problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
2. How many natural numbers from 1 to 2017 have exactly three distinct natural divisors? | Answer: 14.
Solution: Only squares of prime numbers have exactly three divisors. Note that $47^{2}>2017$, so it is sufficient to consider the squares of prime numbers from 2 to 43. There are 14 of them. | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Petya is coming up with a password for his smartphone. The password consists of 4 decimal digits. Petya wants the password not to contain the digit 7, and at the same time, the password should have at least two (or more) identical digits. In how many ways can Petya do this? | # Answer 3537.
Solution: The total number of passwords not containing the digit 7 will be $9^{4}=6561$. Among these, 9x8x7x6=3024 consist of different digits. Therefore, the number of passwords containing identical digits is 6561-3024=3537 passwords. | 3537 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the computer center, there are 200 computers, some of which (in pairs) are connected by cables, a total of 345 cables are used. We will call a "cluster" a set of computers such that a signal from any computer in this set can reach all the others via the cables. Initially, all computers formed one cluster. But one... | Answer: 153.
Solution: Let's try to imagine the problem this way: an evil hacker has cut all the wires. What is the minimum number of wires the admin needs to restore to end up with 8 clusters? Obviously, by adding a wire, the admin can reduce the number of clusters by one. This means that from 200 clusters, 8 can be ... | 153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. At the international StarCraft championship, 100 participants gathered. The game is played in a knockout format, meaning in each match, two players compete, the loser is eliminated from the tournament, and the winner remains. Find the maximum possible number of participants who won exactly two games? | Answer: 49
Solution: Each participant (except the winner) lost one game to someone. There are 99 such participants, so no more than 49 participants could have won 2 games (someone must lose 2 games to them).
We will show that there could be 49. Let's say โ3 won against โ1 and โ2, โ5 - against โ3 and โ4, ... โ99 - aga... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On graph paper, a right-angled triangle with legs equal to 7 cells was drawn (see fig.). Then all the grid lines inside the triangle were outlined. What is the maximum number of triangles that can be found in this drawing? | Answer: 28 triangles
Solution: One of the sides of the triangle must go at an
angle, i.e., lie on the segment BC. If we fix some diagonal segment, the remaining vertex is uniquely determine... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of distinct digits $\Pi, B$, and $\Gamma$, for which the equality $\overline{\Pi B \Gamma}=(\Pi+B+\Gamma) \times(\Pi+B+\Gamma+1)$ holds. | Answer: 156.
Solution: Note that ะ+ะ $\overline{\Pi ะ \Gamma}$ and (ะ + ะ + ะ) should give the same remainder when divided by 9. This is only possible when $ะ+ะ+\Gamma$ is a multiple of 3. Note that $ะ+ะ+\Gamma=9$ - does not work, because (ะ + ะ + $\Gamma) \times(ะ+B+\Gamma+1)=90$-two-digit. By trying $12,15,18,21,24$... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. How many three-digit natural numbers have an even number of distinct natural divisors? | Answer: 878.
Solution: Note that only perfect squares have an odd number of divisors (for non-squares, divisors can be paired with their complements). There are 900 three-digit numbers in total. Among them, the perfect squares are $10^{2}, 11^{2}, \ldots, 31^{2}=961$ ( $32^{2}=1024-$ is a four-digit number). There are... | 878 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. At the international table tennis championship, 200 participants gathered. The game is played in a knockout format, i.e., in each match, two players participate, the loser is eliminated from the championship, and the winner remains. Find the maximum possible number of participants who won at least three matches. | Answer: 66.
Solution: Each participant (except the winner) lost one game to someone. There are 199 such participants, so no more than 66 participants could have won 3 games (someone must lose 3 games to them).
We will show that there could be 66 such participants. Let โ4 win against โ1,2,3; โ7 - against โ4,5,6,... โ1... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. The turtle crawled out of its house and moved in a straight line at a constant speed of 5 m/hour. After an hour, it turned $90^{\circ}$ (right or left) and continued moving, then crawled for another hour, then turned $90^{\circ}$ (right or left) again... and so on. It crawled for 11 hours, turning $90^{\circ}$ at th... | Answer 5 m.
