problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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3 [Divisibility of numbers. General properties]
It is known that $x=2 a^{5}=5 b^{2}>0$, where numbers $a$ and $b$ are integers. What is the smallest possible value of $x$? | The number $x$ is divisible by 2 and 5. $x=2 a^{5}$, so $a$ is divisible by 5. Since $x=5 b^{2}$, $x$ is divisible by $2^{2}$, which means $a$ is also divisible by 2. Assuming $a=2 \cdot 5=10$, we get $x=200000$ and $b=200$.
## Answer
200000. | 200000 | Number Theory | math-word-problem | 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 |
[ [Decimal number system ] $[$ Equations in integers $]$
Find a two-digit number that is equal to the sum of the cube of the number of its tens and the square of the number of its units.
# | Let's denote the number of tens by $x$, and the number of units by $y$. Then $10 x + y = x^3 + y^2$ or $x(10 - x^2) = y(y - 1)$, where $y(y - 1)$ is an even natural number. Therefore, $x$ is even and $10 - x^2 > 0$. Consequently, $x = 2$. Substituting into the equation, we find $y = 4$.
## Answer
24.
Problem | 24 | 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 |
[ Arithmetic progression ]
Author: Shapovalov A.B.
In a $29 \times 29$ table, the numbers $1,2,3, \ldots, 29$ were written, each 29 times. It turned out that the sum of the numbers above the main diagonal is three times the sum of the numbers below this diagonal. Find the number written in the central cell of the tab... | Above (below) the diagonal there are $29 \cdot 14=406$ numbers. It is not difficult to check that the sum of the 406 largest numbers in the table ( $16,17, \ldots, 29$, taken 29 times each) is exactly three times the sum of the 406 smallest numbers ( $1,2, \ldots, 14$, taken 29 times each). Therefore, all the numbers o... | 15 | 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 |
Shapovalov A.V.
Little kids were eating candies. Each one ate 7 candies less than all the others together, but still more than one candy.
How many candies were eaten in total? | Let's choose one of the children - for example, Petya. If we take away 7 candies from all the others, there will be as many left as Petya has. Therefore, twice the number of candies Petya has equals the total number of candies minus seven. The same can be said about any of the children, which means that all the childre... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Prove that for no natural values of $x$ and $y$ the number $x^{8}-x^{7} y+x^{6} y^{2}-\ldots-x y^{7}+y^{8}$ is prime. | $x^{8}-x^{7} y+x^{6} y^{2}-\ldots-x y^{7}+y^{8}=\frac{(x+y)\left(x^{8}-x^{7} y+\ldots-x y^{7}+y^{8}\right)}{x+y}=\frac{x^{9}+y^{9}}{x+y}=\frac{\left(x^{3}+y^{3}\right)\left(x^{6}-x^{3} y^{3}+y^{6}\right)}{x+y}=\left(x^{2}-x y\right.$ $\left.+y^{2}\right)\left(x^{6}-x^{3} y^{3}+y^{6}\right)$
The obtained factors have t... | 19 | Number Theory | proof | 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 |
8,9,10 How many integers exist from 1 to 1000000 that are neither perfect squares, nor perfect cubes, nor fourth powers? | $1000000=1000^{2}=100^{3}=10^{6}$. Therefore, in the specified range, there are exactly 1000 squares and 100 cubes. 10 numbers among them are sixth powers, meaning they are both squares and cubes. All fourth powers are found among the squares. Therefore, the numbers that satisfy the condition are
$1000000-1000-100+10=... | 998910 | Number Theory | 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 |
$[\underline{\text { Divisibility Criteria (other) }}]$
Find all numbers of the form $\overline{13 x y 45 z}$ that are divisible by 792.
# | Apply the divisibility rules for 8, 9, and 11.
## Solution
$792=8 \cdot 9 \cdot 11$. According to the divisibility rule for $8 \mathrm{z}=6$. According to the divisibility rule for $9 x+y=8$ or 17. According to the divisibility rule for 11
$1+x+4+6-3-y-5=x-y+3=0$ or 11. Since $x+y$ and $x-y$ are of the same parity, ... | 1380456 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
There are books on the table that need to be packed. If they are bundled into equal packs of 4, 5, or 6 books, there will always be one extra book left each time, but if they are bundled into packs of 7 books, there will be no extra books left. What is the smallest number of books that can be on the table?
# | Let there be $n$ books. Then $n-1$ is divisible by $60 = \text{GCD}(4, 5, 6)$, and $n$ is divisible by 7. By checking numbers of the form $60k + 1$, we find that the smallest of them that is divisible by 7 is 301.
## Answer
301 books. | 301 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Quadratic equations. Vieta's theorem ] [ Methods for solving problems with parameters ]
For which values of the parameter $a$ is the sum of the squares of the roots of the equation $x^{2}+2 a x+2 a^{2}+4 a+3=0$ the greatest? What is this sum? (The roots are considered with multiplicity.) | The equation has roots when $D / 4=a^{2}-\left(2 a^{2}+4 a+3\right)=-\left(a^{2}+4 a+3\right) \geq 0$, that is, when $-3 \leq a \leq-1$. The sum of the squares of the roots is
$\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(2 a)^{2}-2\left(2 a^{2}+4 a+3\right)=-8 a-6$. This expression takes its maximum value at the small... | 18 | Algebra | 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 |
[ Motion problems ]
The cabins of the ski lift are numbered consecutively from 1 to 99. Igor got into cabin No. 42 at the foot of the mountain and at some point noticed that he was level with cabin No. 13 moving downwards (see figure), and 15 seconds later his cabin was level with cabin No. 12.
