problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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As is known, a quadratic equation has no more than two roots. Can the equation $\$[x \wedge 2]+p x+q=0 \$$ for \$p \ne 0 have more than 100 roots? (\$ \$x^2]\$ denotes the greatest integer not exceeding \$x^2\$.) | Consider, for example, the equation $\$\left[x^{\wedge} 2\right]-100 x+2500=0 \$$. It has 199 roots of the form $\$ 50+\backslash f r a c\{k\}\{100\} \$$ (where \$k = - 99\$, \$- 98, ···, 99\$). Indeed, \$\$lleft[left(50 +|frac\{k\}
$\{100\} \backslash$ right)^2\right]=\left[2500+k+\left(|frac $\{k\}\{100\} \backslash... | 199 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Substitution cipher transformation in the alphabet $A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$, consisting of p different letters, involves replacing each letter of the plaintext with a letter from the same alphabet, with different letters being replaced by different letters. The key for a simple substitution cipher... | The number of different words that can be obtained in the specified encryption process with the initial word СРОЧНО coincides with the smallest cycle number of encryption that gives this initial word.
## Solution
Since different letters are replaced by different ones, different words are obtained when encrypting diff... | 2730 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Classical combinatorics (miscellaneous). ] [ Formulas of abbreviated multiplication (miscellaneous).]
Twenty-five coins are distributed into piles as follows. First, they are arbitrarily divided into two groups. Then any of the existing groups is again divided into two groups, and so on until each group consists of... | The first method. Let's represent the coins as points and connect each pair of points with a segment. We will get 25(25 - 1) : $2=$ 300 segments. Each time we divide one group of coins into two, we will erase all segments connecting points corresponding to coins that ended up in different groups. Suppose at some step w... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tolpygo A.K.
It is known that the number $a$ is positive, and the inequality $1 < x a < 2$ has exactly three solutions in integers. How many solutions in integers can the inequality $2 < x a < 3$ have? | On a half-interval of length 1 on the number line, there is always an integer, so on an interval of length greater than $n$ - there are at least $n$ integers. Therefore, if an interval contains exactly $n$ integers, then its length is greater than $n-1$ (the distance between the extreme integers), but not more than $n+... | 23 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[Percentage and ratio problems]
In the basket, there were no more than 70 mushrooms, among which $52\%$ were white. If you throw out the three smallest mushrooms, then the white ones will make up half.
How many mushrooms are in the basket
# | Let there be $n$ mushrooms in the basket, of which $m$ are white. Since $m = \frac{13}{25} n$, $n$ must be divisible by 25. Additionally, $n < 70$ and $n$ is odd (since $n-3$ is even). The only number that satisfies these conditions is $n=25$.
In this case, there are 13 white mushrooms, and they will become half of th... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ equations in integers ]
The number 1047, when divided by $A$, gives a remainder of 23, and when divided by $A+1$, it gives a remainder of 7. Find $A$.
# | Since 1047 gives a remainder of 23 when divided by $A$, then $1047-23=1024$ is divisible by $A$. Similarly, $1047-7=1040$ is divisible by $A+1$. Since $1024=2^{10}$, then $A=2^{n}$, where $n-$ is a natural number and $n \leq 10$. At the same time, $A>23$, so $n \geq$ 5. Among the numbers $2^{5}+1,2^{6}+1,2^{7}+1,2^{8}+... | 64 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Examples and Counterexamples. Constructions] [ Proof by Contradiction ]
## Auto: Zassavsiy A.A.
In a volleyball tournament, each team met every other team once. Each match consisted of several sets - until one of the teams won three times. If a match ended with a score of $3: 0$ or $3: 1$, the winning team received ... | Evaluation. If there are no more than three teams, then the "Simpletons" won all the matches, which means they have the most points. Contradiction.
If there are four or five teams, then each team will play three or four matches. This means that the "Cunning" won no more than one match and scored a maximum of $5=3+1+1$... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Glibov A. }}$.
Let $n$ be a natural number. We call a sequence $a_{1}, a_{2}, \ldots, a_{n}$ interesting if for each $i=1,2$, $\ldots, n$ one of the equalities $a_{i}=i$ or $a_{i}=i+1$ holds. We call an interesting sequence even if the sum of its terms is even, and odd otherwise. For each odd inter... | Denoting the sum containing the term $2 \cdot 3 \cdot \ldots \cdot n(n+1)$ by $A_{n}$, and the other by $B_{n}$, we will prove the equality $A_{n}-B_{n}=1$ by induction.