Solution: We can consider a coordinate grid with nodes spaced 5 m apart. It is clear that the turtle crawls along
the nodes of this grid. The turtle cannot return to the initial... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Can you use the four arithmetic operations (and also parentheses) to write the number 2016 using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in sequence? | Answer: $1 \cdot 2 \cdot 3 \cdot(4+5) \cdot 6 \cdot 7 \cdot 8: 9=2016$. | 2016 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Anya did not tell Misha how old she is, but she informed him that on each of her birthdays, her mother puts as many coins into the piggy bank as Anya is turning years old. Misha estimated that there are no fewer than 110 but no more than 130 coins in the piggy bank. How old is Anya? | Answer: 15. Solution. Either use the formula for the sum of an arithmetic progression: $110 \leq \frac{1+n}{2} n \leq 130$, or simply calculate the sum "brute force". | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. How many different right-angled triangles exist, one of the legs of which is equal to $\sqrt{2016}$, and the other leg and the hypotenuse are expressed as natural numbers
ANSWER: 12. | Solution. According to the condition $c^{2}-b^{2}=a^{2}=2016$, that is, $(c-b)(c+b)=2^{5} \cdot 3^{2} \cdot 7$. The system $\left\{\begin{array}{l}c-b=n, \\ c+b=k\end{array}\right.$ (here $n-$ is one of the divisors of the number 2016, and $k=\frac{2016}{n}$) has natural solutions $c=\frac{n+k}{2}, b=\frac{k-n}{2}$, if... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. The number $n+2015$ is divisible by 2016, and the number $n+2016$ is divisible by 2015. Find the smallest natural $n$ for which this is possible.
ANS: 4058209. | Solution. According to the condition $\left\{\begin{array}{l}n+2015=2016 m, \\ n+2016=2015 k .\end{array}\right.$ From this, $2016 m-2015 k=-1$. The solution of this equation in integers: $m=-1+2015 p, k=-1+2016 p$. Therefore, $n+2015=2016(-1+2015 p)=-2016+2016 \cdot 2015 p$, which means $n=-2015-2016+2016 \cdot 2015 p... | 4058209 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Kolya is twice as old as Olya was when Kolya was as old as Olya is now. And when Olya is as old as Kolya is now, their combined age will be 36 years. How old is Kolya now?
ANS: 16 years. | Solution: Let $x$ be Kolya's current age, $y$ be Olya's age. We can set up the system $\mathrm{x}=2(y-(x-y)) ; x+(x-y)+y+(x-y)=36$. Solving it: $x=16, y=12$. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. How many different right-angled triangles exist, one of the legs of which is equal to $\sqrt{1001}$, and the other leg and the hypotenuse are expressed as natural numbers
ANSWER: 4. | Solution: Let's write down the Pythagorean theorem: $a^{2}+1001=b^{2}$. From this, we get $(b-a)(b+a)=1001=7 \times 11 \times 13$. We can represent 1001 as the product of two factors $1 \times 1001=7 \times 143=11 \times 91=13 \times 77$ - the first factor must be smaller - there are 4 options in total. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find the largest natural number that cannot be represented as the sum of two composite numbers.
OTBET: 11 | Solution: Even numbers greater than 8 can be represented as the sum of two even numbers greater than 2. And odd numbers greater than 12 can be represented as the sum of 9 and an even composite number. By direct verification, we find that 11 cannot be represented in this way. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $f(x)=x^{2}+p x+q$ where $p, q$ are some coefficients. By how much can the maximum value of the function $g(x)=|f(x)|$ differ from the minimum value of this function on the interval $[2 ; 6]$?
ANSWER: by 2. | Solution: For $f(x)=x^{2}+p x+q$ the difference between the maximum and the minimum value is at least 4 (this can be shown graphically). By choosing $q$,
we find that the maximum value of the modulus of the minimum differs by no more than 2. Example: $f(x)=(x-4)^{2}-2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. How many five-digit numbers of the form $\overline{a b 16 c}$ are divisible by 16? $(a, b, c-$ arbitrary digits, not necessarily different).