How long will it take ... | Since Igor first caught up with cabin No. 13 and then with cabin No. 12, the numbering goes in the direction of the lift's movement. Let's assume the distance between adjacent cabins is one unit. Then the distance along the cable between cabins No. 42 and No. 12 is 69 units: 57 units to cabin No. 99 and another 12 unit... | 17 | Logic and Puzzles | 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 |
Kuzneiov A.
A square box of candies is divided into 49 equal square cells. In each cell, there is a chocolate candy - either black or white. In one sitting, Sasha can eat two candies if they are of the same color and lie in adjacent cells, either by side or by corner. What is the maximum number of candies that Sasha c... | Evaluation. For the layout shown in the left image, none of the 16 black candies can be eaten, and out of the 33 white candies, no more than 32 can be eaten due to parity.
Algorithm. We will... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## [ equations in integers ] Decompositions and partitions $\quad]$ [ GCD and LCM. Mutual simplicity ]
Ostap Bender organized a giveaway of elephants to the population in the city of Fux. 28 union members and 37 non-members showed up for the giveaway, and Ostap distributed the elephants equally among all union members... | Let Bender somehow distributed elephants between two specified groups. To implement another way of distribution, he would have to take away an equal number of elephants from each member of one group and distribute them equally among the members of the other group. Therefore, the total number of elephants transferred fr... | 2072 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Pairing and Grouping; Bijections] $[$ Evenness and Oddness $]$
Given a 29-digit number $X=\overline{a_{1}} \overline{-} \overline{-} \overline{29}\left(0 \leq a_{k} \leq 9, a_{1} \neq 0\right)$. It is known that for every $k$ the digit $a_{k}$ appears in the representation of this number $a_{30-k}$ times (for example... | Let's divide all numbers into pairs $\left(a_{k}, a_{30-k}\right)$. Notice that if among these pairs there is a pair $(a, b)$, then pairs $(a, c)$ and $(c, a)$, where $c \neq b$, cannot occur, otherwise the digit $a$ should appear in the number $X$ $b$ times on one side and $c$ times on the other. Therefore, if the pai... | 201 | Combinatorics | 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 |
Fomin D.
On the board, the numbers $1, 1/2, 1/3, \ldots, 1/100$ are written. We choose two arbitrary numbers $a$ and $b$ from those written on the board, erase them, and write the number
$a + b + ab$. We perform this operation 99 times until only one number remains. What is this number? Find it and prove that it does... | If $a_{1}, a_{2}, \ldots, a_{n}$ are numbers written on the board, then the value $\left(1+a_{1}\right)\left(1+a_{2}\right) \ldots\left(1+a_{n}\right)$ does not change under a permissible operation. Indeed, if $a$ and $b$ are replaced by $a+b+a b$, then the factors not containing $a$ and $b$ do not change, and the prod... | 100 | Algebra | 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 |
In the corners $B$ and $C$ of triangle $ABC$, two circles with radii 2 and 3 are inscribed, touching the angle bisector of angle $A$ of the triangle.
Find this bisector, if the distance between the points where the circles touch $BC$ is 7. | Let a circle of radius 2 with center $O_{1}$ touch the bisector $A D$ and side $B C$ of triangle $A B C$ at points $M$ and $K$ respectively, and a circle of radius 3 with center $O_{2}$ touch $A D$ and $B C$ at points $N$ and $L$ respectively.
Denote $\angle O_{1} D K=\alpha$. Since $D O_{1}$ and $D O_{2}$ are bisecto... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
How many times in a day do the hour and minute hands of a correctly running clock form an angle of $90^{\circ}$?
# | The angle between the hour and minute hands changes uniformly and continuously.
## Solution
Let midnight (the start of the day) be the initial moment of time \( t = 0 \). In one hour, the minute hand completes one full rotation, i.e., \( 360^\circ \), while the hour hand completes \( 30^\circ \). At time \( t \) (whe... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system]
A 1992-digit number is written. Each two-digit number formed by adjacent digits is divisible by 17 or 23. The last digit of the number is 1. What is the first?
# | Let's consider all two-digit numbers divisible by 17 or 23. Recall problem 41.
## Solution
List all two-digit numbers divisible by 17 or 23. These are 17, 34, 51, 68, 85, 23, 46, 69, 92. The last digits of all these numbers are different, so the number can be uniquely restored. The last digit is 1, which means the co... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Arithmetic. Mental arithmetic, etc.] [Theory of algorithms (other).]
There are two hourglasses - one for 7 minutes and one for 11 minutes. An egg needs to be boiled for 15 minutes. How can you measure this time using the available hourglasses?
# | Notice that with the help of two different hourglasses, you can measure not only the time equal to their "sum," but also the time equal to their "difference."
