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1. Find the largest natural number consisting of distinct digits such that the product of the digits of this number is 2016. | Answer: 876321.
Solution: Factorize the number 2016. $2016=2^{5} \cdot 3^{2} \cdot 7$. To make the number as large as possible, it should contain the maximum number of digits. Notice that the number must include the digit 1. Therefore, the number should consist of the digits $1,2,3,6,7,8$.
Grading Criteria. 20 points... | 876321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Ten consecutive natural numbers are written on the board. What is the maximum number of them that can have a digit sum equal to a perfect square? | Answer: 4.
Solution: Note that the sums of the digits of consecutive natural numbers within the same decade are consecutive natural numbers. Since there are 10 numbers, they span two decades. Also note that among ten consecutive natural numbers, there can be no more than 3 perfect squares, and three perfect squares ca... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find all real roots of the equation
$(x+1)^{5}+(x+1)^{4}(x-1)+(x+1)^{3}(x-1)^{2}+(x+1)^{2}(x-1)^{3}+(x+1)(x-1)^{4}+(x-1)^{5}=0$ | Solution. Multiply both sides of the equation by $(x+1)-(x-1)$ (this factor equals 2). Use the formula $a^{6}-b^{6}=(a-b)\left(a^{5}+a^{4} b+a^{3} b^{2}+a^{2} b^{3}+a b^{4}+b^{5}\right)$.
$$
\begin{gathered}
(x+1)^{6}-(x-1)^{6}=0 \\
(x+1)^{6}=(x-1)^{6}
\end{gathered}
$$
The equation
$$
x+1=x-1
$$
has no solutions. ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Four points $A, B, C, D$ are on a plane. It is known that $A B=1, B C=$ $2, C D=\sqrt{3}, \angle A B C=60^{\circ}, \angle B C D=90^{\circ}$. Find $A D$. | Solution. Let's construct the diagram. Let the line $CD$ intersect the line $AB$ at point $O$
(according to the condition, these lines are not parallel). There are two possible positions for... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The parabolas in the figure are obtained by shifting the parabola $f(x)=x^{2}$ and are arranged such that the points of their intersections with the $O X$ axis respectively coincide in pairs, and all vertices lie on the same straight line. There are a total of 2020 parabolas. The length of the segment on the $O X$ a... | Solution. Note that the arrangement of parabolas can be changed by shifting the entire set to the left or right (the lengths of the segments on the $O X$ axis, enclosed between the roots of the parabolas, do not change with the shift). The intersection point of the line drawn through the vertices of the parabolas will ... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Three numbers are stored in a computer's memory. Every second, the following operation is performed: each number in this triplet is replaced by the sum of the other two. For example, the triplet $(1 ; 3 ; 7)$ turns into $(10 ; 8 ; 4)$. What will be the difference between the largest and smallest number in the triple... | 5. Answer: 19. Solution. Let the initial triplet be ( $a ; b ; c$). Since all numbers in the triplet are distinct, we will assume that $a<b<c$. In the next second, the triplet of numbers will look like this: $(b+c$; $a+c ; a+b)$. The largest number in this triplet is $b+c$, and the smallest is $a+b$. Their difference i... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. Find the largest four-digit number in which all digits are different and which is divisible by any of its digits (don't forget to explain why it is the largest). | Answer: 9864.
Firstly, the desired number cannot have the form $\overline{987 a}$, because divisibility by the digit 7 would mean that $a$ is 0 or 7. This means the desired number is smaller. Secondly, consider numbers of the form $\overline{986 a}$. Divisibility by the digit 9 would mean that $9+8+6+a=a+23$ is divisi... | 9864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. To some number $\kappa$, the sum of its digits was added and the result was 2014. Provide an example of such a number. | Answer: 1988 or 2006.
Grading Criteria:
+ a correct example is provided (one is sufficient)
$\pm$ a correct example is provided along with an incorrect one
- the problem is not solved or solved incorrectly | 1988 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.2. The Wolf, the Hedgehog, the Chizh, and the Beaver were dividing an orange. The Hedgehog got twice as many segments as the Chizh, the Chizh got five times fewer segments than the Beaver, and the Beaver got 8 more segments than the Chizh. Find out how many segments were in the orange, if the Wolf only got the peel. | Answer: 16 segments.
