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1. The head of the fish weighs as much as the tail and half of the body, the body weighs as much as the head and the tail together. The tail weighs 1 kg. How much does the fish weigh?
|
Answer: 8 kg.
Solution 1. The body weighs as much as the head and tail, i.e., two tails and half the body. This means that half the body weighs as much as two tails, i.e., the body weighs 4 kg. Then the head weighs $1+2=3$ kg, and the whole fish weighs $4+3+1=8$ kg.
Solution 2. Let $\Gamma, T, X$ be the weight of the head, body, and tail, respectively. Then, according to the condition, $\Gamma=T / 2+X, T=\Gamma+X$. From this, $\Gamma=(\Gamma+X) / 2+X$, i.e., $\Gamma=3 X$. Therefore, the fish weighs $\Gamma+T+X=3 X+(3 X+X)+X=8 X=8$ kg.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The sum of the minuend, subtrahend, and difference is 555. Can the minuend be an integer? If yes, provide an example; if no, explain why.
|
# Answer. Hem.
Solution. Since the sum of the subtrahend and the difference is equal to the minuend, then the sum of the minuend, subtrahend, and difference is equal to twice the minuend, i.e., the minuend is 555/2 - a non-integer.
|
555
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In a psychiatric hospital, there is a chief doctor and many lunatics. During the week, each lunatic bit someone (possibly even themselves) once a day. At the end of the week, it turned out that each of the patients had two bites, and the chief doctor had a hundred bites. How many lunatics are there in the hospital
|
Answer: 20 madmen.
Solution: Let there be n madmen in the hospital. Then, by the end of the week, on the one hand, 7n bites were made, and on the other hand, $2 n+100$. i.e.
$7 n=2 n+100$, from which $n=20$.
|
20
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Real numbers $x, y, z$ (non-zero) satisfy the equation: $x+y=z$. Find all possible values that the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}$ can take.
|
1. Answer: 3.
Solution. Transform the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}=$
$=\frac{z}{x}-\frac{y}{x}+\frac{z}{y}-\frac{x}{y}+\frac{y}{z}+\frac{x}{z}=\frac{z-y}{x}+\frac{z-x}{y}+\frac{y+x}{z}$.
Using the condition $x+y=z$ and the derived formulas $y=z-x$ and $x=$ $z-y$, we get:
$\frac{z-y}{x}+\frac{z-x}{y}+\frac{y+x}{z}=\frac{x}{x}+\frac{y}{y}+\frac{z}{z}=3$.
Grading criteria:
Correct answer with justification provided: 7 points.
Only the correct answer provided without justification: 0 points.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. For different numbers $a$ and $b$, it is known that $\frac{a}{b}+a=\frac{b}{a}+b$. Find $\frac{1}{a}+\frac{1}{b}$.
|
Answer: -1.
Solution. The given equality can be written as $\frac{a}{b}-\frac{b}{a}=b-a$, from which $\frac{a^{2}-b^{2}}{a b}=b-a$ or $\frac{(a-b)(a+b)}{a b}=b-a$. Since the numbers $a$ and $b$ are different, we can divide both sides of the equation by $a-b$, after which we get: $\frac{a+b}{a b}=-1$. This is the required value, since $\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$.
## Grading Criteria:
+ correct answer and complete solution provided
$\pm$ correct answer and generally correct calculations provided, but no justification for the possibility of dividing by $(a-b)$
干 correct reasoning provided, but a computational error was made
- only the answer provided
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $K$ are marked respectively such that $\angle B A M = \angle C K M = 30^{\circ}$. Find $\angle A K D$.
|
Answer: $75^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle B M A=\angle C M K=60^{\circ}$, and then $\angle A M K=60^{\circ}$ (see Fig. 8.5a). Further reasoning can be done in different ways.
First method. Let $A H$ be the perpendicular from vertex $A$ to $M K$. Then the right triangles $A M B$ and $A M H$ are equal by hypotenuse and acute angle, from which $A H=A B$. Using this equality, we get that the right triangles $A K H$ and $A K D$ are equal by hypotenuse and leg. Therefore, $\angle A K D=\frac{1}{2} \angle M K D=\frac{150^{\circ}}{2}=75^{\circ}$.
In the given reasoning, it is used that point $H$ lies on the segment $M K$, and not on its extension beyond point $K$. This is indeed the case, otherwise $A H$ would intersect side $C D$ at point $X$, but then $A H>A X>A D$, which contradicts the equality $A H=A D$. For the lack of this explanation, the student's grade should not be reduced.
Second method. The diagonal $C A$ of the square is the bisector of the internal angle of triangle $C M K$, and the ray $M A$ is the bisector of its external angle, so vertex $A$ is the center of the excircle of this triangle. Therefore, $K A$ is also the bisector of the external angle of triangle $C M K$, so $\angle A K D=\frac{1}{2} \angle M K D=75^{\circ}$.
Third method. Extend the segment $K M$ to intersect the line $A B$ at point $P$ (see Fig. 8.5b). Then $\angle P M B=\angle C M K=\angle A M B$. Therefore, the right triangles $P M B$ and $A M B$ are equal (by leg and acute angle), then $P B=A B$, that is, $A P=2 a$, where $a$ is the side of the given square, and $P M=A M$.
By the property of the leg opposite the $30^{\circ}$ angle in a right triangle, $A M=2 B M$ and $M K=2 M C$. Therefore, $P K=P M+M K=2(B M+M C)=2 B C=2 a$.
Thus, triangle $A P K$ is isosceles with an angle of $30^{\circ}$ at vertex $P$, so its angle at the base is $75^{\circ}$. Since $\angle M K D=150^{\circ}$ and $\angle M K A=75^{\circ}$, then $\angle A K D=75^{\circ}$.
Grading criteria:
+ The correct answer and a fully justified solution are provided
A
Fig. 8.5 b
- Only the answer is provided
|
75
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.6. Sasha drew a square of size $6 \times 6$ cells and alternately colors one cell at a time. After coloring the next cell, he writes down the number of colored cells adjacent to it. After coloring the entire square, Sasha adds up the numbers written in all the cells. Prove that no matter in what order Sasha colors the cells, he will end up with the same sum. (Cells are considered adjacent if they share a common side.)
|
Solution. Consider all unit segments that are common sides for two cells. There are exactly sixty such segments - 30 vertical and 30 horizontal. If a segment separates two shaded cells, we will say that it is "painted." Note that when Sasha writes a number in a cell, he indicates the number of segments that were not painted before shading this cell, but are now painted. When Sasha starts summing, all 60 segments are painted, so the sum of the numbers that Sasha writes will always be 60.
Note that the "materialization" of the specified sum can be described in different ways, for example, instead of a painted segment, one can talk about pairs of adjacent shaded cells and m. n.
Also note that this problem undoubtedly has a brute-force solution, but it is physically impossible to present it in the work.
## Grading criteria:
+ a complete and justified solution is provided
- the number 60 is mentioned but not justified (including by incomplete enumeration or demagogic arguments like "it will always be so...")
|
60
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. In a confectionery store, three types of candies are sold: caramels for 3 rubles, toffees for 5 rubles, and chocolates for 10 rubles. Varya wanted to buy exactly 8 candies of each type and took 200 rubles with her. In the morning, she saw announcements in the store: “When paying for three chocolates, get a free toffee at the cash register.” and “When paying for three toffees, get a free caramel at the cash register.” How much money will Varya have left?
|
Answer: 72 rubles.
Solution. Since Varya will buy more than six but less than nine chocolates, she will get two free caramels. Then she will need to buy another 6 caramels, for which she will get two free toffees, after which she will need to buy another 6 toffees. Then she will spend a total of $8 \cdot 10 + 6 \cdot 5 + 6 \cdot 3 = 128$ rubles. In this case, she will have $200 - 128 = 72$ rubles left.
|
72
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. During a math test, Oleg was supposed to divide a given number by 2, and then add 6 to the result. But he hurried and instead multiplied the given number by 2, and then subtracted 6 from the result. Nevertheless, he got the correct answer. What number was given to Oleg
|
Answer: 8.
Solution. Since Oleg multiplied the number by 2 instead of dividing it, he got a result that is four times greater than the required one. This means that the difference between these two results is three times greater than the required result. However, according to the condition, this difference is equal to $6+6=12$. Therefore, the required result is $12: 3=4$. Then the guessed number is $4 \cdot 2=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In an album, a grid rectangle $3 \times 7$ is drawn. Robot Igor was asked to trace all the lines with a marker, and it took him 26 minutes (the robot draws lines at a constant speed). How many minutes will it take him to trace all the lines of a grid square $5 \times 5 ?$
|
Answer: 30 minutes.
Solution. Note that the grid rectangle $3 \times 7$ consists of four horizontal lines of length 7 and eight vertical lines of length 3. Then the total length of all lines is $4 \cdot 7 + 8 \cdot 3 = 52$. It turns out that Igor spends $26 : 52 = \frac{1}{2}$ minutes on the side of one cell.
The rectangle $5 \times 5$ consists of six vertical and six horizontal lines of length 5. Thus, the total length of the lines is $6 \cdot 5 + 6 \cdot 5 = 60$. Then it will take Igor $60 \cdot \frac{1}{2} = 30$ minutes to outline it.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. A messenger was riding a horse to deliver a message to Ilya Muromets. At some point, he noticed that Ilya Muromets had passed by him (and continued walking in the opposite direction). After 10 seconds (when the horse stopped), the messenger dismounted and ran to catch up with Ilya. How many seconds will it take for the messenger to deliver the message if he runs twice as fast as Ilya Muromets, but five times slower than the galloping horse?
|
Answer: 110.
Solution. Let Ilya Muromets walk at a speed of $x$ meters per second. Then the messenger's speed is $2x$, and the horse's speed is $10x$. Therefore, 10 seconds after the meeting, the distance between the messenger and the horse will be $10 \cdot x$ (Ilya Muromets has walked) $+10 \cdot 10 x$ (the horse has traveled to a complete stop) $=110 x$. After this, the closing speed of the messenger and Ilya Muromets is $2x-x=x$. Thus, the messenger will deliver the message in 110 seconds.
|
110
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. In the kingdom, there live counts, dukes, and marquises. One day, each count dueled with three dukes and several marquises. Each duke dueled with two counts and six marquises. Each marquise dueled with three dukes and two counts. It is known that all counts dueled with an equal number of marquises. How many marquises did each count duel with?
|
Answer: With 6 marquises.
Solution. Let there be $x$ counts, $y$ dukes, and $z$ marquises in the kingdom. Each count fought with three dukes, so there were $3 x$ duels between counts and dukes. But each duke fought with two counts, meaning there were $2 y$ such duels. Therefore, $3 x=2 y$.
Each duke fought with six marquises. Thus, there were $6 y$ duels between dukes and marquises. Each marquis fought with three dukes, so there were $3 z$ such duels. Therefore, $6 y=3 z$, from which $z=2 y=3 x$.
Each marquis fought with two counts, so there were $2 z=6 x$ duels between counts and marquises. Therefore, each count fought with six marquises.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. On an island, there live knights who always tell the truth, and liars who always lie. One day, 15 natives, among whom were both knights and liars, stood in a circle, and each said: "Of the two people standing opposite me, one is a knight, and the other is a liar." How many of them are knights?
|
Answer: 10 knights.
Solution. Consider any knight. He tells the truth, which means that opposite him stand a knight and a liar. One of the people opposite the found liar is the initial knight, so next to him stands another knight. Opposite this new knight stands the previously found liar and another person, who must be a knight. Repeating these considerations, we see that in the circle, there are two knights, then one liar, then again two knights, then another liar, and so on. This means that the liars make up a third of the total number of participants. Therefore, the number of liars is $15 \div 3 = 5$, and the number of knights is $15 - 5 = 10$.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Some squares of the table are mined. Each number written in a square shows the number of mines in the squares adjacent to the given square. (See the figure. Adjacent are squares that have at least one common point; a square with a number is not mined). In how many ways can the mines be placed in the table? Justify your answer.
#
|
# Answer: 14.
Solution. The mine adjacent to the number 1 is in the second, third, or fourth column.
1 case. If the mine is in the second column, it has three possible positions, and it is adjacent to the number two in the first column, which means the second mine adjacent to this two is in the first column, and it has two possible positions. In this case, we get $2 \cdot 3=6$ possible mine placements.
2 case. If the mine is in the fourth column, due to symmetry, we also get 6 possible placements.
3 case. If the mine is in the third column, it has two possible positions in this column. The mines adjacent to the two must be in the same column as the two. In this case, we get two possible placements.