Base case. $A_{1}-B_{1}=2-1=1$.
Inductive step. Represent the sum $A_{n}$ as $A^{\prime}+A^{\prime \prime}$, where $A^{\prime}$ contains all terms o... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
We have a three-digit number $\overline{a b c}$, take $\overline{c b a}$ and subtract the smaller from the larger. We get the number $\overline{a_{1}^{1}} \bar{b}_{\overline{1}}^{\bar{c}} \overline{\overline{1}} \overline{\text {, }}$, perform the same operation with it. Prove that at some step we will get either the n... | If the number $\overline{a_{1}} \overline{\overline{1}} \overline{\overline{1}} \bar{c}_{\overline{1}}^{\overline{1}}$ is non-zero, then it has the following properties:
1) $b_{1}=9$
2) $\overline{a_{1}^{1}} \overline{b_{1}} \overline{1} c_{1}^{-}$ is divisible by 9.
Therefore, $\overline{a_{1}} \overline{\overline{1... | 495 | Number Theory | proof | Yes | Yes | olympiads | false |
\left.\begin{array}{l}{[\text { Systems of exponential equations and inequalities] }} \\ {[\quad \text { Monotonicity and boundedness }}\end{array}\right]
Solve the system in positive numbers:
$$
\begin{cases}x^{y} & =z \\ y^{z} & =x \\ z^{x} & =y\end{cases}
$$ | First, note that if one of the unknowns is equal to one, then the others are also equal to one. Indeed,
Answer
$x=y=z=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ggrmanova T.A.
Tanya sequentially wrote down numbers of the form $\{n \wedge 7-1\}$ for natural numbers $n=2,3, \dots$ and noticed that for $n=8$ the resulting number is divisible by 337. For what smallest $n>1$ will she get a number divisible by $2022?$ | Let the natural number \$ \$ \$ be such that \$n $\wedge 7-1$ \$ is divisible by \$2022=21cdot 3lcdot 337\$. Then \$n 2 and on 3 , so $\$ n \$-$ is an odd number, having a remainder of 1 when divided by 3. In addition, $\$ n \wedge 7-1$ is divisible by 337. Note that if two numbers are congruent modulo 337 (i.e., they ... | 79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
[ Divisibility of numbers. General properties ] $[$ Classical combinatorics (other). $]$
How many natural numbers less than a thousand are not divisible by either 5 or 7? | First, we will strike out from the set of numbers $1,2, \ldots, 999$ the numbers that are multiples of 5; their quantity is $[999 / 5]=199$. Then, from the same set of numbers $1,2, \ldots, 999$, we will strike out the numbers that are multiples of 7; their quantity is $[999 / 7]=142$. In this process, numbers that are... | 686 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A steamship traveled from Nizhny Novgorod to Astrakhan in 5 days, and back in 7 days. How many days do rafts float from Nizhny Novgorod to Astrakhan
# | Let the distance from Nizhny to Astrakhan be $S$. Then the speed of the steamboat downstream is $S / 5$, and upstream it is $S / 7$. Therefore, the speed of the current (and thus the rafts) is $1 / 2(S / 5 - S / 7) = S / 35$.
## Answer
35 days. | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal number system ] $[$ Equations in integers $]$
Find a four-digit number that is a perfect square and such that the first two digits are the same as each other and the last two digits are also the same. | Let $a$ be the first and second digits, $b$ be the third and fourth. Then the given number is equal to $11(b+100 a)$, so $b+100 a=11 x^{2}$ for some natural number $x$. Moreover, $100 \leq b+100 a \leq 908$, hence $3 \leq x \leq 9$.
Calculating the squares of the numbers $33,44, \ldots, 99$, we find that exactly one o... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find a four-digit number that, when divided by 131, gives a remainder of 112, and when divided by 132, gives a remainder of 98.