Solution. First method. Let the number of orange segments given to Chizh be $x$, then Hedgehog received $2x$ segments, and Beaver received $5x$ segments (Wolf - 0 segments). Knowing that Beaver received 8 more segments than Chizh, we set up the equation: $5x - x = 8$. The solution to this equation... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.3. In a seven-story building, domovoi (Russian house spirits) live. The elevator travels between the first and the last floors, stopping at every floor. On each floor, starting from the first, one domovoi entered the elevator, but no one exited. When the thousandth domovoi entered the elevator, it stopped. On which f... | Answer: on the fourth floor.
Solution. First, let's find out how many housekeepers ended up in the elevator after the first trip from the first to the seventh floor and back, until the elevator returned to the first floor. One housekeeper entered on the first and seventh floors, and on all other floors, two housekeepe... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Find the maximum possible area of a quadrilateral in which the product of any two adjacent sides is 1. | 1. Find the maximum possible area of a quadrilateral for which the product of any two adjacent sides is 1.
OTBET: 1.
SOLUTION.
Let the quadrilateral have sides $a, b, c, d$. Then $a b=b c=c d=d a=1$. From the equality $a b=b c$, it follows that $a=c$, and from the equality $b c=c d$, we get that $b=d$. Therefore, th... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.7. In a magic shop, for 20 silver coins you can buy an invisibility cloak and get 4 gold coins as change. For 15 silver coins you can buy an invisibility cloak and get 1 gold coin as change. How many silver coins will you get as change if you buy an invisibility cloak for 14 gold coins? | Answer: 10.
Solution. In the first case, compared to the second, by paying 5 extra silver coins, one can receive 3 extra gold coins in change. Therefore, 5 silver coins are equivalent to 3 gold coins.
In the second case, by paying 15 silver coins (which is equivalent to $3 \cdot 3=9$ gold coins), one can get the cloa... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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?
. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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
$$
\begin{aligned}
& f(1)-f(0)=2 \cdot 0+3 \\
& f(2)-f(1)=2 \cdot 1+3 \\
& f(3)-f(2)=2 \cdot 2+3
\end{aligned}
$$
$$
f(2018)-f(2017)=2 \cdo... | 4076361 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A car left point A for point B, which are 10 km apart, at 7:00. After traveling $2 / 3$ of the way, the car passed point C, from which a cyclist immediately set off for point A. As soon as the car arrived in B, a bus immediately set off from there in the opposite direction and arrived in point A at 9:00. At what dis... | Solution. Let $v_{a}$ be the speed of the car, $v_{\varepsilon}$ be the speed of the cyclist, and $v_{a \varepsilon}$ be the speed of the bus. From the problem statement, we derive the following system of equations:
$$
\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 |
4. In city $\mathrm{N}$, only blondes and brunettes live. Blondes always lie, while brunettes always tell the truth. Every day, the residents dye their hair the opposite color. On one Monday in October, everyone born in the fall was asked: "Were you born this month?" - and 200 residents answered "yes," while no one ans... | 4. Answer. 0.
Solution. Note that on Monday and Friday of the same week, the residents have the same hair color: Monday - Tuesday - Wednesday - Thursday - Friday. Then, if the same people answer differently on Monday and Friday, it means that the month has changed, and the brunettes will answer "no" to the same questi... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. A tournament of dodgeball was held at school. In each game, two teams competed. 15 points were awarded for a win, 11 for a draw, and no points for a loss. Each team played against each other once. By the end of the tournament, the total number of points scored was 1151. How many teams were there? | 5. Answer: 12 teams.
Solution. Let there be $\mathrm{N}$ teams. Then the number of games was $\mathrm{N}(\mathrm{N}-1) / 2$. For each game, a total of 15 or 22 points are scored. Therefore, the number of games was no less than 53 $(1151 / 22)$ and no more than $76(1151 / 15)$.
Note that if there were no more than 10 ... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. In a basket, there are 41 apples: 10 green, 13 yellow, and 18 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What is... | Answer: (a) 13 apples. (b) 39 apples.
Solution. (a) Alyona can take all 13 yellow apples from the basket, for example, if the first 13 apples turn out to be yellow. Since at no point do the yellow apples become fewer than the red ones, Alyona can indeed do this.