## Solution
Start both hourglasses simultaneously. When the 7-minute hourglass runs out, flip it and let it run for 4 more minutes until the 11-minute hourglas... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Residue arithmetic (other).]
$a, b, c$ are integers, and $a+b+c$ is divisible by 6. Prove that $a^{3}+b^{3}+c^{3}$ is also divisible by 6.
# | | | $\left[\begin{array}{l}\text { Division with remainder } \\ \text { Problem } \underline{30604} \text { Topics: }\end{array}\right]$ | Difficulty: 3- |
| :---: | :---: | :---: |
| $[\underline{\text { Classical combinatorics (other) }}]$ | Classes: 7,8,9 | |
How many natural numbers $n$, less than 10000, exist f... | 2858 | Number Theory | proof | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
In a row, 37 numbers are written such that the sum of any six consecutive numbers is 29. The first number is 5. What can the last number be?
# | The sum of the first 36 numbers is $6 \cdot 29$, the sum of the last 36 numbers is also $6 \cdot 29$. Therefore, the last number is equal to the first.
## Answer
5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
After Clive assembled and wound his clock (see problem $\underline{32798}$), setting it by his grandfather's, it started running backward. How many times a day will it show the correct time? | The hour hands on Clive's clock and the correct clock will first coincide at the moment of setting the "correct" time. The next time this will happen is 6 hours after setting, when each of them will have made half a rotation. Of course, the minute hands will coincide then as well. And so on.
## Answer
4 times. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
For many years, every day at noon, a mail steamship departs from Le Havre to New York and at the same time a steamship from New York departs for Le Havre of the same company. Each of these steamships is at sea for exactly seven days, and they follow the same route.
How many steamships of its company will a steamship t... | At the time when a steamship departs from Le Havre, a steamship that left New York 7 days ago arrives in Le Havre.
## Solution
Let's take the departure time of a certain steamship P from Le Havre as the initial time and mark all the steamships it encounters. At the time when steamship P departs from Le Havre, a steam... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
From the ten-digit number 2946835107, five digits were erased. What is the largest number that could result from this?
# | It is more advantageous for larger digits to remain at the beginning of the number.
## Solution
As a result of the deletion, some five-digit number remains. If in the number 2946835107, the first digit is not deleted, then the resulting five-digit number will start with a two and, consequently, will be less than 9851... | 98517 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the minimum value of the fraction $\frac{x^{2}-1}{x^{2}+1}$.
# | $\frac{x^{2}-1}{x^{2}+1}=1-\frac{2}{x^{2}+1}$. The right side is the smallest when the denominator of the fraction is the smallest, that is, when $x=0$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Evenness and Oddness ] [ Counting in Two Ways ]
In the cells of a $10 \times 10$ square table, numbers from 1 to 100 are arranged. Let $S_{1}, S_{2}, \ldots, S_{10}$ be the sums of the numbers in the columns of the table.
Could it be that among the numbers $S_{1}, S_{2}, \ldots, S_{10}$, every two consecutive numb... | If the condition were met, then in the sequence $S_{1}, S_{2}, \ldots, S_{10}$, even and odd numbers would strictly alternate.
## Solution
If $S_{i}$ and $S_{i+1}$ differ by 1, then these two numbers have different parity, meaning that in the sequence $S_{1}, S_{2}, \ldots, S_{10}$, even and odd numbers strictly alte... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Problems on percentages and ratios ]
The Wolf, Hedgehog, Chizh, and Beaver were dividing an orange. The Hedgehog got twice as many segments as Chizh, Chizh got five times fewer segments than the Beaver, and the Beaver got 8 more segments than Chizh. Find how many segments were in the orange, if the Wolf only got the... | Let's assume the number of orange segments that Chizh received is one part, then Hedgehog received two parts, and Beaver received five parts. Beaver received 4 parts more than Chizh, which is 8 segments. Therefore, one part is 2 segments. Since there are 8 parts in total, the number of segments is 16.
## Problem | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3-$ [Methods for solving problems with parameters] The quadratic trinomial $y=a x^{2}+b x+c$ has no roots and $a+b+c>0$. Determine the sign of the coefficient $c$.
# | 
The quadratic polynomial has no roots, which means its graph does not intersect the x-axis. Since \( y(1) = a + b + c > 0 \), the graph is located in the upper half-plane (see the figure), t... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Inequalities Problems. Case Analysis]
Long, long ago, nine identical books cost 11 rubles and some kopecks, and thirteen such books cost 15 rubles and some kopecks.
How much did one book cost
# | From the first condition, it follows that one book cost more than ${ }^{11} / 9=1.222 \ldots$ rubles, and from the second - that it cost less than ${ }^{16} / 15=1.2307 \ldots$ rubles.
## Answer
1 rub. 23 kop. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[
The Tale of the Dead Princess and the Seven Bogatyrs. Once, returning home in the evening, the bogatyrs gave the princess their prey - 29 gray ducks. Each brother shot at least one duck. All of them hunted a different number of ducks: the older the brother, the more game he shot. What was the catch of the eldest bro... | Let's consider the seven smallest natural numbers.
