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7.4. Yura was walking down the road and met a tractor pulling a long pipe. Yura decided to measure the length of the pipe. For this, he walked along it "against the direction of the tractor" and counted 20 steps. After that, he walked along the pipe "in the direction of the tractor" and counted 140 steps. Knowing that ... | Answer: 35 m.
Let's say that in the time it takes Yura to make 20 steps, the train travels $x$ meters. Denoting the length of the train by $L$, we get: $20=L-x, 140=L+7 x$. From this, $L=(140+20 \cdot 7): 8=35$. | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,065 |
7.5. Vasya had 101 real coins. One of them was replaced with a fake one (it differs in weight, but it is unknown whether it is lighter or heavier than a real coin). Vasya wants to find as many real coins as possible in one weighing on a balance scale without weights. What should he do and how many real coins will he be... | Answer: 50 coins.
Divide the coins into piles of $25 ; 25 ; 51$ and place the piles of 25 coins on the scales. If the scales are in balance, then these piles contained genuine coins, and we have found 50 genuine coins. If the scales are not in balance, then the genuine coins are in the third pile, and we have found 51... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,066 |
1. Can the sum of 2017 consecutive natural numbers be the 2017th power of a natural number? | Answer. Yes.
Solution. The sum of 2017 consecutive natural numbers is $n+(n+1)+\cdots+(n+2016)=2017 n+\frac{2016 \cdot 2017}{2}=2017 \cdot(n+1008)$. If $n+1008=2017^{2016}$, then the sum is the 2017th power of the number 2017, and the condition is satisfied (for $n=2017^{2016}-1008$).
Comment. In the absence of a sol... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,067 |
2. The graph of the linear function $y=k x+k+1(k>0)$ intersects the $O x$ axis at point $A$, and the $O y$ axis at point $B$ (see the figure). Find the smallest possible value of the area of triangle $A B O$. | Answer: 2.
Solution. The abscissa of point $A$ of intersection with the $O x$ axis: $0=k x+k+1 ; x=$
$-\left(1+\frac{1}{k}\right)$. The ordinate of point $B$ of intersection with the $O y$ ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,068 |
3. In triangle $ABC$, the bisector $BD$ is drawn. It is known that the center of the circumscribed circle of triangle $ABC$ coincides with the center of the circle inscribed in triangle $BCD$. Find the angles of triangle $ABC$. | Answer. $\angle C=36^{\circ}, \angle A=\angle B=72^{\circ}$.
Solution. Let the common center of the circles be $I$, and the points of tangency of the inscribed circle in triangle $BCD$ with sides $CD$, $BC$, and $BD$ be $E$, $F$, and $G$ respectively (see figure). Segments $IE$ and $IF$ are the perpendicular bisectors... | \angleC=36,\angleA=\angleB=72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,069 |
4. Find all natural $n$ for which both statements are true:
1) the greatest prime divisor of the number $n$ is equal to $\sqrt{n}$,
2) the greatest prime divisor of the number $(n+72)$ is equal to $\sqrt{n+72}$. | Answer: 49 and 289.
Solution. Let the mentioned prime divisors be $p_{1}$ and $p_{2}$. Then $n=p_{1}^{2}, n+72=p_{2}^{2}$, from which $\left(p_{2}+p_{1}\right)\left(p_{2}-p_{1}\right)=72$. Consider the factorizations of the number 72 into natural factors: $72=m \cdot k$ $(m>k)$. Then $p_{1}=\frac{m-k}{2}, p_{2}=\frac{... | 49289 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,070 |
5. The number 24 is written on the board. Lёsha and Vitya take turns changing the number: either by appending one digit to the end, or by erasing the last digit of the previous number, or by rearranging the digits of the previous number in any order (the number cannot start with zero). It is not allowed to leave the nu... | Answer: Lёsha will win.
Solution. Lёsha will win. The second number he gets by appending 0 at the end. Now the board shows 240. Next, if Vitya, by his move, transforms the number 240 into the number $B$, Lёsha again transforms $B$ into 240. This does not contradict the rules, since the number 240 is divisible by $4, 6... | Lёshawillwin | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,071 |
11.2. In the quarry, there are 120 granite slabs weighing 7 tons each and 80 slabs weighing 9 tons each. A railway platform can load up to 40 tons. What is the minimum number of platforms required to transport all the slabs? | Answer: 40 platforms
It is impossible to load 6 slabs onto one platform, even if they weigh 7 tons each. Therefore, at least $200 / 5=40$ platforms are needed. Forty platforms are sufficient: on each platform, you can load 3 slabs weighing 7 tons and 2 slabs weighing 9 tons.
Comment. The necessity of at least 40 plat... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,072 |
11.3. The fractional part of a positive number, its integer part, and the number itself form an increasing geometric progression. Find all such numbers. | Answer. The number is unique $\frac{\sqrt{5}+1}{2}$.
Let $a$ be the desired number, $b$ its integer part, $c$ its fractional part, and $q$ the common ratio of the progression. Then $a=b+c$ and $c q^{2}=c q+c$. Given $c>0$, dividing by $c$, we get the quadratic equation $q^{2}-q-1=0$, which has two roots, one of which ... | \frac{\sqrt{5}+1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,073 |
11.4. Vasya has three cans of paint of different colors. In how many different ways can he paint a fence consisting of 10 boards so that any two adjacent boards are of different colors, and he uses all three colors of paint? | Answer: 1530 ways.
First, let's calculate the number of ways to paint the fence so that any two adjacent boards are painted in different colors. The first board can be painted with any of the three colors, the second with one of the two remaining colors. The third board can be painted with one of the two colors differ... | 1530 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,074 |
11.5. All edges of the tetrahedron are equal to $12 \mathrm{~cm}$. Can it be placed in a box that has the shape of a rectangular parallelepiped with sides $9 \mathrm{~cm}, 13$ cm and 15 cm $?$ | Answer: Yes.
