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Problem 5. A tennis tournament was held among 20 schoolchildren. Each participant played one match every day; in total, over 19 days, each played exactly once with all the others. There was only one tennis court at the school, so the matches were played in sequence. Immediately after their first win in the tournament, ... | Answer: Also on the eleventh.
Solution. On the first day, there were 10 matches, and thus 10 T-shirts were given out. The eleventh T-shirt was only given out on the eleventh day of the tournament, which means that Petya and nine other participants did not win a single match in the first ten days of the tournament. Thi... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,745 |
Problem 6. A row of new recruits stood facing the sergeant. On the command "left," some turned left, while the rest turned right. It turned out that six times more soldiers were looking at the back of their neighbor than in the face. Then, on the command "about face," everyone turned in the opposite direction. Now, sev... | Answer: 98.
Solution. Let's assume that the sergeant lined up the soldiers between two posts. After the first command, each recruit either looks at the back of the neck of the neighbor or at the face, except for the two soldiers at the ends, who can look at the posts.
If a soldier is looking at the back of the neck o... | 98 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,746 |
Problem 1. Today is 17.02.2008. Natasha noticed that in the record of this date, the sum of the first four digits equals the sum of the last four. When will such a coincidence happen for the last time this year?
$[3$ points] (N. M. Netrosova) | Answer: December 25, 2008.
Solution. It is not difficult to verify that such a coincidence will not occur again in the remaining days of the year. | December25,2008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,747 |
Problem 2. The she-rabbit bought seven drums of different sizes and seven pairs of sticks of different lengths for her seven bunnies. If a bunny sees that both its drum is larger and its sticks are longer than those of one of its brothers, it starts to drum loudly. What is the maximum number of bunnies that can start d... | Answer: 6 baby rabbits.
Solution: Not all baby rabbits can play the drum, as the baby rabbit that gets the smallest drum will not play it. On the other hand, if the same baby rabbit is also given the shortest drumsticks, then all the other baby rabbits will play the drum. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,748 |
Problem 3. There were several whole cheese heads on the warehouse. At night, rats came and ate 10 heads, and everyone ate equally. Several rats got stomachaches from overeating. The remaining 7 rats the next night finished off the remaining cheese, but each rat could eat only half as much cheese as the night before. Ho... | Answer: 11 cheese heads
Solution: Let the total number of rats be $k$ ( $k>7$ ), then each rat ate $10 / k$ cheese heads on the first night. On the second night, each rat ate half as much, that is, $5 / k$ cheese heads. Thus, the 7 rats ate a total of $35 / k$ cheese heads. This is an integer. The only divisor of the ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,749 |
Problem 6. Vasilisa the Wise decided to lock Koschei in a straight corridor divided by three passages into four rooms, with a fat, tired guard leaning against one of the walls in each passage. Every time Koschei moves from one room to another, the guard moves to the opposite wall and leans against it. If all the guards... | Answer: Yes, it can.
Solution. Suppose, for example, Vasilisa placed Koschei in the northernmost room and positioned the guards as follows: to the west - to the east - to the west wall ("W-E... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,752 |
Task 1. A number was multiplied by the sum of its digits and the result was 2008. Find this number.
$[4$ points] (I. V. Yashchenko) | Answer: 251.
Solution: The desired number is a divisor of the number 2008. Let's factorize the number 2008 into prime factors: $2008=2 \cdot 2 \cdot 2 \cdot 251$. List all divisors of the number $2008: 1,2,4,8,251,502,1004,2008$. By finding the sum of the digits of each of them, we notice that the condition of the pro... | 251 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,753 |
Task 2. In the Water Supply Cup football tournament, the teams "Pump", "Filter", "Pump", and "Gate" participated. Each team played against each other once (3 points were awarded for a win, 1 for a draw, and 0 for a loss). The team "Pump" scored the most points, and the team "Gate" scored the least. Could it have been t... | Answer: Yes, it could. An example is provided in the table:
| team | P | F | N | S | p |
| :--- | :---: | :---: | :---: | :---: | :---: |
| Pump | - | 1 | 1 | 3 | 5 |
| Filter | 1 | - | 3 | 0 | 4 |
| Pump | 1 | 0 | - | 3 | 4 |
| Gate | 0 | 3 | 0 | - | 3 |
Notes. 1. Since "Pump" overtook "Filter," and "Filter" overtoo... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,754 |
Problem 3. Dima lives in a nine-story building. He descends from his floor to the first floor by elevator in 1 minute. Due to his small stature, Dima cannot reach the button for his floor. Therefore, when going up, he presses the button he can reach, and then walks the rest of the way. The entire journey upwards takes ... | Answer. Dima lives on the seventh floor.
First solution. Consider the part of the journey that Dima travels down by elevator and up on foot. On the one hand, the walk takes twice as long, and on the other, it is 10 seconds longer. Therefore, he traveled this part by elevator in 10 seconds and walked it in 20 seconds. ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,755 |
Task 5. Sergei cut out two identical figures from cardboard. He placed them with an overlap on the bottom of a rectangular box. The bottom turned out to be completely covered. A nail was hammered into the center of the bottom. Could the nail have pierced one piece of cardboard and not the other?
[8 points] (S. V. Mark... | Answer. Yes, he could.
Solution. Let the cardboard pieces initially lie one on top of the other with their edges coinciding. If the cardboard pieces are square, by moving one along the other and overlapping them, it is easy to form a rectangle, but its center will lie inside both (see Fig. $5, a$). This can be correct... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,756 |
Problem 6. Vasya stood at the bus stop for some time. During this time, one bus and two trams passed by. After some time, Spy came to the same stop. While he was sitting there, 10 buses passed by. What is the minimum number of trams that could have passed during this time? Both buses and trams run at equal intervals, a... | Answer: 4 trams.
Solution. First, let's provide an example where four trams passed during the Spy's observation time: suppose buses run at 9:00, 10:00, ..., and trams run with an interval of 1 hour and 58 minutes - at 10:01, 11:59, 13:57, 15:55, 17:53, 19:51, 21:49, ... Then Vasya could have been standing from 10:01 t... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,757 |
Problem 3. Bus stop $B$ is located on a straight highway between stops $A$ and $C$. After some time following departure from $A$, the bus found itself at a point on the highway such that the distance from this point to one of the three stops is equal to the sum of the distances to the other two. After the same amount o... | Answer: 3 hours.
