All pairwise sums can be divided into the following groups: 1) $a+b, c+d$, 2) $a+c, b+d$, 3) $a+d, b+c$. The sum of the numbers in each group is 890. This is only possible if the lost card has the number $890-360=530$, since 620 and 270, 390 and 500 form pairs.
|
530
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Variant 1.
101 natural numbers are written in a circle. It is known that among any 3 consecutive numbers, there is at least one 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 this number and its neighbor, and divide the remaining 99 into 33 sets of 3 consecutive numbers. In each such set, there is at least one even number. Thus, the total number of even numbers is no less than $1+33=34$. Such a situation is possible. Number the numbers in a circle. The even numbers can be those with numbers $1,4,7, \ldots, 100$
|
34
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 8. Variant 1.
Identical coins are laid out on a table in the shape of a hexagon. If they are laid out so that the side of the hexagon consists of 2 coins, then 7 coins are enough, and if the side consists of 3 coins, then a total of 19 coins are required. How many coins are needed to build a hexagon with a side consisting of 10 coins?
|
Answer: 271.
Solution.
Method 1.
We will divide the coins into layers (contours) from the center. The first layer contains 1 coin, the second layer contains 6, and so on. Notice that each new layer contains 6 more coins than the previous one (if we remove the coins at the vertices, we get exactly as many coins as there were in the previous layer). Therefore, the total number of coins can be calculated using the formula $1+6+12+18+24+30+36+42+48+54=271$.
Method 2.
Let the side of the hexagon contain $n$ coins. Consider two opposite sides. Each of them consists of $n$ coins. To each of these sides, there are 2 more sides attached. But we have already counted one coin from each of these sides, so there are $n-1$ coins left on each of these sides, and two coins are counted twice. In total, there will be $2n + 4(n-1) - 2 = 6n - 6$. It remains to notice that each new layer of coins is built around the existing one. Then the total will be: $19 + (6 \cdot 4 - 6) + (6 \cdot 5 - 6) + (6 \cdot 6 - 6) + (6 \cdot 7 - 6) + (6 \cdot 8 - 6) + (6 \cdot 9 - 6) + (6 \cdot 10 - 6) = 19 + 18 + 24 + 30 + 36 + 42 + 48 + 54 = 271$.
|
271
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. There were 10 chatterboxes sitting in a circle. At first, one of them told one joke, the next one clockwise told two jokes, the next one told three, and so on around the circle, until one of them told 100 jokes at once. At this point, the chatterboxes got tired, and the next one clockwise told 99 jokes, the next one told 98, and so on around the circle, until one of them told just one joke, and everyone dispersed. How many jokes did each of these 10 chatterboxes tell in total?
|
Answer: 1000 jokes.
Solution. Let's number the chatterboxes from 1 to 10 clockwise, starting with the one who told the first joke. Then for any pair of chatterboxes with numbers $k$ and $k+1 (1 \leq k \leq 9)$, the $(k+1)$-th chatterbox initially tells one more joke per round than the $k$-th, for 10 rounds. After the chatterboxes get tired, the $(k+1)$-th chatterbox tells one less joke than the $k$-th, also for 10 rounds. Thus, the first and second told the same number of jokes, the second and third - the same number, and so on. Therefore, all of them told the same number of jokes - each one-tenth of the total. The total number of jokes told is $1+2+\cdots+99+100+99+\cdots+2+1=(1+99)+(2+98)+\cdots+(99+1)+100=100 \cdot 100=10000$. Each of the 10 chatterboxes told $\frac{10000}{10}=1000$ jokes.
Comment. It is proven that each chatterbox told one-tenth of the total number of jokes - 4 points. The total number of jokes told is found - 3 points. For potentially useful ideas and approaches in the absence of a solution - 2-3 points.
|
1000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In triangle $ABC$, the bisector $BD$ was drawn, and in triangles $ABD$ and $CBD$ - the bisectors $DE$ and $DF$ respectively. It turned out that $EF \| AC$. Find the angle $DEF$.
|
Answer: $45^{\circ}$.
Solution. Let segments $B D$ and $E F$ intersect at point $G$. From the condition, we have $\angle E D G = \angle E D A = \angle D E G$, hence $G E = G D$. Similarly, $G F = G D$. Therefore, $G E = G F$, which means $B G$ is the bisector and median, and thus also the altitude in triangle $B E F$. From this, $D G$ is the median and altitude, and thus also the bisector in triangle $E D F$, from which $\angle D E G = \angle E D G = \angle F D G = \angle G F D$. Since the sum of the four angles mentioned in the last equality is $180^{\circ}$ degrees, each of them is $45^{\circ}$ degrees.
Comment. A complete and justified solution - 7 points. A generally correct reasoning with minor gaps or inaccuracies - up to 5 points.
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In a football tournament where each team played against each other once, teams A, B, C, D, and E participated. For a win, a team received 3 points, for a draw 1 point, and for a loss 0 points. In the end, it turned out that teams A, B, C, D, and E each had 7 points. What is the maximum number of points that team $\mathrm{E}$ could have?
|
Answer: 7 points.
Solution: In a match where one of the teams won, the teams together score 3 points, in a match that ended in a draw - 2 points. Since 7 is not divisible by 3, the team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament. There were a total of 15 matches, as is easily verified. Therefore, all teams together scored no more than $2 \cdot 3 + 3 \cdot 12 = 42$ points. Of these, teams A, B, C, D, and E scored 35 points. Therefore, team F scored no more than $42 - 35 = 7$ points. How it could have scored exactly 7 points is shown in the table on the right.
Comment: A realization is proposed - 3 points, an estimate is made - 4 points, points are summed. In the absence of a solution, 2-3 points for potentially useful ideas and approaches. Answer without justification - 0 points.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. How many different triangles can be formed from: a) 40 matches; b) 43 matches? Solution. We need to find the number of triples of natural numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ such that $\mathrm{x} \leq \mathrm{y} \leq \mathrm{z}$, $\mathrm{x}+\mathrm{y}+\mathrm{z}=40$ and $\mathrm{x}+\mathrm{y}>\mathrm{z}$. From these inequalities, it follows that $\mathrm{z}$ can take values satisfying the inequalities $14 \leq z \leq 19$. If $z=19$, then $x+y=21$, and $x \leq y \leq 19$. Therefore, $11 \leq y \leq 19$, and we have 9 triangles with $z=19$. Similarly, we establish that the number of triangles for which $\mathrm{z}=18,17,16,15,14$ is respectively $8,6,5,3,2$, and in total we have 33 triangles. Similarly, we find that the number of triangles with a perimeter of 43 is 44.
|
Answer: a) $33 ;$ b) 44.
|
33
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 2. Solve the equation
$$
\sqrt{3 x-2-x^{2}}+\sqrt{x^{2}-4 x+3}=\sqrt{2}(1-\sqrt{x})
$$
|
Solution. Solving the system of inequalities
$$
\left\{\begin{array}{c}
3 x-2-x^{2} \geq 0 \\
x^{2}-4 x+3 \geq 0
\end{array}\right.
$$
we obtain that the domain of the function on the left side of the equation is $\{1\}$. The domain of the function on the right side of the equation is the numerical ray $[0 ;+\infty)$. Therefore, the domain of the equation is $\{1\}$. By verification, we find that $x=1$ is a root of the equation.
Answer: 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.6. First solution. Divide all the coins into two parts of 20 coins each and weigh them. Since the number of counterfeit coins is odd, one of the piles will weigh more. This means that there is no more than one counterfeit coin in it. Divide it into two piles of 10 coins and weigh them. If the scales are in balance, then all 20 weighed coins are genuine. If one of the pans outweighs, then there are 10 genuine coins on that pan, and among the other 10 coins, there is exactly one counterfeit. Divide these 10 coins into three piles of 4, 4, and 2 coins. In the third weighing, compare the two piles of 4 coins. If they balance, then all 8 coins are genuine and we have found 18 genuine coins. If one of the piles outweighs, then there are 4 genuine coins in it, the other pile contains the counterfeit coin, and the 2 set aside coins are genuine. In total, 16 genuine coins have been found.
|
The second solution. Divide all the coins into five equal piles, each containing 8 coins, and number them. Place the 1st and 2nd piles on one side of the scales, and the 3rd and 4th piles on the other.
Consider the first case - the scales balance. Then either there is one fake coin on each side, or all the coins being weighed are genuine. Then we will weigh the 1st and 2nd piles. If they balance, then all 16 coins are genuine. If one of the piles outweighs the other, then it contains 8 genuine coins. The third weighing will compare the 3rd and 4th piles to determine the next 8 genuine coins.
Now consider the second case - the scales do not balance. Let's assume for definiteness that the 1st and 2nd piles outweigh the others, then among them there is no more than one fake coin. The second weighing will compare the 1st and 2nd piles. If they balance, then all 16 coins are genuine. If one of the piles outweighs the other, then it contains 8 genuine coins, and the other has exactly one fake coin. Consequently, the 3rd and 4th piles contain exactly two fake coins, and the 5th pile contains 8 genuine coins. Thus, a total of 16 genuine coins have been found.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Postman Pechkin is riding a bicycle along a highway. He noticed that every 4.5 kilometers, a suburban bus overtakes him, and every 9 minutes, a suburban bus passes him in the opposite direction. The interval of bus movement in both directions is 12 minutes. At what speed is Pechkin riding?
|
Answer: 15 km/h
Solution: Let x (km/h) and y (km/h) be the speeds of the cyclist and the bus, respectively. Since the interval between bus movements is 12 minutes (1/5 hour), the distance between two consecutive buses is y/5 km. Therefore, at the moment the cyclist meets a bus, the distance between the cyclist and the next oncoming bus is y/5 km. Since their meeting will occur in 9 minutes (3/20 hour), and the closing speed is (x+y), we have $3(x+y)/20=y/5$. The time between two consecutive overtakes is 4.5/x, on the other hand, it is equal to $1/5 + 4.5/y$. Thus, we have the system of equations $3(x+y)=4y$ and $4.5/x=1/5+4.5/y$, from which $x=15$ km/h.
Grading criteria: Correct solution - 7 points, the solution process is correct but the answer is wrong due to an arithmetic error - **5** points, the system is set up but the subsequent solution is incomplete or incorrect - **3** points, in all other cases - 0 points.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given a square $A B C D$. Point $N$ lies on side $A D$, such that $A N$ : $N D=2: 3$, point $F$ lies on side $C D$ and $D F: F C=1: 4$, point $K$ lies on side $A B$, such that $A K: K B=1: 4$. Find the angle $K N F$.
|
Solution. Mark point $E$ on side $B C$ such that $B E: E C=2: 3$. Let $M-$ be the midpoint of $K E$. Then $\triangle K M N-$ is an isosceles right triangle and $\triangle M N F-$ is an isosceles triangle. Therefore, $\angle K N F=$ $\angle K N M+\angle M N F=45^{\circ}+90^{\circ}=135^{\circ}$.
|
135
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1 The number 100 was divided by some number less than 50, with a remainder of 6. By what number could the division have occurred?
|
Let $x$ be the number by which they divided, $m$ be the quotient: $100 = m \cdot x + 6$ (since 6 is the remainder, then $x > 6$, and $m \geq 2$, since $x < 50$) $m \cdot x = 94 = 2 \cdot 47$. From this, it is clear that the numbers $m$ and $x$ are exactly the numbers 2 and 47, i.e., $m = 2$, $x = 47$. (since the number 94 can only be represented as a product of two natural numbers in four ways: $1 \cdot 94, 94 \cdot 1, 2 \cdot 47, 47 \cdot 2$, then all ways except $2 \cdot 47$ are eliminated).
