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[ Text problems (miscellaneous) ] [Arithmetic. Mental calculation, etc.]
In response to a question about the age of his children, the mathematician answered:
- My wife and I have three children. When our first child was born, the total age of the family members was 45 years, a year ago, when the third child was born ... | A year ago, the total age of the children was 11 years, which means the parents' total age was 59 years. And on the day of the first child's birth, this sum was 45. Therefore, the time between these two events is (59 - 45) : $2=7$ years. Thus, the first child was 7 years old a year ago, and the second was 4.
## Answer... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
One degree on the Celsius scale is equal to 1.8 degrees on the Fahrenheit scale, while $0^{\circ}$ Celsius corresponds to $32^{\circ}$ Fahrenheit.
Can a temperature be expressed by the same number of degrees both in Celsius and Fahrenheit? | From the condition, it follows that the temperature in Fahrenheit is expressed through the temperature in Celsius as follows: $T_{F}=1.8 T_{C}+32^{\circ}$.
If $T_{F}=T_{C}$, then $0.8 T_{C}+32=0$, that is, $T_{C}=-40$.
## Answer
Can.
| -40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bverlov S.L.
Petya is coloring 2006 points, located on a circle, using 17 colors. Then Kolya draws chords with endpoints at the marked points such that the endpoints of any chord are of the same color and the chords do not have any common points (including common endpoints). Kolya wants to draw as many chords as possi... | Note that $2006=17 \cdot 118$; therefore, there will be two colors in which at least $2 \cdot 118=236$ points are painted in total.
We will prove by induction on $k$ that through $2 k-1$ points of two colors, it is always possible to draw $k-1$ non-intersecting chords with same-colored endpoints.
Base case $(k=1,2)$ ... | 117 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Folkpor
In a convex quadrilateral $A B C D: A C \perp B D, \angle B C A=10^{\circ}, \angle B D A=20^{\circ}, \angle B A C=40^{\circ}$. Find $\angle B D C$. | Let $K$ and $M$ be the points of intersection of the line $CB$ with the line $AD$ and the circumcircle of triangle $ACD$, respectively (see figure). Then
$\angle MDA = \angle MCA = 10^{\circ}$, so $DM$ is the bisector of angle $KDB$. Also note that $\angle ABD = 50^{\circ}$, $\angle CBD = 80^{\circ}$, hence $\angle KB... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the abscissa of the point of intersection.
# | First method. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be simplified to: $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, therefore $x = 1$.
Second method. Notice that $x = 1$ is a solution to the problem: when $x = 1$, both given l... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc. ] [ Problems on percentages and ratios ]
Author: Raskina I.V.
Children went to the forest to pick mushrooms. If Anya gives half of her mushrooms to Vitya, all the children will have the same number of mushrooms, and if instead Anya gives all her mushrooms to Sasha, Sasha will ha... | Let Anya give half of her mushrooms to Vitya. Now all the kids have the same number of mushrooms (this means Vitya had no mushrooms of his own). For Sasha to get all of Anya's mushrooms, he needs to take the mushrooms from Vitya and Anya. Then he will have the mushrooms of three kids - Vitya's, Anya's, and his own. The... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The sum of the three smallest distinct divisors of some number $A$ is 8. How many zeros can the number $A$ end with?
# | The number 8 can be represented as the sum of three different natural numbers in two ways: $8=1+2+5=$ $1+3+4$. The numbers 1, 3, and 4 cannot be the three smallest divisors of the number $A$: if $A$ is divisible by 4, then it is also divisible by 2. Therefore, the three smallest divisors of $A$ are 1, 2, and 5. Thus, $... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (miscellaneous). ] [ Systems of linear equations ]
A calf weighs as much as a kid together with a piglet. And a piglet together with a calf weighs as much as a lamb together with a kid.
How much does the piglet weigh, if the lamb weighs 30 kg? | A lamb and a kid weigh as much as a kid and two piglets. Therefore, a piglet weighs half as much as a lamb.
## Answer
15 kg. | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Galoikin A.i.
Nine digits: $1,2,3, \ldots, 9$ are written in some order (so that a nine-digit number is formed). Consider all triples of consecutive digits, and find the sum of the corresponding seven three-digit numbers. What is the maximum possible value of this sum? | Let $\overline{a b c d e f g h i}-$ be a nine-digit number. From it, the sum $S=\overline{a b c}+\overline{b c d}+\overline{c d e}+\overline{d e f}+\overline{f f g}+\overline{f g h}+\overline{g h i}=$ $100 a+110 b+111(c+d+e+f+g)+11 h+i$ is formed. The sum $S$ reaches its maximum when the larger digits enter it with the... | 4648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Proovvoov v.V.
A chess king has traversed the entire $8 \times 8$ board, visiting each square exactly once and returning to the starting square on the last move.
Prove that he made an even number of diagonal moves. | At each non-diagonal move, the color of the field the king stands on changes; at a diagonal move - it does not change. Since the king has toured the entire board and returned, the color of the field changed from white to black as many times as from black to white, which means the king made an even number of non-diagona... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
Booin D.A.
The whole family drank a full cup of coffee with milk, and Katya drank a quarter of all the milk and a sixth of all the coffee. How many people are in the family? | Let $n$ be the number of cups (the number of people in the family), and $x$ be the amount of milk consumed (in cups). Then the amount of coffee consumed is $n-x$. Katya drank one cup of coffee with milk, which consisted of one quarter of all the milk $(x / 4)$ and one sixth of all the coffee $((n-x) / 6)$. We get
$$
\... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Galochkina A.i.
The digits 1, 2, 3,..., 9 are arranged in a circle in some arbitrary order. Every three consecutive digits, when read clockwise, form a three-digit number. Find the sum of all nine such numbers. Does this sum depend on the order in which the digits are arranged?
# | In our numbers, each digit appears exactly once in each of the hundreds, tens, and units places.
## Solution
A three-digit number, where the hundreds place is digit $a$, the tens place is digit $b$, and the units place is digit $c$, is equal to $100a + 10b + c$. (For example, $394 = 3 \cdot 100 + 9 \cdot 10 + 4$.) Lo... | 4995 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7,8 Authors: Spivak A.V., Yanenko I.V.