Solution. At both moments in time mentioned in the problem, the sum of distances will obviously be the distance from the bus to the farthest stop from it. This cannot be $B$, as it is closer than $C$. Therefore, these were $C$ (before the bus had traveled halfway from $A$ to $C$) and $A$ (after this m... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,760 |
Problem 4. The teacher wrote a two-digit number on the board and asked Dima in turn whether it is divisible by $2?$ by 3? by $4?$... by 9? Dima answered all eight questions correctly, and the number of "yes" and "no" answers was equal.
a) Can you now answer at least one of the teacher's questions correctly without kno... | Answer. a) Yes, on the first. b) No, without knowing the number, it is impossible to guarantee an answer.
Solution. a) We will show that the written number is even. If it were odd, then the answers to the questions about divisibility by 2, 4, 6, and 8 would be "no," and then, accordingly, the answers to the questions ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,761 |
Problem 5. Six mathematicians went fishing. Together they caught 100 fish, and each caught a different number of fish. After fishing, they noticed that any one of them could distribute all their fish to the other five fishermen so that each of the remaining five would have an equal number of fish. Prove that one fisher... | Solution. After one fisherman distributes his fish, the others should each have $100: 5=20$ fish. This means that each caught no more than 20 fish. Suppose Fisherman Ivan has exactly 20 fish. When another mathematician distributes his fish, Ivan does not receive anything, but everyone ends up with the same amount. Ther... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,762 |
Problem 6. Cut a $9 \times 9$ square grid into three figures of equal area such that the perimeter of one of the parts is equal to the sum of the perimeters of the other two.
$[8$ points] (M.A. Evdokimov) | Answer. Examples are shown in the figures. Other solutions are possible.
Comment. Let's show how one can come up with the required dissection. It is clear that each figure should contain $9 \... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,763 |
Problem 2. Doctor Aibolit wants to visit a cow, a wolf, a bug, and a worm. All four live along the same straight road. In the morning, eagles are ready to deliver Aibolit to the first patient, and in the evening, they will pick him up from the last one, but the three intermediate transitions he will have to make on foo... | Answer. The figure shows 4 possible options (the segment is marked in 1 km increments; we draw the scheme so that the cow is to the left of the worm):
Comment. No matter where the eagles del... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,765 |
Problem 5. In quadrilateral $A B C D$, it is known that $A B=$ $=B C=C D, \angle A=70^{\circ}$ and $\angle B=100^{\circ}$. What can the angles $C$ and $D$ be equal to? $\quad[8$ points] (M.A.Volchkevich) | Answer: $60^{\circ}$ and $130^{\circ}$ or $140^{\circ}$ and $50^{\circ}$.
First Solution. Draw segment $B E$ such that point $E$ lies on $A D$, and angle $A B E$ is $40^{\circ}$. Then $\angle A E B = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ}$, hence triangle $A B E$ is isosceles, $A B = B E$. Consider triangl... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,766 |
Task 1. The year 2009 has the following property: by swapping the digits of the number 2009, it is impossible to obtain a smaller four-digit number (numbers do not start with zero). In which year will this property reoccur for the first time? | Answer. In 2022.
Solution. In the years $2010, 2011, \ldots, 2019$ and in 2021, the year number contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be decreased by rearranging the digits. | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,767 |
Problem 3. In the park, there were lindens and maples. Maples among them were $60 \%$. In spring, lindens were planted in the park, after which maples became $20 \%$. In autumn, maples were planted, and maples became $60 \%$ again. By what factor did the number of trees in the park increase over the year?
[6 points] (... | Answer: 6 times.
First solution. Before the planting, lindens constituted $2 / 5$, and maples $-3 / 5$ of all the trees in the park. By summer, the number of maples did not change, but they began to constitute $1 / 5$ of all the trees. Therefore, the total number of trees in the park increased threefold. At the same t... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,769 |
Problem 4. If an octopus has an even number of legs, it always tells the truth. If it has an odd number, it always lies. One day, a green octopus said to a dark blue one:
- I have 8 legs. And you only have 6.
- I have 8 legs, - the dark blue one was offended. - And you only have 7.
- The dark blue one really has 8 le... | Answer: The striped octopus has 8 legs.
Solution: If the purple octopus is telling the truth, then it has an even number of legs. But in this case, it cannot say that it has 9 legs. Therefore, the purple octopus is lying. So, the dark blue octopus does not have 8 legs. But the dark blue octopus says that it has 8 legs... | The\striped\octopus\has\8\legs | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,770 |
Task 5. A curious tourist wants to stroll through the streets of the Old Town from the station (point $A$ on the map) to their hotel (point $B$). The tourist wants their route to be as long as possible, but it is not interesting for them to visit the same intersection twice, and they do not do so. Draw the longest poss... | Solution. One of the possible routes of the tourist is shown in the figure. By following this path, the tourist will walk 34 streets (a street is defined as a segment between two adjacent intersections). We-
 A miserly knight keeps his gold coins in six chests. One day, while counting them, he noticed that if he opens any two chests, he can redistribute the coins in them equally between these two chests. He also noticed that if he opens any 3, 4, or 5 chests, he can also redistribute the coins in them so that ... | Answer. a) Yes b) No.
a) First solution. Divide the chests into three pairs. The total number of coins in each pair of chests is even, so the number of coins in all six chests is even as well. Now divide the chests into two triplets. The number of coins in each triplet is divisible by three, so the total number of coi... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,772 |
Problem 1. Pete and Vasya live in neighboring houses (see plan). Vasya lives in the fourth entrance. It is known that for Pete to run to Vasya by the shortest path (not necessarily along the sides of the cells), it does not matter which side he runs around his house. Determine which entrance Pete lives in.