# | Let $N$ be the desired number. According to the condition, $N=131 k+112=132 l+98$, where $k$ and $l$ are natural numbers. From this, $131(k-l)=l-14$. Denoting $k-l=t$, we have $l=131 t+14, N=132 \cdot 131 t+132 \cdot 14+98$. Since the number $132 \cdot 131$ is a five-digit number, only $t=0$ fits, giving $N=132 \cdot 1... | 1946 | 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 |
[ Arithmetic. Mental arithmetic, etc. ] $[$ Word problems (other). ]
Snow White cut out a large square from batiste and put it in a chest. The First Dwarf came, took out the square, cut it into four squares, and put all four back into the chest. Then the Second Dwarf came, took out one of the squares, cut it into four... | How many squares does each dwarf "add"?
## Solution
Each dwarf takes 1 square from the chest and puts in 4 - that is, adds 3 squares. Therefore, after the
Seventh Dwarf leaves, there will be $1+3 \cdot 7=22$ squares in the chest.
## Answer
22 squares. | 22 | 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 |
$\underline{\text { Folklore }}$
Find the number of solutions in natural numbers of the equation $\left[{ }^{x} / 10\right]=\left[{ }^{x} / 11\right]+1$. | 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 |
Authors: Fomin D., Smirnov S.K.
On the first day of a certain month, there were 10 types of goods in the store, each at the same price per unit. After that, every day each good either doubles or triples in price. On the first day of the following month, all prices turned out to be different. Prove that the ratio of th... | Each item is assigned a number - how many days in a month this item doubled in price. All prices on the first day of the following month are different, so these numbers for all items are different. Among 10 different integers, the smallest differs from the largest by at least 9. Therefore, there are two items whose pri... | 27 | Number Theory | proof | 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 |
Frankin B.P.
Natural numbers $a<b<c$ are such that $b+a$ is divisible by $b-a$, and $c+b$ is divisible by $c-b$. The number $a$ is written with 2011 digits, and the number $b-2012$ is written with 2012 digits. How many digits does the number $c$ have? | According to the condition, the number $2 a=(b+a)-(b-a)$ is divisible by $b-a$. Therefore, $b-a \leq 2 a$, which means $b \leq 3 a$. Similarly, $c \leq 3b$. Thus, $c \leq 9 a < 10 a$, so in the notation with no more than 2012 digits (but not less, since $c \geq b$).
## Answer
2012 digits. | 2012 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folkpor }}$
It is known that the expressions $4 k+5$ and $9 k+4$ for some natural values of $k$ are simultaneously perfect squares. What values can the expression $7 k+4$ take for the same values of $k$? | Let $4 k+5=m^{2}$ and $9 k+4=n^{2}$, where $m$ and $n$ are some natural numbers. Then $9 m^{2}-4 n^{2}=29=(3 m-2 n)(3 m + 2 n)$.
Thus, $3 m-2 n=1, 3 m+2 n=29$, from which $m=5, n=7, k=5$, and $7 k+4=39$.
## Answer
39.
Send a comment | 39 | Number Theory | 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 |
On each of the three axes, there is one rotating gear and one stationary pointer. The gears are connected sequentially. The first gear has 33 teeth, the second has 10, and the third has 7. On each tooth of the first gear, in clockwise order, one letter of the Russian alphabet is written in alphabetical order:
## А Б В... | Carefully determine the stopping moments after the start of encryption. For this, assign the ordinal number to each letter of the Russian alphabet.
## Solution
Determine the stopping moments after the start of encryption. Number the letters of the Russian alphabet: А - 0, Б - 1, and so on. The letters of the word to ... | 515355128523864354 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Isosceles triangles $A B C(A B=B C)$ and $A_{1} B_{1} C_{1} \quad\left(A_{1} B_{1}=B_{1} C_{1}\right)$ are similar and $A C: A_{1} C_{1}=5: \sqrt{3}$.
Vertices $A_{1}$ and $B_{1}$ are located on sides $A C$ and $B C$, respectively, and vertex $C_{1}$ is on the extension of side
$A B$ beyond point $B$, and $A_{1} B_{1... | Prove that $C_{1} A_{1} \perp A C$ and formulate a trigonometric equation relative to the angle at the base of the isosceles triangle $A B C$.