Nataша and Inna bought identical boxes of tea bags. It is known that one tea bag is enough for two or three cups of tea. This box lasted Nataша for 41 cups of tea, and Inna for 58. How many tea bags were in the box? | Note that there could not have been fewer than 20 tea bags in the box: if there were at least 19, then Inna would not have been able to drink more than $19 \cdot 3=57$ cups, but she drank 58. On the other hand, there could not have been more than 20 tea bags: if there were at least 21, then Natasha could not have drunk... | 20 | Other | math-word-problem | Yes | Yes | olympiads | false |
3a To bake a hundred pancakes, it takes Mom 30 minutes, and Anya 40 minutes. Andryusha is ready to eat 100 pancakes in an hour. Mom and Anya bake pancakes without stopping, while Andryusha continuously eats them. After how much time from the start of this process will there be exactly a hundred pancakes on the table? | First method. Mom bakes a hundred pancakes in half an hour, so in two hours she will bake 400 pancakes. Anya bakes a hundred pancakes in forty minutes, so in two hours she will bake 300 pancakes. Andryusha will eat two hundred pancakes in these two hours. Therefore, in two hours there will be $400+300-200=500$ pancakes... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Inclusion-Exclusion Principle] [Arithmetic. Mental calculation, etc.]
In a group of 50 children, some know all the letters except "r", which they simply skip when writing, while others know all the letters except "k", which they also skip. One day, the teacher asked 10 students to write the word "кот" (cat), 18 other... | The word "крот" was not written correctly by anyone, because no one can write both the letter "p" and the letter "к" at the same time. The word "рот" or "кот" should have been written by $10+18=28$ people. Note that only
| the words "рот", "кот", and "от" were written. The first two words were written 15 times each, s... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Petya is playing a shooting game. If he scores less than 1000 points, the computer will add $20 \%$ of his score. If he scores from 1000 to 2000 points, the computer will add $20 \%$ of the first thousand points and $30 \%$ of the remaining points. If Petya scores more than 2000 points, the computer will add $20 \%$ of... | Clearly, Petya scored more than 1000 points (otherwise his result would not exceed 1200) and less than 2000 (otherwise the result would not be less than 2500). Let's discard 1200 points (the first thousand plus the prize for it). The remaining 1170 points constitute $130\%$ of the points Petya scored in the second thou... | 470 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Tiling with domino bones and tiles $]$ [ Text problems (other). $]$
A square $8 \times 8$ was cut into squares $2 \times 2$ and rectangles $1 \times 4$. As a result, the total length of the cuts was 54.
How many figures of each type were obtained? | In an $8 \times 8-64$ square, there are 64 cells, and each of the resulting figures has 4 cells. Therefore, there are 16 figures in total.
Let's find the sum of the perimeters of all the resulting figures. Since each cut boundary is part of the perimeter of two figures, we add to the perimeter of the square twice the ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A chess player played 20 games in a tournament and scored 12.5 points. How many more games did he win than he lost?
# | If a chess player had an equal number of wins and losses, he would have scored 10 points (the same as if he had drawn all 20 games). Since in reality he scored 2.5 points more, five draws need to be replaced by five wins, meaning the chess player has 5 more wins than losses.
## Answer
By 5 games. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
The sequence $a_{n}$ is defined by the condition: $a_{n+1}=a_{n}-a_{n-1}$. Find $a_{100}$, if $a_{1}=3, a_{2}=7$. | Note that $a_{n+3}=a_{n+2}-a_{n+1}=-a_{n}$. Therefore, $a_{n+6}=a_{n}$, which means the terms of the sequence repeat with a period of 6. Since 100 divided by 6 leaves a remainder of 4, we have $a_{100}=a_{4}=-a_{1}$.
## Answer
$-3$. | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { Prime numbers and their properties }} \\ {[\underline{\text { Arithmetic of residues (other) })]}}\end{array}\right.$
a) $p, p+10, p+14$ - prime numbers. Find $p$.
b) $p, 2 p+1,4 p+1$ - prime numbers. Find $p$. | Consider the remainders when dividing by 3. One of these numbers is divisible by 3.
## Answer
$p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest number that gives the following remainders: 1 - when divided by 2, 2 - when divided by 3, 3 - when divided by 4, 4 - when divided by 5, 5 - when divided by 6.
# | Note that this number, increased by 1, is divisible by 2, 3, 4, 5, 6.
## Answer
$59=\operatorname{LCM}(2,3,4,5,6)-1$. | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Numerical inequalities. Comparisons of numbers. ] [ Exponential functions and logarithms (other). ]
How many digits does the number $2^{1000}$ have? | $\lg 2=0.30102 \ldots$ Therefore, $10^{301}<2^{1000}<10^{302}$.
## Answer
302 digits. | 302 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Prove that for any $n$
a) $7^{2 n}-4^{2 n}$ is divisible by 33;
b) $3^{6 n}-2^{6 n}$ is divisible by 35. | a) $7^{2 n}-4^{2 n}$ is divisible by $7^{2}-4^{2}=33$.
b) First method. $3^{6 n}-2^{6 n}$ is divisible by $3^{6}-2^{6}=665=19 \cdot 35$.
Second method. $3^{6 n}-2^{6 n}=(-2)^{6 n}-2^{6 n}=0(\bmod 5), 27^{2 n}-8^{2 n}=(-1)^{2 n}-1^{2 n}=0(\bmod 7)$.
$a \equiv 68(\bmod 1967), \quad a \equiv 69(\bmod 1968)$. Find the re... | 5 | Number Theory | proof | Yes | Yes | olympiads | false |
[ [Motion Problems ]
Author: Shnol D. $\cdot$.
Daddy, Masha, and Yasha are walking to school together. While Daddy takes 3 steps, Masha takes 5 steps. While Masha takes 3 steps, Yasha takes 5 steps. Masha and Yasha counted that together they made 400 steps. How many steps did Daddy take? | While dad makes 9 steps, Masha makes 15 steps, Yasha - 25, and Masha with Yasha together - 40. Since they together walked 400 steps, which is $10 \cdot 40$, then dad will walk 10 times more - 90 steps.
## Answer
90 steps. | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9,10 Can integers be written in the cells of a $4 \times 4$ table so that the sum of all the numbers in the table is positive, while the sum of the numbers in each $3 \times 3$ square is negative? | The central square of size $2 \times 2$ is contained in each square of size $3 \times 3$. Place the number -9 in one of the cells of the central square, and fill the rest of the cells in this table with ones. Then the sum of all numbers in the table is $15+(-9)=6$, and the sum of the numbers inside any $3 \times 3$ squ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Multiplied several natural numbers and got 224, and the smallest number was exactly half of the largest.
How many numbers were multiplied?
# | $224=2^{5} \cdot 7$. Consider the two numbers mentioned in the condition: the smallest and the largest. If one of them is divisible by 7, then the other must also be divisible by 7. But 224 is not divisible by $7^{2}$, so both of these numbers must be powers of two. From the condition, it also follows that these are tw... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Raskin M.A.