Summing up the results, we get a total of $6+6+2=14$.
Criteria. All cases of mine placement are provided, but it is not shown that there are no others: 4 points.
|
14
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. A family of four octopuses came to a shoe store (each octopus has 8 legs). The father octopus already had half of his legs booted, the mother octopus had only 3 legs booted, and their two sons had 6 legs booted each. How many boots did they buy if they left the store fully booted?
|
Answer: 13.
Solution. The daddy octopus had half of his legs booted, that is, 4 legs. Thus, 4 legs were not booted.
The mommy octopus had 3 legs booted, meaning 5 legs were not booted.
Each of the two sons had 6 legs booted, meaning 2 legs were not booted.
Thus, a total of $4+5+2+2=13$ boots were bought.
## Criteria
## 3 p. Only the correct answer is provided.
4 6. The correct answer is provided, along with the calculations used to obtain it.
|
13
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task No. 1.1
## Condition:
On the potion-making exam, each student at Hogwarts School had to brew 4 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same speeds?
|
# Answer: 36
## Exact match of the answer -1 point
## Solution.
Let's calculate the potion-making speed of each of the three students, measured in potions per hour. Hermione made 4 potions in half an hour, so in one hour she would make twice as many potions, which is 8. Therefore, her speed is 8 potions per hour. Harry made 4 potions in 40 minutes, which means 1 potion every 10 minutes, so in 60 minutes he would make 6 potions, giving him a speed of 6 potions per hour. As for Ron, he made 4 potions in one hour, so his speed is 4 potions per hour. Now, let's assume the three students work together for one hour. Hermione would make 8 potions in this hour, Harry would make 6, and Ron would make 4, so together they would make $8+6+4=18$ potions in this hour. Therefore, if they work together for 2 hours, they would make twice as many potions, which is $18 \cdot 2=36$ potions - giving us the desired answer.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 1.3
## Condition:
At the Potions exam, each student of Hogwarts School had to brew 8 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron needed 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same speeds?
|
Answer: 72
Exact match of the answer -1 point
Solution by analogy with task №1.1
#
|
72
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 1.4
## Condition:
At the potion-making exam, each student from Hogwarts School had to brew 10 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same speeds?
|
Answer: 90
Exact match of the answer -1 point
Solution by analogy with task №1.1
#
|
90
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 5.1
## Condition:
In a certain language $\mathrm{N}$, letters denote only 10 consonants and 5 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables in an allowable manner. For example, the Russian word "кошка" (cat) could be a word in language $\mathrm{N}$, as it can be broken down into syllables as "кош-ка" (ko-sh-ka), whereas the word "гроза" (storm) could not exist in language $\mathrm{N}$ because it starts with two consonants that cannot be separated into syllables. How many eight-letter words are there in language $\mathrm{N}$?
|
Answer: 43750000
Exact match of the answer -1 point
## Solution.
Note that the number of vowels cannot be more than half of the total number of letters in the word, that is, more than 4, since in each syllable with a vowel there is at least one consonant. Also, the number of vowels cannot be less than $1 / 3$ of the number of letters in the word, since the number of consonants is at most twice the number of vowels. This means that 2 or fewer vowels in an eight-letter word is not possible. This means that the possible number of vowels is 4 or 3. In the first case, there is only one possible syllable division: cg-cg-cg-cg, and in the second case, there are three possible divisions: cgc-cgc-cg, cgcsc-cgc, cg-cgc-cgc (c - consonant, g - vowel).
Let's calculate the number of words for each of the four syllable divisions:
cg-cg-cg-cg: $10 * 5 * 10 * 5 * 10 * 5 * 10 * 5=10^{4} * 5^{4}$
cgc-cgc-cg: $10 * 5 * 10 * 10 * 5 * 10 * 10 * 5=10^{5} * 5^{3}$
cgc-cg-cgc: $10 * 5 * 10 * 10 * 5 * 10 * 5 * 10=10^{5} * 5^{3}$
cg-cgc-cgc: $10 * 5 * 10 * 5 * 10 * 10 * 5 * 10=10^{5} * 5^{3}$
The total number of eight-letter words in the language $\mathrm{N}$ is the sum of these four numbers, that is, $10^{4} * 5^{4}+3 * 10^{5} * 5^{3}=43750000$.
|
43750000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 5.2
## Condition:
In a certain language $\mathrm{N}$, letters denote only 10 consonants and 8 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables in an acceptable way. For example, the Russian word "кошка" (cat) could be a word in language $\mathrm{N}$, as it can be broken down into syllables as "кош-ка" (ko-sh-ka), whereas the word "гроза" (storm) could not exist in language $\mathrm{N}$ because it starts with two consonants that cannot be separated into syllables. How many eight-letter words are there in language $\mathrm{N}$?
|
Answer: 194560000
Exact match of the answer -1 point
Solution by analogy with task №5.1
#
|
194560000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 5.4
## Condition:
In a certain language $\mathrm{N}$, letters denote a total of 20 consonants and 3 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables in an allowable manner. For example, the Russian word "кошка" (cat) could be a word in language $\mathrm{N}$, as it can be broken down into syllables as "кош-ка" (ko-sh-ka), whereas the word "гроза" (storm) could not exist in language $\mathrm{N}$ because it starts with two consonants that cannot be separated into syllables. How many eight-letter words are there in language $\mathrm{N}$?
|
Answer: 272160000
Exact match of the answer -1 point
Solution by analogy with task №5.1
#
|
272160000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 6.1
## Condition:
Given a triangle $\mathrm{ABC}$, in which $\mathrm{AB}=5$. The median $\mathrm{BM}$ is perpendicular to the bisector $\mathrm{AL}$. Find $\mathrm{AC}$.
|
Answer: 10
Exact match of the answer -1 point
Solution.
Consider triangle ABM: since the bisector drawn from vertex A is perpendicular to side $\mathrm{BM}$, triangle ABM is isosceles.
Therefore, $\mathrm{AB}=\mathrm{AM}=\mathrm{MC}$. Hence, $\mathrm{AC}=10$.
#
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task No. 7.1
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid on it - increasing the price, but only by a factor of two, - that is, the new price can only be twice the last one offered; the participant after whom no one dares to raise the price by another factor of two gets the painting; if no one bids on the painting, that is, no one offers a price higher than the initial one, the painting remains unsold. For each of his paintings, Ivan Konstantinovich decided to set the starting price at 1000 rubles. How many paintings did he sell at the Broken Auction if all his paintings were bought for different prices, and his total revenue from all the sold paintings amounted to 250000 rubles?
|
Answer: 6
Exact match of the answer -1 point
## Solution.
It is claimed that each painting by Ivan Konstantinovich was sold for $1000 \times 2^{\text {x }}$ rubles, where $\mathrm{x}$ is some natural number. Indeed, initially, the painting's price is 1000 rubles, and each participant doubles its price, meaning the price is multiplied by 2. Therefore, when the painting reaches the last participant and is sold, its price is multiplied by 2 several times (possibly once) and becomes $1000 \times 2 \times \ldots \times 2$, or $1000 \times 2^{x}$ rubles. Here, $\mathrm{x}$ cannot be 0 because the painting cannot be sold for 1000 rubles if no one doubles the price; the painting would remain unsold, and the problem only concerns sold paintings.
Thus, Ivan Konstantinovich's total revenue is the sum of several such numbers of the form $1000 \times 2^{\mathrm{a}}$, where the exponent of 2 is some natural number. This means it is 1000 multiplied by some sum of powers of 2. Since all prices, according to the problem, were different, no power of 2 in this sum is repeated. We need to represent the number $250000 / 1000=250$ as a sum of different powers of 2, and the number of terms in such a decomposition will show us the number of paintings sold. Let's do this: $250=128+64+32+16+8+2 \Rightarrow 6$ terms $\Rightarrow$ 6 paintings were sold.
The uniqueness of such a decomposition can be explained using simple reasoning about what can and cannot be in the decomposition: there are no powers of 2 greater than 128 in the sum, as otherwise the sum would be at least 256, but it should be 250. Moreover, 128 must be included, otherwise the maximum sum would be $64+32+16+8+4+2=126$, since all numbers in the sum are different (as the prices are different by condition). It remains to make up the difference of 250$128=122$. Similarly, it can be shown that 64, 32, 16, and 8 must be included in the sum. After that, only 2 remains to be added, which is exactly the number 2.
#
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task No. 7.2
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid on it - increasing the price, but only by a factor of two, - that is, the new price can only be twice the last one offered; the participant after whom no one dares to raise the price by another factor of two gets the painting; if no one bids on the painting, that is, no one offers a price higher than the initial one, the painting remains unsold. For each of his paintings, Ivan Konstantinovich decided to set the starting price at 1000 rubles. How many paintings did he sell at the Broken Auction if all his paintings were bought for different prices, and his total revenue from all the sold paintings amounted to 300000 rubles?
|
Answer: 4
Exact match of the answer - 1 point
Solution by analogy with task №7.1
#
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 7.3
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid on it - increasing the price, but only by two times, - that is, the new price can only be twice the last one offered; the participant after whom no one dares to raise the price by another two times gets the painting; if no one bids on the painting, that is, no one offers a price higher than the initial one, the painting remains unsold. For each of his paintings, Ivan Konstantinovich decided to set the starting price at 1000 rubles. How many paintings did he sell at the Broken Auction if all his paintings were bought for different prices, and his total revenue from all the sold paintings amounted to 350000 rubles?
|
Answer: 6
Exact match of the answer - 1 point
Solution by analogy with task №7.1
#
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task No. 8.1
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=74^{\circ}$. On ray $\mathrm{BC}$, segment $\mathrm{CD}=\mathrm{CB}$ is laid out. Then, from point $\mathrm{D}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, drawn from vertex $\mathrm{B}$, and the intersection point is $\mathrm{F}$. What is the measure of angle CDF? Express your answer in degrees.
|
Answer: 37
## Exact match of the answer -1 point
## Solution.
Let $\mathrm{F}^{\prime}$ be the midpoint of segment $\mathrm{AC}$. Then $\mathrm{F}^{\prime} \mathrm{C}=0.5 \mathrm{AC}=\mathrm{BC}=\mathrm{CD}=0.5 \mathrm{BD}$. From this, it follows that triangle BDF' is a right triangle, since its median is equal to half the side to which it is drawn. Therefore, the angle $\mathrm{DF}^{\prime} \mathrm{B}=90^{\circ}$, and $\mathrm{DF}^{\prime}$ is perpendicular to the median of angle B. This means that point $\mathrm{F}^{\prime}$ coincides with F. Note that triangle FCD is isosceles, since segments FC and CD are equal, and therefore the angles CFD and CDF, as the angles at the base, are equal. Angle C is the exterior angle for triangle FCD, so $\angle \mathrm{C}=\angle \mathrm{CFD}+\angle \mathrm{CDF}=2 \angle \mathrm{CDF}$. Therefore, $\angle \mathrm{CDF}=0.5 \angle \mathrm{C}=37^{\circ}$.
|
37
|
Geometry
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math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem № 8.2
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=106^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CX}=$ CB is laid out. Then, from point $\mathrm{X}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, drawn from vertex $\mathrm{B}$, and the intersection point is $\mathrm{Y}$. What is the measure of angle СХҮ? Express your answer in degrees.
|
Answer: 53
Exact match of the answer -1 point
Solution by analogy with task №8.1
#
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53
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 8.3
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=46^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CM}=$ CB is marked. Then, from point M, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, drawn from vertex $\mathrm{B}$, and the intersection point is N. What is the measure of angle CMN? Express your answer in degrees.
|
Answer: 23
Exact match of the answer - 1 point
Solution by analogy with task №8.1
#
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23
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. In triangle $ABC$, the median $BM$ was drawn. It turned out that $\angle ABM=40^{\circ}$, $\angle MBC=70^{\circ}$. Find the ratio $AB: BM$. Justify your answer.
