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8.1. Masha and Olya bought many identical pens for the new school year at the store. It is known that one pen costs an integer number of rubles, more than 10. Masha bought pens for exactly 357 rubles, and Olya - for exactly 441 rubles. How many pens did they buy in total?
|
8.1. Let a pen cost $r$ rubles, and the numbers 357 and 441 are divisible by $d$. Since the greatest common divisor of the numbers $357=3 \cdot 7 \cdot 17$ and $441=3^{2} \cdot 7^{2}$ is $3 \cdot 7$, then 21 is also divisible by $r$. Since $r>10$, then $r=21$. Therefore, the total number of pens bought is $\frac{357}{21}+$ $\frac{441}{21}=17+21=38$
Answer: 38.
|
38
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. In 8th grade class "G", there are enough underachievers, but Vovochka studies the worst of all. The pedagogical council decided that either Vovochka must correct his twos by the end of the quarter, or he will be expelled. If Vovochka corrects his twos, then the class will have $24 \%$ of underachievers, and if he is expelled, the underachievers will become $25 \%$. What percentage of underachievers are there in 8 "G" now?
|
8.2. Let there be $n$ students in the class now. According to the condition,
$$
0.24 n = 0.25(n-1)
$$
i.e., $0.01 n = 0.25$. Therefore, $n = 25$. One person constitutes $4\%$ of 25, so there are now $24 + 4 = 28\%$ of underachievers.
Answer: $28\%$.
|
28
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. How many natural numbers less than 1000 are divisible by 4 and do not contain the digits $1,3,4,5,7,9$ in their notation?
|
8.4. The desired numbers are written only with the digits $0,2,6,8$.
There is exactly one single-digit number that satisfies the condition, which is the number 8.
There are six two-digit numbers, which are $20,28,60,68,80,88$.
The desired three-digit numbers can end with the following 8 combinations of digits: $00,08,20,28,60,68,80,88$. In each of these variants, the first place can be occupied by any of the three digits $2,6,8$. Thus, we get $3 \cdot 8=24$ desired three-digit numbers.
Therefore, the total number of such numbers is $1+6+24=31$.
Answer: 31
|
31
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. At a round table, 2015 people are sitting, each of them is either a knight or a liar. Knights always tell the truth, and liars always lie. Each person was given a card with a number on it; all the numbers on the cards are different. After looking at their neighbors' cards, each person said: "My number is greater than the number of each of my two neighbors." After this, $k$ of the people said: "My number is less than the number of each of my two neighbors." For what largest $k$ could this have happened?
(O. Podlipsky)
|
Answer: 2013.
Solution. Let $A$ and $B$ be the people who received the cards with the largest and smallest numbers, respectively. Since both of them said the first phrase, $A$ is a knight, and $B$ is a liar. However, if they had said the second phrase, $A$ would have lied, and $B$ would have told the truth; this is impossible. Therefore, $A$ and $B$ cannot say the second phrase, and $k \leqslant 2013$.
We will show that a situation where the remaining 2013 people can say the second phrase is possible. Suppose the people sitting at the table received (in a clockwise direction) cards with the numbers $1, 2, 3, \ldots$, 2015; with the card numbered 2015 going to a knight, and the rest to liars. Then the first phrase can be said by everyone, and the second phrase by everyone except those with cards numbered 1 and 2015.
Remark. There are other examples of card distributions where 2013 people can say the second phrase.
Comment. Only the answer - 0 points.
An example of seating arrangement is provided, where $k$ can be equal to $2013-3$ points.
Only proved that $k \leqslant 2013-4$ points.
Only proved that $k \leqslant 2014-1$ point.
It is claimed that the people with the largest and smallest numbers cannot say the second phrase, but this claim is not justified - 1 point is deducted.
|
2013
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. Several married couples came to the New Year's Eve party, each of whom had from 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families and took them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the necessary trio of people. How many children could there be in total at this party?
(S. Volchonkov)
|
Answer: 33.
Solution: Let there be $p$ married couples and $d$ children at the party (from the condition, $d \leqslant 10 p$). Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and with a fixed choice of mother, a father could be chosen from one of the $p-2$ remaining couples. Therefore, the total number of trios is $d \cdot(p-1)(p-2)=3630$. Since $d \leqslant 10 p$, we get $3630 \leqslant 10 p^{3}$, that is, $p^{3} \geqslant 363>7^{3}$. Thus, $p \geqslant 8$.
Next, the number $3630=2 \cdot 3 \cdot 5 \cdot 11^{2}$ has two divisors $p-1$ and $p-2$, differing by 1. If one of these divisors is divisible by 11, then the other gives a remainder of 1 or 10 when divided by 11. Then it is coprime with 11, and thus divides $2 \cdot 3 \cdot 5=30$ and is at least 10. It is easy to see that this divisor can only be 10; then $p-2=10, p-1=11$ and $d=3630 / 110=33$.
If, however, neither of the numbers $p-2$ and $p-1$ is divisible by 11, then the number $2 \cdot 3 \cdot 5=30$ is divisible by their product; this contradicts the fact that $p \geqslant 8$.
Comment. The equality $d(p-1)(p-2)=3630$ was justified - 4 points.
|
33
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. How many six-digit numbers exist in which four consecutive digits form the number $2021?$
|
Solution. Consider three types of six-digit numbers: $\overline{a b 2021}, \overline{a 2021 b}, \overline{2021 a b}$ (the bar denotes the decimal representation of the number, where $a, b$ - are digits).
In the first case, $\overline{a b 2021}$, $a$ can be any digit from 1 to 9, and $b$ can be any digit from 0 to 9. These digits are chosen independently. Therefore, we get a total of $9 \cdot 10 = 90$ options.
In the second case, $\overline{a 2021 b}$, the reasoning is repeated, $a$ can be any digit from 1 to 9, and $b$ can be any digit from 0 to 9. These digits are chosen independently. Therefore, we again get $9 \cdot 10 = 90$ options. We check that no number from the first group coincides with a number from the second group. They, in particular, differ by the digit in the hundreds place.
In the third case, $\overline{2021 a b}$, the digits $a$ and $b$ are chosen independently and arbitrarily from 0 to 9. In this case, we have 100 options. No number from the third group coincides with the previous options, they differ by the digit in the hundreds place.
The total number of options is $90 + 90 + 100 = 280$.
Answer. 280 numbers.
Leningrad Region
## All-Russian School Olympiad in Mathematics Municipal Stage 2021-2022 academic year
7th grade
Grading Criteria
| Problem 1 | Score | For what it is given |
| :---: | :---: | :--- |
| | 7 | Fully correct solution. A correct diagram is provided. Checking the conditions of the problem for the correct diagram is not required. |
| | | | |
| | Incorrect solution. | |
| Score | For what it is given | |
| :---: | :---: | :--- |
| 7 | Complete solution. Correct groups of exchanges are obtained. | |
| 5 | Correct solution. It is not specified how the condition that there are no whistles left after all exchanges is used. As a result, the division of exchanges into groups is not justified. | |
| | 3 | A sequence of exchanges is described. The uniqueness of the answer does not follow from the reasoning. |
| 1 | It is written that 50 exchanges were made. Further, it is not written or is incorrect. | |
| 0 | Incorrect solution and (or) incorrect answer. | |
| Problem 3 | Score | For what it is given |
| :---: | :---: | :---: |
| | 7 | Complete solution, the correct answer is obtained. |
| | 3 | It is noted that for a simple odd $p$, the number $n-1$ will be odd, and $p-1$ - even. Further progress is absent. |
| | 1 | A check is provided that for $n$, being a power of two, the condition of the problem is satisfied. Proof of the statement formulated in the condition is absent. |
| | 0 | Incorrect solution and (or) incorrect answer. |
| Score | For what it is given |
| :---: | :--- | :--- |
| 7 | Fully correct solution. A correct example is provided and it is written that all conditions are met. |
| 3 | A correct example is provided, but it is not shown that all conditions are met. Possibly, a partial check is performed (for example, the possibility of walking all paths twice). It is not proven that it is impossible to walk all paths once. |
| 1 | A suitable diagram is provided, the conditions are not checked. |
| 0 | Incorrect solution and (or) incorrect answer. |
| Score | For what it is given |
| :---: | :---: | :--- |
| | Fully correct solution. It is proven that Petya will win with the correct strategy. The possibility of each necessary move is shown. |
| | A winning strategy is described. There is no proof of why Petya can always make the necessary move. |
| | A description is provided of the moves Petya and Vasya will make. It is not assumed that Vasya can choose the number of pebbles at his discretion, and Petya can adapt to his choice. |
| 0 | Incorrect solution and (or) incorrect answer. |
| Problem 6 | Score | For what it is given |
| :---: | :---: | :---: |
| | 7 | Fully correct solution. Correct answer. |
| | 6 | Correct solution. Correct answer. The difference from 7 points is that it does not provide a check that no two options for a six-digit number are counted twice. |
| | 5 | Correct solution. Incorrect answer due to placing the digit 0 in the first position. |
| | 1 | Partially correct solution, in which only one placement of the pair of unknown digits is allowed (for example, they are placed only at the end). |
| | 0 | Incorrect solution and (or) incorrect answer that does not meet the criteria for 1 and 5 points. |
|
280
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. This figure contains $9^{2}-4 \cdot 3=69$ cells. Figure 2 shows how a gardener can plant 60 apple trees. We will prove that it is impossible to plant more than 60 apple trees.
Figure 2
First, note that apple trees can definitely be planted on 28 blue cells (these cells receive light through the fence). This leaves 41 cells. Let $x$ be the number of empty cells among these (the "light windows" depicted in yellow in Figure 2). Thus, 41 - x apple trees must be illuminated through $x$ light windows. However, each light window can illuminate no more than four apple trees. Therefore, we arrive at the inequality
$41-x \geq 4 x$, the solution to which is $x \geq 8.2$. Since $x$ is a natural number,
$$
x \geq 9
$$
Thus, there must be at least 9 light windows. Consequently, the number of apple trees cannot exceed $69-9=60$. The statement is proven.
|
Answer: the maximum number of apple trees is 60.
|
60
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
1.1 A courtyard table tennis tournament among 15 players is held according to certain rules. In each round, two players are randomly selected to compete against each other. After the round, the loser receives a black card. The player who receives two black cards is eliminated from the competition. The last remaining player is declared the champion. In table tennis, there are no ties. How many rounds were there in the courtyard tournament if the champion lost exactly once?
|
Answer: 29
Solution. In each match, there is always exactly one loser. Since 14 players were eliminated, there were a total of $14 \cdot 2+1=29$ losses.
|
29
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1 Seven pirates were dividing five identical treasure chests. They agreed that five of them would take one chest each, while the others would receive a fair compensation equal to the value of a chest. Each of the recipients of a chest paid 10000 piastres into a common fund, after which the money was distributed among the remaining pirates. What was the value of one chest?
|
Answer: 35000
Solution. Two pirates received 50000 piastres, so each share amounted to 25000. Since the chests were divided fairly, the total amount of treasure was 25000 for each of the seven, totaling 175000. Then 175000 is the value of five chests, meaning each chest was valued at 35000 piastres.
|
35000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1 Let's say that number A hides number B if you can erase several digits from A to get B (for example, the number 123 hides the numbers 1, 2, 3, 12, 13, and 23). Find the smallest natural number that hides the numbers 2021, 2120, 1220, and 1202.
|
# Answer: 1201201
Solution. Notice that the number contains at least two twos and one zero. If there are exactly two twos, then the zero must stand both between them and after them, but then there must be at least two zeros. Therefore, only on twos and zeros, we need 4 digits (either two twos and two zeros, or three twos and one zero).
If there is one one, then before it should be $2,0,2$, and after it - two more twos and a zero. At the same time, the zero stands both between the twos and after them (1220 and 1202), so we need one more two or one more zero, making it 8 digits in total.
Suppose there are two ones. If there are exactly two twos, then we need a one after these twos (2021), between them (2120), and before them (1220), which leads to a contradiction. Therefore, there must be at least three twos. If there are 6 digits in total, then the zero is the only one. Then it stands after two twos and before one. From the number 1220, we get that one of the ones stands before the first two twos and the zero, and from the number 2021, we get that the other one stands after the last two. Then we definitely get the arrangement - 122021, but alas, from it, we cannot get the number 2120. Therefore, one zero is not enough, and we already have 7 digits.
Suppose there are three ones. We need 4 more digits for twos and zeros, so we have 7 digits in total.
Suppose there are 7 digits. The first digit is not zero, so it must be 1. If the next digit is 0, then it is useless (in our numbers, 0 appears only after twos). Two ones in a row or two zeros make no sense, as in each of our numbers, 0 or 1 appear only once. Therefore, the second most significant digit is 2, followed by 0, and then 1. 1201... We have three more digits, one of which is definitely a two. From the number 1220, we understand that somewhere after it stands a zero, and from the number 2021, we understand that after it stands another one. The smallest such variant is 201, and it fits.
|
1201201
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1 In the example of addition and subtraction, the student replaced the digits with letters according to the rule: identical letters are replaced by identical digits, different letters are replaced by different digits. From how many different examples could the record $0<\overline{\overline{Б A}}+\overline{\text { БА }}-\overline{\text { ЯГА }}<10$ be obtained?
|
# Answer: 31
Solution. The sum of two two-digit numbers is no more than 199, so $\overline{\text { YAG }}$ is a three-digit number starting with 1, Y $=1$. Let's look at the last digit in each number, A. It is added twice and subtracted once, so the value of the expression is $\mathrm{A}$, and $\mathrm{A} \neq 0$. $\mathrm{A} \neq 1$, since Y = 1. In addition, B $\geq 5, B+B=\overline{Y \Gamma}$. If B $=5$, then A can be any digit except 1, 0, and 5 (7 options). Further, if B $=6,7,8,9$, then A can be any digit except 0, 1, B, and $\Gamma$, and $\Gamma$ is determined as the units digit of the number 2B (it is not equal to 1, 0, or B). Thus, in these cases, there are 6 options left. In total, there are $7+4 \cdot 6=7+24=31$ options.
|
31
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. Palm oil has increased in price by $10 \%$. Due to this, the cheese of one of the manufacturers has increased in price by $3 \%$. What is the percentage of palm oil in the cheese of this manufacturer?
|
Answer: $30 \%$.