## Answer: 47.
|
47
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 9.1
All three-digit numbers are written in a row: $100101102 \ldots 998$ 999. How many times in this row does a zero follow a two?
## Number of points 7 Answer: 19
#
|
# Solution
Since a three-digit number cannot start with zero, the two followed by a zero cannot be in the units place of any three-digit number in the sequence. Let's assume the two is in the tens place of a three-digit number. Then the zero following it is in the units place of the same number, i.e., the number ends in 20. There are 9 such numbers: 120, 220, ..., 920. Finally, if the two followed by a zero is in the hundreds place, then the corresponding three-digit number starts with 20. There are 10 such numbers: 200, 201, ..., 209. Therefore, the zero will follow the two 19 times in total.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. Harry, Ron, and Hermione wanted to buy identical waterproof cloaks. However, they lacked the money: Ron was short by a third of the cloak's price, Hermione by a quarter, and Harry by one fifth of the cloak's price. When the price of the cloak dropped by 9.4 sickles during a sale, the friends pooled their savings and bought three cloaks, spending all their money. How many sickles did one cloak cost before the price reduction?
|
Answer: 36 sickles.
Solution. Let the initial cost of the cloak be $x$ sickles, then Ron had $\frac{2}{3} x$ sickles, Hermione had $\frac{3}{4} x$ sickles, and Harry had $\frac{4}{5} x$ sickles. During the sale, the cloak cost $(x-9.4)$ sickles, and three cloaks cost $3(x-9.4)$ sickles. Since the friends bought three cloaks, spending all their money, we have $\frac{2}{3} x+\frac{3}{4} x+\frac{4}{5} x=3(x-9.4)$. Solving this equation, we get: $x=36$.
Grading criteria.
"+" A correct and justified solution and the correct answer are provided
“士” The equation is correctly and justifiedly formulated, but the solution is incomplete or contains a computational error
"耳" The correct answer is provided and it is shown that it satisfies the condition
“耳” Only the correct answer is provided
“-" An incorrect solution is provided or it is missing
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. Each of the thirteen dwarfs is either a knight, who always tells the truth, or a liar, who always lies. One day, all the dwarfs in turn made the statement: “Among the statements made previously, there are exactly two more false ones than true ones.” How many knights could there have been among the dwarfs?
|
Answer: 6
Solution. The first two statements are obviously false, as there were fewer than two statements made before each of them. The third statement is true, as there were 2 false statements and zero true statements made before it. The fourth statement is false, as it adds one true statement to the two false ones, while the fifth statement is true again.
Reasoning similarly, we find that all gnomes making an even-numbered statement were lying, while those making an odd-numbered statement were telling the truth. Thus, the knights are the gnomes who spoke under the numbers: $3,5,7,9,11$ and 13.
Evaluation criteria.
"+" A complete and well-reasoned solution with the correct answer is provided
“士” A generally correct reasoning with minor gaps or inaccuracies and the correct answer is provided
"干" Only the numbers of the gnomes who told the truth are correctly indicated, but any explanations are missing
“-" Only the answer is provided
“-" An incorrect solution is provided or it is missing
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. At each vertex of a cube lives a number, not necessarily positive. All eight numbers are distinct. If a number is equal to the sum of the three numbers living in the adjacent vertices, then it is happy. What is the maximum number of happy numbers that can live at the vertices of the cube?
|
Answer: 8.
Solution. See, for example, Fig. 7.5a. It is easy to verify that each vertex of the cube contains a lucky number.
There are other examples as well. Let's understand how they are structured (this was not required of the olympiad participants). Let's denote the numbers at the vertices of the cube (see Fig. 7.5b). First, write down two equations for the opposite vertices of the lower square: $a=a_{1}+b+d$ (1); $c=c_{1}+b+d$ (2). Subtracting equation (2) from equation (1), we get: $a-c=a_{1}-c_{1}$ (3).
Now, write down similar equations for two vertices of the vertical edge: $b=b_{1}+a+c$ (4); $b_{1}=b+a_{1}+c_{1}$ (5). Substituting $b_{1}$ from equation (5) into equation (4), we get: $a+c=-a_{1}-c_{1}$ (6). Combining equations (3) and (6) into a system, we get that $a=-c_{1}$; $c = -a_{1}$. Thus, the numbers in opposite vertices of the cube must be opposite (similar equations for the other two pairs of numbers follow from the symmetry of the cube).
Therefore, to construct any example, it is sufficient to choose one vertex of the cube and denote the numbers in the adjacent vertices, for example, as $x$, $y$, and $z$. To make the chosen vertex lucky, place the number $x+y+z$ in it. It is only required that the absolute values of the numbers $x$, $y$, $z$, and $x+y+z$ be pairwise distinct. Then, place the numbers $-x$, $-y$, $-z$, and $-x-y-z$ in the opposite vertices, respectively. It is easy to verify that all eight numbers are lucky.
Grading criteria.
“+" Correct answer and correct example provided
“-" Only the answer is provided
“-” Incorrect example or no example provided
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. In the numerical example АБВ $+9=$ ГДЕ, the letters А, Б, В, Г, Д, and Е represent six different digits. What digit is represented by the letter Д?
|
Answer: 0.
Solution: In the addition, the second digit of the first addend АБВ has changed (Д instead of Б). This could only happen if 1 was carried over from the units place to the tens place during the addition. However, the first digit also changed (Г instead of А). This means that 1 was also carried over from the tens place to the hundreds place during the addition. This is only possible if Б $+1=10$. Therefore, $Б=9$, and then $Д=0$.
Note: The puzzle has solutions, for example, $194+9=203$.
Comment: The score is not reduced if the correct answer is given, it is stated that $Б=9$, $Д=0$, but detailed explanations are not provided.
|
0
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.2. Lines parallel to the sides of a square form a smaller square, the center of which coincides with the center of the original square. It is known that the area of the cross formed by the smaller square (see the figure on the right) is 17 times the area of the smaller square. How many times is the area of the original square larger than the area of the smaller square?
|
Answer. 81 times.
Solution. Let the square have dimensions of 1 cm $\times 1$ cm, and the larger square have dimensions of $n$ cm $\times n$ cm. Then the area of the cross is $(2 n-1)$ cm $^{2}$ (the vertical column has dimensions $n \times 1$, the horizontal row has dimensions $1 \times n$, and the area of the square is counted twice). From the equation $2 n-1=17$, we get that $n=9$. Therefore, the area of the square is $9 \times 9=81$ cm $^{2}$.
Comment. The correct answer was obtained by trial and error - 5 points.
|
81
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.5. In the room, there are 10 people - liars and knights (liars always lie, and knights always tell the truth). The first said: "In this room, there is at least 1 liar." The second said: "In this room, there are at least 2 liars." The third said: "In this room, there are at least 3 liars." And so on,
up to the tenth, who said: "In this room, everyone is a liar." How many liars could there be among these 10 people?
|
Answer: 5.
Solution: Let there be $k$ liars in the room. Then the first $k$ people told the truth (and thus were knights), while the remaining $(10-k)$ lied (and were liars). Therefore, $k=10-k$, from which $k=5$.
Comment: The answer is obtained by considering an example -3 points.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On the island, there live knights who always tell the truth and liars who always lie. In the island's football team, there are 11 people. Player number 1 said: "In our team, there are as many knights as there are liars." Player number 2 said: "In our team, the number of knights and the number of liars differ by one," and so on. Player number 11 said: "In our team, the number of knights and the number of liars differ by ten." How many knights are in the team, and if there are any, what are their numbers? Find all possible answers to this question.
|
4. Answer: There are either no knights at all, or there is only one and he plays under number 10. The two answers cannot both be true, as they contradict each other. This means there can be no more than one true answer. If there is no true answer, then the team consists entirely of liars, in which case indeed none of the answers are true. If there is 1 knight and 10 liars in the team, then the answer of the player under number 10 is true, and the rest of the answers are false.
Criteria: correct solution - 7 points. Stated but not explained why there cannot be two correct answers, no other errors - 5 points. The case of no knights in the team is lost - 2 points. Only noted that there may be no knights - 1 point. Only the answer - 0 points.
|
0
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2 Out of 24 matches, a figure in the form of a $3 \times 3$ square is laid out (see figure), the side length of each small square is equal to the length of a match. What is the smallest number of matches that can be removed so that there are no whole $1 \times 1$ squares left, formed from matches.
|
Answer: 5 matches.
Solution: Estimation: It is impossible to manage with four matches, as by removing a match, we "ruin" no more than two squares (each match is a side of one or two adjacent squares), but initially, we have 9 small squares. Example for 5 matches:
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. From the natural numbers $1,2, \ldots, 101$, a group of numbers is chosen such that the greatest common divisor of any two numbers in the group is greater than two. What is the maximum number of numbers that can be in such a group?
|
Answer: 33. Solution: Estimation. Let's divide the numbers $1,2, \ldots, 101$ into 34 sets: $A_{0}=\{1,2\}, A_{1}=\{3,4,5\}$, $A_{2}=\{6,7,8\}, \ldots, A_{33}=\{99,100,101\}$ (i.e., $A_{k}$ for $k \geq 1$ consists of three numbers $3 k, 3 k+1,3 k+2$). In the desired group of numbers, there cannot be a number from $A_{0}$ (otherwise, the GCD of two numbers from the group, one of which belongs to $A_{0}$, would be $\leq 2$), and from each $A_{k}$ for $k \geq 1$, no more than one number can be in the group, since otherwise, if the group includes adjacent numbers, their GCD $=1$, and if the group includes numbers $3 k$ and $3 k+2$, their GCD $\leq 2$ (since their difference is 2). Therefore, the total number of numbers in the desired group is no more than 33 (no more than one from each $A_{k}$ for $k \geq 1$). Example. Consider the group of numbers $\{3,6,9, \ldots, 99\}$: this group contains 33 numbers, and all of them are divisible by 3.
|
33
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. In a convex pentagon $P Q R S T$, angle $P R T$ is half the size of angle $Q R S$, and all sides are equal. Find angle $P R T$.
|
Answer: $30^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle P R Q+\angle T R S=\angle P R T(*)$.
First method. We use the "folding" method. Reflect triangle $P Q R$ symmetrically relative to line $P R$, and triangle $T S R$ - relative to line $T R$ (see Fig. 11.5a). From the equality (*) and the equality $R Q=R S$, it follows that the images of points $Q$ and $S$ are the same point $O$.
Note that triangle $T O P$ is equilateral. Moreover, $O R=O P=O T$. Therefore, $O$ is the center of the circumscribed circle of triangle $P R T$. Then $\angle P R T = 0.5 \angle P O T = 30^{\circ}$.