Consider a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side length $a$. The tetrahedron $A C B_{1} D_{1}$ has all edges equal to $\sqrt{2} a$. Therefore, a tetrahedron with an edge length of 12 can fit into a cube with a side length of $6 \sqrt{2}$. Since $6 \sqrt{2}<9$, the cube, along with the tetrahedr... | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,075 |
Task 4.1. Koschei has three chests.
- On the first one, it is written: “Here lie gold coins.”
- On the second one, it is written: “Here lie silver coins.”
- On the third one, it is written: “Here lie gold or silver coins.”
One of the chests he filled with only gold coins, another with only silver coins, and the remai... | Answer: In the first chest lie silver coins, in the second - gold, in the third - copper.
Solution. Since the inscription on the third chest is incorrect, it can only contain copper coins. In the first chest, neither gold nor copper coins lie, so there must be silver coins. Then, in the second chest lie gold coins. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,076 |
Problem 4.2. Vika wrote the numbers from 1 to 9 in the cells of a $3 \times 3$ table such that the sum of any two numbers in adjacent cells is less than 12. The hooligan Andrey erased all the even numbers: $2, 4, 6,$ and $8$.
Help Vika restore where each number was.
| $A$ | $\mathbf{1}$ | $\mathbf{9}$ |
| :--- | :---... | Answer: $A=8, B=6, C=4, D=2$.
Solution. Since the sum of the numbers 9 and $D$ is less than 12, $D$ can only be 2. Since the sum of the numbers 7 and $C \neq 2$ is less than 12, $C$ can only be 4. Since the sum of the numbers $C=4$ and $B$ is less than 12, and $B$ is 6 or 8, $B$ can only be 6. Then the remaining numbe... | A=8,B=6,C=4,D=2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,077 |
Problem 4.3. All seats at King Arthur's round table are numbered clockwise. The distances between adjacent seats are the same.
One day, King Arthur sat at seat number 10, and Sir Lancelot sat directly opposite him at seat number 29. How many seats are there in total at the round table? | Answer: 38.
Solution. Along one side of the table between Arthur and Lancelot are seats numbered $11,12, \ldots, 28$ - a total of exactly 18 seats. Since these two are sitting directly opposite each other, there are also 18 seats on the other side of the table. Therefore, the total number of seats at the table is $18+... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,078 |
Problem 4.6. Fourth-grader Vasya goes to the cafeteria every school day and buys either 9 marshmallows, or 2 meat pies, or 4 marshmallows and 1 meat pie. Sometimes Vasya is so busy socializing with classmates that he doesn't buy anything at all. Over 15 school days, Vasya bought 30 marshmallows and 9 meat pies. How man... | Answer: 7.
Solution. Let's consider all the options for how many times Vasya could have bought 9 marshmallows - it is clear that this could not have happened more than three times.
Suppose Vasya never bought 9 marshmallows. Then he bought 30 marshmallows in packs of 4 on some days, but 30 is not divisible by 4. Contr... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,081 |
Problem 4.7. At the festival of namesakes, 45 Alexanders, 122 Borises, 27 Vasily, and several Gennady arrived. At the beginning of the festival, all of them stood in a row so that no two people with the same name stood next to each other. What is the minimum number of Gennady that could have arrived at the festival? | Answer: 49.
Solution. Since there are a total of 122 Boris, and between any two of them stands at least one non-Boris (Alexander/Vasily/Gennady), there are at least 121 non-Boris. Since there are a total of 45 Alexanders and 27 Vasilies, there are at least $121-45-27=49$ Gennadys.
Note that there could have been exac... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,082 |
Problem 4.8. A merchant has 6 bags weighing 13, 15, 16, 17, 21, and 24 kg. One bag is filled with turnips, and each of the remaining bags is either onions or carrots. The merchant knows that the total weight of the carrots is twice the total weight of the onions. In which bag can the turnips be? List all possible optio... | Answer: 13 or 16.
Solution. The total weight of onions is a whole number of kilograms, and carrots weigh twice as much. Therefore, the total weight of onions and carrots is a whole number of kilograms divisible by 3, and they are in all bags except one. The total weight of all bags is $13+15+$ $16+17+21+24=106$ kg, so... | 13or16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,083 |
Problem 5.1. Dasha calls a natural number special if four different digits are used to write it. For example, the number 3429 is special, while the number 3430 is not special.
What is the smallest special number greater than 3429? | Answer: 3450.
Solution. Note that all numbers of the form $343 \star$ and $344 \star$ are not special. And the next number after them, 3450, is special. | 3450 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,084 |
Problem 5.2. At first, the magical island was divided into three counties: in the first county lived only elves, in the second - only dwarves, in the third - only centaurs.
- During the first year, each county where non-elves lived was divided into three counties.
- During the second year, each county where non-dwarve... | Answer: 54.
Solution. Initially, there was 1 county of each kind.
After the first year, there was 1 elven county, 3 dwarf counties, and 3 centaur counties.
After the second year, there were 4 elven counties, 3 dwarf counties, and 12 centaur counties.
After the third year, there were 24 elven counties, 18 dwarf coun... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,085 |
Problem 5.3. In five of the nine circles in the picture, the numbers 1, 2, 3, 4, 5 are written. Replace the digits $6, 7, 8, 9$ in the remaining circles $A, B, C, D$ so that the sums of the four numbers along each of the three sides of the triangle are the same.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,088 |
Problem 5.6. There are 4 absolutely identical cubes, each of which has 6 dots marked on one face, 5 dots on another, ..., and 1 dot on the remaining face. These cubes were glued together to form the figure shown in the image.
How many dots are on the four left faces?
, one can get the cloa... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,090 |
Problem 5.8. Each of the 33 bogatyrs (Russian epic heroes) either always lies or always tells the truth. It is known that each bogatyr has exactly one favorite weapon: a sword, spear, axe, or bow.
One day, Ded Chernomor asked each bogatyr four questions:
- Is your favorite weapon a sword?
- Is your favorite weapon a ... | Answer: 12.
Solution. Note that each of the truth-telling heroes answers affirmatively to only one question, while each of the lying heroes answers affirmatively to exactly three questions. Let the number of truth-telling heroes be $x$, and the number of lying heroes be $-(33-x)$. Then the total number of affirmative ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,091 |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,092 |
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?