To solve problem 9.2 Solution: Extend the median $BM$ beyond point $M$ and lay off a segment $MD = BM$ (see the figure). In quadrilateral $ABCD$, the diagonals bisect each other, so this quadrilateral is a parallelogram. Then $\angle ABC + \angle BCD = 180^{\circ}$, from which $\angle BCD = 70^{\circ} = \angle MBC = \angle DBC$. This means that triangle $BCD$ is isosceles, as is the congruent triangle $BAD$. Therefore,
$$
AB: BM = AB: \frac{BD}{2} = AB: \frac{AB}{2} = 2
$$
|
Answer: $A B: B M=2$.
| in progress | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| An example of a triangle satisfying the conditions of the problem is provided | 0 points |
| Incorrect solution or its absence | 0 points |
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. After the bankruptcy of the company "Horns and Hooves," 17 horns, 2 hooves, and one weight remained. All this wealth was divided equally by weight between Pankovskiy and Balaganov, with the weight entirely going to Balaganov. The horns and hooves were not cut into pieces. Each horn is heavier than each hoof and lighter than the weight by the same amount. How many horns and hooves did Pankovskiy get? Provide all possible options and prove that there are no others.
|
Solution: Let one hoof weigh $k$, and one horn weigh $k+\delta$ (all weights in the same units of measurement, for example, in puds). Then, according to the condition, the weight of the weight (the iron piece) is $k+2\delta$, and the total weight of the divided property is $20k+19\delta$. Each person received $10k+9.5\delta$. Balaganov took the weight, and with horns and hooves, he gathered $9k+7.5\delta$. This weight is less than the weight of 9 horns but more than the weight of 7 horns and 2 hooves. Therefore, Balaganov took fewer than 9 but more than 7 horns, which means 8.8 horns weigh $8k+8\delta$, and for hooves, Balaganov has $k-0.5\delta$ left, which is less than the weight of one hoof. This means all hooves and all other horns went to Panikovsky. Incidentally, we established that $k-0.5\delta=0$, meaning the horn is 3 times heavier than the hoof, and the weight is 5 times heavier than the hoof.
Answer: 9 horns and 2 hooves.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and justified answer with the derived relationship between the weights of the horn, hoof, and weight | 7 points |
| Proven that all hooves went to Panikovsky | 2 points |
| Proven that Balaganov could only have 9 horns | 2 points |
| The problem is solved for specifically chosen weights of horns, hooves, and the weight | 0 points |
| Correct answer without justification | 0 points |
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Variant 1.
Masha is distributing tennis balls into identical boxes. If she uses 4 boxes, there is still room for 8 more balls in the last box, and if she uses 3 boxes, 4 balls will not fit into the boxes. How many balls is one box designed to hold?
|
Answer: 12
Solution.
Method 1
Notice that if the balls are in 3 boxes, there are 4 tennis balls left. Take these 4 balls and put them in the fourth box. According to the condition, there will still be room for 8 more balls in this box. This means that the box can hold a total of 12 balls.
Method 2.
Let the box hold $x$ balls. Then $3 x+4=4 x-8$. From this, $x=12$.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Variant 1.
In the number, two digits were swapped, and as a result, it increased by more than 3 times. The resulting number is 8453719. Find the original number.
|
Answer: 1453789.
Solution. When the number 8453719 is reduced by at least 3 times, the result is a number less than 3000000. Therefore, 8 was swapped with a digit less than 3, and the only such digit is 1.
|
1453789
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 5. Option 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to April 27?
|
Answer: 23
Solution: In the period from April 10 to April 30, there are exactly 21 days. Dividing this period into three weeks: from April 10 to April 16, from April 17 to April 23, and from April 24 to April 30, we get exactly one weekend (Monday) in each of the three weeks. Therefore, in the second statement, Kolya told the truth, and in the first statement, he was wrong.
The period from April 1 to April 20 can be divided into two weeks: from April 1 to April 7, and from April 8 to April 14, and a six-day period (from April 15 to April 20). In the two weeks, there are 12 working days, so in the period from April 15 to April 20, there could be either six working days or five. Since Kolya was wrong here, there were five. That is, in the period from April 15 to April 20, there was definitely a weekend, as well as in the period from April 22 to April 27 (since it is obtained by shifting the previous period exactly by 7 days).
In total, from April 1 to April 21, the cafe was open $3 \cdot 6=18$ days, and in the period from April 22 to April 27, another $6-1=5$ days, that is, a total of $18+5=23$ days.
|
23
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Variant 2.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to April 13?
|
Answer: 11
Option 3.
The cafe "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "the cafe was open for 18 days from April 1 to April 20" and "the cafe was also open for 18 days from April 10 to April 30." It is known that he was wrong once. How many days was the cafe open from April 8 to April 20?
Answer: 11
#
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Variant 1.
On a grid paper, a right triangle with sides 6 and 10 is drawn (see figure). Find the total length of the horizontal grid lines inside this triangle.
|
Answer: 27.
Solution. Extend the triangle to form a rectangle.
It is divided into two identical triangles, so the total length of the horizontal lines in the rectangle is twice the desired total length. In the rectangle, there are 9 lines, each with a length of 6. The desired total length is $\frac{9 \cdot 6}{2}=27$.
|
27
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Variant 1.
Petya wrote 9 consecutive natural numbers on the board. Kolya calculated their sum and got the answer 43040102. It turned out that he made a mistake only in the first digit of the sum. What should the first digit be
|
Answer: 8.
Solution: Let the consecutive natural numbers be $a-4, a-3, \cdots, a+3, a+4$. The sum of nine consecutive numbers is divisible by 9, indeed, this sum is equal to $(a-4)+(a-3)+\cdots+(a+3)+(a+4)=9a$. Therefore, the desired sum must be divisible by 9. This means that the sum of the digits of the obtained answer must also be divisible by 9. The sum of all digits except the first one is 10. Therefore, the first digit must be 8. A suitable number would be 83040102.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 8. Variant 1.
101 natural numbers are written in a circle. It is known that among any 5 consecutive numbers, there will be at least two even numbers. What is the minimum number of even numbers that can be among the written numbers?
|
Answer: 41.
Solution. Consider any 5 consecutive numbers. Among them, there is an even number. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there are at least two even numbers. Thus, the total number of even numbers is no less than $1+2 \cdot 20=41$. Such a situation is possible. Number the numbers in a circle. The even numbers can be those with numbers $1,3,6$, $8,11, \ldots, 98,101$ (the number with number 1 is even, and in each set of 5 numbers with numbers $5 \mathrm{k}+2$, $5 \mathrm{k}+3,5 \mathrm{k}+4,5 \mathrm{k}+5,5 \mathrm{k}+6$, the numbers with numbers $5 \mathrm{k}+3$ and $5 \mathrm{k}+6$ are even).
|
41
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 3 blue points. And on any segment with endpoints at blue points, containing 2 blue points inside, there are at least 2 red points. What is the maximum number of blue points that can be on a segment with endpoints at red points, not containing other red points?
|
# Answer. 3.
Solution. Note that on a segment with endpoints at red points, not containing other red points, there cannot be 4 blue points. Indeed, in this case, between the outermost blue points there will be 2 blue points, which means there will be at least 2 more red points. Therefore, there are no more than 3 blue points on such a segment.
We will show that 3 blue points can lie on a segment with endpoints at red points, not containing other red points. Suppose the points are arranged on a line in the following order: 2 red - 3 blue - 2 red - 3 blue - 2 red. Then all conditions are satisfied, and there is a segment with 3 blue points.
Comment. It is proven that there are no more than 3 blue points between neighboring red points - 4 points.
A correct example with 3 blue points is provided - 3 points.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. In the castle, there are 25 identical square rooms arranged in a $5 \times 5$ square. In these rooms, 25 people—liars and knights (liars always lie, knights always tell the truth)—have settled, one person per room. Each of these 25 people said: "At least one of the rooms adjacent to mine is occupied by a liar." What is the maximum number of liars that could be among these 25 people? Rooms are considered adjacent if they share a wall.
|
Answer: 13 liars.
Solution. Note that liars cannot live in adjacent rooms (otherwise, they would be telling the truth). Let's select one corner room, and divide the remaining rooms into 12 pairs of adjacent rooms. Then, in each pair of rooms, there can be no more than one liar. Therefore, the total number of liars cannot exceed $12+1=13$. Consider a chessboard coloring of the rooms in black and white (corner rooms are black). If we place 13 liars in the "black" rooms and 12 knights in the "white" rooms, the condition of the problem will be satisfied (all liars will lie, and all knights will tell the truth).
Comment. A correct answer without justification - 0 points.
An example of placing 13 liars and 12 knights - 2 points.
Only the statement that liars cannot live in adjacent rooms - 1 point.
Proven that there are no more than 13 liars - 5 points.
|
13
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.2. Petya runs down from the fourth floor to the first floor 2 seconds faster than his mother rides the elevator. Mother rides the elevator from the fourth floor to the first floor 2 seconds faster than Petya runs down from the fifth floor to the first floor. How many seconds does it take Petya to run down from the fourth floor to the first floor? (The lengths of the stair flights between all floors are the same).
|
Answer: 12 seconds.
Solution. Between the first and fourth floors, there are 3 flights, and between the fifth and first floors, there are 4. According to the problem, Petya runs 4 flights in 2 seconds longer than it takes his mother to ride the elevator, and 3 flights in 2 seconds less than his mother. Therefore, Petya runs one flight in 4 seconds. Then from the fourth floor to the first (i.e., 3 flights), Petya runs down in $4 \times 3 = 12$ seconds.
## Grading Criteria.
7 points. Correct answer with a complete solution.
5 points. Explained that one flight takes 4 seconds, and the answer is 4 seconds.
3 points. Correct reasoning under the assumption that the distance from the fifth floor to the first is 1.25 times the distance from the fourth floor to the first, and the answer is 16 seconds. Only the answer $-\mathbf{0}$ points.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3. Three people are playing table tennis, with the player who loses a game giving way to the player who did not participate in it. In the end, it turned out that the first player played 10 games, the second - 21. How many games did the third player play?
|
Answer: 11 games.
Solution: According to the condition, the second player played 21 games, so there were at least 21 games in total. Out of any two consecutive games, the first player must participate in at least one, which means there were no more than \(2 \cdot 10 + 1 = 21\) games. Therefore, a total of 21 games were played, and the second player participated in each of them. In 10 games, he met the first player, and in the remaining 11 games, he met the third player.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. Let x be some natural number. Among the statements:
$2 \times$ greater than 70
x less than 100
$3 x$ greater than 25
x not less than 10
x greater than 5
three are true and two are false. What is x
|
Answer: 9.
Solution
The first statement is equivalent to x > 35, and the third statement is equivalent to $\mathrm{x}>8$, since x is a natural number. The first statement is false, because if it were true, then the third, fourth, and fifth statements would also be true, i.e., there would be at least 4 true statements. Therefore, x does not exceed 35. From this, it follows that the second statement is true. But then the fourth statement is false (otherwise, the third and fifth statements would also be true, and again we would have at least 4 true statements). Therefore, statements 3 and 5 are true. And from the falsity of the fourth statement and the truth of the third statement, it immediately follows that $x=9$.
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. The magpie-crow was cooking porridge and feeding her chicks. The third chick received as much porridge as the first two combined. The fourth one received as much as the second and third. The fifth one received as much as the third and fourth. The sixth one received as much as the fourth and fifth. And the seventh one got nothing - the porridge ran out! It is known that the fifth chick received 10 grams of porridge. How much porridge did the magpie-crow cook?
|
Answer: 40 g.
Solution. Let the first chick receive $m$ g of porridge, and the second chick $-n$ g. Then the third chick received $m+n$ (g), the fourth chick $-m+2n$ (g), the fifth chick $-2m+3n$ (g), and the sixth chick $-3m+5n$ (g). Therefore, the total amount of porridge was: $m+n+(m+n)+(m+2n)+(2m+3n)+(3m+5n)=8m+12n=4(2m+3n)$ (g). This is four times the amount that the fifth chick received, so the magpie cooked $10 \cdot 4=40$ (g) of porridge.
## Grading Criteria:
+ a complete and justified solution is provided
✓ the correct answer is obtained by considering specific examples
✓ two variables are reasonably introduced and expressions for the amount of porridge received by each chick are written (possibly with an error), but the solution is not completed to the correct answer
- only the answer is provided
- the problem is not solved or is solved incorrectly
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. $A B C D$ - a convex quadrilateral. It is known that $\angle C A D=\angle D B A=40^{\circ}, \angle C A B=60^{\circ}$, $\angle C B D=20^{\circ}$. Find the angle $C D B$.
|
Answer: $30^{\circ}$.
Solution. Since $\angle C A B=60^{\circ}, \angle A B C=\angle A B D+\angle D B C=60^{\circ}$, triangle $A B C$ is equilateral (see Fig. 8.3a). We can reason in different ways from here.
