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Rabbits sawed several logs. They made 10 cuts and got 16 chunks. How many logs did they saw?
# | Recall problem 4.
## Solution
From each log, you get one more chunk than the number of cuts made. Since there are 6 more chunks, it means there were 6 logs.
## Answer
6 logs. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
$[\quad$ Invariants $]$
The rabbits are sawing the log again, but now both ends of the log are secured. Ten middle pieces fell, while the two end pieces remained secured. How many cuts did the rabbits make? | How many logs did the hares receive
## Solution
The hares received 12 logs - 10 fallen and 2 secured. Therefore, there were 11 cuts.
## Answer
The problem | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
How many times longer is the staircase to the fourth floor of a house compared to the staircase to the second floor of the same house? | Think about how many floors you need to go up to reach the second floor?
## Solution
To get to the 2nd floor, you need to go up 1 floor, and to get to the 4th floor, you need to go up three floors. Therefore, the answer is: 3 times (and not 2, as it might seem at first).
## Answer
3 times. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
For the book, 100 rubles were paid, and it remains to pay as much as it would remain to pay if it had been paid as much as it remains to pay. How much does the book cost? | This task is very similar to task 2, only even more complicated.
## Solution
If you read the condition carefully, you will understand that the 100 rubles paid are the "first half" of the book's cost. This means the book costs 200 rubles.
## Answer
The book costs 200 rubles. | 200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
Malvina asked Buratino to multiply a number by 4 and add 15 to the result, but Buratino multiplied the number by 15 and then added 4, and yet the answer was correct. What was the number? | Try to form an equation
## Solution
Let the unknown number be x and write the equation: $4 x + 15 = 15 x + 4$. Solving this equation, we get: $11 = 11 x$, or $x = 1$.
## Answer
This number is 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
A customer walks into a store, picks out an item costing 20 rubles, and gives the seller a hundred-ruble note. The seller looks - no change. He goes to the neighboring department, gets the hundred-ruble note changed. He gives the customer the item and the change. The customer le... | Note: The seller would not have suffered any loss if it were not for the fake 100-ruble note.
## Solution
The seller gave the buyer goods and change totaling 100 rubles, and also paid 100 rubles to the second seller, but he had already received 100 rubles from the second seller. So the total loss is 100 rubles. It ca... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
The kids brought a full basket of mushrooms from the forest. In total, 289 mushrooms were collected, and each basket had the same number of mushrooms. How many kids were there?
# | Notice that the total number of mushrooms collected is equal to the product of the number of children and the number of mushrooms in each basket.
## Solution
The total number of mushrooms collected is equal to the product of the number of children and the number of mushrooms in each basket. Let's represent the number... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Arithmetic. Mental arithmetic, etc.]
The tower clock strikes three times in 12 seconds. How long will it take for it to strike six times?
# | What is the interval between two adjacent strikes? Recall Problem 4.
## Solution
Since 3 strikes of the clock take 12 seconds, the interval between two consecutive strikes is 6 seconds. From the first strike to the second - 6 seconds, and from the second to the third - also 6 seconds. Six strikes, however, occur with... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
On the mailbox, it is written: "Mail is collected five times a day from 7 to 19 hours." Indeed, the first time the postman collects the mail is at 7 in the morning, and the last time is at 7 in the evening. At what intervals are the letters taken out of the box? | How many intervals will there be between mail pickups? Recall problem 4.
## Solution
If the mail is picked up 5 times within the specified time, there will be 4 intervals, i.e., the duration of one interval will be 3 hours.
## Answer
Every 3 hours. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Tournaments and tournament tables]
As is known, cup football matches are played according to the Olympic system: the loser is eliminated, and in case of a draw, a rematch is played. In that year, there were no rematches, and 75 teams participated in the matches. How many matches were played for the cup? # | Note: after a loss, the team is eliminated.
## Solution
Since one team is eliminated after each game, a total of 74 matches were played.
## Answer
74 matches. | 74 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the last digit of the product of all odd numbers from 1 to 99? And from 1 to 199?
# | Notice that among the numbers involved in the product, there are those ending in 5.
## Solution
The product of any sequence of numbers, among which there are numbers ending in 5, will end in either 0 (if there is at least one even number in the sequence) or 5 (if all numbers are odd). In both cases here, there are no... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
A full milk barrel weighs 34 kg, and one filled to half - 17.5 kg. How much does the empty barrel weigh?
# | How much greater is the doubled weight of a bucket filled halfway compared to the weight of a full bucket?
## Solution
The weight of the bucket is equal to the difference between the doubled weight of the bucket filled halfway (i.e., the weight of the contents + twice the weight of the bucket) and the weight of a ful... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
Liza is 8 years older than Nastya. Two years ago, she was three times as old as Nastya. How old is Liza? | Think about how old the girls were two years ago.
## Solution
Two years ago, Lisa was also 8 years older than Nastya. And if at that time she was also 3 times older, then Nastya was 4 years old, and Lisa was 12. So, Lisa is 14 years old now.
## Answer
14 years old. | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ]
Try to find all natural numbers that are 5 times larger than their last digit.
# | Think about what the last digit of the desired number could be.
## Solution
When multiplied by 5, the last digit did not change, so it must have been 0 or 5. If the last digit was 0, then the entire number would be 0, but we are looking for natural numbers. Therefore, the last digit must have been 5. And the entire n... | 25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ [Decimal numeral system ]
Find the largest six-digit number, in which each digit, starting from the third, is equal to the sum of the two preceding digits.
# | Notice that when the number of digits in two numbers is the same, the larger one is the one with the larger first digit.
## Solution
If the first letter was $a$, and the second $-b$, then the third will be $(a+b)$, the fourth $-(a+2b)$, the fifth $-(2a+3b)$, and the sixth $-(3a+5b)$. We need to choose the maximum pos... | 303369 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ]
Find the largest number in which each digit, starting from the third, is equal to the sum of the two preceding digits.
# | Notice, of the two numbers, the larger one is the one with more digits.
## Solution
At first glance, this problem seems very similar to the previous one, but the solution is quite different. A number will be larger the more digits it has. And the more digits there will be, the smaller the first two digits. Let's chec... | 10112358 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ]
Find a two-digit number that is 5 times the sum of its digits.