.
By the evening of September 8, Yura realized that th... | Answer: 12.
Solution. In the first 3 days, Yura solved $91-46=45$ problems. Let's say on September 7th, he solved $z$ problems, then on September 6th, he solved $(z+1)$ problems, and on September 8th, he solved $(z-1)$ problems. We get that $45=(z+1)+z+(z-1)=3 z$, from which $z=15$.
Since $91=16+15+14+13+12+11+10$, Y... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,094 |
Problem 6.4. Find any solution to the puzzle
$$
\begin{array}{r}
\mathrm{ABA} \\
+\mathrm{ABC} \\
\mathrm{ACC} \\
\hline 1416
\end{array}
$$
where $A, B, C$ are three different non-zero digits.
Enter the values of the digits $A, B, C$.
Instead of the letter $A$ should stand the digit:
Instead of the letter $B$ sho... | Answer: $A=4, B=7, C=6$.
Solution. Suppose $A \leqslant 3$. Then $1416=\overline{A B A}+\overline{A B C}+\overline{A C C}<400+400+400=1200$ - contradiction.
Suppose $A \geqslant 5$. Then $1416=\overline{A B A}+\overline{A B C}+\overline{A C C} \geqslant 500+500+500=1500-$ contradiction.
Therefore, $A=4$. Considering... | A=4,B=7,C=6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,095 |
Problem 6.5. Masha told Sasha that she had thought of two different natural numbers greater than 11. Then she told Sasha their sum.
- Sasha, after thinking, said: "I don't know the numbers you thought of."
- To which Masha replied: "At least one of them is even."
- Then Sasha said: "Now I definitely know the numbers y... | Answer: 12 and 16.
Solution. The sum of two different natural numbers, both greater than 11, is a natural number not less than $12+13=25$.
Suppose Masha told Sasha that the sum of the numbers is 25. Then he would have immediately understood that these are the numbers 12 and 13. Contradiction.
Suppose Masha told Sash... | 1216 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,096 |
Problem 6.6. Ksyusha runs twice as fast as she walks (both speeds are constant).
On Tuesday, when she left home for school, she first walked, and then, when she realized she was running late, she started running. The distance she walked was twice the distance she ran. In the end, it took her exactly 30 minutes to get ... | # Answer: 24.
Solution. Let the distance from home to school be $3 S$, Ksyusha's walking speed be $v$, and her running speed be $-2 v$ (distance is measured in meters, and speed in meters per minute). Then on Tuesday, Ksyusha walked a distance of $2 S$ and ran a distance of $S$. And on Wednesday, she walked a distance... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,097 |
Problem 6.8. In the country of Dragonia, there live red, green, and blue dragons. Each dragon has three heads, each of which always tells the truth or always lies. Moreover, each dragon has at least one head that tells the truth. One day, 530 dragons sat around a round table, and each of them said:
- 1st head: “To my ... | Answer: 176.
Solution. Consider an arbitrary red dragon. To the right of this dragon, at least one head must tell the truth. Note that the 1st and 3rd heads cannot tell the truth (since there is a red dragon to the left), so the 2nd head must tell the truth, and to the right of this dragon, there must be a blue dragon... | 176 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,099 |
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | In\\A\there\is\the\\6,\in\B-2,\in\C-4,\in\D-5,\in\E-3,\in\F-8,\in\G-7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,100 |
Problem 7.2. Petya bought himself shorts for football at the store.
- If he had bought shorts with a T-shirt, the cost of the purchase would have been twice as much.
- If he had bought shorts with cleats, the cost of the purchase would have been five times as much.
- If he had bought shorts with shin guards, the cost ... | Answer: 8.
Solution. Let the shorts cost $x$. Since the shorts with a T-shirt cost $2x$, the T-shirt also costs $x$. Since the shorts with boots cost $5x$, the boots cost $4x$. Since the shorts with shin guards cost $3x$, the shin guards cost $2x$. Then, if Petya bought shorts, a T-shirt, boots, and shin guards, his p... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,101 |
Problem 7.3. Lёsha gathered friends to play hide and seek. He surveyed Andrey, Borya, Vasya, Gena, and Denis and found out the following.
- If Andrey goes to play hide and seek, then Borya will go too, but Vasya will not.
- If Borya goes to play hide and seek, then Gena or Denis will also go.
- If Vasya does not go to... | Answer: With Borya, Vasya, and Denis.
Solution. Regardless of whether Andrey goes to play hide and seek, Borya will definitely go. Then Vasya will also go to play hide and seek (if Vasya didn't go, then Borya wouldn't go either). Then Andrey will not go to play hide and seek (if Andrey went, then Vasya wouldn't go). T... | Borya,Vasya,Denis | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,102 |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
=9 \cdot(18+S)$. Solving this linear equation, we get $S=13.5$. | 13.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,104 |
Problem 7.7. A circle of 130 trees has been planted: birches and lindens (both species are present). On each tree, there is a sign that reads: "Two different trees grow next to me." It is known that among all the trees, the statement is false on all lindens and on exactly one birch. How many birches could have been pla... | Answer: 87.
Solution. Let's divide all the trees into alternating groups of consecutive birches and consecutive lindens (by the condition, there are groups of both types).
Suppose there exists a group of at least 2 lindens. Then the truth would be written on the outermost of them (since it is between a birch and a li... | 87 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,106 |
Problem 7.8. The king and queen had three sons and several daughters (at least one). On September 1st of a certain year, the king and queen noticed that they were both 35 years old, and moreover, the total age of their children also amounted to 35 years. On September 1st several years later, the king and queen noticed ... | Answer: 7 or 9.
Solution. Let the king and queen have $d \geqslant 1$ daughters. Let also $n$ years have passed between the two described moments in time.
Initially, the difference between the total age of the parents and the total age of the children was $35+35-35=35$ years, and after $n$ years, it became 0. Since e... | 7or9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,107 |
Problem 8.1. In front of a pessimist and an optimist, there are glasses (the glasses are identical). Each of them was given water in their glass such that the pessimist's glass turned out to be $60\%$ empty, while the optimist's glass, on the contrary, was $60\%$ full. It turned out that the amount of water in the pess... | Answer: 230.