## Solution
The youngest brother cannot shoot fewer than 1 duck, the next one fewer than 2, the next one fewer than 3, and finally, the oldest brother cannot shoot fewer than 7 ducks. This means that the minimum total number of ducks shot by the brothers is \(1+2+\ld... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7,8 |
48 blacksmiths need to shoe 60 horses. What is the minimum time they will spend on the work if each blacksmith spends 5 minutes on one horseshoe
# | Notice that four blacksmiths cannot shoe one horse at the same time.
## Solution
It will take no less than $60 \cdot 4 \cdot 5: 48=25$ minutes. For this, it is necessary that none of the blacksmiths are idle. It is sufficient to show that four blacksmiths can shoe five horses in 25 minutes. For this, arrange the five... | 25 | Other | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (other).]
The students of class 5A had a total of 2015 pencils. One of them lost a box with five pencils, and instead bought a box containing 50 pencils. How many pencils do the students of class 5A have now?
# | $2015-5+50=2060$.
Answer
2060 pencils. | 2060 | Other | math-word-problem | Yes | Yes | olympiads | false |
Kazicinat T. T.
Grisha has 5000 rubles. In the store, chocolate bunnies are sold at 45 rubles each. To carry the bunnies home, Grisha will have to buy several bags at 30 rubles each. No more than 30 chocolate bunnies can fit in one bag. Grisha bought the maximum possible number of bunnies and enough bags to carry all ... | A full bag of rabbits costs $30+30 \cdot 45=1380$ rubles. Grisha can buy a maximum of 3 full bags, and he will have $5000-3 \cdot 1380=860$ rubles left. With this money, Grisha can buy a fourth (empty) bag and 18 rabbits to put in it. In the end, Grisha will have $860-30-18 \cdot 45=20$ rubles left.
## Answer
20 rubl... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Note that if we flip the sheet on which the digits are written, the digits $0,1,8$ will not change, 6 and 9 will swap places, and the others will lose their meaning. How many nine-digit numbers exist that do not change when the sheet is flipped? | All such numbers should consist of digits that make sense when the sheet is flipped, and the number is determined by its first five digits. Note that the fifth digit cannot be 6 or 9, and the first digit cannot be 0. Therefore, for the first digit, we have 4 options, for the second, third, and fourth digits - 5 each, a... | 1500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
A natural number $a$ was increased by 1, and its square increased by 1001. What is $a$? | Let's write the equation for $a$: $(a+1)^{2}-a^{2}=1001$, from which $a=500$. | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
In the basket, there are 30 russulas and boletus. Among any 12 mushrooms, there is at least one russula, and among any 20 mushrooms, there is at least one boletus. How many russulas and how many boletus are in the basket?
# | Since out of 12 mushrooms there is at least one russula, then there are no more than 11 boletus, and out of any 20 mushrooms there is at least one boletus, which means there are no more than 19 russulas. Since there are 30 mushrooms in total, then there are 11 boletus and 19 russulas. | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Decimal numeral system] Riddles $\quad]$
Top student Polikarp and underachiever Kolya were forming the smallest 5-digit number consisting of different even digits. Polikarp formed his number correctly, while Kolya made a mistake. However, it turned out that the difference between Kolya's number and the correct answer... | Think about what the first two digits of Polycarp's number and the last two digits of Kolya's number should be.
## Solution
If we proceed as in the previous problem, Polycarp should have formed the number 02468, but the first digit cannot be zero, so Polycarp formed the number 20468. Let's try to find Kolya's number.... | 20486 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ]
Top student Polycarp has formed a huge number by writing down natural numbers from 1 to 500 in a row:
123...10111213...499500. Poor student Kolya erased the first 500 digits of this number. What do you think, with which digit does the remaining number start?
# | Note: Kolya was erasing digits, while Polycarp was writing down numbers - single-digit, two-digit, and three-digit.
## Solution
Out of the 500 digits erased by Kolya, 9 digits will be used for single-digit numbers, leaving 491 digits. For two-digit numbers, $90 \times 2 = 180$ digits will be used, leaving 311 digits.... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
Gena went to the shooting range with his dad. They agreed that Gena would make 5 shots and for each hit, he would get the right to make 2 more shots. In total, Gena made 17 shots. How many times did he hit the target?
# | How many shots did Gena earn by hitting the target?
## Solution
It was predetermined that Gena would have 5 shots, and the additional 12 shots were earned by hitting the target - 2 shots for each hit. Therefore, the number of hits was 6.
## Answer
6 times. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Decimal numeral system]
How many two-digit numbers exist where the digit in the tens place is greater than the digit in the units place?
# | The problem can be solved by simple enumeration.
1) If the units digit is 0, then the tens digit can take values from 1 to 9.
2) If the units digit is 1, then the tens digit can take values from 2 to 9.
3) If the units digit is 2, then the tens digit can take values from 3 to 9.
$\cdots$
9) If the units digit is 8, ... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Percentage and ratio problems ]
After the crisis, all prices increased by $25 \%$. By what percentage fewer goods can be bought with the same salary?
# | An increase in prices by $25 \%$ means that the new price of a product is equal to the old price multiplied by $5 / 4$. Therefore, with the previous salary, one can buy
$1:{ }^{5} / 4=4 / 5$ of the old quantity of goods, which is $20 \%$ less.