Masha has two-ruble and five-ruble coins. If she takes all her two-ruble coins, she will be 60 rubles short to buy four pies. If all five-ruble coins - she will be 60 rubles short for five pies. And in total, she is 60 rubles short to buy six pies. How much does a pie cost? | If Masha takes all her two-ruble and five-ruble coins, she will be short of $60+60=120$ rubles for $4+5=9$ pies. On the other hand, she will be short of 60 rubles for 6 pies. That is, three pies cost 60 rubles.
## Answer
]
Nadia baked pies with raspberries, blueberries, and strawberries. The number of pies with raspberries was half of the total number of pies; the number of pies with blueberries was 14 less than the number of pies with raspberries. And the number of pies with strawberries was half the number of p... | Since the number of raspberry pies is half of the total number, the number of blueberry and strawberry pies together is the same as the number of raspberry pies (see the diagram). At the same time, there are 14 fewer blueberry pies than raspberry pies, so these 14 pies are strawberry pies. Then the number of raspberry ... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Sergey and Misha, while walking in the park, stumbled upon a glade surrounded by lindens. Sergey walked around the glade, counting the trees. Misha did the same, but started from a different tree (although he went in the same direction). The tree that was 20th for Sergey was 7th for Misha, and the tree that was 7th for... | Between the first and second mentioned trees, Misha counted another $94-7-1=86$ trees. And Sergei between the second tree and the first tree counted $20-7-1=12$ trees. Therefore, around the clearing, there are $86+$ $12=98$ trees and two more mentioned trees, that is, exactly 100 trees.
## Answer
100. | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The dandelion blooms in the morning, flowers yellow for two days, turns white on the third morning, and by evening it has scattered. Yesterday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and today there are 15 yellow and 11 white.
a) How many yellow dandelions were there on the meadow the da... | a) All dandelions that were yellow the day before yesterday have turned white yesterday or today. Therefore, there were 14 + $11=25$.
b) Out of the yellow dandelions from yesterday, 11 turned white today, and the rest $20-11$ = 9 will turn white tomorrow.
## Answer
a) 25 ;
b) 9. | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The arithmetic mean of ten different natural numbers is 15. Find the greatest value of the greatest of these numbers.
# | Evaluation. The sum of these numbers is 150. Since all numbers are distinct, the sum of the nine smallest of them is no less than $1+2+\ldots+9=45$. Therefore, the largest number cannot be greater than 105.
Example: $(1+2+\ldots+9+105): 10=15$.
## Answer
105. | 105 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
When it is noon in Bratsk - it is 6 a.m. in Gusev, and 2 p.m. in Komsomolsk-on-Amur. When it is noon in Zlatoust - it is 6 p.m. in Elizovo, and 9 a.m. in Gusev. What time is it in Komsomolsk-on-Amur when it is noon in Elizovo? | When it is noon in Elizovo, it is 3 o'clock in Gusev (from the second condition). And when it is 3 o'clock in Gusev, it is 11 o'clock in Komsomolsk-on-Amur (from the first condition).
## Otвет
11 o'clock in the morning. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (other).]
In a seven-story building, house spirits live. The elevator travels between the first and the last floors, stopping at each floor. On each floor, starting from the first, one house spirit entered the elevator, but no one exited. When the thousandth house spirit entered the elevator, the eleva... | Let's find out how many house elves ended up in the elevator for a trip from the first to the seventh floor and back, until the elevator returned to the first floor. One house elf entered on the first and seventh floors, and two house elves entered on all other floors. Thus, 12 house elves enter the elevator in one tri... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Text problems (other)]
Forty gold and forty silver coins were distributed among three empty chests, with both types of coins in each chest. In the first chest, there were 7 more gold coins than silver ones, and in the second chest, there were 15 fewer silver coins than gold ones. Which type of coin is more in the thi... | In the first two chests, the total number of gold coins is $7+15=22$ more than the total number of silver coins. Since initially, the number of gold and silver coins was equal, in the third chest, there are 22 fewer gold coins than silver.
## Answer
There are 22 more silver coins. | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Prove that if the expression $\left(x^{2}-x+1\right)^{2014}$ is expanded and like terms are combined, then some coefficient of the resulting polynomial will be negative.
# | Let's find the coefficient of $x$ in the obtained polynomial. Similar terms with the literal part $x$ are formed by multiplying 2014 identical brackets in the following way: in one of the brackets, the term - $x$ is taken, and in the remaining brackets, the term 1 is taken. Therefore, the coefficient of $x$ will be equ... | -2014 | Algebra | proof | Yes | Yes | olympiads | false |
[The Fundamental Theorem of Arithmetic. Prime Factorization]
## Authors: Raskina I.V., Fedumin L.E.
A hunter told a friend that he saw a wolf with a meter-long tail in the forest. That friend told another friend that a wolf with a two-meter-long tail had been seen in the forest. Passing on the news further, ordinary ... | Each storyteller either doubled or tripled the length of the tail, so 864 is the product of a certain number of twos and threes. Since
$864=2^{5} \cdot 3^{3}$, the wolf's tail was "lengthened" by five ordinary people and three creative individuals.
## Answer
5 ordinary people and 3 creative individuals. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Folkpor
Three pirates were dividing a bag of coins. The first took $3 / 7$ of all the coins, the second took $51 \%$ of the remainder, after which the third received 8 coins less than the second. How many coins were in the bag? | Since the second pirate took $51\%$ of the coins remaining after the first, the third pirate received $49\%$ of this amount. Therefore, 8 coins represent $2\%$ of the coins remaining after the first pirate. Thus, the share of the second and third pirates is $8 \cdot 50 = 400$ coins, which is $\frac{4}{7}$ of their tota... | 700 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Inequality problems. Case analysis]
In a certain school, a council of 5 students was elected in each of the 20 classes. Petya turned out to be the only boy elected to the council along with four girls. He noticed that in 15 other classes, more girls were elected than boys, although in total, the number of boys and gi... | In total, 5$\cdot$20 = 100 people were selected to the council. Half of them are girls, which is 50. If more girls were selected from a class than boys, then at least three girls were selected. Therefore, in 15 classes, at least 45 girls were selected. Four more girls were selected from Petya's class. Since Petya is th... | 19 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
To repair the propeller, Karlson needs to buy three blades and one screw. In the store, blades cost 120 tugriks each and screws cost 9 tugriks each. However, after a purchase of at least 250 tugriks, a $20 \%$ discount is given on all subsequent purchases. Will Karlson be able to repai... | Let the first purchase be two blades and two screws, costing $2(120+9)=258$ tugriks. Since the cost of the purchase is more than 250 tugriks, Carlson can buy the third blade with a $20\%$ discount, spending $120 \cdot 0.8=96$ tugriks. Therefore, in total, Carlson will spend $258+96=354$ tugriks.
