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2. A road 28 kilometers long was divided into three unequal parts. The distance between the midpoints of the extreme parts is 16 km. Find the length of the middle part.
|
Answer: 4 km.
Solution. The distance between the midpoints of the outermost sections consists of half of the outer sections and the entire middle section, i.e., twice this number equals the length of the road plus the length of the middle section. Thus, the length of the middle section $=16 * 2-28=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In five 15-liter buckets, there are 1, 2, 3, 4, and 5 liters of water respectively. It is allowed to triple the amount of water in any container by pouring water from one other container (if there is not enough water to triple the amount, then it is not allowed to pour from this bucket). What is the maximum amount of water that can be collected in one bucket using such actions?
|
# Answer
The answer depends on the interpretation of the condition -- whether it is allowed to pour NOT all the contents of the bucket (essentially -- whether it is possible to measure OUT ONE LITER of water)
A) If it is not allowed, then the answer is 9 liters
B) If it is allowed, then the answer is 12 liters
## Solution:
Variant $A$ )
We will show how to collect 9 liters in one of the buckets:
$1,2,3,4,5=>1,6,3,0,5=>1,0,9,0,5$.
## Variant B)
We will show how to collect 12 liters in one of the buckets:
$1,2,3,4,5=>1,6,3,4,1=>1,0,9,4,1=>1,0,1,12,1$
In any case, it is necessary to prove that this is the maximum number (that it is impossible to get MORE).
Let the maximum number of liters be $n>=9$.
Consider the last operation with this bucket (at least one operation was performed
--- otherwise $n<=5$ ).
Since $n$ is the maximum number of liters, the last operation could not have involved pouring from this bucket
(otherwise, there would have been more before), i.e., it was filled, so $n$ is a multiple of 3.
There are 15 liters in all the buckets combined.
Note that after each step, there is a non-empty bucket, the number of liters in which is a multiple of three. (solution)
Assume that $n=15$.
Then, on the previous step, there were exactly two non-empty buckets, one with 5 and the other with 10 liters.
But neither of these numbers is a multiple of 3. Contradiction with (*)
## Variant A)
Assume that $n=12$.
Then, on the previous step, there should have been two non-empty buckets: 4 and 8 liters.
Then, due to condition (*), the remaining 3 liters should have been in one bucket.
But then, one step earlier, the amount of water in the buckets should have been 1, 2, 4, 8, which contradicts the condition (*).
Thus, $n=9$, an example of how to get 9 liters is provided above.
Variant B) Thus, $n=12$, an example of how to get 12 liters is provided above.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter Г on the board, at least one colored cell is found?
|
Solution: Consider a $2 \times 3$ rectangle. In it, obviously, a minimum number of cells need to be colored. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown in the figure.
Answer: 12.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Two painters are painting a 15-meter corridor. Each of them moves from the beginning of the corridor to its end and starts painting at some point until the paint runs out. The first painter has red paint, which is enough to paint 9 meters of the corridor; the second has yellow paint, which is enough for 10 meters.
The first painter starts painting when he is two meters from the beginning of the corridor; and the second finishes painting when he is one meter from the end of the corridor. How many meters of the corridor are painted with exactly one layer?
|
Answer: 5 meters.
Solution. The first painter starts painting 2 meters from the beginning of the corridor and finishes at $2+9=11$ meters from the beginning of the corridor.
The second painter finishes painting 1 meter from the end of the corridor, which is 14 meters from the beginning of the corridor, and starts at $14-10=4$ meters from the beginning of the corridor (Fig. 1).
Fig. 1: to the solution of problem 2
Thus, there are two sections of the corridor that are painted with one layer:
- the section from 2 meters (from the beginning of the corridor) to 4 meters (from the beginning of the corridor) was painted only by the first painter (length of the section 2 meters);
- the section from 11 meters (from the beginning of the corridor) to 14 meters (from the beginning of the corridor) was painted only by the second painter (length of the section 3 meters).
This results in $3 \mathrm{m}+2 \mathrm{m}=5 \mathrm{m}$.
## Criteria
3 p. The correct answer is obtained.
4 6. The correct answer and justification are present.
The justification is considered to be a "picture" showing the sections painted by the painters.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. As is known, balance scales come to equilibrium when the weight on both pans is the same. On one pan, there are 9 identical diamonds, and on the other, 4 identical emeralds. If one more such emerald is added to the diamonds, the scales will be balanced. How many diamonds will balance one emerald? The answer needs to be justified.
|
Answer: 3 diamonds.
Solution. From the condition of the problem, it follows that 9 diamonds and 1 emerald weigh as much as 4 emeralds. Thus, if we remove one emerald from each side of the scales, the equality will be preserved, that is, 9 diamonds weigh as much as 3 emeralds. This means that 3 diamonds weigh as much as 1 emerald.
## Criteria
1 p. The correct answer is obtained.
4 p. The correct answer and justification are present.
If the work mentions (or illustrates) three cases of balance on the scales:
- 9 diamonds and 1 emerald balance 4 emeralds,
- 9 diamonds balance 3 emeralds,
- 3 diamonds balance 1 emerald,
then this is considered a correct justification.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On the faces $BCD, ACD, ABD$, and $ABC$ of the tetrahedron $ABCD$, points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are marked, respectively. It is known that the lines $AA_{1}, BB_{1}, CC_{1}$, and $DD_{1}$ intersect at point $P$, and $\frac{AP}{A_{1}P}=\frac{BP}{B_{1}P}=\frac{CP}{C_{1}P}=\frac{DP}{D_{1}P}=r$. Find all possible values of $r$.
|
Solution. Let V be the volume* of tetrahedron ABCD. We introduce the consideration of the partition of the original tetrahedron ABCD into tetrahedra PBCD, PACD, PABD, and PABC. Then for the volumes of the specified tetrahedra, the following is true:
$$
V=V_{\mathrm{PBCD}}+V_{\mathrm{PACD}}+V_{\mathrm{PABD}}+V_{\mathrm{PABC}}
$$
Note that if $\mathrm{H}$ is the height of tetrahedron $\mathrm{ABCD}$, dropped from vertex A to the base BCD, and $\mathrm{h}$ is the height of tetrahedron PBCD, dropped from vertex P to the base BCD, then $\frac{H}{h}=\frac{A A_{1}}{A_{1} P}$. Due to this observation and the condition $r=\frac{A P}{A_{1} P}=\frac{B P}{B_{1} P}=\frac{C P}{C_{1} P}=\frac{D P}{D_{1} P}$, we have
$$
\frac{V}{V_{\mathrm{PBCD}}}=\frac{\mathrm{A} A_{1}}{A_{1} P}=\frac{A P+A_{1} P}{A_{1} P}=1+\frac{A P}{A_{1} P}=r+1
$$
Similar relations are valid for the other tetrahedra, so
$$
\frac{v}{V_{\mathrm{PBCD}}}=\frac{v}{V_{\mathrm{PACD}}}=\frac{v}{V_{\mathrm{PABD}}}=\frac{v}{V_{\mathrm{PABC}}}=r+1 \text { or }
$$
$V=(r+1) V_{\mathrm{PBCD}}=(r+1) V_{\mathrm{PACD}}=(r+1) V_{\mathrm{PABD}}=(r+1) V_{\mathrm{PABC}}$.
Then
$$
r+1=\frac{(r+1) V_{\mathrm{PBCD}}+(r+1) V_{\mathrm{PACD}}+(r+1) V_{\mathrm{PABD}}+(r+1) V_{\mathrm{PABC}}}{V_{\mathrm{PBCD}}+V_{\mathrm{PACD}}+V_{\mathrm{PABD}}+V_{\mathrm{PABC}}}
$$
or
$$
r+1=\frac{4 V}{V_{\mathrm{PBCD}}+V_{\mathrm{PACD}}+V_{\mathrm{PABD}}+V_{\mathrm{PABC}}}=\frac{4 V}{V}=4
$$
Thus, $r=3$.
*Note that for solving this problem, it is not necessary to know the formula for the volume of a tetrahedron. It is sufficient to introduce a conditional quantity $V=S h$, and use its additivity (a known fact from mathematics and physics courses).
## Recommendations for evaluating solutions.
A correct solution for a specific case - 2 points.
The partition of the original tetrahedron ABCD into tetrahedra PBCD, PACD, PABD, and PABC is presented, and relations of the form $\frac{V}{V_{\text {PBCD }}}=\frac{\mathrm{AA}_{1}}{A_{1} P}=\frac{A P+A_{1} P}{A_{1} P}=1+\frac{A P}{A_{1} P}=r+1$ are obtained - no less than 3 points.
Note that the problem can be solved using the vector method.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 7.1. (7 points)
Find the value of the expression
$$
\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \ldots\left(1+\frac{1}{2 m}\right)\left(1-\frac{1}{2 m+1}\right)
$$
|
Answer: 1.
Solution: Notice that
$$
\begin{array}{r}
\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=\frac{3}{2} \cdot \frac{2}{3}=1,\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\frac{5}{4} \cdot \frac{4}{5}=1, \ldots \\
\left(1+\frac{1}{2 m}\right)\left(1-\frac{1}{2 m+1}\right)=\frac{2 m+1}{2 m} \cdot \frac{2 m}{2 m+1}=1
\end{array}
$$
Therefore, the value of the expression is 1.
#
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 7.2. (7 points)
Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon and, without stopping, arrived: one - in $B$ at 4 PM, and the other - in $A$ at 9 PM. At what hour was dawn that day?
|
Answer: at 6 o'clock.
Solution: Let $x$ be the number of hours from dawn to noon. The first pedestrian walked $x$ hours before noon and 4 after, the second - $x$ before noon and 9 after. Note that the ratio of times is equal to the ratio of the lengths of the paths before and after the meeting point, so $\frac{x}{4}=\frac{9}{x}$. From this proportion, we find that $x=6$.
Indeed, let the speed of the first pedestrian (walking from $A$ to $B$) be $V_{1}$, and the speed of the second pedestrian (walking from $B$ to $A$) be $V_{2}$. Then before the meeting point, the first pedestrian walked $V_{1} \cdot x$ hours, and the second $-V_{2} \cdot x$. After noon, the first pedestrian walked $V_{1} \cdot 4$, and the second $-V_{2} \cdot 9$. But after noon, the first pedestrian walked as much as the second did before noon. And vice versa, after noon, the second pedestrian walked as much as the first did before noon. We have:
$$
\begin{aligned}
& V_{1} \cdot x=V_{2} \cdot 9, \frac{V_{1}}{V_{2}}=\frac{9}{x} \\
& V_{1} \cdot 4=V_{2} \cdot x, \frac{V_{1}}{V_{2}}=\frac{x}{4}
\end{aligned}
$$
From this, $\frac{x}{4}=\frac{9}{x}$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 7.5. (7 points)
In a family, there are six children. Five of them are older than the youngest by 2, 6, 8, 12, and 14 years, respectively. How old is the youngest if the ages of all the children are prime numbers?
|
Answer: 5 years.
Solution. First, let's check the prime numbers less than six. It is obvious that the number we are looking for is odd. The number 3 does not satisfy the condition because $3+6=9-$ is not a prime number. The number 5 satisfies the condition because $5+2=7, 5+6=11, 5+8=13, 5+12=17, 5+14=19$, which means the ages of all the children are indeed prime numbers.
Let's prove that this answer is unique. For this, consider the remainders when the numbers added to the age of the youngest child are divided by 5: the remainder of 2 is 2, the remainder of 6 is 1, the remainder of 8 is 3, the remainder of 12 is 2, and the remainder of 14 is 4. Thus, all remainders from 1 to 4 are present. Therefore, if the age of the youngest child is not a multiple of five, adding a number with a remainder that complements it to five will result in a number that is a multiple of five and greater than five, which is a composite number.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The second term of an infinite decreasing geometric progression is 3. Find the smallest possible value of the sum $A$ of this progression, given that $A>0$.
|
Answer: 12.
Solution. Let the first term of the progression be $a$, and the common ratio be $q$. The sum of the progression $A$ is $\frac{a}{1-q}$. From the condition, we have $3=a q$, from which $a=3 / q$. Therefore, we need to find the minimum value of $A=\frac{3}{q(1-q)}$. Note that from the condition it follows: $0<q<1$. The number $A$ takes its minimum value when the denominator of the fraction $q(1-q)$ is maximized, but $\left(-q^{2}+q\right)$ has a maximum point at $q=1 / 2$, and the maximum value of $q(1-q)$ is $1 / 4$. Then the minimum possible value of $A=3 / 1 / 4=12$.
Comment. The sum $A$ is represented as a function of one variable - 3 points.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.6. The sheriff believes that if the number of bandits he catches on a certain day is a prime number, then he is lucky. On Monday and Tuesday, the sheriff was lucky, and starting from Wednesday, the number of bandits he caught was equal to the sum of the number from the day before yesterday and twice the number from yesterday. What is the maximum number of consecutive days the sheriff could have been lucky this week? Justify your answer.
|
Solution: Let's say the sheriff caught 7 bandits on Monday and 3 on Tuesday. Then on Wednesday, Thursday, and Friday, he caught 13, 29, and 71 bandits, respectively. All these numbers are prime, and the sheriff has been lucky for five days in a row.
We will show that the sheriff cannot be lucky for six days in a row. Note that by Wednesday, this number is no less than $2 \cdot 2 + 3 = 7$, and it only increases from there. Therefore, it is sufficient to show that on Wednesday, Thursday, Friday, or Saturday, there will be a day when the number of bandits caught is divisible by 3.