Solution. Let's assume that the cheese cost 100 conditional rubles per kilogram. Then it increased by 3 rubles. Since this happened due to the increase in the price of palm oil, 3 rubles is $10 \%$ of the cost of palm oil in the cheese, meaning the cost of palm oil in a kilogram of cheese is 30 rubles. Therefore, palm oil constitutes $30 \%$.
Comment. Correct answer without justification - 0 points.
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. Three cyclists leave the city. The speed of the first one is 12 km/h, the second one is 16 km/h, and the third one is 24 km/h. It is known that the first one was cycling exactly when the second and third were resting (standing still), and that at no time did two cyclists cycle simultaneously. It is also known that in 3 hours, all three covered the same distance. Find this distance.
|
Answer: 16 km
Solution: From the condition, it follows that each of the cyclists was riding at the time when the other two were standing, and, moreover, at any given moment, one of the cyclists was riding (the first one rode when the second and third were standing). Since they all traveled the same distance, and the ratio of their speeds is $3: 4: 6$, the time of movement for the first cyclist is 24 parts, for the second - 18 parts, and for the third - 12 parts of time. In total, they rode for 3 hours $=180$ minutes. Therefore, one part of time is $10 / 3$ minutes. Consequently, the first one rode for 80 minutes, which is $4 / 3$ hours. During this time, he traveled 16 km.
Comment: A correct answer without justification - 0 points.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Usually, we write the date in the format of day, month, and year (for example, 17.12.2021). In the USA, however, it is customary to write the month number, day number, and year in sequence (for example, 12.17.2021). How many days in a year cannot be determined unequivocally by its writing?
|
Answer: 132.
Solution. Obviously, these are the days where the date can be the number of the month, that is, it takes values from 1 to 12. There are such days $12 \times 12=144$. But the days where the number matches the month number are unambiguous. There are 12 such days. Therefore, the number of days sought is $144-12=132$.
|
132
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Does there exist a pair of unequal integers $a, b$, for which the equality
$$
\frac{a}{2015}+\frac{b}{2016}=\frac{2015+2016}{2015 \cdot 2016}
$$
holds? If such a pair does not exist, justify it. If such a pair does exist, provide an example.
|
# Solution.
Multiply the equation by the number $2015 \cdot 2016$. We get the following equation
$$
2016 a + 2015 b = 2015 + 2016
$$
Rewrite the equation in the following form
$$
2016(a-1) = 2015(1-b)
$$
Since the numbers 2016 and 2015 are coprime, then $(a-1)$ is divisible by 2015, i.e., there exists an integer $k$ for which the equality $a-1 = 2015 k \Rightarrow 1-b = 2016 k \Rightarrow a = 2015 k + 1; b = 1 - 2016 k$ holds.
Substituting the values $a = 2015 k + 1; b = 1 - 2016 k$ for any integer $k$ into the original equation, we get a true identity.
For $k = 1, a = 2016, b = -2015$.
For $k = -1, a = -2014, b = 2017$.
For $k = 2, a = 4031, b = -4031$.
For $k = -2, a = -4029, b = 4033$.
Other various examples can also be represented
One could have guessed
$$
\begin{aligned}
& \frac{2016}{2015} + \frac{-2015}{2016} = \frac{2015 + 2016}{2015 \cdot 2016} \\
& \frac{2016^2 - 2015^2}{2015 \cdot 2016} = \frac{2015 + 2016}{2015 \cdot 2016}
\end{aligned}
$$
## Recommendations for checking.
A correct example is provided - 7 points
|
4031
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In the parliament of a certain state, there are 2016 deputies, who are divided into 3 factions: "blues," "reds," and "greens." Each deputy either always tells the truth or always lies. Each deputy was asked the following three questions:
1) Are you a member of the "blues" faction?
2) Are you a member of the "reds" faction?
3) Are you a member of the "greens" faction?
1208 deputies answered affirmatively to the first question, 908 deputies answered affirmatively to the second question, and 608 deputies answered affirmatively to the third question. In which faction are there more deputies who lie than those who tell the truth, and by how many?
|
# Solution.
Let the number of deputies telling the truth in the "blue," "red," and "green" factions be $r_{1}, r_{2},$ and $r_{3}$ respectively, and the number of deputies lying in the "blue," "red," and "green" factions be $l_{1}, l_{2},$ and $l_{3}$ respectively.
According to the problem:
$\left\{\begin{array}{l}r_{1}+r_{2}+r_{3}+l_{1}+l_{2}+l_{3}=2016, \\ r_{1}+l_{2}+l_{3}=1208, \\ r_{2}+l_{1}+l_{3}=908, \\ r_{3}+l_{1}+l_{2}=608 .\end{array}\right.$
Let $l_{1}-r_{1}=a, l_{2}-r_{2}=b, l_{3}-r_{3}=c$.
Subtracting the third equation from the second, we get $b-a=300$. Or $a=b-300$.
Subtracting the fourth equation from the third, we get $c-b=300$. Or $c=b+300$.
Subtracting the first equation from the sum of the second, third, and fourth equations, we get
$l_{1}+l_{2}+l_{3}=708$. From this, $r_{1}+r_{2}+r_{3}=1308$.
Then $a+b+c=-600$. Or $3 b=-600$.
Thus, $b=-200, a=-500, c=100$. That is, there are 100 more liars in the "green" faction.
Answer: There are 100 more people lying in the "green" faction.
## Recommendations for checking.
The mathematical model of the problem is correctly formulated - 3 points.
|
100
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given an acute-angled triangle $A B C$. The feet of the altitudes $B M$ and $C N$ have perpendiculars $M L$ to $N C$ and $N K$ to $B M$. Find the angle at vertex $A$, if the ratio $K L: B C=3: 4$.
|
# Solution.
We will prove that triangles $N A M$ and $C A B$ are similar with a similarity coefficient of $\cos A$. The ratio of the leg $A M$ to the hypotenuse $A B$ is $\cos A$. Similarly, $A M / A C=\cos A$.
Since angle $A$ is common to both triangles
and the corresponding sides are proportional, the triangles are similar with $k=\cos A$. Therefore, the proportion also holds for the third pair of sides of the similar triangles
$$
\frac{M N}{B C}=\cos A
$$
In quadrilateral ANPM, two opposite right angles exist, so $\angle N P M=180^{\circ}-\angle A \Rightarrow \angle M P L=\angle A \Rightarrow \angle K M L=\angle P M L=90^{\circ}-\angle A$
Two right angles subtend the same hypotenuse, hence they are inscribed in the same circle with radius $R=M N / 2$. By the Law of Sines, the following equalities hold
$$
\frac{K L}{\sin \left(90^{\circ}-A\right)}=2 R=2 N O=M N
$$
$$
\frac{K L}{\cos A}=M N=B C \cos A \Rightarrow \frac{K L}{B C}=\cos ^{2} A \Rightarrow \cos ^{2} A=\frac{3}{4} \Rightarrow \cos A=\frac{\sqrt{3}}{2}
$$
Therefore, $\mathrm{A}=30^{\circ}$
Answer: $\mathrm{A}=30^{\circ}$
## Recommendations for checking.
Do not deduct points for the absence of a proof of the similarity of triangles NAM and CAB. Consider it a known fact.
Do not add points for the presence of only a proof of the similarity of triangles NAM and CAB, i.e., 0 points.
If only the equality of angles $\angle M P L=\angle A$ or $\angle K M L=\angle P M L=$ $90^{\circ}-\angle A$ is proven, 1 point is awarded.
If the circle circumscribed around quadrilateral ANPM is present, 1 point is added.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. What is the minimum number of factors that need to be crossed out from the number 99! (99! is the product of all numbers from 1 to 99) so that the product of the remaining factors ends in 2?
|
Answer: 20.
Solution. From the number 99! all factors that are multiples of 5 must be removed, otherwise the product will end in 0. There are 19 such factors in total.
The product of the remaining factors ends in 6. Indeed, the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9$ ends in 6, and similar products in each subsequent decade also end in 6. Therefore, it is sufficient to remove one more factor, for example, 8. After this, the product of the remaining factors will end in 2.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Sasha marked several cells in an $8 \times 13$ table such that in any $2 \times 2$ square, there was an odd number of marked cells. Then he marked a few more cells, as a result of which in each $2 \times 2$ square, there became an even number of marked cells. What is the smallest total number of cells that Sasha could have marked?
|
Answer: 48.
Solution. See example in the figure (the digit 1 is in the cells marked the first time, the digit 2 - the second time)
Estimate. In the table, 24 independent 2x2 squares can be placed. In the first round, at least one cell in each of them was marked. Since each of them ended up with an odd number of marked cells, and after the second round, the number of marked cells became even, at least one cell in each of them was marked again in the second round. In total, we get a minimum of 48 marked
| | 1 | | 1 | | 1 | | 1 | | 1 | | 1 | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
| | 1 | | 1 | | 1 | | 1 | | 1 | | 1 | |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
| | 1 | | 1 | | 1 | | 1 | | 1 | | 1 | |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
| | 1 | | 1 | | 1 | | 1 | | 1 | | 1 | |
| | 2 | | 2 | | 2 | | 2 | | 2 | | 2 | |
cells.
|
48
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. From Zlatoust to Miass, a "GAZ", a "MAZ", and a "KAMAZ" set off simultaneously. The "KAMAZ", having reached Miass, immediately turned back and met the "MAZ" 18 km from Miass, and the "GAZ" - 25 km from Miass. The "MAZ", having reached Miass, also immediately turned back and met the "GAZ" 8 km from Miass. What is the distance from Zlatoust to Miass?
|
Answer: 60 km.
Solution. Let the distance between the cities be $x$ km, and the speeds of the trucks: "GAZ" $-g$ km/h, "MAZ" - $m$ km/h, "KAMAZ" $-k$ km/h. For each pair of vehicles, we equate their travel time until they meet. We get $\frac{x+18}{k}=\frac{x-18}{m}, \frac{x+25}{k}=\frac{x-25}{g}$ and $\frac{x+8}{m}=\frac{x-8}{g}$. Let's rewrite these equations differently: $\frac{x+18}{x-18}=\frac{k}{m}, \frac{x-25}{x+25}=\frac{g}{k}$ and $\frac{x+8}{x-8}=\frac{m}{g}$. Multiplying them term by term, we get: $\frac{x+18}{x-18} \cdot \frac{x-25}{x+25} \cdot \frac{x+8}{x-8}=1$.
Transform the obtained equation, considering that the denominator of each fraction is not zero: $x^{3}+x^{2}+(18 \cdot 8-18 \cdot 25-8 \cdot 25) x-18 \cdot 8 \cdot 25=x^{3}-x^{2}+(18 \cdot 8-18 \cdot 25-8 \cdot 25) x+18 \cdot 8 \cdot 25 \Leftrightarrow 2 x^{2}=2 \cdot 18 \cdot 8 \cdot 25$. Since $x>0$, then $x=60$.
## Grading Criteria:
+ a complete and well-reasoned solution is provided
$\pm$ the equation with one variable (the unknown) is correctly derived, but an error is made in solving it
干 the three proportions are correctly written, but there is no further progress
$\mp$ only the correct answer is provided
- the problem is not solved or is solved incorrectly
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. Square $A B C D$ and isosceles right triangle $A E F$ $\left(\angle A E F=90^{\circ}\right)$ are positioned such that point $E$ lies on segment $B C$ (see figure). Find the angle $D C F$.
---
The square $A B C D$ and the isosceles right triangle $A E F$ with $\angle A E F = 90^{\circ}$ are arranged so that point $E$ lies on the segment $B C$. Determine the measure of angle $D C F$.
|
Answer: $45^{\circ}$.
Solution. First method. Let $P$ be the foot of the perpendicular dropped from point $F$ to line $B C$ (see Fig. 9.5a). Since $\angle F E P=90^{\circ}-\angle B E A=\angle E A B$, the right triangles $F E P$ and $E A B$ are congruent (by hypotenuse and acute angle). Therefore, $P F=B E$. Moreover, $B E=B C-C E=$ $=A B-C E=E P-C E=P C$. Thus, $P F=P C$, which means triangle $C P F$ is a right isosceles triangle. Hence, $\angle F C P=45^{\circ}$, and therefore $\angle D C F=45^{\circ}$.
Fig. 9.5a
Fig. 9.5b
Second method. Draw the diagonal $A C$. Since $\angle E C A=\angle E F A=45^{\circ}$, the quadrilateral $E F C A$ is cyclic (see Fig. 9.5b). Then $\angle A C F=\angle A E F=90^{\circ}$. Therefore, $\angle D C F=\angle A C F-$ $-\angle A C D=90^{\circ}-45^{\circ}=45^{\circ}$.
## Grading Criteria:
+ A complete and well-reasoned solution is provided
$\pm$ A generally correct reasoning is provided, with minor inaccuracies or gaps
- Only the answer is provided or the answer is obtained from a specific example
- The problem is not solved or is solved incorrectly
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. In a box, there are white and blue balls, with the number of white balls being 8 times the number of blue balls. It is known that if you pull out 100 balls, there will definitely be at least one blue ball among them. How many balls are there in the box?
|
Answer: 108.
Solution: Since the number of white balls in the box is 8 times the number of blue balls, the total number of balls in the box is divisible by 9. Since 100 balls are drawn from it, the smallest possible number of balls in the box $n=108$, of which 12 are blue and 96 are white. Since there are fewer than a hundred white balls, there will be blue balls in any set of a hundred balls. If $n>108$, then $n \geq 117$ (since $n$ is divisible by 9). In this case, the number of white balls will be greater than or equal to 104, and there will be sets of a hundred balls that do not contain any blue balls.
|
108
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. In the Banana Republic, parliamentary elections were held in which all residents participated. All those who voted for the "Mandarin" party love mandarins. Among those who voted for other parties, $90\%$ do not like mandarins (the rest do like them). What percentage of the votes did the "Mandarin" party receive in the elections, if exactly $46\%$ of the republic's residents love mandarins?
|
Answer: $40 \%$.