# | Note: The desired number must be divisible by 5.
## Solution
A number that is 5 times the sum of its digits must be divisible by 5. Therefore, it ends in 0 or 5. However, it cannot end in 0, because in that case it would be 10 times the sum of its digits. Thus, the desired number can be written in the form $10a + 5$.... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental arithmetic, etc.]
Ten people wanted to found a club. For this, they need to collect a certain amount of membership fees. If there were five more organizers, each of them would have to contribute 100 dollars less. How much did each person contribute? | Try to represent the condition of the problem as a system of equations. Think about how to solve this problem without forming a system of equations.
## Solution
Let's denote the amount of the entry fee by $x$. Then we can form the equation $10 x=15(x-100)$, solving which, we determine $x=300$ dollars. We could also s... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Rational Equations]
Solve the equation:
$$
x+\frac{x}{x}+\frac{x}{x+\frac{x}{x}}=1
$$ | On the domain of definition, the equation can be reduced to $\mathrm{x}+1+\mathrm{x} /(\mathrm{x}+1)=1$. Multiply both sides of the equation by $\mathrm{x}+1$. After simplification, we get: $\mathrm{x}^{2}+2 \mathrm{x}=0$, that is, $\mathrm{x}=0$ or $\mathrm{x}=-2$. The root of the equation is only $\mathrm{x}=$ $-2$.
... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A cube with a side of 1 m was sawn into cubes with a side of 1 cm and laid in a row (in a straight line). What length did the row turn out to be
# | We get $100 \times 100 \times 100=1000000$ (cm) or 10000 m $=10$ km. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$2-$
Find the numbers equal to twice the sum of their digits.
# | If the first digit of a two-digit number is $a$, and the second digit is $b$, then the number itself is $10a + b$.
## Solution
It is easy to notice that there are no single-digit numbers greater than zero with the required property. Let's try to find a solution among two-digit numbers. If the first digit of a two-dig... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Percentage and ratio problems ]
During the year, the prices of strudels were raised twice by $50 \%$, and before New Year's, they started selling them at half price.
How much does one strudel cost now if it cost 80 rubles at the beginning of the year?
# | After the first price increase, the strudel cost $80 \cdot 1.5 = 120$ rubles; after the second price increase $120 \cdot 1.5 = 180$ rubles; finally, after a two-fold price reduction, it cost $180 : 2 = 90$ rubles.
## Answer
90 rubles. | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ [ divisibility rules for 3 and 9 ]
It is known that $35!=10333147966386144929 * 66651337523200000000$. Find the digit replaced by the asterisk.
The number 35! is divisible by 9.
# | Since 35! is divisible by 9, the sum of the digits of this number is also divisible by 9. It is not difficult to calculate that the sum of the digits (excluding the asterisk) of the written number gives a remainder of 3 when divided by 9. Therefore, the digit replaced by the asterisk is 6.
## Answer
6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Podlipsky o.k.
Oleg drew an empty $50 \times 50$ table and wrote a number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers writte... | Evaluation. Suppose that among the rational numbers there is 0 and it is written at the top side of the table. Let along the left side of the table be written $x$ irrational and $50-x$ rational numbers. Then along the top side are written $50-x$ irrational and $x$ rational numbers. Note that the product of a non-zero r... | 1275 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
In a sequence of natural numbers, each number, except the first, is obtained by adding to the previous one its largest digit.
What is the maximum number of consecutive odd terms that can occur in the sequence? | Let $a_{1}, \ldots, a_{5}$ be consecutive odd terms of a sequence. The last digits of these numbers are odd, and the largest digits of the numbers $a_{1}, \ldots, a_{4}$ are even and, therefore, not the last. When moving to the next number, the largest digit of the previous number does not change (otherwise, it would i... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalovo $A$.
On the table were two decks, each with 36 cards. The first deck was shuffled and placed on top of the second. Then, for each card in the first deck, the number of cards between it and the same card in the second deck was counted (i.e., how many cards between the sevens of hearts, between the queens of... | Let's number the cards from top to bottom in order. In the upper deck, the numbers are from 1 to 36, and in the lower deck, the numbers are from 37 to 72. Let the $i$-th card of the upper deck coincide with the $k_{i}$-th card of the lower deck ($i=1,2, \ldots, 36$). Between them lies $k_{i}-i-1$ cards, so the desired ... | 1260 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ The Extreme Principle (Miscellaneous). ] [ Cauchy's Inequality $]$ [ Pairing and Grouping; Bijections ]
a) What is the maximum number of edges in a 30-vertex graph that contains no triangles?
b) What is the maximum number of edges in a 30-vertex graph that contains no complete subgraph of four vertices? | Select the vertex of the highest degree.
## Solution
a) Estimate. Let's choose a vertex $A$ of the highest degree $n$ and consider the subgraph $G$ formed by the vertices to which edges from $A$ lead. It is clear that there are no edges in this subgraph. Each vertex not in $G$ has a degree no greater than $n$, and th... | 225 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}{[\text { Decimal numeral system }]} \\ {\left[\begin{array}{l}\text { Case analysis }\end{array}\right]} \\ {[\underline{\text { Divisibility rules for 11 }}]} \\ {[\underline{\text { Divisibility rules for 3 and 9 }}]}\end{array}\right]$
A two-digit number is appended to the right of a certain... | Let $a$ denote the first natural number, and $b$ and $c$ the two-digit numbers written after it. Let $x = a + b + c$. By the condition,
$10^{4} a + 100 b + c = x^{3}$.
If $x \geq 100$, then $x^{3} \geq 10^{4} x = 10^{4}(a + b + c) > 10^{4} a + 100 b + c$, which means the equation has no solutions.
Therefore, $x$ is ... | 91125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Systems of linear equations]
Seven coins (weighing 1, 2, ... 7 grams) are laid out in a row in some order. For each coin (except the outermost ones), the sum of the weights of its neighbors is known.
What is the maximum number of coins whose weight can be guaranteed to be known?