## Answer
$20 \%$ less. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
What is the value of the expression $\left(10^{2}+11^{2}+12^{2}+13^{2}+14^{2}\right) / 365$ ? | Notice that one can simply "calculate directly," i.e., compute each of the squares, add them all up, and finally divide; or one can recall the formula for the square of a sum and make some transformations.
## Solution
Let's outline the steps that can quite easily be done mentally.
$$
10^{2}+11^{2}+12^{2}+13^{2}+14^{... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Continue the sequence: $2,6,12,20,30, \ldots$ | Try to compare the value of a number with the position number it occupies.
## Solution
The number standing at position $n$ is increased by the number $2 n+2$.
## Answer
42 | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
Grandpa is twice as strong as Grandma, Grandma is three times stronger than Granddaughter, Granddaughter is four times stronger than Puss, Puss is five times stronger than Cat, and Cat is six times stronger than Mouse. Grandpa, Grandma, Granddaughter, Puss, and Cat together with... | How many Mice does a Cat replace? And a Granddaughter?
## Solution
A Cat replaces 6 Mice, a Puppy replaces $5 \times 6$ Mice, a Granddaughter replaces $4 \times 5 \times 6$ Mice, a Grandma replaces $3 \times 4 \times 5 \times 6$ Mice, and a Grandpa replaces $2 \times 3 \times 4 \times 5 \times 6$ Mice. In total, it r... | 1237 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ [Decimal numeral system ]
From the number 1234567...5657585960, remove 100 digits so that the remaining number is: a) the smallest; b) the largest.
# | a) Since we need to find the smallest number, we cross out the 100 largest digits so that the number starts with zeros. The original number contains 111 digits, so 11 digits will remain: $00000123450=123450$.
b) The largest number starts with the largest number of nines (there can be no more than five), which are from... | 123450 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can a black and a white rook be placed on a chessboard so that they do not attack each other
# | The white rook can be placed on any of the 64 squares of the board, and from each of them, it attacks 15 squares (including the square it stands on). There remain 49 squares on which the black rook can be placed. Thus, the white and black rooks can be arranged in $64 \cdot 49=3136$ ways.
## Answer
3136 ways. | 3136 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
- My phone rang.
- Who's calling?
- The Elephant.
Then the Crocodile called, then the Bunny, then the Monkeys, then the Bear, then the Herons... So, the Elephant, Crocodile, Bunnies, Monkeys, Bear, Herons, and I all have phones. Every two phone devices are connected by a wire. How many wires are needed for this?
# | Notice, each wire connects two devices.
## Solution
There are a total of 7 telephones, each connected to six others. Therefore, there are 42 connections in total. Since a wire represents 2 connections, a total of 21 wires were needed.
## Answer
21 wires. | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Word Problems ]
Chuk and Gek were decorating the Christmas tree. To prevent them from fighting, their mother gave each of them the same number of branches and the same number of toys. Chuk tried to hang one toy on each branch, but he was short of one branch. Gek tried to hang two toys on each branch, but one branch ... | Try to do as Chuk did - hang one toy on each branch.
## Solution
Let's try to do as Chuk did - hang one toy on each branch, then one toy will be left over. Now, take two toys - one that is left over, and another one from one of the branches. If we now hang these toys as the second ones on the branches that still have... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ]
In the competition, 50 shooters participated. The first scored 60 points; the second - 80; the third - the arithmetic mean of the points of the first two; the fourth - the arithmetic mean of the points of the first three. Each subsequent shooter scored the arithmetic mea... | Do you remember that if you add a number equal to the arithmetic mean of a group of numbers to that group, the arithmetic mean of the new group will be equal to the arithmetic mean of the initial group.
## Solution
The third player scored $(60+80): 2=70$ points. Each subsequent player also scored 70 points: if you ad... | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A bus ticket will be considered lucky if between its digits you can insert the signs of the four arithmetic operations and parentheses in the right places so that the value of the resulting expression equals 100. Is the ticket N 123456 lucky?
# | $1+(2+3+4) \cdot(5+6)=100$. There are other solutions (try to find at least one more).
## Answer
Yes.
Author: $\underline{\text { Yashchenko I.V. }}$
Place parentheses so that the equation is correct:
$$
1-2 \cdot 3+4+5 \cdot 6 \cdot 7+8 \cdot 9=1995
$$
## Answer
$(1-2) \cdot 3+(4+5 \cdot 6 \cdot 7+8) \cdot 9=19... | 1995 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Galoycin A.i.
In the basket, there are 30 mushrooms - russulas and boletus. It is known that among any 12 mushrooms, there is at least one russula, and among any 20 mushrooms - at least one boletus. How many russulas and how many boletus are in the basket?
# | There are no more than 19 russulas.
## Solution
Since among any 12 mushrooms, at least one is a russula, there are no more than 11 boletus. Since among any 20 mushrooms, at least one is a boletus, there are no more than 19 russulas. Since there are 30 mushrooms in total in the basket, there are exactly 11 boletus and... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Khachaturyan A.v.
A kilogram of beef with bones costs 78 rubles, a kilogram of boneless beef costs 90 rubles, and a kilogram of bones costs 15 rubles. How many grams of bones are there in a kilogram of beef?