ANSWER: 90. | Solution: Note that the first digit does not affect divisibility, hence, a=1,..,9. On the other hand, divisibility by 8 implies that c=0 or 8. If c=0, then $b$ must be even, and if $c=8$ - odd. In both cases, we get 5 options, from which the total number is $9 *(5+5)=90$. | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Philatelist Andrey decided to distribute all his stamps equally into 3 envelopes, but it turned out that one stamp was extra. When he distributed them equally into 5 envelopes, 3 stamps were extra; finally, when he distributed them equally into 7 envelopes, 5 stamps remained. How many stamps does Andrey have in tota... | Solution. If the desired number is $x$, then the number $x+2$ must be divisible by 3, 5, and 7, i.e., it has the form $3 \cdot 5 \cdot 7 \cdot p$. Therefore, $x=105 p-2$. Since by the condition $150<x \leq 300$, then $p=2$. Therefore, $x=208$. | 208 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Solve the equation in natural numbers $2 n-\frac{1}{n^{5}}=3-\frac{2}{n}$
ANSWER: $n=1$. | Solution: $2 n=3-\frac{2}{n}+\frac{1}{n^{5}} \leq 3$, only $\mathrm{n}=1$ fits
Lomonosov Moscow State University
## School Olympiad "Conquer Sparrow Hills" in Mathematics
Final Stage Tasks for the 2015/2016 Academic Year for 9th Grade | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest natural number $N$ such that $N+2$ is divisible (without remainder) by 2, $N+3$ by 3, ..., $N+10$ by 10.
ANSWER: 2520. | Solution: Note that $N$ must be divisible by $2,3,4, \ldots, 10$, therefore, N= LCM $(2,3,4, . ., 10)=2^{3} \times 3^{2} \times 5 \times 7=2520$. | 2520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Captain Jack Sparrow found a cave with a pirate treasure. In it, there are 6 chests, and the treasure is only in one of them, while the other chests are home to venomous snakes, ready to attack anyone who disturbs their peace.
On the first chest, it says โThe treasure is in the third chest.โ
On the second, โThe tr... | Solution: Let's create a $6 \times 6$ table. In the ั-th row and ั-th column, we will place a cross if the ั-th statement is true when the treasure is in the ั-th chest:
| | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | | | X | | | |
| 2 | X | X | | | | |
| 3 | X | X... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. $A B C D E F$ - a regular hexagon, point O - its center. How many different isosceles triangles with vertices at the specified seven points can be constructed? Triangles that differ only in the order of vertices are considered as
.
$ and $f(x, y)$ in this chain, there are no other solutions to the equation.
Suppose such a solution $(\alpha, \beta)$ does exist, then $(x, y)<(\alpha, \beta)<f(x, y)$. Apply the mapping $g$, which is the inverse of $f: g(x, y)=(3 x-2 y,-4 x+3 y)$, to all parts of this inequal... | Answer: 696.
V-2. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 15-story hotel by the sea. Know-it-all noticed that the sum of all room numbers from the first to his own, inclusive, is twice the sum of all room numbers from the first to the one where Don't-know-it stayed... | 539 | Number Theory | proof | Yes | Yes | olympiads | false |
1. Place the numbers $1,2,3,4,5,6,7,8$ and 9 in the nine cells of the figure shown in the diagram, so that the sum of the numbers in each column, starting from the second, is 1 more than in the previous one. It is sufficient to find at least one such arrangement. In your answer, indicate the number in the first column.... | Answer: 7.
Solution: For now, we will not pay attention to the order of numbers in one column.
The sum of the given numbers is 45. Let $x$ be the number in the bottom-left cell. Then $5x + 10 = 45$, from which $x = 7$. Therefore, the sum of the numbers in the second column is $8 = 5 + 3 = 6 + 2$. If the second column... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. A certain 4-digit number is a perfect square. If you remove the first digit from the left, it becomes a perfect cube, and if you remove the first 2 digits, it becomes a fourth power of an integer. Find this number. | Answer: 9216.