## Solution
Let $A_{1} B_{1}=B_{1} C_{1}=1, \angle C=\gamma$. Then $A B=\frac{5}{\sqrt{3}}, \angle C_{1} A_{1} C=\angle C_{1} A_{1} B_{1}+\angle B_{1} A_{1} C=\gamma+\left(90^... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
a) The "Tower of Hanoi" puzzle consists of eight disks arranged in decreasing size on one of three pegs. The goal is to move the entire tower to another peg, moving only one disk at a time and never placing a larger disk on a smaller one. Prove that the puzzle has a solution. What method would be optimal (in terms of t... | a) Let the minimum number of steps to move a tower of $n$ disks be $K_{n}$. Express $K_{n+1}$ in terms of $K_{n}$. Consider a tower of $n+1$ disks. We can divide the entire procedure into three stages.
Stage 1 - from the beginning to the first move of the bottom disk. By the end of this stage, the tower of $n$ disks s... | 4373 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
What digits should be placed instead of the asterisks so that the number 454** is divisible by 2, 7, and 9?
# | Let these be the digits $a$ and $b$. For the number to be even, the digit $b$ must be even, and for the number to be divisible by 9, $a+b$ must equal 5 or 14. We get five options - 45414, 45432, 45450, 45468, 45486, - of which only 45486 is divisible by 7.
Answer
45486. | 45486 | 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 |
Aakhnov $H . X$.
On the board, the expression $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}$ is written, where $a, b, c, d, e, f$ are natural numbers. If the number $a$ is increased by 1, then the value of this expression increases by 3. If in the original expression the number $c$ is increased by 1, then its value... | The product of the denominators can be equal to 60. One of the possible examples: $\frac{20}{3} \cdot \frac{15}{4} \cdot \frac{12}{5}$.
Evaluation. Let the value of the original expression be $A$. Then, after the first operation, the product will take the value $\frac{a+1}{a} \cdot A=A+3$, from which
$A=3a$. This mea... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Akopyan E.
At the beginning of the year, there were 25 students in the 7th grade. After seven new students joined, the percentage of excellent students increased by 10 (if it was $a \%$ at the beginning of the year, it is now $(a+10)\%$). How many excellent students are there in the class now? | Let there be $x$ excellent students in the class, and $y$ more excellent students came. From the condition, we get the equation: ${ }^{x+y} / 32-x / 25=$ 0.1. Eliminating the denominators, we get: $25 y-7 x=80$. Therefore, $7 x=5(5 y-16)$. From this, it is clear that $5 y-16$ is divisible by 7. Therefore, $(5 y-16)+21=... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
To H.X.К a natural number $N$ was added its largest divisor, less than $N$, and the result was a power of ten. Find all such $N$.
# | Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=m p$, where $p$ is the smallest prime divisor of the number $N$. We have
$m(p+1)=N+m=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is odd, and therefore $m$ is odd. Since $10^{k}$ is divisi... | 75 | Number Theory | 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 |
Evdokimov M.A.
By the river, there lives a tribe of Mumbo-Jumbo. One day, with urgent news to the neighboring tribe, a young warrior Mumbo and a wise shaman Yumbo set out simultaneously. Mumbo ran at a speed of 11 km/h to the nearest raft storage and then sailed on a raft to the neighboring tribe. Yumbo, on the other ... | Let the habitat of the Mumbo-Jumbo tribe be denoted by $O$, the storage that Mumbo ran to by $M$, and the storage that Jumbo went to by $U$. Clearly, $M$ is located upstream from $O$, while $U$ is downstream. Let the distances from $O$ to $M$ and $U$ be $x$ and $y$ km, respectively $(x < y)$, and the speed of the river... | 26 | Algebra | 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 |
[ Trigonometric equations ] [ Divisibility of numbers. General properties ]
Authors: Begun $\underline{\text { A.V. }}, \underline{\text { Goryashin D.V. }}$
What is the maximum number of factors of the form $\sin \frac{n \pi}{x}$ that can be crossed out in the left-hand side of the equation
$\sin \frac{\pi}{x} \sin... | Note that the factor $\sin n \pi / x$ becomes zero only when $n \pi / x = \pi k$, that is, $x = n / k$, where $k \in \mathbf{Z}$. Among these values, the natural numbers will only be the number $n$ itself and all its divisors.