# | Let $x$ kg be the weight of bones in a kilogram of beef, then the weight of "pure" beef in it is (1 - $x$) kg. Thus,
$$
15 x+90(1-x)=78
$$
from which $x=0.16$.
## Answer
160 grams. | 160 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Yatsenko I.V.
Vanya thought of a simple three-digit number, all digits of which are different.
What digit can it end with if its last digit is equal to the sum of the first two? | Use the divisibility rules for 2, $2,3$.
## Solution
Obviously, the last digit is greater than 1. A three-digit prime number cannot end in an even digit (i.e., $0,2,4,6$ or 8), nor in the digit 5. If the last digit is 3 or 9, then the sum of all the digits of the number, which is twice the last digit, is divisible by... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ GCD and LCM. Mutual simplicity ]
Candies "Sweet Mathematics" are sold in boxes of 12, and candies "Geometry with Nuts" - in boxes of 15.
What is the smallest number of boxes of candies of both types that need to be bought so that there are an equal number of each type of candy?
# | $\operatorname{HOK}(12,15)=60$.
## Answer
5 boxes of "Sweet Mathematics" and 4 boxes of "Geometry with Nuts".
Send a comment | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Prove the equality $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$. | Let $u=\sqrt[3]{2+\sqrt{5}}, v=\sqrt[3]{2-\sqrt{5}}$. Then $u^{3}+v^{3}=4, u v=\sqrt[3]{4-5}=-1$. From the equality $(u+v)^{3}$ $=u^{3}+v^{3}+3 u v(u+v)$ it is clear that $u+v$ is a root of the equation $x^{3}+3 x-4=0$. This equation has an obvious root $x=1$, and it has no other roots (see problem $\underline{61252}$ ... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
Bogdanov I.I.
Papa Carlo has 130 planks. From 5 planks, he can make a toy windmill, from 7 planks - a steamboat, and from 14 planks - an airplane. An airplane costs 19 gold, a steamboat - 8 gold, and a windmill - 6 gold. What is the maximum amount of gold Papa Carlo can earn? | For one airplane, the same 14 planks are required as for two steamships, but the airplane generates more revenue. Therefore, it makes no sense to build more than one steamship. From 15 planks, you can build three mills (revenue of 18 gold) or an airplane (revenue of 19 gold, and the extra plank won't hurt). Therefore, ... | 172 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Biinkov Add:
A thick issue of the newspaper costs 30 rubles, and a thin one is cheaper. A discount of the same percentage is applied to all newspapers for pensioners, so they buy a thin issue of the same newspaper for 15 rubles. It is known that in any case, the newspaper costs a whole number of rubles. How much does ... | Let the cost of a thick newspaper for pensioners be $30 k$ rubles $(k<1)$, and the full cost of a thin newspaper be $x$ rubles $(x<30)$. Then $k x=15,30 k x=450$. By the condition, $30 k$ and $x$ are integers, so $x$ is a divisor of the number $450=2 \cdot 3^{2} \cdot 5^{2}$, and $15<x<30$. Therefore, $x=25$ or $x=18$.... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3+$ [ Examples and counterexamples. Constructions]
In a square table of size $100 \times 100$, some cells are shaded. Each shaded cell is the only shaded cell either in its column or in its row. What is the maximum number of cells that can be shaded
# | Example. We will color all the cells of one row and all the cells of one column, except for their common cell. In this case, the condition of the problem is satisfied and exactly 198 cells are colored.
Evaluation. For each colored cell, we will highlight the line (row or column) in which it is the only colored one. In... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shnol D.e.
In the morning, a dandelion blooms, it flowers yellow for three days, on the fourth morning it turns white, and by the evening of the fifth day, it withers. On Monday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and on Wednesday - 15 yellow and 11 white. How many white dandelions w... | A dandelion that has bloomed can be white on the fourth and fifth day. This means that on Saturday, the dandelions that bloomed on Tuesday or Wednesday will be white. Let's determine how many there are. The dandelions that were white on Monday had flown away by Wednesday, and 20 yellow ones had definitely survived unti... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Two trains, each with 20 identical cars, were moving towards each other on parallel tracks at constant speeds. Exactly 36 seconds after the first cars of the trains met, passenger Vova, sitting in the fourth car, passed passenger Oleg from the oncoming train, and another 44 seconds later, the last cars of the trains ha... | Let's assume that the first train (in which Vova was traveling) is stationary. Then the other train was moving relative to the first, covering a distance equal to $20+20=40$ carriages in $36+44=80$ seconds, which means it was traveling at a speed of 0.5 carriages per second.
From the moment the first carriages of the ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Inequality problems. Case analysis]
A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar?
... | Let there be $x$ bugs in the first jar, then in the second jar - no less than $x+1$ bug, in the third jar - no less than $x+2$ bugs, ..., in the tenth jar no less than $x+9$ bugs. Therefore, the total number of bugs is no less than $10 x+45$. Considering that a total of 150 bugs were distributed, we get $x \leq 10$.
