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Avor: frankin 5.
A natural number $\$ \mathrm{k} \$$ is called interesting if the product of the first $\$ \mathrm{k} \$$ prime numbers is divisible by $\$ \mathrm{k} \$$ (for example, the product of the first two prime numbers is $2 \cdot 3=6$, and 2 is an interesting number).
What is the maximum number of interesti... | Evaluation. Obviously, a number that is a multiple of 4 is not interesting.
Examples. $1,2,3 ; 5,6,7$.
## Answer
3 numbers. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9,10
In the tournament, each chess player scored half of all their points in matches against the participants who took the last three places.
How many people in total participated in the tournament
# | For brevity, let's call the players who took the last three places "bad," and all others "good." The bad players played three games among themselves, and in these games, a total of three points were scored. According to the condition, this is half of all the points scored by the bad players; therefore, in games with th... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3 [Chessboards and Chess Pieces]
Avoris A. Anjanis
What is the minimum number of cells that need to be marked on a chessboard so that
1) among the marked cells, there are no adjacent cells (having a common side or a common vertex),
2) adding any one cell to these cells would violate point 1?
# | Evaluation. Let's divide the board into nine parts (see the figure). In each of them, a cell must be marked, otherwise the black cell contained in it can be added.
Example. The nine black c... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Senderov V.A.
1999 numbers stand in a row. The first number is 1. It is known that each number, except the first and the last, is equal to the sum of its two neighbors.
Find the last number.
# | Let's denote our numbers as $a_{1}, a_{2}, \ldots, a_{1999}$. By adding the equations $a_{n+1}=a_{n+2}+a_{n}$ and $a_{n+2}=a_{n+3}+a_{n+1}$, we get $a_{n+3}+a_{n}=0$, or $a_{n+3}=-a_{n}$. From this, $a_{n+6}=-a_{n+3}=a_{n}$, which means the sequence has a period of 6. Therefore, $a_{1999}=a_{6 \cdot 333+1}=a_{1}=1$.
#... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shaovalov A.v.
For which $n>2$ can the integers from 1 to $n$ be arranged in a circle so that the sum of any two adjacent numbers is divisible by the next number in the clockwise direction? | Just like in solving the problem $\underline{98454}$, we prove that there is no more than one even number. Therefore, there are no more than three numbers in total. The numbers 1, 2, 3 can be placed in any order.
## Answer
Only when $n=3$ | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
There are six children in the family. Five of them are respectively 2, 6, 8, 12, and 14 years older than the youngest, and the age of each child is a prime number.
How old is the youngest? | Can the age of the youngest child be an even number?
## Solution
The age of the youngest child cannot be an even number, as otherwise the ages of the older children would not be prime numbers. It cannot end in $1,3,7,9$ - otherwise the age of one of the older children would be divisible by 5. The only prime number th... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3Among the unseen beasts that left tracks on unknown paths, there was a herd of one-headed Thirty-four-legged creatures and three-headed Dragons. In total, the herd has 286 legs and 31 heads. How many legs does a three-headed Dragon have?
# | Think about how many Thirty-four-legged creatures can be in a herd.
## Solution
In the herd, there are 286 legs, which means that the number of Thirty-four-legged creatures cannot be more than 8, since $34 \cdot 9 > 286$. In the herd, there are 31 heads, and each Dragon has three heads, so the number of Thirty-four-l... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Kaibhanov A.K. }}$
The auditorium has the shape of a regular hexagon with a side length of 3 meters. In each corner, there is a snore meter that determines the number of sleeping students within a distance of no more than 3 meters. How many sleeping students are there in the auditorium if the sum o... | 
Each student is "seen" by 2, 3, or 6 snorometers (see fig). Therefore, 7 can be broken down into a sum of terms, each of which is 2, 3, or 6. It is easy to see that 7 can be represented as suc... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kuyna A.K.
Among any five nodes of a regular grid paper, there will definitely be two nodes, the midpoint of the segment between which is also a node of the grid paper. What is the minimum number of nodes of a grid made of regular hexagons that need to be taken so that among them there will definitely be two, the midp... | Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.
Proof of the lemma. Introduce the o... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
|
| $[$ Decimal numeral system $]$ | | |
Find the last digit of the number $7^{7}$.
# | $7^{2} \equiv-1(\bmod 10)$, so, $7^{4} \equiv 1(\bmod 10). 7^{7} \equiv(-1)^{7}=-1(\bmod 4)$, therefore $7^{7^{7}} \equiv 7^{3}=343(\bmod 10)$.