Solution: Only 16 and 81 are two-digit fourth powers. But 81 does not work, since no three-digit cube ends in $81\left(5^{3}=125,7^{3}=343,9^{3}=729\right)$. But 16 is the ending of $6^{3}=216$. Next, we look for a perfect square that ends in 216. | 9216 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. How many natural numbers from 1 to 2015 inclusive have a sum of digits that is a multiple of 5? | Answer: 402.
Solution: Note that among ten numbers of the form $\overline{a 0}, \ldots, \overline{a 9}$, exactly two numbers have a sum of digits that is a multiple of five. Thus, among the numbers from 10 to 2009, there are exactly 200 such tens, and therefore, 400 such numbers. Considering also the numbers 5 and 201... | 402 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Drop perpendiculars $D D_{1}, D D_{2}, D D_{3}$ from point $D$ to the planes $S B C$, $S A C$, and $S A B$ respectively. Let $D D_{1}=x, D D_{2}=y, D D_{3}=z$. According to the condition, we form the system of equations
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5 \\
x^{2}+z^{2}=13 \\
x^{2}+y^{2}=10
\end{array}\right.
... | Answer: 27.
Answer to option 17-2: 108.
Answer to option $17-3: 27$.
Answer to option $17-4: 108$.
[^0]: ${ }^{1}$ This equality can be proven by expressing $B C^{2}$ from two triangles $B A C$ and $B D C$ using the planimetric cosine theorem. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $x, y, z$ be the number of students in the categories of biology, physics, and chemistry, respectively. Then, according to the problem, we get the system of equations:
$$
\left\{\begin{array} { l }
{ 5 x = 2 ( y + z ) , } \\
{ 7 z = 3 ( x + y ) . }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ 5 x - 2 y =... | Answer: 29. Answer to option: 4-2: 11.
# | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The solution can only exist if $a \in\left\{-\frac{\pi}{12}\right\} \cup\left(0 ; \frac{\pi}{12}\right]$, since otherwise the left side of the equation is either undefined or strictly positive. When $a=-\frac{\pi}{12}$, the equation becomes $2|x-16|=0$. Therefore, when $a=-\frac{\pi}{12}$, $x=16$. If $a \in\left(0 ;... | Answer: $x=16$ when $a=-\frac{\pi}{12}$. For other $a$, there are no solutions. Answer to option $5-2: x=-16$ when $a=\frac{\pi}{12}$. For other $a$, there are no solutions.
Lomonosov Moscow State University
## Olympiad "Conquer Sparrow Hills"
Option $6-1$ (Nizhny Novgorod) | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let's introduce the function $g(x)=f(x)+(x-2021)^{2}-4$. For this function, the conditions $g(2019)=g(2020)=g(2021)=g(2022)=g(2023)=0$ are satisfied, meaning that the function $g(x)$ has 5 roots. Since it is a polynomial of the 5th degree, it has no other roots. Therefore,
$$
g(x)=(x-2019)(x-2020)(x-2021)(x-2022)(x... | Answer: -125. Answer to option: 7-2: -115. 7-3: 115. 7-4: 125. | -125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest possible value of $\left|2015 m^{5}-2014 n^{4}\right|$, given that $m, n$ are natural numbers. | Answer: 0.
Solution: Let's find $N=2014^{x} \cdot 2015^{y}$ such that $m^{5}=2014^{x-1} \cdot 2015^{y}$ and $n^{4}=2014^{x} \cdot 2015^{y-1}$. For this, $x$ and $y-1$ must be multiples of 4, and $x-1$ and $y-5$ must be multiples of 5. For example, $x=16$ and $y=5$ work. Then, if we take $m=2014^{3} \cdot 2015$ and $n=... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Find the largest three-digit number that is divisible by the sum of its digits and in which the first digit matches the third, but does not match the second. | Answer: 828.