Let's divide the factors of the form $\sin { }^{n \pi} / x$ on the left side of the equation... | 1007 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In each cell of a chessboard, there are two cockroaches. At a certain moment, each cockroach crawls to an adjacent (by side) cell, and the cockroaches that were in the same cell crawl to different cells. What is the maximum number of cells on the board that can remain free after this? | Evaluation. We will color 20 cells of the board as shown in the left figure. Each colored cell has the following property: no matter which two adjacent cells the cockroaches crawl into from it, the cockroaches will not be able to reach these cells from any other colored cell. Therefore, after all the cockroaches have m... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Senderov V.A. }}$
A thousand non-overlapping arcs are placed on a circle, and two natural numbers are written on each of them. The sum of the numbers on each arc is divisible by the product of the numbers on the next arc in the clockwise direction. What is the greatest possible value of the largest... | Let's prove that the numbers on the circle do not exceed 2001.
Lemma 1. Let \( x \) and \( y \) be natural numbers. If \( xy = x + y \), then \( x = y = 2 \), and if \( x > 0, y > 0, x \leq y \), then the sum \( x + y \) decreases as \( x \) increases.
Proof. This follows from the decrease of the function \( f(x) = x... | 2001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9,10,11 |
Authors: Agakhanov N.Kh., Bogdanov I.I.
In how many ways can the numbers $2^{0}, 2^{1}, 2^{2}, \ldots, 2^{2005}$ be divided into two non-empty sets $A$ and $B$ such that the equation $x^{2}-S(A) x+S(B)=0$, where $S(M)$ is the sum of the numbers in set $M$, has an integer root? | If $x_{1} \leq x_{2}$ are the roots of the equation, then $x_{1}, x_{2} \in \mathbf{N}$ and $x_{1}+x_{2}=S(A), x_{1} x_{2}=S(B)$, therefore $\left(x_{1}+1\right)\left(x_{2}+1\right)=S(B)+$ $S(A)+1=1+2+4+\ldots+2^{2005}+1=2^{2006}$. Hence, $x_{1}+1=2^{k}, x_{2}+1=2^{2006-k}$, where $k$ can take values $1,2, \ldots, 1003... | 1003 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can you place on a chessboard so that they do not attack each other
a) two rooks; b) two kings; c) two bishops; d) two knights; e) two queens?
All figures are of the same color. | In part a) the answer is half of that in problem 102877, and in part b) - of that in problem 30327, since in those calculations each position was counted twice. In the other parts, the calculations are similar.
## Answer
a) $\frac{64 \cdot 49}{2}=1568 ;$ b) $\frac{4 \cdot 60+24 \cdot 58+36 \cdot 55}{2}=1806$ ways; c)... | 1848 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Gapochkin A.i.
How many integers from 1 to 1997 have a sum of digits that is divisible by 5? | Consider 10 consecutive numbers, starting from a number ending in zero and ending with a number ending in nine. The sums of the digits of these numbers also form 10 consecutive numbers, so among them, exactly two numbers are divisible by 5. The numbers from 0 to 1999 are divided into 200 such tens, therefore, among the... | 399 | Number Theory | 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 |
[ Periodicity and Aperiodicity $]$ [ Classical Combinatorics (other)]
In an old manor, the house is surrounded by tall trees - pines, spruces, and birches. There are a total of 96 trees. These trees have a strange property: of the two trees growing two apart from any conifer, one is coniferous and the other is deciduo... | Notice that the condition is imposed on trees of the same "parity".
## Solution
Let's mentally remove half of the trees - those planted every other one. Then 48 trees will remain, and the condition will become as follows: of the two trees growing next to a coniferous one, one is coniferous and the other is a birch, a... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Tolkpy A.K. }}$
A circle is divided into seven arcs such that the sum of any two adjacent arcs does not exceed $103^{\circ}$.
Name the largest number $A$ such that in any such division, each of the seven arcs contains at least $A^{\circ}$. | Evaluation. The sum of six consecutive arcs does not exceed $309 \|$, therefore, the length of the seventh arc is not less than $51^{\circ}$. Example: divide the circle into arcs of $51^{\circ}, 52^{\circ}, 51^{\circ}, 52^{\circ}, 51^{\circ}, 52^{\circ}, 51^{\circ}$.
## Answer
$A=51$. | 51 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A.K. Tolpygo
Ten numbers are written around a circle, their sum is 100. It is known that the sum of any three consecutive numbers is not less than 29.
Indicate the smallest number $A$ such that in any such set of numbers, each number does not exceed $A$. | Evaluation. Consider some single number $d$, and divide the other nine numbers into three pairs of adjacent numbers. The sum of these nine numbers is no less than 87, so $d \leq 13$. Therefore, in any set that meets the conditions, each number does not exceed 13.