## [5 point... | Answer. In the sixth entrance.
Solution. The shortest path from point $A$ to Vasya's entrance is the segment $A D$. The shortest path from point $B$ to Vasya's entrance is the path along segment $B C$, and then along segment $C D$. Since triangles $A E D$ and
![](https://cdn.mathpix.com/cropped/2024_05_06_37ba496d27b0... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,773 |
Problem 2. On each of the two gardens, Grandpa planted the same number of turnips. If Granddaughter enters the garden, she pulls out exactly $1 / 3$ of the turnips present at that moment. If Doggy enters, she pulls out $1 / 7$ of the turnips, and if Mousey enters, she pulls out only $1 / 12$ of the turnips. By the end ... | Answer. Yes, she did.
First solution. Each time after granddaughter visits the garden, 2/3 of the turnips that were there before remain, after the visit of the Ladybug, -6/7, and after the visit of the Mouse - 11/12. Therefore, the number of turnips in the garden cannot become a multiple of 7 after the visit of any of... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,774 |
Problem 3. The underwater king has octopuses with six, seven, or eight legs in his service. Those with 7 legs always lie, while those with 6 or 8 legs always tell the truth. Four octopuses met. The blue one said: "We have 28 legs in total," the green one: "We have 27 legs in total," the yellow one: "We have 26 legs in ... | Answer: The green octopus has 6 legs, while the others have 7 legs each.
Solution: Since the octopuses contradict each other, there are two possible cases: either all the octopuses are lying, or exactly one of them is telling the truth.
If all the octopuses are lying, then each of them has 7 legs. Therefore, together... | Thegreenoctopushas6legs,whiletheothershave7legseach | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,775 |
Problem 4. A miserly knight keeps gold coins in 77 chests. Once, while counting them, he noticed that if he opened any two chests, he could redistribute the coins lying in them equally between these two chests. Then he noticed that if he opened any 3, or any 4, ..., or any 76 chests, he could also redistribute the coin... | Answer. Yes: all the coins can be evenly distributed among all the chests.
Solution. Divide the chests into 11 groups of 7 chests each. The total number of coins in each group of chests must be divisible by 7, which means the total number of coins in all 77 chests is also divisible by 7.
Now divide the chests into 7 ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,776 |
Problem 6. Using any number of coins with denominations of $1, 2, 5$, and 10 rubles, as well as (free) parentheses and the four arithmetic operation signs, form an expression with a value of 2009, spending as little money as possible.
[28 rubles - 3 points; 26 rubles - 5 points;
24 rubles - 7 points; 23 rubles - 10 p... | Answer. The number 2009 can be obtained for 23 rubles in the following way: $2009=(2 \cdot(2+1)+1)(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot(2+1) \cdot(2+1)-1)$.
Solution. Let's explain how this problem could be solved. Note that the expression can be formed from arbitrary numbers, replacing each with the sum of the cor... | 2009=(2\cdot(2+1)+1)(2\cdot2\cdot2\cdot2\cdot2\cdot(2+1)\cdot(2+1)-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,778 |
Problem 1. Two sums are written on the board:
$$
\begin{array}{r}
1+22+333+4444+55555+666666+7777777+ \\
+88888888+999999999
\end{array}
$$
$9+98+987+9876+98765+987654+9876543+$
$+98765432+987654321$
Determine which of them is greater (or if they are equal).
$[4$ points] (G. Galperin) | Answer: The sums are equal.
Solution. Let's write both sums in a column, and for better clarity, the second one in reverse order. In both sums, the digits from 1 to 9 will be added in the units place, the digits from 2 to 9 in the tens place, the digits from 3 to 9 in the hundreds place, and so on. The digit obtained ... | 1097393685 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,779 |
Problem 2. Vasya solved problems for 10 days - at least one problem each day. Each day (except the first), if the weather was cloudy, he solved one more problem than the previous day, and if it was sunny, one less problem. In the first 9 days, Vasya solved 13 problems. What was the weather like on the tenth day? [5 poi... | Answer: Cloudy.
Solution. Consider any two consecutive days. On each day, at least one problem is solved, but it cannot be exactly one on both days, so at least three problems are solved over these two days. Thus, over the first 8 days, Vasya solved at least \(4 \cdot 3 = 12\) problems. If he solved at least two probl... | Cloudy | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,780 |
Problem 3. One hundred carousel seats are arranged in a circle at equal intervals. Each is painted yellow, blue, or red. Seats of the same color are placed consecutively and numbered $1,2,3, \ldots$ clockwise. The blue seat № 7 is opposite the red № 3, and the yellow № 7 is opposite the red № 23. Find how many yellow, ... | Answer. Yellow 34, blue 20, red 46.
Solution. There are 19 seats between the 3rd and 23rd red seats, so there are also 19 seats between the 7th blue and the 7th yellow seats. This includes the first six yellow seats, and thus $19-6=13$ blue seats with numbers greater than 7. Therefore, the total number of blue seats i... | Yellow34,blue20,red46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,781 |
Problem 5. The figure "violinist" attacks the cell to the left on the side (with the elbow) and the cell to the upper right diagonally (with the bow), if he is right-handed, and, conversely, the right cell on the side and the upper left cell diagonally, if he is left-handed (all violinists are facing us). Place as many... | Solution. Placing 32 violinists is not difficult: for example, you can fill four columns every other one (it doesn't matter whether they are right-handed or left-handed). However, it is possible to place more. An example with 34 violinists is shown in the figure.
| $\mathrm{p}$ | $\Lambda$ | | | $\mathrm{p}$ | $\Lam... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,783 |
Task 6. Koschei imprisoned a crowd of captives and told them: “Tomorrow you will face a trial. I will choose several of you (whomever I wish, but at least three), seat you around a round table in some order (whichever I desire), and stick a piece of paper with a figure on each of your foreheads. The figures may repeat,... | Solution. Since there are different numbers of figurines, one figurine is used more than the others. Let's call it the main figurine. If figurines of any other shape are at least two fewer, then every prisoner can see which figurine is the main one. If everyone draws it, there will be those who guess correctly.
Suppos... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,784 |
Problem 1. Anya calls a date beautiful if all 6 digits of its notation are different. For example, 19.04.23 is a beautiful date, while 19.02.23 and 01.06.23 are not. How many beautiful dates are there in 2023? 3 $[4$ points] (M. Evdokimov) | Answer: 30.