Assume the opposite, that on all these days, the number of bandits caught is not divisible by 3. In particular, the numbers of bandits caught on Wednesday and Thursday do not divide by 3; let's denote them as $A$ and $B$ respectively. Then the number of bandits caught on Friday is $A + 2B$. There are two possible situations.
1) The remainders of $A$ and $B$ when divided by 3 are the same. Then the number $A + 2B = A + B + B$ is the sum of three numbers with the same remainder when divided by 3, so it is divisible by 3. Contradiction.
2) The remainders of $A$ and $B$ when divided by 3 are different: one is 1, the other is 2. Then the number $A + B$ is divisible by 3, so the number $A + 2B = (A + B) + B$ has the same remainder when divided by 3 as the number $B$. Therefore, the numbers of bandits caught on Thursday and Friday have the same remainder when divided by 3. Reasoning as in point 1, we get that the number of bandits caught on Saturday is divisible by 3. Again, a contradiction. The proof is complete.
Answer: 5 days.
Note: The mathematical model of the problem is as follows: Given a sequence $\left\{a_{n}\right\}$ of natural numbers, for all natural $n$ satisfying the relation $a_{n} + 2a_{n+1} = a_{n+2}$. We need to determine the maximum number of consecutive terms in this sequence that can be prime numbers.
The key point in the solution is the assertion that among any four terms of this sequence, there is at least one that is divisible by 3. This fact can be proven in different ways, including by considering all remainders when dividing $a_{1}$ and $a_{2}$ by 3. Of course, the enumeration must be complete.
From the solution, it also follows that to get five consecutive prime numbers, one of the numbers $a_{1}$ or $a_{2}$ must be equal to 3.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| Justified that the sheriff cannot be lucky for 6 days in a row, but no justification that he can be lucky for 5 days in a row | 4 points |
| The idea of considering remainders when dividing by 3 (or any other number divisible by 3) | 2 points |
| Example where the sheriff is lucky for exactly 5 days without proof that a larger number is impossible | 1 point |
| Answer without justification or incorrect answer, as well as examples where the sheriff is lucky for fewer than 5 consecutive days | 0 points |
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. 9 knights and liars stood in a row. Each said that there is exactly one liar next to him. How many liars are there among them, if knights always tell the truth, and liars always lie?
|
Answer: 3 liars.
Sketch of the solution. Consider the partition of all into groups of people of the same type standing in a row, with people of different types in adjacent groups. In such a group, there can only be one liar. If there are at least two, the extreme liars tell the truth. There are no more than two knights in a group. If there are at least three, the middle knights lie. A group of two knights cannot stand at the edge, otherwise the edge knight lies. A group of one knight can only stand at the edge, otherwise, there are two liars around the knight, and he lies.
Based on these statements, we get that there are two possible arrangements: LRRRLRRRL and RLRRLRRRL. The first one is impossible, as there is a group of two knights at the edge.
Criteria: Answer without justification: 1 point.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Find the smallest integer $x$ that satisfies the inequality $\frac{100}{|x|}>x^{2}+1$.
|
Answer: -4.
If $|x| \geq 5$, then $\frac{100}{|x|} \leq \frac{100}{5}=20 \leq 17 = (-4)^{2} + 1$ we verify the correctness of the answer.
Comment. The correct answer without a rigorous justification up to 2 points. The inequality is solved, but the wrong answer is chosen 2 points.
|
-4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. At a drama club rehearsal, 8 people gathered. Some of them (honest people) always tell the truth, while the others always lie. One of those present said: "There is not a single honest person here." The second said: "There is no more than one honest person here." The third said: "There is no more than two honest people here," and so on up to the eighth, who said: "There is no more than seven honest people here." How many honest people were there among those gathered?
|
3. The standard evaluation methodology for solutions is provided below.
| Points | Correctness (Incorrectness) of the Solution |
| :---: | :--- |
| 7 | Fully correct solution. |
| $6-7$ | Correct solution, but with minor flaws that do not significantly affect the solution. |
| $5-6$ | The solution is generally correct. However, it contains errors or omits cases that do not affect the logical reasoning. |
| $3-4$ | In cases where the solution to the problem is divided into two equally important parts—solution of one of the parts. |
| $2-3$ | Auxiliary statements that help in solving the problem have been proven. |
| $0-1$ | Individual cases have been considered in the absence of a solution. |
| 0 | The solution is incorrect, and there is no progress. |
| 0 | The solution is absent. |
It is important to note that any correct solution is awarded 7 points. Points should not be deducted for a solution being too long, or for a correct solution that differs from the one provided in the criteria or from other solutions known to the jury. At the same time, any solution, no matter how long, that does not contain useful progress should be awarded 0 points.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Two-headed and seven-headed dragons came to a meeting. At the very beginning of the meeting, one of the heads of one of the seven-headed dragons counted all the other heads. There were 25 of them. How many dragons in total came to the meeting?
|
Answer: 8 dragons. Solution. Subtract the 6 heads belonging to the seven-headed dragon from the 25 heads counted by the seven-headed dragon. 19 heads remain. The remaining dragons cannot all be two-headed (19 is an odd number). There can only be one more seven-headed dragon (if there were two, an odd number of heads would remain for the two-headed dragons. And for three seven-headed dragons, there would not be enough heads ($7 \cdot 3=21>19$)). Subtract the 7 heads of this single dragon from the 19 heads, and we get the total number of heads belonging to the two-headed dragons. Therefore, the number of two-headed dragons: (19 - 7) $: 2=6$. In total: $6+1+1=8$ dragons.
Grading criteria: Full solution - $\mathbf{7}$ points. Only answer - $\mathbf{2}$ points.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Vovodya is running on a circular track at a constant speed. There are two photographers standing at two points on the track. After the start, Vovodya was closer to the first photographer for 2 minutes, then closer to the second photographer for 3 minutes, and then closer to the first photographer again. How long did it take Vovodya to run the entire circle?
|
Solution. Let's divide both arcs of the circle between the photographers in half. The halves of the arcs adjacent to the second photographer make up half the distance. Vasya ran this half in 3 minutes, so he will run the entire circle in 6 minutes.
Answer: in 6 minutes.
Grading criteria: Full solution - 7 points. Only answer - $\mathbf{2}$ points.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. There are pan scales without weights and 11 visually identical coins, among which one may be counterfeit, and it is unknown whether it is lighter or heavier than the genuine coins (genuine coins have the same weight). How can you find at least 8 genuine coins in two weighings?
|
Solution. Let's divide the coins into three piles of three coins each. Compare pile 1 and pile 2, and then compare pile 2 and pile 3. If all three piles weigh the same, then all the coins in them are genuine, and we have found 9 genuine coins. Otherwise, one of the piles differs in weight from the others, and the counterfeit coin can only be in it. Then the genuine coins are in the other two piles plus the two remaining coins that we did not weigh.
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. A round table was sat at by 10 people - liars and knights. Liars always lie, and knights always tell the truth. Each of them was given a coin. Then each of them passed their coin to one of their two neighbors. After that, each one said: “I have more coins than my right neighbor.” What is the maximum number of knights that could have been sitting at the table?
|
Answer: 6.
Solution: After the coins are passed, each person sitting at the table can have 0, 1, or 2 coins. Note that 3 knights cannot sit in a row. Indeed, let knights $A, B, C$ sit next to each other, with $B$ sitting to the right of $A$, $C$ to the right of $B$, and $D$ to the right of $C$. If $A$ has $x$ coins, $B$ has $y$ coins, $C$ has $z$ coins, and $D$ has $t$ coins, then the inequality $x > y > z > t$ would hold, which is impossible for the numbers 0, 1, 2. Therefore, among any 3 people sitting in a row, there is a liar. Choose any liar sitting at the table (such a person exists because there are 3 people sitting in a row), and divide the remaining people into 3 groups of 3 people each sitting next to each other. In each of these groups, there is a liar. Thus, there are at least $1 + 3 = 4$ liars at the table, and therefore no more than 6 knights. We will show that 6 knights could be sitting at the table. Suppose they are seated as follows: -K-K-L-L-K-K-L-K-K-L-. And the knights sitting next to each other exchange coins, while the liars give their coins to the people sitting to their right. Then the number of coins each person has will be -K(2)-K(1)-L(0)-L(1)-K(2)-K(1)-L(0)-K(2)-K(1)-L(0)-. And the knights will tell the truth, while the liars will lie.
Comment: It is proven that no more than 6 knights can sit at the table - 5 points. It is proven that 6 knights can sit at the table - 2 points.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. On the island of knights and liars, each resident was asked about each of the others: is he a knight or a liar. In total, 42 answers of "knight" and 48 answers of "liar" were received. What is the maximum number of knights that could have been on the island? Justify your answer. (It is known that knights always tell the truth, while liars always lie.)
|
Solution: Each of the $n$ residents gave $n-1$ answers; in total, there were $n(n-1)$ answers, which, according to the problem, equals $42+48=90$. Hence, $n=10$, meaning there are 10 residents on the island. Let the number of knights be $x$, then the number of liars is $10-x$. Answers of "liar" arise in two cases: when a knight speaks about a liar (there are $x(10-x)$ such cases) and vice versa (the same number of cases). We have the equation $2 x(10-x)=48$, from which $x=6$ or $x=4$. Both options are possible.
Note: Instead of the equation $2 x(10-x)=48$, the equation $x(x-1)+(10-x)(9-x)=42$ (by the number of "knight" answers) or even an equation of the type $\frac{2 x(10-x)}{x(x-1)+(10-x)(9-x)}=\frac{48}{42}$ might arise. All of them are equally valid.
Answer: 6 knights.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| A correct approach to the solution resulted in an incorrect answer due to calculation errors | 6 points |
| A correct equation (system of equations) that fully describes the problem condition; this equation (system) is not solved | 4 points |
| The number of residents on the island is correctly found (in the absence of further progress) | 3 points |
| It is shown that the number of knights could have been 4 or 6 (at least one of these numbers) and it is not proven that there are no other options | 1 point |
| Answer without justification | 0 points |
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. A tourist goes on a hike from $A$ to $B$ and back, and completes the entire journey in 3 hours and 41 minutes. The route from $A$ to $B$ first goes uphill, then on flat ground, and finally downhill. Over what distance does the road run on flat ground, if the tourist's speed is 4 km/h when climbing uphill, 5 km/h on flat ground, and 6 km/h when descending, and the distance $\mathrm{AB}$ is 9 km?
|
Solution. Let $x$ km of the path be on flat ground, then $9-x$ km of the path (uphill and downhill) the tourist travels twice, once (each of the ascent or descent) at a speed of 4 km/h, the other at a speed of 6 km/h, and spends $(9-x) / 4 + (9-x) / 6$ hours on this part. Since the tourist walks $2x / 5$ hours on flat ground, and the round trip takes 3 hours and 41 minutes, then $2x / 5 + (9-x) / 4 + (9-x) / 6 = 221 / 60$, from which $x=4$ km.
Answer: 4 km.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2-0. The number $n$ is such that $8n$ is a 100-digit number, and $81n$ is a 102-digit number. What can the second digit from the beginning of $n$ be?
|
Answer: 2
Solution. Since $8 n10^{101}$ (equality here is obviously impossible), it means $n>123 \cdot 10^{97}$. Therefore, the second digit from the beginning of the number $n$ is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-0. A certain quadratic trinomial $x^{2}-p x+q$ has integer roots $x_{1}$ and $x_{2}$. It turns out that the numbers $x_{1}$, $x_{2}$, and $q$ form a decreasing arithmetic progression. Find the sum of all possible values of $x_{2}$.
|
Answer: -5
Solution. By Vieta's theorem $q=x_{1} x_{2}$. Then $2 x_{2}=x_{1}+x_{1} x_{2}$, from which $x_{1}=\frac{2 x_{2}}{1+x_{2}}$. Since the roots are integers, $2 x_{2}$ is divisible by $1+x_{2}$. But $2+2 x_{2}$ is divisible by $1+x_{2}$, which means 2 is also divisible by $1+x_{2}$. Therefore, $x_{2}=-3,-2,0$ or 1. In these cases, $x_{1}$ takes the values $3,4,0$ and 1. Since $x_{2}<x_{1}$, only the first two options are suitable, from which the sum of possible values of $x_{2}$ is -5.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. For numbers $a, b, c, d$, it is known that $a^{2}+b^{2}=1, c^{2}+d^{2}=1, a c+b d=0$. Calculate $a b+c d$.
|
4. Consider the equality: $(a c+b d)(a d+b c)=0$, since $a c+b d=0$.
We get $a^{2} c d+b^{2} c d+c^{2} a b+d^{2} a b=0$ or $\left(a^{2}+b^{2}\right) c d+\left(c^{2}+d^{2}\right) a b=0$.