Solution. Let $x \%$ be the percentage of votes for the party "Mandarin". Then the percentage of votes for the other parties is $(100-x) \%$. One tenth of $(100-x) \%$ love mandarins, so we get the equation $x+(100-x) / 10=46$. Solving this, we find that $x=40$.
## Criteria.
5 points. A correct equation is correctly justified.
|
40
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. The sum of ten natural numbers is 1001. What is the greatest value that the GCD (greatest common divisor) of these numbers can take?
|
Answer: 91.
Solution. Example. Consider nine numbers equal to 91, and the number 182. Their sum is 1001.
Estimate. We will prove that the GCD cannot take a value greater than 91. Note that $1001=7 \times 11 \times 13$. Since each term in this sum is divisible by the GCD, the GCD is a divisor of the number 1001. On the other hand, the smallest term in the sum (and thus the GCD) is no greater than 101. It remains to note that 91 is the largest divisor of the number 1001 that satisfies this condition.
## Grading criteria:
+ fully justified solution
$\pm$ correct answer provided, estimate is proven, but no example is given
干 correct answer and an example of ten numbers are provided
- only the answer is provided
- the problem is not solved or is solved incorrectly
|
91
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. Misha and Masha had the same multi-digit integer written in their notebooks, ending in 9876. Masha placed a plus sign between the third and fourth digits from the right, while Misha placed a plus sign between the fourth and fifth digits from the right. To the surprise of the schoolchildren, both resulting sums turned out to be the same. What number was originally written by the schoolchildren? Provide all possible answers and prove that there are no others.
|
Solution: Let the written number have the form $\overline{x 9876}$, where $x$ is also some natural number. Then Misha got the sum $x+9876$, and Masha got the sum $10 x+9+876$. From the equality $x+9876=10 x+9+876$, we find that $x=999$.
Answer: 9999876 and there is no other number.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| the correct answer is given and its uniqueness is proven | 7 points |
| the answer is incorrect only due to an arithmetic error | 6 points |
| the condition is correctly written as a Diophantine equation, but the latter is not solved | 5 points |
| cases of $n$-digit numbers are correctly analyzed for specific natural values of $n$ | |
| the correct answer is given, and the check that the number 9999876 fits the condition of the problem AND/OR there is an unsuccessful attempt to write the condition of the problem in general as an equation in integers | 2 points |
| the correct answer without justification | 1 point |
|
9999876
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Answer. 190. Solution 1. Among the lines intersecting $y=20-x$, there is the line $y=x$. Moreover, the picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(19 ; 19)$, $180: 9=20$ lines are drawn, of which 19 intersect the line $y=20-x$. For each point on the line $y=20-x$, the sum of the coordinates is 20, so the sum of the abscissas and ordinates of all points is $20 \cdot 19=380$, then the sum of the abscissas is half of this and equals 190.
|
Solution 2. Through the point $(19 ; 19)$, $180: 9=20$ lines are drawn, of which 19 intersect the line $y=20-x$. Let the line $y=x$ intersect the line $y=20-x$ at point $A$, then $x_{A}=10$. The remaining 18 lines are divided into pairs, intersecting the line $y=20-x$ at symmetric points $B$ and $C$ (triangles $M A B$ and $M A C$ are equal by a leg and an acute angle). Then $x_{B}+x_{C}=2 x_{A}=20$. The sum of the abscissas of all intersection points is $10+9 \cdot 20=190$.
Grading criteria. Correct solution - 7 points. An error in calculating the number of lines passing through the point (19;19): 360:9=40, all other reasoning is correct - 4 points. It is justified that the picture is symmetric with respect to the line $y=x$, but there is no further progress - 2 points. In all other cases - 0 points.
|
190
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the digits of a natural number $a$ is to the sum of the digits of the number $2a$ as 19 to 9. Prove that the decimal representation of the number $a$ has at least 29 digits.
|
Solution. From the condition, it follows that the sum of the digits of the number $2a$ is divisible by 9. Therefore, the number $2a$ itself, and consequently the number $a$, and the sum of its digits, are divisible by 9. In addition, the sum of the digits of the number $a$ is divisible by 19. Thus, it is divisible by 171. Let it be 171n. If there were no carries across the digits when adding $a+a$ in a column, the sum of the digits of $2a$ would be 342n. In reality, it equals $\frac{9 \cdot 17 n}{19}=81$ n. Therefore, the sum of the digits is reduced by $261 n$ due to the carries. Each carry reduces the sum by 9. Thus, there were $\frac{261 n}{9}=29 n$ carries, which means the number of digits in the sum is at least 29.
|
29
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. On a plane, 100 points are marked. It turns out that on two different lines a and b, there are 40 marked points each. What is the maximum number of marked points that can lie on a line that does not coincide with a and b?
|
Answer: 23. Solution: Lines $a$ and $b$ have no more than one common point. If such a point exists and is marked, then together on lines $a$ and $b$ there are 79 marked points; otherwise, there are 80. Therefore, outside lines $a$ and $b$, there are no more than 21 marked points.
Take an arbitrary line $c$ that does not coincide with $a$ and $b$. On it, there are no more than 21 points not lying on $a$ and $b$, and, in addition, the points of its intersection with $a$ and $b$ can be marked - a total of no more than 23 points.
Mark the vertices of some triangle $ABC$. On its sides $a=BC$ and $b=AC$, mark another 38 points each, and on side $c=AB$, mark another 21 points. We have constructed an example where exactly 23 points lie on line $c$, which completes the solution.
- See also problem $7-2$.
- For an answer without justification - 0 points. An answer with a correct example but without justification of the estimate - 3 points, the justification of the estimate is evaluated from the remaining 4 points.
|
23
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. A row of 101 numbers is written (the numbers are not necessarily integers). The arithmetic mean of all the numbers without the first one is 2022, the arithmetic mean of all the numbers without the last one is 2023, and the arithmetic mean of the first and last numbers is 51. What is the sum of all the written numbers?
|
Answer. 202301. Solution. Let the sum of all numbers be $S$, the first number be $a$, and the last number be $b$. According to the conditions, $(S-a) / 100=2022, \quad(S-b) / 100=2023$, $(a+b) / 2=51$, from which we get $S-a=2022 \cdot 100, S-b=2023 \cdot 100, a+b=51 \cdot 2$. Adding these three equations, we get $2 S=4045 \cdot 100+51 \cdot 2$, from which $S=4045 \cdot 50+51=202301$.
- For an answer without justification - 0 points.
|
202301
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In an online store, two types of New Year's gifts are sold. The first type of gift contains a toy, 3 chocolate, and 15 caramel candies and costs 350 rubles. The second type of gift contains 20 chocolate and 5 caramel candies and costs 500 rubles. Eugene wants to buy an equal number of caramel and chocolate candies, and these candies are only sold in the above-mentioned gift sets. What is the smallest non-zero amount of money he will have to spend?
|
Answer: 3750 rubles.
Solution. Consider integer values $m$ and $n$ - the quantities of purchased gift sets of candies of the 1st and 2nd types, respectively. These quantities must satisfy the conditions of the problem: the total number of caramel and chocolate candies in them is the same, and this number is the smallest possible under the conditions. Then, satisfying the conditions of the problem, we form an equation for the unknown quantities: $3 m + 20 n = 15 m + 5 n$
solving which as an equation in integers, we get the ratio $5 n = 4 m$,
from which we obtain a non-zero solution $n = 4 k, m = 5 k, k \in \mathbb{N}$. The smallest solution will be for $k=1$. That is, $m=5, n=4$. Then, the smallest cost of such a quantity of gifts will be $350 m + 500 n = 350 \cdot 5 + 500 \cdot 4 = 3750$ rubles.
Remarks on evaluation. Answer without justification - 0 points. Guessing the values of the quantities of gift sets without justifying the minimality - 2 points. Equation for the quantities of gift sets and candies is formed and the solutions are analyzed - 3 points.
|
3750
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The school dance studio calculated that this year they have already performed the dance "Circle Dance" 40 times, and in each performance, exactly 10 people participated, and any two dancers performed together no more than once. Prove that the studio has at least 60 dancers.
|
Solution. First solution. According to the problem, any two dancers could meet at no more than one performance. We will consider such two dancers as a pair. Consider any performance: from 10 participants of the performance, no more than $\frac{10 \cdot 9}{2}=45$ pairs can be formed (the number of "handshakes," $C_{10}^{2}=45$). Since there were 40 performances in total, from all their participants, no fewer than $45 \cdot 40=1800$ pairs can be formed. Consider a group of 60 dancers. From 60 people, no more than $\frac{60 \cdot 59}{2}=30 \cdot 59=1770$ pairs can be formed, which is less than 1800. Thus, the group has no fewer than 60 dancers.
Second solution. By contradiction. Suppose the studio group consists of $N$ dancers, where $N\frac{400}{60}>6$, which means there is at least one dancer who participated in no fewer than seven performances. Then, according to the problem, in these performances, he met with all their participants exactly once, that is, with no fewer than $7 \cdot 9=63$ different dancers. Contradiction, as there were no more than 60 in total.
Remarks on evaluation. Answer without justification - 0 points. Proof on specific cases - 0 points.
|
60
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
3. Points $O$ and $I$ are the centers of the circumcircle and incircle of triangle $ABC$, and $M$ is the midpoint of the arc $AC$ of the circumcircle (not containing $B$). It is known that $AB=15, BC=7$, and $MI=MO$. Find $AC$.
|
Answer: $A C=13$.
Solution. (Fig. 5). First, we will prove that $M I=M A$ (trident lemma).
Indeed, the external angle $A I M$ of triangle $A I B$ is equal to the sum of angles $B A I$ and $A B I$, and since $A I$ and $B I$ are angle bisectors, $\angle A I M=\frac{1}{2} \angle A+\frac{1}{2} \angle B$. Angle $I A M$ is equal to the sum of angles $I A C$ and $C A M$. But $\angle I A C=\frac{1}{2} \angle A$, and $\angle C A M=\angle C B M=\frac{1}{2} \angle B$ - as inscribed angles.
From this, it follows that $\angle I A M=\frac{1}{2} \angle A+\frac{1}{2} \angle B=\angle A I M$, and therefore, triangle $A M I$ is isosceles, $M I=M A$. By the condition $M O=M I$, so by the trident lemma $A O=M O=M I=M A$. Thus, triangle $A O M$ is equilateral and $\angle A O M=60^{\circ}$. Since the central angle $A O M$ is twice the inscribed angle $A B M$, we have $\frac{1}{2} \angle B=30^{\circ}$, that is, $\angle B=60^{\circ}$.
By the cosine theorem, $A C^{2}=A B^{2}+B C^{2}-2 \cdot A B \cdot B C \cos 60^{\circ}=13^{2}$.
Criteria. Answer without explanation - 0 points. Proved that angle $B$ is $60^{\circ}-5$ points. Points are not deducted for the absence of proof of the trident lemma. Full solution - 7 points.
|
13
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The magpie-crow was cooking porridge, feeding her chicks. The third chick received as much porridge as the first two combined. The fourth chick received as much as the second and third. The fifth chick received as much as the third and fourth. The sixth chick received as much as the fourth and fifth. The seventh chick 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 a grams of porridge, the second - b grams, Then the others received $\mathrm{a}+\mathrm{b}, \mathrm{a}+2 \mathrm{~b}, 2 \mathrm{a}+3 \mathrm{~b}, 3 \mathrm{a}+5 \mathrm{~b}, 0$ grams respectively. In total, they received $8 \mathrm{a}+12 \mathrm{~b}$ grams of porridge, which is 4 times more than what the fifth one received.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.5. Anton guessed a three-digit number, and Lesha is trying to guess it. Lesha sequentially named the numbers 109, 704, and 124. Anton noticed that each of these numbers matches the guessed number in exactly one digit place. What number did Anton guess?
|
Answer: 729.
Solution. Note that the first and third numbers have a common hundreds digit 1, the first and second have a common tens digit 0, and the second and third have a common units digit 4.
Suppose the first digit of the guessed number is 1. Then the first and third numbers no longer have common digits with the guessed number in the same place, so the second digit is neither 0 nor 2, and the third digit is neither 4 nor 9. But then the second number cannot have a common digit with the guessed number in the same place. This leads to a contradiction, so the first digit is not 1.
Reasoning similarly, we find that the second digit is not 0 (otherwise the third number would not have a common digit with the guessed number in the same place), and the third digit is not 4 (otherwise the first number would not have a common digit with the guessed number in the same place).
Thus, the second number can only have the common hundreds digit 7 with the guessed number, the third number can only have the common tens digit 2, and the first number can only have the common units digit 9, i.e., the guessed number is 729.
|
729
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zone - ring or central field - a certain number of points is awarded.)
|
Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38): 2=34
$$
|
34
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.8. In a grove, there are trees of four species: birches, firs, pines, and aspens. There are a total of 100 trees. It is known that among any 85 trees, there will be trees of all four species. Among what minimum number of any trees in this grove will there definitely be trees of at least three species?
|
Answer: 69.
Solution. Suppose there are no more than 15 birches in the grove. Then there are at least 85 other trees, and according to the problem's condition, among them, there must be trees of all four types. This is a contradiction. Therefore, there must be at least 16 birches in the grove. Similarly, we can conclude that there are at least 16 trees of each type.
We will prove that among any 69 trees, there will definitely be three different types of trees. Suppose this is not the case, and among some 69 trees, there are only two different types. Then among the remaining 31 trees, there are trees of the other two types, but as we have proven earlier, each type is represented by at least 16 trees. This is a contradiction.