Example. A set containing the number 13 exists: 10, 10,... | 13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Bershin A.
The sequence $\left\{x_{n}\right\}$ is defined by the conditions: $x_{n+2}=x_{n}-1 / x_{n+1} \quad$ for $n \geq 1$.
Prove that there will be a zero among the terms of the sequence. Find the index of this term. | Note that if some term of this sequence is equal to 0, then the subsequent terms are undefined, as the formula for determining the first of them contains division by 0. All subsequent considerations apply to the terms of the sequence up to and including this zero (or to an infinite sequence if there is no zero).
We ha... | 1845 | Algebra | proof | 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.
 Find this rule.
b) Find all natural numbers that transform into themselves (according to this rule).
c) Prove that the number $2^{1991}$ will become a single-digit number after several transformations. | a) The law can be guessed, noticing, for example, that while the number is a single digit, it doubles, and then - apparently not. And the fact that 10 turns into 2 suggests that it is not the number itself that is doubled, but the sum of its digits. Thus, the sought law has been discovered: "Double the sum of the digit... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Spivak A.V.
If for a number $x$ we calculate the sum of its digits and repeat this process with the resulting number two more times, we will get three more numbers. Find the smallest $x$ for which all four numbers are different, and the last of them is 2. | If $a<b$, then the smallest number with the sum of digits $a$ will be less than the smallest number with the sum of digits $b$.
## Solution
First, let's prove that if $a<b$, then the smallest number with the sum of digits $a$ will be less than the smallest number with the sum of digits $b$. Let $B$ be the smallest nu... | 2999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Motion Problems ]
Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon and, without stopping, arrived: one - in $B$ at 4 PM, and the other - in $A$ at 9 PM. At what hour of the day was dawn? | After noon, the first pedestrian walked as much as the second did before noon.
## Solution
Let $x$ be the number of hours from dawn to noon. The first pedestrian walked $x$ hours before noon and 4 after, while the second walked $x$ hours before noon and 9 after. The ratio of the times is equal to the ratio of the len... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Five football teams held a tournament - each team played against each other once. 3 points were awarded for a win, 1 point for a draw, and no points for a loss. Four teams scored 1, 2, 5, and 7 points respectively. How many points did the fifth team score?
# | Each team played 4 games. It is clear that the first team drew once and lost the rest of the games. The second team has two draws and two losses. The third team could not have scored five points with draws alone, so it must have won once, in addition to having two draws and one loss. The fourth team won twice (otherwis... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A.K. Tolpygo
Petya's bank account contains 500 dollars. The bank allows only two types of transactions: withdrawing 300 dollars or adding 198 dollars.
What is the maximum amount Petya can withdraw from his account if he has no other money?
# | Since 300 and 198 are divisible by 6, Petya will only be able to withdraw an amount that is a multiple of 6 dollars. The maximum number that is a multiple of 6 and does not exceed $500 is 498.
Let's show how to withdraw 498 dollars. Perform the following operations: $500-300=200, 200+198=398, 398-$ $300=98, 98+198=296... | 498 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Case Analysis $]$ [ Proof by Contradiction ]
In the cells of a $3 \times 3$ table, numbers are arranged such that the sum of the numbers in each column and each row is zero. What is the smallest number of non-zero numbers that can be in this table, given that this number is odd? | Example.
| 0 | -1 | 1 |
| :---: | :---: | :---: |
| -1 | 2 | -1 |
| 1 | -1 | 0 |
Evaluation. We will prove that it is impossible to get by with fewer non-zero numbers.
If the table contains exactly one non-zero number, then the sum of the numbers in the row containing this number is not zero. Suppose the table conta... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Problems on Mixtures and Concentrations]
From two pieces of alloys (with different lead content) 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 |
[ Classical combinatorics (miscellaneous).] $[\quad$ Case enumeration $\quad]$
Find all odd natural numbers greater than 500 but less than 1000, for each of which the sum of the last digits of all divisors (including 1 and the number itself) is 33. | For an odd number, all its divisors are odd. Since the sum of their last digits is odd, there must be an odd number of divisors. If a number has an odd number of divisors, then it is a square of a natural number (see problem $\underline{30365}$). Let's consider all the squares of odd numbers within the given range: $23... | 729 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
People enter the metro from the street evenly. After passing through the turnstiles, they find themselves in a small hall before the escalators. The doors just opened, and the hall in front of the escalators was initially empty, with only one escalator operating for descent. One escalator could not handle the crowd, so... | When one escalator is turned on, 1/12 of the hall is filled in a minute. When two escalators are turned on, 1/1 of the hall is filled in a minute. We can reason in different ways.