Fig. 8.3a
Fig. 8.3b
First method. In triangle $A B D: \angle A B D=40^{\circ}, \angle B A D=\angle B A C+\angle C A D=100^{\circ}$, so $\angle B D A=180^{\circ}-(40^{\circ}+100^{\circ})=40^{\circ}$. Therefore, triangle $A B D$ is isosceles (see Fig. 8.3a). Thus, $A B=B C=C A=A D$, so triangle $C A D$ is also isosceles. Then $\angle A D C=\angle A C D=(180^{\circ}-\angle C A D): 2=70^{\circ}, \angle C D B=\angle C D A-\angle B D A=70^{\circ}-40^{\circ}=30^{\circ}$.
Second method. Draw a circle with center $A$ and radius $A B=A C$ and prove that it contains point $D$. Suppose this is not the case, then the circle intersects ray $A D$ at some point $E$, different from $D$ (see Fig. 8.3b). By the inscribed angle theorem, $\angle C B E=\frac{1}{2} \angle C A E=20^{\circ}$, so rays $B E$ and $B D$ coincide. Therefore, points $E$ and $D$ coincide, which is a contradiction. Since the circle passes through point $D$, $\angle C D B=\frac{1}{2} \angle C A B=30^{\circ}$.
It is also possible to use a similar solution where the isosceles nature of triangle $C A D$ is first proven, and then the circle with center $A$ and radius $A C=A D$ is used. Note that attempting to use the auxiliary circle immediately (without proving the equality of any two segments) is unlikely to be successful.
## Grading Criteria:
+ A complete and well-reasoned solution is provided
$\pm$ A generally correct reasoning is provided, with minor inaccuracies or gaps
干 An auxiliary circle with center $A$ containing points $B, C$, and $D$ is drawn, but it is not proven that it exists, and the correct answer is obtained
- Only the answer is provided
- The problem is not solved or is solved incorrectly
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. Twelve chairs are arranged in a row. Sometimes a person sits on one of the free chairs. In this case, exactly one of his neighbors (if they were there) stands up and leaves. What is the maximum number of people that can be sitting simultaneously, if initially all the chairs were empty?
|
Answer: 11.
Solution. Evaluation. Note that it is impossible to occupy all chairs simultaneously, as at the moment when a person sits on the last unoccupied chair, one of his neighbors will stand up. Therefore, the number of people sitting simultaneously cannot exceed 11.
Example. We will show how to seat 11 people. Number the chairs from 1 to 12. The first chair is easy to occupy. The second chair can be occupied in two stages. In the first stage, a person sits on the third chair, and in the second stage, we seat a person on the second chair, and the person on the third chair stands up. We proceed similarly: if chairs numbered from 1 to \( k \) are occupied, we first seat a person on the chair numbered \( k+2 \), and then seat a person on the chair numbered \( k+1 \), thereby freeing the chair numbered \( k+2 \). After this operation is performed for all \( k \) from 1 to 10, the chairs numbered from 1 to 11 will be occupied, and the twelfth chair will be free.
## Evaluation Criteria:
+ A complete and justified solution is provided
± The correct answer is provided and it is explained how to seat 11 (a correct example is given), but it is not explained why there cannot be 12 sitting (no evaluation)
干 The correct answer is provided and it is explained why there cannot be 12 sitting, but the seating example for 11 is not provided or is incorrect
- Only the answer is provided
- The problem is not solved or is solved incorrectly
8.5. Inside an equilateral triangle \( ABC \), a point \( M \) is marked. Prove that it is possible to choose a point \( C_{1} \) on side \( AB \), a point \( A_{1} \) on side \( BC \), and a point \( B_{1} \) on side \( AC \) such that the lengths of the sides of triangle \( A_{1} B_{1} C_{1} \) are equal to the segments \( MA, MB \), and \( MC \).
Solution. Mark points \( C_{1} \) on side \( AB \), \( A_{1} \) on side \( BC \), and \( B_{1} \) on side \( AC \) such that \( M C_{1} \parallel BC, M A_{1} \parallel AC, M B_{1} \parallel AB \) (see Fig. 8.5). Then the segments \( M A_{1}, M B_{1} \), and \( M C_{1} \) will divide the given triangle into three trapezoids. From the parallelism, it follows that each angle at the larger base of these trapezoids is \( 60^{\circ} \), so these trapezoids are isosceles. Consequently, in each trapezoid, the diagonals are equal: \( B_{1} C_{1} = MA \), \( A_{1} C_{1} = MB \), \( A_{1} B_{1} = MC \), which is what we needed to prove.
Evaluation Criteria:
Fig. 8.5
+ A complete and justified solution is provided
± A correct construction is provided, but it is not explained (or explained incorrectly) why the sides of the resulting triangle are the desired ones
- The problem is not solved or is solved incorrectly
The presence of any reasoning about the number of ways to mark the desired points in the solution does not affect the evaluation result.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (7 points) In a three-digit number, the first digit (hundreds place) was increased by 3, the second digit by 2, and the third digit by 1. As a result, the number increased by 4 times. Provide an example of such an original number.
Answer: 107.
|
Solution. The answer can be found in the following way. Let $x$ be the desired number. Then the condition of the problem immediately leads to the equation $x+321=4x$, the only solution of which is $x=107$.
Criteria. Correct answer, even without any comments: 7 points.
Incorrect answer: 0 points.
|
107
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (7 points) A movie ticket cost 300 rubles. When the price was reduced, the number of visitors increased by 50 percent, and the cinema's revenue increased by 35 percent. How many rubles does one ticket cost now?
|
Answer: 270.
Solution. Let the price of the new ticket be $s$ rubles. Suppose the initial number of visitors was $N$, and after increasing by $50 \%$ it became $1.5 N$. Then, according to the condition, the current revenue of the cinema $1.5 N \cdot s$ is $35 \%$ more than $N \cdot 300$, from which we have $1.5 N s = 1.35 \cdot N \cdot 300$, and $s=270$.
Criteria. Any correct solution: 7 points.
Only the correct answer is provided: 1 point.
|
270
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) Given an arithmetic progression. The sum of its first 10 terms is 60, and the sum of its first 20 terms is 320. What can the 15th term of this progression be?
|
Answer: 25.
Solution. Let the first term of the sequence be $a$, and the common difference be $b$. Then the sum of the first 10 terms is $a+(a+b)+\ldots+(a+9b)=$ $10a+45b$. The sum of the first 20 terms is $a+(a+b)+\ldots+(a+19b)=$ $20a+190b$. According to the problem, $10a+45b=60, 20a+190b=320$. Solving the system, we find $a=-3, b=2$. Then the 15th term is $a+14b=25$.
Criteria. Any correct solution: 7 points.
A generally correct solution containing arithmetic errors that do not affect the solution process: 5 points.
Only the answer: 1 point.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (7 points) Consider the equation $\sin ^{3}(x)+\cos ^{3}(x)=-1$. How many solutions does it have on the interval $[0 ; 6 \pi]$?
Answer: 6.
|
Solution. From the main trigonometric identity, we have $\sin ^{2}(x)+$ $\cos ^{2}(x)=1$. Adding this to the given equality, we get
$$
0=\sin ^{2}(x) \cdot(1+\sin (x))+\cos ^{2}(x) \cdot(1+\cos (x))
$$
In this expression, all factors are non-negative, so both terms $\sin ^{2}(x)(1+\sin (x))$ and $\cos ^{2}(x)(1+\cos (x))$ are equal to 0.
Case 1. Let $\sin (x)=0$. Then $\cos (x) \neq 0$, so $\cos (x)=-1$. Therefore, $x=\pi+2 \pi k$.
Case 2. Let $\sin (x) \neq 0$. Then $\sin x=-1$, which implies $\cos (x)=0$. Therefore, $x=\frac{3 \pi}{2}+2 \pi n$.
It is easy to see that all numbers of the form $\pi+2 \pi k$ and $\frac{3 \pi}{2}+2 \pi n$ are roots of the original equation. In the required interval, three roots of the first form fall, where $k=0,1,2$ and three roots of the second form, where $n=0,1,2$. In total, we have $3+3=6$ roots.
Criteria. Any correct solution: 7 points.
General formulas for the roots are obtained, but the number of roots on the interval $[0 ; 6 \pi]$ is calculated incorrectly: 5 points.
Only the answer, even with the full set of roots presented: 1 point.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. On a circle, 2012 points are marked, dividing it into equal arcs. From these, $k$ points are chosen and a convex $k$-gon is constructed with vertices at the chosen points. What is the largest $k$ for which it could turn out that this polygon has no parallel sides?
|
Answer. For $k=1509$.
Solution. Let $A_{1}, A_{2}, \ldots, A_{2012}$ be the marked points in the order of traversal (we will assume that $A_{2013}=A_{1}, A_{2014}=$ $A_{2}$). We will divide them into quadruples of points $\left(A_{1}, A_{2}, A_{1007}, A_{1008}\right)$, $\left(A_{3}, A_{4}, A_{1009}, A_{1010}\right), \ldots, \left(A_{1005}, A_{1006}, A_{2011}, A_{2012}\right)$. If among the selected $k$ points, all points of some quadruple $\left(A_{2 i-1}, A_{2 i}, A_{2 i+1005}, A_{2 i+1006}\right)$ are found, then in the resulting polygon, there will be two sides $A_{2 i-1} A_{2 i}$ and $A_{2 i+1005} A_{2 i+1006}$, which are symmetric relative to the center of the circle and therefore parallel. This is impossible; hence, in each of the 503 quadruples, no more than three vertices will be marked, that is, $k \leqslant 503 \cdot 3=1509$.
It remains to provide an example of a 1509-gon without parallel sides with vertices at the marked points. For example, the polygon $A_{1} A_{2} \ldots A_{1006} A_{1008} A_{1010} \ldots A_{2012}$ (its vertices are all points with numbers from 1 to 1006 and all points with even numbers from 2008 to 2012) fits. Indeed, the sides $A_{2012} A_{1}, A_{1} A_{2}, \ldots, A_{1005} A_{1006}$ lie on one side of the diameter $A_{2012} A_{1006}$ and therefore cannot be parallel; similarly, the sides $A_{1006} A_{1008}, \ldots, A_{2010} A_{2012}$ are pairwise non-parallel. Finally, the small diagonal $A_{j} A_{j+2}$ of a regular 2012-gon is not parallel to its sides; hence, no two sides of the form $A_{i} A_{i+1}$ and $A_{j} A_{j+2}$ can be parallel.
|
1509
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the problem: octopuses with an even number of legs always lie, while octopuses with an odd number of legs always tell the truth. Five octopuses met, each having between 7 and 9 legs.
The first said: "We have 36 legs in total";
The second: "We have 37 legs in total";
The third: "We have 38 legs in total";
The fourth: "We have 39 legs in total";
The fifth: "We have 40 legs in total".
How many legs did they actually have?
|
Solution. All answers are different, so one is telling the truth or all are lying. If all were lying, they would have 8 legs, i.e., a total of 40, which matches the last answer, a contradiction.
Therefore, four are lying, one is telling the truth, and the one telling the truth has an odd number of legs since $4 \cdot 8 = 32$, the truthful octopus can only have 7 legs, otherwise, there would be no correct answer. Thus, the fourth octopus with seven legs is telling the truth, and the others are liars with 8 legs.
Answer: 39 legs.
|
39
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Two cars started from the same point on a circular track 150 km long, heading in opposite directions. After some time, they met and continued moving in the same directions. Two hours after the start, the cars met for the second time. Find the speed of the second car if the speed of the first car is 60 km/h.
|
Solution. Let the speed of the first car be $x$ km/h, then their closing speed will be $(x+60)$ km/h. Obviously, in two hours they traveled two laps, i.e., 300 km, hence $2 \cdot(x+60)=300$, from which we get that $x=90($ km/h).
Answer: 90 km/h
|
90
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. At the railway platform, many people gathered early in the morning waiting for the train. On the first train, one-tenth of all those waiting left, on the second train, one-seventh of the remaining left, and on the third train, one-fifth of the remaining left. How many passengers were initially on the platform if 216 passengers remained after the departure of the third train?
|
Solution. Solving from the end, we get that before the departure of the third train, there were $216: 6 \cdot 5=270$ passengers on the platform, before the departure of the second - $270: 6 \cdot 7=315$ passengers, before the departure of the first - $315: 9 \cdot 10=350$ passengers.
Answer: 350 passengers.
|
350
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. On the board, there are $n$ different integers. The product of the two largest is 77. The product of the two smallest is also 77. For what largest $n$ is this possible?
(R. Zhenodarov, jury)
|
Answer. For $n=17$.
Solution. The numbers $-11, -7, -6, -5, \ldots, 6, 7, 11$ provide an example for $n=17$.
Assume that there are at least 18 such numbers. Then, at least 9 of them will have the same sign (all positive or all negative). Among these 9 numbers, the absolute values of the two largest will be at least 8 and 9, respectively. Then their product cannot be equal to 77.