Solution: The digits 2 and 3 are already used in the year number, so we need to consider only the months 01, 04, 05, 06, 07, 08, 09, and 10. Each of these month numbers contains a 0, so in a beautiful date, there will be no day number starting with 0, 2, or 3, and there will also be no days 10, 11, 12, and... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,786 |
Problem 4. Two squares are arranged as shown in the figure, the marked segments are equal. Prove that triangle $B D G$ is isosceles.
$[6$ points]
 | Solution 1. Consider triangles $D E G$ and $D E B$. They share the side $D E$, and the sides $E G$ and $E B$ are equal (as two sides of the square). It remains to prove that the angles $D E G$ and $D E B$ are equal, - then the specified triangles will be equal (by two sides and the angle between them), and therefore, t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,788 |
Problem 5. In the 7th grade, there are 100 students, some of whom are friends with each other. On September 1, they organized several clubs, each founded by three students (each club had its own). Every day, students who were friends with at least three members of a club joined that club. By February 19, all students i... | Answer: Yes, it could.
Solution. Divide the seventh graders into two groups of 50 students each. Let all students in each group be friends with everyone, and let three students $A, B$, and $C$ from the first group have three different friends each in the second group, and no one else be friends with anyone else.
. For Sanya to now get all of Anya's mushrooms, he needs to take the mushrooms from Vitya and Anya. Then he will have the mushr... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,791 |
Problem 3. In the morning, a dandelion blooms, it flowers yellow for two days, on the third day in the morning it turns white, and by evening it sheds its seeds. Yesterday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and today there are 15 yellow and 11 white.
a) How many yellow dandelions wer... | Answer. a) 25 yellow dandelions; b) 9 white dandelions.
Solution. a) All dandelions that were yellow the day before yesterday have turned white yesterday or today. Therefore, there were $14+11=25$.
b) Out of the yellow dandelions from yesterday, 11 turned white today, and the remaining $20-11=9$ will turn white tomor... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,793 |
Problem 5. Mom baked pies - three with rice, three with cabbage, and one with cherries - and laid them out in a circle on a plate (see fig.). Then she put the plate in the microwave to warm them up. On the outside, all the pies look the same.
. Prove that the shaded triangle
; therefore, its sides $A B$ and $B C$ are equal. But its sides $A B$ and $A C$ are also equal (as corresponding sides of equal paper triangles); therefore, triangle $A B C$ is equilateral.
Comments. 1. The so... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,798 |
Task 3. Replace the letters in the word MATHEMATICA with digits and addition and subtraction signs so that a numerical expression equal to 2014 is obtained. (Identical letters represent identical digits or signs, different letters represent different digits or signs. It is sufficient to provide an example.) [6 points] ... | Answer. $183+1839-8=2014$ M A T E M A T I K A
Comment. To find the solution (and also to prove its uniqueness), it is useful to think about which of the letters are operation signs.
It is clear that A cannot be a sign (as the last symbol of the expression). Then, there must be at least one sign between the first two ... | 183+1839-8=2014 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,799 |
Problem 4. In the morning, a dandelion blooms, it flowers yellow for three days, on the fourth day in the morning it turns white, and by the evening of the fifth day, it withers. On Monday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and on Wednesday there were 15 yellow and 11 white. How many... | Answer: 6 white dandelions.
Solution: A blooming dandelion is white on the fourth and fifth day. Therefore, on Saturday, the dandelions that bloomed on Tuesday or Wednesday will be white. Let's determine how many there are.
14 dandelions that were white on Monday had all their seeds dispersed by Wednesday, while 20 y... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,800 |
Problem 6. On the board, two numbers are written: 2014 and 2015. Petya and Vasya take turns, with Petya starting. In one move, they can
- either decrease one of the numbers by its non-zero digit or by a non-zero digit of the other number;
- or divide one of the numbers by two if it is even.
The player who first write... | Answer: Petya can win.
Solution. Let Petya first replace 2015 with 2014, and with each subsequent move, he will equalize the numbers (he can always do this by repeating Vasya's move with the number Vasya did not change):
 were either seals or otters, and Igor remembered that 5 were either otters or sea l... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,803 |
Task 2. Find some solution to the rebus $\Phi / \mathrm{E}+$ $+\mathrm{BP} / \mathrm{A} / \mathrm{J}=1$. Different letters correspond to different digits. The line denotes division. [5 points] (E. Akopyan) | Answer. There are different options, for example:
$$
2 / 4+79 / 158=1, \quad 6 / 8+35 / 140=1, \quad 4 / 5+72 / 360=1
$$
and others. | 2/4+79/158=1,\quad6/8+35/140=1,\quad4/5+72/360=1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,804 |
Problem 3. Vanya assembled a $3 \times 3 \times 3$ cube from red and blue blocks of size $1 \times 1 \times 3$. Then he started drawing what he had created. When Tanya arrived, Vanya had only colored 8 out of 27 cells on the visible surface of the drawn cube (see the image below).
$ and a square $AKLM$ are positioned as shown in the figure. Point $S$ on $AB$ is such that $AS = SL$. Find the measure of angle $SLB$.
[8 points] (L. Ponov) | Answer: $90^{\circ}$.
Solution. Consider triangles $A K S$ and $L K S$. They are equal by three sides.
Therefore, angles $K A S$ and $K L S$ are equal.
In the isosceles triangle $A B C$, ang... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,806 |
Problem 6. The tsar has 7 bags of gold coins, each containing 100 coins. The tsar remembers that in one bag all the coins weigh 7 g, in the second 8 g, in the third 9 g, in the fourth 10 g, in the fifth 11 g, in the sixth 12 g, and in the seventh 13 g, but he does not remember which is which. The tsar informed the cour... | Answer. a, b) Yes, it can.
Solution. a) Let's take one coin from each of the bags, except the specified one, and place them on the left pan, and put 8 coins from the specified bag on the right pan.
Then, if the coins in the specified bag weigh the minimum (7 g), the right pan will be lighter: $8+9+$ $+10+11+12+13>8 \... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,807 |
Problem 2. a) Fill in each circle with a non-zero digit so that the sum of the digits in the two top circles is 7 times less than the sum of the other digits, and the sum of the digits in the two left circles is 5 times
 See the diagram.