Second method. We will prove that $Q P T S$ is a parallelogram (see Fig. 11.5b). Indeed, using the equality of the angles at the bases of the isosceles triangles $P Q R$ and $R S T$ and the equality (*), we get: $\angle Q P T + \angle P T S = \angle Q P R + \angle R P T + \angle R T P + \angle S T R = \angle P R Q + \angle T R S + (180^{\circ} - \angle P R T) = 180^{\circ}$.
Thus, $P Q \| S T$ and $P Q = S T$ (by condition), so $Q P T S$ is a parallelogram. Then $Q S = P T$, hence triangle $Q R S$ is equilateral. Therefore, $\angle P R T = 0.5 \angle Q R S = 30^{\circ}$.
Grading criteria.
"+" A complete and justified solution is provided.
"±" A generally correct solution is provided, containing minor gaps or inaccuracies (for example, the fact that the images of points $Q$ and $S$ coincide under the symmetries is used but not justified).
"Ғ" The correct answer is obtained based on the fact that $Q P T S$ is a parallelogram, but this is not proven.
"-" Only the answer is provided or the answer is obtained by considering a regular pentagon.
"-" An incorrect solution is provided or no solution is provided.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.6. A stack consists of 300 cards: 100 white, 100 black, and 100 red. For each white card, the number of black cards lying below it is counted; for each black card, the number of red cards lying below it is counted; and for each red card, the number of white cards lying below it is counted. Find the maximum possible value of the sum of the three hundred resulting numbers.
|
Answer: 20000.
Solution. First method. The number of different permutations of cards is finite. Therefore, their arrangement with the largest indicated sum exists (possibly not unique).
Let the cards lie in such a way that this sum is maximal. Without loss of generality, we can assume that the top card is white. Then, in this arrangement, pairs of cards CB, BK, and WB cannot lie consecutively from top to bottom, otherwise, the sum can be increased by swapping them in such pairs (symmetric pairs do not increase the desired sum when swapped). Therefore, the cards must lie as follows (from top to bottom): WW...WBB...BBKK...KKWW...W...
The length of each subsequent series of cards of the same color cannot be less than the length of the previous series. Indeed, if, for example, in the arrangement with the largest sum, there is a fragment ...WWWBBK..., then the K card can be moved to the top: ...KWWWBB..., increasing the sum. Since the number of cards of each color is the same, the lengths of all series must be the same (otherwise, there will be more cards of the color that ends up at the bottom than cards of another color). Then the series of the same color can be rearranged "in a cycle," without changing the sum, that is, to get such an arrangement of cards: 100 white cards on top, 100 black cards below them, and 100 red cards at the bottom. Therefore, the desired sum is $100 \cdot 100 + 100 \cdot 100 = 20000$.
Second method. Let the number of cards of each of the three colors be $n$. Using the method of mathematical induction, we will prove that for the specified sum $S$, the inequality $S \leq 2 n^{2}$ holds.
Base of induction. For $n=1$, by enumeration, we verify that $S \leq 2$. Induction step: Suppose the inequality is true for $n$ cards of each color. We will prove that it is true if the number of cards of each color is $n+1$. Consider how the sum $S$ can increase if one card of each color is added. Without loss of generality, we can assume that the white card is added to the very top of the stack, and the added black and red cards are the topmost among the cards of their color. Let there be $b$ previously lying black cards above the first red card from the top, and $w$ previously lying white cards above the first black card from the top. Then the white card adds $n+1$ to the sum (considering all black cards lying below it), the black card adds $n+1$ (considering all red cards lying below it) and $w$, due to it lying below $w$ old white cards, and the red card adds no more than $n-w$ due to white cards lying below it and $b$ due to it lying below $b$ old black cards. In total, $S \leq 2 n^{2} + n + 1 + n + 1 + w + n - w + b = 2 n^{2} + 3 n + b + 2$. Considering that $b \leq n$, we get: $S \leq 2 n^{2} + 4 n + 2 = 2(n+1)^{2}$.
Thus, the statement is proved for all natural $n$. For $n=100$, we get that $S \leq 2 \cdot 100^{2} = 20000$. This value is achieved, for example, with the following arrangement: 100 white cards on top, 100 black cards below them, and 100 red cards at the bottom.
Grading criteria.
“+" A complete and well-reasoned solution is provided
“士” A generally correct solution is provided, containing minor gaps or inaccuracies
“Ғ” The correct answer is obtained based on the assumption that the lengths of all monochromatic series of cards are the same, but this is not proven
"Ғ" The solution contains correct ideas on how to maximize the sum by rearranging the cards, but the solution is not completed or contains errors
"-" Only the answer is provided or the answer is obtained by considering only specific cases
“-" An incorrect solution is provided or it is absent
|
20000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a right triangle $ABC$ with angle $\angle A=60^{\circ}$, a point $N$ is marked on the hypotenuse $AB$, and the midpoint of segment $CN$ is marked as point $K$. It turns out that $AK=AC$. The medians of triangle $BCN$ intersect at point $M$. Find the angle between the lines $AM$ and $CN$.
|
Solution. By the property of a right triangle with an angle of $30^{\circ}$, we get that $A B: A C=2: 1$. Therefore, $A B: A K=2: 1$. By the property of medians in a triangle, $\mathrm{BM}: \mathrm{MK}=2: 1$. Then, by the converse of the angle bisector theorem, we get that $A M$ is the bisector of angle $\angle \mathrm{BAK}$. Let $\angle \mathrm{B}=\bar{\alpha}=$. Then $\angle \mathrm{K} M=\alpha$,
$\angle \mathrm{KAC}=60^{\circ}-2 \alpha$. Since triangle KAC is isosceles, $\angle \mathrm{KCA}=60^{\circ}+\alpha$. By the theorem of the sum of angles in a triangle, we get that the angle between lines $A M$ and $C N$ is $180^{\circ}-\left(60^{\circ}-\alpha+60^{\circ}+\alpha\right)=60^{\circ}$.
Criteria. If the solution is incorrect - 0 points. If it is proven that triangle OXY is isosceles - 3 points. If the solution is correct - 7 points.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Vitya Pereperepkin always calculates percentages incorrectly in surveys: he divides the number of people who answered in a certain way by the number of all the others. For example, in the survey "What is your name?", conducted among 7 Anyas, 9 Ols, 8 Yuls, Vitya counted $50 \%$ Yuls.
Vitya conducted a survey in his school: what kind of triangle is one with sides $3,4,5$? According to his calculations, $5 \%$ answered "acute-angled", $5 \%$ - "obtuse-angled", $5 \%$ - "such a triangle does not exist", $50 \%$ - "right-angled", and the remaining $a \%$ - "it depends on the geometry". What is $a$?
|
Answer: $a=110$ (in the $2-nd$ variant $a=104$). When Vitya calculates that some part of the students constitutes $5 \%$ of the entire class, in reality, it constitutes $5 / 100=1 / 20$ of the remaining students, $1 / 21$ of the entire class. Similarly, Vitya's $50 \%$ is one third of the class. Therefore, all those who answered in the first four ways constitute a share of $1 / 21+1 / 21+1 / 21+1 / 3=10 / 21$ of the entire class. Thus, the remaining students constitute $11 / 10$ of all except them, that is, according to Vitya's opinion, $110 \%$.
|
110
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In a checkers tournament, students from 10th and 11th grades participated. Each player played against every other player once. A player received 2 points for a win, 1 point for a draw, and 0 points for a loss. There were 10 times more 11th graders than 10th graders, and together they scored 4.5 times more points than all the 10th graders. How many points did the most successful 10th grader score?
|
Answer: 20.
Solution: 1. Let $a$ be the number of tenth graders who participated in the tournament, earning $b$ points. Then, $10a$ eleventh graders played, earning $4.5b$ points. In each match, 2 points are played for, and a total of $11a$ players play $\frac{11 a(11 a-1)}{2}$ matches. Therefore, from the condition of the problem, we have the equation $11 a(11 a-1)=5.5 b$, from which it follows that $b=2 a(11 a-1)$.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Two given quadratic trinomials $f(x)$ and $g(x)$ each have two roots, and the equalities $f(1)=g(2)$ and $g(1)=f(2)$ hold. Find the sum of all four roots of these trinomials.
|
Answer: 6.
First solution. Let $f(x)=x^{2}+a x+b, g(x)=x^{2}+$ $+c x+d$. Then the conditions of the problem can be written as
$$
1+a+b=4+2 c+d \quad \text { and } \quad 4+2 a+b=1+c+d
$$
Subtracting the second equation from the first, we get $-3-a=3+c$, which means $a+c=-6$. By Vieta's theorem, $-a$ is the sum of the roots of the first quadratic, and $-c$ is the sum of the roots of the second quadratic, from which the required result follows.
Second solution. Consider the auxiliary quadratic polynomial $h(x)=g(3-x)$ (it is also a monic polynomial!). Then $h(x)-f(x)$ is a linear polynomial with roots 1 and 2; hence, it is identically zero, i.e., $f(x)=g(3-x)$. Therefore, if $x_{0}$ is a root of $f(x)$, then $3-x_{0}$ is a root of $g(x)$, and the sum of these two roots is 3. Similarly, the sum of the other roots of these polynomials is also 3.
Comment. Only the correct answer - 0 points.
Only the correct answer with an example of two polynomials satisfying the conditions of the problem - 1 point.
Proved that $a+c=-6$ (in the notation of the first solution) - 3 points.
Proved that $f(x)=g(3-x)$ - 3 points.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. Each of the 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some integer. Then the first said: “My number is greater than 1”, the second said: “My number is greater than 2”, \ldots, the tenth said: “My number is greater than 10”. After that, all ten, speaking in some order, said: “My number is less than 1”, “My number is less than 2”, \ldots, “My number is less than 10” (each said exactly one of these ten phrases). What is the maximum number of knights that could have been among these 10 people?
(O. Podlipsky)
|
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 number is less than 1," "My number is less than 2," ..., "My number is less than 10." Therefore, there could not have been more than eight knights.
We will show that there could have been 8 knights. Suppose the first knight thought of the number 2, the second -3, ..., the eighth -9, and the liars thought of the numbers 5 and 6. Then the $k$-th knight could have said the phrases "My number is greater than $k$" and "My number is less than $k+2$," while the liars could have said the phrases: one - "My number is greater than 9" and "My number is less than 1," and the other - "My number is greater than 10" and "My number is less than 2."
Note. The example given above ceases to be valid if the liars think of numbers outside the interval $[1; 10]$, as then some of their statements become true.
Comment. Proved that there are no more than $9-$ 0 points.
Proved that there are no more than 8 knights (or, equivalently, at least two liars -3 points.
Provided an example showing that there could have been 8 knights, with a correct indication of which person said which phrase - 4 points.