Solution. The pessimist's glass is $40 \%$ full, while the optimist's is $60 \%$ full. The pessimist has $20 \%$ less water than the optimist, which is $\frac{1}{5}$ of the total volume of the glass. Since this difference is 46 milliliters according to the problem, the total volume of the glass is $46 \cd... | 230 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,108 |
Problem 8.2. On the board, 23 signs are drawn - some pluses and some minuses. If any 10 of them are chosen, there will definitely be at least one plus among them. If any 15 of them are chosen, there will definitely be at least one minus among them. How many pluses are there in total? | Answer: 14.
Solution. Since among any 10 signs there is a plus, the number of minuses on the board is no more than 9 (otherwise, we could choose 10 minuses).
Since among any 15 signs there is a minus, the number of pluses on the board is no more than 14 (otherwise, we could choose 15 pluses).
Then the total number o... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,109 |
Problem 8.3. In the grove, there are 140 chameleons - blue and red. One day, several blue chameleons changed their color to red. As a result, the number of blue chameleons decreased by 5 times, and the number of red chameleons increased by 3 times. How many chameleons changed their color? | Answer: 80.
Solution. Let the number of blue chameleons become $x$. Then initially, there were $5 x$ blue chameleons. Accordingly, the number of red chameleons initially was $140-5 x$. Then the number of red chameleons became $3 \cdot(140-5 x)$. Since the total number of chameleons remained the same, we get the equati... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,110 |
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 $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,111 |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,112 |
Problem 8.6. Vasya thought of three natural numbers with a sum of 1003. Calculating their product, Vasya noticed that it ends with $N$ zeros. What is the largest value that
$N$ can take? | Answer: 7.
Solution. The product of the three thought-of numbers could end with 7 zeros, for example, if these were the numbers $625, 250, 128$. Indeed, $625+250+128=1003$ and
$$
625 \cdot 250 \cdot 128=5^{4} \cdot\left(2 \cdot 5^{3}\right) \cdot 2^{7}=2^{8} \cdot 5^{7}=2 \cdot 10^{7}=20000000
$$
Suppose there exist... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,113 |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$ m/s, he would cover the same distance to school 2.5 times faster. This means that $\frac{v+2}{v}=2.5$, from which we find $v=\frac{4}{3}$. If he had initially run at a speed of $(v+4)$ m/s, he would have arrived at school $\fr... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,116 |
Problem 9.2. In the vertices of a cube, numbers $1,2, \ldots, 8$ are written in some order. It turned out that on three faces of the cube, the following condition is met: one of the numbers at the vertices is equal to the sum of the other three.
From the vertex with the number 6, three edges emanate. What three number... | Answer: There are three triples: $(2,3,5),(3,5,7),(2,3,7)$.
Solution. Let's call a face of the cube beautiful if one of the numbers on it is equal to the sum of the other three. Let's call a number large if on a beautiful face it is equal to the sum of the other three numbers. Clearly, the large numbers can only be $6... | (2,3,5),(3,5,7),(2,3,7) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,117 |
Problem 9.3. The teacher wrote an arithmetic example on the board: the product of two mixed fractions equals 43. Then he replaced the denominator of the first fraction with $X$, and the integer part of the second fraction with $Y$. The resulting expression is
$$
5 \frac{1}{X} \cdot Y \frac{1}{2}=43
$$
Restore the exa... | Answer: $X=17, Y=8$.
Solution. Since $5\frac{43}{5}>Y \frac{1}{2} \geqslant \frac{43}{5.5}=\frac{86}{11}>7.5
$
The number $Y$ is an integer, so $Y=8$.
Then $5 \frac{1}{X}=\frac{43}{8.5}=\frac{86}{17}=5 \frac{1}{17}$, so $X=17$. | X=17,Y=8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,118 |
Problem 9.4. In city $\mathrm{N}$, there are exactly three monuments. One day, a group of 42 tourists arrived in this city. Each of them took no more than one photo of each of the three monuments. It turned out that any two tourists together had photos of all three monuments. What is the minimum number of photos that a... | Answer: 123.
Solution. Let's number the three monuments. Note that the photo of the first monument is missing from no more than one tourist (otherwise, we could choose two tourists who, in total, have photos of no more than two monuments), so at least $42-1=41$ photos of the first monument were taken. Similarly, at le... | 123 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,119 |
Problem 9.5. On the base $AC$ of isosceles triangle $ABC (AB = BC)$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,120 |
Problem 9.7. On the coordinate plane, point $A$ is marked. On the $O x$ axis, point $B$ is marked, and on the $O y$ axis, point $C$ is marked.
It is known that the equations of the lines $A B, B C, A C$ in some order have the form $y=a x+4$, $y=2 x+b$ and $y=\frac{a}{2} x+8$ for some real numbers $a$ and $b$.
Find th... | Answer: 13 or 20.
Solution. Let the graph of the function $y=a x+4$ be denoted by $l_{1}$, the function $y=2 x+b-$ by $l_{2}$, and the function $y=\frac{a}{2} x+8-$ by $l_{3}$.
From the condition, it follows that $l_{1}, l_{2}, l_{3}$ are pairwise intersecting lines (each of the points $A, B, C$ belongs to some two o... | 13or20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,122 |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,126 |
Problem 10.4. Given a quadratic trinomial $P(x)$. It is known that the equations $P(x)=x-2$ and $P(x)=1-x / 2$ have exactly one root each. What is the discriminant of $P(x) ?$ | Answer: $-1 / 2$.
Solution. Let $P(x)=a x^{2}+b x+c, a \neq 0$. Consider the first quadratic equation and its discriminant:
$$
\begin{gathered}
a x^{2}+b x+c=x-2 \\
a x^{2}+x(b-1)+(c+2)=0 \\
D_{1}=(b-1)^{2}-4 a(c+2)
\end{gathered}
$$
Now consider the second quadratic equation and its discriminant:
$$
a x^{2}+b x+c=... | \frac{-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,127 |
Problem 10.5. Greedy Vovochka has 25 classmates. For his birthday, he brought 200 candies to the class. Vovochka's mother, to prevent him from eating them all himself, told him to distribute the candies so that any 16 of his classmates would collectively have at least 100 candies. What is the maximum number of candies ... | Answer: 37.