Since $a^{2}+b^{2}=1$ and $c^{2}+d^{2}=1$ we obtain $a b+c d=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. How many solutions in integers \(x, y\) does the equation \(6 x^{2}+2 x y+y+x=2019\) have?
|
Answer. 4 solutions. Solution. Express $y$ from the given equation: $y=\frac{2019-6 x^{2}-x}{2 x+1}$. Dividing $6 x^{2}+x$ by $2 x+1$ with a remainder, we get $6 x^{2}+x=(3 x-1)(2 x+1)+1$. Thus, the expression for $y$ will take the form: $y=\frac{2019-(3 x-1)(2 x+1)-1}{2 x+1}=\frac{2018}{2 x+1}-3 x+1$. Therefore, for $y$ to be an integer, 2018 must be divisible by $2 x+1$. Since $2 x+1$ is an odd number, and 2018:2=1009 is a prime number, $2 x+1$ can take 4 values: $1, -1, 1009, -1009$. Then the corresponding values of $x$ and $y$ are: (0;2019), (-1;-2014), (504;-1509), (-505;1514).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.3. Irina did poorly in math at the beginning of the school year, so she had 3 threes and 2 twos in her journal. But in mid-October, she pulled herself together and started getting only fives. What is the minimum number of fives Irina needs to get so that her average grade is exactly 4?
|
Answer: 7.
Solution. Let Irina need to get $x$ fives. Then the sum of her grades will be $3 \cdot 3+2 \cdot 2+5 x=5 x+13$, and the number of grades $-3+2+x=x+5$.
We get the equation
$$
\begin{aligned}
(5 x+13) & :(x+5)=4 \\
5 x+13 & =4(x+5) \\
5 x+13 & =4 x+20 \\
x & =7
\end{aligned}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.8. A boastful fisherman says the same phrase every day: "Today I caught more perch than I did the day before yesterday (2 days ago), but less than I did 9 days ago." What is the maximum number of days in a row that he can tell the truth?
|
Answer: 8.
Solution. First, let's provide an example where he tells the truth for 8 days in a row. The numbers of perch he caught on consecutive days are indicated:
$$
2020202020202061 \underbrace{728394105}_{\text {truth }} \text {. }
$$
Now let's prove that he could not have told the truth for 9 days in a row. Suppose he did. Let's denote the truthful days as the 3rd, ..., 11th, the previous day as the 2nd, and the day before that as the 1st.
Then on the 10th day, he caught more than on the 8th, which is more than on the 6th, 4th, and 2nd. At the same time, on the 11th day, he caught less than on the 2nd, since it was exactly 9 days ago. Therefore, on the 11th day, he caught less than on the 10th.
On the other hand, on the 11th day, the fisherman caught more than on the 9th, which is more than on the 7th, 5th, 3rd, and 1st. At the same time, on the 10th day, he caught less than on the 1st. This means that on the 10th day, he caught less than on the 11th, which contradicts the previously obtained conclusion.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. Borya found the smallest prime number \( p \) such that \( 5 p^{2} + p^{3} \) is a square of some natural number. What number did Borya find?
|
Answer: 11.
Solution: Since $5 p^{2}+p^{3}=p^{2}(5+p)$, the original number is a perfect square if and only if the number $5+p$ is a perfect square. Since $p$ is a prime number, $p+5 \geq 7$. It is sufficient to verify that if $p+5=9$, then $p=4$ is not a prime; if $p+5=16$, then $p=11$ satisfies the condition, and this value of $p$ will be the smallest.
Note: The problem can also be solved by sequentially checking prime numbers starting from $p=2$. Criteria: Only the answer (including verification) - 0 points. Correct answer with a correct solution (including finding by checking all (!) primes $p$ up to 11) - 7 points. The problem is reduced to the fact that $(5+p)$ is a prime number, but not completed or further with an error - 3 points.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 9.4. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars always lie, and truth-tellers always tell the truth. What is the minimum number of liars in the presidium for the described situation to be possible? (Two members of the presidium are neighbors if one of them sits to the left, right, in front of, or behind the other).
|
Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The figure shows how eight liars can be seated in the presidium so that the condition of the problem is satisfied.
Comments. Without an example of seating the liars - 5 points.
#
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
#
|
# Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. Then the emeralds are in the two remaining boxes, and there are a total of $5+7=12$.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
|
Answer: 7.
Solution. Since $A B C D$ is a square, then $A B=B C=C D=A D$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle A B L$ to $90^{\circ}$. Then the right triangles $A B K$ and $C B L$ are equal by the acute angle and the leg $A B=B C$ (Fig. 1). Therefore, $A K=C L=6$. Then
$$
L D=C D-C L=A D-C L=(K D-A K)-C L=K D-2 \cdot C L=19-2 \cdot 6=7
$$
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given the functions $f(x)=x^{2}+4 x+3$ and $g(x)=x^{2}+2 x-1$. Find all integer solutions to the equation $f(g(f(x)))=g(f(g(x)))$.
|
4. Answer. $x=-2$. Solution. Let's represent the functions as $f(x)=x^{2}+4 x+3=(x+2)^{2}-1$ and $g(x)=x^{2}+2 x-1=(x+1)^{2}-2 . \quad$ Then $\quad f(g(x))=\left((x+1)^{2}-2+2\right)^{2}-1=(x+1)^{4}-1$, $g(f(x))=\left((x+2)^{2}-1+1\right)^{2}-2=(x+2)^{4}-2$. Performing similar operations, we get $g(f(g(x)))=(x+1)^{8}-2$ and $f(g(f(x)))=(x+2)^{8}-1$. Thus, the equation is
$$
\begin{gathered}
(x+1)^{8}-(x+2)^{8}=1 \\
\left((x+1)^{4}-(x+2)^{4}\right)\left((x+1)^{4}+(x+2)^{4}\right)=1
\end{gathered}
$$
Obviously, the second bracket can only be equal to 1, which means the first bracket is also equal to 1. This system has a unique solution $x=-2$.
Grading criteria. Correct solution - 7 points, all function compositions are correctly written but the solution is incomplete or incorrect - 2 points, the solution is incorrect or only the answer - $\mathbf{0}$ points.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Sasha chose a natural number $N>1$ and wrote down in ascending order all its natural divisors: $d_{1}<\ldots<d_{s}$ (so that $d_{1}=1$ and $d_{s}=N$). Then, for each pair of adjacent numbers, he calculated their greatest common divisor; the sum of the resulting $s-1$ numbers turned out to be $N-2$. What values could $N$ take?
(A. Kuznetsov)
|
Answer: $N=3$.
Solution. Note immediately that $d_{s+1-i}=N / d_{i}$ for all $i=$ $=1,2, \ldots, s$.
The number $d_{i+1}-d_{i}$ is divisible by the GCD $\left(d_{i}, d_{i+1}\right)$, so the GCD $\left(d_{i}, d_{i+1}\right) \leqslant d_{i+1}-d_{i}$. For $i=1, \ldots, s-1$, let $r_{i}=\left(d_{i+1}-d_{i}\right)-$ GCD $\left(d_{i}, d_{i+1}\right) \geqslant 0$. According to the condition,
$$
\left(d_{2}-d_{1}\right)+\left(d_{3}-d_{2}\right)+\ldots+\left(d_{s}-d_{s-1}\right)=d_{s}-d_{1}=N-1
$$
and
GCD $\left(d_{1}, d_{2}\right)+$ GCD $\left(d_{2}, d_{3}\right)+\ldots+$ GCD $\left(d_{s-1}, d_{s}\right)=N-2$.
Subtracting the second equation from the first, we get $r_{1}+\ldots+r_{s-1}=$ $=1$. This means that $r_{k}=1$ for some $k$, while all other $r_{i}$ are zero.
Thus, $1=\left(d_{k+1}-d_{k}\right)-$ GCD $\left(d_{k}, d_{k+1}\right)$. The right, and therefore the left side of this equation, is divisible by GCD $\left(d_{k}, d_{k+1}\right)$, so GCD $\left(d_{k}, d_{k+1}\right)=1$ and $d_{k+1}-d_{k}=2$. This is possible only if both numbers $d_{k}$ and $d_{k+1}$ are odd.
Since $d_{k}$ and $d_{k+1}$ are two consecutive divisors of the number $N$, then $\frac{N}{d_{k+1}}$ and $\frac{N}{d_{k}}$ are also two consecutive divisors of the number $N$. Therefore, if $\frac{N}{d_{k+1}}=d_{m}$, then $\frac{N}{d_{k}}=d_{m+1}$. In this case,
$$
\begin{aligned}
\operatorname{GCD}\left(d_{m}, d_{m+1}\right)= & \frac{N}{\operatorname{LCM}\left(d_{k}, d_{k+1}\right)}=\frac{N \cdot \operatorname{GCD}\left(d_{k}, d_{k+1}\right)}{d_{k} d_{k+1}}=1$, which is possible only when $k=m$ (and, consequently, $s=2 k)$.
Thus, $d_{k+1}=\frac{N}{d_{k}}$, meaning the number $N=d_{k} d_{k+1}$ is odd. But then $d_{s-1} \leqslant \frac{N}{3}$, from which GCD $\left(d_{s-1}, d_{s}\right) \leqslant d_{s-1} \leqslant \frac{N}{3}$. Therefore, $1 \geqslant r_{s-1} \geqslant \frac{2 N}{3}-\frac{N}{3}=\frac{N}{3}$, i.e., $N \leqslant 3$. Since $N>1$, we get the only possible value $N=3$, which, as is easy to verify, satisfies the condition.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Two-digit numbers are written on the board. Each number is composite, but any two numbers are coprime. What is the maximum number of numbers that can be written?
|
Answer: four numbers.
Solution. Evaluation. Since any two of the written numbers are coprime, each of the prime numbers 2, 3, 5, and 7 can appear in the factorization of no more than one of them. If there are five or more numbers on the board, then all prime factors in the factorization of some of them must be at least 11. But this composite number is at least 121. This contradicts the condition that all written numbers are two-digit. Therefore, there are no more than four numbers on the board.
Example. 25, 26, $33, 49$.
Other examples exist.
Grading criteria.
“+” - a complete and justified solution is provided
“ $\pm$ " - a correct estimate is provided, but the example is not given or is incorrect
"耳” - only the correct answer and correct example are provided
“-" - only the answer is provided
“-" - the problem is not solved or is solved incorrectly
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In volleyball competitions, where there are no ties, 5 teams participate. All teams played against each other. The team that took 1st place won all their matches, and the teams that took 2nd and 3rd place each won exactly two matches. In the case of equal points, the position is determined by the result of the match between the teams. How many victories did the team that took last place achieve? Determine who won against whom.
|
Solution. Let's denote a team by a point. If team A won against team B, we draw an arrow from A to B.
The number of outgoing arrows equals the number of wins. In total, 10 arrows can be drawn between five points, i.e., 10 matches were played. The first, second, and third teams won a total of 8 matches. This leaves 2 wins for the fourth and fifth teams. Since only the second and third teams won 2 matches each, the fourth and fifth teams won 1 match each. Since in the case of equal points, the higher position is determined by a win in the head-to-head match, the second team beat the third, and the fourth beat the fifth.
The third team won 2 matches, so it beat the fourth and fifth teams.
The fifth team won its only match against the second team.
One arrow remains.
Comment. A correct answer alone is worth 3 points. Subsequent points are awarded based on the completeness of the justification of the answer.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. The following numbers are written on the wall: $1,3,4,6,8,9,11,12,16$. Four of these numbers were written by Vova, four numbers were written by Dima, and one number was simply the house number of the local police officer. The police officer found out that the sum of the numbers written by Vova is three times the sum of the numbers written by Dima. Calculate the house number.
|
Solution. The sum of all numbers on the wall is $1+3+4+6+8+9+11+12+16=70$. Let $n$ be the sum of the numbers written by Dima, then $3n$ is the sum of the numbers written by Vova, $4n$ is the sum of the numbers written by both, $70-4n-$ is the house number of the district police officer, which when divided by 4 leaves a remainder of 2. The only such number among those given is 6.
Answer: 6.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Masha left the house for school. A few minutes later, Vanya ran out of the same house to school. He overtook Masha at one-third of the way, and when he arrived at school, Masha still had half of the way left to go. How many times faster does Vanya run compared to how Masha walks?
|
Solution. At one third of the way, Masha and Vanya were at the same time. After that, Vanya ran $2 / 3$ of the way, while Masha walked $1 / 2 - 1 / 3 = 1 / 6$ of the way in the same time. This means that he runs $2 / 3$ : $1 / 6 = 4$ times more in the same amount of time than Masha.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In a football tournament, 17 teams participate, and each team plays against each other exactly once. A team earns 3 points for a win. For a draw, 1 point is awarded. The losing team gets no points. What is the maximum number of teams that can accumulate exactly 10 points?
|
Answer: 11.
Sketch of the solution. Estimation. Let $n$ teams have scored exactly 10 points each. The total points include all points scored by these teams in matches against each other (at least 2) and possibly in matches against other teams: $10 n \geq 2(n-1) n / 2$. Hence, $n \leq 11$.
Example: All eleven teams played to a draw against each other and lost to the rest.
Criteria. Answer only, without justification: 1 point.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given three non-zero real numbers $a, b, c$ such that the equations: $a x^{2}+b x+c=0, b x^{2}+c x+a=0, c x^{2}+a x+b=0$ each have two roots. How many of the roots of these equations can be negative?
|
Answer: 2.
Sketch of the solution.
If the numbers $a, b, c$ are replaced by their opposites, then the "new" equations will have the same set of roots as the original ones. There are two possible cases: the numbers $a, b, c$ are of the same sign; among them, there are both positive and negative numbers.