Comment. Proving that $n \leqslant 17-$ is worth 5 points.
Providing an example for $n=17-2$ points.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.1. Find any solution to the puzzle
$$
\overline{A B}+A \cdot \overline{C C C}=247
$$
where $A, B, C$ are three different non-zero digits; the notation $\overline{A B}$ represents a two-digit number composed of the digits $A$ and $B$; the notation $\overline{C C C}$ represents a three-digit number consisting only of the digit $C$. Write the three-digit number $\overline{A B C}$ as your answer.
|
Answer: 251.
Solution. The number $\overline{C C C}$ is divisible by 111 and is less than 247, so $A \cdot \overline{C C C}$ is either 111 or 222. In the first case, we get that $\overline{A B}=247-111=136$, which is impossible. In the second case, $\overline{A B}=247-222=25$, that is, $A=2, B=5$, and therefore, $C=1$.
|
251
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.2. At a ball, princesses and knights gathered - a total of 22 people. The first princess danced with seven knights, the second with eight knights, the third with nine knights, ..., the last danced with all the knights present. How many princesses were at the ball?
|
Answer: 8.
Solution. Note that the number of knights the princess danced with is 6 more than her number. Let there be $x$ princesses in total, then the last one has the number $x$ and danced with all the knights, and there are $x+6$ of them in total. We get that there were $x+$ $(x+6)=2x+6=22$ people at the ball, which means there were 8 princesses.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.3. Dima, Misha, and Yura decided to find out who among them is the most athletic. For this, they held 10 competitions. The winner received 3 points, the second place 1 point, and the third place received nothing (in each competition, there was a first, second, and third place). In total, Dima scored 22 points, and Misha scored 8 points. How many points did Yura score?
|
Answer: 10.
Solution. In each competition, the boys in total received $3+1+0=4$ points. For all competitions, they scored $4 \cdot 10=40$ points. Dima and Misha in total scored $22+8=30$ points, so the remaining $40-30=10$ points were scored by Yura.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.4. On her birthday, Katya treated her classmates with candies. After giving out some candies, she noticed that she had 10 more candies left than Artem received. After that, she gave everyone one more candy, and it turned out that all the children in the class (including Katya) had the same number of candies. How many classmates does Katya have?
|
Answer: 9.
Solution. Initially, the number of candies Kati and Artyom had differed by 10. When Kati gave everyone one more candy, the number of candies Artyom had increased by 1, and the number of candies Kati had decreased by the number of her classmates, and they ended up with the same amount. This means that 10 is the number of her classmates plus 1. Therefore, Kati has 9 classmates.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.5. Cat Matroskin, Uncle Fyodor, Postman Pechkin, and Sharik sat down at a round table. In front of each of them was a plate with 15 sandwiches. Every minute, three of them ate a sandwich from their own plate, while the fourth ate a sandwich from their neighbor's plate. After 5 minutes of the meal, there were 8 sandwiches left in Uncle Fyodor's plate. What is the minimum number of sandwiches that could have remained on Cat Matroskin's plate?
|
# Answer: 7
Solution. We will call the sandwiches eaten from a neighbor's plate stolen. Note that in 5 minutes, exactly 5 sandwiches were stolen.
At the same time, 7 sandwiches disappeared from Uncle Fyodor's plate, of which he himself ate no more than 5, meaning that at least 2 were stolen. From this, it is clear that no more than 3 sandwiches were stolen from Matroskin's plate, that is, no more than 8 sandwiches in total disappeared from there. Therefore, at least 7 sandwiches remained.
Note that exactly 7 sandwiches could remain if Uncle Fyodor and Matroskin always ate sandwiches from their own plates, while Sharik sat between them and ate 2 sandwiches from Uncle Fyodor's plate and 3 sandwiches from Matroskin's plate.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Does there exist a four-digit natural number with distinct non-zero digits that has the following property: if this number is added to the same number written in reverse order, the result is divisible by $101$?
|
1. Answer. It exists.
For example, the number 1234 works. Indeed, $1234+4321=5555=101 \cdot 55$.
|
1234
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other?
|
2. Answer: 8.
Note that it is impossible to arrange all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor, the card with the number 1. Therefore, both cards 5 and 7 must be at the ends, and the card with the number 1 must be adjacent to each of them, which is impossible. It is possible to select 8 cards and arrange them in a row according to the requirements of the problem, for example: $9,3,6,2,4,8,1,5$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Petya bought one cupcake, two muffins, and three bagels, Anya bought three cupcakes and a bagel, and Kolya bought six muffins. They all paid the same amount of money for their purchases. Lena bought two cupcakes and two bagels. How many muffins could she have bought for the same amount she spent?
|
3. Answer. 5 cupcakes.
The total cost of Petya and Anya's purchases is equal to the cost of two of Kolya's purchases. If we denote $x, y$, and $z$ as the costs of a cake, a cupcake, and a bagel respectively, we get the equation: $(x+2 y+3 z)+(3 x+z)=12 y, \quad$ from which it follows that $\quad 4 x+4 z=10 y$, \quad that is, \quad $2 x+2 z=5 y$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. What is the minimum number of L-shaped corners consisting of 3 cells that need to be painted in a $6 \times 6$ square of cells so that no more L-shaped corners can be painted? (Painted L-shaped corners should not overlap.)
|
5. Answer. 6.
Let the cells of a $6 \times 6$ square be painted in such a way that no more corners can be painted.
Then, in each $2 \times 2$ square, at least 2 cells are painted, otherwise, a corner in this square can still be painted. By dividing the $6 \times 6$ square into 9 $2 \times 2$ squares, we get that at least $9 \cdot 2 = 18$ cells are painted. Therefore, at least 6 corners are painted. The figure below shows how to paint 6 corners so that no more corners can be painted.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. Three pirates divided the diamonds they had obtained during the day in the evening: twelve each for Bill and Sam, and the rest went to John, who couldn't count. At night, Bill stole one diamond from Sam, Sam stole one from John, and John stole one from Bill. As a result, the average weight of Bill's diamonds decreased by one carat, Sam's decreased by two carats, and John's increased by four carats. How many diamonds did John get?
|
Answer: 9 diamonds.
Solution. The first method (arithmetic). Note that the number of diamonds each pirate has did not change overnight. Since Bill has 12 diamonds, and their average weight decreased by 1 carat, the total weight of his diamonds decreased by 12 carats. Similarly, Sam also has 12 diamonds, and their average weight decreased by 2 carats, so the total weight of his diamonds decreased by 24 carats. Since the total weight of the diamonds of Bill and Sam decreased by 36 carats, the total weight of John's diamonds increased by the same 36 carats. Since the average weight of John's diamonds increased by 4 carats, he had $36: 4=9$ diamonds.
The second method (algebraic). Let John have $x$ diamonds. Let the average weight of the diamonds that Bill received be denoted by $b$, Sam's by $s$, and John's by $d$. Then the total weight of the diamonds that Bill had was $12 b$, Sam's was $-12 s$, and John's was $-x d$.
In the morning, the number of diamonds each had did not change, but the average weight of the diamonds became: for Bill - $(b-1)$, for Sam - $(s-2)$, for John - $(d+4)$. The total weight of the diamonds became: for Bill $12(b-1)$, for Sam $-12(s-2)$, for John $-x(d+4)$. Since the total weight of the diamonds of the three pirates did not change, we have $12 b+12 s+x d=12(b-1)+12(s-2)+x(d+4)$. Expanding the brackets and combining like terms, we get: $4 x-36=0$, so $x=9$.
## Grading criteria:
+ a complete and well-reasoned solution is provided
$\pm$ a generally correct solution is provided, but an arithmetic error is made
干 the equation is correctly set up, but not solved or solved incorrectly
干 the correct answer is obtained from a specific example
- only the answer is provided
- the problem is not solved or solved incorrectly
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.1. The numbers $x, y, z$ are such that $x \in[-3,7], y \in[-2,5], z \in[-5,3]$.
(a) (1 point) Find the smallest possible value of the quantity $x^{2}+y^{2}$.
(b) (3 points) Find the smallest possible value of the quantity $x y z - z^{2}$.
|
# Answer:
(a) (1 point) 0.
(b) (3 points) -200.
Solution. (a) Note that $x^{2} \geqslant 0$ and $y^{2} \geqslant 0$, so $x^{2}+y^{2} \geqslant 0$. The value $x^{2}+y^{2}=0$ is possible when $x=0, y=0$.
(b) Note that $|x y z|=|x||y||z| \leqslant 7 \cdot 5 \cdot 5$ and $z^{2} \leqslant 5^{2}$, so $x y z-z^{2} \geqslant-7 \cdot 5 \cdot 5-5^{2}=-200$. The value $x y z-z^{2}=200$ is possible when $x=7, y=5, z=-5$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.2. Given a rectangle $A B C D$. A line passing through vertex $A$ and point $K$ on side $B C$ divides the entire rectangle into two parts, the area of one of which is 5 times smaller than the area of the other. Find the length of segment $K C$, if $A D=60$.
|
Answer: 40.
Fig. 1: to the solution of problem 8.2
Solution. Draw a line through $K$ parallel to $AB$. Let it intersect side $AD$ at point $L$ (Fig. 1), then $ABKL$ and $DCKL$ are rectangles. Let $S_{ABK}=S$, then $S_{KLA}=S$. It is obvious that $S_{AKCD}=5S$ and $S_{LKCD}=5S-S=4S$.
The ratio of the areas of rectangles $ABKL$ and $DCKL$ with the common side $KL$ is equal to the ratio of the sides $BK$ and $CK$. Therefore, $\frac{BK}{CK}=\frac{2S}{4S}=\frac{1}{2}$, from which we get $KC=\frac{2}{3}BC=$ $\frac{2}{3} \cdot 60=40$
|
40
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.3. In a row, there are 127 balls, each of which is either red, green, or blue. It is known that
- there is at least one red, at least one green, and at least one blue ball;
- to the left of each blue ball, there is a red ball;
- to the right of each green ball, there is a red ball.
(a) (1 point) What is the maximum number of red balls that can lie in the row?
(b) (3 points) What is the minimum number of red balls that can lie in the row?
|
# Answer:
(a) (1 point) 125.
(b) (3 points) 43.
Solution. (a) Among 127 balls, there is at least 1 green and at least 1 blue, so there are no more than 125 red balls. Note also that there can be exactly 125 if the leftmost ball is blue, the rightmost ball is green, and all 125 balls between them are red.
(b) Suppose there are 3 consecutive balls, none of which are red. If the middle ball among them is blue, then to the left of it lies a non-red ball, and if the middle ball is green, then to the right of it lies a non-red ball. This is a contradiction, so among any 3 consecutive balls, there is at least 1 red ball.
Suppose among the two leftmost balls, there is no red ball. If the leftmost ball among them is green, then to the left of it there is no red ball, and if the leftmost ball is blue, then to the right of it there is a non-red ball. This is a contradiction, so among the two leftmost balls, there is at least 1 red ball. Similarly, among the two rightmost balls, there is at least 1 red ball.
All balls, except for the two leftmost and two rightmost, can be divided into \(\frac{127-4}{3}=41\) groups of three consecutive balls. Therefore, the total number of red balls is no less than \(41+1+1=43\). There can be exactly 43 red balls, for example, like this: RBRGRBRGR...RBRGR.
|
125
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.7. Given a right isosceles triangle $ABC$ with a right angle at $A$. A square $KLMN$ is positioned as shown in the figure: points $K, L, N$ lie on sides $AB, BC, AC$ respectively, and point $M$ is located inside triangle $ABC$.
Find the length of segment $AC$, if it is known that $AK=7, AN=3$.
|
Answer: 17.
Fig. 2: to the solution of problem 8.7
Solution. Mark a point $H$ on the segment $B K$ such that $L H \perp B K$ (Fig. 2). Triangle $B H L$ is a right isosceles triangle, $H B=H L$. Note that $\angle L K H=90^{\circ}-$
$\angle A K N=\angle A N K$, and also $K L=K N$. Right triangles $L K H$ and $K N A$ are equal by an acute angle and the hypotenuse, so $K H=A N=3$ and $L H=A K=7$. Therefore,
$$
A C=A B=A K+K H+H B=7+3+7=17
$$
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar?
|
Answer: 16.
Solution. If the first jar contains no less than 11 beetles, then the second jar contains no less than 12, the third jar no less than 13, ..., and the tenth jar no less than 20. And in all jars, there are no less than $11+12+\ldots+20=155$ beetles. Contradiction. Therefore, the first jar contains no more than 10 beetles.