If the example provided does not fully describe the situation (for example, it is not specified what numbers the liars thought of, or it is not clearly indicated who said which phrase) - out of 4 points for the example, no more than 2 points are given.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. Each face of a cube $1000 \times$ $\times 1000 \times 1000$ is divided into $1000^{2}$ square cells with side 1. What is the maximum number of these cells that can be painted so that no two painted cells share a side?
$$
\text { (S. Dolgikh) }
$$
|
Answer. $3 \cdot 1000^{2}-2000=$ $=2998000$ cells.
Solution. Consider an arbitrary coloring that satisfies the condition. Divide all the cells on the surface into "edges" as shown in Fig. 2 - 500 edges around each of the eight vertices (one of the edges is marked in gray). Then, in the $k$-th edge, counting from the vertex, there will be $S_{k}=6 k-3$ cells. Since no two colored cells can be adjacent, in this edge there will be no more than $\left[\frac{S_{k}}{2}\right]=3 k-2=\frac{S_{k}-1}{2}$ colored cells. Summing over all 4000 edges and noting that their total area is $6 \cdot 1000^{2}$, we get that the total number of colored cells does not exceed $\frac{6 \cdot 1000^{2}-4000}{2}=$ $=3 \cdot 10^{6}-2000$.
It remains to provide an example showing that so many cells can be colored. Let's call two opposite faces of the cube the top and bottom, and the others - side faces. On each of the side faces, half of the cells can be marked in a checkerboard pattern. After that, on the top and bottom faces, half of the cells can also be colored in all rows except the two outermost, leaving them empty - see Fig. 3, where two side faces and the top face are visible. It is easy to see that with such a coloring, in each edge there will be the maximum possible number of colored cells. (Instead of checking each edge, one can note that the entire surface is divided into strips $1 \times 100$,
Fig. 2
Fig. 3
four of which are empty, and in each of the others, exactly half of the cells are colored.)
Remark. There are other optimal examples. In particular, in the given example, the coloring of the top face can be changed as follows: divide the top face into 4 triangles with diagonals, and in each of them, color the cells in a checkerboard pattern (so that the coloring of this triangle agrees with the coloring of the adjacent side face).
Comment. Only the answer - 0 points.
Only provided a correct example of coloring $N=$ $=3 \cdot 1000^{2}-2000$ cells - 1 point.
Only proved that in any coloring satisfying the conditions, there are no more than $N$ colored cells - 5 points.
|
2998000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. For what least natural $n$ do there exist integers $a_{1}, a_{2}, \ldots, a_{n}$ such that the quadratic trinomial
$$
x^{2}-2\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} x+\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$
has at least one integer root?
(P. Kozlov)
|
Answer. For $n=6$.
Solution. For $n=6$, we can set $a_{1}=a_{2}=a_{3}=a_{4}=1$ and $a_{5}=a_{6}=-1$; then the quadratic trinomial from the condition becomes $x^{2}-8 x+7$ and has two integer roots: 1 and 7. It remains to show that this is the smallest possible value of $n$.
Suppose the numbers $a_{1}, a_{2}, \ldots, a_{n}$ satisfy the condition of the problem; then the discriminant of the quadratic trinomial from the condition, divided by 4, must be a perfect square. It is equal to
$$
d=\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{4}-\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$
Then the number $d$ is odd and is a square, so it gives a remainder of 1 when divided by 8.
Rewrite the above equality as
$$
d+1+a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}=\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{4}
$$
and consider it modulo 8. It is not hard to check that the fourth powers of integers give only remainders 0 and 1 when divided by 8, so the right-hand side of the equality gives a remainder of 0 or 1. The left-hand side, however, is comparable to $1+1+k$, where $k$ is the number of odd numbers among $a_{i}$. Therefore, $n \geqslant k \geqslant 6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-4-1. In the figure, $O$ is the center of the circle, $A B \| C D$. Find the degree measure of the angle marked with a «?».
|
Answer: $54^{\circ}$.
Solution variant 1. Quadrilateral $A D C B$ is an inscribed trapezoid in a circle. As is known, such a trapezoid is isosceles, and in an isosceles trapezoid, the angles at the base are equal: $\angle B A D=\angle C B A=63^{\circ}$. Triangle $D O A$ is isosceles ($O A$ and $O D$ are equal as radii), so its base angles $D A O$ and $A D O$ are equal. Therefore,
$$
\angle D O A=180^{\circ}-\angle D A O-\angle A D O=180^{\circ}-63^{\circ}-63^{\circ}=54^{\circ}
$$
|
54
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.1. In a notebook, a triangular grid is drawn (see figure). Tanya placed integers at the nodes of the grid. We will call two numbers close if they are in adjacent nodes of the grid. It is known that
- the sum of all ten numbers is 43;
- the sum of any three numbers such that any two of them are close is 11.
What is the central number?
|
Answer: 10.
Solution. Let's denote the numbers by variables as shown in the figure.
Then
\[
\begin{gathered}
a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}=c_{1}+c_{2}+c_{3}=11 \\
\left(a_{1}+a_{2}+a_{3}\right)+\left(b_{1}+b_{2}+b_{3}\right)+\left(c_{1}+c_{2}+c_{3}\right)+x=43
\end{gathered}
\]
From this, we get that \( x=10 \).
An example of a suitable arrangement of numbers is shown below.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.2. The least common multiple of four pairwise distinct numbers is 165. What is the maximum value that the sum of these numbers can take?
|
Answer: 268.
Solution. Since 165 is the least common multiple of four numbers, these numbers are divisors of 165. To maximize the sum of these numbers, it is sufficient to take the four largest divisors of 165. If one of them is the number 165 itself, then the LCM will definitely be equal to it.
Then the maximum sum will be
$$
165+\frac{165}{3}+\frac{165}{5}+\frac{165}{11}=165+55+33+15=268
$$
|
268
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.3. The teacher wrote a fraction on the board, where the numerator and the denominator are natural numbers. Misha added 30 to the numerator of the given fraction and wrote the resulting fraction in his notebook, while Lesha subtracted 6 from the denominator of the fraction written on the board and also wrote the resulting fraction in his notebook. The fractions recorded by the boys turned out to be equal to the same number. What is this number?
|
Answer: 5.
Solution. Let $\frac{a}{b}$ be the original fraction. Then Misha wrote down the fraction $\frac{a+30}{b}$ in his notebook, and Lёsha wrote down $-\frac{a}{b-6}$.
Let's write the equation
$$
\frac{a+30}{b}=\frac{a}{b-6}
$$
Transforming it, we get
$$
\begin{gathered}
(a+30)(b-6)=a b \\
a b+30 b-6 a-180=a b ; \\
30 b=6 a+180 \\
\frac{a+30}{b}=5
\end{gathered}
$$
This is the value of the fraction obtained by Misha. It should also have been obtained by Lёsha.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.4. Given a cyclic quadrilateral $A B C D$. It is known that $\angle A D B=48^{\circ}, \angle B D C=$ $56^{\circ}$. Inside triangle $A B C$, a point $X$ is marked such that $\angle B C X=24^{\circ}$, and ray $A X$ is the angle bisector of $\angle B A C$. Find the angle $C B X$.
|
Answer: $38^{\circ}$.
Solution. Angles $B D C$ and $B A C$ are equal since they subtend the same arc. Similarly, angles $A D B$ and $A C B$ are equal. Then
$$
\angle A C X=\angle A C B-\angle X C B=\angle A D B-\angle X C B=48^{\circ}-24^{\circ}=24^{\circ}=\angle X C B .
$$
We obtain that $C X$ is the bisector of angle $A C B$, so $X$ is the point of intersection of the angle bisectors of triangle $A B C$.
Now it is easy to find the required angle:
$$
\angle C B X=\frac{\angle A B C}{2}=\frac{180^{\circ}-\angle B A C-\angle A C B}{2}=\frac{180^{\circ}-56^{\circ}-48^{\circ}}{2}=38^{\circ}
$$
|
38
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.5. On the board, the graph of the function $y=x^{2}+a x+b$ is drawn. Yulia drew two lines parallel to the $O x$ axis on the same drawing. The first line intersects the graph at points $A$ and $B$, and the second line intersects the graph at points $C$ and $D$. Find the distance between the lines if it is known that $A B=5, C D=11$.
|
Answer: 24.
Solution. Let the first line have the equation $y=s$, and the second line have the equation $y=t$. Then the distance between the lines is $(t-s)$.
The length of segment $A B$ is equal to the absolute value of the difference of the roots of the equation $x^{2}+a x+b=s$. We can express the difference of the roots using the formula for solving the quadratic equation $x^{2}+a x+(b-s)=0$:
$$
\frac{-a+\sqrt{a^{2}-4(b-s)}}{2}-\frac{-a-\sqrt{a^{2}-4(b-s)}}{2}=\sqrt{a^{2}-4(b-s)}=5
$$
from which we derive
$$
a^{2}-4(b-s)=25
$$
Similarly, we obtain
$$
a^{2}-4(b-t)=121
$$
Subtract the first equation from the second and we get
$$
121-25=\left(a^{2}-4(b-t)\right)-\left(a^{2}-4(b-s)\right)=4(t-s)
$$
We have $4(t-s)=96$, that is, $t-s=24$.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.6. On a line, two red points and several blue points are marked. It turned out that one of the red points is contained in exactly 56 segments with blue endpoints, and the other - in 50 segments with blue endpoints. How many blue points are marked?
|
Answer: 15.
Solution. Let there be $a$ blue points to the left of the first red point, and $b$ blue points to the right; $c$ blue points to the left of the second red point, and $d$ blue points to the right. Then $a b=56, c d=50$. Additionally, $a+b=c+d$ - the number of blue points.
Notice that among the numbers $c$ and $d$, exactly one is even, since the number 50 is divisible by 2 but not by 4. This means that among the numbers $a$ and $b$, exactly one is also even. There are exactly two ways to represent 56 as the product of an even number and an odd number: $56=8 \cdot 7$ and $56=56 \cdot 1$.
In the first case, $c=10, d=5$ work. As an example, you can place 5 blue points on the line, then a red one, then 3 more blue points, then another red one, and finally 7 more blue points.
In the second case, the total number of blue points is 57. Then it only remains to understand that the following system of equations has no integer solutions:
$$
\left\{\begin{array}{l}
c+d=57 \\
c d=50
\end{array}\right.
$$
One can check that the quadratic trinomial $t^{2}-57 t+50=0$, whose roots by Vieta's theorem should be the numbers $c$ and $d$, does not have a pair of natural solutions (for example, it is sufficient to verify that its values at $t=0$ and $t=1$ have different signs, meaning one of the roots lies in the interval $(0 ; 1))$.
Alternatively, one can simply enumerate the factorizations of the number 50 into $c$ and $d$.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Pantelej and Gerasim received 20 grades each in November, and Pantelej received as many fives as Gerasim received fours, as many fours as Gerasim received threes, as many threes as Gerasim received twos, and as many twos as Gerasim received fives. At the same time, their average grade for November is the same. How many twos did Pantelej receive in November?
|
Solution. Let's add one to each of Gerasim's grades. His total score will increase by 20. On the other hand, it will become greater than Pantelej's total score by four times the number of Pantelej's twos.