Solution. Among all 25 classmates, select 16 people with the smallest number of candies given. Note that among them, there is a person who received no less than 7 candies (otherwise, if they all received no more than 6 candies, then in total they received no more than $16 \cdot 6=96$ candies, which is less... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,128 |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,129 |
Problem 10.7. A square board $30 \times 30$ was cut along the grid lines into 225 parts of equal area. Find the maximum possible value of the total length of the cuts. | Answer: 1065.
Solution. The total length of the cuts is equal to the sum of the perimeters of all figures, minus the perimeter of the square, divided by 2 (each cut is adjacent to exactly two figures). Therefore, to get the maximum length of the cuts, the perimeters of the figures should be as large as possible.
The ... | 1065 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,130 |
Problem 10.8. A natural number $1 \leqslant n \leqslant 221$ is called lucky if the remainder of 221
divided by $n$ is divisible by the quotient (in this case, the remainder can be equal to 0). How many lucky numbers are there? | Answer: 115.
Solution. Let for some successful number $n$ the quotient be $k$, and the remainder be $k s$, by definition it is non-negative and less than the divisor $n$ (from the condition it follows that $k$ is a natural number, and $s$ is a non-negative integer.) Then
$$
221=n k+k s=k(n+s)
$$
Since $221=13 \cdot ... | 115 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,131 |
Problem 11.1. Petya wrote down ten natural numbers in a row as follows: the first two numbers he wrote down arbitrarily, and each subsequent number, starting from the third, was equal to the sum of the two preceding ones. Find the fourth number if the seventh is 42 and the ninth is 110. | Answer: 10.
Solution. From the condition, it follows that the eighth number is equal to the difference between the ninth and the seventh, i.e., $110-42=68$. Then the sixth is $68-42=26$, the fifth is $42-26=16$, and the fourth is $26-16=10$.
Remark. In fact, the numbers on the board are the doubled Fibonacci numbers. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,132 |
Problem 11.2. In the store, there are 9 headphones, 13 computer mice, and 5 keyboards. In addition, there are 4 sets of "keyboard and mouse" and 5 sets of "headphones and mouse". In how many ways can you buy three items: headphones, a keyboard, and a mouse? Answer: 646. | Solution. Let's consider the cases of whether any set was purchased.
- Suppose the set "keyboard and mouse" was purchased, then headphones were added to it. This results in exactly $4 \cdot 9=36$ ways to make the purchase.
- Suppose the set "headphones and mouse" was purchased, then a keyboard was added to it. This re... | 646 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,133 |
Problem 11.3. The teacher wrote the number 1818 on the board. Vasya noticed that if a multiplication sign is written between the hundreds and tens digits, the value of the resulting expression is a perfect square $\left(18 \times 18=324=18^{2}\right)$. What is the next four-digit number after 1818 that has the same pro... | Answer: 1832.
Solution. Since we need to find the nearest four-digit number, let's try to find it in the form $\overline{18 a b}$. Then the number $18 \cdot \overline{a b}=3^{2} \cdot(2 \cdot \overline{a b})$ must be a perfect square. From this, it follows that $2 \cdot \overline{a b}$ must also be a perfect square. C... | 1832 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,134 |
Problem 11.4. It is known that $\frac{7}{13}+\sin \varphi=\cos \varphi$ for some real $\varphi$. What is $\sin 2 \varphi$? | Answer: $\frac{120}{169}$.
Solution. From the condition, it follows that $\frac{7}{13}=\cos \varphi-\sin \varphi$. Squaring this equality, we get $\frac{49}{169}=\cos ^{2} \varphi+\sin ^{2} \varphi-2 \sin \varphi \cos \varphi=1-\sin 2 \varphi$. Therefore, $\sin 2 \varphi=1-\frac{49}{169}=\frac{120}{169}$. | \frac{120}{169} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,135 |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
, and different numbers have different small divi... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,137 |
Problem 11.7. An archipelago consists of $N \geqslant 7$ islands. Any two islands are connected by no more than one bridge. It is known that from each island, no more than 5 bridges lead, and among any 7 islands, there are definitely two connected by a bridge. What is the largest value that $N$ can take? | Answer: 36.
Solution. There can be 36 islands in the archipelago, for example, as follows: they form 6 groups of 6 islands each, and two islands are connected by a bridge if and only if they are in the same group. It is clear that from each island, exactly 5 bridges lead out, and among any 7 islands, there are necessa... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,138 |
Problem 11.8. In the quadrilateral pyramid $S A B C D$
- the lateral faces $S A B, S B C, S C D, S D A$ have areas 9, 9, 27, 27 respectively;
- the dihedral angles at the edges $A B, B C, C D, D A$ are equal;
- the quadrilateral $A B C D$ is inscribed, and its area is 36.
Find the volume of the pyramid $S A B C D$. | Answer: 54.
Solution. Let the angle between the lateral face and the base of the pyramid be denoted as $\alpha$.
Fig. 10: to the solution of problem 11.8
Draw the height $S H$ to the base ... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,139 |
1. Find all pairs of real numbers (x, y) for which the equality $\sqrt{x^{2}+y^{2}-1}=x+y-1$ holds. | Answer: All pairs of numbers $(1, t),(t, 1)$, where $t$ is any non-negative number.
Hint. By squaring and factoring, we get that $\mathrm{x}=1$ or $\mathrm{y}=1$. The square root is non-negative, so the sum $x+y-1 \geq 0$. Substituting the possible values of the variables into this inequality, we get that the other va... | (1,),(,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,140 |
2. The numbers from 1 to 8 are arranged in a circle. A number is called large if it is greater than its neighbors, and small if it is less than its neighbors. Each number in the arrangement is either large or small. What is the smallest possible sum of the large numbers? | Answer: 23.