In the tenth jar, there are at least 9 more beetles than in the first jar, which means 9 beetles are no more than half the number of beetles in the tenth jar, so the first jar must contain no fewer than 9 beetles. If there are 9, then the tenth jar can only have 18. But $9+10+\ldots+18=135$. Contradiction. Therefore, the first jar contains exactly 10 beetles, and the tenth jar contains a maximum of 20. Since there are eleven numbers from 10 to 20 and $10+11+\ldots+20=165$, the variant with 15 beetles in a jar is missing, and the following are present: $10,11,12,13,14,16,17,18,19,20$. Therefore, the sixth jar contains 16 beetles.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 8.2 (7 points)
The shares of the company "Nu-i-Nu" increase in price by 10 percent every day. Businessman Borya bought shares of the company for 1000 rubles every day for three days in a row, and on the fourth day, he sold them all. How much money did he make from this operation?
|
Solution: $1000 \cdot 1.1^{3}+1000 \cdot 1.1^{2}+1000 \cdot 1.1-3 \cdot 1000=1331+1210+1100-3000=641$.
| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Correct approach with arithmetic error | 4 |
| Correct answer without justification | 0 |
## Answer: 641 rubles
#
|
641
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 8.5 (7 points)
Find the smallest natural number that is divisible by $48^{2}$ and contains only the digits 0 and 1.
|
Solution:
$48^{2}=2^{8} \cdot 3^{2}$. For a number to be divisible by $2^{8}$, it must end with at least 8 zeros according to the divisibility rule, because otherwise, with fewer zeros ($n \leq 7$), it would have the form $1 \ldots .1 \cdot 10^{\text {n }}$ and would only be divisible by the n-th power of two, but we need divisibility by the 8th power of two. For a number to be divisible by $3^{2}=9$, it must have a sum of digits that is a multiple of 9, meaning the number must contain a multiple of 9 ones, i.e., at least 9. Therefore, our number must have at least 9 ones and end with at least 8 zeros, and the smallest such number is 11111111100000000 (9 ones and 8 zeros), which is divisible by both $2^{8}$ and $3^{2}$, and thus, due to their mutual simplicity, by their product $2^{8} \cdot 3^{2}=48^{2}$.
| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Necessity of 8 zeros derived | 3 |
| Divisibility by 9 shown | 3 |
| Incorrect solution | 0 |
Answer: 11111111100000000 (9 ones and 8 zeros)
|
11111111100000000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.1. In a store, there are 20 items, the costs of which are different natural numbers from 1 to 20 rubles. The store has decided to have a promotion: when buying any 5 items, one of them is given as a gift, and the customer chooses which item to receive for free. Vlad wants to buy all 20 items in this store, paying as little as possible. How many rubles will he need? (Each of the 20 items is sold in 1 copy.)
|
Answer: 136.
Solution. Vlad can take advantage of the offer no more than 4 times, so he will get no more than 4 items for free. The total cost of these 4 items does not exceed $17+18+$ $19+20$ rubles. Therefore, the rubles Vlad needs are not less than
$$
(1+2+3+\ldots+20)-(17+18+19+20)=1+2+3+\ldots+16=\frac{16 \cdot 17}{2}=136
$$
We will show that 136 rubles will definitely be enough for him. He can make purchases of items with the following costs: $(1,2,3,4,17),(5,6,7,8,18),(9,10,11,12,19),(13,14,15,16,20)$. If Vlad takes the last item for free in each purchase, he will spend exactly 136 rubles.
|
136
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.2. Vanya thought of two natural numbers, the product of which equals 7200. What is the greatest value that the GCD of these numbers can take?
|
Answer: 60.
Solution. Since each of these numbers is divisible by their GCD, their product is divisible by the square of this GCD. The greatest exact square that divides the number $7200=2^{5} \cdot 3^{2} \cdot 5^{2}$ is $3600=\left(2^{2} \cdot 3 \cdot 5\right)^{2}$, so the GCD of the two numbers in question does not exceed 60. The GCD can be equal to 60 if the two numbers in question are 60 and 120.
|
60
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.3. Four cities and five roads are arranged as shown in the figure. The lengths of all roads are whole numbers of kilometers. The lengths of four roads are indicated in the figure. How many kilometers is the length of the remaining one?
|
# Answer: 17.
Solution. We will use the triangle inequality: in any non-degenerate triangle, the sum of any two sides is strictly greater than the remaining one.
Let $x$ km be the unknown length. From the left triangle, we see that $x < 10 + 8 = 18$. But if $x \leqslant 16$, then in the right triangle, the triangle inequality does not hold: $x + 5 \leqslant 21$. Since $x$ is an integer less than 18 and greater than 16, it equals 17.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.4. A prime number $p$ is such that the number $p+25$ is the seventh power of a prime number. What can $p$ be? List all possible options.
|
Answer: 103
Solution. We will use the fact that the only even prime number is 2.
- Let $p=2$, then $p+25=27$, which is not a seventh power. Contradiction.
- Let $p>2$, then $p$ is odd, and $p+25$ is even. Since $p+25$ is even and is a seventh power of a prime number, this prime number must be 2. Therefore, $p+25=2^{7}=128$, from which we get $p=103$.
|
103
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.5. On an island, there live knights who always tell the truth, and liars who always lie.
One day, 80 residents of the island gathered, each wearing a T-shirt with a number from 1 to 80 (different residents had different numbers). Each of them said one of two phrases:
- "Among those gathered, at least 5 liars have a T-shirt number greater than mine."
- "Among those gathered, at least 5 liars have a T-shirt number less than mine."
What is the smallest number of knights that could have been among these 80 residents?
|
Answer: 70.
Solution. Suppose there are at least 11 liars. Arrange the numbers on their T-shirts in ascending order and select the liar with the 6th number. Then he must be telling the truth, as there are at least 5 liars with a smaller number and at least 5 liars with a larger number. Thus, there are no more than 10 liars, i.e., there are at least 70 knights.
We will show that there could be exactly 70 knights. For example, let the knights be in T-shirts numbered $1-70$, and the liars in T-shirts numbered $71-80$. All knights and liars with numbers 76 - 80 said the first phrase, while liars with numbers $71-75$ said the second phrase. It is clear that all conditions of the problem are satisfied.
|
70
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.6. Given an obtuse triangle $ABC$ with an obtuse angle $C$. On its sides $AB$ and $BC$, points $P$ and $Q$ are marked such that $\angle ACP = CPQ = 90^\circ$. Find the length of the segment $PQ$, if it is known that $AC = 25, CP = 20, \angle APC = \angle A + \angle B$.
|
Answer: 16.
Solution. Since $\angle P C B+\angle P B C=\angle A P C=\angle P A C+\angle P B C$, we obtain $\angle P C B=\angle P A C$. Note that the right triangles $P A C$ and $Q C P$ are similar by the acute angle, and
$$
\frac{25}{20}=\frac{A C}{C P}=\frac{P C}{P Q}=\frac{20}{P Q}
$$
from which we find $P Q=\frac{20 \cdot 20}{25}=16$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.7. Given a quadratic trinomial $P(x)$, whose leading coefficient is 1. On the graph of $y=P(x)$, two points with abscissas 10 and 30 are marked. It turns out that the bisector of the first quadrant of the coordinate plane intersects the segment between them at its midpoint. Find $P(20)$.
|
Answer: -80.
Solution. The midpoint of this segment has coordinates $\left(\frac{10+30}{2}, \frac{P(10)+P(30)}{2}\right)$. Since it lies on the bisector of the first quadrant, i.e., on the line $y=x$, these coordinates are equal. From this, we get $P(10)+P(30)=40$.
Since $P(x)$ is a monic polynomial, it can be written as $P(x)=x^{2}+a x+b$. Then the condition $P(10)+P(30)=40$ can be rewritten as $100+10 a+b+900+30 a+b=40$, which implies $40 a+2 b=-960$ and $20 a+b=-480$.
Therefore, $P(20)=400+20 a+b=400-480=-80$.
|
-80
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.8. In an $8 \times 12$ table, some $N$ cells are black, and the rest are white. In one operation, it is allowed to paint three cells forming a three-cell corner to white (some of them could already be white before repainting). It turned out that it is impossible to make the entire table white in fewer than 25 such operations. Find the smallest possible value of $N$.
|
Answer: 27.
Solution. Divide the $8 \times 12$ table into 24 squares of $2 \times 2$ (Fig. 9a).
(a)
(b)
Fig. 9: to the solution of problem 9.8
First, we will show how to color 27 cells black in the required manner. In each $2 \times 2$ square, color the top-left cell, and in the bottom-right square, also color all the other cells (Fig. 9b). Note that no three-cell corner can simultaneously cover black cells from two different $2 \times 2$ squares. Therefore, it will take at least 2 operations for the bottom-right square, and another 23 operations for the remaining squares, i.e., a total of at least 25 operations.
Now, we will show that if there are 26 black cells, the table can be guaranteed to be made completely white in no more than 24 operations (if there are initially fewer than 26 black cells, some white cells can be considered black). For this, it is sufficient to either once repaint 3 black cells to white, or twice repaint 2 black cells to white. Indeed, after such actions, the number of available operations will be no less than the remaining number of black cells, so it will be enough to repaint at least one black cell each time.
Since there are a total of 26 black cells, and there are 24 $2 \times 2$ squares, either there is a square with 3 black cells, or there are at least two squares with 2 black cells each (otherwise, there would be no more than $24+1=25$ black cells). By repainting the corresponding three-cell corners, and then repainting at least one black cell each time, we will achieve the required result in no more than 24 operations.
|
27
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. The graph of the quadratic function $y=a x^{2}+c$ intersects the coordinate axes at the vertices of an equilateral triangle. What is the value of ac?
|
Answer: -3.
Solution. Since the graph intersects the OX axis at two points, the numbers a and c have different signs. The points of intersection are: $A\left(\sqrt{-\frac{c}{a}} ; 0\right) ; B\left(-\sqrt{-\frac{c}{a}} ; 0\right)$. The graph intersects the OY axis at point $C(0 ; c)$. Then the side $AB$ of the equilateral triangle is $2 \sqrt{-\frac{c}{a}}$, and its height is $|c|$. Since the side $a$ and the height $h$ of an equilateral triangle are related by $h=\frac{a \sqrt{3}}{2}$, we have $|c|=\frac{\sqrt{3}}{2} \cdot 2 \sqrt{-\frac{c}{a}}$, from which $c^{2}=-3 \cdot \frac{c}{a}$, so $a c=-3$.
There are other ways to obtain the equation relating a and c, for example, by using that $A B=A C$.
Grading criteria.
“+” A complete and well-reasoned solution is provided
“士” A generally correct reasoning is provided, containing minor gaps or inaccuracies (e.g., only considering $s > 0$)
“Ғ” Correct reasoning, but a computational error is made
“ๆ” Only the correct answer is provided
“-” The problem is not solved or is solved incorrectly
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.6. The sides of the base of a brick are 28 cm and 9 cm, and the height is 6 cm. A snail crawls in a straight line along the faces of the brick from one vertex of the lower base to the opposite vertex of the upper base. The horizontal and vertical components of its velocity, $v_{x}$ and $v_{y}$, are related by the equation $v_{x}^{2}+4 v_{y}^{2}=1$ (for example, on the top face $v_{y}=0$ cm/min, so $v_{x}=v=1$ cm/min). What is the minimum time the snail can spend on its journey?
|
Answer: 35 min.
Fig. $10.6 \mathrm{a}$
Solution. Let the given brick be a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D$, where $A B-=9 \mathrm{~cm}, B C=28 \mathrm{~cm}, A A^{\prime}=6 \mathrm{~cm}$.
First method. Increase the height of the brick and the vertical component of the snail's speed by two times. We get a brick $A B C D E F G H$, on which the snail's speed will always be the same: $v=\sqrt{V_{x}^{2}+\left(2 V_{y}\right)^{2}}=1 \mathrm{~cm} /$ min, and the minimum time of movement is the same as on the original brick.
Suppose the snail crawls from vertex $B$ to vertex $F$. Consider the development of three faces of this brick: the base $A B C D$ and two adjacent vertical faces with a common edge $D F$. Since the snail's speed is constant, the minimum time will be when the path length is the shortest, that is, when moving along the diagonal of one of the two rectangles: BCFE or $A B G F$ (see Fig. 10.6a). By the Pythagorean theorem, we get that $B F=35$ cm; $B F=41$ cm. Therefore, it is more advantageous to move along the diagonal $B F$, then at a speed of 1 cm/min, the minimum time will be 35 minutes.