Solution. b) If the sum of the digits in the two upper circles is 7 times less than the sum of the remaining digits, then it is 8 times less than the sum of all five digits. Reasoning similarly, we get that
 | Answer: 45.
Solution. Let's set the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed the monkeys no more than one tangerine each, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. Let's show how 45 monkeys can be made happy:
5 mo... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,812 |
Problem 6. Yura drew a rectangle (along the grid lines) on graph paper and painted a picture on it. After that, he drew a frame one cell wide around the picture (see figure). It turned out that the area of the picture is equal to the area of the frame.
. Then, in the small frame, as in the large one, there will be four corner cells (they are shaded), and each sid... | 3\times10or4\times6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,813 |
Problem 2. In an aquarium, there are three types of fish: gold, silver, and red. If the cat eats all the gold fish, the number of fish will be 1 less than $2 / 3$ of the original number. If the cat eats all the red fish, the number of fish will be 4 more than $2 / 3$ of the original number. Which fish - gold or silver ... | Answer: There are 2 more silver fish.
Solution: From the first condition, there is 1 more goldfish than a third. From the second condition, there are 4 fewer red fish than a third. Therefore, there are 3 more silver fish than a third. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,815 |
Problem 4. The Smeshariki live on the shores of a pond in the shape of an equilateral triangle with a side length of 600 meters. Krosh and Barash live on one shore, 300 meters apart from each other. In the summer, it is 900 meters for Losharik to walk to Krosh, and 900 meters for Barash to walk to Nyusha. Prove that in... | Solution. Let Krosh live at a distance $x$ from the nearest corner of the pond, $A K=x$. Then the distance from Barash to his corner of the pond is $B B=600-300-x=300-x$. Now, by the condition, $B J=900-B K=300+x$ (note that since 900 is exactly half the perimeter of the pond, it doesn't matter which of the two paths L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,816 |
Problem 5. There is a set of two cards: 1 and 2. In one operation, it is allowed to form an expression using the numbers on the cards, arithmetic operations, and parentheses. If its value is a non-negative integer, it is issued on a new card. (For example, having cards 3, 5, and 7, you can form the expression 7 [ 5/3 a... | Answer. a) For example
$[1+2=3 ; \quad 3+2=5 ; \quad 3-2-11=0 ; \quad 201[5=2015$
or
$[1+2=3 ; \quad[1]=[13 ; \quad 3[1=31 ; \quad([2+3) \cdot 13 \cdot 31=2015$.
b) $[1+2=3 ; \quad 3 \cdot 2-1=63 ; \quad(63+2) \cdot 31=2015$.
Comments. 1. To solve the problem, it is useful to first factorize 2015 into prime factor... | 2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,817 |
Problem 6. Petya wrote 25 numbers in the cells of a $5 \times 5$ square. It is known that their sum is 500. Vasya can ask him to name the sum of the numbers in any cell and all its side neighbors. Can Vasya find out, in a few such questions, what number is written in the central cell?
$[8$ points] (E. V. Bakayev) | Answer: Yes.
Solution: By asking questions about the 6 cells marked on the left diagram, Vasya can learn the sum of all numbers except \( L \) and \( R \). By subtracting this sum from 500, he can find \( L+R \). Similarly, he can find \( U+D \). After that, Vasya needs to learn the sum of the numbers in the central c... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,818 |
Task 1. A $3 \times 3$ square is filled with digits as shown in the figure on the left. It is allowed to move along the cells of this square, transitioning from one cell to an adjacent one (by side), but it is not allowed to visit any cell more than once.
| 1 | 8 | 4 |
| :--- | :--- | :--- |
| 6 | 3 | 9 |
| 5 | 7 | 2 ... | Answer. The largest number that can be obtained is -573618492 (see fig.).
Comments. 1. Let's explain how the problem could be solved (this was not required of the participants).
Notice that a number larger than the one given
| 1 | 8 | 4 |
| :---: | :---: | :---: |
| 6 | 3 | 9 |
| 5 | 7 | 2 |
in the problem can be o... | 573618492 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,819 |
Problem 4. On each of the two branches of the river, a kilometer before their confluence, there is a pier, and another pier is located two kilometers downstream from the confluence (see figure). A boat traveled from one of the piers to another (unknown which) in 30 minutes, and from another to the third in 18 minutes. ... | Answer: Either 24 or 72 minutes.
Solution. Let's denote the piers in the order of visit as $A, B$, and $C$. Let's think about which of them could have been after the confluence.
- Pier $B$ could not have been after the confluence. Indeed, otherwise the boat would have traveled 3 km downstream to it, and the same 3 km... | 24or72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,821 |
Problem 5. Vasya wrote a correct statement:
"In this phrase, $1 / 3$ of all digits are the digit 3, and $1 / 2$ of all digits are the digit 1."
And Kolya wrote the phrase:
"In this phrase, $1 / \ldots$ of all digits are the digit *, the fractions of digits * and * are equal and are $1 / \ldots$, and the fraction of ... | Answer. "In this phrase, $1 / 2$ of all digits are the digit 1, the shares of digits 2 and 5 are the same and equal to $1 / 5$, and the share of all other digits is 1/10" (or "…the shares of digits 0 and 2 are the same…" or "…the shares of digits 0 and 5 are the same…").
Solution. Let's explain how the problem could b... | "Inthisphrase,1/2ofalldigitsthedigit1,thesharesofdigits25thesameequalto1/5,theshareofallotherdigitsis1/10" | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,822 |
Problem 6. Having defeated Koschei, Ivan demanded gold to ransom Vasilisa from the bandits. Koschei led him to a cave and said:
"In the chest lie gold ingots. But you cannot simply take them: they are enchanted. Put one or several into your bag. Then I will put one or several from the bag back into the chest, but it m... | Answer. a) 13; b) 13.
Solution. Ivan will act in such a way that each time Kashchey's move will be the only one possible: all other numbers have either already appeared in previous moves or are too large - Ivan does not have that many ingots at that moment. We will record the moves of the game as follows: the number o... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,823 |
Problem 3. Vanya assembled a $3 \times 3 \times 3$ cube from red and blue blocks of size $1 \times 1 \times 3$. Then he started drawing what he had created. When Tanya arrived, Vanya had only colored 8 out of 27 cells on the visible surface of the drawn cube (see the image below).