First case. Due to the initial observation, we can assume that they are positive. And due to the cyclic symmetry of the coefficients of the equations, we can assume that $0<a \leq b, 0<a \leq c$. Multiplying these inequalities, we get $\mathrm{a}^{2} \leq \mathrm{bc}<4 \mathrm{bc}$. From this inequality, it follows that the discriminant of the equation $c x^{2}+a x+b=0$ is negative, and the equation has no roots. This case is impossible.
Second case. Due to the initial observation and the cyclic symmetry of the coefficients of the equations, we can assume that $0<a, 0<c, b<0$. From these inequalities, it follows that in the equations $b x^{2}+c x+a=0$, $c x^{2}+a x+b=0$, the product of the roots is negative, and thus exactly one root is negative. In the equation $a x^{2}+b x+c=0$, the product of the roots is positive and the sum is positive, meaning both roots are positive. In the end, we have two negative roots.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. Two cyclists, Andrey and Boris, are riding at a constant and identical speed along a straight highway in the same direction, so that the distance between them remains constant. There is a turnoff to a village ahead. At some point in time, the distance from Andrey to the turnoff was equal to the square of the distance from Boris to the same turnoff. When each of them had traveled another 1 km, the distance from Andrey to the turnoff became three times the distance from Boris to the turnoff. What is the distance between the cyclists?
|
Answer: 2 or 0 km.
Solution: Let the first mentioned distances be $a$ and $b$, then $a=b^{2}$. When each of them had traveled another kilometer, the remaining distance to the turn was $a-1$ and $b-1$ km, respectively, so $a-1=3(b-1)$, which means $b^{2}-1=3(b-1),(b-1)(b+1)=3(b-1)$, from which $b=1$ or $b=2$. When $b=1$, we have $a=1$, so $a-b=0$. When $b=2$, we have $a=4$, so $a-b=2$.
Comment: A complete and justified solution - 7 points. Both equations are correctly set up, but the answer is not obtained - 3 points. The answer is found by selecting numbers that satisfy the condition, but it is not shown that other answers are impossible - 2 points, if only part of the answer is found by selection - 1 point. The solution is correctly started, but there is no significant progress - 1 point. Errors in transformations are made in an otherwise correct solution - deduct 2 points per error. Only the answer is provided - 0 points. The problem is not solved or is solved incorrectly - 0 points.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. It is known that for real numbers $a$ and $b$ the following equalities hold:
$$
a^{3}-3 a b^{2}=11, \quad b^{3}-3 a^{2} b=2
$$
What values can the expression $a^{2}+b^{2}$ take?
|
Answer: 5.
Solution. We have
$$
\left(a^{2}+b^{2}\right)^{3}=a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}=\left(a^{3}-3 a b^{2}\right)^{2}+\left(b^{3}-3 a^{2} b\right)^{2}=11^{2}+2^{2}=125
$$
From this, $a^{2}+b^{2}=5$.
Comment. A correct and justified solution - 7 points. The correct answer obtained based on a simple example satisfying the condition ( $a=-1, b=2$ ), but not proven that the same answer for other pairs - 3 points. Some progress - 2 points. The solution is started, but the progress is insignificant - 1 point. Only the answer is provided - 0 points. The problem is not solved or solved incorrectly - 0 points.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (7 points) The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the abscissa of the point of intersection.
#
|
# Solution.
Method 1. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be reduced to $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, so $x = 1$.
Method 2. Notice that $x = 1$ is a solution to the problem, because when $x = 1$, both given linear functions take the same value $y = k + b$. Since their graphs intersect, meaning these lines have exactly one common point, there are no other solutions.
Answer. $x = 1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. On a circle, $2 N$ points are marked ($N$ is a natural number). It is known that through any point inside the circle, no more than two chords with endpoints at the marked points pass. We will call a matching a set of $N$ chords with endpoints at the marked points such that each marked point is the endpoint of exactly one of these chords. We will call a matching even if the number of points where its chords intersect is even, and odd otherwise. Find the difference between the number of even and odd matchings.
(V. Shmarov)
|
Answer. 1.
First solution. We will prove by induction on $N$ that there is one more even matching than odd. For $N=1$, the statement is obvious: there is only one matching, and it is even. Now we will prove the statement for $2N$ points, assuming it is true for $2(N-1)$ points. Let the marked points be $A_{1}, A_{2}, \ldots, A_{2N}$ in the order of traversal of the circle clockwise.
Lemma. Suppose a chord $A_{1}A_{i}$ is involved in the matching. Then for even $i$, it intersects an even number of chords, and for odd $i$, an odd number.
Proof. Let the chord $A_{1}A_{i}$ intersect exactly $k$ chords. Consider the points $A_{2}, \ldots, A_{i-1}$; exactly $k$ of them are the endpoints of chords intersecting $A_{1}A_{i}$ (one endpoint of each chord). The remaining $i-2-k$ points are paired with points connected by chords that do not intersect $A_{1}A_{i}$. Thus, the number $i-2-k$ is even, meaning that the numbers $i$ and $k$ have the same parity. The lemma is proved.
Now we will divide all matchings into $2N-1$ groups $\Pi_{2}, \ldots, \Pi_{2N}$: the group $\Pi_{i}$ will contain those matchings in which point $A_{1}$ is connected to $A_{i}$. Now, remove the chord $A_{1}A_{i}$ from each matching in $\Pi_{i}$; we will get all possible matchings on the remaining $2N-2$ points. By the induction hypothesis, among them, there is one more even matching than odd. At the same time, if $i$ is even, then according to the lemma, the parity of the matching does not change when removed, and if $i$ is odd, it changes. Thus, in each of the $N$ sets $\Pi_{2}, \ldots, \Pi_{2N}$, there is one more even matching than odd, and in each of the $N-1$ sets $\Pi_{3}, \ldots, \Pi_{2N-1}$, there is one more odd than even. In total, there are more even matchings than odd by $N-(N-1)=1$, which is what we needed to prove.
Second solution. We will provide another proof of the induction step.
Let the marked points be $A_{1}, \ldots, A_{2N}$. Consider all matchings in which $A_{2N-1}$ and $A_{2N}$ are connected by a chord. This chord does not intersect any other. Thus, by removing it from each of the considered matchings, we will get all matchings on the points $A_{1}, \ldots, A_{2N-2}$, and the parity of each of them will be preserved. By the induction hypothesis, among our matchings, there is one more even than odd.
To complete the proof, it is sufficient to show that among all other matchings, there are equally many even and odd. Consider any of them; suppose it contains chords $A_{2N-1}A_{i}$ and $A_{2N}A_{k}$. Now, "swap" the points $A_{2N-1}$ and $A_{2N}$, that is, replace the chords with $A_{2N}A_{i}$ and $A_{2N-1}A_{k}$. In this case, if the original chord intersected with some of the others, then the new chord will also intersect with it. On the other hand, if the chords $A_{2N-1}A_{i}$ and $A_{2N}A_{k}$ did not intersect, then the new chords will intersect, and vice versa. Thus, to each remaining even matching, we have associated an odd one, and vice versa; different matchings, obviously, correspond to different ones. Therefore, there are equally many remaining even and odd matchings, which is what we needed to prove.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.6. Petya chose a natural number $a>1$ and wrote down fifteen numbers $1+a, 1+a^{2}, 1+a^{3}, \ldots, 1+a^{15}$ on the board. Then he erased several numbers so that any two remaining numbers are coprime. What is the maximum number of numbers that could remain on the board?
(O. Podlipsky)
|
# Answer. 4 numbers.
Solution. First, we will show that there cannot be more than four such numbers. Note that if $k$ is odd, then the number $1+a^{n k}=1^{k}+\left(a^{n}\right)^{k}$ is divisible by $1+a^{n}$. Next, each of the numbers $1,2, \ldots, 15$ has one of the forms $k, 2 k, 4 k, 8 k$, where $k$ is odd. Thus, each of the numbers $1+a, 1+a^{2}, 1+a^{3}, \ldots, 1+a^{15}$ is divisible by either $1+a$, or $1+a^{2}$, or $1+a^{4}$, or $1+a^{8}$. Therefore, if we take at least five numbers, then among them there will be two that are divisible by the same number greater than 1; hence, they will not be coprime. Thus, there cannot be more than four remaining numbers.
It remains to show that four numbers could remain. Indeed, if $a=2$, then we can leave the numbers $1+2=3$, $1+2^{2}=5$, $1+2^{4}=17$, and $1+2^{8}=257$. All of them are pairwise coprime.
Remark. It can be shown that for any even $a$, the numbers $1+a, 1+a^{2}, 1+a^{4}, 1+a^{8}$ will be pairwise coprime.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. In triangle $A B C$, the median $A M$ is perpendicular to the bisector $B D$. Find the perimeter of the triangle, given that $A B=1$, and the lengths of all sides are integers.
|
Answer: 5.
Solution. Let $O$ be the point of intersection of the median $A M$ and the bisector $B D$. Triangles $A B O$ and $M B O$ are congruent (by the common side $B O$ and the two adjacent angles). Therefore, $A B = B M = 1$ and $C M = 1$, since $A M$ is a median. Thus, $A B = 1, B C = 2$. By the triangle inequality, $A C = B C - A B = 1$, and since the length of $A C$ is given to be an integer, $A C = 2$. Therefore, the perimeter of the triangle is 5.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. It is known that the quadratic trinomials $x^{2}+p x+q$ and $x^{2}+q x+p$
have different real roots. Consider all possible pairwise products of the roots of the first quadratic trinomial with the roots of the second (there are four such products in total). Prove that the sum of the reciprocals of these products does not depend on $p$ and $q$.
|
Let $x_{1}, x_{2}$ be the roots of the first quadratic polynomial and $x_{3}, x_{4}$, then we need to prove that $\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{2} x_{4}}$ does not depend on $p$ and $q$. Transform the given expression:
$\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{2} x_{4}}=\frac{1}{x_{1}}\left(\frac{1}{x_{3}}+\frac{1}{x_{4}}\right)+\frac{1}{x_{2}}\left(\frac{1}{x_{3}}+\frac{1}{x_{4}}\right)=$
$=\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)\left(\frac{1}{x_{3}}+\frac{1}{x_{4}}\right)=\left(\frac{x_{1}+x_{2}}{x_{1} x_{2}}\right)\left(\frac{x_{3}+x_{4}}{x_{3} x_{4}}\right)$
By Vieta's formulas, we have: $\left\{\begin{array}{c}x_{1}+x_{2}=-p \\ x_{1} x_{2}=q\end{array}\right.$ and $\left\{\begin{array}{c}x_{3}+x_{4}=-q \\ x_{3} x_{4}=p\end{array}\right.$
Then $\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{2} x_{4}}=\frac{-p(-q)}{q p}=1$.
|
1
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
3. Solve the equation in integers $x^{4}-2 y^{4}-4 z^{4}-8 t^{4}=0$.
## Answer: $x=y=z=t=0$
|
Note that $x$ is even. Let $x=2x_{1}$, then we get
$4x_{1}^{4}-8y_{1}^{4}-z^{4}-2t^{4}=0$. Therefore, $z=2z_{1}$ and $2x_{1}^{4}-4y_{1}^{4}-8z_{1}^{4}-t^{4}=0$. Similarly, $t=2t_{1}$ and $x_{1}^{4}-2y_{1}^{4}-4z_{1}^{4}-8t_{1}^{4}=0$ we obtain the original equation, but for variables $x_{1}, y_{1}, z_{1}, t_{1}$, so we can repeat the same reasoning and so on. We get that each variable is divisible by any power of 2. Therefore, $x=y=z=t=0$.
## Recommendations for checking
1) Noted that $x$ is even - 1 point.
2) Using correct reasoning, the equation
$x_{1}^{4}-2y_{1}^{4}-4z_{1}^{4}-8t_{1}^{4}=0$, where $x=2x_{1}, y=2y_{1}, z=2z_{1}, t=2t_{1}$, but no corresponding conclusion - 4 points.
3) Only the answer is provided without proof that there are no other solutions - 0 points.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. It is known that in the pyramid $A B C D$ with vertex $D$, the sum $\angle A B D+\angle D B C=\pi$. Find the length of the segment $D L$, where $L$ is the base of the bisector $B L$ of triangle $A B C$, if it is known that
$$
A B=9, B C=6, A C=5, D B=1 .
$$
|
Answer: 7.
Solution. Let point $M$ lie on the line, outside the segment $B C$, beyond point $B$. From the condition, we have the equality of angles $\angle A B D=\angle D B M$. Then the projection of line $B D$ onto the plane $A B C$ is the bisector $B K$ of angle $A B M$. The bisectors $B K$ and $B L$ are perpendicular, as the bisectors of the internal and external angles. Also, line $B L$ is perpendicular to the height of the pyramid $D H$. Therefore, $B L \perp D B K$ and, consequently, $B L \perp D B$, and triangle $L B D$ is a right triangle with $\angle B=\pi / 2$. Similarly, consider the case when point $M$ is beyond point $C$. From the ratio $C L / L A=B C / B A=2 / 3$, we find that $L C=2, A L=3$. Since $B L^{2}=A B \cdot B C-A L \cdot L C$, then $B L^{2}=48$. By the Pythagorean theorem, we get $D L^{2}=D B^{2}+B L^{2}=49$. Answer: 7.
## Grading Criteria.
Guessed the solution: 0 points.
Found the length of the bisector: 1 point.