# | Make sure that the weights of the second, fourth, and sixth coins can be expressed through known weights.
## Solution
Let the weights of the coins in the order of their arrangement be: $x_{1}, x_{2}, \ldots, x_{7}$. From the condition, we have:
$x_{1}+x_{3}=a_{2}, x_{2}+x_{4}=a_{3}, x_{3}+x_{5}=a_{4}, x_{4}+x_{6}=a_... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ equations in integers ]
A combination ( $x, y, z$ ) of three natural numbers, lying in the range from 10 to 20 inclusive, is an unlocking combination for a code lock if
$3 x^{2}-y^{2}-7 z=99$. Find all the unlocking combinations. | Note that $3 x^{2}-y^{2} \equiv 1(\bmod 7)$. There are 8 such pairs of remainders: $(1,3),(1,4),(2,2),(2,5),(5,2),(5,5),(6,3),(6,4)$. Considering the range of values for $x$ and $y$, we get 19 possible pairs: $(15,10),(15,17),(16,16),(16,12),(16,19)$, $(12,16),(19,16),(12,12),(12,19),(19,12),(19,19),(13,10),(13,17),(20... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Divisibility rules (etc.).]
Find the smallest number, the representation of which consists only of zeros and ones, that is divisible by 225. | $225=9 \cdot 25$. For a number to be divisible by 25, it should end with two zeros. For divisibility by 9, there should be at least 9 ones in its representation.
## Oтвет
11111111100. | 11111111100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
|
For what x and y is the number $\overline{x x y y}$ a square of a natural number?
# | The number $\overline{x x y y}$ is divisible by 11. If it is a perfect square, then the number $\overline{x 0 y}$ is also divisible by 11.
## Answer
$88^{2}=7744$. | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
For what value of $a$ does the polynomial $P(x)=x^{1000}+a x^{2}+9$ divide by $x+1$? | $P(-1)=1+a+9=a+10$. By the theorem of Bezout, this number should equal zero.
## Answer
For $a=-10$. | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Bezout's Theorem. Factorization]
For what values of the parameter $a$ does the polynomial $P(x)=x^{n}+a x^{n-2}(n \geq 2)$ divide by $x-2$? | $P(2)=2^{n}+a \cdot 2^{n-2}=2^{n-2}(4+a)$.
## Answer
For $a=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Prime numbers and their properties ] [ Examples and counterexamples. Constructions]
Find the smallest natural number $n$, for which the following condition is satisfied: if the number $p-$ is prime and $n$ is divisible by $p-1$, then $n$ is divisible by $p$. | Let $n$ satisfy this condition. Since $n$ is divisible by $1=2-1$, it must be divisible by 2, but then it is divisible by $3=2+1$, by
$7=2 \cdot 3+1$ and by $43=2 \cdot 3 \cdot 7+1$. Therefore, $n$ must be divisible by $1806=2 \cdot 3 \cdot 7 \cdot 43$. Hence, the minimum $n$ (if it exists) is no less than 1806.
On t... | 1806 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the remainder when the number $10^{10}+10^{10^{2}}+10^{10^{3}}+\ldots+10^{10^{10}}$ is divided by 7.
# | $10^{6} \equiv 1(\bmod 7)$, since $10^{3}+1$ is divisible by 7, and $10^{k} \equiv 4(\bmod 6)$ for $k \geq 1$, since the number $9 \ldots . .96$ is even and divisible by 3. Therefore,
## Methods for solving problems with parameters
$$
\text { }
$$
-5(8 l+7)=13$, so the GCD(5l+6, $8 l+7$) is 13 or 1. When $l=4$, we get the fraction $26 / 39$, which is indeed reducible by 13.
## Answer
Only by 13. | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Solve the equation $x^{2 y-1}+(x+1)^{2 y-1}=(x+2)^{2 y-1}$ in natural numbers.
# | $x^{2 y-1} \equiv 1(\bmod x+1)$. Since the number $2 y-1$ is odd, then $x^{2 y-1}=-1(\bmod x+1)$. Therefore, $0=(x+2)^{2 y-1}-x^{2 y-1}$ $-(x+1)^{2 y-1}=1+1=2(\bmod x+1)$, that is, $x+1=2$. Hence, $1+2^{2 y-1}=3^{2 y-1} \Leftrightarrow 2 y-1=1 \Leftrightarrow y=1$.
## Answer
$x=y=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let's consider the sum of the digits of all numbers from 1 to 1000000 inclusive. For the resulting numbers, we will again consider the sum of the digits, and so on, until we get a million single-digit numbers. Which are more numerous among them - ones or twos? | The sum of the digits of a number gives the same remainder when divided by 9 as the number itself. Therefore, units are obtained from numbers of the form $9k+1$, and twos from numbers of the form $9k+2$. All numbers from 1 to 999999 are divided into consecutive nines, each of which contains one number of each kind. The... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9 | |
In a square table of $9 \times 9$ cells, 9 cells are marked at the intersections of the second, fifth, and eighth rows with the second, fifth, and eighth columns. How many ways are there to get from the bottom-left cell to the top-right cell, moving only through unmarked cells upwards or to the right? | Let $a_{i, j}$ be the number of paths from the bottom-left cell to the cell ( $i, j$ ), not passing through the marked cells. Note that if the cell ( $i, j$ ) is not marked, then $a_{i, j}=a_{i-1, j}+a_{i, j-1}$ (and if it is marked, then $a_{i, j}=0$ ). Now it is not difficult to calculate these numbers for each cell ... | 678 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.I
In a row, 10 integers are written. The second row is constructed as follows: under each number $A$ in the first row, a number is written that is equal to the count of numbers in the first row that are greater than $A$ and are located to the right of $A$. The third row is constructed similarly from the seco... | a) In all rows starting from the second, all numbers are non-negative integers. In the second row, the tenth number (the rightmost) is 0, as there is nothing to the right of it. In the third row, the tenth number is also 0, and the ninth number is also 0, because in the second row, the ninth number is non-negative, and... | 10 | Combinatorics | proof | Yes | Yes | olympiads | false |
[ Fibonacci words ]
Find the number of words of length 10, consisting only of the letters "a" and "b" and not containing two consecutive "b" letters in their sequence.