Second method. Notice that the snail must crawl 6 cm vertically in any case, so we can assume that it uniformly crawled from vertex $D$ to vertex $B'$ and has already reached the top face of the brick. In this case, it ended up at one of the points $A_{1}$ or $A_{2}$ at a distance $a$ from vertex $D_{1}$ (see Fig. 10.6b).
Let it have spent $t$ minutes on this path, then multiplying both sides of the equation $v_{x}^{2}+4 v_{y}^{2}=1$ by $t^{2}$, we get: $t^{2} v_{x}^{2}+4 t^{2} v_{y}^{2}=t^{2}$. In this case, $a=t v_{x}$ and $6=t v_{y}$, so $a^{2}+4 \cdot 36=t^{2}$, from which $t=\sqrt{a^{2}+144}$.
Consider two possible routes on the top face: $A_{1} B^{\prime}=\sqrt{28^{2}+\left(9-a_{1}\right)^{2}}$ or $A_{2} B^{\prime}=\sqrt{9^{2}+\left(28-a_{2}\right)^{2}}$. The speed of its movement on the top face is 1 cm/min, so the time is numerically equal to the length of the path traveled. Find the values of $a$ for which the functions $f_{1}(a)=\sqrt{a^{2}+144}+\sqrt{28^{2}+(9-a)^{2}}$ and $f_{2}(a)=\sqrt{a^{2}+144}+\sqrt{9^{2}+(28-a)^{2}}$ take their minimum values.
For the first function, the geometric interpretation is the sum of the distances on the coordinate plane between the points $(0 ; 0)$ and $(12 ; a)$ and between the points $(12 ; a)$ and $(40 ; 9)$. This sum takes the minimum value if the point $(12 ; a)$ lies on the segment with endpoints at the other two points. Therefore, $\frac{40}{12}=\frac{9}{a}$, that is, $a_{1}=2.7$, $f_{1}(2.7)=\sqrt{2.7^{2}+144}+\sqrt{28^{2}+6.3^{2}}=12.3+28.7=41$.
For the second function, the geometric interpretation is the sum of the distances between the points $(0 ; 0)$ and $(12 ; a)$ and between the points $(12 ; a)$ and $(21 ; 28)$. Reasoning similarly, we get: $\frac{21}{12}=\frac{28}{a}$, then $a_{2}=16, f_{2}(16)=\sqrt{16^{2}+144}+\sqrt{9^{2}+12^{2}}=20+15=35$.
Therefore, the minimum time of movement is 35 minutes.
Students who already know how to compute derivatives of functions can act differently in the final part of the solution.
Find the derivatives of the obtained functions:
$$
f_{1}^{\prime}(a)=\frac{a}{\sqrt{a^{2}+144}}+\frac{a-9}{\sqrt{28^{2}+(9-a)^{2}}} u
$$
$f_{2}^{\prime}(a)=\frac{a}{\sqrt{a^{2}+144}}+\frac{a-28}{\sqrt{9^{2}+(28-a)^{2}}} . \quad$ Zeros of the derivatives: $a_{1}=2.7 ; \quad a_{2}=16 . \quad$ Since the derivatives take negative values to the left of these values of $a$ and positive values to the right, these points are the only points of minimum of the functions. Due to the continuity of each function on its domain, the minimum values of these functions will be achieved at these points. These values are: $f_{1}(2.7)=\sqrt{2.7^{2}+144}+\sqrt{28^{2}+6.3^{2}}=12.3+28.7=41$ and $f_{2}(16)=\sqrt{16^{2}+144}+\sqrt{9^{2}+12^{2}}=20+15=35$.
Grading criteria. «+» A complete and justified solution is provided
«士» The correct answer is obtained justifiably, based on the consideration of only one movement option
«Ғ» The answer 41 minutes is obtained justifiably
«Ғ» The time functions are obtained justifiably, but there is no further progress or it contains errors
«-» Only the answer is provided
«-» The problem is not solved or is solved incorrectly
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given $n>2$ natural numbers, among which there are no three equal, and the sum of any two of them is a prime number. What is the largest possible value of $n$?
|
1. Answer: 3. Note that the triplet $1,1,2$ satisfies the condition. Suppose there are more than 3 numbers. Consider any 4 of them a, b, c, d. Among these four numbers, there cannot be two even numbers, otherwise their sum would be an even prime greater than two. Therefore, at least 3 of them must be odd. Their pairwise sums are even primes. Then all three considered numbers must be 1, which violates the condition.
Grading recommendations:
only answer - 0 points
example for three numbers provided - 2 points.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A circle is circumscribed around a right triangle $\mathrm{ABC}$ with hypotenuse $\mathrm{AB}$. On the larger leg $\mathrm{AC}$, a point $\mathrm{P}$ is marked such that $\mathrm{AP}=\mathrm{BC}$. On the arc $\mathrm{ACB}$, its midpoint $\mathrm{M}$ is marked. What can the angle $\mathrm{PMC}$ be equal to?
|
5. Answer: 90 degrees
Consider triangles АРМ and
ВСМ. In them, ВС=АР by condition,
АМ=ВМ - chords subtending equal
arcs, angles МВС and МАС are equal,
since they subtend the same arc.
Therefore, the triangles are equal.
Thus, angles ВМС and АМР are equal. Since angle АМВ is 90 degrees, then angle $\mathrm{PMC}=\mathrm{AMB}-\mathrm{AMP}+\mathrm{BMC}=90$.
Grading recommendations: only answer - 0 points.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (7 points) The kids were given the task to convert the turtle's speed from centimeters per second to meters per minute. Masha got an answer of 25 m/min, but she thought there were 60 cm in a meter and 100 seconds in a minute. Help Masha find the correct answer.
|
Answer: 9 m/min.
Solution. The turtle covers a distance of 25 Machine "meters" in one Machine "minute," meaning it crawls $25 \cdot 60$ centimeters in 100 seconds. Therefore, the speed of the turtle is $\frac{25 \cdot 60}{100}=15$ cm/sec. Thus, in 60 seconds, the turtle will crawl $15 \cdot 60$ centimeters, which is $\frac{15 \cdot 60}{100}=9$ meters.
Criteria. Any correct solution: 7 points.
Correctly found the speed in centimeters per second, but the subsequent part is not done or done with an error: 3 points.
Only the answer without a solution: 1 point.
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) At a certain moment, Anya measured the angle between the hour and minute hands of her clock. Exactly one hour later, she measured the angle between the hands again. The angle turned out to be the same. What could this angle be? (Consider all cases.)
|
Answer: $15^{\circ}$ or $165^{\circ}$.
Solution. After 1 hour, the minute hand remains in its original position. During this time, the hour hand has turned $30^{\circ}$. Since the angle has not changed, the minute hand must bisect one of the angles between the positions of the hour hand (either the $30^{\circ}$ angle or the supplementary angle of $330^{\circ}$).
Therefore, the hour hand was either at $15^{\circ}$ earlier or at $165^{\circ}$ later.
Criteria. Any correct solution: 7 points.
Both correct answers given without justification or with incorrect justification: 3 points.
One of the correct answers given: 1 point.
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (7 points) Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon (i.e., exactly at 12 o'clock) and, without stopping, arrived: one at $B$ at 4 PM, and the other at $A$ at 9 PM. At what time was dawn that day?
|
Answer: at 6 AM.
Solution. Let's denote the meeting point as $C$. Let $x$ be the number of hours from dawn to noon.
The speed of the first pedestrian on segment $A C$ is $A C / x$, and on segment $B C$ it is $B C / 4$. Since his speed is constant, we have $\frac{A C}{x}=\frac{B C}{4}$, which can be rewritten as $\frac{A C}{B C}=\frac{x}{4}$.
Similarly for the second pedestrian: the equality of speeds on segments $B C$ and $A C$ results in the ratio $\frac{B C}{x}=\frac{A C}{9}$, which we can rewrite as $\frac{A C}{B C}=\frac{9}{x}$.
We get $\frac{x}{4}=\frac{9}{x}$, and by the property of proportions, $x^{2}=36, x=6$. Dawn was 6 hours before noon, i.e., at 6 AM.
Criteria. Any correct solution: 7 points.
The time interval from dawn to the meeting is correctly found, but the time of dawn is not found or is found with an error: 5 points.
Only the answer without a solution: 1 point.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (7 points) Determine the number of points at which 10 lines intersect, given that only two of them are parallel and exactly three of these lines intersect at one point.
|
Answer: 42.
Solution. Let's number the lines so that lines 1, 2, and 3 intersect at one point (denote this point as $X$). List all possible pairs of lines (1 and 2, 1 and 3, 1 and $4, \ldots, 8$ and 9, 8 and 10, 9 and 10) and their points of intersection. There are a total of 45 pairs of lines (there are 9 pairs of the form 1 and $\ell$, 8 pairs of the form 2 and $\ell$, and so on; $9+8+7+6+5+4+3+2+1=45$). According to the problem, exactly two lines are parallel. Therefore, a total of 44 points of intersection will be listed. In this case, all points of intersection of the lines except $X$ will be listed exactly once, and point $X$ will appear three times: for the pairs of lines 1 and 2, 1 and 3, 2 and 3. Erase two extra letters $X$ from the list of points of intersection. There will be exactly 42 points left, and this time all points of intersection will be counted exactly once.
Criteria. Any correct solution: 7 points.
Correctly calculated the number of pairs of lines and provided the correct answer: 2 points.
Considered only special cases or provided the correct answer without explanation: 1 point.
|
42
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. There are three acrobat brothers. Their average height is 1 meter 74 centimeters. The average height of two of these brothers: the tallest and the shortest - 1 meter 75 centimeters. What is the height of the middle brother? Justify your answer.
|
Answer: 1 meter 72 centimeters.
Solution. Since the average height of all three is 1 meter 74 centimeters, the total height of all three is 5 meters 22 centimeters. The average height of the two brothers is 1 meter 75 centimeters, so their combined height is 3 meters 50 centimeters. Therefore, the height of the middle brother is 1 meter 72 centimeters.
## Criteria
2 p. Correct solution approach, but an arithmetic error was made.
4 p. The correct answer and justification are provided.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. Let's call a three-digit number interesting if at least one of its digits is divisible by 3. What is the maximum number of consecutive interesting numbers that can exist? (Provide an example and prove that it is impossible to have more consecutive numbers.)
|
Answer: 122.
Solution. The numbers $289,290, \ldots, 299,300, \ldots, 399,400, \ldots, 409,410$ are interesting (recall that 0 is divisible by 3), and there are 122 of them in total. Let's prove that a larger number is not possible.
Suppose we managed to find a larger number of consecutive interesting numbers; choose 123 consecutive ones from them.
Let's call a hundred consecutive numbers, all of which have the same hundreds digit that is divisible by 3, an interesting hundred. Note that before any interesting hundred, there are only 11 interesting numbers ending in $89,90, \ldots, 99$, and the 12th number ends in 88 and is not interesting. Similarly, after an interesting hundred, there are also only 11 interesting numbers ending in $00, \ldots, 09,10$, and the 12th number ends in 11 and is also not interesting.
If our sequence of 123 numbers intersects with some interesting hundred, then it contains at least 12 numbers either before or after this hundred. Consequently, at least one number in it is not interesting.
If our sequence of 123 numbers does not intersect with an interesting hundred, then it contains at least one number ending in 55 (as well as any other combination of digits). But this number is not interesting, as none of its digits are divisible by 3.
## Criteria
0 6. Incorrect answer and incorrect (or missing) example.
16 . Correct answer provided.
1 b. Correct example provided, but the count of numbers in it is incorrect.
2 6. Correct example and answer provided, but no justification that a larger number of numbers is impossible.
4 b. Correct answer, correct example, and justification provided.
|
122
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. In a country, there are 100 cities. Between any two cities, there is either no connection, or there is an air route, or there is a railway (both air routes and railways cannot exist simultaneously). It is known that if two cities are connected to a third city by railway, then there is an air route between them, and if two cities are connected to a third city by air routes, then there is a railway between them. Due to a natural disaster, all flights in the country have been canceled. The government has decided to place humanitarian aid centers in some cities, such that from each city, it is possible to reach such a center. Prove that at least 20 such centers need to be opened.
|
Solution. Note that no city is connected by railway to more than two other cities. Indeed, suppose some three cities are connected by railway to one. Then all of them are connected to each other by air routes, which is impossible by the condition of the problem. Therefore, each city is connected by railway to no more than two others. Then the cities connected to each other by railway form a chain (possibly closed). We will prove that each such chain contains no more than 5 vertices. Let the cities $A, B, C, D$ and $E$ stand sequentially in the chain. Since $A$ and $C$ are connected to $B$ by railways, there is an air route between $A$ and $C$. Similarly, there is an air route between $C$ and $E$. Then there is a railway between $A$ and $E$, and thus the cities are connected in a circle and are not connected to any other city. Therefore, each chain contains no more than 5 cities. To be able to reach a humanitarian center from each city, it is necessary to open one in each such chain. This means that at least $100: 5=20$ centers need to be built, as required.