## Answer
3.
Prove that the sum of the squares of five consecutive natural numbers is not a perfect square.
## Solution
First method. The remainder of the division by 4 o... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
}
For which $n>3$ can a set of weights with masses $1,2,3, \ldots, n$ grams be divided into three equal-mass piles?
# | Six weights, whose masses are six consecutive numbers, can easily be divided into three equal mass piles.
## Solution
Let's call a number $n$ good if a set of weights with masses $1,2,3, \ldots, n$ can be divided into three equal mass piles. For a good $n$, the total mass of the weights, which is $1 / 2 n(n+1)$, is d... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the remainder from dividing the polynomial $P(x)=x^{81}+x^{27}+x^{9}+x^{3}+x$ by
a) $x-1$;
b) $x^{2}-1$.
# | a) $P(1)=5$.
b) First method. Let $P(x)=(x^2-1)Q(x) + ax + b$. Substituting $x=-1$, we get $b-a=-5$; substituting $x=1$, we get $a+b=5$.
From this, $b=0, a=5$.
Second method. $P(x)=x(x^{80}+x^{26}+x^{8}+x^{2}+1)$. By substituting $t=x^2$, the second factor transforms into the polynomial $Q(t)=t^{40}+t^{13}+t^{4}+t+1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For what positive value of $p$ do the equations $3 x^{2}-4 p x+9=0$ and $x^{2}-2 p x+5=0$ have a common root?
# | The common root of the given equations must also be a root of the equation $\left(3 x^{2}-4 p x+9\right)-3\left(x^{2}-2 p x+5\right)=0 \quad \Leftrightarrow$ $2 p x-6=0$. Therefore, it equals $3 / p$. Substituting, for example, into the second equation, we get $9 / p^{2}=1$, from which $p=3$.
## Answer
For $p=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3+ [Inequalities. Case Analysis] Pinocchio buried two ingots - a golden one and a silver one - on the Field of Wonders. On good weather days, the golden ingot increases by $30\%$, and the silver ingot by $20\%$. On bad weather days, the golden ingot decreases by $30\%$, and the silver ingot by $20\%$. After a week, it ... | Increasing a number by $20 \%$ is equivalent to multiplying it by 1.2, and decreasing a number by $20 \%$ is equivalent to multiplying it by 0.8 (for $30 \%$ - by 1.3 and 0.7, respectively). Therefore, the result does not depend on the sequence of good and bad weather, but only on the number of good and bad days.
Afte... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Berlov S.L. }}$.
A natural number $n$ is called good if each of its natural divisors, increased by 1, is a divisor of the number $n+1$.
Find all good natural numbers. | Clearly, $n=1$ satisfies the condition. Also, all odd prime numbers satisfy it: the divisors of such a number $p$, increased by 1, are 2 and
$p+1$; both of them divide $p+1$.
On the other hand, any number $n$ that satisfies the condition has a divisor 1; hence, $n+1$ is divisible by $1+$ 1, which means $n$ is odd.
S... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## [ Non-Convex Polygons Division with Remainder $\quad]$
A secret base is surrounded by a transparent winding fence in the shape of a non-convex polygon, with a swamp outside. A straight power line with 36 poles runs through the swamp, some of which are outside the base, and some are inside. (The power line does not... | A straight power line crosses the territory of the base in several segments. When the spy is at the end of one of these segments, he counts the poles. Consider one of them ($A B$). When the spy is at point $A$, he counts the poles on one side of $A B$, and when he is at point $B$, he counts the poles on the other side ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Frankin B.R.
On three red and three blue cards, six positive numbers are written, all of them are different. It is known that on the cards of one color, the pairwise sums of some three numbers are written, and on the cards of the other color - the pairwise products of the same three numbers. Can these three numbers al... | All the sought numbers are distinct, otherwise some numbers on the cards would be equal. Since their pairwise products are positive, all numbers have the same sign, and since their pairwise sums are positive, this sign is positive.
Let the sought numbers be \( x < y < z \).
First method. For the cards of each color, ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov P.G
In the cells of a $4 \times 4$ table, numbers are written such that the sum of the neighbors of each number is 1 (cells are considered neighbors if they share a side).
Find the sum of all the numbers in the table.