Answer: 5 twos.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Each pair of numbers $A$ and $B$ is assigned a number $A * B$. Find $2021 * 1999$, if it is known that for any three numbers $A, B, C$ the identities are satisfied: $A * A=0$ and $A *(B * C)=(A * B)+C$.
|
Solution.
$A *(A * A)=A * 0=A *(B * B)=A * B+B$,
$A *(A * A)=(A * A)+A=0+A=A$, then $A * B+B=A$.
Therefore, $A * B=A-B$. Hence, $2021 * 1999=2021-1999=22$.
Answer: 22.
|
22
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Masha and Olya bought many identical pens for the new school year at the store. It is known that one pen costs an integer number of rubles, more than 10. Masha bought pens for exactly 357 rubles, and Olya - for exactly 441 rubles. How many pens did they buy in total?
|
8.1. Let a pen cost $r$ rubles, and the numbers 357 and 441 are divisible by $d$. Since the greatest common divisor of the numbers $357=3 \cdot 7 \cdot 17$ and $441=3^{2} \cdot 7^{2}$ is $3 \cdot 7$, then 21 is also divisible by $r$. Since $r>10$, then $r=21$. Therefore, the total number of pens bought is $\frac{357}{21}+$ $\frac{441}{21}=17+21=38$
Answer: 38.
|
38
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. In 8th grade class "G", there are enough underachievers, but Vovochka studies the worst of all. The pedagogical council decided that either Vovochka must correct his twos by the end of the quarter, or he will be expelled. If Vovochka corrects his twos, then the class will have $24 \%$ of underachievers, and if he is expelled, the underachievers will become $25 \%$. What percentage of underachievers are there in 8 "G" now?
|
8.2. Let there be $n$ students in the class now. According to the condition,
$$
0.24 n = 0.25(n-1)
$$
i.e., $0.01 n = 0.25$. Therefore, $n = 25$. One person constitutes $4\%$ of 25, so there are now $24 + 4 = 28\%$ of underachievers.
Answer: $28\%$.
|
28
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. In triangle $\mathrm{ABC}$, $\mathrm{AC}=1$, $\mathrm{AB}=2$, $\mathrm{O}$ is the point of intersection of the angle bisectors. A segment passing through point O and parallel to side $\mathrm{BC}$ intersects sides $\mathrm{AC}$ and $\mathrm{AB}$ at points K and M, respectively. Find the perimeter of triangle $\mathrm{AKM}$.
|
8.3. $\angle \mathrm{KCO}=\angle \mathrm{BCO}=\angle \mathrm{KOC}$ (alternate interior angles). Therefore, OK = KC, and similarly BM = OM. Then
$$
A K+A M+K M=A K+K C+A M+B M=3 .
$$
Answer: 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. How many natural numbers less than 1000 are divisible by 4 and do not contain the digits $1,3,4,5,7,9$ in their notation?
|
8.4. The desired numbers are written only with the digits $0,2,6,8$.
There is exactly one single-digit number that satisfies the condition, which is the number 8.
There are six two-digit numbers, which are $20,28,60,68,80,88$.
The desired three-digit numbers can end with the following 8 combinations of digits: $00,08,20,28,60,68,80,88$. In each of these variants, the first place can be occupied by any of the three digits $2,6,8$. Thus, we get $3 \cdot 8=24$ desired three-digit numbers.
Therefore, the total number of such numbers is $1+6+24=31$.
Answer: 31
|
31
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. At a round table, 2015 people are sitting, each of them is either a knight or a liar. Knights always tell the truth, and liars always lie. Each person was given a card with a number on it; all the numbers on the cards are different. After looking at their neighbors' cards, each person said: "My number is greater than the number of each of my two neighbors." After this, $k$ of the people said: "My number is less than the number of each of my two neighbors." For what largest $k$ could this have happened?
(O. Podlipsky)
|
Answer: 2013.
Solution. Let $A$ and $B$ be the people who received the cards with the largest and smallest numbers, respectively. Since both of them said the first phrase, $A$ is a knight, and $B$ is a liar. However, if they had said the second phrase, $A$ would have lied, and $B$ would have told the truth; this is impossible. Therefore, $A$ and $B$ cannot say the second phrase, and $k \leqslant 2013$.
We will show that a situation where the remaining 2013 people can say the second phrase is possible. Suppose the people sitting at the table received (in a clockwise direction) cards with the numbers $1, 2, 3, \ldots$, 2015; with the card numbered 2015 going to a knight, and the rest to liars. Then the first phrase can be said by everyone, and the second phrase by everyone except those with cards numbered 1 and 2015.
Remark. There are other examples of card distributions where 2013 people can say the second phrase.
Comment. Only the answer - 0 points.
An example of seating arrangement is provided, where $k$ can be equal to $2013-3$ points.
Only proved that $k \leqslant 2013-4$ points.
Only proved that $k \leqslant 2014-1$ point.
It is claimed that the people with the largest and smallest numbers cannot say the second phrase, but this claim is not justified - 1 point is deducted.
|
2013
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. We will call a natural number interesting if the sum of its digits is a prime number. What is the maximum number of interesting numbers that can be among five consecutive natural numbers?
(
#
|
# Answer: 4.
Solution. Among five consecutive natural numbers, there can be 4 interesting numbers. For example, the numbers 199, 200, 201, 202, 203 (with digit sums 19, 2, 3, 4, and 5) will work.
Now, let's prove that all 5 numbers cannot be interesting. Among our five numbers, there are three that lie within the same decade. Then their digit sums are consecutive numbers; hence, they cannot all be prime simultaneously.
Remark 1. There are other examples of five consecutive numbers, four of which are interesting. Moreover, the number divisible by 10 can also be in a different position, as in the set 197, 198, 199, 200, 201. However, in any such example, the digit sums of two of the numbers in the set must be 2 and 3.
Remark 2. One can also prove that five consecutive numbers cannot all be interesting in a different way. Among our numbers, at least two have even digit sums (even if there is a transition "across a decade"). But both of these sums cannot equal 2, since the digit sum 2 is only found in numbers of the form $10 \ldots 010 \ldots 0$ and $20 \ldots 0$, and two such numbers cannot appear among 5 consecutive natural numbers.
Comment. Only the answer - 0 points.
Providing an example of five consecutive numbers, four of which are interesting - 3 points.
Providing only an example of five consecutive numbers, among which fewer than 4 are interesting - 0 points.
Proving that there do not exist five consecutive interesting numbers - 3 points.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. Several married couples came to the New Year's Eve party, each of whom had from 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families and took them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the necessary trio of people. How many children could there be in total at this party?
(S. Volchonkov)
|
Answer: 33.
Solution: Let there be $p$ married couples and $d$ children at the party (from the condition, $d \leqslant 10 p$). Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and with a fixed choice of mother, a father could be chosen from one of the $p-2$ remaining couples. Therefore, the total number of trios is $d \cdot(p-1)(p-2)=3630$. Since $d \leqslant 10 p$, we get $3630 \leqslant 10 p^{3}$, that is, $p^{3} \geqslant 363>7^{3}$. Thus, $p \geqslant 8$.
Next, the number $3630=2 \cdot 3 \cdot 5 \cdot 11^{2}$ has two divisors $p-1$ and $p-2$, differing by 1. If one of these divisors is divisible by 11, then the other gives a remainder of 1 or 10 when divided by 11. Then it is coprime with 11, and thus divides $2 \cdot 3 \cdot 5=30$ and is at least 10. It is easy to see that this divisor can only be 10; then $p-2=10, p-1=11$ and $d=3630 / 110=33$.
If, however, neither of the numbers $p-2$ and $p-1$ is divisible by 11, then the number $2 \cdot 3 \cdot 5=30$ is divisible by their product; this contradicts the fact that $p \geqslant 8$.
Comment. The equality $d(p-1)(p-2)=3630$ was justified - 4 points.
|
33
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. At the New Year's celebration, schoolchildren organized an exchange game: if they were given five tangerines, they would exchange them for three crackers and a candy, and if they were given two crackers, they would exchange them for three tangerines and a candy. Father Frost played this game with them several times and received a total of 50 candies. Initially, he only had a bag of tangerines, and after all the exchanges, he had no crackers left. How many tangerines did Father Frost give to the children?
|
Solution. Ded Moroz conducted 50 exchanges, as he was given 50 candies. At the end of the game, he had no crackers left, meaning he exchanged all of them back. For every two exchanges of tangerines for crackers (ten given - six received), he conducted three exchanges of crackers for tangerines (six given - nine received). Therefore, every five exchanges, Ded Moroz irreversibly gave away one tangerine. All exchanges can be divided into ten such groups. Therefore, Ded Moroz gave away 10 tangerines in total.
Answer. Ded Moroz gave away 10 tangerines.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. How many six-digit numbers exist in which four consecutive digits form the number $2021?$
|
Solution. Consider three types of six-digit numbers: $\overline{a b 2021}, \overline{a 2021 b}, \overline{2021 a b}$ (the bar denotes the decimal representation of the number, where $a, b$ - are digits).
In the first case, $\overline{a b 2021}$, $a$ can be any digit from 1 to 9, and $b$ can be any digit from 0 to 9. These digits are chosen independently. Therefore, we get a total of $9 \cdot 10 = 90$ options.
In the second case, $\overline{a 2021 b}$, the reasoning is repeated, $a$ can be any digit from 1 to 9, and $b$ can be any digit from 0 to 9. These digits are chosen independently. Therefore, we again get $9 \cdot 10 = 90$ options. We check that no number from the first group coincides with a number from the second group. They, in particular, differ by the digit in the hundreds place.
In the third case, $\overline{2021 a b}$, the digits $a$ and $b$ are chosen independently and arbitrarily from 0 to 9. In this case, we have 100 options. No number from the third group coincides with the previous options, they differ by the digit in the hundreds place.
The total number of options is $90 + 90 + 100 = 280$.
Answer. 280 numbers.