First method. The difference 1/12 - 1/30 = 1/20 shows what part of the hall one escalator empties in a minute. When a third escalator is tu... | 500 | 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 |
Proizvolov V.V.
Given fifty different natural numbers, twenty-five of which do not exceed 50, and the rest are greater than 50 but do not exceed 100. Moreover, no two of them differ exactly by 50. Find the sum of these numbers. | Subtract 50 from each number that is greater than 50. We will get 50 different numbers, that is, numbers from 1 to 50. Their sum is $1+2+\ldots+50=25 \cdot 51$, and the sum of the original numbers is $-25 \cdot 51+25 \cdot 50=25 \cdot 101$.
## Answer
2525. | 2525 | 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.
. We will denote a row (column) by the same color as the rooks standing on it.
Since there are more than 16 rooks, there are at least three... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Solve the equation: $2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}=\frac{10}{x-4}$. | Since the left, and therefore the right, part of the equation takes only positive values, then $x>4$. On the interval $(4,+\infty)$, 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 |
Folklore
A rope was folded in half, then again in half, and once more in half, and then all layers of the rope were cut in one place.
What could have been the length of the rope, if it is known that some two of the resulting pieces had lengths of 9 meters and 4 meters? | 
When cutting, three different types of pieces could be formed (see fig.), and the length of one type of piece is exactly twice that of another. Let's denote these lengths as \( x, 2x \), and... | 52 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Berroov S.L.
A five-digit number is called non-decomposable if it cannot be factored into the product of two three-digit numbers.
What is the maximum number of consecutive non-decomposable five-digit numbers?
# | The smallest number that can be represented as the product of two three-digit numbers is $100 \cdot 100=10000$. The next such number is:
$100 \cdot 101=10100$, so the numbers $10001,10002, \ldots, 10099$ - are non-decomposable. Thus, there are 99 consecutive non-decomposable five-digit numbers.
More than 99 non-decom... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Doctor Aibolit distributed 2006 miraculous tablets to four sick animals. The rhinoceros received one more than the crocodile, the hippopotamus received one more than the rhinoceros, and the elephant received one more than the hippopotamus. How many tablets will the elephant have to eat?
# | While the animals haven't eaten the medicine, let's take one pill from the rhinoceros, two from the hippopotamus, and three from the elephant. Now all four have an equal amount. We took 6 pills, so there are 2000 left - 500 for each. We took 3 pills from the elephant, so Aybolit prescribed 503 pills to the elephant.
#... | 503 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
In an example of adding two numbers, the first addend is less than the sum by 2000, and the sum is greater than the second addend by 6.
Restore the example.
# | If from the sum of two numbers you subtract one of the addends, you get the other addend. From the condition, it follows that the second addend is 2000, and the first is -6.
## Answer
$6+2000=2006$. | 2006 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ostap Bender and Kisa Vorobyaninov divided the revenue from selling elephants to the population between themselves. Ostap thought: if I had taken 40% more money, Kisa's share would have decreased by 60%. And how would Vorobyaninov's share change if Ostap had taken 50% more money? | $40 \%$ of Ostap's money equals $60 \%$ of Kisa's money, so Ostap has $1.5$ times more money. Therefore, an increase in Ostap's share by $n \%$ decreases Kisa's share by $1.5 n \%$.
## Answer
It would decrease by $75 \%$. | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Zlobin C.A.
Two different numbers $x$ and $y$ (not necessarily integers) are such that $x^{2}-2000 x=y^{2}-2000 y$. Find the sum of the numbers $x$ and $y$. | We can rewrite the given equation as follows: $2000(x-y)=(x-y)(x+y)$. From this, $x+y=2000$ (since $x \neq y$, we can cancel out $x-y$).