Solution: Let this number be $\overline{a b a}=100 a+10 b+a$, where $a \neq b$. It must be divisible by $2 a+b$, so $101 a+10 b-10(2 a+b)=81 a$ is also divisible by $2 a+b$.
Since we need to find the largest such number, consider $a=9$. Then $81 a=729=3^{6}$, i.e., all divisors are powers of three, so $1... | 828 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the equation in natural numbers
$$
a b c + a b + b c + a c + a + b + c = 164
$$
In your answer, specify the product $a b c$. | Answer: 80.
Solution: $(a+1) \times(b+1) \times(c+1)=a b c+a b+b c+a c+a+b+c+1=$ $165=3 \times 5 \times 11$, therefore, $a=2, b=4$ and $c=10$. Note that the solution is unique up to the permutation of $a, b$ and $c$, since $3,5,11$ are prime numbers.
## 2013/2014 Academic Year CRITERIA FOR DETERMINING WINNERS AND PRI... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. On the board in the laboratory, two numbers are written. Every day, the senior researcher Pyotr Ivanovich erases both numbers from the board and writes down their arithmetic mean and harmonic mean ${ }^{2}$. In the morning of the first day, the numbers 1 and 2 were written on the board.
Find the product of the nu... | # Solution:
The product of the numbers on the board does not change.
Indeed, $\frac{a+b}{2} \times \frac{2}{\frac{1}{a}+\frac{1}{b}}=a b$. Therefore, the desired product is 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.6. In a certain company, there are 100 shareholders, and any 66 of them own no less than $50 \%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? | Solution. Let M be the shareholder owning the largest percentage of shares - x percent of shares. Divide the other 99 shareholders into three groups A, B, and
C, each with 33 shareholders. Let them own a, b, c percent of shares, respectively. Then
$$
2(100-x)=2(a+b+c)=(a+b)+(b+c)+(c+a) \geq 50+50+50
$$
That is, $x \... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. A ticket to the Historical Museum costs 300 rubles for an adult and 60 rubles for a schoolchild, while pensioners can visit the museum for free. There is also a
"family ticket" for two adults and two children, which costs 650 rubles. What is the minimum amount in rubles that a family, including a father, a mother... | Solution. All possible options are presented in the table.
| Grandmother | Father | Mother | 1st child | 1st child | 1st child | 1st child | Total cost, RUB |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| free | "adult" ticket 300 | "adult" ticket 300 | "child" 60 | "child" 60 | "child" 60 | "chi... | 770 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.4. Is it possible to measure out exactly 4 liters of water into the larger of the two containers, using only a 3-liter and a 5-liter container, from a tap? If so, provide an example of how to do it. | Solution. You can, fill up 5 liters, pour 3 liters into the second container, then pour out this water, leaving 2 liters in the 5-liter container. Transfer these 2 liters to the 3-liter container. Fill the 5-liter container to the top and top up the second container (exactly 1 liter), then you will have 4 liters left i... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. When one of two integers was increased 1996 times, and the other was reduced 96 times, their sum did not change. What can their quotient be | Solution. Let the first number be x, and the second y. Then the equation $1996 x+\frac{y}{96}=x+y$ must hold, from which we find that 2016x=y. Therefore, their quotient is 2016 or $\frac{1}{2016}$.
Answer: 2016 or $\frac{1}{2016}$.