Leningrad Region
## All-Russian School Olympiad in Mathematics Municipal Stage 2021-2022 academic year
7th grade
Grading Criteria
| Problem 1 | Score | For what it is given |
| :---: | :---: | :--- |
| | 7 | Fully correct solution. A correct diagram is provided. Checking the conditions of the problem for the correct diagram is not required. |
| | | | |
| | Incorrect solution. | |
| Score | For what it is given | |
| :---: | :---: | :--- |
| 7 | Complete solution. Correct groups of exchanges are obtained. | |
| 5 | Correct solution. It is not specified how the condition that there are no whistles left after all exchanges is used. As a result, the division of exchanges into groups is not justified. | |
| | 3 | A sequence of exchanges is described. The uniqueness of the answer does not follow from the reasoning. |
| 1 | It is written that 50 exchanges were made. Further, it is not written or is incorrect. | |
| 0 | Incorrect solution and (or) incorrect answer. | |
| Problem 3 | Score | For what it is given |
| :---: | :---: | :---: |
| | 7 | Complete solution, the correct answer is obtained. |
| | 3 | It is noted that for a simple odd $p$, the number $n-1$ will be odd, and $p-1$ - even. Further progress is absent. |
| | 1 | A check is provided that for $n$, being a power of two, the condition of the problem is satisfied. Proof of the statement formulated in the condition is absent. |
| | 0 | Incorrect solution and (or) incorrect answer. |
| Score | For what it is given |
| :---: | :--- | :--- |
| 7 | Fully correct solution. A correct example is provided and it is written that all conditions are met. |
| 3 | A correct example is provided, but it is not shown that all conditions are met. Possibly, a partial check is performed (for example, the possibility of walking all paths twice). It is not proven that it is impossible to walk all paths once. |
| 1 | A suitable diagram is provided, the conditions are not checked. |
| 0 | Incorrect solution and (or) incorrect answer. |
| Score | For what it is given |
| :---: | :---: | :--- |
| | Fully correct solution. It is proven that Petya will win with the correct strategy. The possibility of each necessary move is shown. |
| | A winning strategy is described. There is no proof of why Petya can always make the necessary move. |
| | A description is provided of the moves Petya and Vasya will make. It is not assumed that Vasya can choose the number of pebbles at his discretion, and Petya can adapt to his choice. |
| 0 | Incorrect solution and (or) incorrect answer. |
| Problem 6 | Score | For what it is given |
| :---: | :---: | :---: |
| | 7 | Fully correct solution. Correct answer. |
| | 6 | Correct solution. Correct answer. The difference from 7 points is that it does not provide a check that no two options for a six-digit number are counted twice. |
| | 5 | Correct solution. Incorrect answer due to placing the digit 0 in the first position. |
| | 1 | Partially correct solution, in which only one placement of the pair of unknown digits is allowed (for example, they are placed only at the end). |
| | 0 | Incorrect solution and (or) incorrect answer that does not meet the criteria for 1 and 5 points. |
|
280
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. This figure contains $9^{2}-4 \cdot 3=69$ cells. Figure 2 shows how a gardener can plant 60 apple trees. We will prove that it is impossible to plant more than 60 apple trees.
Figure 2
First, note that apple trees can definitely be planted on 28 blue cells (these cells receive light through the fence). This leaves 41 cells. Let $x$ be the number of empty cells among these (the "light windows" depicted in yellow in Figure 2). Thus, 41 - x apple trees must be illuminated through $x$ light windows. However, each light window can illuminate no more than four apple trees. Therefore, we arrive at the inequality
$41-x \geq 4 x$, the solution to which is $x \geq 8.2$. Since $x$ is a natural number,
$$
x \geq 9
$$
Thus, there must be at least 9 light windows. Consequently, the number of apple trees cannot exceed $69-9=60$. The statement is proven.
|
Answer: the maximum number of apple trees is 60.
|
60
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
10-1. Piglet has balloons of five colors. He managed to arrange them in a row in such a way that for any two different colors in the row, there will always be two adjacent balloons of these colors. What is the minimum number of balloons Piglet could have?
|
Answer: 11 balls.
Solution. Consider the balls of color $a$. Their neighbors must be balls of all 4 other colors. But one ball can have no more than two neighbors, so there must be at least 2 balls of color $a$. This is true for each of the 5 colors, so there must be at least 10 balls in total.
Notice that some ball of color $b$ is at the end of the row. It has only one neighbor, so two balls of this color have no more than 3 neighbors. Therefore, to meet the conditions, we need to add at least one more ball of color $b$. Thus, we need at least 11 balls in total.
We will show that 11 balls are sufficient. Let $a, b, c, d, e$ be the colors of the balls. They can be arranged, for example, as follows: $a b c d e c a d b e a$.
Criteria. Only the answer - 0 points. Estimation (that there are no fewer than 11 balls) - 4 points. Example of the required arrangement - 3 points.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-5. From 80 identical Lego parts, several figures were assembled, with the number of parts used in all figures being different. For the manufacture of the three smallest figures, 14 parts were used, and in the three largest, 43 were used in total. How many figures were assembled? How many parts are in the largest figure?
|
Answer: 8 figurines, 16 parts.
Solution. Let the number of parts in the figurines be denoted by $a_{1}43$, so $a_{n-2} \leq 13$.
Remove the three largest and three smallest figurines. In the remaining figurines, there will be $80-14$ - 43 $=23$ parts, and each will have between 7 and 12 parts. One figurine is clearly insufficient, and three would be too many (7 $+8+9=24$). Therefore, 23 parts form 2 figurines. This is possible, and in only one way: $23=11+12$.
We have $43=13+14+16$ - the only decomposition with $a_{6} \geq 13$.
Criteria. Only the answer - 0 points. Only correct estimates for $a_{3}$ and $a_{n-3}-3$ points. Only a justified answer for the number of figurines - 5 points. Complete solution - 7 points.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1 A courtyard table tennis tournament among 15 players is held according to certain rules. In each round, two players are randomly selected to compete against each other. After the round, the loser receives a black card. The player who receives two black cards is eliminated from the competition. The last remaining player is declared the champion. In table tennis, there are no ties. How many rounds were there in the courtyard tournament if the champion lost exactly once?
|
Answer: 29
Solution. In each match, there is always exactly one loser. Since 14 players were eliminated, there were a total of $14 \cdot 2+1=29$ losses.
|
29
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1 Two ants are running along a circle towards each other at constant speeds. While one of them runs 9 laps, the other runs 6. At the point where the ants meet, a red dot appears. How many red dots will there be on the circle?
#
|
# Answer: 5
Solution. The ratio of the speeds of the ants is $\frac{9}{6}=\frac{3}{2}$. Without loss of generality, assume the circumference of the circle is 5, and let the ants meet at some moment at point 0. Place points $1, 2, 3,$ and 4 clockwise so that they, together with point 0, divide the circle into 5 equal parts. If the slower ant runs counterclockwise, they will meet again at point 2. Now we can consider this point as the starting point, and thus they will meet next at point 4, and then at point 1 (since $4+2=6$, but the circumference is 5). The next meeting will be at point 3, then 0, after which the process will repeat, and there will be a total of 5 points.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1 Seven pirates were dividing five identical treasure chests. They agreed that five of them would take one chest each, while the others would receive a fair compensation equal to the value of a chest. Each of the recipients of a chest paid 10000 piastres into a common fund, after which the money was distributed among the remaining pirates. What was the value of one chest?
|
Answer: 35000
Solution. Two pirates received 50000 piastres, so each share amounted to 25000. Since the chests were divided fairly, the total amount of treasure was 25000 for each of the seven, totaling 175000. Then 175000 is the value of five chests, meaning each chest was valued at 35000 piastres.
|
35000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1 Let's say that number A hides number B if you can erase several digits from A to get B (for example, the number 123 hides the numbers 1, 2, 3, 12, 13, and 23). Find the smallest natural number that hides the numbers 2021, 2120, 1220, and 1202.
|
# Answer: 1201201
Solution. Notice that the number contains at least two twos and one zero. If there are exactly two twos, then the zero must stand both between them and after them, but then there must be at least two zeros. Therefore, only on twos and zeros, we need 4 digits (either two twos and two zeros, or three twos and one zero).
If there is one one, then before it should be $2,0,2$, and after it - two more twos and a zero. At the same time, the zero stands both between the twos and after them (1220 and 1202), so we need one more two or one more zero, making it 8 digits in total.
Suppose there are two ones. If there are exactly two twos, then we need a one after these twos (2021), between them (2120), and before them (1220), which leads to a contradiction. Therefore, there must be at least three twos. If there are 6 digits in total, then the zero is the only one. Then it stands after two twos and before one. From the number 1220, we get that one of the ones stands before the first two twos and the zero, and from the number 2021, we get that the other one stands after the last two. Then we definitely get the arrangement - 122021, but alas, from it, we cannot get the number 2120. Therefore, one zero is not enough, and we already have 7 digits.
Suppose there are three ones. We need 4 more digits for twos and zeros, so we have 7 digits in total.
Suppose there are 7 digits. The first digit is not zero, so it must be 1. If the next digit is 0, then it is useless (in our numbers, 0 appears only after twos). Two ones in a row or two zeros make no sense, as in each of our numbers, 0 or 1 appear only once. Therefore, the second most significant digit is 2, followed by 0, and then 1. 1201... We have three more digits, one of which is definitely a two. From the number 1220, we understand that somewhere after it stands a zero, and from the number 2021, we understand that after it stands another one. The smallest such variant is 201, and it fits.
|
1201201
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1 In the example of addition and subtraction, the student replaced the digits with letters according to the rule: identical letters are replaced by identical digits, different letters are replaced by different digits. From how many different examples could the record $0<\overline{\overline{Б A}}+\overline{\text { БА }}-\overline{\text { ЯГА }}<10$ be obtained?
|
# Answer: 31
Solution. The sum of two two-digit numbers is no more than 199, so $\overline{\text { YAG }}$ is a three-digit number starting with 1, Y $=1$. Let's look at the last digit in each number, A. It is added twice and subtracted once, so the value of the expression is $\mathrm{A}$, and $\mathrm{A} \neq 0$. $\mathrm{A} \neq 1$, since Y = 1. In addition, B $\geq 5, B+B=\overline{Y \Gamma}$. If B $=5$, then A can be any digit except 1, 0, and 5 (7 options). Further, if B $=6,7,8,9$, then A can be any digit except 0, 1, B, and $\Gamma$, and $\Gamma$ is determined as the units digit of the number 2B (it is not equal to 1, 0, or B). Thus, in these cases, there are 6 options left. In total, there are $7+4 \cdot 6=7+24=31$ options.
|
31
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. Palm oil has increased in price by $10 \%$. Due to this, the cheese of one of the manufacturers has increased in price by $3 \%$. What is the percentage of palm oil in the cheese of this manufacturer?
|
Answer: $30 \%$.
Solution. Let's assume that the cheese cost 100 conditional rubles per kilogram. Then it increased by 3 rubles. Since this happened due to the increase in the price of palm oil, 3 rubles is $10 \%$ of the cost of palm oil in the cheese, meaning the cost of palm oil in a kilogram of cheese is 30 rubles. Therefore, palm oil constitutes $30 \%$.
Comment. Correct answer without justification - 0 points.
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. Three cyclists leave the city. The speed of the first one is 12 km/h, the second one is 16 km/h, and the third one is 24 km/h. It is known that the first one was cycling exactly when the second and third were resting (standing still), and that at no time did two cyclists cycle simultaneously. It is also known that in 3 hours, all three covered the same distance. Find this distance.
|
Answer: 16 km
Solution: From the condition, it follows that each of the cyclists was riding at the time when the other two were standing, and, moreover, at any given moment, one of the cyclists was riding (the first one rode when the second and third were standing). Since they all traveled the same distance, and the ratio of their speeds is $3: 4: 6$, the time of movement for the first cyclist is 24 parts, for the second - 18 parts, and for the third - 12 parts of time. In total, they rode for 3 hours $=180$ minutes. Therefore, one part of time is $10 / 3$ minutes. Consequently, the first one rode for 80 minutes, which is $4 / 3$ hours. During this time, he traveled 16 km.