Determined that triangle $L B D$ is a right triangle: 4 points.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.4. In a $3 \times 4$ rectangle, natural numbers $1,2,3, \ldots, 12$ were written, each exactly once. The table had the property that in each column, the sum of the top two numbers was twice the bottom number. Over time, some numbers were erased. Find all possible numbers that could have been written in place of $\star$.
| 6 | | | |
| :--- | :--- | :--- | :--- |
| 4 | | | |
| | 8 | 11 | $\star$ |
Answer, variant 1. 11.
Answer, variant 2. 8.
Answer, variant 3. 11.
Answer, variant 4. 2.
|
Solution 1. According to the condition, the number 2 is written in the bottom-left cell. The sum of the two unknown numbers in the second column is 10, and in the third column, it is 16. Therefore, the sum of the numbers in the first three columns is $6+15+24$, and the sum of all numbers in the table is $1+2+\ldots+12=78$. Thus, the sum of the numbers in the fourth column is $78-6-15-24=33$. From the condition, it follows that the number at $\star$ is one-third of 33, which is 11.
Remark. The arrangement of numbers described in the condition is indeed possible. For example, such an arrangement:
| 1 | 6 | 9 | 10 |
| :--- | :--- | :--- | :--- |
| 3 | 4 | 7 | 12 |
| 2 | 5 | 8 | 11 |
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.4. 10 children stood in a circle. Each thought of an integer and told it to their clockwise neighbor. Then each loudly announced the sum of their number and the number of the counterclockwise neighbor. The first said "10", the next clockwise - "9", the next clockwise - "8", and so on, the ninth said "2". What number did the remaining child say?
Answer, option 1.5.
Answer, option 2. 9.
Answer, option 3. 14.
Answer, option 4. 6.
|
Solution option 1. The sum of all ten thought-of numbers will be obtained if we add what the first child said to what the third, fifth, seventh, and ninth children said. Similarly, we add what the second, fourth, sixth, eighth, and tenth children said. Again, we get the sum of all ten thought-of numbers, so the equation $1+3+5+7+9=2+4+6+8+x$ must hold, where $x$ is the number named by the tenth child. From this, $x=5$.
Remark. Where was the condition that the numbers thought of are integers used? Nowhere. The solution works for non-integer numbers as well.
|
5
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
# Task 9.1
For which natural numbers $n$ is the expression $n^{2}-4 n+11$ a square of a natural number?
## Number of points 7
|
Answer:
for $n=5$
## Solution
Let $n^{2}-4 n+11=t^{2}$.
Note that $n^{2}-4 n+4=(n-2)^{2}$ is also the square of some integer $r=n-2$, which is less than $t$.
We get that $t^{2}-r^{2}=(t+r)(t-r)=7$.
The numbers $(t+r)$ and $(t-r)$ are natural and the first is greater than the second.
Thus, $(t+r)=7$, and $(t-r)=1$.
Solving this system, we get $t=4, r=3$, which gives $n=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In a beach soccer tournament, 17 teams participate, and each team plays against each other exactly once. Teams are awarded 3 points for a win in regular time, 2 points for a win in extra time, and 1 point for a win on penalties. The losing team does not receive any points. What is the maximum number of teams that can accumulate exactly 5 points?
|
Answer: 11.
Sketch of the solution. Estimation. Let $n$ teams have scored exactly 5 points each. The total points include all points scored by these teams in matches against each other (at least 1) and possibly in matches against other teams: $5 n \geq(n-1) n / 2$. Hence, $n \leq 11$.
Example: Place eleven teams in a circle, and let each team win against the five teams following them in a clockwise direction by penalty. The remaining teams lost to them.
Criteria. Example without estimation: 2 points. Estimation without example: 4 points.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Find the number of roots of the equation
$$
|x|+|x+1|+\ldots+|x+2018|=x^{2}+2018 x-2019
$$
(V. Dubinskaya)
|
Answer: 2.
Solution. For $x \in(-2019,1)$ there are no roots, since on the given interval the left side is non-negative, while the right side is negative.
For $x \in[1, \infty)$ all absolute values are resolved with a positive sign, so the equation will take the form $g(x)=0$, where $g(x)=x^{2}-x-2009+(1+2+\ldots+2018)$. Since $g(1)<0$, this quadratic equation has a single root on the interval $[1, \infty)$.
Since the graphs of the functions on the left and right sides are symmetric with respect to the line $x=-1009$ (i.e., $f(x)=f(-2018-x)$), there are as many roots on the interval $(-\infty,-2019]$ as there are on the interval $[1,+\infty)$, i.e., exactly one root. In total, the given equation has two roots.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Initially, there are 111 pieces of clay of the same mass on the table. In one operation, you can choose several groups with the same number of pieces and in each group, combine all the clay into one piece. What is the minimum number of operations required to get exactly 11 pieces, any two of which have different masses?
(I. Bogdanov)
|
Answer. In two operations.
Solution. Let the mass of one original piece be 1. If in the first operation there are $k$ pieces in each group, then after it each piece will have a mass of 1 or $k$; therefore, it is impossible to obtain eleven pieces of different masses in one operation.
We will show that the required result can be achieved in two operations. In the first operation, select 37 groups of 2 pieces each; after the operation, there will be 37 pieces with masses of 1 and 2. In the second operation, select 9 groups of 8 pieces each: in the $i$-th group $(1 \leqslant i \leqslant 9)$ there will be $i-1$ pieces of mass 2 and $9-i$ pieces of mass 1. Then two pieces of masses 1 and 2 will remain unused, and from the $i$-th group a piece of mass $9-i+2(i-1)=7+i$ will be obtained. Thus, 11 pieces with masses $1,2,8,9, \ldots, 16$ will be obtained, as required.
Remark. It can be shown that the method presented is the only possible one.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-2. In square $A B C D$, a segment $C E$ is drawn such that the angles shown in the diagram are $7 \alpha$ and $8 \alpha$. Find the value of angle $\alpha$ in degrees. If necessary, round the answer to 0.01 or write the answer as a common fraction.
|
Answer: $9^{\circ}$.
Solution. In triangle $D F E$, the angles are $7 \alpha, 8 \alpha$ and $45^{\circ}$.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha + 8 \alpha + 45^{\circ} = 180^{\circ}$, from which $\alpha = \frac{1}{15} \cdot 135^{\circ} = 9^{\circ}$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Solve the inequality $\sqrt{2 x^{2}-8 x+6}+\sqrt{4 x-x^{2}-3}<x-1$
|
Note that all solutions to the original inequality exist if the expressions under the square roots are non-negative. These inequalities are simultaneously satisfied only under the condition $x^{2}-4 x+3=0$. This equation has two roots, 1 and 3. Checking shows that the original inequality has a unique solution 3.
Answer: 3.
#
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. Solve the equation $2021 x^{2021}-2021+x=\sqrt[2021]{2022-2021 x}$. (7 points)
#
|
# Solution
The function $f(x)=2021x^{2021}-2021+x$ is increasing, while the function $g(x)=\sqrt[2021]{2022-2021x}$ is decreasing. Therefore, the equation $f(x)=g(x)$ has no more than one root. However, it is obvious that $f(1)=g(1)$.
Answer: $x=1$.
| criteria | points |
| :--- | :---: |
| correct solution | 7 |
| Proven that there is no more than one root. | 4 |
| Root found, but not proven that there are no other roots | 2 |
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Four non-zero numbers are written on the board, and the sum of any three of them is less than the fourth number. What is the smallest number of negative numbers that can be written on the board? Justify your answer.
|
Solution: Let the numbers on the board be $a \geqslant b \geqslant c \geqslant d$. The condition of the problem is equivalent to the inequality $a+b+c < d$ for optimality | 7 points |
| There is a proof that there are no fewer than three negative numbers (in the absence of an example with three numbers) | 3 points |
| An example of a set with three negative numbers is provided (in the absence of a proof of its optimality) | 1 point |
| Correct answer without justification (or with incorrect justification) | 0 points |
| Examples of sets with four negative numbers | not evaluated |
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. At the exchange office, only the following operations can be performed:
1) exchange 2 gold coins for three silver coins and one copper coin;
2) exchange 5 silver coins for three gold coins and one copper coin.
Nikolai had only silver coins. After several visits to the exchange office, he had fewer silver coins, no gold coins appeared, but he got 50 copper coins. By how much did the number of silver coins decrease for Nikolai? Justify your answer. Note that exchanging money at the bank is not an equivalent operation, meaning that with each exchange, the purchasing power of the coins Nikolai had slightly decreased.
|
Solution: As a result of each operation, Nikolai acquires exactly 1 copper coin, which means there were exactly 50 operations in total. Of these, some (let's say \(a\)) were of the first type, and the rest \(50-a\) were of the second type. On operations of the first type, Nikolai spent \(2a\) gold coins, and on operations of the second type, he earned \(3(50-a)\) gold coins. Since he has no gold coins left, \(2a = 3(50-a)\), from which \(a = 30\). This means that on operations of the first type, Nikolai received \(30 \cdot 3 = 90\) silver coins, and on operations of the second type, he lost \((50-30) \cdot 5 = 100\) such coins. Thus, the number of silver coins decreased by \(100 - 90 = 10\) coins.
Answer: By 10 coins.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| Correct solution path, but the answer is incorrect due to arithmetic errors | 6 points |
| Correctly and justifiedly found the number of operations of one type (either the first or the second) | 5 points |
| Proven that the ratio of the number of operations of the 1st and 2nd types is 3:2 AND/OR the total number of operations of both types is justified | 3 points |
| The problem is correctly solved under the assumption of equal value of all exchanges (the general case is not investigated) | 2 points |
| A specific example of exchanges is provided, showing that the answer 10 is possible, but its uniqueness is not justified | 1 point |
| Correct answer without justification (with incorrect justification) | |
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find all natural numbers $n \geq 2$ such that $20^{n}+19^{n}$ is divisible by $20^{n-2}+19^{n-2}$.
|
Solution.
Consider the expression
$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)
$$
By the condition, it is divisible by $20^{n-2}+19^{n-2}$. On the other hand,
$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)=20^{n-2}\left(20^{2}-19^{2}\right)=20^{n-2} \cdot 39
$$
Note that $20^{n-2}$ and $20^{n-2}+19^{n-2}$ are coprime, since no prime divisor of $20^{n-2}$ is a divisor of $19^{n-2}$, and the expressions $20^{n-2}$ and $20^{n-2}+19^{n-2}$ do not have
common prime divisors. Therefore, in the product $20^{n-2} \cdot 39$, only the second factor can be divisible by $20^{n-2}+19^{n-2}$. For $n-2>1$, the expression $20^{n-2}+19^{n-2}$ exceeds 39, so it cannot be its divisor. It remains to check by substitution $n=2,3$.
For $n=2$, $20^{0}+19^{0}=2$, the odd number $20^{2}+19^{2}=761$ is not divisible by 2. For $n=3$, $20^{1}+19^{1}=39,20^{3}+19^{3}=39 \cdot\left(20^{2}-20 \cdot 19+19^{2}\right)$ is divisible by $20^{1}+19^{1}=39$. Answer. $n=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. At a round table, 10 people are sitting, some of them are knights, and the rest are liars (knights always tell the truth, while liars always lie). It is known that among them, there is at least one knight and at least one liar. What is the maximum number of people sitting at the table who can say: "Both of my neighbors are knights"? (A false statement is one that is at least partially not true.)
#
|
# Answer. 9.
Solution. Note that all 10 could not have said such a phrase. Since at the table there is both a knight and a liar, there will be a liar and a knight sitting next to each other. But then this knight does not have both neighbors as knights. If, however, at the table there are 9 liars and 1 knight, then each of these 9 liars could say the phrase "Both of my neighbors are knights," since each liar has a liar among their neighbors.
Comment. It has been proven that all 10 people could not have said the required phrase - 4 points.
It has been shown that with a certain seating arrangement, 9 people could have said the required phrase - 3 points.
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3. What is the minimum number of L-shaped corners consisting of 3 cells that need to be painted in a $5 \times 5$ square so that no more L-shaped corners can be painted? (Painted L-shaped corners should not overlap.)
|
Answer: 4.
Solution: Let the cells of a $5 \times 5$ square be painted in such a way that no more corners can be painted. Consider the 4 corners marked in Fig. 7. Since none of these corners can be painted, at least one cell in each of these corners must be painted. Note that one corner cannot paint cells of two marked corners. Therefore, at least 4 corners must be painted.
Fig. 7
Fig. 8
Fig. 8 shows how to paint 4 corners so that no more corners can be painted.
Comment: It is proven that the number of painted corners is not less than $4-4$ points.
A correct example with 4 painted corners is drawn 3 points.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime
|
Answer: 5.
Solution: Note that an odd semiprime number can only be the sum of two and an odd prime number.
We will show that three consecutive odd numbers $2n+1$, $2n+3$, and $2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assuming the contrary, we get that the numbers $2n-1$, $2n+1$, and $2n+3$ are prime, and all are greater than 3. But one of these three numbers is divisible by 3. Contradiction.
Note that among any six consecutive numbers, there are three consecutive odd numbers; therefore, there cannot be more than five consecutive semiprime numbers. Five consecutive numbers can be semiprimes; for example, $30=17+13, 31=29+2, 32=19+13, 33=31+2$, $34=23+11$.
Remark: There are other examples.
Comment: It is proven that there cannot be more than 5 consecutive semiprime numbers (greater than 25) - 4 points.