## Answer
He can.
Se... | 354 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Each of the artists in the creative association "Patience and Labor" has their own work schedule. Six of them paint one picture every two days, another eight artists - one picture every three days, and the rest never paint pictures. From September 22 to 26, they painted a total of 30 pictures. How many pictures will th... | Let's see how many paintings the artists will create from September 22 to 27 inclusive. Each of the six artists, who paint one painting every two days, will create three paintings (one for each pair of days 22-23, 24-25, and 26-27), and each of the eight artists, who paint one painting every three days, will create two... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Identical Transformations ]
It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-\right.$ $\left.a^{2}\right)$ take? | From the condition, it follows that $a^{2}-b^{2}=c-b, b^{2}-c^{2}=a-c$ and $c^{2}-a^{2}=b-a$. Therefore, $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+$ $c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0$.
## Answer
0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10,11
Does there exist a natural number $n$, greater than 1, such that the value of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number? | For example, $n=2^{8}=256$. Indeed, $\sqrt{n \sqrt{n \sqrt{n}}}=n^{7 / 8}$. For $n=2^{8}$, the value of this expression is $2^{7}=$ 128.
## Answer
It exists. | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Divisibility of numbers. General properties ]
Find the smallest natural $n$ for which $(n+1)(n+2)(n+3)(n+4)$ is divisible by 1000. | For any natural $n$, the given product is divisible by 8, since among any four consecutive natural numbers, one is divisible by 4 and another by 2. Therefore, it is sufficient to find the smallest $n$ for which the given product is divisible by $125=5^{3}$. Since only one of the factors can be divisible by 5, $n$ is th... | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Text problems (other) ]
At each of the four math club sessions, 20 schoolchildren were present. Nine students attended exactly three out of these four sessions, five students - exactly two sessions, and three students only attended one session. How many schoolchildren attended all sessions? | Let's fill in the "attendance record" for this club. In total, for the specified four sessions, there will be $20 \cdot 4=80$ attendance marks. Each student who attended three sessions is marked three times in the record, so nine such students are marked a total of $9 \cdot 3=27$ times. Similarly, those who attended tw... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (other) ]
In the thirtieth kingdom, there are two exchange offices. The first one gives 3000 tugriks for a ruble but charges a 7000 tugriks commission for the exchange, while the second one gives only 2950 tugriks for a ruble but does not charge any commission. A tourist noticed that it makes no differ... | Since the tourist does not care which exchange office to use, before the commission is deducted, in the first exchange office, for all the rubles he intends to exchange, the tourist will receive 7000 tugriks more. Since for each ruble the tourist receives 50 tugriks more in the first office than in the second, he inten... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Motion problems ]
Sasha and Ilya had to run 600 meters. But Sasha ran the first half of the time and walked the second half. And Ilya ran the first half of the distance and walked the second half. They started and finished at the same time. Both of them walk at a speed of 5 km/h. At what speed did Ilya run, if Sasha... | Since Sasha runs twice as fast as he walks, he ran twice the distance he walked, which is 400 m. Since the boys walk at the same speed, they walked the last 200 m together. Therefore, while Sasha ran 400 m, Ilya ran 300 m and walked 100 m. While Ilya walked 100 m, Sasha ran 200 m. And while Sasha ran $400-200=200$ m, I... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ]
In a test, each of the 25 students received one of the grades "3", "4", or "5". By how many more were there fives than threes, if the sum of all grades is 106?
# | Let $a$ students received a grade of 3, $b$ students received a grade of 4, and $c$ students received a grade of 5. From the problem, we have $a+b+c=25$ and $3a+4b+5c=106$.
Multiply both sides of the first equation by 4: $4a+4b+4c=100$. Now subtract the obtained equation from the second equation, then $c-a=6$.
## Ans... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
Prosperous Mole in autumn dug up 8 bags of grain. For each winter month, he needs either 3 bags of grain or 1 bag of grain and 3 bags of millet. Mole can trade with other moles 1 bag of grain for 2 bags of millet. But his burrow cannot hold more than 12 bags, and in winter Mole ... | A mole can exchange 3 bags of grain for 6 bags of millet, then in its burrow there will be 11 bags: 5 bags of grain and 6 bags of millet. For one month, it will spend 3 bags of grain, and for each of the other two - 1 bag of grain and 3 bags of millet. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Text problems (other)]
Pilyulkin, Guslya, and Toropyzhka are decorating a New Year's tree. It is known that Guslya hung twice as many ornaments on the tree as Pilyulkin, Pilyulkin hung 15 fewer ornaments than Toropyzhka, and Toropyzhka hung twice as many as Guslya and Pilyulkin combined. How many ornaments are decora... | The first method. Suppose Pilulkin hung one part of all the toys, then Guslya hung two such parts, and together they hung three parts of the toys (see the figure). Therefore, Toropyzhka hung 6 parts, which is 5 parts more than what Pilulkin hung. Since these 5 parts consist of 15 toys, one part is three toys. Thus, the... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Ya.
Does there exist a number whose square, in its decimal representation, contains the sequence of digits "2018"? | Indeed,
$$
\left(5 \cdot 10^{6}+2018\right)^{2}=25 \cdot 10^{12}+2018 \cdot 10^{7}+4072324
$$
so the square of this number contains the digit sequence "2018".
## Answer
Yes, for example, 5002018. | 5002018 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
When the barrel is empty by $30 \%$, it contains 30 liters more honey than when it is filled to $30 \%$. How many liters of honey are in a full barrel? | The first method. If the barrel is empty by $30 \%$, it means it is filled to $70 \%$, so 30 liters make up $40 \%$ of its volume. Therefore, $10 \%$ of the volume of the barrel is 7.5 liters, and the entire volume is 75 liters.
The second method. Let the volume of the barrel be $x$ liters. According to the problem: $... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
What angle do the hour and minute hands form at 4:12?
# | The first method. At 12:00, each of the hands is pointing vertically upwards. Let's find the angles that each of the hands forms with this position at 4 hours and 12 minutes.
The hour hand turns $30^{\circ}$ every hour, and in 12 minutes, it will turn an additional $30^{\circ} \div 5 = 6^{\circ}$. Thus, it will form a... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a pack of 20 cards: blue, red, and yellow. There are six times fewer blue cards than yellow ones, and fewer red cards than yellow ones. What is the minimum number of cards you need to pull out without looking to ensure that there is at least one red card among them?
# | If there is one blue card, then there are six yellow cards, and red cards: $20-1-6=13$. But then there are more red cards than yellow, which contradicts the condition.