Comment: A correct answer without justification - 0 points.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. 16 travelers, each of whom is either a liar or a knight (liars always lie, knights always tell the truth), settled into 3 rooms of a hotel. When everyone gathered in their rooms, Basil, who was staying in the first room, said: "There are more liars than knights in this room right now. Although no - there are more knights than liars in this room right now." After that, Basil went to the second room and said the same two statements there. And then he went to the third room and said the same two statements there as well. How many knights could there have been among these 16 travelers?
|
Answer: 9 knights.
Solution: Since Vasily's statements contradict each other, Vasily is a liar. Therefore, both of Vasily's statements (about each room) are false, and in each room (when he was there) the number of liars and knights was equal. This means that in each room, without Vasily, there was one more knight than liar. Therefore, among the 15 remaining travelers, there are 3 more knights than liars, which means there are 9 knights and 6 liars among them.
Comment: Correct answer without justification - 0 points.
Showing that Vasily is a liar - 1 point.
Proving that in each room, without Vasily, there is one more knight than liar - 4 points.
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. It is known that the numbers EGGPLANT and FROG are divisible by 3. What is the remainder when the number CLAN is divided by 3? (Letters represent digits, the same letters represent the same digits, different letters represent different digits).
Answer: 0
|
Solution. By the divisibility rule for 3, the sums B+A+K+L+A+Z+A+N and Z+A+B+A are divisible by 3, and therefore the difference of these sums $K+N+A+H$ is also divisible by 3, which by the rule means that the number KLAN is divisible by 3, hence the remainder is 0.
Criteria. Only the answer - 0 points
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Usually, we write the date in the format of day, month, and year (for example, 17.12.2021). In the USA, however, it is customary to write the month number, day number, and year in sequence (for example, 12.17.2021). How many days in a year cannot be determined unequivocally by its writing?
|
Answer: 132.
Solution. Obviously, these are the days where the date can be the number of the month, that is, it takes values from 1 to 12. There are such days $12 \times 12=144$. But the days where the number matches the month number are unambiguous. There are 12 such days. Therefore, the number of days sought is $144-12=132$.
|
132
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. Find the value of the expression $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}$, if $a+b+c=0$.
|
Answer: 3.
Solution. $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+(a+b+c)(3 a b+3 b c+3 c a)-3 a b c$. Using the condition $a+b+c=0$, we get: $a^{3}+b^{3}+c^{3}=3 a b c$. Then $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}=\frac{a^{3}+b^{3}+c^{3}}{a b c}=\frac{3 a b c}{a b c}=3$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Does there exist a pair of unequal integers $a, b$, for which the equality
$$
\frac{a}{2015}+\frac{b}{2016}=\frac{2015+2016}{2015 \cdot 2016}
$$
holds? If such a pair does not exist, justify it. If such a pair does exist, provide an example.
|
# Solution.
Multiply the equation by the number $2015 \cdot 2016$. We get the following equation
$$
2016 a + 2015 b = 2015 + 2016
$$
Rewrite the equation in the following form
$$
2016(a-1) = 2015(1-b)
$$
Since the numbers 2016 and 2015 are coprime, then $(a-1)$ is divisible by 2015, i.e., there exists an integer $k$ for which the equality $a-1 = 2015 k \Rightarrow 1-b = 2016 k \Rightarrow a = 2015 k + 1; b = 1 - 2016 k$ holds.
Substituting the values $a = 2015 k + 1; b = 1 - 2016 k$ for any integer $k$ into the original equation, we get a true identity.
For $k = 1, a = 2016, b = -2015$.
For $k = -1, a = -2014, b = 2017$.
For $k = 2, a = 4031, b = -4031$.
For $k = -2, a = -4029, b = 4033$.
Other various examples can also be represented
One could have guessed
$$
\begin{aligned}
& \frac{2016}{2015} + \frac{-2015}{2016} = \frac{2015 + 2016}{2015 \cdot 2016} \\
& \frac{2016^2 - 2015^2}{2015 \cdot 2016} = \frac{2015 + 2016}{2015 \cdot 2016}
\end{aligned}
$$
## Recommendations for checking.
A correct example is provided - 7 points
|
4031
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In the parliament of a certain state, there are 2016 deputies, who are divided into 3 factions: "blues," "reds," and "greens." Each deputy either always tells the truth or always lies. Each deputy was asked the following three questions:
1) Are you a member of the "blues" faction?
2) Are you a member of the "reds" faction?
3) Are you a member of the "greens" faction?
1208 deputies answered affirmatively to the first question, 908 deputies answered affirmatively to the second question, and 608 deputies answered affirmatively to the third question. In which faction are there more deputies who lie than those who tell the truth, and by how many?
|
# Solution.
Let the number of deputies telling the truth in the "blue," "red," and "green" factions be $r_{1}, r_{2},$ and $r_{3}$ respectively, and the number of deputies lying in the "blue," "red," and "green" factions be $l_{1}, l_{2},$ and $l_{3}$ respectively.
According to the problem:
$\left\{\begin{array}{l}r_{1}+r_{2}+r_{3}+l_{1}+l_{2}+l_{3}=2016, \\ r_{1}+l_{2}+l_{3}=1208, \\ r_{2}+l_{1}+l_{3}=908, \\ r_{3}+l_{1}+l_{2}=608 .\end{array}\right.$
Let $l_{1}-r_{1}=a, l_{2}-r_{2}=b, l_{3}-r_{3}=c$.
Subtracting the third equation from the second, we get $b-a=300$. Or $a=b-300$.
Subtracting the fourth equation from the third, we get $c-b=300$. Or $c=b+300$.
Subtracting the first equation from the sum of the second, third, and fourth equations, we get
$l_{1}+l_{2}+l_{3}=708$. From this, $r_{1}+r_{2}+r_{3}=1308$.
Then $a+b+c=-600$. Or $3 b=-600$.
Thus, $b=-200, a=-500, c=100$. That is, there are 100 more liars in the "green" faction.
Answer: There are 100 more people lying in the "green" faction.
## Recommendations for checking.
The mathematical model of the problem is correctly formulated - 3 points.
|
100
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. At present, the exchange rates of the US dollar and the euro are as follows: $D=6$ yuan and $E=7$ yuan. The People's Bank of China determines the yuan exchange rate independently of market conditions and adheres to a policy of approximate equality of currencies. One bank employee proposed the following scheme for changing the exchange rate to the management. Over one year, the exchange rates $D$ and $E$ are allowed to change according to the following four rules. Either change $D$ and $E$ to the pair $(D+E, 2D \pm 1)$, or to the pair $(D+E, 2E \pm 1)$. Moreover, it is forbidden for the exchange rates of the dollar and euro to be equal at any time.
For example: From the pair $(6,7)$, after one year, the following pairs can be made: $(13,11)$, $(11,13)$, $(13,15)$, or $(15,13)$. What is the smallest value that the difference between the larger and smaller of the resulting exchange rates can take after 101 years?
|
# Solution.
Note that if $D$ and $E$ have different parity, their sum is odd, and if they have the same parity, their sum is even. The second number in the pair is always odd. Initially, we have a pair $О Н$ (Odd, Odd).
After one year
After two years $\quad \mathrm{H}+\mathrm{H}=$ О, Н. And so on.
That is, after an odd number of years, there will be a pair of odd numbers.
Since they do not coincide, the smallest difference can only be 2.
We will also prove that after an even number of years, a difference of 1 can be achieved. We will prove this fact by mathematical induction.
For $n=1$, i.e., after one year $(13,11)$, the difference is 2. For $n=2(24,13 \cdot 2$ - $1=25)$. The difference is 1. Suppose for $n=2 k$ the difference is 1, i.e., $(2 m ; 2 m-1) \Rightarrow(4 m-$ $1 ; 4 m-3)$, the difference is 2. If $n=2 k-1$, the difference, by the induction hypothesis, is 2, then for $n=2 k$ from two courses $(2 m ; 2 m+2) \Rightarrow(4 m+2 ; 4 m+3)$. The difference is 1. Therefore, after 101 years, the smallest difference can be 2.
Answer: 2.
## Recommendations for checking.
Proved only the fact of the alternation of the parity of the difference of the courses - 2 points
Answer without justification - 0 points
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given an acute-angled triangle $A B C$. The feet of the altitudes $B M$ and $C N$ have perpendiculars $M L$ to $N C$ and $N K$ to $B M$. Find the angle at vertex $A$, if the ratio $K L: B C=3: 4$.
|
# Solution.
We will prove that triangles $N A M$ and $C A B$ are similar with a similarity coefficient of $\cos A$. The ratio of the leg $A M$ to the hypotenuse $A B$ is $\cos A$. Similarly, $A M / A C=\cos A$.
Since angle $A$ is common to both triangles
and the corresponding sides are proportional, the triangles are similar with $k=\cos A$. Therefore, the proportion also holds for the third pair of sides of the similar triangles
$$
\frac{M N}{B C}=\cos A
$$
In quadrilateral ANPM, two opposite right angles exist, so $\angle N P M=180^{\circ}-\angle A \Rightarrow \angle M P L=\angle A \Rightarrow \angle K M L=\angle P M L=90^{\circ}-\angle A$
Two right angles subtend the same hypotenuse, hence they are inscribed in the same circle with radius $R=M N / 2$. By the Law of Sines, the following equalities hold
$$
\frac{K L}{\sin \left(90^{\circ}-A\right)}=2 R=2 N O=M N
$$
$$
\frac{K L}{\cos A}=M N=B C \cos A \Rightarrow \frac{K L}{B C}=\cos ^{2} A \Rightarrow \cos ^{2} A=\frac{3}{4} \Rightarrow \cos A=\frac{\sqrt{3}}{2}
$$
Therefore, $\mathrm{A}=30^{\circ}$
Answer: $\mathrm{A}=30^{\circ}$
## Recommendations for checking.
Do not deduct points for the absence of a proof of the similarity of triangles NAM and CAB. Consider it a known fact.
Do not add points for the presence of only a proof of the similarity of triangles NAM and CAB, i.e., 0 points.
If only the equality of angles $\angle M P L=\angle A$ or $\angle K M L=\angle P M L=$ $90^{\circ}-\angle A$ is proven, 1 point is awarded.
If the circle circumscribed around quadrilateral ANPM is present, 1 point is added.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. What is the minimum number of factors that need to be crossed out from the number 99! (99! is the product of all numbers from 1 to 99) so that the product of the remaining factors ends in 2?
|
Answer: 20.
Solution. From the number 99! all factors that are multiples of 5 must be removed, otherwise the product will end in 0. There are 19 such factors in total.
The product of the remaining factors ends in 6. Indeed, the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9$ ends in 6, and similar products in each subsequent decade also end in 6. Therefore, it is sufficient to remove one more factor, for example, 8. After this, the product of the remaining factors will end in 2.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Sasha marked several cells in an $8 \times 13$ table such that in any $2 \times 2$ square, there was an odd number of marked cells. Then he marked a few more cells, as a result of which in each $2 \times 2$ square, there became an even number of marked cells. What is the smallest total number of cells that Sasha could have marked?
|
Answer: 48.