An example of 5 consecutive semiprime numbers (greater than 25) is provided - 3 points.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. Kuzya cut a convex paper 67-gon along a straight line into two polygons, then similarly cut one of the two resulting polygons, then one of the three resulting ones, and so on. In the end, he got eight $n$-gons. Find all possible values of $n$.
|
Answer: $n=11$.
Solution. A straight-line cut can be of three types: from side to side, from a vertex to a side, and from a vertex to a vertex. Therefore, after one cut, the total number of sides of the polygons increases by 4, 3, or 2, respectively. Kuzya made 7 cuts, so the number of sides added is at least 14 but no more than 28. Therefore, the total number of sides of the eight $n$-gons is between 81 and 95. Among the integers in this range, only the number 88 is divisible by 8 without a remainder, so $n=88 \div 8=11$.
Grading Criteria.
“+” A complete and well-reasoned solution is provided
“土” A generally correct reasoning is provided, but it contains minor gaps or inaccuracies (for example, it is not explained how the total number of sides changes in each of the three cases)
“干” The correct answer is obtained, but not all cases are considered
“干” The correct answer is provided and it is only verified that it satisfies the condition, but there is no proof of the absence of other answers
“-” Only the answer is provided
“-” The problem is not solved or is solved incorrectly
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.4. Lev Alex decided to count the stripes on Marty the zebra (black and white stripes alternate). It turned out that there is one more black stripe than white ones. Alex also noticed that all white stripes are of the same width, while black stripes can be wide or narrow, and there are 7 more white stripes than wide black ones. How many narrow black stripes does Marty have?
|
# Answer: 8.
Solution. First, let's look only at the wide black stripes. There are 7 fewer of them than white ones. If we add the narrow black stripes to the wide black ones, we get all the black stripes, which are 1 more than the white ones. Therefore, to find the number of narrow black stripes, we need to first "compensate" for the 7 white stripes - the number by which the white ones were greater, and then add one more. This gives us $7+1=8$.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one corner, the second carpet $6 \times 6$ - in the opposite corner, and the third carpet $5 \times 7$ - in one of the remaining corners (all dimensions are in meters).
Find the area of the part of the hall covered by carpets in three layers (give the answer in square meters).
|
Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 units from the bottom side.
The first carpet intersects this rectangle horizontally between the 5th and 8th meters from the left side of the square room, and vertically between the 4th and 6th meters from the top side. In the end, this results in a rectangle $2 \times 3$, the area of which is 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.3. In the class, there are 29 students: some are excellent students and some are troublemakers. Excellent students always tell the truth, while troublemakers always lie.
All students in this class sat around a round table.
- Several students said: “There is exactly one troublemaker next to me.”
- All other students said: “There are exactly two troublemakers next to me.”
What is the minimum number of troublemakers that can be in the class?
|
Answer: 10.
Solution. If there were three straight-A students in a row, the middle one would definitely lie. Therefore, among any three consecutive people, there must be at least one troublemaker.
Choose an arbitrary troublemaker. Assign this person the number 29, and number all the people following them clockwise from 1 to 28. Since in each of the non-overlapping groups $(1,2,3),(4,5,6), \ldots,(25,26,27),(29)$ there is at least one troublemaker, there must be at least $\frac{27}{3}+1=10$ troublemakers in total.
Note also that there could have been exactly 10 troublemakers. Again, numbering the people clockwise from 1 to 29, let the students with numbers $3,6,9,12,15,18,21,24$, 27, 29 be the troublemakers, and the students with all other numbers be straight-A students. In this case, all straight-A students except the one with number 28 said the first phrase, while the student with number 28 and all the troublemakers said the second phrase. It is easy to see that all the conditions of the problem are satisfied.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.4. Points $D$ and $E$ are marked on sides $A C$ and $B C$ of triangle $A B C$ respectively, such that $A D=E C$. It turns out that $B D=E D, \angle B D C=\angle D E B$. Find the length of segment $A C$, given that $A B=7$ and $B E=2$.
|
Answer: 12.
Solution. Note that triangles $D E C$ and $B D A$ are equal. Indeed, $D E=B D, E C=$ $D A$ and $\angle D E C=180^{\circ}-\angle B E D=180^{\circ}-\angle B D C=\angle B D A$. From this, it follows that $D C=A B=7$ and $\angle D C E=\angle B A D$ (Fig. 3). From the equality of these angles, it follows that triangle $A B C$ is isosceles, $7=A B=B C=2+E C$, from which we get $5=E C=A D$ and $A C=A D+D C=5+7=12$.
Fig. 3: to the solution of problem 8.4
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
|
Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
Notice that triangle $A B D$ is equal to triangle $E B D$ by three sides: $B D$ is a common side, $A D=D E, A B=B E$ from the equality of triangles $A B C$ and $E B D$. Then $\angle D A B=$ $\angle B E D=\angle B A C$ and $\angle A B D=\angle D B E=\angle A B E=\frac{1}{3} \cdot 360^{\circ}=120^{\circ}$.
Since $A B=B E$, triangle $A B E$ is isosceles with an angle of $120^{\circ}$, so $\angle B A E=\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=30^{\circ}$. Therefore,
$$
\angle B A C=\angle D A B=\angle D A E-\angle B A E=37^{\circ}-30^{\circ}=7^{\circ}
$$
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.1. On an island, there live red, yellow, green, and blue chameleons.
- On a cloudy day, either one red chameleon changes its color to yellow, or one green chameleon changes its color to blue.
- On a sunny day, either one red chameleon changes its color to green, or one yellow chameleon changes its color to blue.
In September, there were 18 sunny and 12 cloudy days. As a result, the number of yellow chameleons increased by 5. By how much did the number of green chameleons increase?
|
Answer: 11.
Solution. Let $A$ be the number of green chameleons on the island, and $B$ be the number of yellow chameleons. Consider the quantity $A-B$. Note that each cloudy day it decreases by 1, and each sunny day it increases by 1. Since there were $18-12=6$ more sunny days than cloudy days in September, the quantity $A-B$ increased by 6 over this period. Since $B$ increased by 5, $A$ must have increased by $5+6=11$.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.3. Lёsha cut a cube $n \times n \times n$ into 153 smaller cubes. Moreover, all the cubes, except one, have an edge length of 1. Find $n$.
|
Answer: 6.
Solution. Let the remaining cube have an edge of $s$. Obviously, the numbers $n$ and $s$ are natural, and $n>s$.
The difference in the volumes of the cubes with edges $n$ and $s$ is equal to the total volume of the unit cubes. Therefore, $n^{3}-s^{3}=152$.
Since $n^{3}>152>5^{3}$, then $n>5$. Note that $n=6$ satisfies the equation above, corresponding to $s=4$, since $6^{3}-4^{3}=152$. However, $n=7$ does not satisfy the equation, as the number $7^{3}-152=191$ is not a perfect cube. Neither do all $n \geqslant 8$ satisfy the equation, since otherwise
$$
152=n^{3}-s^{3} \geqslant n^{3}-(n-1)^{3}=3 n^{2}-3 n+1=3 n(n-1)+1 \geqslant 3 \cdot 8 \cdot 7+1>152
$$
Clearly, the only possible value of $n=6$ is valid: a cube $6 \times 6 \times 6$ can be cut into a cube $4 \times 4 \times 4$ and 152 unit cubes $1 \times 1 \times 1$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 10.5. Vika has 60 cards with numbers from 1 to 60. She wants to divide all the cards into pairs so that the modulus of the difference of the numbers in all pairs is the same. How many ways are there to do this?
|
# Answer: 8.
Solution. Let $d$ be the absolute difference between the numbers. It is clear that the number $d$ is a natural number.
It is clear that the number 1 must be paired with $d+1$, the number 2 must be paired with $d+2, \ldots$, the number $d$ must be paired with $2d$. Therefore, the first $2d$ natural numbers are paired with each other.
Considering the number $2d+1$, we get that it must be paired with $3d+1$, the number $2d+2$ must be paired with $3d+2, \ldots$, the number $3d$ must be paired with $4d$. Therefore, the next $2d$ natural numbers are also paired with each other.
Continuing in this manner, we get that the set of all numbers $\{1,2, \ldots, 60\}$ should be divided into non-overlapping groups of $2d$ numbers (which in turn are divided into $d$ pairs). Therefore, 60 is divisible by $2d$, which is equivalent to 30 being divisible by $d$. The number 30 has exactly 8 natural divisors: $1,2,3,5,6,10,15,30$. It is clear that all of them are suitable, as each number $d$ among them corresponds to $\frac{30}{d}$ groups of the form $\{2kd+1,2kd+2, \ldots, 4kd\}$, which are obviously divided into pairs with the absolute difference $d$.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
|
Answer: -6.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the condition, it is clear that $x_{1} < 0$ and $x_{2} > 0$. Since $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $f(x)$, by Vieta's theorem, we have $x_{1} \cdot x_{2} = 12b$, from which we get $b = \frac{x_{1} \cdot x_{2}}{12} < 0$.
Let $H$ be the point with coordinates $(3 ; 0)$ (Fig. 11). Clearly, in the isosceles triangle $A T C$, the segment $T H$ is the height, and therefore it is also the median. Thus, $3 - x_{1} = A H = H C = x_{2} - 3$, from which we get $x_{1} = 6 - x_{2}$.
Let $M$ be the point with coordinates $(0, 3)$. Since $T H = T M = 3$ and $T A = T B$, the right triangles $A T H$ and $B T M$ are equal by the leg and hypotenuse. Therefore, $3 - x_{1} = H A = M B = 3 - b$, that is, $x_{1} = b = \frac{x_{1} \cdot x_{2}}{12}$ (by Vieta's theorem), from which we find $x_{2} = 12$. Finally, $x_{1} = 6 - x_{2} = 6 - 12 = -6$ and $b = x_{1} = -6$.
Another solution. As in the previous solution, let the abscissas of points $A$ and $C$ be $x_{1}$ and $x_{2}$, respectively; we will also use the fact that point $B$ has coordinates $(0 ; b)$. Immediately, we understand that $O A = |x_{1}| = -x_{1}$, $O C = |x_{2}| = x_{2}$, and $O B = |b| = -b$.
Let's find the second intersection of the circle with the y-axis, let this be point $D$ with coordinates $(0 ; d)$ (Fig. 12). The chords $A C$ and $B D$ of the circle intersect at the origin $O$; from the properties of the circle, we know that $O A \cdot O C = O B \cdot O D$. We get $-x_{1} \cdot x_{2} = -b \cdot d$, from which, replacing $x_{1} \cdot x_{2}$ with $12b$ by Vieta's theorem, we get $d = 12$.
Fig. 12: to the solution of problem 10.7
It remains to note that triangle $BTD$ is isosceles, and the midpoint of its base, point $M$, has coordinates $(0 ; 3)$. Reflecting point $D(0 ; 12)$ relative to it, we get $B(0 ; -6)$.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.2. On the plate, there were 15 doughnuts. Karlson took three times more doughnuts than Little Man, and Little Man's dog Bibbo took three times fewer than Little Man. How many doughnuts are left on the plate? Explain your answer.
|
Answer: 2 doughnuts are left.
From the condition of the problem, it follows that Little One took three times as many doughnuts as Bimbo, and Karlson took three times as many doughnuts as Little One. We can reason in different ways from here.
First method. If Bimbo took one doughnut, then Little One took three doughnuts, and Karlson took nine doughnuts, so together they took $1+3+9=13$ doughnuts. Thus, two doughnuts are left on the plate.
If Bimbo had taken two doughnuts or more, then Little One would have taken six doughnuts or more, and Karlson would have taken 18 doughnuts or more, which is impossible.
Second method. Let Bimbo take $x$ doughnuts, then Little One took $3x$ doughnuts, and Karlson took $3 \cdot 3x = 9x$ doughnuts. Then together they took $x + 3x + 9x = 13x$ doughnuts. Since the number $13x$ must not exceed 15, the only possible natural value for $x$ is 1. Therefore, together they ate 13 doughnuts, and two doughnuts are left on the plate.
+ a complete and justified solution
$\pm$ the correct answer is provided, it is shown that it satisfies the condition, and there is a statement that it is the only one
$\pm$ the number of doughnuts each person ate is justified, but the explicit answer to the question is not provided
干 the correct answer is provided, it is shown that it satisfies the condition, but there is no mention of its uniqueness
干 only the correct answer is provided
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.5. Ladybugs gathered on a clearing. If a ladybug has 6 spots on its back, it always tells the truth, and if it has 4 spots, it always lies, and there were no other ladybugs on the clearing. The first ladybug said: "Each of us has the same number of spots on our backs." The second said: "Together, we have 30 spots on our backs." "No, together we have 26 spots on our backs," - the third one objected. "Exactly one of these three told the truth," - each of the other ladybugs stated. How many ladybugs gathered on the clearing?
|
Answer: 5 ladybugs.
If the first ladybug is telling the truth, then the second and third should also be telling the truth, as they should have the same number of spots as the first. However, the second and third ladybugs contradict each other, so at least one of them is lying, which means the first ladybug is also lying.
Assume that each of the first three ladybugs lied, then all the others lied as well, since none of these three told the truth. This means all the ladybugs are liars, so each ladybug should have four spots on its back. But in this case, it turns out that the first ladybug actually told the truth, which cannot be. Therefore, the first three ladybugs cannot lie simultaneously, so either the second or the third told the truth, and the other two are liars.