Now, let's provide an example where among some 68 trees, there are no more than two types (i.e., 68 trees may "not be enough"). Suppose in the grove, there are 34 birches, 34 firs, 16 pines, and 16 aspens. From the previous reasoning, we understand that among any 85 trees, there will be trees of all four types (since among the missing 15 trees, no type can be entirely absent). However, if we take all the birches and firs, a total of 68 trees, then among them, there will be no trees of three types.
## 5th grade
|
69
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.6. Vanya received three sets of candies as a New Year's gift. In the sets, there are three types of candies: lollipops, chocolate, and jelly. The total number of lollipops in all three sets is equal to the total number of chocolate candies in all three sets, as well as the total number of jelly candies in all three sets. In the first set, there are an equal number of chocolate and jelly candies, and 7 more lollipops. In the second set, there are an equal number of lollipops and chocolate candies, and 15 fewer jelly candies. How many candies are in the third set, if it is known that there are no lollipops there?
|
Answer: 29.
Solution. There are more lollipops than jelly candies in the first set by 7, and in the second set by 15. Since the total number of each type of candy is the same in all sets, and there are 0 lollipops in the third set, there must be $7+15=22$ jelly candies in the third set.
Similarly, there are more lollipops than chocolate candies in the first set by 7, and in the second set by 0. Since the total number of each type of candy is the same in all sets, and there are 0 lollipops in the third set, there must be 7 chocolate candies in the third set.
Therefore, the total number of candies in the third set is $22+7+0=29$.
|
29
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.3. Three merchants: Foma, Yerema, and Julius met in Novgorod. If Foma gives Yerema 70 gold coins, then Yerema and Julius will have the same amount of money. If Foma gives Yerema 40 gold coins, then Foma and Julius will have the same amount of money. How many gold coins should Foma give Yerema so that they both have the same amount of money?
|
Answer: 55.
Solution. From the first condition, it follows that Yuliy has 70 more coins than Yeremy. From the second condition, it follows that Foma has 40 more coins than Yuliy. Therefore, Foma has $40+70=110$ more coins than Yeremy. For them to have an equal amount of money, Foma must give Yeremy half of this difference, which is $\frac{110}{2}=55$ coins.
|
55
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.4. In a coastal village, 7 people go fishing every day, 8 people go fishing every other day, 3 people go fishing every three days, and the rest do not fish at all. Yesterday, 12 people went fishing, and today, 10 people are fishing. How many people will go fishing tomorrow?
|
Answer: 15.
Solution. Let's calculate how many times in total they fished yesterday and today. 7 people who fish every day fished 2 times each, i.e., a total of 14 times. 8 people who fish every other day fished exactly 1 time each, i.e., a total of 8 times. Therefore, these 15 people fished a total of $14+8=22$ times over the past two days.
According to the problem, they fished $12+10=22$ times in total over the past two days. Therefore, the fishermen who fish every three days were not present yesterday and today, and they will all come to fish tomorrow. Additionally, 7 people who fish every day will come tomorrow, and $12-7=5$ fishermen who fish every other day will also come. Thus, a total of $3+7+5=15$ people will come to fish tomorrow.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
|
Answer: 29.
Solution. The side of the largest square (with vertex $A$) is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the length of segment $C D$, which is 5. And its side is greater than the side of the smallest square by the length of segment $E F$, which is 13. In total, the side of the largest square is greater than the side of the smallest square by the length of segment $G H$, which is $11+5+13=29$.
|
29
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.6. Four girls and eight boys came to take a class photo. The children approach the photographer in pairs and take a joint photo. Among what minimum number of
photos will there definitely be either a photo of two boys, or a photo of two girls, or two photos of the same children?
|
Answer: 33.
Solution. Suppose at some point there is neither a photo of two boys, nor a photo of two girls, nor two photos of the same children. Then, on each photo, there is a boy and a girl, and on different photos, there are different pairs. However, the total number of possible pairs consisting of a boy and a girl is $4 \cdot 8=32$, and each of these pairs can be captured in no more than one photo. Therefore, among any 33 photos, there must be either a photo of people of the same gender, or two photos of the same pair of people.
At the same time, 32 photos may not be enough: they can capture different pairs consisting of a boy and a girl.
|
33
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.3. The pages in a book are numbered as follows: the first sheet is two pages (with numbers 1 and 2), the second sheet is the next two pages (with numbers 3 and 4), and so on. The hooligan Petya tore out several consecutive sheets from the book: the first torn page has the number 185, and the number of the last torn page consists of the same digits but in a different order. How many sheets did Petya tear out?
|
Answer: 167.
Solution. Since any leaf ends with a page number that is an even number, the number of the last torn-out page is either 158 or 518. But 158 does not work, since 158 < 185. Therefore, the last page ends with the number 518. Now let's calculate the number of torn-out pages. Among the pages from 1 to 518, the pages from 1 to 184 were not torn out. Therefore, 518 - 184 = 334 pages were torn out. There are half as many leaves, which is 334 / 2 = 167.
|
167
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
|
Answer: 52.
Solution. Inside the square, we will write the length of its side. Let the sides of the two squares be $a$ and $b$, and we will sequentially calculate the lengths of the sides of the squares.
The sum of the lengths of the sides of the two squares adjacent to the left side of the rectangle is equal to the sum of the lengths of the sides of the two squares adjacent to the right side of the rectangle. We get the equation
$$
\begin{aligned}
(2 a+b)+(3 a+b) & =(12 a-2 b)+(8 a-b) \\
5 a+2 b & =20 a-3 b \\
b & =3 a
\end{aligned}
$$
Thus, to minimize the perimeter of the rectangle, we need to choose $a=1$, $b=3$. It is easy to check that with these values, the rectangle will have dimensions $11 \times 15$, and its perimeter will be 52.
|
52
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.6. The distance between cities A and B is an integer number of kilometers. Along the road between the cities, there is a signpost every kilometer: on one side is the distance to city A, and on the other side is the distance to city B. Slava walked from city A to city B. During his journey, Slava calculated the GCD of the numbers on each signpost. It turned out that among the calculated GCDs, only the numbers 1, 3, and 13 appeared. What is the distance between the cities?
|
Answer: 39.
Solution. Suppose we are standing next to a sign with numbers $x$ and $y$ written on it. If $\text{GCD}(x, y) = d$, then $(x + y) \vdots d$, i.e., the distance between the cities is divisible by all the calculated GCDs.
Now suppose the distance between the cities (let's call it $S$) is divisible by some natural number $d$. Then at a distance of $d$ kilometers from city A, the numbers on the sign are $d$ and $S - d$, and the GCD of these numbers is exactly $d$.
From all this, we conclude that 1, 3, and 13 are the complete list of divisors of the distance between the cities (excluding the distance itself). Therefore, the distance is 39 kilometers.
|
39
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.8. Carlson and Little Man have several jars of jam, each weighing an integer number of pounds.
The total weight of all the jars of jam that Carlson has is 13 times the total weight of all the jars that Little Man has. Carlson gave Little Man the jar with the smallest weight (from those he had), after which the total weight of his jars turned out to be 8 times the total weight of Little Man's jars.
What is the maximum number of jars of jam that Carlson could have had initially?
|
Answer: 23.
Solution. All variables in the solution will be natural numbers, since the weights of all jars are integers by condition.
Let Little have a total of $n$ pounds of jam initially, then Karlson had $13 n$ pounds of jam. Let Karlson give away his smallest jar with $a$ pounds of jam. Then, by condition, $13 n-a=8(n+a)$, i.e., $5 n=9 a$. Therefore, $9 a$ is divisible by 5, which means $a$ is divisible by 5, i.e., $a=5 k$ for some $k$. Then $5 n=9 \cdot 5 k$, and $n=9 k$. Consequently, initially, Karlson had several jars with a total of $13 \cdot 9 k=117 k$ pounds of jam, with the smallest jar weighing $5 k$ pounds. But then the number of jars Karlson had is no more than $\frac{117 k}{5 k}=23.4$. Therefore, there are no more than 23 jars.
Karlson could have had 23 jars. For example, let him have 1 jar weighing 7 pounds and 22 jars weighing 5 pounds each, and Little have 1 jar weighing 9 pounds. Then all conditions are met: $117=13 \cdot 9$ and $117-5=8 \cdot(9+5)$.
## 8th grade
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.1. A square was cut into four equal rectangles, and from them, a large letter P was formed, as shown in the figure, with a perimeter of 56.
What is the perimeter of the original square
|
Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
From this, we get the equation
$$
\begin{gathered}
28 x=56 \\
x=2
\end{gathered}
$$
The perimeter of the square is $16 x=32$.
|
32
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.2. The numbers from 1 to 9 were placed in the cells of a $3 \times 3$ table such that the sum of the numbers on one diagonal is 7, and on the other - 21. What is the sum of the numbers in the five shaded cells?
|
Answer: 25.
Solution. Note that 7 can be represented uniquely as the sum of numbers from 1 to 9 - this is $1+2+4=7$.
Let's look at the other diagonal with a sum of 21. The largest possible value of the sum in it is $9+8+4=21$ (since the number in the central cell is no more than 4). Therefore, it must contain the numbers $9,8,4$.
Thus, the number 4 is in the central cell, and the numbers 1, 2, 8, and 9 are at the corners. Now it is not difficult to find the sum of the numbers in the shaded cells: $3+4+5+6+7=25$.
|
25
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.5. At the ball, ladies and gentlemen arrived - in total less than 50 people. During the first dance, only a quarter of the ladies were not invited to dance, and 2/7 of the total number of gentlemen did not invite anyone. How many people came to the ball? (For the dance, a certain gentleman invites a certain lady.)
|
Answer: 41.
Solution. Let the number of ladies be $n$, and the number of gentlemen be $m$. We will count the number of pairs who danced. On one hand, it is $\frac{3}{4} n$, and on the other hand, $\frac{5}{7} m$. By equating, we find the ratio
$$
\frac{n}{m}=\frac{20}{21}
$$
Since the fraction on the right side is irreducible, we get that $n=20 k$ and $m=21 k$ for some natural number $k$. It remains to note that the total number of people $20 k +$ $21 k=41 k$ is less than 50, so only $k=1$ fits, and the total number of people at the ball is $20+21=41$.
|
41
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
|
Answer: 17.
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute-angled ($\angle ABD = \angle CBD < 90^\circ$, $\left.\angle BAD = \angle ADB = \frac{180^\circ - \angle ABD}{2} < 90^\circ\right)$, point $H$ lies on the segment $AB$.
Notice that the right triangles $BDH$ and $BDC$ are equal by the common hypotenuse $BD$ and the acute angle at vertex $B$. Therefore, $BH = BC$ and $DH = CD$.
Now, notice that the right triangles $ADH$ and $EDC$ are also equal by the hypotenuse $AD = ED$ and the leg $DH = CD$. Therefore, $EC = AH$.
Thus, $BD = BA = BH + AH = BC + EC = (7 + 5) + 5 = 17$.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.7. For three real numbers $p, q$, and $r$, it is known that
$$
p+q+r=5, \quad \frac{1}{p+q}+\frac{1}{q+r}+\frac{1}{p+r}=9
$$
What is the value of the expression
$$
\frac{r}{p+q}+\frac{p}{q+r}+\frac{q}{p+r} ?
$$
|
Answer: 42.
Solution. Multiply the two given equalities, we get
$$
5 \cdot 9=\frac{p+q+r}{p+q}+\frac{p+q+r}{q+r}+\frac{p+q+r}{p+r}=\left(1+\frac{r}{p+q}\right)+\left(1+\frac{p}{q+r}\right)+\left(1+\frac{q}{p+r}\right)
$$
Subtract 3 from both sides of the equation, we get
$$
\frac{r}{p+q}+\frac{p}{q+r}+\frac{q}{p+r}=42
$$
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.8. Masha wrote on the board in ascending order all natural divisors of some number $N$ (the very first divisor written is 1, the largest divisor written is the number $N$ itself). It turned out that the third from the end divisor is 21 times greater than the second from the beginning. What is the largest value that $N$ can take?
|
Answer: 441.
Solution. The second divisor from the beginning is the smallest prime divisor of the number $N$, let's denote it as $p$. The third divisor from the beginning is either $p^{2}$ or the second largest prime divisor of the number $N$, let's denote it as $q$.
Case 1. The third divisor from the beginning is $p^{2}$. Then the third divisor from the end is $\frac{N}{p^{2}}$. According to the problem,
$$
\begin{aligned}
& \frac{N}{p^{2}}=21 p \\
& N=21 p^{3}
\end{aligned}
$$
It is clear that 3 and 7 are divisors of the number $N$, so $p \leqslant 3$. If $p=2$, then the third largest divisor of the number $N$ is 3; if $p=3$, then the third largest divisor of the number $N$ is no more than 7, i.e., not $3^{2}$. Contradiction.
Case 2. The third divisor from the beginning is $q$. Then the third divisor from the end is $\frac{N}{q}$. According to the problem,
$$
\begin{aligned}
\frac{N}{q} & =21 p \\
N & =21 p q
\end{aligned}
$$
It is clear that 3 and 7 are divisors of the number $N$, so $p \leqslant 3, q \leqslant 7$. From this, we get that $N \leqslant 441$. It is not difficult to check that this number satisfies the condition.
## 9th grade
|
441
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.3. To 30 palm trees in different parts of an uninhabited island, a sign is nailed.
- On 15 of them, it is written: "Exactly under 15 signs, treasure is buried."
- On 8 of them, it is written: "Exactly under 8 signs, treasure is buried."
- On 4 of them, it is written: "Exactly under 4 signs, treasure is buried."
- On 3 of them, it is written: "Exactly under 3 signs, treasure is buried."
It is known that only the signs under which there is no treasure are true.
Under what minimum number of signs can the treasure be buried?
|
Answer: 15.
Solution. Suppose the treasure is not buried under at least 16 plaques. Then there are two plaques with different inscriptions under which there is no treasure. According to the condition, the inscriptions on both should be true, but they contradict each other. Contradiction.