If there are two blue cards, then there are 12 yellow cards, and red cards: $20-2-12=6$. This case satisfies the condition. Then, if you draw no more t... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Motion problems ]
A cyclist rode from point A to point B, where he stayed for 30 minutes, and then returned to A. On the way to B, he overtook a pedestrian, and 2 hours later met him on the return trip. The pedestrian arrived in B at the same time the cyclist returned to A. How much time did it take the pedestrian t... | The first method. The distance that the pedestrian covers in 2 hours, we will take as a unit. Then the cyclist covers this same distance in 30 minutes. From the moment of their first meeting, the pedestrian has walked one unit, while the cyclist has traveled three units (he rested for half an hour in point B). Therefor... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find the last digit of the number $2^{50}$.
# | $2^{50}=(16)^{3} \cdot 4.6$ any power of 6 ends in 6, so $2^{50}$ ends with the same digit as $6 \cdot 4=24$.
## Answer
4. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Periodicity and Non-periodicity] [
Find the last digit of the number $7^{1988}+9^{1988}$.
# | $7^{1988}+9^{1988}=49^{994}+9^{1988} \equiv(-1)^{994}+(-1)^{1988}=2(\bmod 10)$.
## Answer
2. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
| Reverse Process |
| :---: | :---: |
| | [ Arithmetic. Mental calculation, etc.] |
| | [ Word problems (miscellaneous). [ Iterations |
John had a full basket of trampolines. First, he met Anna and gave her half of his trampolines and another half trampoline. Then he met Banna and gave her half of the remaining tra... | Notice that before meeting Vanna, John had one tremponch left, as half of this amount was half a tremponch. Before meeting Banna, he had 3 tremponchs, because half of this amount was one and a half tremponchs, that is, one and a half. Similarly, we get that initially there were 7 tremponchs.
## Answer
7 tremponchs. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest $k$ such that $k$! is divisible by 2040.
# | $2040=2^{3} \cdot 3 \cdot 5 \cdot 17$
## Answer
$k=17$. | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal number system ] [ Divisibility of numbers. General properties ]
Find the last two digits in the decimal representation of the number $1!+2!+\ldots+2001!+2002!$. | From some point, factorials start to be divisible by 100.
## Solution
Notice that 10! is divisible by 100, and \( n! \) for \( n > 10 \) will also be divisible by 100. Therefore, the last two digits of the number \( 1! + 2! + \ldots + 2001! + 2002! \) are the same as the last two digits of the number \( 1! + 2! + \ld... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Fomin S.B.
A natural number $n$ is written in the decimal system. It is known that if any digit is included in this representation, then $n$ is divisible by this digit (0 does not appear in the representation). What is the maximum number of different digits that this representation can contain? | If the digit 5 is included in the number, then the number must end in 5. Thus, it is odd and, consequently, contains only odd digits. Therefore, it cannot have more than five digits. If 5 does not appear in the decimal representation of the number, then it can include all the other 8 digits. Here is an example: 1471963... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.v.
A number was multiplied by the sum of its digits and the result was 2008. Find this number.
# | The sought number is a divisor of the number $2008=2^{3} \cdot 251$. Let's list all divisors of 2008: $1,2,4,8,251$, 502, 1004, 2008. By finding the sum of the digits of each of them, we notice that the condition of the problem is only satisfied for the number 251 $(2008=251 \cdot(2+5+1))$.
## Answer
251. | 251 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arrangement in ascending (descending) order. $]$
[ Pigeonhole Principle (other). $\quad]$
The sum of 123 numbers is 3813. Prove that from these numbers, 100 can be chosen with a sum of no less than 3100. | The sum of the 100 largest numbers is not less than 100/123 of the sum of all numbers, that is, not less than ${ }^{100} / 123 \cdot 3813=3100$. | 3100 | Combinatorics | proof | Yes | Yes | olympiads | false |
Sergei and Lena have several chocolate bars, each weighing no more than 100 grams. No matter how they divide these chocolate bars, the total weight of the chocolate bars for one of them will not exceed 100 grams. What is the greatest possible total weight of all the chocolate bars? | It is possible to divide the chocolates "almost equally" - such that the difference in weight between the chocolates given to Lena and Sergei does not exceed 100 grams.
## Solution
First, give all the chocolates to Sergei and then start giving one chocolate at a time to Lena. Initially, the weight of the chocolates w... | 300 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7,8,9 |
| $[\underline{\text { Modular Arithmetic (other) }}]$ | | |
Find all prime numbers that are equal to the sum of two prime numbers and the difference of two prime numbers. | The given prime number $p$ is odd, so the sum and difference involve numbers of different parity. Thus, $p=q+2=r-2$. From this, it is clear that the numbers give different remainders when divided by 3, meaning one of them is divisible by 3, and since it is prime, it equals 3.
## Answer
5.
=42$. This means there are 42 students in total; 6 of them received fives; 14 received fours; 21 received threes. Therefore... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The Wolf with the Three Little Pigs wrote a detective novel "Three Little Pigs-2," and then, together with Little Red Riding Hood and her grandmother, a cookbook "Little Red Riding Hood-2." The publishing house paid the royalties for both books to the pig Naf-Naf. He took his share and handed over the remaining 2100 go... | For the book "The Three Little Pigs-2," each author is supposed to receive a quarter of the royalties. However, since Naf-Naf has already taken his share, the Wolf is entitled to $1 / 3$ of the remainder. For the book "Little Red Riding Hood-2," he is also due 1⁄3 of the royalties. Therefore, in total, he should receiv... | 700 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Dasha and Tanya live in the same entrance. Dasha lives on the 6th floor. Leaving Dasha's place, Tanya went up instead of down as she needed to. Reaching the top floor, Tanya realized her mistake and went down to her floor. It turned out that Tanya walked one and a half times more than if she had gone down immediately. ... | From Dasha to Tanya is no more than 5 floors, so the extra path was no more than two floors, and the path upwards was no more than one. Therefore, Tanya went up exactly one floor.
## Answer
7 floors. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
. How many pairs $(m, n)$ of positive integers with $m<n$ fulfill the equation
$$
\frac{3}{2008}=\frac{1}{m}+\frac{1}{n} ?
$$
## Answer: 5. | Let $d$ be the greatest common divisor of $m$ and $n$, and let $m=d x$ and $n=d y$. Then the equation is equivalent to
$$
3 d x y=2008(x+y) \text {. }
$$
The numbers $x$ and $y$ are relatively prime and have no common divisors with $x+y$ and hence they are both divisors of 2008. Notice that $2008=8 \cdot 251$ and 251... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. Consider a set $A$ of positive integers such that the least element of $A$ equals 1001 and the product of all elements of $A$ is a perfect square. What is the least possible value of the greatest element of $A$ ?
Answer: 1040. | We first prove that max $A$ has to be at least 1040 .