(b) We will show that Alyona can take all the apples ex... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. A polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-8$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smallest... | Answer: (a) -24. (b) 6.
Solution. (a) Since the graph of $G(x)$ is symmetric with respect to the line $x=-8$, the points at which it takes the same value must be divided into pairs of symmetric points, except possibly one point that lies on the axis of symmetry. Therefore, the middle of the five given points should li... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. On a horizontal floor, there are three volleyball balls with a radius of 18, each touching the other two. Above them, a tennis ball with a radius of 6 is placed, touching all three volleyball balls. Find the distance from the top point of the tennis ball to the floor. (All balls are spherical.)
Fig. 9: to the solution of problem 11.4
Solution. Let the centers of the volleyballs be denoted by $A, B, C$, and the center of the tennis ball by $D$; the top point of the tenn... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Point $M$ is the midpoint of the base $B C$ of trapezoid $A B C D$. A point $P$ is chosen on the base $A D$. Ray $P M$ intersects ray $D C$ at point $Q$. The perpendicular to base $A D$, drawn through point $P$, intersects segment $B Q$ at point $K$.
It is known that $\angle K Q D=64^{\circ}$ and $\angle... | Answer: $39^{2}$.
Fig. 11: to the solution of problem 11.6
Solution. From the condition, it follows that $\angle B K D=\angle K Q D+\angle K D Q=64^{\circ}+38^{\circ}=102^{\circ}$.
Let the... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Real numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=11 \\
b^{2}+b c+c^{2}=11
\end{array}\right.
$$
(a) (1 point) What is the smallest value that the expression $c^{2}+c a+a^{2}$ can take?
(b) (3 points) What is the largest value that the expression $c^{2}+c a+a^{2}$ can take... | Answer: (a) 0; (b) 44.
Solution. (a) Note that if $a=c=0$, and $b=\sqrt{11}$, then the given system of equalities is satisfied, and $c^{2}+c a+a^{2}$ equals 0. On the other hand,
$$
c^{2}+c a+a^{2}=\frac{c^{2}}{2}+\frac{a^{2}}{2}+\frac{(c+a)^{2}}{2} \geqslant 0
$$
Thus, the smallest value of the expression $c^{2}+c ... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.1.1. In a basket, there are 41 apples: 10 green, 13 yellow, and 18 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What ... | Answer: (a) 13 apples. (b) 39 apples. | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.1.2. In a basket, there are 38 apples: 9 green, 12 yellow, and 17 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What i... | Answer: (a) 12 apples. (b) 36 apples. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.1.3. In a basket, there are 35 apples: 8 green, 11 yellow, and 16 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What i... | Answer: (a) 11 apples. (b) 33 apples. | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.1.4. In a basket, there are 44 apples: 11 green, 14 yellow, and 19 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What ... | Answer: (a) 14 apples. (b) 42 apples. | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.3.1. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-8$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -24 . (b) 6. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.3.3. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-7$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -21 . (b) 6. | -21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.3.4. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-6$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -18 . (b) 6. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.7.4. Let $N$ be the least common multiple of ten different natural numbers $a_{1}<a_{2}<a_{3}<\ldots<a_{10}$.
(a) (2 points) What is the smallest value that $N / a_{6}$ can take?
(b) (2 points) Identify all possible values of $a_{6}$ in the interval $[1 ; 2000]$, for which the value of $N / a_{1}$ can take... | Answer: (a) 5. (b) $504,1008,1512$. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. $\left(225-4209520: \frac{1000795+(250+x) \cdot 50}{27}\right)=113$
4209520: $\frac{1000795+(250+x) \cdot 50}{27}=112$
$\frac{1000795+(250+x) \cdot 50}{27}=37585$
$1000795+(250+x) \cdot 50=1014795$
$250+x=280$
$x=30$ | Answer: $\boldsymbol{x}=30$.