Therefore, the treasure is not buried under more than 15 plaques. Thus, there are at least 15 plaques under which there is treasure. It is not difficult to come up with a suitable example: let the treasures be buried only under plaques with one of the following three phrases.
- "The treasure is buried under exactly 8 plaques."
- "The treasure is buried under exactly 4 plaques."
- "The treasure is buried under exactly 3 plaques."
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
|
Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =(A+B)-(B+C+D)+(D+E+F)-(F+G)= \\
& =12^{2}-9^{2}+7^{2}-3^{2}=103
\end{aligned}
$$
|
103
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.5. Buratino has many coins of 5 and 6 soldi, more than 10 of each type. Coming to the store and buying a book for $N$ soldi, he realized that he could not pay for it without receiving change. What is the greatest value that the natural number $N$ can take if it is no more than 50?
|
Answer: 19.
Solution. It is easy to check that with $N=19$ change is necessary.
Notice that numbers from 20 to 24 do not meet the condition, as they can be paid for without change: $N=20=4 \cdot 5, 21=3 \cdot 5+6, 22=2 \cdot 5+2 \cdot 6, 23=5+3 \cdot 6, 24=4 \cdot 6$.
It is clear that then numbers from 25 to 50 also do not meet the condition, as each of them is the sum of one or more terms equal to 5, as well as a number from 20 to 24, which is the sum of terms, each equal to 5 and 6.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.6. At a ball, 29 boys and 15 girls arrived. Some boys danced with some girls (no more than once in each pair). After the ball, each person told their parents how many times they danced. What is the maximum number of different numbers the children could have mentioned?
|
Answer: 29.
Solution. The largest possible number that could have been named is 29 (in the case where there is a girl who danced with all the boys), and the smallest is 0. Thus, we have that the number of different numbers named is no more than 30.
We will prove that it cannot be exactly 30. Suppose the opposite, that all numbers from 0 to 29 were named. Then the numbers from 16 to 29 were named only by girls (this is already 14 numbers). The number 0 was definitely not named by a boy, since there is a girl who danced with all the boys. This means that the girls must have named exactly fifteen numbers: 0, 16, 17, 18, ..., 29.
At the same time, the number 15 was not named by any of the girls, since it is already known which numbers they named. If the number 15 was named by a boy, it would mean that he danced with all the girls. But there is a girl who named the number 0, who did not dance with anyone. Contradiction.
Now we will show how 29 different numbers could have been named. Let's number the girls from 1 to 15 and the boys from 1 to 29. Let boy $i$ dance with girl $j$ if and only if $i \geqslant j$.
We can illustrate this example with a $15 \times 29$ table: we will assign a column of the table to each boy and a row to each girl; if a boy danced with a girl, we will color the cell at the intersection of the corresponding row and column. In Fig. 4, above each column is written how many cells are colored in it; to the left of each row is written how many cells are colored in it.
Fig. 4: example for the solution to problem 9.6
|
29
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
|
Answer: 16.
Solution. Divide the $5 \times 5$ square without the central cell into four $2 \times 3$ rectangles, and each of these into two $1 \times 3$ rectangles.
We get 8 rectangles $1 \times 3$, the sum of the numbers in each of which is 23. Since the sum of all the numbers is 200, we find the number in the central cell as $200 - 23 \cdot 8 = 16$.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 10.4. In the central cell of a $21 \times 21$ board, there is a chip. In one move, the chip can be moved to an adjacent cell by side. Alina made 10 moves. How many cells can the chip end up in?
|
Answer: 121.
Solution. We will paint the entire board in a checkerboard pattern so that the central cell of the board is black. With each move of the chip to an adjacent cell, the color of the cell on which the chip is standing will change. After an odd number of moves, the chip will always be on a white cell, and after an even number of moves - on a black cell. Therefore, after 10 moves, the chip will definitely be on a black cell.
We will show that to all black cells that can be reached in no more than 10 moves, it is possible to reach them exactly in 10 moves. Consider an arbitrary black cell $A$ that can be reached in fewer than 10 moves from the center. Since by the time the chip reaches cell $A$, an even number of moves, less than 10, has been made, one can simply move the chip to an adjacent cell and back until exactly 10 moves have been made. Therefore, we need to count the number of black cells that can be reached in no more than 10 moves.
In 0 moves, you can only reach the starting cell, in 2 moves you can reach the starting cell and $4 \cdot 2=8$ new cells, in 4 moves you can reach the cells that have already been reached and $4 \cdot 4=16$ new cells, and so on. We get that the number of cells that can be reached in no more than 10 steps is $1+8+16+24+\ldots+40=$ $1+8(1+2+3+4+5)=121$
Remark. The centers of the cells that can be reached in $k \leqslant 10$ steps form the vertices of a square grid rotated by $45^{\circ}$ (Fig. 6). From this, it is clear that the total number of such cells is $11^{2}$.
|
121
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.7. Oleg has four cards, on each of which natural numbers are written on both sides (a total of 8 numbers are written). He considers all possible quadruples of numbers, where the first number is written on the first card, the second on the second, the third on the third, and the fourth on the fourth. Then, for each quadruple, he writes down the product of the numbers in his notebook. What is the sum of the eight numbers on the cards if the sum of the sixteen numbers in Oleg's notebook is $330 ?$
|
Answer: 21.
Solution. Let the numbers on one card be $a$ and $b$, on another card - $c$ and $d$, on the third card - $e$ and $f$, and on the fourth card - $g$ and $h$. According to the problem, the sum of 16 terms of the form $a c e g + a c e h + \ldots + b d f h$ equals 330. Note that this sum is also obtained by expanding all the brackets in the expression $(a+b)(c+d)(e+f)(g+h)$.
Therefore, $(a+b)(c+d)(e+f)(g+h)=330=2 \cdot 3 \cdot 5 \cdot 11$. Since all numbers are natural, each bracket is greater than 1. Hence, the brackets are equal to the numbers $2, 3, 5$, and 11 in some order. Then their sum is $a+b+c+d+e+f+g+h=2+3+5+11=21$.
|
21
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
|
Answer: 35.
Solution. Let's start with the assumption that the quadrilateral KMDС is inscribed (several proofs of this fact will be proposed below).
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\angle M D C}{2}=45^{\circ}$. The angle $A M K$ as an external angle for triangle $K D M$ is equal to the sum of angles $M K D$ and $K D A$, so the required angle КDA is $80^{\circ}-45^{\circ}=35^{\circ}$.
Fig. 8: to the solution of problem 10.8
Let's provide the first possible proof of the inscribed nature of quadrilateral KMDC. Consider triangle $M K C$ and its circumscribed circle. Note that point $D$ lies on the bisector of angle $M K C$, and is equidistant from vertices $M$ and $C$ (Fig. 8). However, the bisector of an angle of a non-isosceles triangle and the perpendicular bisector of its opposite side, as is known, intersect at the midpoint of the "smaller" arc of the circumscribed circle of the triangle. In other words, $D$ is the midpoint of the arc $M C$ of the circumscribed circle of triangle $M K C$, not containing point $K$. It should also be noted that $M K \neq K C$ (otherwise triangles $K M D$ and $K C D$ would be equal, but $\angle K M D>90^{\circ}>\angle K C D$ ).
Let's provide the second possible proof of the inscribed nature of quadrilateral KMDC. It will use the fourth criterion for the equality of triangles: if two sides and an angle not between them are equal in two triangles, then these triangles are either equal or the sum of the other two angles not between them is $180^{\circ}$. The fourth criterion is satisfied for triangles $M D K$ and $C D K (M D=D C, D K$ - common, $\angle M K D=\angle C K D)$. However, angles $K M D$ and $K C D$ are not equal (again, the first is obtuse, and the second is acute), so their sum is $180^{\circ}$, which are the opposite angles of quadrilateral KMDC. Therefore, it is inscribed.
## 11th grade
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.1. Inside a circle, 16 radii of the circle and 10 concentric circles, whose centers coincide with the center of the circle, are drawn. Into how many regions do the radii and circles divide the circle?
|
Answer: 176.
Solution. 10 circles divide the circle into 10 rings and one smaller circle, a total of 11 parts. 16 radii divide each of the 11 parts into 16 more. In total, there are $11 \cdot 16=176$ regions.
|
176
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.2. Along a road, 25 poles are standing in a row. Sometimes a goldfinch lands on one of the poles, and immediately a goldfinch flies away from one of the adjacent poles (if there was at least one goldfinch on the adjacent poles at that moment). Also, no more than one goldfinch can sit on each pole.
Initially, there are no birds on the poles. What is the maximum number of goldfinches that can be on the poles at the same time?
|
Answer: 24.
Solution. First, we will show that it is impossible to occupy all the poles. Suppose this happened. Consider the last tit that sat down. Since it occupied the last unoccupied pole, there must have been an occupied pole next to it. Consequently, the tit sitting on that pole must have flown away. Contradiction.
Now, let's provide an example of how the tits could occupy 24 poles. For convenience, let's number all the poles in order. Let the first tit sit on the first pole. Suppose all poles from 1 to $k$ are occupied, where $k \leqslant 23$. We will show how to achieve a situation where all poles from 1 to ( $k+1$ ) are occupied. By sequentially changing $k$ from 1 to 23, we will obtain a situation where the first 24 poles are occupied.
Suppose all poles from 1 to $k$ are occupied, then the next tit sits on the $(k+2)$-th pole (where $k+2 \leqslant 25$), and the next tit after that sits on the $(k+1)$-th pole, and the tit from the $(k+2)$-th pole flies away. We obtain a situation where all poles from 1 to ( $k+1$ ) are occupied.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.3. A natural number $n$ is called interesting if $2 n$ is a perfect square, and $15 n$ is a perfect cube. Find the smallest interesting number.
Answer: 1800.
|
Solution. Let's factorize the number $n$ into prime factors. For a number to be a square, it is necessary that in this factorization all prime numbers appear in even powers, and for a number to be a cube, it is necessary that all prime numbers appear in powers divisible by 3.
Let's see what power of two divides $n$. First, this power is odd, since $2 n$ is a perfect square. Second, this power is divisible by 3, since $15 n$ is a perfect cube. Therefore, the minimum power of two is 3.
Now let's see what power of three divides $n$. First, this power is even, since $2 n$ is a perfect square. Second, this power gives a remainder of 2 when divided by 3, since $15 n$ is a perfect cube. Therefore, the minimum power of three is 2.
Similarly, for five, we get that its minimum power is 2.
Therefore, $n$ is divisible by $2^{3} 3^{2} 5^{2}=1800$, i.e., $n \geqslant 1800$. It is not difficult to check that $n=$ 1800 satisfies all the conditions of the problem.
|
1800
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.4. Senya has three straight sticks, each 24 centimeters long. Senya broke one of them into two pieces so that he could form the outline of a right-angled triangle with the two pieces of this stick and the two whole sticks. How many square centimeters is the area of this triangle?
|
Answer: 216.
Solution. Let Senya break one of the sticks into parts of lengths $a$ and $24-a$, which we will call small, while sticks of length 24 will be called large. From four sticks, Senya formed a triangle, so one of the sides consists of two sticks, and the other two sides consist of one stick each.
Let's see which two sticks can be put together. If we put together two small sticks, we get an equilateral triangle, but we need a right triangle. If we put together two large sticks into one side, it will be greater than the sum of the other two sides (which contradicts the triangle inequality). The only remaining option is to combine a large and a small stick, and the other two sides will be just a small stick and a large stick (and these will be the legs, as they are shorter than the first side).
Fig. 9: to the solution of problem 11.4
Thus, the lengths of the sides of the right triangle are $a, 24$, and $48-a$ (Fig. 9). From the Pythagorean theorem, we get $a^{2}+24^{2}=(48-a)^{2}$. Transforming this equation, we find that $a=18$. Therefore, the area of the triangle is $\frac{18 \cdot 24}{2}=216$.
|
216
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.5. At the call of the voivode, 55 soldiers arrived: archers and swordsmen. All of them were dressed either in golden or black armor. It is known that swordsmen tell the truth when wearing black armor and lie when wearing golden armor, while archers do the opposite.
- In response to the question "Are you wearing golden armor?", 44 people answered affirmatively.
- In response to the question "Are you an archer?", 33 people answered affirmatively.
- In response to the question "Is it Monday today?", 22 people answered affirmatively.
How many archers in golden armor came at the voivode's call?
|
Answer: 22.
Solution. To the first question, affirmative answers will be given by archers in gold armor and archers in black armor, that is, all archers.
To the second question, affirmative answers will be given by archers in gold armor and swordsmen in gold armor, that is, all soldiers in gold armor.
To the third question, affirmative answers will be given either by swordsmen in gold armor and archers in black armor (if it is not Monday), or by swordsmen in black armor and archers in gold armor (if it is Monday). In the first case (if it is not Monday), we will sum the number of affirmative answers to all three questions. This will result in double the number of swordsmen in gold armor plus double the number of archers, i.e., an even number, but according to the condition, it equals \(22+33+44=99\) and is odd, which is a contradiction. Therefore, to the third question, affirmative answers were given by swordsmen in black armor and archers in gold armor.
Now, let's sum the number of affirmative answers. This will result in triple the number of archers in gold armor plus the number of all others (once each). Then, if we subtract the total number of soldiers from this number, we get double the number of archers in gold armor, which needs to be divided by two:
\[
\frac{22+33+44-55}{2}=22
\]
|
22
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3. Find the length of the segment $O D_{1}$.
|
Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ to this face, then point $X$ is the center of $\omega$ (point $O$ is equidistant from all points of $\omega$, so the projection of $O$ onto the plane of $\omega$ is also equidistant from them). Let $Y$ be an arbitrary point on $\omega$ (Fig. 10). Triangle $OXY$ is a right triangle; by the problem's condition, $X Y=3$ and $O Y=10$. By the Pythagorean theorem, we get $O X^{2}=10^{2}-3^{2}=91$.