As $1001=13 \cdot 77$ and $13 \nmid 77$, the set $A$ must contain a multiple of 13 that is greater than $13 \cdot 77$. Consider the following cases:
- $13 \cdot 78 \in$. But $13 \cdot 78=13^{2} \cdot 6$, hence $A$ must also contain some greater multiple of 13 .
- ... | 1040 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. For a positive integer $n$, let $S(n)$ denote the sum of its digits. Find the largest possible value of the expression $\frac{S(n)}{S(16 n)}$.
## Answer: 13 | It is obvious that $S(a b) \leq S(a) S(b)$ for all positive integers $a$ and $b$. From here we get
$$
S(n)=S(n \cdot 10000)=S(16 n \cdot 625) \leq S(16 n) \cdot 13 \text {; }
$$
so we get $\frac{S(n)}{S(16 n)} \leq 13$.
For $n=625$ we have an equality. So the largest value is 13 . | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. For an upcoming international mathematics contest, the participating countries were asked to choose from nine combinatorics problems. Given how hard it usually is to agree, nobody was surprised that the following happened:
- Every country voted for exactly three problems.
- Any two countries voted for different sets... | Certainly, the 56 three-element subsets of the set $\{1,2, \ldots, 8\}$ would do. Now we prove that 56 is the maximum. Assume we have a maximal configuration. Let $Y$ be the family of the three-element subsets, which were chosen by the participating countries and $N$ be the family of the three-element subsets, which we... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
. Some $1 \times 2$ dominoes, each covering two adjacent unit squares, are placed on a board of size $n \times n$ so that no two of them touch (not even at a corner). Given that the total area covered by the dominoes is 2008 , find the least possible value of $n$.
Answer: 77 | Following the pattern from the figure, we have space for
$$
6+18+30+\ldots+150=\frac{156 \cdot 13}{2}=1014
$$
dominoes, giving the area $2028>2008$.
The square $76 \times 76$ is not enoug... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $N$ be a positive integer. Two persons play the following game. The first player writes a list of positive integers not greater than 25 , not necessarily different, such that their sum is at least 200 . The second player wins if he can select some of these numbers so that their sum $S$ satisfies the condition $200-... | If $N=11$, then the second player can simply remove numbers from the list, starting with the smallest number, until the sum of the remaining numbers is less than 212. If the last number removed was not 24 or 25 , then the sum of the remaining numbers is at least $212-23=189$. If the last number removed was 24 or 25 , t... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n>1$ such that any prime divisor of $n^{6}-1$ is a divisor of $\left(n^{3}-1\right)\left(n^{2}-1\right)$. | Clearly $n=2$ is such an integer. We will show that there are no others.
Consider the equality
$$
n^{6}-1=\left(n^{2}-n+1\right)(n+1)\left(n^{3}-1\right) \text {. }
$$
The integer $n^{2}-n+1=n(n-1)+1$ clearly has an odd divisor $p$. Then $p \mid n^{3}+1$. Therefore, $p$ does not divide $n^{3}-1$ and consequently $p ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $X$ be a subset of $\{1,2,3, \ldots, 10000\}$ with the following property: If $a, b \in X, a \neq b$, then $a \cdot b \notin X$. What is the maximal number of elements in $X$ ?
Answer: 9901. | If $X=\{100,101,102, \ldots, 9999,10000\}$, then for any two selected $a$ and $b, a \neq b$, $a \cdot b \geq 100 \cdot 101>10000$, so $a \cdot b \notin X$. So $X$ may have 9901 elements.
Suppose that $x_{1}1$ and consider the pairs
$$
\begin{gathered}
200-x_{1},\left(200-x_{1}\right) \cdot x_{1} \\
200-x_{2},\left(20... | 9901 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A lattice point in the plane is a point whose coordinates are both integral. The centroid of four points $\left(x_{i}, y_{i}\right), i=1,2,3,4$, is the point $\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}, \frac{y_{1}+y_{2}+y_{3}+y_{4}}{4}\right)$. Let $n$ be the largest natural number with the following property: There are ... | To prove $n \geq 12$, we have to show that there are 12 lattice points $\left(x_{i}, y_{i}\right)$, $i=1,2, \ldots, 12$, such that no four determine a lattice point centroid. This is guaranteed if we just choose the points such that $x_{i} \equiv 0(\bmod 4)$ for $i=1, \ldots, 6, x_{i} \equiv 1(\bmod 4)$ for
$i=7, \ldot... | 12 | Combinatorics | proof | Yes | Yes | olympiads | false |
Is it possible to select 1000 points in a plane so that at least 6000 distances between two of them are equal?
Answer: Yes. | Let's start with configuration of 4 points and 5 distances equal to $d$, like in this figure:
$(\alpha)$
Now take $(\alpha)$ and two copies of it obtainable by parallel shifts along vectors... | 6075 | Combinatorics | proof | Yes | Yes | olympiads | false |
Find the smallest positive integer $n$ having the property: for any set of $n$ distinct integers $a_{1}, a_{2}, \ldots, a_{n}$ the product of all differences $a_{i}-a_{j}, i<j$ is divisible by 1991 . | Let $S=\prod_{1 \leq i<j \leq n}\left(a_{i}-a_{j}\right)$. Note that $1991=11 \cdot 181$. Therefore $S$ is divisible by 1991 if and only if it is divisible by both 11 and 181 . If $n \leq 181$ then we can take the numbers $a_{1}, \ldots, a_{n}$ from distinct congruence classes modulo 181 so that $S$ will not be divisib... | 182 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. Let $\Gamma$ be a circle in the plane and $S$ be a point on $\Gamma$. Mario and Luigi drive around the circle $\Gamma$ with their go-karts. They both start at $S$ at the same time. They both drive for exactly 6 minutes at constant speed counterclockwise around the track. During these 6 minutes, Luigi makes exactly on... | . Without loss of generality, we assume that $\Gamma$ is the unit circle and $S=(1,0)$. Three points are marked with bananas:
(i) After 45 seconds, Luigi has passed through an arc with a subtended angle of $45^{\circ}$ and is at the point $\left(\sqrt{2} / 2, \sqrt{2} / 2\right.$ ), whereas Mario has passed through an... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
. Distinct positive integers $a, b, c, d$ satisfy
$$
\left\{\begin{array}{l}
a \mid b^{2}+c^{2}+d^{2}, \\
b \mid a^{2}+c^{2}+d^{2}, \\
c \mid a^{2}+b^{2}+d^{2}, \\
d \mid a^{2}+b^{2}+c^{2},
\end{array}\right.
$$
and none of them is larger than the product of the three others. What is the largest possible number of pr... | At first we note that the given condition is equivalent to $a, b, c, d \mid a^{2}+b^{2}+c^{2}+d^{2}$.