Criteria. Correct solution - 7 points. $3 a$ for each computational error minus 1 point. Error in the algorithm for solving the equation - 0 points | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. 11088 is $110\%$ of the containers in December compared to November. This means that in November, $11088: 1.10=10080$ containers were manufactured, which is $105\%$ compared to October. This means that in October, $10080: 1.05=9600$ containers were manufactured, which is $120\%$ compared to September. Therefore, in ... | Answer: in September 8000, in October 9600, in November 10080 containers. | 8000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In one lyceum, $76 \%$ of the students have at least once not done their homework, and $\frac{5}{37}$ sometimes forget their second pair of shoes. Find the number of students in the lyceum, if it is more than 1000 but less than 2000. | Solution. Since $76 \%=\frac{76}{100}=\frac{19}{25}$, and the numbers 25 and 37 are coprime, the number of students is a multiple of $25 \cdot 37$, i.e., $925 \mathrm{k}$, where k is a natural number. Since $1000<925 k<2000$, then $\mathrm{k}=2$, and the number of students $925 \cdot 2$ $=1850$.
Answer. 1850
Recommen... | 1850 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Denis housed chameleons that can change color only to two colors: red and brown. Initially, the number of red chameleons was five times the number of brown chameleons. After two brown chameleons turned red, the number of red chameleons became eight times the number of brown chameleons. Find out how many chameleons D... | Solution. Let $t$ be the number of brown chameleons Denis had. Then the number of red chameleons was $5t$. From the problem statement, we get the equation $5 t+2=8(t-2)$. Solving this, we find $t=6$. Therefore, the total number of chameleons is $6 t$, which is 36.
Answer. 36
Recommendations for checking. Only the cor... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. An odd six-digit number is called "simply cool" if it consists of digits that are prime numbers, and no two identical digits stand next to each other. How many "simply cool" numbers exist? | Solution. There are four single-digit prime numbers in total - 2, 3, 5, 7. We will place the digits starting from the least significant digit. In the units place, three of them can stand $-3, 5, 7$. In the tens place, there are also three out of the four (all except the one placed in the units place). In the hundreds p... | 729 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the $3000-$th year, the World Hockey Championship will be held under new rules: 12 points will be awarded for a win, 5 points will be deducted for a loss, and no points will be awarded for a draw. If the Brazilian national team plays 38 matches in this championship, scores 60 points, and loses at least once, how ... | Solution. Let Brazil win in x matches and lose in y matches. We form the equation $12 x-5 y=60$. We see that $12 \mathrm{x} \vdots 12$ and $60 \vdots 12$. GCD(5, 12)=1, i.e., $y \vdots 12$. Possible: a) $y=12$. Then we get the equation $12 x-60=60$. Thus, $x=10$. This is possible. b) $y=24$. We get the equation: $12 x-... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Three boys and 20 girls stood in a row. Each child counted the number of girls who were to the left of them, the number of boys who were to the right of them, and added the results. What is the maximum number of different sums that the children could have obtained? (Provide an example of how such a number could be o... | Solution. Let's consider how the number changes when moving from left to right by one person. If the adjacent children are of different genders, the number does not change. If we move from a girl to a girl, the number increases by 1, and if from a boy to a boy, it decreases by 1. Thus, the smallest number in the row co... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.3. Find all natural $k$ such that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first).
(V. Senderov) | Answer. $k=1$.
Solution. Let $n \geqslant 2$, and $2=p_{1}<p_{2}<\ldots<p_{k}$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation $(*)$ would be divisible by $q$, which is impossible. Therefore, $a>p_{k}$.
Without loss of generality, we can ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. Petya wrote ten natural numbers on the board, none of which are equal. It is known that among these ten numbers, three can be chosen that are divisible by 5. It is also known that among the ten numbers written, four can be chosen that are divisible by 4. Can the sum of all the numbers written on the board be less ... | Answer. It can.
Solution. Example: $1,2,3,4,5,6,8,10,12,20$. In this set, three numbers $(5,10,20)$ are divisible by 5, four numbers $(4,8,12,20)$ are divisible by 4, and the total sum is 71.
Remark. It can be proven (but, of course, this is not required in the problem), that in any example satisfying the problem's c... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. On the board, a certain natural number $N$ was written nine times (one under the other). Petya appended a non-zero digit to the left or right of each of the 9 numbers; all the appended digits are different. What is the maximum number of prime numbers that could result among the 9 obtained numbers?
(I. Efremov) | Answer: 6.