Criteria: Full solution - 7 points; correct answer without solution 1 point. | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The function $f(x)$ is such that for all natural $n>1$ there exists a prime divisor $p$ of the number $n$ such that
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$
It is known that $f(1001)=1$. What is $f(1002) ?$ | Solution. Note that for any prime number p, the value $f(p)=f(1)-$ $f(p)$. Therefore, $f(p)=\frac{f(1)}{2}$ for any prime number. For prime numbers p and q, we get that either $\mathrm{f}(\mathrm{pq})=\mathrm{f}(\mathrm{p})-\mathrm{f}(\mathrm{q})=0$, or $\mathrm{f}(\mathrm{pq})=\mathrm{f}(\mathrm{q})-\mathrm{f}(\mathrm... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On a line, 3025 points are marked. The midpoints of every two of the marked points are painted green, blue, or red. Prove that the number of points painted in one of the colors on the line is at least 2016. | Solution. Let's start by considering three points. Obviously, for three points on a line, the midpoints of each pair of them are three different points. Consider the extreme point on the line. The midpoints between it and the two nearest points to it are two points that are not midpoints of any other points. If we remo... | 2016 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. When one of two integers was increased 1996 times, and the other was reduced 96 times, their sum did not change. What can their quotient be? | Solution. Let the first number be x, and the second y. Then the equation $1996 x + \frac{y}{96} = x + y$ must hold, from which we find that $2016 x = y$. Therefore, their quotient is 2016 or $\frac{1}{2016}$.
Answer: 2016 or $\frac{1}{2016}$.
Criteria: Full solution - 7 points; correct answer without solution 1 point... | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. At noon, a "Moskvich" left point A for point B. At the same time, "Zhiguli" left point B for point A on the same road. An hour later, the "Moskvich" was halfway between A and the "Zhiguli". When will it be halfway between the "Zhiguli" and B? (The speeds of the cars are constant and differ by less than a factor of... | Solution. Let the speeds of the "Moskvich" and "Zhiguli" be u and v, respectively. From the problem statement, it follows that if the speed of the "Moskvich" were $2 \mathrm{u}$, then its meeting with the "Zhiguli" (traveling at speed v) would occur one hour after the start of the journey. From this, it follows that if... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. The faces of a cube are labeled with six different numbers from 6 to 11. The cube was rolled twice. The first time, the sum of the numbers on the four side faces was 36, the second time - 33. What number is written on the face opposite the one where the digit $10$ is written? | Solution: The sum of the numbers on all faces is
$$
6+7+8+9+10+11=51
$$
On the first roll, the sum of the numbers on the top and bottom faces is 51 $-36=15$, on the second roll $-51-33=18$. Therefore, the sum on the third pair of opposite faces is $51-15-18=18$. The sum of 18 can be obtained in two ways: $11+7$ or $1... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.6. Ostap Bender put new tires on the car "Gnu Antelope". It is known that the front tires of the car wear out after 25,000 km, while the rear tires wear out after 15,000 km (the tires are the same both in the front and in the rear, but the rear ones wear out more). After how many kilometers should Ostap Bender swap t... | Solution. Let Ostap Bender swap the tires after x kilometers. Then the rear tires have used up [x/15000] of their resource, and the front tires [x/25000]. After the swap, they can work for another
$$
25000 \cdot\left(1-\frac{x}{15000}\right) \text { and } 15000 \cdot\left(1-\frac{x}{25000}\right)
$$
kilometers, respe... | 9375 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.3. To number the pages of a book, a total of 1392 digits were used. How many pages are in this book? | Solution. The first nine pages will require 9 digits, and for the next 90 pages, 2 digits are needed for each page, which means 2 * 90 digits are required. Let the book have x pages, then the pages with three digits will be x - 99, and the digits on them will be 3 * (x - 99). We get the equation: $9 + 2 \cdot 90 + 3 \c... | 500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. There are 30 logs with lengths of 3 and 4 meters, the total length of which is 100 meters. How many cuts can be made to saw the logs into logs of 1 meter length? (Each cut saws exactly one log.)
# | # Solution.
First solution. Glue all the logs into one 100-meter log.
To divide it into 100 parts, 99 cuts are needed, 29 of which have already been made.
Second solution. If there were $m$ three-meter logs and $n$ four-meter logs,
then $m+n=30, 3m+4n=100$, from which $m=20, n=10$. Therefore, $20 \cdot 2 + 10 \cdot... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.1. On the Island of Knights and Liars, knights always tell the truth, while liars always lie. One day, a traveler interviewed seven residents of the island.
- I am a knight, - said the first.