It is possible that three of the given numbers are primes, for example for $a=2, b=3, c=13$ and $d=26$. In this case $2^{2}+3^{2}+13^{2}+26^{2}=13 \cdot 66$ which is divisible by all four given numbers. Furthermore we... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. Find all real numbers $x, y, z$ so that
$$
\begin{aligned}
x^{2} y+y^{2} z+z^{2} & =0 \\
z^{3}+z^{2} y+z y^{3}+x^{2} y & =\frac{1}{4}\left(x^{4}+y^{4}\right)
\end{aligned}
$$ | Answer: $x=y=z=0$.
$y=0 \Longrightarrow z^{2}=0 \Longrightarrow z=0 \Longrightarrow \frac{1}{4} x^{4}=0 \Longrightarrow x=0 . x=y=z=0$ is a solution, so assume that $y \neq 0$. Then $z=0 \Longrightarrow x^{2} y=0 \Longrightarrow x=0 \Longrightarrow \frac{1}{4} y^{4}=0$, which is a contradiction. Hence $z \neq 0$. Now ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
. Let $n>2$ be a given positive integer. There are $n$ guests at Georg's bachelor party and each guest is friends with at least one other guest. Georg organizes a party game among the guests. Each guest receives a jug of water such that there are no two guests with the same amount of water in their jugs. All guests now... | Answer: 2 .
If there are guests $1,2, \ldots, n$ and guest $i$ is friends with guest $i-1$ and $i+1$ modulo $n$ (e.g. guest 1 and guest $n$ are friends). Then if guest $i$ has $i$ amount of water in their jug at the start of the game, then only guest 1 and $n$ end up with a different amount of water than they started ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
. A mason has bricks with dimensions $2 \times 5 \times 8$ and other bricks with dimensions $2 \times 3 \times 7$. She also has a box with dimensions $10 \times 11 \times 14$. The bricks and the box are all rectangular parallelepipeds. The mason wants to pack bricks into the box filling its entire volume and with no br... | Answer: 24.
Let the number of $2 \times 5 \times 8$ bricks in the box be $x$, and the number of $2 \times 3 \times 7$ bricks $y$. We must figure out the sum $x+y$. The volume of the box is divisible by 7 , and so is the volume of any $2 \times 3 \times 7$ brick. The volume of a $2 \times 5 \times 8$ brick is not divis... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
. Let $n \geqslant 1$ be a positive integer. We say that an integer $k$ is a fan of $n$ if $0 \leqslant k \leqslant n-1$ and there exist integers $x, y, z \in \mathbb{Z}$ such that
$$
\begin{aligned}
x^{2}+y^{2}+z^{2} & \equiv 0 \quad(\bmod n) ; \\
x y z & \equiv k \quad(\bmod n) .
\end{aligned}
$$
Let $f(n)$ be the ... | Answer: $f(2020)=f(4) \cdot f(5) \cdot f(101)=1 \cdot 1 \cdot 101=101$.
To prove our claim we show that $f$ is multiplicative, that is, $f(r s)=f(r) f(s)$ for coprime numbers $r, s \in \mathbb{N}$, and that
(i) $f(4)=1$,
(ii) $f(5)=1$,
(iii) $f(101)=101$.
The multiplicative property follows from the Chinese Remain... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can we paint 16 seats in a row, each red or green, in such a way that the number of consecutive seats painted in the same colour is always odd?
Answer: 1974. | Let $g_{k}, r_{k}$ be the numbers of possible odd paintings of $k$ seats such that the first seat is painted green or red, respectively. Obviously, $g_{k}=r_{k}$ for any $k$. Note that $g_{k}=r_{k-1}+g_{k-2}=g_{k-1}+g_{k-2}$, since $r_{k-1}$ is the number of odd paintings with first seat green and second seat red and $... | 1974 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A $100 \times 100$ table is given. For each $k, 1 \leq k \leq 100$, the $k$-th row of the table contains the numbers $1,2, \ldots, k$ in increasing order (from left to right) but not necessarily in consecutive cells; the remaining $100-k$ cells are filled with zeroes. Prove that there exist two columns such that the su... | Observe that the sum of numbers in the first column is at most $1 \cdot 100=100$, the sum in the first and second columns is at most $1 \cdot 100+2 \cdot 99$, the sum in the first, second and third columns is at most $1 \cdot 100+2 \cdot 99+3 \cdot 98$, etc. But the sum of all nonzero numbers equals $\sum_{i=1}^{100} i... | 1917 | Combinatorics | proof | Yes | Yes | olympiads | false |
On a $16 \times 16$ torus as shown all 512 edges are colored red or blue. A coloring is good if every vertex is an endpoint of an even number of red edges. A move consists of switching the color of each of the 4 edges of an arbitrary cell. What is the largest number of good colorings such that none of them can be conve... | 
Answer: 4. Representatives of the equivalence classes are: all blue, all blue with one longitudinal red ring, all blue with one transversal red ring, all blue with one longitudinal and one ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The integers from 1 to $n$ are written, one on each of $n$ cards. The first player removes one card. Then the second player removes two cards with consecutive integers. After that the first player removes three cards with consecutive integers. Finally, the second player removes four cards with consecutive integers. Wha... | Answer: $n=14$.
At first, let's show that for $n=13$ the first player can ensure that after his second move no 4 consecutive numbers are left. In the first move he can erase number 4 and in the second move he can ensure that numbers 8,9 and 10 are erased. No interval of length 4 is left.
If $n=14$ the second player c... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $x_{1}=1$ and $x_{n+1}=x_{n}+\left\lfloor\frac{x_{n}}{n}\right\rfloor+2$ for $n=1,2,3, \ldots$, where $\lfloor x\rfloor$ denotes the largest integer not greater than $x$. Determine $x_{1997}$. | Answer: $x_{1997}=23913$.
Note that if $x_{n}=a n+b$ with $0 \leqslant b<n$, then
$$
x_{n+1}=x_{n}+a+2=a(n+1)+b+2 \text {. }
$$
Hence if $x_{N}=A N$ for some positive integers $A$ and $N$, then for $i=0,1, \ldots, N$ we have $x_{N+i}=A(N+i)+2 i$, and $x_{2 N}=(A+1) \cdot 2 N$. Since for $N=1$ the condition $x_{N}=A ... | 23913 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A rectangle can be divided into $n$ equal squares. The same rectangle can also be divided into $n+76$ equal squares. Find all possible values of $n$. | Answer: $n=324$.