![](https://cdn.mathpix.com/cropped/2... | Answer. See the figure on the right.
Solution. First, let's determine how the blocks are arranged in the cube. The block that passes through the blue cell on the front face is bounded below and to the right by red $1 \times 1 \times 1$ cubes, so it passes through the center
![](https://cdn.mathpix.com/cropped/2024_05... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,824 |
Task 1. "This is for you to see a bit too early,"- said Baba Yaga to her 33 students and commanded: "Close your eyes!" The right eye was closed by all the boys and a third of the girls. The left eye was closed by all the girls and a third of the boys. How many students still saw what they were not supposed to see yet? ... | Answer: 22 students.
Solution: What is seen is still too early, two-thirds of the girls saw with their right eye, and two-thirds of the boys saw with their left eye. In total, then, one eye was not closed by two-thirds of all students - 22 people. | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,825 |
Task 3. Before the football match between the teams "North" and "South," five predictions were given:
a) there will be no draw;
b) goals will be scored in "South's" net;
c) "North" will win;
d) "North" will not lose;
![](https://cdn.mathpix.com/cropped/2024_05_06_ad0910d2ff8aaa6a488fg-1.jpg?height=315&width=310&to... | Answer: 1:2.
Solution. Since exactly 3 forecasts are correct, exactly 2 forecasts are incorrect.
First, let's determine which team won. Suppose "North" won. Then 4 forecasts («a», «b», «v» and «g») would be correct, which contradicts the condition.
Suppose the match ended in a draw; then forecasts «a», «v», and «d» ... | 1:2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,827 |
Problem 4. Find all solutions to the puzzle
$$
\mathrm{G}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}+\mathrm{OH}=\mathrm{Mb} .
$$
(Same letters encode the same digits, different letters encode different digits.)
[5 points] (D. E. Schnol) | Answer. $0+12+12+12+12+12+12+12+12=96$.
Solution. Since $8 \cdot \mathrm{OH}<100$, then $\mathrm{OH} \leqslant 12$, which means $\mathrm{O}=1$, and Н is either 0 or 2. But $\mathrm{H}$ cannot be 0, because then Я and Ы would represent the same digit. Therefore, $\mathrm{H}=2$. Thus, $8 \cdot \mathrm{OH}=8 \cdot 12=96$... | 0+12+12+12+12+12+12+12+12=96 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,828 |
Task 5. A dragon imprisoned six dwarfs in a cave and said: “I have seven caps of seven colors of the rainbow. Tomorrow morning I will blindfold you and put a cap on each of you, and hide one cap. Then I will remove the blindfolds, and you will be able to see the caps on the heads of the others, but I will not allow you... | Solution. Each dwarf can see all the hats except for two: their own and the hidden one. They need to agree on which of the two colors to name. This can be done, for example, as follows.
Let's number the colors from 1 to 7 (for example, in the same order as the colors of the rainbow) and arrange them in a circle in adv... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,829 |
Problem 6. A wooden block was cut into eight smaller blocks with three cuts. On Fig. 5, the surface area of seven blocks is indicated. What is the surface area of the invisible block?
$[8$ points] (A. V. Shapovalov) | Answer: 22.
Solution: For each small block, the surface of the cuts constitutes half of its entire surface. We will only consider this. We will color the small blocks in black and white as shown in Fig. 6 (the invisible block is black). Then, every two identical rectangles touching on the cut are of different colors. ... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,830 |
Problem 4. The magician taught Kashtanka to bark as many times as he showed her secretly. When Kashtanka correctly answered how much two times two is in this way, he hid a delicious cake in a suitcase with a combination lock and said:
- The eight-digit code for the suitcase is the solution to the puzzle УЧУЙ $=\kappa ... | Answer. a) No. b) Yes, УЧУЙ $=2021$.
Solution. a) Note that КЕ and КС represent different numbers, but swapping them does not change the product УЧУЙ. Therefore, for each solution to the puzzle, there is a paired solution where the digits corresponding to $\mathrm{E}$ and $\mathrm{C}$ are swapped. Thus, it is impossib... | 2021 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,834 |
Problem 5. Given an equilateral triangle $ABC$. On side $AB$, point $K$ is marked, and on side $BC$ - points $L$ and $M$ ( $L$ lies on segment $BM$ ) such that $KL=KM, BL=2, AK=3$. Find $CM$.
[7 points]
. Each person can either be in the group or not, so there are two options for each... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,836 |
Task 1. a) Fill in the cells with four different digits so that the product of the fractions equals $\frac{20}{21}$:
$$
\frac{\square}{\square} \times \frac{\square}{\square}=\frac{20}{21}
$$
Solve this problem for the other three arithmetic operations:
b) division
c) subtraction;
d) addition.
$[4$ points] (A. V.... | Answer.
$$
\frac{4}{3} \times \frac{5}{7}=\frac{20}{21}, \quad \frac{4}{3}: \frac{7}{5}=\frac{20}{21}, \quad \frac{9}{7}-\frac{1}{3}=\frac{20}{21}, \quad \frac{4}{6}+\frac{2}{7}=\frac{20}{21}
$$
There are other examples. Interestingly, in the last problem, it is not possible to use irreducible fractions. | \frac{4}{3}\times\frac{5}{7}=\frac{20}{21},\quad\frac{4}{3}:\frac{7}{5}=\frac{20}{21},\quad\frac{9}{7}-\frac{1}{3}=\frac{20}{21},\quad\frac{4}{6}+\frac{2}{7}=\frac{20}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,837 |
Problem 2. a) Malvina divided each face of a $2 \times 2 \times 2$ cube into unit squares and asked Buratino to write crosses in some squares and zeros in the others such that each square shares a side with exactly two crosses and two zeros.
 Let's imagine that the cube is made up of eight unit cubes and look at the shaded cube in the picture. No matter what Buratino writes on its bottom face, Malvina's requirement will be violated. If there is a cross, then the zero on its front face will be adjacent to
. If the figure is cut along all horizontal grid lines, 20 strips 1 cell wide will be obtained. How many strips will be ob... | Answer: 21.