Solution. Let $S$ be the sum of the digits of the number $N$. Then the sums of the digits of the obtained numbers will be $S+1, S+2, \ldots, S+9$. Three of these sums will be divisible by 3. By the divisibility rule for 3, the corresponding three numbers on the board will also be divisible by 3. Since these... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.4. In the company, some pairs of people are friends (if $A$ is friends with $B$, then $B$ is also friends with $A$). It turned out that among any 100 people in the company, the number of pairs of friends is odd. Find the largest possible number of people in such a company.
(E. Bakayev) | Answer: 101.
Solution: In all solutions below, we consider the friendship graph, where vertices are people in the company, and two people are connected by an edge if they are friends.
If the graph is a cycle containing 101 vertices, then any 100 vertices will have exactly 99 edges, so such a company satisfies the con... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The numbers $1,2,3, \ldots, 29,30$ were written in a row in a random order, and partial sums were calculated: the first sum $S_{1}$ equals the first number, the second sum $S_{2}$ equals the sum of the first and second numbers, $S_{3}$ equals the sum of the first, second, and third numbers, and so on. The last sum $... | Answer: 23.
Solution: Evaluation: adding an odd number changes the parity of the sum, there are 15 odd numbers, so the parity of the sums changes at least 14 times. Therefore, there will be at least 7 even sums, and thus no more than 23 odd sums.
Implementation: arrange the numbers as follows: 1, then all even number... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
113. In the forest, there grew pines, cedars, and larches, and there were cones on all the trees, and they were equal in number. A gentle breeze blew, and several cones fell to the ground. It turned out that $11\%$ of the cones fell from each pine, $54\%$ from each cedar, and $97\%$ from each larch. At the same time, e... | 11.3. Let $m$ be the number of cones on each tree, $s, k, l-$ the number of pines, cedars, and deciduous trees in the forest. Then the condition of the problem corresponds to the equation $0.3 m(s+k+l)=0.11 m s+0.54 m k+0.97 m l$. This is equivalent to $19(s+k+l)=43 k+86 l$. The right side of this expression is divisib... | 43 | Number Theory | proof | Yes | Yes | olympiads | false |
115. In how many non-empty subsets of the set $\{1,2,3, \ldots, 10\}$ will there be no two consecutive numbers? | 115. 143.
Let $A_{n}$ denote the set of non-empty subsets of the set $\{1,2,3, \ldots, n\}$ that do not contain two consecutive numbers, and let $a_{n}$ be the number of such subsets. Clearly, the set $A_{1}$ consists of the subset $\{1\}$, and the set $A_{2}$ consists of the subsets $\{1\}$ and $\{2\}$. Thus, $a_{1}... | 143 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.3. Lёsha colors cells inside a $6 \times 6$ square drawn on graph paper. Then he marks the nodes (intersections of the grid lines) to which the same number of colored and uncolored squares are adjacent. What is the maximum number of nodes that can be marked? | Answer: 45.
## Reasoning.
Evaluation. Each grid node belongs to one, two, or four squares.
The corner vertices of the original square are adjacent to only one small square each, so Lёsha will not be able to mark them. Therefore, the maximum number of marked nodes does not exceed $7 \cdot 7-4=45$.
, where \( N > 5 \). What is the smallest value that \( N \) can have?
(O. Dmitriev)
# | # Answer: 14.
Solution. The number $N$ can equal 14, as shown, for example, by the quartet of numbers $4, 15, 70, 84$. It remains to show that $N \geqslant 14$.
Lemma. Among the pairwise GCDs of four numbers, there cannot be exactly two numbers divisible by some natural number $k$.
Proof. If among the original four ... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.4. All cells of a $100 \times 100$ square table are numbered in some order with numbers from 1 to 10000. Petya colors the cells according to the following rules. Initially, he colors $k$ cells at his discretion. Then, on each move, Petya can color one more uncolored cell with number $a$ if at least one of the two con... | Answer. $k=1$.
Solution. First, let's prove the following statement.
Lemma. For any two cells $A$ and $B$, there exists a cell $C$ such that by coloring it, one can then color both $A$ and $B$ (possibly $C$ coincides with $A$ or $B$).
Proof. We can assume that the number $a$ of cell $A$ is less than the number $b$ o... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.2. On the board, the expression $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}$ is written, where $a, b, c, d, e, f$ are natural numbers. If the number $a$ is increased by 1, then the value of this expression increases by 3. If in the original expression the number $c$ is increased by 1, then its value increases b... | # Answer: 60.