Let $a b=n$ and $c d=n+76$, where $a, b$ and $c, d$ are the numbers of squares in each direction for the partitioning of the rectangle into $n$ and $n+76$ squares, respectively. Then $\frac{a}{c}=\frac{b}{d}$, or $a d=b c$. Denote $u=\operatorname{gcd}(a, c)$ ja $v=\operatorname{gcd}(b, d)$, then ther... | 324 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a forest each of $n$ animals $(n \geqslant 3)$ lives in its own cave, and there is exactly one separate path between any two of these caves. Before the election for King of the Forest some of the animals make an election campaign. Each campaign-making animal visits each of the other caves exactly once,
uses only the... | Answer: b) 4 .
a) As each campaign-making animal uses exactly $n$ paths and the total number of paths is $\frac{n(n-1)}{2}$, the number of campaign-making animals cannot exceed $\frac{n-1}{2}$. Labeling the caves by integers $0,1,2, \ldots, n-1$, we can construct $\frac{n-1}{2}$ non-intersecting campaign routes as fol... | 4 | Combinatorics | proof | Yes | Yes | olympiads | false |
Twelve cards lie in a row. The cards are of three kinds: with both sides white, both sides black, or with a white and a black side. Initially, nine of the twelve cards have a black side up. The cards 1-6 are turned, and subsequently four of the twelve cards have a black side up. Now cards $4-9$ are turned, and six card... | Answer: there are 9 cards with one black and one white side and 3 cards with both sides white.
Divide the cards into four types according to the table below.
| Type | Initially up | Initially down |
| :---: | :---: | :---: |
| $A$ | black | white |
| $B$ | white | black |
| $C$ | white | white |
| $D$ | black | black... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, you see three half-circles. The circle $C$ is tangent to two of the half-circles and to the line $P Q$ perpendicular to the diameter $A B$. The area of the shaded region is $39 \pi$, and the area of the circle $C$ is $9 \pi$. Find the length of the diameter $A B$.
^{2}-r^{2}-s^{2}\right)-9 \pi
$$
hence $r s=48$. Let $M$ be the midpoint of the diameter $A B, N$ be the midpoint of $P B, O$ be the centre of the circle $C$, and let $F$ be the orthogonal p... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The graph of the function $f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ (where $n>1$ ), intersects the line $y=b$ at the points $B_{1}, B_{2}, \ldots, B_{n}$ (from left to right), and the line $y=c(c \neq b)$ at the points $C_{1}, C_{2}, \ldots, C_{n}$ (from left to right). Let $P$ be a point on the line $y=c$, to ... | Let the points $B_{i}$ and $C_{i}$ have the coordinates $\left(b_{i}, b\right)$ and $\left(c_{i}, c\right)$, respectively, for $i=1,2, \ldots, n$. Then we have
$$
\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P=\frac{1}{b-c} \sum_{i=1}^{n}\left(b_{i}-c_{i}\right)
$$
The numbers $b_{i}$ and $c_{i}$ are the ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Using each of the eight digits $1,3,4,5,6,7,8$ and 9 exactly once, a three-digit number $A$, two twodigit numbers $B$ and $C, B<C$, and a one-digit number $D$ are formed. The numbers are such that $A+D=B+C=143$. In how many ways can this be done? | From $A=143-D$ and $1 \leq D \leq 9$, it follows that $134 \leq A \leq 142$. The hundreds digit of $A$ is therefore 1 , and the tens digit is either 3 or 4 . If the tens digit of $A$ is 4 , then the sum of the units digits of
$A$ and $D$ must be 3 , which is impossible, as the digits 0 and 2 are not among the eight dig... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Consider the sequence $a_{k}$ defined by $a_{1}=1, a_{2}=\frac{1}{2}$,
$$
a_{k+2}=a_{k}+\frac{1}{2} a_{k+1}+\frac{1}{4 a_{k} a_{k+1}} \quad \text { for } k \geq 1
$$
Prove that
$$
\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}}+\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{98} a_{100}}<4
$$ | Note that
$$
\frac{1}{a_{k} a_{k+2}}a_{k}+\frac{1}{2} a_{k+1}
$$
which is evident for the given sequence. Now we have
$$
\begin{aligned}
\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}} & +\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{98} a_{100}} \\
& <\frac{2}{a_{1} a_{2}}-\frac{2}{a_{2} a_{3}}+\frac{2}{a_{2} a_{3}}-\frac{2... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
Consider a grid of $25 \times 25$ unit squares. Draw with a red pen contours of squares of any size on the grid. What is the minimal number of squares we must draw in order to colour all the lines of the grid?
Answer: 48 squares. | Consider a diagonal of the square grid. For any grid vertex $A$ on this diagonal denote by $C$ the farthest endpoint of this diagonal. Let the square with the diagonal $A C$ be red. Thus, we have defined the set of 48 red squares (24 for each diagonal). It is clear that if we draw all these squares, all the lines in th... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $m=30030=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ and let $M$ be the set of its positive divisors which have exactly two prime factors. Determine the minimal integer $n$ with the following property: for any choice of numbers from $M$, there exist three numbers $a, b, c$ among them satisfying $a \cdot b \cdot c=... | Taking the 10 divisors without the prime 13 shows that $n \geq 11$. Consider the following partition of the 15 divisors into five groups of three each with the property that the product of the numbers in every group equals $m$.
$$
\begin{array}{ll}
\{2 \cdot 3,5 \cdot 13,7 \cdot 11\}, & \{2 \cdot 5,3 \cdot 7,11 \cdot ... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the smallest number of circles of radius $\sqrt{2}$ that are needed to cover a rectangle
(a) of size $6 \times 3$ ?
(b) of size $5 \times 3$ ?
Answer: (a) Six circles, (b) five circles. | (a) Consider the four corners and the two midpoints of the sides of length 6 . The distance between any two of these six points is 3 or more, so one circle cannot cover two of these points, and at least six circles are needed.
On the other hand one circle will cover a $2 \times 2$ square, and it is easy to see that si... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let the medians of the triangle $A B C$ meet at $M$. Let $D$ and $E$ be different points on the line $B C$ such that $D C=C E=A B$, and let $P$ and $Q$ be points on the segments $B D$ and $B E$, respectively, such that $2 B P=P D$ and $2 B Q=Q E$. Determine $\angle P M Q$.
Answer: $\angle P M Q=90^{\circ}$. | Draw the parallelogram $A B C A^{\prime}$, with $A A^{\prime} \| B C$. Then $M$ lies on $B A^{\prime}$, and $B M=\frac{1}{3} B A^{\prime}$. So $M$ is on the homothetic image (centre $B$, dilation $1 / 3$ ) of the circle with centre $C$ and radius $A B$, which meets $B C$ at $D$ and $E$. The image meets $B C$ at $P$ and... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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