O... | 16 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[ Average values ]
Three pirates divided the diamonds they had obtained during the day in the evening: twelve each for Bill and Sam, and the rest for John, who could not count. At night, Bill stole one diamond from Sam, Sam stole one from John, and John stole one from Bill. As a result, the average weight of Bill's di... | The number of diamonds each pirate had did not change over the night. Since Bill had 12 diamonds, and their average weight decreased by 1 carat, the total weight of his diamonds decreased by 12 carats. Similarly, the total weight of Sam's diamonds decreased by 24 carats. Since the total weight of Bill's and Sam's diamo... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A set of several numbers, none of which are the same, has the following property: the arithmetic mean of some two numbers from this set is equal to the arithmetic mean of some three numbers from the set and is equal to the arithmetic mean of some four numbers from the set. What is the smallest possible number of number... | Let $C\left(a_{1}, \ldots, a_{k}\right)$ be the arithmetic mean of the numbers $\left(a_{1}, \ldots, a_{k}\right)$. Note that adding a number to the set that is different from its arithmetic mean changes the original arithmetic mean of the set.
Suppose that $(a, b, c, d)$ is a set of four numbers satisfying the condit... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shapovaoov A.B.
A monkey becomes happy when it eats three different fruits. What is the maximum number of monkeys that can be made happy with 20 pears, 30 bananas, 40 peaches, and 50 tangerines? | Let's set the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed each monkey no more than one tangerine, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. Here's how we can satisfy 45 monkeys:
5 monkeys eat a pear, a banana, and a t... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Authors: Vysokiy I. Raskina I.V.
In the aquarium, there are three types of fish: gold, silver, and red. If the cat eats all the gold fish, the number of fish will be one less than $2 / 3$ of the original number. If the cat eats all the red fish, the number of fish will be four more than $2 / 3$ of the original number.... | From the first condition, it is clear that the number of goldfish is 1 more than a third. From the second condition, it follows that the number of red fish is 4 less than a third. Therefore, the number of silver fish is 3 more than a third.
## Answer
The number of silver fish is 2 more
Authors: Baksueva E.V., Khacha... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Garkovv S.B.
The sequence $\left(a_{n}\right)$ is such that $a_{n}=n^{2}$ for $1 \leq n \leq 5$ and for all natural $n$ the equality $a_{n+5} + a_{n+1} = a_{n+4} + a_{n}$ holds. Find $a_{2015}$. | $a_{n+4}+a_{n}=a_{n+3}+a_{n-1}=\ldots=a_{5}+a_{1}=25+1=26$. Therefore, $a_{n}=26-a_{n+4}=26-\left(26-a_{n+8}\right)=a_{n+8}$, which means the sequence $\left(a_{n}\right)$ is periodic with a period of 8. Since the remainder of 2015 divided by 8 is 7, $a_{2015}=a_{7}=26-a_{3}=17$ | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Out of 101 Dalmatians, 29 have a spot only on the left ear, 17 have a spot only on the right ear, and 22 Dalmatians have no spots on their ears.
How many Dalmatians have a spot on the right ear
# | Subtract from 101 the number of Dalmatians that have a spot only on their left ear, and the number of those that have no spots at all:
$101-22-29=50$.
## Answer
50 Dalmatians. | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Percentage and Ratio Problems ]
Karlson ate $40 \%$ of the cake for breakfast, and Little One ate 150 g. For lunch, Fräulein Bock ate $30 \%$ of the remainder and another 120 g, and Matilda licked the remaining 90 g of crumbs from the cake. What was the initial weight of the cake?
# | $90+120=210$ g remained after Freken Bok ate $30 \%$ of the remainder. Since Freken Bok ate $30 \%$ of the remainder, 210 g is $70 \%$ of the remainder.
$210: 0.7=300$ g was before Freken Bok started her lunch. $300+150=450$ g was before Little Man started eating. Since Karlson ate $40 \%$ of the cake, 450 g is $60 \%... | 750 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$:$ Tolinho A.K.
How many different ways are there to break down the number 2004 into natural addends that are approximately equal? There can be one or several addends. Numbers are considered approximately equal if their difference is no more than 1. Ways that differ only in the order of the addends are considered the... | Each method is uniquely determined by the number of summands. Indeed, let 2004 be split into $n$ summands, $r$ of which are equal to $q+1$, and the rest are equal to $q(0 \leq r<n)$. Then $2004=q n+r$. Thus, the numbers $q$ and $r$ are the quotient and remainder of the division of 2004 by $n$, they are uniquely determi... | 2004 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
On the first horizontal line of the chessboard, there are 8 black queens, and on the last - 8 white queens. In what minimum number of moves can the white queens exchange places with the black ones? White and black take turns, moving one queen per move. | The rook on the squares b1 and b8, which moved first, could not move to the opposite rank with that move, meaning these two rooks together made at least three moves. Similarly, pairs of rooks standing on the files c, ..., g each made at least three moves. The first of the four corner rooks to move did not land on the s... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
In a small town, there is only one tram line. It is a circular line, and trams run in both directions. There are stops called Circus, Park, and Zoo on the loop. The journey from Park to Zoo via Circus is three times longer than the journey not via Circus. The journey from Circus to Zoo via Park is half... | Let's get on the tram at the Zoo stop and travel through the Circus to the Park, and then, without leaving the tram, return to the Zoo. The second part of the journey is three times shorter than the first, meaning the first part takes up three-quarters of the full circle, and the second part takes up one-quarter. Mark ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$3+$
Avor: Bakayeva.v.