Instructions. Adjacent numbers cannot be of the same type, so larger and smaller numbers alternate, and there are four of each. 8 is large. 7 is also large, since a small number must be less than two numbers, and seven is less than only one. 1 and 2 are small. 3 and 4 cannot both be large, as there are onl... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,141 |
3. On 20 cards, one digit was written on each card such that each digit was written exactly twice. One card with the digit one was lost. Can two numbers be formed using all the remaining 19 cards such that their ratio is $2018 ?$ | Answer: No.
Instructions. Proof by contradiction. Note that the sum of the numbers, the ratio of which is 2018, is divisible by 3: $m + 2018m = 2019m$, and 2019 is divisible by 3. Any number gives the same remainder when divided by 3 as the sum of all its digits (the criterion of equal remainder when divided by 3), so... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,142 |
5. The circle inscribed in the right triangle ABC touches the legs CA and CB at points P and Q, respectively. The line PQ intersects the line passing through the center of the inscribed circle and parallel to the hypotenuse at point N. M is the midpoint of the hypotenuse. Find the measure of angle MCN. | Answer: $90^{\circ}$.
Instructions. Let the center of the inscribed circle be denoted by I. The points of intersection of the line,
parallel to the hypotenuse and passing through I, with th... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,143 |
Task 1. The height of the Eiffel Tower in Paris is 324 m, its weight is 8000 t, it is made entirely of steel. What will be the height of an exact model of the tower weighing 1 kg made of the same steel? | Answer: 1.62 m. Solution: The linear dimensions of similar figures are related as the cube roots of their volumes. Since the materials of the tower and the model are the same, this ratio coincides with the ratio of the cube roots of the weights of the model and the tower, which is $1: 200$. Therefore, the height of the... | 1.62 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,144 |
Problem 2. There are 22 batteries, 15 of which are charged and 7 are discharged. The camera works with three charged batteries. You can insert any three batteries into it and check if it works. How can you guarantee to turn on the camera in 10 such attempts? | Solution. Let's number the batteries: $1,2, \ldots, 22$. The first six tests will involve inserting batteries into the camera as follows: $1,2,3 ; 4,5,6 ; \ldots, 16,17,18$. If at least one of these groups turns on the camera, everything is fine. If not, then among the first 18 batteries, there are at least 6 discharge... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,145 |
Problem 5. Inside triangle ABC, two points are given. The distances from one of them to the lines AB, BC, and AC are 1, 3, and 15 cm, respectively, and from the other - 4, 5, and 11 cm. Find the radius of the circle inscribed in triangle ABC. | Answer: 7 cm. First solution. Let $M_{1}$ and $M_{2}$ be the first and second given points, and let point $O$ be such that point $M_{2}$ is the midpoint of segment $O M_{1}$. Drop perpendiculars $M_{1} N_{1}, M_{2} N_{2}$, and $O N_{3}$ to line $A B$. Then segment $M_{2} N_{2}$ will be the midline of trapezoid $O N_{3}... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,147 |
7.1 Little Dima has two more teeth than little Yulia. Boxer Kolya has as many teeth as Dima and Yulia combined, but half as many as biologist Vanya. In total, they have 64 teeth. How many teeth does each of them have? | Solution: Dima has $x$ teeth, Yulia has $x-2$, Kolya has $x+x-2=2 x-2$, Vanya has $2 \cdot(2 x-2)=4 x-4$. In total, $x+x-2+2 x-2+4 x-4=8 x-8=64 . x=9$.
Number of teeth: D -9, Y $-7, K-16$, V -32. | D=9,Y=7,K=16,V=32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,148 |
7.2 In the class, there are 33 students, and the sum of their ages is 430 years. Prove that there are 20 students in the class whose sum of ages is greater than 260 years. | Solution: Let $V_{1} \geq V_{2} \geq \ldots \geq V_{20} \geq V_{21} \geq \ldots \geq V_{33}$ be the ages of the students, $A=V_{1}+V_{2}+\ldots+V_{20}, \quad B=V_{21}+\ldots+V_{33}$. According to the condition, $A+B=430$. It is clear that $A \geq 20 V_{20}, \quad B \leq 13 V_{20}, \quad$ hence $B \leq \frac{13}{20} A$.... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,149 |
11.5. Initially, 10 ones are written on the board. Grisha and Gleb play a game, taking turns. On his turn, Grisha squares 5 of the numbers on the board. On his turn, Gleb chooses several (possibly none) of the numbers on the board and increases each of them by 1. If a number divisible by 2023 appears on the board withi... | Answer. Grisha wins.
Solution. Note that $2023=7 \cdot 17^{2}$. Grisha will divide the numbers on the board into two groups of 5 and will square the numbers from the first group and the second group alternately. It is easy to see that the squares of integers not divisible by 7, when divided by 7, can only give remaind... | Grishawins | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,150 |
11.6. Plane $\alpha$ intersects the edges $A B, B C, C D$ and $D A$ of tetrahedron $A B C D$ at points $X, Y, Z$ and $T$ respectively. It turns out that points $Y$ and $T$ lie on circle $\omega$, constructed on segment $X Z$ as its diameter. Point $P$ is marked in plane $\alpha$ such that lines $P Y$ and $P T$ are tang... | Solution. From the problem statement, we immediately obtain that $\angle X Y Z=90^{\circ}=\angle X T Z$. Let $Q$ be the intersection point of the lines $X Y$ and $Z T$, and $R$ be the intersection point of the lines $Z Y$ and $X T$ (see Fig. 7). Without loss of generality, we can assume that point $Z$ lies on the segme... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,151 |
11.8. In a country with $N$ cities, there are $N(N-1)$ one-way roads: one road from $X$ to $Y$ for each ordered pair of cities $X \neq Y$. Each road has a maintenance cost. For a given $k=1, \ldots, N$, consider all ways to select $k$ cities and $N-k$ roads such that from each city, it is possible to reach some selecte... | Solution. Networks consisting of $N-k$ roads are called $k$-networks hereafter. Consider the undirected graph formed by the roads of a $k$-network. In it, there are no more than $k$ connected components, since each has a designated city. On the other hand, the number of components is at least $k$, since there are no mo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,153 |
11.6. The following functions are written on the board: $x+1, x^{2}+1, x^{3}+1, x^{4}+1$. It is allowed to add new functions to the board that are obtained from the written functions using the operations of subtraction and multiplication. Show how to obtain a non-zero function that takes non-negative values for positiv... | Solution. For example, $\left(x^{4}+1\right)(x+1)-\left(x^{4}+1\right)=$ $=x\left(x^{4}+1\right)$ is suitable.