# | Let $\mathrm{a}_{\mathrm{n}}$ denote the number of words of length n, consisting only of the letters "a" and "b" and not containing two consecutive "b" letters. In the sequence $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}$, express each subsequent term in terms of the previous ones.
## Solution
Le... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Inequality problems. Case analysis]
A student passed 31 exams over 5 years of study. In each subsequent year, he passed more exams than in the previous year, and in the fifth year, he passed three times as many exams as in the first year. How many exams did he pass in the fourth year?
# | Try to determine the number of exams taken in the first (and, respectively, fifth) year.
## Solution
It is easy to determine that three exams were taken in the first year and, accordingly, nine in the fifth year. There are two possible distributions of the number of exams in the other years: $4+7+8$ and $5+6+8$. Thus... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{lll}{[} & \text { Evenness and Oddness } & \text { Processes and Operations }\end{array}\right]$ [ Examples and Counterexamples. Constructions ]
The number 123456789 is written on the board. Two adjacent digits are chosen, and if neither of them is 0, 1 is subtracted from each digit, and the chos... | When performing each operation, the parity of each digit does not change.
## Solution
Notice that when performing each operation, the parity of the digit in each position does not change. Indeed, initially, we had the number 123456789, which is a number of the form OEOEOEOEO (O stands for an odd digit, and E stands f... | 101010101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Trigonometric Inequalities]
Find the maximum value of the expression $\sin x \sin y \sin z + \cos x \cos y \cos z$. | What is the greatest value of the expression $\sin x \sin y+\cos x \cos y$?
## Solution
When $x=y=z=0$, the expression $\sin x \sin y \sin z+\cos x \cos y \cos z$ equals 1. We will show that a value greater than 1 cannot be achieved. The expression $\sin x \sin y \sin z+\cos x \cos y \cos z_{\text {does not }}$ excee... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
A group of tourists was supposed to arrive at the train station at 5 o'clock. By this time, a bus was supposed to come from the tourist base to pick them up. However, arriving at the train station at 3:10, the tourists started walking to the tourist base. Meeting the bus on the road, they got on and arrived at the tour... | Tourists saved 20 minutes, during which the bus would have traveled the distance they walked twice.
Therefore, on the way to the station, the bus saved 10 minutes, meaning it met the tourists at 4:50. This means the tourists walked the distance from the station to the meeting point in 100 minutes, which is 10 times sl... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
a) A traveler stopped at an inn, and the owner agreed to accept rings from a golden chain the traveler wore on his wrist as payment for lodging. However, he set a condition that the payment should be daily: each day the owner should have one more ring than the previous day. The closed chain contained 11 rings, and the ... | a) It is enough to cut two rings so that pieces of three and six rings are separated. On the third day, the traveler gives the piece of three rings and receives two rings as change, and on the sixth day, the piece of six rings and receives five rings as change.
b) Arrange the resulting pieces of the chain (not counting... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How many (non-degenerate) triangles with integer side lengths and a perimeter of 100 exist?
# | We need to find the number of triples $(a, b, c)$ of natural numbers where $a \leq b \leq c, a+b+c=100, a+b>c$.
It is clear that $c$ can take values from 34 to 49. For each of these values, $a+b=100-c$, so $b$ can take values from
$\frac{1}{2}(100-c)$ to $c$, more precisely, from $50-\frac{c}{2}$ to $c$ for even $c$ ... | 208 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Chess boards and chess pieces]
What is the maximum number of kings that can be placed on a chessboard so that no two of them attack each other
# | A chessboard can be divided into 16 squares of $2 \times 2$. No more than one king can be placed in each of them.
On the other hand, one king can be placed in each: for example, in the upper left corners of each of these squares.
## Answer
16 kings. | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Greatest Common Divisor (GCD) and Least Common Multiple (LCM). Mutual simplicity ]
A wheel with a radius of 18 rolls around a circle with a radius of 40. A nail is driven into the wheel, which leaves marks on the circle each time it hits it. How many such marks will the nail leave on the circle? How many times will ... | Let $l=4 \pi$. The length of the large circle is 201, and the small one is $9 l$. The nail will hit the previously marked point when the small circle rolls over an arc of length GCD$(20,9) l=180 l$, which means it will roll 9 times around the large circle. During this time, it will leave $180: 9=20$ marks.
## Answer
... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A disappointed investor in the "Neftalmazinvest" fund tore a share into 8 pieces. Not satisfied with this, he tore one of the pieces into 8 again, and so on.
Could he have ended up with 2002 pieces? | Each tear adds 7 pieces, so the number of pieces always gives a remainder of 1 when divided by 7. And 2002 is divisible by 7.
## Task
| 2002 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all three-digit numbers that are 12 times the sum of their digits.
# | According to the condition, the number is divisible by 3. Therefore, the sum of its digits is also divisible by 3, which means the number itself is divisible by 9. In addition, it is divisible by 4. Therefore, we need to look for numbers that are divisible by 36. The sum of the digits of a three-digit number does not e... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Ordinary fractions ]
How many representations does the fraction $\frac{2 n+1}{n(n+1)}$ admit as a sum of two positive fractions with denominators $n$ and $n+1$? | $\frac{2 n+1}{n(n+1)}=\frac{1}{n}+\frac{1}{n+1}$. Since both terms are the smallest fractions with the specified denominators, there are no other representations
## Answer
One. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kolya Vasin was walking after school for five hours. First, he walked along a horizontal road, then climbed a hill, and finally returned to the starting point via the old route. His speed was 4 km/h on the horizontal section of the road, 3 km/h while climbing the hill, and 6 km/h while descending the hill. What distanc... | For each section of the mountain road on the way there, Kolya spent twice as much time as on the way back.
Therefore, his average speed on the "mountain" road was $(2 \cdot 3 + 1 \cdot 6) : 3 = 4$ km/h, which is equal to the speed on the flat road. Consequently, he covered $4 \cdot 5 = 20$ km.