# | Let's divide all cells into 6 groups (in the figure, cells of each group are marked with their own symbol). Each group consists of all neighbors of a single cell, so the sum of the numbers in it is 1. Therefore, the sum of all numbers is 6.
 weighing 6 and 12 kg, pieces of equal mass were cut. Each of the cut pieces was melted with the remainder of the other piece, after which the percentage of lead in both alloys became the same. What are the masses of each ... | In each of the received pieces, the ratio of the first alloy to the second should be the same (namely $1: 2$, since the second alloy is twice as much). This means the cut piece constitutes $1 / 3$ of 12 kg (and $2 / 3$ of 6 kg).
## Answer
4 kg. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ [Divisibility of numbers. General properties]
Authors: Binkov A.d,, Raskina I.v.
Several whole heads of cheese were stored in the warehouse. At night, rats came and ate 10 heads, and all of them ate equally. Several rats got sick from overeating. The remaining seven rats the next night finished off the remaining ch... | Let there be $k$ rats in total ($k>7$), then each rat ate $10 / k$ pieces of cheese on the first night. On the second night, each rat ate half as much, i.e., $5 / k$ pieces. Seven rats thus ate $35 / k$ pieces. This is an integer.
The only divisor of the number 35 that exceeds 7 is the number 35 itself. Therefore,
$3... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A pedestrian set out from point $A$ to point $B$. At the same time, a cyclist set out from $B$ to $A$. After one hour, the pedestrian found himself exactly halfway between point $A$ and the cyclist. Another 15 minutes later, they met, and each continued on their way.
How much time did the pedestrian spend on the journ... | Let one hour after the start of the movement, the pedestrian be at point $C$, and the cyclist at point $D$. Then the segment $A C$ is equal to the segment $C D$ (see figure). Let them meet at point $E$ after 15 minutes.
$, the function
$f(x)=2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}$ is increasing, while the function $g(x)=\frac{10}{x-4}$ is decreasing, so the equation $f(x)=g(x)$ has no more than one root.
It rem... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A sowing plot for rye has a rectangular shape. As part of the restructuring of the collective farm lands, one side of the plot was increased by $20\%$, and the other side was reduced by $20\%$. Will the rye yield change as a result, and if so, by how much? | Let $a$ and $b$ be the original sides of the rectangle. The new area is equal to $1.2 a \cdot 0.8 b = 0.96 a b$, which means the area has decreased by $4\%$.
## Answer
It will decrease by $4\%$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Find the largest natural $n$, for which $n^{200}<5^{300}$. | Let's rewrite the inequality as: $\left(n^{2}\right)^{100}<\left(5^{3}\right)^{100}$. Therefore, it is sufficient to find the greatest natural solution to the inequality $n^{2}<125$.
## Answer
$n=11$. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Formulas for abbreviated multiplication (other).]
Calculate: $\frac{(2001 \cdot 2021+100)(1991 \cdot 2031+400)}{2011^{4}}$ | $2001 \cdot 2021+100=(2011-10)(2011+10)+100=2011^{2}-10^{2}+100=2011^{2}, 1991 \cdot 2031=2011^{2}-20^{2}+400=$ 20112. Thus, the numerator is equal to the denominator.
## Answer
1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental arithmetic, etc. ]
[Mathematical logic (other)]
The rabbits sawed several logs. They made 10 cuts and got 16 chunks. How many logs did they saw? | Recall the problem $\underline{89914}$ - the problem about how the rabbits made 10 cuts to split one log.
## Solution
From each log, you get one more piece than the number of cuts. Since there are 6 more pieces, it means there were 6 logs. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ]
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 $x = 1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Word problems ]
In a pet store, they sell large and small birds. A large bird costs twice as much as a small one. One lady bought 5 large birds and 3 small ones, while another bought 5 small birds and 3 large ones. In this case, the first lady paid 20 rubles more. How much does each bird cost?
# | Try to express the difference in purchases of two ladies "in small birds".
## Solution
The first lady paid for her purchase as if for 13 small birds (let's recall that a large bird is twice as expensive as a small one), while the second lady paid as if for 11 small birds. That is, the difference in purchases is 2 sma... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ [Evenness and Oddness]
All the dominoes were laid out in a chain. At one end, there were 5 dots. How many dots are at the other end?
# | The number five on half of the dominoes appears 8 times. Inside the chain, all fives are paired. The unpaired five is at one end. Therefore, the remaining five is at the other end.
## Answer
5 points. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Find the remainder of $4^{18}+5^{17}$ when divided by 3.