Remark. There are other examples.
Comment. Any correct example -7 points. | x(x^{4}+1) | Algebra | proof | Yes | Yes | olympiads | false | 12,154 |
11.7. Can all natural numbers be colored in two colors so that the sum of any two different numbers of the same color is never a power of two?
(D. Khramtsov) | Answer. Possible.
First solution. We will paint numbers of the form $2^{s}(4 k+1)$, where $s, k=0,1, \ldots$, in the first color, and numbers of the form $2^{s}(4 k+3)$, where $(s, k=0,1, \ldots)$, in the second color. We will show that no two numbers of the first color can sum to a power of two; the reasoning for num... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,155 |
11.8. It is known that for some $x$ and $y$, the sums $\sin x + \cos y$ and $\sin y + \cos x$ are positive rational numbers. Prove that there exist natural numbers $m$ and $n$ such that $m \sin x + n \cos x$ is a natural number.
(N. Agakhanov) | Solution. Let $\sin x + \cos y = a$ and $\sin y + \cos x = b$. Then $\cos y = a - \sin x$ and $\sin y = b - \cos x$. Squaring these equations and adding them, we get, by the fundamental trigonometric identity, $1 = a^2 + b^2 - 2a \sin x - 2b \cos x + 1$, which simplifies to $2a \sin x + 2b \cos x = a^2 + b^2$. Let $N$ ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,156 |
11.9. Three spheres touch each other externally at points $A, B$, and $C$, and also touch the plane $\alpha$ at points $D, E$, and $F$. Prove that the radius of the circumcircle of triangle $A B C$ is less than the radius of the circumcircle of triangle $D E F$.
(A. Kuznetsov) | The first solution. Let the given spheres be denoted by $\Omega_{1}, \Omega_{2}$, and $\Omega_{3}$, such that they touch the plane $\alpha$ at points $D, E$, and $F$ respectively. Let $C$ be the point of tangency of the spheres $\Omega_{1}$ and $\Omega_{2}$. Denote by $\beta$ the plane perpendicular to $\alpha$ and con... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,157 |
11.10. Petya thought of two polynomials $f(x)$ and $g(x)$, each of the form $a x^{2} + b x + c$ (i.e., the degree of each polynomial does not exceed 2). On each turn, Vasya names a number $t$, and Petya tells him (at his discretion) one of the values $f(t)$ or $g(t)$ (without specifying which one he reported). After $n... | Answer. When $n=8$.
Solution. We will call a polynomial of the form $a x^{2}+b x+c$ simply a polynomial, and the graph of such a polynomial - simply a graph. We will use the following well-known lemma.
Lemma. Through any three points $\left(a_{i}, b_{i}\right)(i=1,2,3)$ with different abscissas, there passes exactly ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,158 |
1. Find all natural numbers $n$ such that the value of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number less than 2217.
Answer: $n=3^{8}, n=2^{8}$. | Solution. Transform the expression $\sqrt{n \sqrt{n \sqrt{n}}}=\left(n\left(n(n)^{\frac{1}{2}}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=$
$\left(n\left(n^{\frac{3}{2}}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(n(n)^{\frac{3}{4}}\right)^{\frac{1}{2}}=\left(n^{\frac{7}{4}}\right)^{\frac{1}{2}}=n^{\frac{7}{8}}$.
... | n=3^{8},n=2^{8} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,159 |
2. It is known that $\operatorname{tg} \alpha$ and $\operatorname{tg} 3 \alpha$ are integers. Find all possible values of $\operatorname{tg} \alpha$.
Answer: $-1 ; 0$ or 1. | Solution. Let $\operatorname{tg} \alpha=n$, and $\operatorname{tg} 3 \alpha=k$. Using the formula for the tangent of a triple angle (or deriving it), we get: $k=\frac{3 n-n^{3}}{1-3 n^{2}}$.
Transform the equation: $k\left(3 n^{2}-1\right)=n\left(n^{2}-3\right)$.
Let $|n|>1$.
Thus, $\left(3 n^{2}-1\right)$ and $n$ a... | -1,0,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,160 |
3. Do there exist two quadratic trinomials $a x^{2}+b x+c$ and $(a+1) x^{2}+(b+1) x+(c+1)$ with integer coefficients, each of which has two integer roots?
Om веm: No. | Solution. Suppose such trinomials exist, then by Vieta's theorem:
$\left\{\begin{array}{c}x_{1}+x_{2}=-\frac{b}{a}, \\ x_{1} x_{2}=\frac{c}{a} ;\end{array}\right.$ and $\left\{\begin{array}{c}x_{3}+x_{4}=-\frac{b+1}{a+1}, \\ x_{3} x_{4}=\frac{c+1}{a+1} .\end{array}\right.$
Rewrite the systems in a convenient form:
$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,161 |
5. A teacher fills in the cells of a class journal of size $7 \times 8$ (7 rows, 8 columns). In each cell, she puts one of three grades: 3, 4, or 5. After filling in the entire journal, it turns out that in each row, the number of threes is not less than the number of fours and not less than the number of fives, and in... | Answer: 8 fives.
## Solution.
First step.