Thus, each of the other ladybugs told the truth.
Therefore, there are two ladybugs with 4 spots on their backs, and several ladybugs with 6 spots on their backs, and in total, the spots on the backs of all the ladybugs are either 30 or 26.
1) If there are 30 spots, then $30 - 2 \cdot 4 = 22$, which is not divisible by 6, so this case is impossible.
2) If there are 26 spots, then $(26 - 2 \cdot 4) : 6 = 3$. This means there are $2 + 3 = 5$ ladybugs on the meadow.
+ A complete and justified solution
$\pm$ Correctly and justifiedly determined who lied and who told the truth, but the answer is not found or is incorrect
$\mp$ Provided the correct answer and shown that it satisfies the condition
- Only the answer is provided
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (7 points) Cheburashka and Gena ate a cake. Cheburashka ate twice as slowly as Gena, but started eating a minute earlier. In the end, they each got an equal amount of cake. How long would it take Cheburashka to eat the cake alone?
|
Answer. In 4 minutes.
Solution.
First method. If Cheburashka eats twice as slowly as Gena, then to eat the same amount of cake as Gena, he needs twice as much time. This means that the time Cheburashka ate alone (1 minute) is half of the total time it took him to eat half the cake. Thus, he ate half the cake in 2 minutes, and the whole cake in 4 minutes.
Second method. Let Gena eat the whole cake in $x$ minutes, then Cheburashka needs $2x$ minutes for the whole cake. Each of them got half the cake, so Gena ate for $0.5x$ minutes, and Cheburashka for $x$ minutes. From the condition, it follows that $0.5x + 1 = x$, from which $x = 2$. Therefore, Cheburashka will eat the cake in $2 \cdot 2 = 4$ minutes.
## Grading criteria.
- Any complete correct solution - 7 points.
- The equation is correctly set up and solved, or correct reasoning is provided, but the answer is to a different question - 6 points.
- The solution considers a specific mass of the cake - 2 points.
- The equation is set up correctly but solved incorrectly - 2 points.
- The correct answer is provided, and it is verified that it satisfies the problem's condition - 1 point.
- Only the answer is provided - 0 points.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. There are seven cards on the table. In one move, it is allowed to flip any five cards. What is the minimum number of moves required to flip all the cards?
|
9.1. Answer. 3 moves.
Obviously, one move is not enough. After two moves, there will be at least three cards that have been flipped twice, which means these cards will be in their original position.
Let's provide an example of flipping all the cards in three moves. Number the cards from 1 to 7 and flip cards numbered $1,2,3,4,5$ on the first move, $-1,3,4,5,6$ on the second move, and $-1,3,4,5,7$ on the third move.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. In the country, there are 20 cities. An airline wants to organize two-way flights between them so that from any city, it is possible to reach any other city with no more than $k$ transfers. At the same time, the number of air routes from any city should not exceed four. What is the smallest $k$ for which this is possible?
## 9th grade
## Second day
|
9.4. Answer. $k=2$.
Note that at least two transfers will be required. Indeed, from an arbitrary city $A$ without a transfer, one can reach no more than 4 cities, and with exactly one transfer - no more than $4 \cdot 3=12$ cities (since one of the flights from each of these cities leads back to $A$). Therefore, if using no more than one transfer, one can reach no more than 16 other cities from any city, but it is required
Fig. 4
Fig. 5
- to reach 19. In Fig. 5, it is shown how flights can be organized so that there are no more than two transfers. The diagram is symmetric, so it is sufficient to show how to reach from the first city to any other. From it, without a transfer, one can reach cities $2,3,4,5$. Then with one transfer to cities $6,7,8$ (from 5), 9 (from 2), 13 (from 3), and 17 (from 4). And with two transfers - to cities 10, 11, 12 (from 9), to 14, 15, 16 (from 13), to $18,19,20$ (from 17).
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Each of the 5 brothers owns a plot of land. One day, they pooled their money, bought a neighbor's garden, and divided the new land equally among themselves. As a result, Andrey's plot increased by $10 \%$, Boris's plot - by $\frac{1}{15}$, Vladimir's plot - by $5 \%$, Grigory's plot - by $4 \%$, and Dmitry's plot - by $\frac{1}{30}$. By what percentage did the total area of their land increase as a result?
|
Answer: By $5 \%$.
Solution. Let A, B, V, G, D be the areas of the plots of each brother, respectively. Then, according to the problem, $\frac{1}{10} \mathrm{~A}=\frac{1}{15} \mathrm{~B}=\frac{1}{20} \mathrm{~V}=\frac{1}{25} \mathrm{~G}=\frac{1}{30} \mathrm{~D}$ (*). Denoting $\mathrm{A}=x$, we find: $\mathrm{B}=1.5 x, \mathrm{~V}=2 x, \mathrm{~G}=2.5 x$, D $=3 x$. Therefore, the total area of their plots was $10 x$, and increased by $\frac{1}{10} \mathrm{~A}=0.1 x$ for each, that is, by $0.5 x$ in total. Thus, the increase amounted to $\frac{0.5 x}{10 x} \cdot 100 \% = 5 \%$.
Comment. The relationship (*) is obtained - 2 points. The correct answer is obtained in the case of a particular example - 2 points. If arithmetic errors are made in the correct solution process, deduct $1-2$ points.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The participants of the Olympiad left 9 pens in the office. Among any four pens, at least two belong to the same owner. And among any five pens, no more than three belong to the same owner. How many students forgot their pens, and how many pens does each student have?
|
Answer. There are three students, each owning three pens.
Solution. No student owned more than three pens, as otherwise the condition "among any five pens, no more than three belonged to one owner" would not be met. There are a total of 9 pens, so there are no fewer than 3 students. On the other hand, among any four pens, there are two pens belonging to one student, so there are fewer than 4 students. Therefore, there are three students, and each forgot no more than three pens, with a total of 9 pens. This means each student forgot 3 pens.
Comment. A complete and justified solution - 7 points. Only proved that no student owned more than three pens - 1 point. Only proved that there are no fewer than 3 students - 2 points. Only proved that there are fewer than 4 students - 2 points. Proved that there are three students - 5 points. Provided the answer without justification or with incorrect justifications - 0 points.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter $\Gamma$, there is at least one colored cell?
|
Answer: 12.
Solution: Consider a $2 \times 3$ rectangle. It is obvious that at least 2 cells need to be colored in it. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown on the right.
Comment: A correct example with 12 colored cells - 3 points. An estimate that fewer than 12 cells are not enough - 4 points. The estimate should not rely on an example. Reasoning such as: "in my example, it is impossible to reduce the number of colored cells" is not an estimate.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's number, if it is known that after he ran away, 3 people remained in the line? (After each command, one or several players ran away, after which the line closed, and there were no empty spaces between the remaining players.)
|
Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
|
Answer: 3.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals are bisected by their intersection point $L$, it is a parallelogram (in particular, $AC = DX$). Therefore, $DX \parallel AC$. Since $AC \parallel ED$ by the problem's condition, the points $X, D, E$ lie on the same line.
Since $AC \parallel EX$, then $\angle EAX = \angle CAX = \angle AXE$, i.e., triangle $AEX$ is isosceles, $EA = EX$. Then
$$
ED = EX - XD = EA - AC = 15 - 12 = 3
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. Option 1.
In the village, 7 people live. Some of them are liars who always lie, while the rest are knights (who always tell the truth). Each resident said whether each of the others is a knight or a liar. A total of 42 answers were received, and in 24 cases, a resident called a fellow villager a liar. What is the minimum number of knights that can live in the village?
|
Answer: 3.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of "knight-liar," the phrase "He is a liar" will be said twice. Since this phrase was said 24 times in total, there are 12 pairs of "knight-liar," from which we can determine that there are 3 knights and 4 liars, or vice versa.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Variant 1.
A line parallel to the leg $A C$ of the right triangle $A B C$ intersects the leg $B C$ at point $K$, and the hypotenuse $A B$ at point $N$. On the leg $A C$, a point $M$ is chosen such that $M K=M N$. Find the ratio $\frac{A M}{M C}$, if $\frac{B K}{B C}=14$.
|
Answer: 7.
Solution.
Drop the altitude $M H$ from point $M$ in the isosceles triangle $M N K$. Then $M H K C$ is a rectangle and $M C=K H=H N$. Let $M C=K H=H N=y$. Let $K B=x$, then $C K=C B-K B=4 x-x=3 x$. From the similarity of triangles $N K B$ and $A C B$, it follows that $\frac{A C}{C B}=\frac{N K}{K B}$. Therefore, $\frac{A C}{4 x}=\frac{2 y}{x}$. Hence, $A C=8 y$. We get that $\frac{A M}{M C}=\frac{8 x-x}{x}=7$.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task No. 1.1
## Condition:
Five friends - Masha, Nastya, Irina, Olya, and Anya - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Irina: I was the first in line!
Olya: No one was after me.
Anya: Only one person was after me.
Masha: There were five people in front of me.
Nastya: I was standing next to Irina.
The girls are friends, so they don't lie to each other. How many people were between Masha and Nastya?
|
Answer: 3
Exact match of the answer - 1 point
## Solution.
From the statements of Irina and Olya, it is clear that they were first and last, respectively. Since there was only one person after Anya, it was Olya. Nastya stood next to Irina, but she could not have stood in front of her, so Nastya was second. This means that Masha stood somewhere between Nastya and Anya, and since there were five people in front of her, two of whom were Irina and Nastya, there were exactly three people between Masha and Nastya.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task No. 1.2
## Condition:
Five friends - Katya, Polina, Alyona, Lena, and Svetlana - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Polina: I stood next to Alyona.
Alyona: I was the first in line!
Lena: No one was after me.
Katya: There were five people in front of me.
Svetlana: Only one person was after me.
The girls are friends, so they don't lie to each other. How many people were between Katya and Polina?
|
# Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
#
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 1.3
## Condition:
Five friends - Sasha, Yulia, Rita, Alina, and Natasha - meet in the park every day after buying ice cream from the little shop around the corner. One day, the girls had a conversation.
Sasha: There were five people in front of me.
Alina: There was no one after me.
Rita: I was the first in line!
Natasha: There was only one person after me.
Yulia: I was standing next to Rita.
The girls are friends, so they don't lie to each other. How many people were there between Sasha and Yulia?
|
Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
#
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task No. 1.4
## Condition:
Five friends - Kristina, Nadya, Marina, Liza, and Galia - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Kristina: There were five people in front of me.
Marina: I was the first in line!
Liza: No one was behind me.
Nadya: I was standing next to Marina.
Galia: Only one person was behind me.
The girls are friends, so they don't lie to each other. How many people were between Kristina and Nadya?
|
# Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
#
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 3.2
## Condition:
Artyom makes watches for a jewelry store on order. Each watch consists of a bracelet and a dial. The bracelet can be leather, metal, or nylon. Artyom has round, square, and oval dials. Watches can be mechanical, quartz, or electronic.
Artyom is only satisfied when the watches are arranged in a row from left to right on the display according to the following rules:
- There must be mechanical watches with a round dial;
- Next to the watches with a round dial, there must be electronic watches to the right and quartz watches to the left;
- The three watches in the row must have different mechanisms, bracelets, and dials.
How many ways are there to make Artyom happy?
|
Answer: 12
Exact match of the answer -1 point
Solution by analogy with task №3.1
#
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Given a sequence $x_{n}$ such that $x_{1}=1, x_{2}=2, x_{n+2}=\left|x_{n+1}\right|-x_{n}$. Find $x_{2015}$.
|
5. Answer. ${ }^{x_{2015}}=0$.
We will show that the given sequence is periodic with a period of 9. Let's find
$x_{1}=1, x_{2}=2, x_{3}=1, x_{4}=-1, x_{5}=0, x_{6}=1, x_{7}=1, x_{8}=0, x_{9}=-1, x_{10}=1, x_{11}=2, \ldots$.
Since the sequence is completely determined by any two consecutive terms, we have obtained that 9 is the period of this sequence. This means that for any $k, r \in N$, the equality $x_{9 k+r}=x_{r}$ holds.
To find $x_{2015}$, we divide the number 2015 by 9 with a remainder: 2015=9$\cdot$223+8. Then $x_{2015}=x_{8}=0$.
## Evaluation Criteria and Organization of Work Checking
Mathematical Olympiad tasks are creative and allow for several different solutions. Additionally, partial progress in problems (for example, considering an important case, proving a lemma, finding an example, etc.) is evaluated. Thus, when calculating the final scores for a problem, the jury takes into account all these cases, as well as possible logical and arithmetic errors in the solutions.
The checking of works at the Mathematical Olympiad is carried out in two stages. In the first stage, the jury checks the works without assigning scores, using a so-called "plus-minus" system. The sign is assigned according to the table provided below. The preliminary evaluation in the "plus-minus" system may be slightly adjusted after discussing the criteria and classifying the cases.
| Correctness of Solution | |
| :---: | :--- |
| + | Complete correct solution |
| .+ | Correct solution. There are minor flaws that do not significantly affect the solution |
| $\pm$ | The solution is generally correct. However, the solution contains significant errors or cases are omitted, which do not affect the logical reasoning |
| $+/ 2$ | One of two (more complex) significant cases is correctly considered, or in a problem of the type "estimate + example," the estimate is correctly obtained |
| $\mp$ | Auxiliary statements that help in solving the problem are proven |
| .- | Important individual cases are considered in the absence of a solution |
| - | The solution is incorrect, and there is no progress |
| 0 | The solution is absent |
Sometimes the grade "++" is given to note a correct and elegant solution.