Similarly, we find the squares of the distances from point $O$ to the planes $A_{1} B_{1} C_{1} D_{1}$ and $A D D_{1} A_{1}$, which are both equal to $10^{2}-1^{2}=99$.
By the spatial Pythagorean theorem, the square of the length of segment $O D_{1}$ is equal to the sum of the squares of the distances from point $O$ to the three faces containing point $D_{1}$. Therefore, $O D_{1}^{2}=$ $91+99+99=289$, from which $O D_{1}=17$.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 11.7. For what least natural $x$ is the expression
$$
\sqrt{29+\sqrt{x}}+\sqrt{29-\sqrt{x}}
$$
an integer?
|
Answer: 400.
Solution. If we square the given expression, we get $58+2 \sqrt{29^{2}-x}$. If the initial expression was an integer, then this number is a perfect square. We need to find the smallest $x$, i.e., we need to make this expression the largest possible perfect square. Notice that $58+2 \sqrt{29^{2}-x}<58+2 \sqrt{29^{2}}=116$. The largest square less than 116 is 100. Solving the equation $58+2 \sqrt{29^{2}-x}=100$, we get $x=400$. It is not difficult to verify that when $x=400$, the original expression is indeed an integer.
|
400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. There were 2013 empty boxes. In one of them, 13 new boxes (not nested inside each other) were placed. Thus, there became 2026 boxes. In one of them, 13 new boxes (not nested inside each other) were placed again, and so on. After several such operations, there became 2013 non-empty boxes. How many boxes are there in total? Answer: 28182.
|
# Solution.
After each operation, the number of non-empty boxes increases by 1. Initially, there were no non-empty boxes. Therefore, a total of 2013 operations were performed. With each operation, 13 new boxes are added. Thus, we have
$2013 + 13 \times 2013 = 2013 \times 14 = 28182$.
Recommendations for checking.
Answer only or calculations without comments: 1 point.
|
28182
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. An electronic clock displays time from 00.00.00 to 23.59.59. How much time during a day does the clock display exactly four digit 3s?
|
3. If the display shows ab.cd.mn, then $a \neq 3$, so there are 5 cases where one of the digits $b, c, d, m, n$ is not 3, while the others are 3. a) On the display ab.33.33, where $b \neq 3$. There are 21 such sets.
b) On the display a3.c3.33, where c is not 3. Here a can take any of the three values 0, 1, or 2, and c can take any of the five values 0, 1, 2, 4, or 5. There are $3 \cdot 5 = 15$ such sets.
c) On the display a3.3d.33, where $d \neq 3$. Here $a=0,1,2, d=0,1,2,4,5,6,7,8,9$, for a total of $3 \cdot 9 = 27$ sets.
d) a3.33.m3 - 15 sets.
e) a3.33.3n - 27 sets.
In total, $21 + 15 + 27 + 15 + 27 = 105$ sets, each of which is displayed for 1 second.
Answer: 105 seconds.
|
105
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.8. A straight rod 2 meters long was sawn into $N$ sticks, the length of each of which is expressed as an integer number of centimeters. For what smallest $N$ can it be guaranteed that, using all the resulting sticks, one can, without breaking them, form the contour of some rectangle?
(A. Magazinov)
|
Answer. $N=102$.
Solution. First solution. Let $N \leqslant 101$. Cut the stick into $N-1$ sticks of 1 cm each and one stick of $(201-N)$ cm. It is impossible to form a rectangle from this set, as each side of the rectangle is less than half the perimeter, and thus the stick of length $201-N \geqslant 100$ cm cannot be part of any side. Therefore, $N \geqslant 102$.
We will show that for $N=102$, the desired rectangle can be found. Note that among all the sticks, there will be two sticks of 1 cm each. Indeed, if this were not the case, the total length of the sticks would be at least $2 \cdot 101 + 1 = 203$ cm, which is incorrect.
Set aside these two sticks. Let the lengths of the remaining sticks be $a_{1}, a_{2}, \ldots, a_{100} \mathrm{~cm}$, then we have $a_{1}+a_{2}+\ldots+a_{100} = 198$. Among the 100 numbers $A_{1}=a_{1}, A_{2}=a_{1}+a_{2}, A_{3}=a_{1}+a_{2}+a_{3}, \ldots, A_{100}=a_{1}+a_{2}+\ldots+a_{100}$, there will be two that give the same remainder when divided by 99. Let these be $A_{k}$ and $A_{\ell}, k<\ell$. The number $A_{\ell}-A_{k}$ is strictly greater than zero and strictly less than 198, and it is divisible by 99. Thus, $A_{\ell}-A_{k}=99=a_{k+1}+a_{k+2}+\ldots+a_{\ell}$.
Thus, we have found several sticks with a total length of $99 \mathrm{~cm}$. Set these aside as well. The remaining sticks also have a total length of 99 cm. Therefore, we can form a rectangle of $1 \times 99$ cm.
Second solution. We will provide another proof that for $N=102$, it is possible to form a rectangle.
Let the lengths of the sticks in the set, expressed in centimeters, be $a_{1}, a_{2}, \ldots, a_{102}$. We have $a_{1}+a_{2}+\ldots+a_{102}=200$. Consider a circle of length 200 and divide it into 102 red points, forming arcs of lengths $a_{1}, a_{2}, \ldots, a_{102}$. These points are some 102 vertices of a regular 200-gon $T$ inscribed in this circle. The vertices of $T$ are paired into opposite pairs. There are 100 such pairs, and 102 red points, so among the red points, there will be two pairs of opposite points. These two pairs of points divide the circle into two pairs of equal arcs. Thus, we have divided all the sticks into four groups $A, B, C, D$, where the total lengths in groups $A$ and $C$, as well as in groups $B$ and $D$, are equal. Therefore, it is possible to form a rectangle using each group to form one of its sides.
Comment. Only the correct answer -0 points.
An example is provided showing that $N \geqslant 102-1$ point. It is proven that $N=102$ works, but its minimality is not justified -5 points.
|
102
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4 In the thirtieth kingdom, there are three types of coins in circulation: bronze rubles, silver coins worth 9 rubles, and gold coins worth 81 rubles. From the treasury, which contains an unlimited supply of each type of coin, a certain amount was issued with 23 coins, which is less than 700 rubles. Find this amount, given that it is impossible to issue it with a smaller number of coins.
|
Solution. Since the issued amount is less than 700 rubles, the number of gold coins is less than $700: 81$, which means no more than 7. Then the number of bronze and silver coins together is at least $23-7=16$. Note that the number of bronze coins cannot exceed 8: otherwise, we can replace 9 bronze coins with 1 silver coin, and the method of issuing the amount would not be unique. Similarly, the number of silver coins is no more than 8. Therefore, there are exactly 8 of each, and 7 gold coins. The issued amount (in rubles) is $8 \cdot 1 + 8 \cdot 9 + 7 \cdot 81 = 647$.
Answer: 647 rubles.
| INCLUDED IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| The correct answer is provided and it is shown that the amount of 647 rubles can be issued in a unique way. However, the uniqueness of this number is not demonstrated. | 4 points |
| It is proven that the number of bronze (or silver) coins issued is no more than 8 | 3 points |
| Correct but unexplained answer | 1 point |
|
647
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5 In some cells of a $10 \times 10$ table, crosses are placed such that each of them is the only one either in its row or in its column. What is the maximum number of crosses that can be in such a table? Justify your answer.
|
Solution. We will place crosses in all cells of the first row and the first column, excluding the top-left cell - a total of 18 crosses. Then each cross in the first row is unique in its column, and each cross in the first column is unique in its row. Therefore, it is possible to place 18 crosses.
We will show that if we place 19 crosses (or more), some cross will not be unique in its row or column. Assume the opposite. We will call rows and columns simply lines. We will assign to each cross the line in which it is unique (if there are two such lines, then any one of them). Then different crosses are assigned different lines; there are 19 crosses and 20 lines, so all lines, except one, are assigned to some cross, meaning they contain exactly one cross. The only unassigned line contains no more than 10 crosses (more cannot fit on it). The total number of pairs cross - line, where the line contains the cross, will be no more than $19+10=29$. But in reality, there are $19 \cdot 2=38$ - a contradiction.
Answer: 18 crosses.
| PRESENT IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| Correct proof that 19 crosses cannot be placed in the table (in the absence of proof that 19 can) | 4 points |
| Correct example of placing 18 crosses (or proof of the existence of such an example) | 2 points |
| Examples of placing fewer than 18 crosses and/or proofs that there cannot be 20 or more crosses | 0 points |
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On the sides $B C$ and $A D$ of the convex quadrilateral $A B C D$, their midpoints - points $M$ and $N$ respectively - are marked. Segments $M N$ and $A C$ intersect at point $O$, and $M O=O N$. It is known that the area of triangle $A B C$ is 2017. Find the area of quadrilateral $A B C D$.
|
Answer: 4034.
## Solution:
Let $A C$ and $M N$ intersect at point $O$ (see the first figure on the right), $S_{\triangle A B C}=2017=S$.
We will prove that $S_{\triangle A D C}=S_{\triangle A B C}$, then $S_{A B C D}=2 S$.
There are various ways to reason.
First method.
Draw $A M$ and $C N$ (see the first figure). Since $A O$ is the median of triangle $A M N$, then $S_{\triangle А o N}=S_{\triangle А o м}$. Similarly, $S_{\triangle C O N}=S_{\triangle С o м}$.
Therefore, $S_{\triangle A N C}=S_{\triangle A M C}$. Since $A M$ and $C N$ are medians of triangles $A B C$ and $A D C$ respectively, then $S_{\triangle A D C}=2 S_{\triangle A N C}=2 S_{\triangle A M C}=S_{\triangle A B C}=S$.
Second method.
Let $K$ and $L$ be the midpoints of sides $A B$ and $C D$, then $K M L N$ is a parallelogram, so segment $K L$ contains point $O$ (see the second figure).
Since $K L$ is the midline of triangle $A B C$, then $S_{\triangle K O M}=\frac{1}{4} S_{\triangle A B C}=\frac{1}{4} S$.
Similarly, $S_{\triangle N O L}=\frac{1}{4} S_{\triangle A D C}$. Since $S_{\triangle N O L}=S_{\triangle K O M}$, then $S_{\triangle A D C}=S_{\triangle A B C}=S$.
Note that $K M L N$ is the Varignon parallelogram, the area of which is half the area of $A B C D$.
Third method.
We use the fact that in any convex quadrilateral $A B C D$, the midpoints $M$ and $N$ of opposite sides and the midpoints $P$ and $Q$ of the diagonals are the vertices of a parallelogram (see the third figure).
Therefore, the midpoint $O$ of segment $M N$ is also the midpoint of segment $P Q$.
Since in our case point $O$ lies on diagonal $A C$, then the midpoint $Q$ of diagonal $B D$ also lies on $A C$ (see the fourth
figure).
$$
\begin{aligned}
& \text { Then } S_{\triangle A D Q}=S_{\triangle A B Q} \text { and } S_{\triangle C D Q}=S_{\triangle C B Q} \cdot \\
& \text { Therefore, } S_{\triangle A D C}=S_{\triangle A B C}=S .
\end{aligned}
$$
## Criteria:
A correct answer without a correct justification is not scored. For proving the equality of areas of separate elements of the triangle without further correct progressions (without a complete solution), no more than 2 points in total should be given.
|
4034
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a box, there are balls of seven colors. One tenth of the balls are red, one eighth are orange, and one third are yellow. There are 9 more green balls than red ones, and 10 more blue balls than orange ones. There are 8 blue balls in the box. The remaining balls are purple. What is the smallest possible number of purple balls?
|
Solution. Let the total number of balls be x, and the number of violet balls be y. Then
$$
\frac{x}{10}+\frac{x}{8}+\frac{x}{3}+\frac{x}{10}+9+\frac{x}{8}+10+8+y=x
$$
From which $47x / 60 + 27 + y = x$, that is, $y = 13x / 60 - 27$, and the smallest value of y is achieved at the smallest value of x. Since y is an integer, x is a multiple of 60. According to the condition, x must be a multiple of 8, so x is divisible by 120. For x = 120, y = -1, which is impossible. For x = 240, y = 25 and this is the smallest possible number of violet balls.
Answer. 25 balls.
|
25
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.1.1. Dima and Seryozha were picking berries from a raspberry bush, on which 900 berries grew. Dima alternated his actions while picking: he put one berry in the basket and ate the next one. Seryozha also alternated: he put two berries in the basket and ate the next one. It is known that Dima picks berries twice as fast as Seryozha. At some point, the boys had picked all the raspberries from the bush.
Who of them ended up putting more berries in the basket? What will the difference be?
|
Answer: Dima, 100 berries more.
Solution. While Dima picks 6 berries, Seryozha manages to pick only 3. At the same time, Dima puts only 3 out of his 6 berries into the basket, while Seryozha puts only 2 out of his 3.
This means that among every 6+3=9 berries picked, Dima puts exactly 3 into the basket, and Seryozha puts exactly 2. Then, Dima will put a total of $\frac{3}{9} \cdot 900=300$ berries into the basket, and Seryozha will put $\frac{2}{9} \cdot 900=200$ berries. Therefore, Dima will put $300-200=100$ more berries into the basket.
|
100
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Variant 8.1.3. Dima and Seryozha were picking berries from a raspberry bush, on which 450 berries grew. Dima alternated his actions while picking: he put one berry in the basket and ate the next one. Seryozha also alternated: he put two berries in the basket and ate the next one. It is known that Dima picks berries twice as fast as Seryozha. At some point, the boys had picked all the raspberries from the bush.
Who of them ended up putting more berries in the basket? What will the difference be?
|
Answer: Dima, 50 berries more.
|
50
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Variant 8.1.4. Dima and Seryozha were picking berries from a raspberry bush, on which 450 berries grew. Seryozha alternated actions while picking berries: he put one berry in the basket and ate the next one. Dima also alternated: he put two berries in the basket and ate the next one. It is known that Seryozha picks berries twice as fast as Dima. At some point, the boys had picked all the raspberries from the bush.