First solution. Let the value of the original expression be $A$. Then, as a result of the first operation, the product will take the value $\frac{a+1}{a} \cdot A=A+3$, from which $A=3a$. This means that $A$ is a natural number. Moreover, from this equality, it follows that it is divisible by 3. Similarly... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. $\mathrm{ABCD}$ is a rhombus. Points $E$ and $F$ lie on sides $AB$ and $BC$ respectively, such that $\frac{AE}{BE}=\frac{BF}{CF}=5$. Triangle $DEF$ is equilateral. Find the angles of the rhombus. Solution.
The notation is shown in the figure. The situation where angle $D$ of the rhombus is acute is impossible (see ... | Answer. The angles of the rhombus are $60^{\circ}$ and $120^{\circ}$. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.2 About numbers a and b, it is known that the system of equations
$$
\left\{\begin{array}{l}
y^{2}=x^{2}+a x+b \\
x^{2}=y^{2}+a y+b
\end{array}\right.
$$
has no solutions. Find a. | Solution: Since the system has no solutions, in particular, there are no solutions with $x=y$. When $x=y$, both equations of the system are equivalent to the equation $a x+b=0$. This linear equation has no roots only when its slope coefficient $a$ is zero.
## Criteria:
- Points are not deducted for the absence of an ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4 Another participant in the competition for meaningless activity marked the centers of 13 cells in a grid rectangle of size $(N-1) \times(N+1)$ such that the distance between any two marked points is greater than 2. What is the smallest value that $N$ can take? | Solution: We will show that it is impossible to mark cells in a $6 \times 8$ rectangle (and thus in any smaller size) in such a way. Indeed, let's divide the rectangle into $2 \times 2$ squares. In each of them, the pairwise distances between the centers of the cells do not exceed $\sqrt{2}$, so no more than one cell i... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. In pentagon $M N P Q S: \quad M N=N P=P Q=Q S=S M$ and $\angle M N P=2 \angle Q N S$. Find the measure of angle $M N P$. | Answer: $60^{\circ}$
Solution. Since $\angle S N Q=\angle M N S+\angle P N Q$, we can take a point $T$ on the side $S Q$ such that $\angle S N T=\angle M N S=\angle M S N$, i.e., $N T \| M S$.
Then $\angle T N Q=\angle S N Q-\angle S N T=\angle P N Q=\angle N Q P$, i.e., $N T \| P Q$. Therefore, $M S \| P Q$, and sin... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. The teams participating in the quiz need to answer 50 questions. The cost (in integer points) of a correct answer to each question was determined by experts after the quiz, the cost of an incorrect answer - 0 points. The final score of the team was determined by the sum of points received for correct answers. Whe... | Answer: 50.
Solution: We will prove that with 50 teams, such a distribution of points can exist. The example is obvious - let the $k$-th team answer only the $k$-th question. Then, by assigning the costs of the questions as $a_{1}, a_{2}, \ldots, a_{50}$, where $\left\{a_{1}, a_{2}, \ldots, a_{50}\right\}=\{1,2, \ldot... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9-3-1. The numbers from 1 to 217 are divided into two groups: one group has 10 numbers, and the other has 207. It turns out that the arithmetic means of the numbers in the two groups are equal. Find the sum of the numbers in the group of 10 numbers. | Answer: 1090.
Solution Variant 1. The shortest solution to this problem is based on the following statement:
If the arithmetic means of the numbers in two groups are equal, then this number is also equal to the arithmetic mean of all the numbers.
A formal proof of this is not difficult: it is sufficient to denote th... | 1090 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9-4-1. In the figure, $O$ is the center of the circle, $A B \| C D$. Find the degree measure of the angle marked with a «?».
 | Answer: $54^{\circ}$.
Solution variant 1. Quadrilateral $A D C B$ is an inscribed trapezoid in a circle. As is known, such a trapezoid is isosceles, and in an isosceles trapezoid, the angles at the base are equal: $\angle B A D=\angle C B A=63^{\circ}$. Triangle $D O A$ is isosceles ($O A$ and $O D$ are equal as radii... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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