## Answer
2000. | 2000 | 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 |
Shapovvoov A. ·
On a sausage, thin rings are drawn across. If you cut along the red rings, you get 5 pieces; if along the yellow ones, you get 7 pieces; and if along the green ones, you get 11 pieces. How many pieces of sausage will you get if you cut along the rings of all three colors? | Note that the number of parts is always one more than the number of cuts. Therefore, there are 4 red rings, 6 yellow ones, and 10 green ones. Thus, the total number of cuts is $4+6+10=20$, and the number of parts is 21.
## Answer
21 pieces. | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The product of two natural numbers, neither of which is divisible by 10, is 1000. Find their sum.
# | Since $1000 = 5^3 \cdot 2^3$, each of the numbers in their prime factorization can only contain twos and fives. At the same time, both of these factors cannot be present in the factorization of one number, otherwise it would be divisible by 10.
Therefore, one of the numbers is $5^3$, and the other is $2^3$, and their ... | 133 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Find $x^{3}+y^{3}$, given that $x+y=5$ and $x+y+x^{2} y+x y^{2}=24$.
# | Since $x+y+x^{2} y+x y^{2}=x+y+x y(x+y)=(x+y)(x y+1)=24$, using the condition $x+y=5$, we get $x y=3.8$.
We can proceed in different ways:
First method. $x^{3}+y^{3}(x+y)^{3}-3 x y(x+y)=125-3 \times 3.8 \times 5=68$.
Second method. $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=(x+y)\left((x+y)^{2}-3 x y\right)=5(25... | 68 | 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 |
Shaovalov A.V.
Along the path between the houses of Nезнayka and Sineglazka, there were 15 peonies and 15 tulips growing in a row, mixed together. Setting out from home to visit Nезнayka, Sineglazka watered all the flowers in a row. After the 10th tulip, the water ran out, and 10 flowers remained unwatered. The next d... | There were 10 flowers left unwatered, which means that $30-10=20$ flowers were watered. Consider the last tulip watered by Blue-Eyes. There are 5 more tulips growing after it. Therefore, Nезнайка will pick these 5 tulips and finish picking flowers right at the last tulip watered by Blue-Eyes. But this means that all th... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
In a herd consisting of horses, two-humped and one-humped camels, there are a total of 200 humps.
How many animals are in the herd if the number of horses is equal to the number of two-humped camels? . | Let each two-humped camel "share" a hump with a horse. Then each animal will have one hump, so the number of humps is equal to the number of animals in the herd.
## Answer
200. | 200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
Authors: Muroaikin M.Vv, Shapovesovo A.
In six baskets lie pears, plums, and apples. The number of plums in each basket is equal to the number of apples in the other baskets combined, and the number of apples in each basket is equal to the number of pears in the other baskets c... | The total number of plums is 5 times the number of apples, which is 5 times the number of pears, so the total number of fruits is $1+5+25=31$ times the number of pears.
Send a comment | 31 | Number Theory | proof | 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 calculation, etc.]
How many cells does the diagonal cross in a grid rectangle of size $199 \times 991$? | The diagonal intersects $199+991$ - 1 = 1189 cells. | 1189 | Geometry | 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 |
[Text problems $]$
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. | 14 | 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 |
[Problems on percentages and ratios]
In the club, there are girls, but boys make up more than $94\%$ of the composition. What is the minimum number of people that can be in the club? | One girl makes up less than $6 \%$, therefore, in total in the club (100\%) there are more than $100: 6=16.66 \ldots$ people.
## Answer
17 kids. | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Tournament and tournament tables]
30 teams played a tournament in the Olympic system. How many matches were played in total?
# | As many as the eliminated teams.
## Answer
29 matches. | 29 | Combinatorics | 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 |
It is known that $a+b+c=5$ and $ab+bc+ac=5$. What can $a^{2}+b^{2}+c^{2}$ be?
# | $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ac)=5^{2}-2 \cdot 5=15$.
Send a comment | 15 | Algebra | 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 |
On each kilometer of the highway between the villages of Yolki and Palki, there is a post with a sign. On one side of the sign, it says how many kilometers it is to Yolki, and on the other side - to Palki. A Thoughtful Observer noticed that on each post, the sum of these numbers is 13. What is the distance from Yolki t... | Note: On each post, one number shows the distance from the post to Yolkin, and the other number shows the distance from the post to Palkin.
## Solution
On each post, one number shows the distance from the post to Yolkin, and the other number shows the distance from the post to Palkin. Therefore, the distance from Yol... | 13 | Algebra | 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 |
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