Who of them ended up putting more berries in the basket? What will the difference be?
|
Answer: Sergey, 50 berries more.
|
50
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.2.1. Given trapezoid $A B C D(A D \| B C)$. It turns out that $\angle A B D=\angle B C D$. Find the length of segment $B D$, if $B C=36$ and $A D=64$.
|
Answer: 48.
Solution. Since $A D \| B C$, we have $\angle C B D=\angle B D A$. Then triangles $A B D$ and $D C B$ are similar by the first criterion. Therefore, $\frac{64}{B D}=\frac{A D}{B D}=\frac{B D}{B C}=\frac{B D}{36}$, from which we find $B D=\sqrt{64 \cdot 36}=48$.
|
48
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.3.1. As a homework exercise, Tanya was asked to come up with 20 examples of the form $*+*=*$, where $*$ should be replaced with different natural numbers (i.e., a total of 60 different numbers should be used). Tanya loves prime numbers very much, so she decided to use as many of them as possible, while still ensuring the examples are correct. What is the maximum number of prime numbers Tanya can use?
|
# Answer: 41.
Solution. Note that in each example, instead of asterisks, three odd numbers cannot be used, i.e., at least one even number must be used. There is exactly one even prime number, which is 2. Therefore, among the 60 different numbers in the examples, at least 19 even composite numbers will be used, and there can be no more than 41 primes.
As is known, there are infinitely many prime numbers. We will show how examples can be constructed so that exactly 41 prime numbers are involved.
- $2+3=5$ - 3 prime numbers are used;
- $7+11=18$ - 2 prime numbers are used;
- $13+17=30$ - 2 prime numbers are used;
- $19+23=42$ - 2 prime numbers are used;
- in each subsequent example, two consecutive prime numbers are used in the left part.
All 60 numbers in these 20 examples will be different: all odd numbers are different primes, and even numbers, except for 2, are sums that increase when moving from the previous example to the next, and therefore cannot repeat.
|
41
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.4.1. A rectangle was cut into six smaller rectangles, the areas of five of them are marked on the diagram. Find the area of the remaining rectangle.
|
Answer: 101.
Solution. Let's introduce the notation as shown in the figure.
Notice that
$$
2=\frac{40}{20}=\frac{S_{H I L K}}{S_{D E I H}}=\frac{H K \cdot H I}{H D \cdot H I}=\frac{H K}{H D} \quad \text { and } \quad 2=\frac{126}{63}=\frac{S_{A B H G}}{S_{G H K J}}=\frac{B H \cdot G H}{H K \cdot G H}=\frac{B H}{H K}
$$
Therefore, if $H D=x$, then $H K=2 x$, and $B H=4 x$. From this, it follows immediately that $B D=3 x=D K$.
We obtain that the rectangles $B C F D$ and $D F M K$ are equal, as are their areas. Thus,
$$
S_{E F M L}=S_{D F M K}-S_{D E I H}-S_{H I L K}=161-20-40=101
$$
|
101
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.6.1. Even natural numbers $a$ and $b$ are such that $\operatorname{GCD}(a, b) + \operatorname{LCM}(a, b) = 2^{23}$. How many different values can $\operatorname{LCM}(a, b)$ take?
|
Answer: 22.
Solution. Note that LCM $(a, b) \vdots$ GCD $(a, b)$, therefore
$$
2^{23}=\text { GCD }(a, b)+\text { LCM }(a, b) \vdots \text { GCD }(a, b) .
$$
From this, it follows that GCD $(a, b)$ is a natural divisor of the number $2^{23}$.
At the same time, GCD $(a, b) \neq 1$ (since $a$ and $b-$ are even numbers), and GCD $(a, b) \neq 2^{23}$ (since GCD $\left.(a, b)=2^{23}-\operatorname{LCM}(a, b)<2^{23}\right)$. Thus, GCD $(a, b)$ takes one of the values $2,2^{2}, 2^{3}, \ldots, 2^{22}$.
Note that all these values are possible: to make GCD $(a, b)$ equal to $2^{k}$ for $1 \leqslant k \leqslant 22$, it is sufficient to choose $a=2^{k}$ and $b=2^{23}-2^{k}$. In this case, $b \vdots a$, so GCD $(a, b)=a$ and LCM $(a, b)=b$, that is, GCD $(a, b)+$ LCM $(a, b)=a+b=2^{23}$.
|
22
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.8.1. Different positive numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
a^{2}+b c=115 \\
b^{2}+a c=127 \\
c^{2}+a b=115
\end{array}\right.
$$
Find $a+b+c$.
|
Answer: 22.
Solution. Subtract the third equation from the first and transform:
$$
\begin{gathered}
\left(a^{2}+b c\right)-\left(c^{2}+a b\right)=0 \\
a^{2}-c^{2}+b c-a b=0 \\
(a-c)(a+c)+b(c-a)=0 \\
(a-c)(a+c-b)=0
\end{gathered}
$$
By the condition $a \neq c$, therefore $b=a+c$. Now add the first two equations:
$$
\begin{gathered}
\left(a^{2}+b c\right)+\left(b^{2}+a c\right)=115+127 \\
\left(a^{2}+b c\right)+\left(b^{2}+a c\right)=242
\end{gathered}
$$
Substitute $b=a+c$ into the resulting equation:
$$
\begin{aligned}
242 & =a^{2}+b c+b^{2}+a c=a^{2}+(a+c) c+(a+c)^{2}+a c= \\
& =a^{2}+a c+c^{2}+(a+c)^{2}+a c=2(a+c)^{2}
\end{aligned}
$$
from which $(a+c)^{2}=121$. Since the numbers $a$ and $c$ are positive and $b=a+c$, we get that $a+c=11$ and $a+b+c=22$.
|
22
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.9. In a class, there are $m$ students. During September, each of them went to the swimming pool several times; no one went twice on the same day. On October 1st, it turned out that all the numbers of swimming pool visits by the students were different. Moreover, for any two of them, there was definitely a day when the first one was at the pool and the second one was not, and a day when, conversely, the second one was at the pool and the first one was not. Find the maximum possible value of $m$. (September has 30 days.)
|
Answer. $m=28$.
Solution. For each natural $n$, let $X_{n}=$ $=\{1,2, \ldots, n\}$. To each student, we associate the set of all days when he went to the pool (this will be a subset of $X_{30}$). Thus, we have obtained a set of $m$ (according to the condition, non-empty) subsets of $X_{30}$. The condition is equivalent to the fact that all subsets have different numbers of elements, and none of them is contained in another; we will call such a set of subsets $x o r o-$ good. Thus, we need to find the maximum number of sets in a good family of subsets of $X_{30}$.
First, let's prove that such a set cannot contain more than 28 sets. This is obvious if the set contains a 30-element subset, since it contains any other. Therefore, we can assume that the sets in the set can consist only of $1,2, \ldots, 29$ elements (and there are no more than 29 of them). Suppose that in a good set there is a 29-element set $A$ and a 1-element set $B$. Since $B$ is not contained in $A$, they do not intersect. Then any other subset in $X_{30}$ either contains $B$ or is contained in $A$. Therefore, in this case, the good set consists only of two subsets. Finally, if the set does not contain a 1 or 29-element subset, then it already contains no more than 28 sets, which is what we needed.
It remains to provide an example of a good set of 28 subsets in $X_{30}$. For this, we will show by induction on $k \geqslant 2$ that there exists a good set $A_{1}, A_{2}, \ldots, A_{2 k-2}$ of subsets in $X_{2 k}$, and $A_{i}$ contains $i+1$ elements. In the base case $k=2$, the subsets $A_{1}=\{1,2\}$ and $A_{2}=\{1,3,4\}$ work.
Suppose that for some $k$ a required good set $B_{1}, \ldots, B_{2 k-2}$ of subsets in $X_{2 k}$ has already been constructed. Then the required good set of subsets in $X_{2 k+2}$ can be constructed as follows. Let $A_{i+1}=B_{i} \cup\{2 k+2\}$ for $i=1,2, \ldots, 2 k-2$; these sets contain $3,4, \ldots, 2 k$ elements, respectively. Finally, let $A_{1}=\{2 k+1,2 k+2\}$ and $A_{2 k}=\{1,2, \ldots, 2 k+1\}$. It is not difficult to check that they form the required good set. Thus, the induction step is proved.
Remark. The arguments in the second paragraph of the solution show that in a good set of subsets in $X_{n}$ there are no more than $n-2$ sets if $n \geqslant 4$. On the other hand, acting similarly to the second half of the solution, one can construct a good set of $2 k-1$ subsets in $X_{2 k+1}$ for $k \geqslant 1$; the base of induction is provided by the subset $A_{1}=\{1,2\}$.
One can also arrange a direct induction transition that allows constructing a good set of $n-1$ subsets in $X_{n+1}$ from a good set of $n-2$ subsets in $X_{n}$ (for $n \geqslant 5$). For such a transition, the following consideration is useful: if you take the complements of all sets of a good set, you will also get a good set.
Comment. Only the answer - 0 points.
Proved that the number of students in the class does not exceed $29-$ 0 points.
Provided an example with no more than 27 students - 0 points.
Proved only that the number of students in the class does not exceed $28-2$ points
Only provided an example with 28 students - 4 points.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Variant 1. Given two natural numbers. One number was increased by 3, and the other was decreased by 3. As a result, their product increased by 600. By how much will the product decrease if the opposite is done: the first number is decreased by 3, and the second is increased by 3?
|
Answer: 618.
Solution: Let these numbers be $a$ and $b$. Then, according to the condition, $(a-3)(b+3)-ab=600$. Expanding the brackets: $ab+3a-3b-9-ab=600$, so $a-b=203$. We need to find the difference $ab-(a+3)(b-3)=$ $ab-ab+3a-3b+9=3(a-b)+9=3 \cdot 203+9=618$.
|
618
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Variant 1. At the intersection of perpendicular roads, a highway from Moscow to Kazan and a road from Vladimir to Ryazan intersect. Dima and Tolya set out with constant speeds from Moscow to Kazan and from Vladimir to Ryazan, respectively. When Dima passed the intersection, Tolya had 900 meters left to reach it. When Tolya passed the intersection, Dima was 600 meters away from it. How many meters will be between the boys when Tolya has traveled 900 meters from the moment he crossed the intersection?
|
Answer: 1500.
Solution: When Tolya has traveled 900 meters, Dima will have traveled 600 meters, so at the moment when Tolya is 900 meters from the intersection, Dima will be 1200 meters from the intersection. According to the Pythagorean theorem, the distance between the boys is 1500 meters.
|
1500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Variant 1. Petya and Masha take candies from a box in turns. Masha took one candy, then Petya took 2 candies, Masha - 3 candies, Petya - 4 candies, and so on. When the number of candies in the box became less than needed for the next turn, all the remaining candies went to the one whose turn it was to take candies. How many candies did Petya get if Masha got 101 candies
|
Answer: 110.
Solution. Since $1+3+5+7+9+11+13+15+17+19=100101$, then Masha got the last candy. Then Petya took for himself $2+4+6+8+10+12+14+16+18+20=2(1+2+3+4+5+6+7+8+9+10)=110$ candies.
|
110
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Variant 1. An ant, starting from point A, goes $1+\frac{1}{10}$ cm north, then $2+\frac{2}{10}$ cm west, then $3+\frac{3}{10}$ cm south, then $4+\frac{4}{10}$ cm east, then $5+\frac{5}{10}$ cm north, then $6+\frac{6}{10}$ cm west, and so on. After 1000 steps, the ant is at point B. Find the distance between points A and B in centimeters in a straight line. In the answer, write the square of the found distance.
|
Answer: 605000.
Solution. Let's divide 1000 steps into quartets. After each quartet, the ant will move southeast relative to its current position, by a distance equal to the diagonal of a square with side 2.2, i.e., $\sqrt{2.2^{2}+2.2^{2}}=\sqrt{9.68}$. After 250 such quartets, the ant will be at a distance of $250 \cdot \sqrt{9.68}$ from point A. The square of this distance is $250^{2} \cdot 9.68=605000$.
|
605000
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Variant 1. Grisha thought of such a set of 10 different natural numbers that their arithmetic mean is 16. What is the maximum possible value of the largest of the numbers he thought of?
|
Answer: 115.
Solution: The sum of the given numbers is $10 \cdot 16=160$. Since all the numbers are distinct, the sum of the 9 smallest of them is no less than $1+2+\cdots+9=45$. Therefore, the largest number cannot be greater than $160-45=115$. This is possible: $(1+2+\cdots+9+115): 10=16$.
|
115
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Variant 1. Roma decided to create his own multiplication table. The rows correspond to the numbers 12, $13,14, \ldots, 60$, and the columns - to the numbers $15,16,17, \ldots, 40$. In the cells of the table, he wrote the products of the pairs of numbers from the row and column. How many of these products will be even numbers?
|
Answer: 962.
Solution. Note that the product of two numbers is odd if and only if both factors are odd, and even in all other cases. In total, the table contains 49$\cdot$26 products. Note that among the numbers from 12 to 60, there are 24 odd numbers, and among the numbers from 15 to 40, there are 13 odd numbers. Therefore, there will be $24 \cdot 13$ odd products in the table. The remaining $49 \cdot 26-24 \cdot 13=962$ will be even.
|
962
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Variant 1. In triangle $A B C$, the bisector $A L$ and the median $B M$ are drawn. It turns out that $A B=2 B L$. What is the measure of angle $B C A$, if $\angle L M A=127^{\circ}$?
|
Answer: $74^{\circ}$.
Solution.
By the property of the angle bisector, $\frac{A C}{C L}=\frac{A B}{B L}=\frac{2}{1}$, so $A C=2 C L$. Since $M$ is the midpoint of $A C$, then $A M=$ $M C=C L$. In the isosceles triangle $C M L$, the angle $C$ is $180^{\circ}-2 \angle C M L=180^{\circ}-2\left(180^{\circ}-\right.$ $\angle L M A)=2 \angle L M A-180^{\circ}=74^{\circ}$.
|
74
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. For the Day of the Russian Flag, the seller decided to decorate the store window with 10 horizontal strips of fabric in three colors. At the same time, he follows two conditions:
1) strips of the same color should not hang next to each other;
2) each blue strip must hang between a white and a red one.