Solution. See example in the figure (the digit 1 is in the cells marked the first time, the digit 2 - the second time)
Estimate. In the table, 24 independent 2x2 squares can be placed. In the first round, at least one cell in each of them was marked. Since each of them ended up with an odd number of marked cells, and after the second round, the number of marked cells became even, at least one cell in each of them was marked again in the second round. In total, we get a minimum of 48 marked
| | 1 | | 1 | | 1 | | 1 | | 1 | | 1 | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
| | 1 | | 1 | | 1 | | 1 | | 1 | | 1 | |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
| | 1 | | 1 | | 1 | | 1 | | 1 | | 1 | |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
| | 1 | | 1 | | 1 | | 1 | | 1 | | 1 | |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
cells.
|
48
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression
$$
a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right) ?
$$
take?
|
Answer: 0.
Solution. From the condition, it follows that $a^{2}-b^{2}=c-b$, $b^{2}-c^{2}=a-c$, and $c^{2}-a^{2}=b-a$. Therefore, $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0$.
## Grading Criteria:
+ a complete and justified solution is provided
the correct answer is obtained by considering specific examples
- only the answer is provided
- the problem is not solved or is solved incorrectly
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. From Zlatoust to Miass, a "GAZ", a "MAZ", and a "KAMAZ" set off simultaneously. The "KAMAZ", having reached Miass, immediately turned back and met the "MAZ" 18 km from Miass, and the "GAZ" - 25 km from Miass. The "MAZ", having reached Miass, also immediately turned back and met the "GAZ" 8 km from Miass. What is the distance from Zlatoust to Miass?
|
Answer: 60 km.
Solution. Let the distance between the cities be $x$ km, and the speeds of the trucks: "GAZ" $-g$ km/h, "MAZ" - $m$ km/h, "KAMAZ" $-k$ km/h. For each pair of vehicles, we equate their travel time until they meet. We get $\frac{x+18}{k}=\frac{x-18}{m}, \frac{x+25}{k}=\frac{x-25}{g}$ and $\frac{x+8}{m}=\frac{x-8}{g}$. Let's rewrite these equations differently: $\frac{x+18}{x-18}=\frac{k}{m}, \frac{x-25}{x+25}=\frac{g}{k}$ and $\frac{x+8}{x-8}=\frac{m}{g}$. Multiplying them term by term, we get: $\frac{x+18}{x-18} \cdot \frac{x-25}{x+25} \cdot \frac{x+8}{x-8}=1$.
Transform the obtained equation, considering that the denominator of each fraction is not zero: $x^{3}+x^{2}+(18 \cdot 8-18 \cdot 25-8 \cdot 25) x-18 \cdot 8 \cdot 25=x^{3}-x^{2}+(18 \cdot 8-18 \cdot 25-8 \cdot 25) x+18 \cdot 8 \cdot 25 \Leftrightarrow 2 x^{2}=2 \cdot 18 \cdot 8 \cdot 25$. Since $x>0$, then $x=60$.
## Grading Criteria:
+ a complete and well-reasoned solution is provided
$\pm$ the equation with one variable (the unknown) is correctly derived, but an error is made in solving it
干 the three proportions are correctly written, but there is no further progress
$\mp$ only the correct answer is provided
- the problem is not solved or is solved incorrectly
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. Square $A B C D$ and isosceles right triangle $A E F$ $\left(\angle A E F=90^{\circ}\right)$ are positioned such that point $E$ lies on segment $B C$ (see figure). Find the angle $D C F$.
---
The square $A B C D$ and the isosceles right triangle $A E F$ with $\angle A E F = 90^{\circ}$ are arranged so that point $E$ lies on the segment $B C$. Determine the measure of angle $D C F$.
|
Answer: $45^{\circ}$.
Solution. First method. Let $P$ be the foot of the perpendicular dropped from point $F$ to line $B C$ (see Fig. 9.5a). Since $\angle F E P=90^{\circ}-\angle B E A=\angle E A B$, the right triangles $F E P$ and $E A B$ are congruent (by hypotenuse and acute angle). Therefore, $P F=B E$. Moreover, $B E=B C-C E=$ $=A B-C E=E P-C E=P C$. Thus, $P F=P C$, which means triangle $C P F$ is a right isosceles triangle. Hence, $\angle F C P=45^{\circ}$, and therefore $\angle D C F=45^{\circ}$.
Fig. 9.5a
Fig. 9.5b
Second method. Draw the diagonal $A C$. Since $\angle E C A=\angle E F A=45^{\circ}$, the quadrilateral $E F C A$ is cyclic (see Fig. 9.5b). Then $\angle A C F=\angle A E F=90^{\circ}$. Therefore, $\angle D C F=\angle A C F-$ $-\angle A C D=90^{\circ}-45^{\circ}=45^{\circ}$.
## Grading Criteria:
+ A complete and well-reasoned solution is provided
$\pm$ A generally correct reasoning is provided, with minor inaccuracies or gaps
- Only the answer is provided or the answer is obtained from a specific example
- The problem is not solved or is solved incorrectly
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.6. Waiting for customers, a watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). In the process, the seller wrote down on a piece of paper the weights he used. What is the smallest number of different numbers that could have appeared in his notes, if the weight of each weight is an integer number of kilograms?
|
Answer: 6 numbers.
Solution. Let's check that with weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, one can weigh any of the given watermelons. Indeed, $2=1+1 ; 4=3+1 ; 6=5+1 ; 8=7+1 ; 11=10+1$; $12=9+3 ; 13=10+3 ; 14=9+5 ; 15=10+5 ; 16=9+7 ; 17=10+7 ; 18=9+9 ; 19=10+9 ;$ $20=10+10$. Thus, 6 different numbers could have been recorded.
We will show that five types of weights are insufficient for the required weighings. If there are five weights, then, generally speaking, twenty watermelons can be weighed. Specifically: five watermelons can be balanced with single weights, five with double weights, and the remaining $\frac{5 \cdot 4}{2}=10$ watermelons with pairs of different weights. However, each combination of weights must be used exactly once.
Notice that half of the watermelons have an odd mass. Let $k$ be the number of weights with an odd mass, and $(5-k)$ be the number of weights with an even mass. Then the number of ways to weigh a watermelon with an odd mass is exactly $k+k(5-k)=6k-k^2$. However, for no $k=0 ; 1 ; 2 ; 3 ; 4 ; 5$ does this expression equal 10 (this can be verified either by substitution or by solving the quadratic equation $6k-k^2=10$).
Therefore, the seller could not have recorded fewer than 6 numbers.
## Grading criteria:
+ a complete and well-reasoned solution is provided
$\pm$ a generally correct solution is provided, containing minor gaps or inaccuracies
$\pm$ it is proven that fewer than six numbers could not have been recorded, a correct set of six weights is indicated, but it is not shown how exactly to weigh the watermelons
dry it is only proven that five weights are insufficient
Ғ it is not proven that five weights are insufficient, but a correct set of six weights is provided (regardless of whether it is shown how exactly to weigh the watermelons)
- only the answer is provided
- the problem is not solved or is solved incorrectly
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The working day at the enterprise lasted 8 hours. During this time, the labor productivity was as planned for the first six hours, and then it decreased by $25 \%$. The director (in agreement with the labor collective) extended the shift by one hour. As a result, it turned out that again the first six hours were worked with planned productivity, and then it decreased by $30 \%$. By what percentage did the overall labor productivity for the shift increase as a result of extending the working day?
|
2. Answer: by 8 percent. Solution. Let's take 1 as the planned labor productivity (the volume of work performed per hour). Then before the shift extension, workers completed $6+1.5=7.5$ units of work per shift. And after the extension, $8+2.1=8.1$ units. Thus, the overall productivity per shift became $8.1: 7.5 \times 100 \% = 108 \%$ of the initial, meaning it increased by $8 \%$. Grading criteria. Correct solution - 7 points. Initial and final productivity per shift correctly expressed in conditional units, but the percentage ratio is expressed incorrectly - 4 points. In all other cases - 0 points.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In pentagon $A B C D E \quad A B=A E=C D=1, \quad B C+D E=1$, $\angle A B C=\angle A E D=90^{\circ}$. Find the area of the pentagon.
|
6. Answer: 1. Solution. Extend the segment $C B$ beyond point $B$ to segment $B F=D E$, and then draw segments $C A, C E$ and $A F$. It is easy to see that triangles $C D E$ and $A B F$ are equal. Therefore, $C E=A F$ and the areas of the pentagon and quadrilateral $A E C F$ coincide. But triangles $A E C$ and $A F C$ are equal by three sides, and the area of triangle $A F C$ is $1 / 2$. Therefore, the area of quadrilateral $A E C F$ is 1. Grading criteria. Correct solution - 7 points. In all other cases - 0 points.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (7 points) In a building, on all floors in all entrances there is an equal number of apartments (more than one). Also, in all entrances there are an equal number of floors. The number of floors is greater than the number of apartments per floor, but less than the number of entrances. How many floors are there in the building if there are a total of $715$ apartments?
|
Answer: 11.
Solution. Let the number of apartments per floor be K, the number of floors be F, and the number of entrances be E. According to the condition, $1<K<F<E$. The number 715 can be factored into numbers greater than one in only one way: $715=5 \cdot 11 \cdot 13$. Therefore, $K=5, F=11, E=13$.
Criterion. Correct solution: 7 points.
Answer without explanation: 3 points.
Incorrect answer: 0 points.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Anya left the house, and after some time Vanya left from the same place and soon caught up with Anya. If Vanya had walked twice as fast, he would have caught up with Anya three times faster. How many times faster would Vanya have caught up with Anya (compared to the actual time) if, in addition, Anya walked twice as slow
|
Answer: 7.
Solution: Let Ane's speed be $v$, and Vanya's speed be $V$. The distance Ane has walked is proportional to $v$, while Vanya catches up with her at a speed of $u=V-v$. When Vanya doubles his speed, this difference triples, i.e., $u+V=2V-v=3u$. From this, we get $V=2u=2v$. If Vanya's speed doubles and Ane's speed is halved, the new difference in speeds will be $3.5v$, which is 7 times the new speed of Masha. Therefore, Vanya will catch up to her 7 times faster.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. In a box, there are white and blue balls, with the number of white balls being 8 times the number of blue balls. It is known that if you pull out 100 balls, there will definitely be at least one blue ball among them. How many balls are there in the box?
|
Answer: 108.
Solution: Since the number of white balls in the box is 8 times the number of blue balls, the total number of balls in the box is divisible by 9. Since 100 balls are drawn from it, the smallest possible number of balls in the box $n=108$, of which 12 are blue and 96 are white. Since there are fewer than a hundred white balls, there will be blue balls in any set of a hundred balls. If $n>108$, then $n \geq 117$ (since $n$ is divisible by 9). In this case, the number of white balls will be greater than or equal to 104, and there will be sets of a hundred balls that do not contain any blue balls.
|
108
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. Solve the equation $\sqrt{2 x-1}+\sqrt[3]{x}=\sqrt[4]{17-x}$.
|
Answer: 1.
Solution. It is easy to verify that $x=1$ is a root of the equation.
Since the left side of the equation is an increasing function (as the sum of two increasing functions), and the right side is a decreasing function, there are no other roots.
Comment. The correct answer without proof of its uniqueness - 2 points.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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