- Yes, he is a knight, - said the second.
- Among the first two, there are no less than 50% liars, - said the third.
- Among ... | Answer: 5.
Solution: Suppose the first inhabitant is a knight. Then the second is also a knight, while the third and fourth are liars. If, on the other hand, the first is a liar, then the second is also a liar, while the third and fourth are knights. In either case, among the first four, there are exactly two knights ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3.1. Find all pairs of two-digit natural numbers for which the arithmetic mean is $25 / 24$ times greater than the geometric mean. In your answer, specify the largest of the arithmetic means for all such pairs. | Answer: 75.
Solution: Let $a, b$ be the required numbers (without loss of generality, we can assume that $a > b$) and let $\frac{a+b}{2}=25x$ and $\sqrt{ab}=24x$. Then $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}=98x$ and $(\sqrt{a}-\sqrt{b})^2=$ $a+b-2\sqrt{ab}=2x$. From this, it follows that $\sqrt{a}+\sqrt{b}=7(\sqrt{a}-\... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. All natural numbers from 1 to 2017 inclusive were written in a row. How many times was the digit 7 written? | Answer: 602.
Solution. First, consider the numbers from 1 to 2000. Then the digit 7 can be in the third position from the end: numbers of the form $7 * *$ or $17 * *$ - there are 200 such numbers. It can be in the second position: $* 7 *$ or $1 * 7 *$ - there are also 200 such numbers; or the last position: $* * 7$ or... | 602 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. At the vertices of a cube, numbers $\pm 1$ are placed, and on its faces - numbers equal to the product of the numbers at the vertices of that face. Find all possible values that the sum of these 14 numbers can take. In your answer, specify their product. | Answer: -20160.
Solution. It is obvious that the maximum value of the sum is 14. Note that if we change the sign of one of the vertices, the sum of the numbers in the vertices will increase or decrease by 2. On the other hand, the signs of three faces will change. If their sum was $1, -1, 3, -3$, it will become $-1, 1... | -20160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. How many numbers from 1 to 1000 (inclusive) cannot be represented as the difference of two squares of integers | Answer: 250.
Solution. Note that any odd number $2n+1$ can be represented as $(n+1)^{2}-n^{2}$. Moreover, an even number that is a multiple of 4 can be represented as $4n=(n+1)^{2}-(n-1)^{2}$. The numbers of the form $4n+2$ remain. Note that a square can give remainders of 0 or 1 when divided by 4, so numbers of the f... | 250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. The sequence is defined by the relations $a_{1}=1$,
$$
a_{2 n}=\left\{\begin{array}{ll}
a_{n}, & \text { if } n \text { is even, } \\
2 a_{n}, & \text { if } n \text { is odd; }
\end{array} \quad a_{2 n+1}= \begin{cases}2 a_{n}+1, & \text { if } n \text { is even, } \\
a_{n}, & \text { if } n \text { is odd. }\en... | Answer: 5.
Solution: The given rules are easily interpreted in terms of the binary system: if $n$ ends in 0 and 1 is appended to the right, then 1 is appended to the right of $a_{n}$. If $n$ ends in 1 and 0 is appended, then 0 is appended to the right of $a_{n}$. In all other cases, $a_{n}$ does not change (when 0 is ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. On the Island of Knights and Liars, knights always tell the truth, while liars always lie. One day, a traveler interviewed seven residents of the island.
- I am a knight, - said the first.
- Yes, he is a knight, - said the second.
- Among the first two, there are no less than 50% liars, - said the third.
- Among ... | Answer: 5.
Solution: Suppose the first inhabitant is a knight. Then the second is also a knight, while the third and fourth are liars. If, on the other hand, the first is a liar, then the second is also a liar, while the third and fourth are knights. In either case, among the first four, there are exactly two knights ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.1. How many four-digit numbers exist that contain the digit 9 in their notation, immediately followed by the digit 5? | Answer: 279.