Forty children were playing in a circle. Of them, 22 were holding hands with a boy and 30 were holding hands with a girl. How many girls were in the circle? | $22+30=52$, so $52-40=12$ children held hands with both a boy and a girl. Therefore, 30 - 12 = 18 children held hands only with girls. These 18 children held $18 \cdot 2=36$ girls' hands, and another 12 held one girl's hand each, so the girls had a total of $36+12=48$ hands. Therefore, there were $48: 2=24$ girls.
## ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## $[\quad$ Processes and Operations $\quad]$ [ Prime Numbers and Their Properties ] [ Decimal Number System ] $[$ Arithmetic of Residues (other). $]$
Initially, a prime number is displayed on the computer screen. Every second, the number on the screen is replaced by the number obtained from the previous one by adding... | Let the initial number on the screen be 2, then we get the following sequence: $2-5-11-13-17$ - 25. The sixth number is composite, so in this case, it will take 5 seconds. We will prove that in other cases, it will take no more than 5 seconds.
Indeed, if the number on the screen was an odd prime number not ending in 9... | 996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
Cells of a $9 \times 9$ board are painted in a checkerboard pattern with black and white colors (corner cells are white). What is the minimum number of rooks needed to be placed on this board so that all white cells are under attack by these rooks? (A rook attacks all cells in the row and column where i... | Evaluation. A rook attacks no more than two cells of a white diagonal, so four rooks are insufficient.
Example. Let's place the rooks on the squares indicated in the figure.
## Answer
5 rook... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a small Scottish town stood a school where exactly 1000 students studied. Each of them had a locker for their clothes - a total of 1000 lockers, numbered from 1 to 1000. And in this school lived ghosts - exactly 1000 ghosts. Every student, leaving the school, locked their locker, and at night the ghosts began to pla... | If the cabinet number $C$ is not a perfect square, then it has an even number of divisors (see problem 30365). Such a cabinet will change positions an even number of times and will end up closed. If, however, the cabinet number $C$ is a perfect square, then the number of its distinct divisors will be odd, and the cabin... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Examples and counterexamples. Constructions ] [ Partitions into pairs and groups; bijections ] [ Decimal number system ]
A subset $X$ of the set of "two-digit" numbers $00,01, \ldots, 98,99$ is such that in any infinite sequence of digits, there are two adjacent digits that form a number in $X$. What is the smalle... | In the sequence of digits $n, m, n, m, \ldots, n, m, \ldots$, there will be two adjacent digits forming a number from $X$. Therefore, for any two digits $n$ and $m$, either the number $\overline{m n}$ is contained in $X$, or the number $\overline{n m}$ is contained in $X$. In particular, when $n=m$, we get that all num... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
##
Side $A B$ of parallelogram $A B C D$ is equal to $2, \angle A=45^{\circ}$. Points $E$ and $F$ are located on diagonal $B D$, such that
$\angle A E B=\angle C F D=90^{\circ}, B F=\frac{3}{2} B E$.
Find the area of the parallelogram. | Let $E F=x$, then $B E=2 x$. Since the right triangles $A B E$ and $C D F$ are equal, $F D=2 x$, $B D=5 x$.
The height $B K$ of the parallelogram is $\sqrt{2}$. By the Pythagorean theorem, $D K^{2}=25 x^{2}-2, D A^{2}-9 x^{2}=A E^{2}=4-4 x^{2}$, so $A D^{2}=5 x^{2}+4$. Moreover, from the similarity of triangles $E D K... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Auto: Rybnikov and..
A bus traveling a 100 km route is equipped with a computer that shows the predicted remaining time until arrival at the final destination. This time is calculated on the assumption that the average speed of the bus on the remaining part of the route will be the same as on the part already traveled... | Let $s$ km be the distance the bus has traveled $t$ hours after the start of the journey ($2 / 3 \leq t \leq 5^{2} / 3$). Then its average speed over the traveled part of the route is ${ }^{s / t}$. According to the problem, moving at this speed, the bus will travel $100-s$ km in one hour, that is, $s / t=100-s$. From ... | 85 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Auto: : Yashchenko I. I. From Flower City to Sunny City, there is a highway 12 km long. On the second kilometer of this highway, there is a railway crossing, which is closed for three minutes and open for three minutes, and so on. On the fourth and sixth kilometers, there are traffic lights, which are red for two minut... | Nезнайка cannot pass the crossing located at the second kilometer of the highway until three minutes have passed, as well as on the seventh, eighth, and ninth minutes, on the thirteenth to fifteenth minutes, and so on. Graphically, this means that the graph of Nезнайка's movement (a straight line) cannot intersect the ... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Petya's mother also gave him money for pencils. The conditions of the promotional offer are the same as in problem 104109). Petya tried to buy as many pencils as possible and as a result, he was able to buy 12 more pencils than his mother asked for. How many pencils did his mother give money for? | By repeating the reasoning of problem 104109, we can be convinced that if Mom gave Petya money for 49 pencils, then he could buy 61 pencils, which is 12 more pencils than Mom asked for. Therefore, 49 is one of the answers. Similarly, we can check that if Mom had given Petya money for 50 pencils, then he could have boug... | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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