# | $4^{18}+5^{17} \equiv 1^{18}+(-1)^{17}=0(\bmod 3)$.
Answer
0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Residue arithmetic (other).]
Find the remainder of $\left(116+17^{17}\right)^{21} \cdot 7^{49}$ when divided by 8.
# | $\left(116+17^{17}\right)^{21} \cdot 7^{49} \equiv\left(4+1^{17}\right)^{21} \cdot 7 \cdot 49^{24} \equiv 5^{21} \cdot(-1) \cdot 1^{24}=-5 \cdot 25^{10} \equiv-5 \cdot 1^{10} \equiv 3(\bmod 8)$.
Answer
## 3. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Inequalities Problems. Case Analysis]
9 kg of candies cost less than 10 rubles, and 10 kg of the same candies cost more than 11 rubles. How much does 1 kg of these candies cost? | The first condition is equivalent to the statement that 1 kg of candies costs less than $10 / 9$ rubles $= 1$ ruble $11_{1}^{1} / 9$ kopecks. Similarly, the second condition is equivalent to the statement that 1 kg of candies costs more than ${ }^{11} / 10$ rubles = 1 ruble 10 kopecks. Therefore, the answer to the prob... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Rabbits are sawing a log. They made 10 cuts. How many chunks did they get? | Into how many parts is a log divided by the first cut? How does the number of pieces change after each subsequent cut?
# Solution
The number of chunks is always one more than the number of cuts, since the first cut divides the log into two parts, and each subsequent cut adds one more chunk. Answer: 11 chunks.
## Ans... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Decimal numeral system]
How many two-digit numbers have a sum of digits equal to 10?
# | There are nine such numbers: $19,28,37, \ldots 91$.
## Answer
There are 9 numbers. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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.]
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.]
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 |
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 |
[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 |
[ [ 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 |
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 |
[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 |
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 |
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
$$
\text { }
$$
^{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 |
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 |
[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 |
[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 |
[ 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
\[
\begin{aligned} & {\left[\begin{array}{c}\text { Cryptography } \\ {[\underline{\text { Evenness and Oddness }}]}\end{array}\right]}\end{aligned}
\]
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 |
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 |
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 |
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 |
$\stackrel{\text { Ceнгероов V... }}{ }$.
$a, b, c$ 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 |
At the "Lukomorye" station, they sell cards for one, five, and twenty rides. All cards cost a whole number of gold coins. Five cards for one ride are more expensive than one card for five rides, and four cards for five rides are more expensive than one card for twenty rides. It turned out that the cheapest way for 33 b... | Since 5, 20, and 35 are all divisible by 5, the number of tickets bought for a single ride is also divisible by 5. However, it is more cost-effective to replace every five such tickets with one ticket for five rides. Therefore, there is no need to buy tickets for a single ride. For the same reason, it is most cost-effe... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3 Products and factorials $\quad$ K
Determine the highest natural power of the number 2007 that divides 2007! | $2007=3^{2} \cdot 223$. In the prime factorization of the number 2007!, the exponent of the number 3 will be quite large, as the factor 3 is included in the factorization of every third number. The factor 223 is only included in the factorization of numbers of the form 223 p, where $p$ is a natural number not exceeding... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3 [Arithmetic. Mental arithmetic, etc.]
Having walked $4 / 9$ of the length of the bridge, the pedestrian noticed that a car was catching up to him, which had not yet entered the bridge. Then he turned back and met it at the beginning of the bridge. If he had continued his movement, the car would have caught up with h... | From the condition, it follows that the time it takes for the car to approach the bridge is equal to the time it takes for the pedestrian to walk $4 / 9$ of the bridge. Therefore, if the pedestrian continues moving, by the time the car reaches the bridge, he will have walked ${ }^{8} / 9$ of the bridge. This means that... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Factorization] [Exponential Equations]
Find all positive roots of the equation $x^{x}+x^{1-x}=x+1$.
# | Since $x>0$, we have $0=x^{2x}+x-x^{x+1}-x^{x}=x^{x}(x^{x}-1)-x(x^{x}-1)=x(x^{x}-1)(x^{x-1}-1)$. Therefore, $x=1$.
## Answer
$x=1$.