First solution. Let the rectangle occupy $a$ cells vertically and $b$ horizontally, $a+b=50: 2=25$. Similarly, let the dimensions of the hole be $x$ cells vertically and $y$ horizontally, $x+y=32: 2=16$.
. He couldn't do more, and all his shovels broke. The Tsar got angry and put him in prison. Then the younger brother Ivan... | Solution. Ivan can save his brother, earn money, and even (if the tsar keeps his word) receive half the kingdom by planting all 6 types of trees as shown in the figure.
There are no other way... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,841 |
Problem 6. In the showcase of a jewelry store, there are 15 diamonds. Next to them are signs indicating their weights, which are $1, 2, \ldots, 15$ carats. The seller has a balance scale and four weights with masses of $1, 2, 4,$ and 8 carats. The customer is allowed only one type of weighing: place one of the diamonds... | Answer: 800 coins.
Solution. Example.
| Which weights we buy | What we weigh | How many coins we paid |
| :---: | :--- | :---: |
| 1 | $\mathbf{1 = 1}$ | 100 |
| 2 | $1+2=\mathbf{3}, 2=\mathbf{2}$ | 200 |
| 4 | $2+4=\mathbf{6}$ | 300 |
| 1 | $1+2+\mathbf{4}=\mathbf{7}, 1+\mathbf{4}=\mathbf{5}, \mathbf{4}=\mathbf{4}$ ... | 800 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,842 |
Problem 1. Sasha wrote down the numbers from one to one hundred, and Misha erased some of them. Among the remaining numbers, 20 numbers have a one in their notation, 19 numbers have a two in their notation, and 30 numbers have neither a one nor a two. How many numbers did Misha erase?
$[4$ points] (A. V. Shapovalov) | Answer: 33.
Solution: Among the numbers from 1 to 100, the digit 1 appears exactly twenty times: the number 1 itself, ten numbers from 10 to 19, the numbers $21, 31, \ldots, 91$ (eight of them), and the number 100. Therefore, none of these numbers were erased. Similarly, the digit 2 appears exactly nineteen times: the... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,843 |
Problem 3. Senya cannot write some letters and always makes mistakes in them. In the word TETRAHEDRON he would make five mistakes, in the word DODECAHEDRON - six, and in the word ICOSAHEDRON - seven. How many mistakes would he make in the word OCTAHEDRON?
. Assign (draw with arrows) several more one-way flights so that from any city to any other, one could get there with no more than two transfers. Try to make the number of additional flights as small as possible.
 | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,846 |
Problem 5. Around a circular lake, 2019 trees grow at equal intervals: 1009 pines and 1010 spruces. Prove that there must be a tree next to which a pine grows and on the other side, one tree away, another pine also grows. $[8$ points] ( (E. V. Bakayev) | Solution. Let's walk around the lake in a circle and write the letters: A, B, C, then again A, B, C, and so on. There will be $2019: 3=673$ trees with each letter. If there were no more than 336 pines with each letter, then there would be no more than $336 \cdot 3=1008$ pines in total. Since there are 1009 pines, there... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,847 |
Problem 6. Each face of a cube $6 \times 6 \times 6$ is divided into $1 \times 1$ cells. The cube is covered with $2 \times 2$ squares such that each square covers exactly four cells, no squares overlap, and each cell is covered by the same number of squares. What is the maximum value that this identical number can tak... | Answer: 3.
Solution: Estimation. A cell in the corner of a face can be covered in three ways (entirely within the face, with a fold over one edge of the corner, with a fold over the other edge of the corner). Therefore, each cell is covered by no more than three squares.
Example. Consider the usual covering of a cube... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,848 |
Task 2. On two cards, four different digits are written - one on each side of the card. Can it be that every two-digit number that can be formed from these cards is prime? (You cannot flip the digits upside down, i.e., turn a 6 into a 9 and vice versa.)
$[4$ points] (A.V. Shapovalov) | Answer: No, it cannot.
Solution: All two-digit numbers ending in 0, $2, 4, 6$ or 8 are even, and those ending in 5 are multiples of five. Therefore, such numbers cannot be prime, and it makes no sense to write these digits on cards. The remaining digits are 1, 3, 7, and 9. If the digits 3 and 9 are written on differen... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,850 |
Problem 3. Among all the faces of eight identical cubes, one third are blue, and the rest are red. These cubes are used to form a larger cube. Now, among the visible faces of the smaller cubes, exactly one third are red. Prove that it is possible to form a cube that is completely red on the outside from these smaller c... | Solution. Eight cubes have $8 \cdot 6=48$ faces. Of these, $48: 3=16$ are blue. On the faces of the large cube $2 \times 2 \times 2$, we see $6 \cdot 4=24$ faces of the small cubes, of which $24: 3=8$ are red, and the remaining $24-8=16$ are blue. Thus, all 16 blue faces of the small cubes are on the surface of the lar... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,851 |
Problem 5. A group of tourists is dividing cookies. If they evenly distribute two identical packs, one extra cookie will remain. But if they evenly distribute three such packs, 13 extra cookies will remain. How many tourists are in the group? [7 points] (I.V. Raskina) | Answer: 23.
First solution. Distribute three times two packs, 3 * 1 = 3 cookies will remain. But the same six packs of cookies can be distributed differently - three and another three, and then 2 * 13 = 26 cookies will remain. Therefore, 26 - 3 = 23 cookies can be divided equally among the tourists. Since the number 2... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,852 |
Task 6. Koschei the Deathless captured 43 people and took them to an island. Ivan the Tsarevich set out in a two-person boat to rescue them. Koschei told Ivan:
- I'm tired of feeding these freeloaders; let them row back in your boat in good health. Remember: only two can cross from the island to the shore, but one can... | Answer. Yes, there is.
Solution. Ivan can divide the prisoners into twenty pairs and one trio, and order Koschei to tell each of them that everyone except those in the same pair (trio) with him are werewolves. Then the condition will be satisfied, and the prisoners can cross as follows. Let's call the prisoners from t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,853 |
Task 2. A pharmacist has three weights, with which he weighed out 100 g of iodine for one customer, 101 g of honey for another, and 102 g of hydrogen peroxide for a third. He always placed the weights on one pan of the scales and the goods on the other. Could it be that each weight was less than 90 g?