After the first stage of checking, the group of checkers for each problem, analyzing and generalizing the provided solutions, highlights various ways of solving, typical partial progress, and main errors. In accordance with the comparative analysis of different progress, a scale of evaluation criteria is developed.
In the second stage, final scores are assigned for each problem. According to the regulations for conducting school mathematical Olympiads, each problem is scored out of 7 points. The table below provides the scale for converting signs into points.
| Sign | + | .+ | $\pm$ | $+/ 2$ | $\mp$ | .- | - | 0 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Points | 7 | $6-7$ | $5-6$ | 4 | $2-3$ | $0-1$ | 0 | 0 |
The maximum score for completing all tasks is 35.
It is important to note that any correct solution is scored 7 points. Points are not deducted for a solution being too long or for a solution that differs from the one provided in methodological developments or from other solutions known to the jury.
At the same time, any arbitrarily long text of a solution that does not contain useful progress is scored 0 points.
A traditional mistake made by students when solving proof problems is using the statement to be proven as an initial condition or the basis of the proof. For example, in a problem requiring proof that a triangle is isosceles, the proof begins with the words: "Let triangle $A B C$ be isosceles." Such "solutions" are scored 0 points due to a gross logical error.
Each work is evaluated and checked (rechecked) by at least two members of the jury.
After the preliminary results of the Olympiad work checking are published, Participants have the right to review their works, including informing about their disagreement with the assigned scores. In this case, the Chairman of the Olympiad Jury appoints a jury member to reconsider the work. The score for the work can be changed if the Participant's request for a score change is deemed justified. The score change is agreed upon with the Chairman of the Jury and entered into the final table.
A final table is created for each parallel based on the results of the Olympiad.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3. Several different real numbers are written on the board. It is known that the sum of any three of them is rational, while the sum of any two of them is irrational. What is the largest number of numbers that can be written on the board? Justify your answer.
|
Solution: The set of three numbers $\sqrt{2}, \sqrt{3}, -\sqrt{2}-\sqrt{3}$, as is easily verified, satisfies the condition of the problem. We will show that no more than three numbers can be written on the board.
Assume the contrary, and let $a_{i} (i=\overline{1,4})$ be some four numbers from this set. Then the numbers $a_{1}+a_{2}+a_{3}$ and $a_{1}+a_{2}+a_{4}$ are rational, so the number $\left(a_{1}+a_{2}+a_{3}\right)-\left(a_{1}+a_{2}+a_{4}\right)=a_{3}-a_{4}$ is also rational. Similarly, the rationality of the number $a_{3}-a_{2}$ can be proven. Then the number $\left(a_{3}-a_{4}\right)+\left(a_{3}-a_{2}\right)-\left(a_{2}+a_{3}+a_{4}\right)=a_{3}$ is rational. The same can be shown for the rationality of all numbers $a_{i}$. But then the sum of any two of them is rational, which contradicts the condition of the problem.
Answer: Three numbers.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| There is a correct example of three numbers, but there are inaccuracies in the proof (which is basically correct) of its optimality | 5 points |
| Correct proof that 4 numbers cannot exist (in the absence of an example of 3 numbers) | 3 points |
| In a correct (in general) proof that 4 numbers cannot exist, there are inaccuracies and no example of 3 numbers | 2 points |
| An example of three numbers is given, but the proof of its optimality is missing | 1 point |
| Correct answer without justification (or with incorrect justification) | 0 points |
| Proof that there are fewer than 5 (or even more) numbers | not evaluated |
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Plot the graph of the function $y=(4 \sin 4 x-2 \cos 2 x+3)^{0.5}+(4 \cos 4 x+2 \cos 2 x+3)^{0.5}$.
|
Solution:
$\mathbf{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2+4 \sin ^{2} x+3}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x-2+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x+1}$
$\mathrm{y}=2 \sin ^{2} x+1+2 \cos ^{2} x+1, \mathrm{y}=4$.
Answer: The graph of the function will be a straight line $y=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Two very small fleas are jumping on a large sheet of paper. The first jump of the fleas is along a straight line towards each other (their jumps may have different lengths). The first flea first jumps to the right, then up, then to the left, then down, then to the right again, and so on. Each jump is 1 cm longer than the previous one. The second flea first jumps to the left, then up, then to the right, then down, then to the left again, and so on. Each subsequent jump is also 1 cm longer than the previous one. After 100 jumps, the fleas are 3 meters apart. How far apart were the fleas initially?
|
4. Answer: 2 meters. Solution: Let's observe the first four jumps of the first flea: 2 cm more to the left than to the right, and 2 cm more down than up. That is, after four jumps, the first flea moves 2 cm to the left and 2 cm down. Therefore, after 100 jumps, it will move 50 cm to the left and 50 cm down. Similarly, the second flea will move 50 cm to the right and 50 cm down after 100 jumps. Since both fleas have moved down the same amount, the distance between them has increased by $50 \mathrm{~cm} + 50 \mathrm{~cm} = 100 \mathrm{~cm} = 1$ m. Therefore, the initial distance between them was 2 m. Criteria: correct solution - 7 points. Correct reasoning but with an arithmetic error - 5 points. Correct conclusion about the change in position of the flea after a series of four jumps, but no further progress - 2 points. In all other cases, 0 points.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. Find the maximum value of the expression $a+b+c+d-ab-bc-cd-da$, if each of the numbers $a, b, c$ and $d$ belongs to the interval $[0 ; 1]$.
|
Answer: 2.
Solution. The first method. The value 2 is achieved, for example, if $a=c=1, b=d=0$. We will prove that with the given values of the variables $a+b+c+d-ab-bc-cd-da \leqslant 2$.
Notice that $a+b+c+d-ab-bc-cd-da=(a+c)+(b+d)-(a+c)(b+d)$. Let $a+c=x, b+d=y$, then we need to prove that $x+y-xy \leqslant 2$, if $0 \leqslant x \leqslant 2$ and $0 \leqslant y \leqslant 2$.
Indeed, $x+y-xy=(x+y-xy-1)+1=(x-1)(1-y)+1$, where $-1 \leqslant x-1 \leqslant 1$ and $-1 \leqslant 1-y \leqslant 1$. Therefore, $(x-1)(1-y) \leqslant 1$, so $x+y-xy \leqslant 2$.
By introducing new variables, we can reason differently. Fix the variable $y$ and consider the function $f(x) = (1-y)x + y$, where $x \in [0; 2]$. Since it is linear, its maximum value is achieved at one of the endpoints of the interval $[0; 2]$. But $f(0) = y \leqslant 2$ and $f(2) = 2 - y \leqslant 2$, so for all $x \in [0; 2]$ the inequality $f(x) \leqslant 2$ holds.
The second method. The same idea of linearity can be used initially. The given expression can be considered a linear function of one of the variables if the other three variables are fixed.
For example, fix the values of the variables $b, c$, and $d$ and consider the function $f(a) = (1-b-d)a + b + c + d - bc - cd$, where $a \in [0; 1]$. Due to monotonicity, its maximum value is achieved at one of the endpoints of the interval $[0; 1]$. The same situation will arise for similarly defined linear functions of the variables $b$, $c$, and $d$.
Therefore, the maximum value of the original expression can only be achieved when the variables $a, b, c$, and $d$ take one of two values: 0 or 1. Considering the symmetry of the given expression, it is sufficient to check the following cases:
1) if $a=b=c=d=0$ or $a=b=c=d=1$, the value of the expression is 0;
2) if $a=b=c=0, d=1$ or $a=b=c=1, d=0$, the value of the expression is 1;
3) if $a=b=0, c=d=1$ or $a=b=1, c=d=0$, the value of the expression is 1;
4) if $a=c=0, b=d=1$ or $a=c=1, b=d=0$, the value of the expression is 2.
Thus, the maximum value of the given expression is 2.
## Grading Criteria:
+ The correct answer and a fully justified solution are provided
$\pm$ The correct answer and generally correct reasoning are provided, but there are minor gaps or inaccuracies (e.g., the estimate is proven, but an example is missing)
干 The correct line of reasoning is provided, but a computational error is made
$\mp$ The idea of linearity is present, but it is not fully developed
干 The correct answer is provided, and the values of the variables at which it can be achieved are indicated, but the estimate is not conducted
- Only the answer is provided
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. There was a whole number of cheese heads on the kitchen. At night, rats came and ate 10 heads, and everyone ate equally. Several rats got stomachaches from overeating. The remaining seven rats the next night finished off the remaining cheese, but each rat could eat only half as much cheese as the night before. How much cheese was there in the kitchen initially?
|
Solution. Let there be $k$ rats in total $(k>7)$, then each rat ate $\frac{10}{k}$ pieces of cheese on the first night. On the second night, each rat ate half as much, that is, $\frac{5}{k}$ pieces. Then seven rats ate $\frac{35}{k}$ pieces. This is an integer. The only divisor of the number 35 that exceeds 7 is the number 35 itself. Therefore, $\frac{35}{k}=1$, and there were $10+1=$ 11 pieces of cheese in the warehouse before the rat invasion. Answer: 11.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 4 blue points. And on any segment with endpoints at blue points, containing 3 blue points inside, there are at least 2 red points. What is the maximum number of blue points that can be on a segment with endpoints at red points, not containing other red points?
|
Answer: 4.
Solution: Note that on a segment with endpoints at red points, not containing other red points, there cannot be 5 blue points. Indeed, in this case, between the outermost blue points, there would be 3 blue points, which means there would be at least 2 more red points. Therefore, on such a segment, there are no more than 4 blue points.
We will show that 4 blue points can lie on a segment with endpoints at red points, not containing other red points. Suppose the points are arranged on a line in the following order: 2 red - 4 blue - 2 red - 4 blue - 2 red. Then all conditions are satisfied, and there is a segment with 4 blue points.
Comment: It is proven that there are no more than 4 blue points between neighboring red points - 4 points.
A correct example with 4 blue points is provided - 3 points.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. In the castle, there are 9 identical square rooms, forming a $3 \times 3$ square. Nine people, consisting of liars and knights (liars always lie, knights always tell the truth), each occupied one of these rooms. Each of these 9 people said: "At least one of the neighboring rooms to mine is occupied by a liar." What is the maximum number of knights that could be among these 9 people? Rooms are considered neighboring if they share a common wall.
|
Answer: 6 knights.
Solution: Note that each knight must have at least one neighbor who is a liar. We will show that there must be at least 3 liars (thus showing that there are no more than 6 knights). Suppose there are no more than 2 liars, then there will be a "vertical row" of rooms where only knights live. But then each of these knights must have a neighbor who is a liar (and these neighbors are different). Therefore, there must be at least 3 liars.
The diagram below shows how 6 knights and 3 liars could be accommodated.
| $\mathrm{P}$ | L | $\mathrm{P}$ |
| :--- | :--- | :--- |
| $\mathrm{P}$ | $\mathrm{P}$ | L |
| L | $\mathrm{P}$ | $\mathrm{P}$ |
Comment: A correct answer without justification - 0 points.
Example of accommodating 3 liars and 6 knights - 2 points.
Proven that there are no more than 6 knights - 5 points.
Note 1: In the example, the liars should not be neighbors.
Note 2: There are other examples.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. For what largest $k$ can we assert that in any coloring of $k$ cells in black in a white $7 \times 7$ square, there will necessarily remain a completely white $3 \times 3$ square with sides along the grid lines?
|
Answer: 3
Solution. Let's highlight four $3 \times 3$ squares that are adjacent to the corners of the $7 \times 7$ square. These squares do not overlap, so if no more than three cells are shaded, at least one of these $3 \times 3$ squares will remain entirely white. If, however, we shade 4 cells, marked in gray on the diagram, then no white $3 \times 3$ square will remain.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. A pedestrian left point $A$ for point $B$. At the same time, a cyclist left point $B$ for point $A$. After one hour, the pedestrian was three times farther from the cyclist than from point $A$. Another 30 minutes later, they met, after which both continued their journey. How many hours did it take the pedestrian to travel from $A$ to $B$?
|
Answer: 9
Solution. Let the distance from $A$ to $B$ be 1 km, the pedestrian moves at a speed of $x$ km/h, and the cyclist at $y$ km/h. Then in one hour, the pedestrian has walked $x$ km, the cyclist has traveled $y$ km, and the distance between them is $1-x-y$, which should be three times $x$. Therefore, $3 x=1-x-y, y=1-4 x$. In the next half hour, they will together cover $1-x-y$ km. Since their closing speed is $x+y$, we get the equation $\frac{1}{2}(x+y)=1-x-y$, from which $x+y=\frac{2}{3}$. Substituting $y=1-4 x$ into the last equation: $1-3 x=\frac{2}{3}, x=\frac{1}{9}$. Therefore, it will take the pedestrian 9 hours to travel the distance from $A$ to $B$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4. We will call a natural number interesting if it is the product of exactly two (distinct or equal) prime numbers. What is the greatest number of consecutive numbers, all of which are interesting
|
Solution. One of four consecutive numbers is divisible by 4. However, among the numbers divisible by 4, only the number 4 itself is interesting. But the numbers 3 and 5 are not interesting, so four consecutive interesting numbers do not exist. An example of three consecutive interesting numbers: $33,34,35$. Answer: three.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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