## Answer
## Problem | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Folklore
In Italy, they produce clocks where the hour hand makes one revolution in a day, and the minute hand makes 24 revolutions, with the minute hand being longer than the hour hand (in regular clocks, the hour hand makes two revolutions in a day, and the minute hand makes 24). Consider all positions of the two han... | Let some position of the hour hand of ordinary clocks before noon be taken at time $t$ (counting from the beginning of the day; $0 \leq t<12$ : we measure time in hours). Since the hour hand of Italian clocks moves twice as slowly as the hour hand of ordinary clocks, it will coincide with the considered position of the... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Rynbniov i...
In the store, 20 kg of cheese was delivered, and a queue formed. After selling cheese to the next customer, the saleswoman accurately calculates the average weight of the purchase for all the cheese sold and informs how many people the remaining cheese will last if everyone buys exactly this average weig... | Let $s_{k}$ be the average weight of cheese sold to the first $k$ customers. By the condition $20-k s_{k}=10 s_{k}$, hence
$s_{k}=\frac{20}{k+10}$ kg, and after the $k$-th customer, there remains
200
$k+10$
kg of cheese. Since the obtained sequence is decreasing, the seller can make such a statement after each cust... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ equations in integers ]
A shepherd was tending a herd of 100 heads. For this, he was paid 200 r. For each bull, he was paid 20 r, for each cow - 10 r, and for each calf - 1 r.
How many bulls, how many cows, and how many calves are in the herd? | The number of calves is a multiple of 10.
## Solution
From the condition, it follows that the number of calves is a multiple of 10. Let the number of bulls be $a$, cows be $b$, and calves be $10 c$. Then $a+b+10 c=100$ and $20 a+10 b+10 c=200$, which simplifies to $2 a+b+c=20$.
From the first equation, it is clear t... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Yatsenko I.V.
In the elections to the 100-seat parliament, 12 parties participated. Parties that received strictly more than $5\%$ of the voters' votes enter the parliament. Among the parties that entered the parliament, seats are distributed proportionally to the number of votes they received. After the elections, it... | If 10 parties receive exactly 5% of the votes each, and two, including the Party of Mathematics Enthusiasts (PME), receive 25% each, then the representatives of PME will receive exactly 50 seats in the parliament.
Let's prove that PME cannot receive more seats. If 11 parties did not make it into the parliament, they w... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The distance between points $A$ and $B$ is 40 km. A pedestrian left $A$ at 4 o'clock. When he had walked half the distance, he was caught up by a cyclist who had left $A$ at 7:20. An hour after this, the pedestrian met another cyclist who had left $B$ at 8:30. The speeds of the cyclists are the same. Determine the spee... | The second cyclist was on the road for $11 / 6$ hours less than the first (by the time of the second cyclist's meeting with the pedestrian). If he had set out simultaneously with the first cyclist, he would have been halfway through his journey one hour earlier than the moment of his meeting with the pedestrian. Theref... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Golovanov A.S.
The numbers from 1 to 10 are divided into two groups such that the product of the numbers in the first group is divisible by the product of the numbers in the second group.
What is the smallest value that the quotient of the first product divided by the second can have? | Among the numbers from 1 to 10, only the number 7 is divisible by 7. Therefore, it must be included in the first group, and the quotient is no less than 7.
For example, when it equals $7:(3 \cdot 5 \cdot 6 \cdot 7 \cdot 8):(1 \cdot 2 \cdot 4 \cdot 9 \cdot 10)$.
## Answer
7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Percentage and Ratio Problems ]
Joe knows that to convert from pounds to kilograms, you need to divide the weight in pounds by 2 and then decrease the resulting number by $10 \%$. From this, Joe concluded that to convert from kilograms to pounds, you need to multiply the weight in kilograms by 2 and then increase th... | According to the condition, 1 pound equals $0.5 - 0.05 = 0.45$ kg. When converting 0.45 kg to pounds using John's method, he will get $0.9 + 0.09 = 0.99$ pounds, which is $1\%$ less.
## Answer
By $1\%$ less. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In 8th grade class "G", there are enough underachievers, but Vovochka studies the worst of all. The pedagogical council decided that either Vovochka must correct his twos by the end of the quarter, or he will be expelled. If Vovochka corrects his twos, then 24% of the class will be underachievers, and if he is expelled... | Let there be $n$ students in the class. According to the condition, $0.24 n = 0.25(n-1)$, which means $0.01 n = 0.25$. Therefore, $n = 25$. One person constitutes $4\%$ of 25, so there are now $24 + 4 = 28\%$ of underachievers.
## Answer
$28\%$. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$:$ enderovv B.A.
The numbers $a, b, c$ are such that $a^{2}(b+c)=b^{2}(a+c)=2008$ and $a \neq b$. Find the value of the expression $c^{2}(a+b)$. | $(a-b)(ab+ac+bc)=ab(a-b)+(a^2-b^2)c=a^2(b+c)-b^2(a+c)=0$. Since $a \neq b$, then $ab+ac+bc=0$.
Multiplying by $a-c$, we have
$(a-c)(ac+ab+bc)=a^2(b+c)-c^2(a+b)=0$. Therefore, $c^2(a+b)=a^2(b+c)=2008$.
## Answer
2008. | 2008 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the maximum number of cells that can be marked on a chessboard so that from each of them, it is possible to move to any other marked cell in exactly two moves of a chess knight?
# | Let the knight be on a square of a certain color, then after two moves it will be on a square of the same color. Therefore, the marked squares should be of the same color (let's say black).
We will divide all the black squares of the board into eight four-square figures of two types (striped and gray), as shown in the... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Systems of linear equations ] [ Arithmetic. Mental calculation, etc. ]
At the bottom of the lake, springs are bubbling. A herd of 183 elephants could drink the lake dry in 1 day, while a herd of 37 elephants could do it in 5 days.
How many days would it take for one elephant to drink the lake dry? | 37 elephants over five days drink as much as $37 \cdot 5=185$ elephants in one day. The difference of two elephants is explained by the fact that over the extra four days, the springs "fill up" with as much water as two elephants drink in a day. Thus, the springs replenish half of an elephant's daily portion each day. ... | 365 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Nazarov $\Phi$.