## [ Numerical tables and their properties ]
Problem | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Binkov A. D:
After the lesson, the graph of the function $y=k / x$ and five lines parallel to the line $y=k x$ ( $k \neq 0$ ) remained on the board. Find the product of the abscissas of all ten intersection points. | A line parallel to the line $y=k x$ has the equation $y=k x+b$. The abscissas of the points of its intersection with the hyperbola are both roots of the equation $k / x=k x+b \quad \Leftrightarrow \quad k x^{2}+b x-k=0$. The product of the roots of this equation is -1. Multiplying five such products, we get the answer.... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Raskina I.V.
At the exchange office, two types of operations are carried out:
1) give 2 euros - get 3 dollars and a candy as a gift;
2) give 5 dollars - get 3 euros and a candy as a gift.
When the wealthy Pinocchio came to the exchange office, he only had dollars. When he left, he had fewer dollars, no euros appeare... | Since Buratino received 50 candies, he performed exactly 50 operations. During this time, he exchanged all the euros he received back into dollars. Therefore, for every two operations of the second type, there were three operations of the first type, and on these five operations, Buratino lost $2 \cdot 5 - 3 \cdot 3 = ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Find the smallest natural $n$, for which the number $A=n^{3}+12 n^{2}+15 n+180$ is divisible by 23. | $n^{3}+12 n^{2}+15 n+180=n^{2}(n+12)+15(n+12)=(n+12)\left(n^{2}+15\right)$. The smallest value of $n$ for which the first factor is divisible by 23 is 11, and for the second factor, such $n$ is 10. This can be verified by direct enumeration or by noting that
$n^{2}+15 \equiv n^{2}-100=(n-10)(n+10)(\bmod 23)$.
## Answ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the class, there are a teacher and several students. It is known that the teacher's age is 24 years older than the average age of the students and 20 years older than the average age of all present in the class. How many students are in the class?
# | The assumption that all students are of the same age does not affect the average age and, therefore, the answer. Under this assumption, the age of the student is 4 years less than the average age of all present. To "balance" the 20 years "excess" of the teacher, the number of students should be $20: 4=5$.
## Answer
5... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Twelve toddlers went out to the yard to play in the sandbox. Each one who brought a bucket also brought a shovel. Nine toddlers forgot their bucket at home, and two forgot their shovel. By how many fewer toddlers brought a bucket than those who brought a shovel but forgot their bucket? | 3 kids brought a bucket, which means. and a spade. 10 kids brought a spade. So, 7 kids brought a spade without a bucket. $7-3=4$.
## Answer
It's 4. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
A number not less than 10 digits, in its notation only two different digits are used, and identical digits do not stand next to each other.
By what highest power of two can such a number be divisible?
# | Let's cut off the "tail" of the last 8 digits from the number in the condition. The remaining number ends with 8 zeros, so it is divisible by $2^{8}$. The "tail" has the form $\frac{a b a b a b a b}{}=\overline{a b} \cdot 1010101$, where $a$ and $b$ are digits. It is clear that the twos in the prime factorization of th... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a clearing, ladybugs gathered. If a ladybug has six spots on its back, it always tells the truth, and if it has four spots, it always lies, and there were no other ladybugs on the clearing. The first ladybug said: "Each of us has the same number of spots on our backs." The second said: "All of us together have 30 sp... | If the first ladybug tells the truth, then the second and third ones should also tell the truth, as they should have the same number of spots on their backs as the first one. But the second and third ladybugs contradict each other, so at least one of them is lying. Therefore, the first ladybug is also lying.
Suppose e... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Number of divisors and sum of divisors of a number ] equations in integers $\quad]$
Find a natural number $n$, knowing that it has two prime divisors and satisfies the conditions $\tau(n)=6$, $\sigma(n)=28$. | According to the formulas from problem $\underline{60537} n=p^{2} q,\left(p^{2}+p+1\right)(q+1)=28=22 \cdot 7$, where $p$ and $q-$ are prime numbers. From this, $q$ $=3, p=2$.
## Answer
$n=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ $\underline{\text { Modular Arithmetic (other). ] }}$ [Periodicity and Non-periodicity ] Find the remainder of the division by 17 of the number $2^{1999}+1$. | $2^{1999}+1=8 \cdot 16^{499} \equiv 8 \cdot(-1)^{499}=-8 \equiv 9(\bmod 17)$.
## Answer
9. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Prove that the sum $\cos \alpha+\cos \left(72^{\circ}+\alpha\right)+\cos \left(144^{\circ}+\alpha\right)+\cos \left(216^{\circ}+\alpha\right)+\cos \left(288^{\circ}+\alpha\right)$ does not depend on $\alpha$.