In each row, there are no fewer threes than fours, so in the entire journal, there are no fewer threes than fours. In each column, there are no fewer fours than threes, so in the entire journal, there are no fewer fours than threes. Therefore, the number of threes and fours ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,162 |
1. Solve the system of equations $\left\{\begin{array}{l}\sqrt{11 x-y}-\sqrt{y-x}=1, \\ 7 \sqrt{y-x}+6 y-26 x=3\end{array}\right.$ | Solution. Introduce new variables $u=\sqrt{11 x-y}$ and $v=\sqrt{y-x}, u \geq 0, v \geq 0$. Then
$$
\left\{\begin{array} { l }
{ 1 1 x - y = u ^ { 2 } , } \\
{ y - x = v ^ { 2 } }
\end{array} \left\{\begin{array}{l}
x=0.1\left(u^{2}+v^{2}\right) \\
y=0.1\left(u^{2}+11 v^{2}\right)
\end{array}\right.\right.
$$
And, m... | (0.5;1.5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,163 |
2. The function $y=f(x)$ is such that for all values of $x$, the equality $f(x+1)=f(x)+2x+3$ holds. It is known that $f(0)=1$. Find $f(2018)$. | Solution. Rewrite the condition of the problem as $f(x+1)-f(x)=2 x+3$. Substituting sequentially instead of $x$ the numbers $0,1,2, \ldots, 2017$, we get the following equalities
$$
\begin{aligned}
& f(1)-f(0)=2 \cdot 0+3 \\
& f(2)-f(1)=2 \cdot 1+3 \\
& f(3)-f(2)=2 \cdot 2+3
\end{aligned}
$$
$$
f(2018)-f(2017)=2 \cdo... | 4076361 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,164 |
3. Twenty-one girls and twenty-one boys participated in a math competition. Each participant solved no more than six problems. For any girl and boy, there is at least one problem solved by both. Prove that there was a problem solved by at least three girls and at least three boys. | Solution. Suppose there is a problem that was solved by no more than two girls or no more than two boys. We will consider a problem "red" if it was solved by no more than two girls and "black" otherwise (then it was solved by no more than two boys).
Consider a chessboard with 21 rows, each corresponding to a girl, and... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,165 |
4. A car left point A for point B, which are 10 km apart, at 7:00. After traveling $2 / 3$ of the way, the car passed point C, from which a cyclist immediately set off for point A. As soon as the car arrived in B, a bus immediately set off from there in the opposite direction and arrived in point A at 9:00. At what dis... | Solution. Let $v_{a}$ be the speed of the car, $v_{\varepsilon}$ be the speed of the cyclist, and $v_{a \varepsilon}$ be the speed of the bus. From the problem statement, we derive the following system of equations:
$$
\left\{\begin{array}{l}
\frac{20 / 3}{v_{a}}+\frac{20 / 3}{v_{s}}=3 \\
\frac{10}{v_{a}}+\frac{10}{v_... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,166 |
5. Find all values of the parameter $a$, for each of which the sum of the lengths of the intervals that make up the solution to the inequality
$$
\frac{x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-6}{x^{2}+\left(a^{2}+5 a-5\right) x-a^{2}+4 a-6}<0
$$
is not less than 1. | Solution. Let's analyze the graphs of the functions in the numerator and denominator of the fraction. Let $y=f(x)$ be the function of the numerator, and $y=g(x)$ be the function of the denominator. Both of these functions are quadratic trinomials, and their graphs are parabolas. The problem reduces to the inequality $\... | \in(-\infty;2]\cup[3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 12,167 |
1. The number expressing the area of one rectangle made of matches (assuming the length of a match is 1) is one more than the number expressing the perimeter of this rectangle, and the area of the second rectangle is one less than its perimeter. What are the dimensions of these rectangles? | Solution. Let the lengths of the sides of the first rectangle be denoted by x and y. Then, from the condition of the problem, $xy = 2(x + y) + 1$. From this, we get $y = \frac{2x + 1}{x - 2} = 2 + \frac{5}{x - 2}$. For y to be an integer when x is an integer, we need $x - 2 = 1$ or $x - 2 = 5$. In the first case, we ge... | 7\times35\times3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,168 |
2. Find the area of the triangle if it is known that its medians $C M$ and $B N$ are 6 and 4.5 respectively, and $\angle B K M=45^{\circ}$, where $K$ is the point of intersection of the medians. | Solution. The given triangle $ABC$ is shown in the diagram, where the medians $CM$ and $BN$ are 6 and 4.5, respectively, $\angle BKM=45^{\circ}$, and $K$ is the point of intersection of the medians. It is easy to prove that the medians of a triangle divide it into 6 equal-area triangles. Therefore, to find the area of ... | 9\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,169 |
# 3. Solve the system of equations
$$
\left\{\begin{array}{l}
x^{5}+3 x^{4}+5 x^{3}+5 x^{2}+6 x+2=0 \\
x^{3}+3 x^{2}+4 x+1=0
\end{array}\right.
$$ | Solution. Transform the first equation:
$$
\begin{aligned}
& \left(x^{5}+3 x^{4}+4 x^{3}+x^{2}\right)+x^{3}+4 x^{2}+6 x+2=0 \Rightarrow \\
& x^{2}\left(x^{3}+3 x^{2}+4 x+1\right)+\left(x^{3}+3 x^{2}+4 x+1\right)+x^{2}+2 x+1=0
\end{aligned}
$$
Considering that $x^{3}+3 x^{2}+4 x+1=0$, we get $x^{2}+2 x+1=0$. Solving t... | thesystemhasnosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,170 |
# 5. Plot the graph of the function
$$
y=\frac{\sqrt{\frac{1+x^{2}}{2 x}+1}+\sqrt{\frac{1+x^{2}}{2 x}-1}}{\sqrt{\frac{1+x^{2}}{2 x}+1}-\sqrt{\frac{1+x^{2}}{2 x}-1}}
$$ | Solution. $D(y)=\mathbf{R}_{+}$. After simplification, we get:
$$
y=\frac{1+x+|1-x|}{1+x-|1-x|}=\left\{\begin{array}{c}
\frac{1}{x}, \quad \text { for } 0<x<1 \\
x, \quad \text { for } x \geq 1
\end{array}\right.
$$
The graph of the function is shown in the figure.
Grading criteria. Correct example provided - 7 points. In other cases - 0 points. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,173 |
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