15 elephants are standing in a row, each weighing an integer number of kilograms. If you take any elephant except the one on the far right and add twice the weight of its right neighbor, the result is 15 tons (for each of the 14 elephants). Find the weight of each of the 15 elephants. | Number the elephants from left to right from 1 to 15. Denote the weight of the elephant with number $i$ as $5000 + x_{i}$ kg ($i=1, \ldots, 15$). Then $\left(5000 + x_{i}\right) + 2\left(5000 + x_{i+1}\right) = 15000$, which means $x_{i} = -2 x_{i+1}$. Therefore, $x_{1} = -2 x_{2} = 2^{2} x_{3} = -2^{3} x_{4} = \ldots ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc. ] $[$ Word problems (miscellaneous). ]
Authors: Yashchenko I.V., Botin D.A.
Gulliver found himself in the land of Lilliput with 7,000,000 rubles. He immediately spent all his money on bottles of kefir, which cost 7 rubles per bottle (an empty bottle cost 1 ruble at the time). A... | Conduct calculations in "hard" currency - empty bottles. Then there will be no inflation!
## Solution
Note that if all transactions are conducted in hard currency - empty bottles, - there will be no inflation. That is, a bottle of kefir will always cost 7 empty bottles, and the kefir in it - 6 empty bottles. Gulliver... | 1166666 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Frankin B.R. }}$.
A traveler visited a village where every person either always tells the truth or always lies.
The villagers stood in a circle, and each one told the traveler whether the person to their right was truthful. Based on these statements, the traveler was able to uniquely determine wha... | Let $x$ be the fraction of liars. Imagine that all truthful residents became liars, and all liars "reformed." Then the traveler would hear the same thing! Indeed, the truthfulness of any resident has changed, but the truthfulness of the neighbor they are talking about has also changed. But the fraction of truthful peop... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Factorization ] [GCD and LCM. Mutual simplicity ]
Prove that the greatest common divisor of numbers of the form $p^{4}-1$, where $p$ is a prime number greater than 5, is 240. | All numbers of this form are divisible by 240 (see problem 79525). To ensure that this is the greatest common divisor of all such numbers, it is sufficient to provide a pair of such numbers that do not have a greater common divisor, for example $7^{4}-1=48 \cdot 50=240 \cdot 10$ and $11^{4}-1=120 \cdot 122=240 \cdot 61... | 240 | Number Theory | proof | Yes | Yes | olympiads | false |
[Periodicity and Non-periodicity] [Decimal Number System]
Find the last four digits of the number $2^{1965}$. | Note that $5^{n+4} \equiv 5^{n}\left(\bmod 10^{4}\right)$ for $n \geq 4$. Indeed, $5^{n+4}-5^{n}$ is divisible by $5^{4}$ and $5^{4} \equiv 9^{2} \equiv 1(\bmod 16)$.
The number 1965 gives a remainder of 1 when divided by 4, hence $5^{1965} \equiv 5^{5}=3125\left(\bmod 10^{4}\right)$.
## Answer
3125. | 3125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zaytsev C.A.
What is the maximum number of chips that can be placed on the cells of a chessboard so that on each row, column, and diagonal (not only the main ones) there is an even number of chips?
# | Note that on a chessboard, there are 16 diagonals containing an odd number of cells and having no common cells. Therefore, the number of chips cannot be greater than $64-16=48$. A satisfying arrangement of 48 chips is shown in the figure.
 or $n=p^{2} q$, where $p$ and $q$ - are different prime numbers.
In the first case $1+p+p^{2}+p^{3}+p^{4}+p^{5}=3500, p\left(p+p^{2}+p^{3}+p^{4}\right)=3499$. The number 3499 is not divisible by $2,3,5$ and 7, so $p>10$, but in this case
$p\left(p+p^{2}+... | 1996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all such natural numbers $p$, that $p$ and $2 p^{2}+1$ are primes.
# | If $p$ is not divisible by 3, then $2 p^{2}+1$ is divisible by 3.
## Answer
$p=3$.
## Problem | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Integer and fractional parts. Archimedes' principle]
How many solutions in natural numbers does the equation $\left[{ }^{x} / 10\right]=\left[{ }^{x} / 11\right]+1$ have? | Let $x=11 n+r$, where $n \geq 0,0 \leq r \leq 10$. Then $\left[{ }^{x} / 11\right]=n, n+1=\left[{ }^{x} / 10\right]=n+\left[{ }^{n+r} / 10\right]$, that is, $10 \leq n+r<20,10$ $-r \leq n \leq 19-r$. For each $r$ from 0 to 10, we get 10 solutions.
## Answer
110 solutions. | 110 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
It's very boring to look at a black-and-white clock face, so Clive painted the number 12 red at exactly noon and decided to paint the current hour red every 57 hours.
a) How many numbers on the clock face will end up being painted?
b) How many red numbers will there be if Clive paints them every 1913 hours? | a) Since GCD $(12,57)=3$, every third hour will be red: 12, 3, 6, and 9 o'clock.
b) Since GCD $(12,1913)=1$, all numbers on the clock will turn red. (This will happen, however, almost after 3 years.)
## Answer
a) 4 numbers; b) all 12 numbers. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the cells of an $8 \times 8$ table, the numbers 1 and -1 are written such that the product of the numbers in each row, each column, and each diagonal (including the corner cells) is 1. What is the maximum number of negative ones possible?
# | Consider the diagonals containing an odd number of cells.
## Solution
In each row specified in the condition, there must be an even number minus one. Consider all diagonals containing an odd number of cells. There are 16 of them, and no two of them share any cells. On each of them, there must be at least one unit, so... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
If 6 is subtracted from a certain three-digit number, it will be divisible by 7; if 7 is subtracted, it will be divisible by 8; and if 8 is subtracted, it will be divisible by 9.
Determine ... | What will happen if you add one to the desired number?
## Solution
If you add 1 to the desired number, it will be divisible by $7 \cdot 8 \cdot 9=504$. Since the number is a three-digit number, there is only one possibility - it is equal to
$503=7 \cdot 8 \cdot 9-1$.
## Answer
503. | 503 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$[$ Modular arithmetic (other) $]$ [ Decimal number system ]
How many integers from 1 to 2001 have a sum of digits that is divisible by 5? | Prove that in every complete decade there are two numbers with a sum of digits divisible by 5.