## Criteria
1 p. It is proven that each city is connected by railway to no more than two others, or an equivalent statement for air routes.
3 p. It is proven that each chain of railway connections contains no more than 5 cities, or an equivalent statement for air routes.
4 6. A correct proof is provided.
|
20
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.3. Four cities and five roads are arranged as shown in the figure. The lengths of all roads are equal to an integer number of kilometers. The lengths of four roads are indicated in the figure. How many kilometers is the length of the remaining one?
#
|
# Answer: 17.
Solution. We will use the triangle inequality: in any non-degenerate triangle, the sum of any two sides is strictly greater than the remaining one.
Let $x$ km be the unknown length. From the left triangle, we see that $x < 10 + 8 = 18$. But if $x \leqslant 16$, then in the right triangle, the triangle inequality does not hold: $x + 5 \leqslant 21$. Since $x$ is an integer less than 18 and greater than 16, it equals 17.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.6. Given an obtuse triangle $ABC$ with an obtuse angle $C$. On its sides $AB$ and $BC$, points $P$ and $Q$ are marked such that $\angle ACP = CPQ = 90^\circ$. Find the length of the segment $PQ$, if it is known that $AC = 25$, $CP = 20$, and $\angle APC = \angle A + \angle B$.
|
Answer: 16.
Solution. Since $\angle P C B+\angle P B C=\angle A P C=\angle P A C+\angle P B C$, we obtain $\angle P C B=\angle P A C$. Note that the right triangles $P A C$ and $Q C P$ are similar by the acute angle, and
$$
\frac{25}{20}=\frac{A C}{C P}=\frac{P C}{P Q}=\frac{20}{P Q}
$$
from which we find $P Q=\frac{20 \cdot 20}{25}=16$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. 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, the number of knights and the number of liars differ by one." Player number 2 said: "In our team, the number of knights and the number of liars differ by two," and so on. Player number 11 said: "In our team, the number of knights and the number of liars differ by eleven." How many knights are in the team, and if there are any, what are their numbers?
|
3. Answer: the only knight plays under number 9. The solution is that the number of knights cannot be equal to the number of liars, so one of the answers must be correct. Two answers cannot be true, as they contradict each other. Therefore, there is exactly 1 knight and 10 liars in the team, which is stated by the player under number 9.
Criteria: correct solution - 7 points. Failure to indicate that the number of knights in the team cannot be equal to the number of liars - 2 points deducted. Correct answer with an explanation of its validity - 2 points. Only answer - 0 points.
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A three-digit number is called cool if one of its digits is half the product of the other two digits. A three-digit number is called supercool if such digits are in two or three of its positions. How many different supercool numbers exist? (Zero cannot be part of the representation of cool or supercool numbers.)
|
4. Answer: 25 numbers. Solution: let the record of a superclass number include digits a, b, c in some order. Then, two equalities are satisfied: $2 \mathrm{a}=\mathrm{bc} 2 \mathrm{~b}=\mathrm{ac}$. If the numbers a and b are different, then by swapping them, we will violate the existing equality. Therefore, $\mathrm{a}=\mathrm{b}$ and c $=2$. In this case, both equalities are automatically satisfied. This means that a superclass number consists of a two and two identical non-zero digits, and the two can be in any of the three available positions. This gives 24 options plus the number 222.
Criteria: correct solution - 7 points. All calculations are correct, but the equality of two digits is not explained - 3 points. The number 222 is forgotten - 4 points. Only the answer - 0 points. All numbers are listed or described in detail, but the absence of others is not proven - two points.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Variant 1.
In the figure, an example is given of how 3 rays divide the plane into 3 parts. Into what maximum number of parts can 11 rays divide the plane?
|
Answer: 56.
## Solution.
If the $(n+1)$-th ray is drawn so that it intersects all $n$ previous rays, then $n$ intersection points on it divide it into $n+1$ segments. The segment closest to the vertex does not add new parts to the partition, while each of the other $n$ segments ( $n-1$ segments and 1 ray) divides some existing part of the partition into two new parts, i.e., $n$ new parts will be added. (Accordingly, if the ray intersects fewer rays, then fewer new parts will be added to the partition).
1 ray - 1 part.
2 rays - $1+1$.
3 rays - $1+1+2$.
...
$n$ rays - $1+1+2+\ldots+(n-1)=1+n(n-1)/2=(n^2-n+2)/2$.
For $n=11$, the answer is $(121-11+2)/2=56$.
|
56
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Variant 1.
In the "Triangle" cinema, the seats are arranged in a triangular shape: in the first row, there is one seat with number 1, in the second row - seats with numbers 2 and 3, in the third row - 4, 5, 6, and so on (the figure shows an example of such a triangular hall with 45 seats). The best seat in the cinema hall is the one located at the center of the hall, i.e., at the midpoint of the height drawn from the vertex of the triangle corresponding to seat number 1. In the figure, an example of such a triangular hall with 45 seats is shown, where the best seat is number 13.
CINEMA "TRIANGLE"
How many seats are in the cinema hall where the best seat has the number $265?$
|
Answer: 1035
Solution.
1st method.
Note that the number of rows in the cinema cannot be even, otherwise there would be no best seat. Let the total number of rows in the cinema be $2 n+1$, then the best seat is in the $n+1$ row. If we remove this row, the triangle can be divided into 4 parts, and the number of seats in each part will be the same, see the figure.
Thus, the problem reduces to finding the number $n$. We need to find the largest integer $n$ that satisfies the inequality: $1+2+\ldots+n<265$ or $n^{2}+n-530<0$, i.e., $n=22$. Then, in each of the four marked triangles, there will be $1+2+\ldots+22=23 \cdot 11=253$ seats, and in total, in the hall, there will be $253 \cdot 4+23=1035$ seats.
2nd method.
Note that the "height" will consist of the numbers located in the center of the rows with odd numbers.
Let $a_{k}$ be the number of the seat located in the center of the row with number $k$ (where $k$ is odd). To the right and left of it in the given row, there are $(k-1)/2$ seats. Therefore, when transitioning from $a_{k}$ to $a_{k+2}$, $(k-1)/2$ last seats from the row with number $k$, the $(k+1)$-th seat from the row with number $k+1$, and $(k+1)/2$ first seats from the row with number $k+2$ are added.
Thus, $a_{k+2}=a_{k}+(k-1)/2+(k+1)+(k+1)/2+1=a_{k}+2(k+1)$.
$a_{1}=1$
$a_{3}=a_{1}+2 \cdot 2=1+4$
$a_{5}=a_{3}+2 \cdot 4=1+4+8$
$a_{7}=a_{3}+2 \cdot 6=1+4+8+12$
$\cdots$
$a_{k}=1+4+8+12+\ldots+2(k-1)=1+4(1+2+\ldots+(k-1)/2)=1+(k+1)(k-1)/2$,
Using this formula, we will find the row number where the best seat is located.
$$
\begin{gathered}
1+(k+1)(k-1) / 2=265 \\
\left(k^{2}-1\right) / 2=264 \\
k^{2}=264 \cdot 2+1=529 \\
k=\sqrt{529}=23 .
\end{gathered}
$$
The total number of rows is $2 k-1$ (before and after the row with the best seat, there are $k-1$ rows each). Therefore, the total number of seats in the hall is given by the formula
$(1+2 k-1) \cdot(2 k-1) / 2=k \cdot(2 k-1)$. For $k=23$, we get $23 \cdot 45=1035$.
|
1035
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. In triangle $A B C$, points $D$ and $F$ are marked on sides $A B$ and $A C$ respectively such that lines $D C$ and $B F$ are perpendicular to each other and intersect at point $E$ inside triangle $A B C$. It turns out that $A D=D C$ and $D E \cdot E C=F E \cdot E B$. What degree measure can angle $B A C$ have? (7 points)
|
# Solution:
$D E \cdot E C=F E \cdot E B \Rightarrow \frac{D E}{E F}=\frac{E B}{E C}, \angle D E B=\angle F E C=90^{\circ}$, so triangles $\triangle D E B$ and $\triangle F E C$ are similar. From the similarity, we get: $\angle D B E=\angle F C E=\alpha$. (This equality of angles can also be justified by the fact that quadrilateral $D B C F$ is inscribed in a circle, from the condition $D E \cdot E C=F E \cdot E B$)
$\triangle A D C$ is isosceles $(A D=D C) \Rightarrow \angle D A C=\angle D C A(\angle F C E)=\angle D B E=\alpha$.
$\angle B D E$ is the exterior angle for $\triangle A D C \Rightarrow \angle B D E=\angle D C A+\angle D A C=2 \alpha$.
$\angle D B E+\angle B D E=2 \alpha+\alpha=3 \alpha=90^{\circ}$ (sum of the acute angles in a right triangle), hence $\alpha=30^{\circ}$.
Answer: $\angle B A C=30^{\circ}$.
| Criteria | Points |
| :--- | :---: |
| Complete solution of the problem. | 7 |
| Proved that $\angle D B E=\angle F C E$, without further progress. | 4 |
| Determined that quadrilateral $D B C F$ is inscribed in a circle. | 2 |
| Incorrect solution. | 0 |
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. On an $8 \times 8$ battleship game field, a "piglet" figure is placed. What is the minimum number of shots needed to definitely hit one "piglet"? (7 points)
#
|
# Solution:
## Evaluation:
Divide the board into rectangles $4 \times 2$. To ensure that a piglet cannot be placed inside one such figure, at least two shots are required. There are 8 such figures, so at least $8 \cdot 2 = 16$ shots are needed. Example:
Answer: 16 shots.
| Criteria | Points |
| :--- | :---: |
| Complete solution to the problem. | 7 |
| Estimation through division into 2x2 squares, without explaining what to do with the fifth cell + correct answer with an example. | 6 |
| Only estimation, without an example. | 4 |
| -Correct answer with an example. -Only estimation through division into 2x2 squares, without explaining what to do with the fifth cell. | 3 |
| Only the correct answer, without estimation and example. | 0 |
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Let $O$ be the center of the circumcircle of triangle $ABC$, points $O$ and $B$ lie on opposite sides of line $AC$, $\angle AOC = 60^\circ$. Find the angle $AMC$, where $M$ is the center of the incircle of triangle $ABC$.
|
# Solution.
Since points $O$ and $B$ lie on opposite sides of line $A C$, the degree measure of arc $A C$, not containing point $B$, is $360^{\circ}-60^{\circ}=300^{\circ}$. Therefore,
$\angle A B C=(1 / 2) \cdot 300^{\circ}=150^{\circ}$. The sum of the angles at vertices $A$ and $C$ of triangle $A B C$ is $180^{\circ}-150^{\circ}=30^{\circ}$, and since $A M$ and $C M$ are the angle bisectors of triangle $A B C$, the sum of the angles at vertices $A$ and $C$ of triangle $A M C$ is $15^{\circ}$. Therefore, $\angle A M C=180^{\circ}-15^{\circ}=165^{\circ}$.
Answer: $165^{\circ}$.
Instructions. Only the answer -0 points; found $\angle A B C-1$ point.
|
165
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On a chessboard of size $2015 \times 2015$, dominoes are placed. It is known that in every row and every column there is a cell covered by a domino. What is the minimum number of dominoes required for this?
|
Solution Let's start filling the 2015 × 2015 square with dominoes starting from the lower left corner (see figure). The first domino will be placed horizontally, covering two vertical lines and one horizontal line. The next domino will be placed vertically, covering two horizontal lines and one vertical line. In total, these two dominoes will cover 3 horizontal lines and 3 vertical lines (see figure). We will place such small squares along the diagonal of the large square - as many as will fit. There will be $2015: 3=671$ (remainder 2).
In total, there will be 671 squares, which contain $671 \cdot 2$ dominoes and cover all vertical and horizontal lines in the $2013 \times 2013$ square. The remaining $2 \times 2$ square, which is not covered, is covered by two dominoes. Therefore, a total of $671 \cdot 2 +$ $2=1344$ dominoes are required.
Answer: 1344 dominoes.
|
1344
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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