# | Apply the formula for the sum of cosines:
$$
\begin{gathered}
\cos \alpha + \left(\cos \left(72^{\circ} + \alpha\right) + \cos \left(288^{\circ} + \alpha\right)\right) + \left(\cos \left(144^{\circ} + \alpha\right) + \cos \left(216^{\circ} + \alpha\right)\right) = \\
= \cos \alpha + 2 \cos \left(180^{\circ} + \alpha\r... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
Ribamko A.v.
In the lower left corner of a $n \times n$ chessboard, there is a knight. It is known that the minimum number of moves it takes for the knight to reach the upper right corner is equal to the minimum number of moves it takes to reach the lower right corner. Find $n$.
# | Let $n$ be even. In this case, the lower left field has the same color as the upper right, while the lower right has a different color. After each move, the knight lands on a field of the opposite color. Therefore, any path from the lower left corner to the upper right consists of an even number of moves, and to the lo... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The numbers $x, y$ and $z$ are such that $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1$. What values can the expression $\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}$ take? | $\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}+(x+y+z)=\frac{x^{2}}{y+z}+x+\frac{y^{2}}{z+x}+y+\frac{z^{2}}{x+y}+z=$
$=\frac{x^{2}+xy+xz}{y+z}+\frac{y^{2}+yz+yx}{z+x}+\frac{z^{2}+zx+zy}{x+y}=$
$\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)(x+y+z)=x+y+z$.
From this, it is clear that the desired sum i... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a single-round-robin tournament, 10 chess players are participating. After what minimum number of rounds can it happen that a sole winner has already been determined prematurely? (In each round, the participants are divided into pairs. Win - 1 point, draw - 0.5 points, loss - 0). | Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one who has no less than 3 points. Since there are still 3 rounds ahead, the winner is stil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berpow S.l.
Seryozha chose two different natural numbers $a$ and $b$. He wrote down four numbers in his notebook: $a, a+2, b$ and $b+2$. Then he wrote on the board all six pairwise products of the numbers from the notebook. What is the maximum number of perfect squares that can be among the numbers on the board? | Note that no two squares of natural numbers differ by 1: $x^{2}-y^{2}=(x-y)(x+y)$, and the second bracket is greater than one. Therefore, the numbers
$a(a+2)=(a+1)^{2}-1$ and $b(b+2)=(b+1)^{2}-1$ are not squares.
The numbers $a b$ and $a(b+2)$ cannot both be squares; otherwise, their product $a^2 b(b+2)$ would also b... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kuzneuov A.
Sasha chose a natural number $N>1$ and wrote down in ascending order all its natural divisors: $d_{1}<\ldots<d_{s}$ (so that $d_{1}=1$ and $d_{s}=N$). Then for each pair of adjacent numbers, he calculated their greatest common divisor; the sum of the resulting $s-1$ numbers turned out to be
$N-2$. What va... | Note that $d_{s+1-i}=N / d_{i}$ for all $i=1,2,3, \ldots, s$.
The number $d_{i+1}-d_{i}$ is divisible by $\left(d_{i}, d_{i+1}\right)$, so $\left(d_{i}, d_{i+1}\right) \leq d_{i+1}-d_{i}$. For $i=1,2, s-1$, let $r_{i}=\left(d_{i+1}-d_{i}\right)$
$-\left(d_{i}, d_{i+1}\right) \geq 0$. By the condition
$\left(d_{2}-d_{... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Fomin $D$ d.
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 200 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 200 can be balanced by some number of weights from the set, and in a unique way (t... | The correct set must correspond to the factorization of the number 201 (see the solution to problem $\underline{98056}$), and it only factors into two factors: $201=3 \cdot 67$.
## Answer
a) Two weights of 67 grams each and 66 weights of 1 gram each, or 66 weights of 3 grams each and two of 1 gram each.
b) 3 sets.
... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
Find all such prime numbers $p$ that the number $p^{2}+11$ has exactly six distinct divisors (including one and the number itself). | Note that $p^{2}+11=(p-1)(p+1)+12$. If $p \geq 5$ and is prime, then the numbers $p-1$ and $p+1$ are even, and one of them is divisible by 3. Therefore, the product $(p-1)(p+1)$ is divisible by 12, hence $p^{2}+11$ is also divisible by 12, which means it has at least seven divisors (6 divisors of the number 12 and the ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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