## Solution
We will divide all numbers from 1 to 2001 into decades: two incomplete ones, the first of which includes numbers from 1 to 9, and the second includes numbers 2000 and 2001, and 199 complete decades - from 10 to ... | 399 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$[\quad$ The Pigeonhole Principle (miscellaneous). $\quad] \quad$ Complexity: 3 [Chessboards and chess pieces]
What is the maximum number of kings that can be placed on a chessboard so that no two of them attack each other
# | If in a square of four cells there are two kings, then they attack each other.
## Solution
We will divide the board into 16 squares $2 \times 2$.
Estimate. In each of these 16 squares, there can be no more than one king.
Example. Place a king in the lower left corner of each of the 16 squares.
## Answer
16 kings. | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the sum of the coefficients of the even powers in the polynomial that results from the expression $f(x)=(x^{3} - x + 1)^{100}$ after expanding the brackets and combining like terms.
# | What will be the result if we substitute $x=1$ and $x=-1$ into the given expression?
## Solution
If we substitute $x=1$ into the polynomial $f(x)$, we get the sum of all coefficients of $x^{k}$. If we substitute $x=-1$, we get the difference between the sums of coefficients of even and odd powers. Therefore, the sum ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal number system ] [ Divisibility of numbers. General properties ]
Do there exist two consecutive natural numbers, the sum of the digits of each of which is divisible by 7? | If when transitioning from number $n$ to number $n+1$, $k$ digits are carried over, then the sum of the digits changes by $1 - 9k$.
## Solution
For example, the numbers 69999 and 70000.
## Answer
They exist.
What is the minimum sum of digits that a natural number divisible by $99$ can have?
## Hint
Use the divis... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many integers from 1 to 33000 are not divisible by 3 or 5, but are divisible by 11?
# | 11 divides 3000 numbers (see problem $\underline{60438}$ a). Of these, 3000: 3 = 1000 numbers are also divisible by 3, and 3000: 5 = 600 numbers are divisible by 5. 600: 3 = 200 numbers are divisible by 3, 5, and 11. Therefore, 3000 - 1000 - 600 + 200 = 1600 numbers satisfy the condition of the problem.
## Answer
160... | 1600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
By which natural numbers can the fraction $\frac{3 m-n}{5 n+2 m}$ be reduced, given that it is reducible and that the numbers $m$ and $n$ are coprime. | If $3 m-n$ and $5 n+2 m$ are divisible by $d$, then the numbers $17 m=5(3 m-n)+5 n+2 m$ and $17 n=3(5 n+2 m)-2(3 m-n)$ are also divisible by $d$. But $\operatorname{HOK}(17 m, 17 n)=17$. Therefore, $d=17$. This is possible, for example, when $m=1, n=3$ (or when $m=$ $6, n=1)$.
## Answer
Divisible by 17. | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
\[
\begin{aligned} & {\left[\begin{array}{c}\text { Cryptography } \\ {[\underline{\text { Evenness and Oddness }}]}\end{array}\right]}\end{aligned}
\]
Suppose we need to transmit a message consisting of \( n^2 \) zeros and ones. Write it as an \( n \times n \) square table. Append to each row the sum of its elements ... | a) In the extended table, the sum of elements in any column and any row is even. If one element is changed, the sums for one row and one column will become odd. To correct such an error, the element at the intersection of the row and column with odd sums needs to be changed.
b) The minimum number of errors that cannot... | 4 | Combinatorics | proof | Yes | Yes | olympiads | false |
[ [ $\left[\begin{array}{l}\text { Divisibility of numbers. General properties } \\ \text { p }\end{array}\right]$
Find such an $n$ that the number $10^{n}-1$ is divisible by a) 7 ; b) 13 ; c) 91 ; d) 819. | $10^{6}-1$ is divisible by $10^{3}+1=1001=7 \cdot 11 \cdot 13$, and therefore, by 7, and 13, and $91=7 \cdot 13$. In addition, $10^{6}-1=$ 999999 is divisible by 9, and therefore, by
$819=9 \cdot 91$.
## Answer
$n=6$. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Quadratic equations. Vieta's theorem ] [ Methods for solving problems with parameters ]
For what value of the parameter $m$ is the sum of the squares of the roots of the equation $x^{2}-(m+1) x+m-1=0$ the smallest? | The sum of the squares of the roots $\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(m+1)^{2}-2(m-1)=m^{2}+3$ is minimal when $m=0$. Note that in this case, the equation
$x^{2}-x-1=0$ has roots.
## Answer
When $m=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Algebraic equations and systems of equations (other)]
Formulas for abbreviated multiplication (other)
The number $a$ is a root of the equation $x^{11} + x^{7} + x^{3} = 1$. For which natural values of $n$ does the equality $a^{4} + a^{3} = a^{n} + 1$ hold? | By substitution into the equation, we verify that $a \neq 0$ and $a \neq \pm 1$.
From the condition, it follows that $a^{11}+a^{7}+a^{3}=1$. Multiplying both sides of this equality by $a^{4}-1$, we get $a^{15}-a^{3}=a^{4}-1 \Leftrightarrow a^{4}+a^{3}=a^{15}+1$. Comparing the obtained equality with the equality given ... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10,11Auto: Voomenkov C.I.
At the New Year's Eve party, several married couples arrived, each with 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families to take them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the required trio. How m... | Let there be $p$ married couples and $d$ children at the evening. Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and after her selection, a father could be chosen from one of the $p-2$ remaining couples. Therefore, the total number of trios is
$d(p-1)(p-... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovavov A.B.
10 children were standing in a row. The total number of nuts that the girls and boys had was equal. Each child gave one nut to each of the children standing to their right. After this, the girls had 25 more nuts than they originally had. How many girls are in the row? | The first child from the left gave away 9 nuts, which means he had 9 fewer nuts; the second child gave away 8 nuts and received 1, which means he had 7 fewer nuts. Continuing with similar reasoning, we notice that the first five children had 9, 7, 5, 3, and 1 fewer nuts, respectively, while the next five had 1, 3, 5, 7... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
It is known that the remainder of dividing a certain prime number by 60 is a composite number. Which one? | Since $60=2^{2} \cdot 3 \cdot 5$, the remainder cannot be a multiple of the numbers 2, 3, or 5 (otherwise, the original number would have the same property and could not be prime). Since the remainder is less than 60 and is the product of at least two prime factors, it is equal to 7$\cdot$7. Other options are not possi... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
|
| $[\underline{\text { Evenness and Oddness }}]$ | | |
Authors: Akopyan E., Kalinin D. $\underline{\text {. }}$.