In how many ways can he do this?
|
5. Let's call the white and red strips of fabric green. Denote by $T_{n}$ the number of ways to decorate the shop window with $n$ strips of fabric. The first strip can only be a green strip. If the second strip is blue, then the following strips must be such that the first one is green. The total number of ways to hang the strips in this manner will be $T_{n-2}$. If the second strip is also green, then together with it, there will be $n-1$ remaining strips, which can be hung in $T_{n-1}$ ways. Thus, we get:
$$
T_{n}=T_{n-1}+T_{n-2}(*)
$$
Obviously, for $n=1$, $T_{1}=2$ (one strip can be either white or red), and for $n=2$, $T_{2}=2$ (two strips can be either white and red or red and white). Then all subsequent $T_{n}$ are found using the formula (*):
$$
\begin{gathered}
T_{3}=T_{2}+T_{1}=2+2=4, T_{4}=T_{3}+T_{2}=4+2=6, T_{5}=T_{4}+T_{3}=6+4=10, \\
T_{6}=T_{5}+T_{4}=10+6=16, T_{7}=T_{6}+T_{5}=16+10=26, T_{8}=T_{7}+T_{6}=26+16=42, \\
T_{9}=T_{8}+T_{7}=42+26=68, T_{10}=T_{9}+T_{8}=68+42=110 .
\end{gathered}
$$
|
110
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Let $p(x)=2 x^{3}-3 x^{2}+1$. How many squares of integers are among the numbers $p(1), p(2), \ldots, p(2016)$?
|
# Answer: 32.
## Solution:
Notice that $p(x)=(x-1)^{2}(2 x+1)$.
For an integer $x(1 \leq x \leq 2016)$, the number $p(x)=(x-1)^{2}(2 x+1)$ is a perfect square of an integer either when $x=1(p(1)=0)$, or (for $x \geq 2$) when the number $2 x+1$ is a perfect square.
Note that for $x \geq 2$, the inequality holds: $5 \leq 2 x+1 \leq 4033$, and also that $2 x+1$ is odd.
Such squares are the numbers 9 (for $x=4$), 25 (for $x=12$), 49 (for $x=24$), and so on up to $63^{2}=3969$ (for $x=1984$) - that is, we have considered the squares of odd numbers from $3^{2}$ to $63^{2}$ (a total of 31). Together with the previously found square at $x=1$, we get 32 perfect squares among the numbers of the specified form.
|
32
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The letters А, Б, К, М, П, У, Ш were encoded with sequences of zeros and ones (each with its own). Then, in the word ПАПАМАМАБАБУШКА, the letters were replaced with their codes. Could the length of the resulting sequence be shorter than 40 characters, if the sequence can be uniquely decoded?
保留源文本的换行和格式，所以翻译结果如下：
5. The letters А, Б, К, М, П, У, Ш were encoded with sequences of zeros and ones (each with its own). Then, in the word ПАПАМАМАБАБУШКА, the letters were replaced with their codes. Could the length of the resulting sequence be shorter than 40 characters, if the sequence can be uniquely decoded?
|
Answer: She could.
## Solution:
Here is an example of a code table:
| A | Б | К | М | П | У | Ш |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 110 | 1111 | 100 | 101 | 11100 | 11101 |
The word will look like this:
10101010100010001100110111001110111110 - a total of 38 characters. Decoding is unambiguous. To verify this, start from the left edge. The far left can only be the code 101, as no other code starts with the same combination of symbols. Next, similarly, as no code starts with a combination of symbols that matches any other code.
## Comment:
A uniform code (3 symbols per letter) gives 45 characters. Therefore, it is necessary to use codes of variable length for different letters. The more frequently a letter appears, the shorter its code should be. This consideration can help with verification. Be careful.
Participants can justify the unambiguousness of decoding not by sequentially examining the code sequence (from left to right), but by referring to a correct fact about such codes. The given code is an example of a prefix code - the code of any letter is not the beginning of the code of any other letter (Fano's condition), and for it, the theorem of unambiguous decoding is true. A reference to the corresponding theorem can be counted as justification (naturally, provided that it is correctly formulated).
## Instructions for Checking:
Score - either 0 if the answer is incorrect or the code table does not give the desired result, or 4 points if the codes are correct but there is no justification for the unambiguousness of decoding, or 7 points.
|
38
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. In the train, there are 18 identical cars. In some cars, exactly half of the seats are free, in some others - exactly one third of the seats are free, and in the rest, all seats are occupied. At the same time, in the entire train, exactly one ninth of all seats are free. In how many cars are all seats occupied?
|
Answer: in 13 carriages.
Solution. Let the number of passengers in each carriage be taken as a unit. We can reason in different ways.
First method. Since exactly one ninth of all seats in the train are free, this is equivalent to two carriages being completely free. The number 2 can be uniquely decomposed into the sum of thirds and halves: $2=3 \cdot \frac{1}{3}+2 \cdot \frac{1}{2}$. This means that there are free seats in five carriages, so in 13 carriages all seats are occupied.
Second method. Let $x$ be the number of carriages where half of the seats are free, and $y$ be the number of carriages where a third of the seats are free. Then $x \cdot \frac{1}{2}+y \cdot \frac{1}{3}=18 \cdot \frac{1}{9}$. Eliminating the denominators of the fractions, we get: $3 x+2 y=12$. Since $x$ and $y$ are natural numbers, by trial and error, we find that $x=2, y=3$. Therefore, the required number of carriages is: $18-2-3=13$.
Natural solutions to the equation $3 x+2 y=12$ can also be found from considerations of divisibility if one variable is expressed in terms of the other.
Evaluation criteria.
“+” A complete and justified solution is provided
“士” A correct reasoning is provided, but a computational error is made
“さ” A generally correct reasoning and the correct answer are provided, but the uniqueness of the solution to the equation or the uniqueness of the decomposition of 2 into a linear combination of fractions $1 / 2$ and $1 / 3$ is not explained.
“Ғ” The equation is correctly set up, but further reasoning is incorrect or absent
“干” Only the correct answer is provided
“-” The problem is not solved or is solved incorrectly
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. In grandmother's garden, apples have ripened: Antonovka, Grushovka, and White Naliv. If there were three times as many Antonovka apples, the total number of apples would increase by $70 \%$. If there were three times as many Grushovka apples, it would increase by $50 \%$. By what percentage would the total number of apples change if there were three times as many White Naliv apples?
|
Answer: increased by $80 \%$.
Solution. First method. If the amount of each type of apple were three times as much, the total number of apples would increase by $200 \%$. Of this, $70 \%$ is due to the increase in Antonovka, and $50 \%$ is due to the increase in Grushovka. Therefore, the increase due to White Naliv would be $200 \%-70 \%-50 \%=80 \%$.
Second method. Since adding twice the amount of Antonovka results in a $70 \%$ increase, Antonovka makes up 0.35 of all apples. Similarly, Grushovka makes up 0.25 of all apples. Therefore, the share of White Naliv is 0.4. If you add 0.4 twice to a number, the number will increase by $80 \%$.
Grading criteria.
“+” A complete and justified solution is provided
“士” A correct reasoning is provided, but a computational error is made
“Ғ” Only the correct answer is provided or it is obtained by considering a specific example
“-” The problem is not solved or is solved incorrectly
|
80
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.6. In each cell of a $5 \times 5$ square, exactly one diagonal has been drawn. A vertex of a cell is free if it is not the endpoint of any of the drawn diagonals. Find the maximum possible number of free vertices.
|
Answer: 18 vertices.
Solution. Example. See Fig. 7.6a. On each of the six horizontal lines, three vertices are free.
Estimate. The total number of cell vertices: $6 \cdot 6=36$. Let's select nine cells that do not share any vertices (see Fig. 7.6b). They contain all 36 vertices. In each cell, a diagonal is drawn, so in each cell, two
Fig. 7.6b
vertices are not free. Therefore, at least 18 vertices are used, so no more than $36-18=18$ can be free.
Grading Criteria.
“+” A complete and justified solution is provided
“Ғ” The correct answer and a correct example are provided, but the estimate is not made
“干” The correct answer is provided and the estimate is proven, but the example is missing or incorrect
“-” Only the answer is provided
“-” The problem is not solved or is solved incorrectly
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Five consecutive natural numbers were written on the board, and then one number was erased. It turned out that the sum of the remaining four numbers is 2015. Find the smallest of these four numbers.
|
Answer: 502
Solution. If the smallest number is K, then the sum of all numbers is not less than $\mathrm{K}+(\mathrm{K}+1)+(\mathrm{K}+2)+ (\mathrm{K}+3)=4 \mathrm{~K}+6$ and not more than $\mathrm{K}+(\mathrm{K}+2)+(\mathrm{K}+3)+(\mathrm{K}+4)=4 \mathrm{~K}+9$. Therefore, $4 \mathrm{~K}+6 \leq 2015 \leq 4 \mathrm{~K}+9$, from which $\mathrm{K}=502$.
|
502
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. What can the value of the expression $p^{4}-3 p^{3}-5 p^{2}+16 p+2015$ be if $p$ is a root of the equation $x^{3}-5 x+1=0$?
Answer: 2018
|
Solution: $\mathrm{p}^{4}-3 \mathrm{p}^{3}-5 \mathrm{p}^{2}+16 \mathrm{p}+2015=\left(\mathrm{p}^{3}-5 \mathrm{p}+1\right)(\mathrm{p}-3)+2018$. Since $\mathrm{p}$ is a root of the polynomial in the first parenthesis, the entire expression equals 2018.
|
2018
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. There are 400 students in a school. For New Year, each student sent 200 greetings to other students. What is the minimum number of pairs of students who could have greeted each other? Answer: 200.
|
Solution: The total number of greetings was $200 * 400=80000$. And the number of different pairs of students is $-400 * 399 / 2=79800$, which is 200 less than the number of greetings. Therefore, at least 200 pairs had 2 greetings. We can show that exactly 200 pairs can be: Let the 1st student greet everyone from the 2nd to the 201st, the 2nd student greet from the 3rd to the 202nd, ... the 200th student greet everyone from the 201st to the 400th, and then the 201st student greet everyone from the 202nd to the 400th and the 1st, the 202nd student greet from the 203rd to the 400th and from the 1st to the 2nd, and so on. Then the paired greetings will only be between students whose numbers differ by exactly 200.
5. Several schoolchildren have 128 identical tokens among them. They play a game according to the following rule: if someone has no less than half of all the tokens, then each of the others takes from this player as many tokens as they already have (if there are two such players, they choose by lot). Seven such exchanges occurred. Prove that now all 128 tokens are with one schoolchild.
Solution: Note that the number of tokens for those who take tokens doubles. Therefore, after the first transfer of tokens, they had a number of tokens that is a multiple of 2, but then the first player also had a number of tokens that is a multiple of two, since their total number is even. Similarly, after the second transfer of tokens, everyone will have a number of tokens that is a multiple of 4. After the 7th exchange, everyone will have a number of tokens that is a multiple of 128, that is, either 0 or 128.
|
200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. A merchant bought several bags of salt in Tver and sold them in Moscow with a profit of 100 rubles. With all the money earned, he again bought salt in Tver (at the Tver price) and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he spend on the first purchase?
|
Answer: 500 rubles.
First method. Let the cost of a kilogram of salt in Tver be $x$ rubles, and in Moscow $-y$ rubles, and let the merchant buy $a$ kg of salt the first time. Then, according to the condition, $a(y-x)=100$.
The revenue amounted to $ay$ rubles, so the second time the merchant was able to buy $\frac{ay}{x}$ kg of salt. In this case, the profit was $\frac{ay}{x} \cdot y - ay = \frac{ay(y-x)}{x}$ rubles. According to the condition, $\frac{ay(y-x)}{x}=120$.
From the two equations obtained, it follows that $\frac{100y}{x}=120$, that is, $y=\frac{6}{5}x$. Substituting this result into the first equation, we get $ax=500$.
Second method. Let the merchant pay $x$ rubles for the salt during the first purchase in Tver. Then he sold it in Moscow for $x+100$ rubles. The second time he spent $x+100$ rubles in Tver, and received $x+100+120=x+220$ rubles in Moscow. Since the ratio of Moscow and Tver prices did not change, we set up the proportion: $\frac{x}{x+100}=\frac{x+100}{x+220}$. Solving the equation, we get $x=500$.
+ complete justified solution
$\pm$ correct solution with minor gaps or unclear places
干 correct answer obtained based on reasoning with specific numerical data
干 correct answer provided and only verified that it satisfies the condition
- only the answer is provided or the answer with unclear calculations without explanations
If the student assumes that the same number of whole bags was always bought and considers the price of one bag, the grade is not reduced.
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Let $p(x)=2 x^{3}-3 x^{2}+1$. How many squares of integers are among the numbers $p(1), p(2), \ldots$, $p(2016) ?$
#
|
# Answer: 32.
## Solution:
Notice that $p(x)=(x-1)^{2}(2 x+1)$.
For an integer $x(1 \leq x \leq 2016)$, the number $p(x)=(x-1)^{2}(2 x+1)$ is a perfect square of an integer either when $x=1(p(1)=0)$, or (for $x \geq 2$) when the number $2 x+1$ is a perfect square.
Note that for $x \geq 2$, the inequality holds: $5 \leq 2 x+1 \leq 4033$, and also that $2 x+1$ is odd.
Such squares are the numbers 9 (for $x=4$), 25 (for $x=12$), 49 (for $x=24$), and so on up to $63^{2}=3969$ (for $x=1984$) - that is, we have considered the squares of odd numbers from $3^{2}$ to $63^{2}$ (a total of 31). Together with the previously found square at $x=1$, we get 32 perfect squares among the numbers of the specified form.
|
32
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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