[4 points] (A.V.... | Answer: Yes, it could.
Solution: Weights of 49.5 g, 50.5 g, and 51.5 g would work. The first and second weights together weigh 100 g, the first and third weigh 101 g, and the second and third weigh 102 g.
Comment: An example can be found and proven to be unique as follows. First, note that in none of the weighings di... | 49.5,50.5,51.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,854 |
Problem 4. Given a square $ABCD$. On the extension of the diagonal $AC$ beyond point $C$, a point $K$ is marked such that $BK=AC$. Find the angle $BKC$. $[6$ points] ( | Answer: $30^{\circ}$.
Solution: Since the picture is symmetric with respect to the line $A C$, we have $D K=B K$. By the condition, $B K=A C$.
And since the diagonals in a square are equal, $A C=B D$. Thus, in triangle $B K D$ all sides are equal, i.e., it is equilateral, and $\angle B K D=60^{\circ}$. Again, due to ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,855 |
Problem 5. Can the digits $1,2, \ldots, 8$ be placed in the cells of a) the letter Ш; b) the strips (see figure), so that for any cutting of the figure into two parts, the sum of all the digits in one part is divisible by the sum of all the digits in the other? (Cuts can only be made along cell boundaries. Each cell mu... | Solution. Let the sum of the numbers in one of the parts be $x$, in the other $y$, and $y$ is divisible by $x$. Then $x+y$ is also divisible by $x$, and this is the sum of all the numbers, which is $1+2+3+4+5+6+7+8=$ $=36$. Therefore, the smaller of the sums of the parts is a divisor of the number 36. The converse is a... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,856 |
Problem 6. Among 49 schoolchildren, each is acquainted with at least 25 others. Prove that they can be divided into groups of 2 or 3 people such that everyone in a group is acquainted with everyone else in their group.
$[8$ points $]$
(A. V. Shapovalov) | Solution. Let's assume that initially all 49 students are standing in the corridor, and we will gradually let them into the classroom.
We will do this in such a way that at any moment in time, the children in the classroom are divided into the required groups.
Let's say that a student named Fyodor is standing in the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,857 |
Task 1. Electronic clocks show the time: hours, minutes, and seconds, for example, 18:00:00. One day, two digits on the clock went out, and only the digits $2,0,2,2$ (in that exact order) remained. Name the latest moment in the day when this could have happened.
[3 points
(М. А. Калинин)] | Answer: 23:50:22.
Solution. The latest time that electronic clocks can display is 23:59:59. If we look at the time as a six-digit number, two moments of time are compared in the same way as ordinary numbers: the time is greater, the greater the first digit; if the first digits are the same, the second is compared, and... | 23:50:22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,858 |
Problem 3. Find a six-digit number where the first digit is 6 times less than the sum of all digits to the right of it and the second digit is 6 times less than the sum of all digits to the right of it.
$[4$ points] (A. V. Shapovalov) | Answer: 769999.
First solution. The sum of the last five digits of the number, by condition, is divisible by 6 (the quotient is equal to the first digit). The sum of the last four digits of the number is also divisible by 6 (the quotient is equal to the second digit). Therefore, the second digit, as the difference of ... | 769999 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,860 |
Problem 4. Three frogs were sitting on a straight line. At first, one of them jumped, then another, and finally the third. Each frog landed exactly in the middle of the segment between the other two. It turned out that the lengths of two of these three jumps were 60 cm.
a) What could the length of the remaining jump b... | Answer. a) 30 cm or 120 cm; b) 100 cm or 160 cm.
Solution. Regardless of the initial positions of the frogs, after the first jump, the first frog (A) will land exactly in the middle of the segment between the other two.
30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,861 |
Problem 6. The digits from 0 to 9 are encrypted by the letters A, B, C, D, E, F, G, H, I, J in some order. In one question, you can find out the encrypted record of the sum of several different letters. For example, if you ask "A $+\mathrm{B}=$ ?", then in the case where $\mathrm{A}=9, \mathrm{~B}=1, \mathrm{C}=0$, the... | Solution. See problem 3 for 6th grade (p. 4).
Comment. If we do not limit ourselves in the number of questions, then we can find out which letter is coded by which digit even without asking about the total sum of the digits. For example, we can always ask only about pairs of digits. Thus, by sequentially checking the ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,863 |
4. Petya observes two ants crawling along a straight path at constant speeds. After 3 minutes of observation, the distance between the ants was 9 meters, after 5 minutes - 5 meters, after 9 minutes - 3 meters. What was the distance between the ants after 8 minutes of observation?
$[5$ points]
.
![](https://cdn.mathpix.com/cropped/2024_05_06_031fac0a0a7981eac22cg-04.jpg... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,872 |
Problem 4. 13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." The latter said to their right neighbor... | # Answer: 7.
Solution. It is clear that there were both boys and girls at the table. Let's see how the children were seated. After a group of boys sitting next to each other comes a group of girls, then boys again, then girls, and so on (a group can consist of just one person). Groups of boys and girls alternate, so t... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,873 |
Problem 5. Small Island and Big Island have a rectangular shape and are divided into rectangular counties. In each county, a road is laid along one of the diagonals. On each island, these roads form a closed path that does not pass through any point more than once. Here is how Small Island is organized, with a total of... | Answer. Figure 2 provides an example for 9 counties.
Comment. We will show that examples do not exist for 7 counties (or fewer), while at the same time pointing out a property characteristic of all such examples.
All roads can be divided into two types: some roads connect the top-left corner of a county with the bott... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,874 |
Problem 6. Thirty-three bogatyrs (Russian knights) were hired to guard Lake Luka for 240 coins. The cunning Chernomor can divide the bogatyrs into squads of any size (or record all in one squad), and then distribute the entire salary among the squads. Each squad divides its coins equally, and the remainder goes to Cher... | Answer. a) 31 coins; b) 30 coins.
Solution. From each detachment of $N$ bogatyrs, Chernomor will receive at most $N-1$ coins in the best case, since the remainder is less than the divisor. Therefore, he will receive no more than $33-K$ coins in total, where $K$ is the number of detachments. Can Chernomor get 32 coins ... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,875 |
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