Malvina wrote down 2016 common proper fractions in order: $1 / 2,1 / 3,2 / 3,1 / 4,2 / 4,3 / 4, \ldots$ (including reducible ones). Fractions whose value is less than $1 / 2$ she painted red, and the r... | Note that only one fraction with a denominator of 2 is recorded, exactly two fractions with a denominator of 3, exactly three with a denominator of 4, and so on. Since
$2016=64 \cdot 63: 2$, the number 2016 is equal to the sum of all natural numbers from 1 to 63. Therefore, Malvina recorded all fractions with denomina... | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
nine (not necessarily distinct) nine-digit numbers have been formed; each digit has been used in each number exactly once. What is the maximum number of zeros that the sum of these nine numbers can end with
# | Evaluation. Let the sum end with nine zeros. Each of the numbers formed is divisible by 9, since the sum of its digits is divisible by 9. Therefore, their sum is also divisible by 9. The smallest natural number that is divisible by 9 and ends with nine zeros is $9 \cdot 10^{9}$, so the sum of our numbers is no less tha... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6,7 | |
| $[\underline{\text { Prime numbers and their properties }}]$ | | |
## Author: Raskina I.V.
A group of tourists is dividing cookies. If they evenly divide two identical packs, one extra cookie remains. But if they evenly divide three such packs, 13 extra cookies remain. How many tourists are in the group? | Distribute three times two packs, 3 cookies will remain. But the same six packs of cookies can be distributed differently - three and another three, and then $2 \cdot 13=26$ cookies will remain. This means that $26-3=23$ cookies can be divided equally among the tourists. Since the number 23 is prime, this is possible o... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8,9
Let $N$ be an even number that is not divisible by 10. Find the tens digit of the number $N^{20}$. | From the condition, it follows that $N$ is not divisible by 5. Therefore, according to Fermat's Little Theorem, $N^{4} \equiv 1(\bmod 5)$. Therefore, $N^{20}-1=\left(N^{4}-1\right)\left(N^{16}+N^{12}+N^{8}+N^{4}+1\right)$ is divisible by $5^{2}$. Since the number $N^{20}$ is even, it ends in 26 or 76. But a number endi... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10,11
How many different integer solutions does the inequality $|x|+|y|<100$ have? | For a natural number $n$, the equation $|x|+|y|=n$ has exactly $4n$ integer solutions, and for $n=0$ the solution is unique. Therefore, the number of solutions to the original inequality is $1+4(1+2+\ldots+99)=19801$.
## Answer
19801. | 19801 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Karpenkov O.
To open the safe, you need to enter a code - a number consisting of seven digits: twos and threes. The safe will open if there are more twos than threes, and the code is divisible by both 3 and 4. Come up with a code that opens the safe.
# | Use the divisibility rule for 3.
## Solution
Since there are more twos than threes, the number of twos can be 4, 5, 6, or 7. In the first case, the sum of the digits is 17, in the second case 16, in the third case 15, and in the last case 14. According to the divisibility rule for 3, only the third option works.
Thu... | 2222232 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\stackrel{\text { Ceнгероов V... }}{ }$.
$a, b, c$ are the sides of a triangle. Prove the inequality $\frac{a^{2}+2 b c}{b^{2}+c^{2}}+\frac{b^{2}+2 a c}{c^{2}+a^{2}}+\frac{c^{2}+2 a b}{a^{2}+b^{2}}>3$. | Given the triangle inequality $a^{2}>(b-c)^{2}$. Hence, $a^{2}+2 b c>b^{2}+c^{2}$. Therefore, the first term on the left side of the inequality to be proven is greater than 1. The same is true for the other two. Therefore, their sum is greater than 3.
Author: $\underline{\text { Kuznetsov }} \underline{\text { R.M. }}... | 0 | Inequalities | proof | Yes | Yes | olympiads | false |
Zlobin C.A.
The greatest common divisor of natural numbers $m$ and $n$ is 1. What is the greatest possible value of $\gcd(m+2000 n, n+2000 m) ?$ | Let $a=2000 m+n, b=2000 n+m, d=$ GCD $(a, b)$. Then $d$ divides the numbers $2000 a-b=\left(2000^{2}-1\right) m$ and $2000 b-a=\left(2000^{2}-1\right) n$. Since $m$ and $n$ are coprime, $d$ divides $2000^{2}-1$.
On the other hand, for $m=2000^{2}-2000-1, n=1$, we get $a=\left(2000^{2}-1\right)(2000-1), b=2000^{2}-1=d$... | 3999999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
At every kilometer along the highway between the villages of Yolki and Palkino, there is a post with a sign. On one side of the sign, it shows how many kilometers are left to Yolki, and on the other side, how many kilometers are left to Palkino. Borya noticed that on each post, the sum of all the digit... | Let the distance from Yolki no to Palki no be $n$ kilometers. It is clear that $n \geq 10$. Moreover, $n \leq 49$ (otherwise, the sum of the digits on the 49th sign would be greater than 13).
On the tenth sign from Yolki no, one side reads 10, and the other side reads $n-10$, which does not exceed 39. The sum of its d... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Divisibility of numbers. General properties ] [ Examples and counterexamples. Constructions ]
A five-digit number is called indivisible if it cannot be factored into the product of two three-digit numbers.
What is the largest number of consecutive indivisible five-digit numbers? | Note that all numbers divisible by 100 are decomposable. Therefore, there cannot be more than 99 consecutive non-decomposable numbers. On the other hand, between the numbers $100 \cdot 100$ and $100 \cdot 101$, all numbers are decomposable, and there are exactly 99 of them.
## Answer
99.
All coefficients of the pol... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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