Solution. For numbers of the form $95 * *$, the last two digits can be anything - there are $10 \cdot 10=100$ such numbers, and for numbers of the form $* 95 *$ and $* * 95$, the first digit cannot be 0, so there are $10 \cdot 9=90$ of each. The number 9595 was counted twice, so we get 279 numbers. | 279 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. All natural numbers from 1 to 2017 inclusive were written in a row. How many times was the digit 7 written? | Answer: 602.
Solution. First, consider the numbers from 1 to 2000. Then the digit 7 can be in the third position from the end: numbers of the form $7 * *$ or $17 * *$ - there are 200 such numbers. It can be in the second position: $* 7 *$ or $1 * 7 *$ - there are also 200 such numbers; or the last position: $* * 7$ or... | 602 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. How many numbers from 1 to 1000 (inclusive) cannot be represented as the difference of two squares of integers | Answer: 250.
Solution. Note that any odd number $2 n+1$ can be represented as $(n+1)^{2}-n^{2}$. Moreover, an even number that is a multiple of 4 can be represented as $4 n=(n+1)^{2}-(n-1)^{2}$. The numbers of the form $4 n+2$ remain. Note that a square can give remainders of 0 or 1 when divided by 4, so numbers of th... | 250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3.1. Among all integer solutions of the equation $20 x+19 y=2019$, find the one for which the value of $|x-y|$ is minimal. In the answer, write the product $x y$. | Answer: 2623.
Solution. One of the solutions to the equation is the pair $x=100, y=1$. Therefore, the set of all integer solutions is $x=100-19 n, y=1+20 n, n \in \mathbb{Z}$. The absolute difference $|x-y|=$ $|100-19 n-1-20 n|=|99-39 n|$ is minimized when $n=3$, and the corresponding solution is $(x, y)=(43,61)$. We ... | 2623 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4.1. A teacher at a summer math camp took with him for the whole summer several shirts, several pairs of pants, several pairs of shoes, and two jackets. At each lesson, he wore pants, a shirt, and shoes, and he wore a jacket on some lessons. On any two lessons, at least one of the items of his clothing or shoes was dif... | Answer: 126.
Solution: Let the teacher bring $x$ shirts, $y$ pairs of trousers, $z$ pairs of shoes, and 2 jackets. Then he can conduct $3 x y z$ lessons (the number 3 means: 2 lessons in each of the jackets and 1 lesson without a jacket). If he has one more shirt, the number of lessons will increase by $3 y z$. If he ... | 126 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4.4. A teacher at a summer math camp took with him for the whole summer several shirts, several pairs of pants, several pairs of shoes, and two jackets. At each lesson, he wore pants, a shirt, and shoes, and he wore a jacket on some lessons. On any two lessons, at least one of the items of his clothing or shoes was dif... | Answer: 216.
Note. In versions $4.2,4.3,4.4$, answers $252,360,420$ respectively are also counted as correct. | 216 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. It is known that $P(x)$ is a polynomial of degree 9 and $P(k)=2^{k}$ for all $k=1,2,3, \ldots, 10$. Find $P(12)$. | Answer: 4072.
Solution. Let $P(x)$ be a polynomial of degree $n$ and $P(k)=2^{k}$ for all $k=1,2,3, \ldots, n+1$. We will find $P(n+m+2), m=0,1, \ldots$. By the binomial theorem for any $k \in \mathbb{N}$ we have
$$
2^{k}=2 \cdot(1+1)^{k-1}=2 \sum_{i=0}^{k-1} C_{k-1}^{i}=2 \sum_{i=0}^{k-1} \frac{(k-1)(k-2) \ldots(k-i... | 4072 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. The sequences $\left\{x_{n}\right\},\left\{y_{n}\right\}$ are defined by the conditions $x_{1}=11, y_{1}=7, x_{n+1}=3 x_{n}+2 y_{n}$, $y_{n+1}=4 x_{n}+3 y_{n}, n \in \mathbb{N}$. Find the remainder of the division of the number $y_{1855}^{2018}-2 x_{1855}^{2018}$ by 2018. | Answer: 1825.
Solution. For all $n \in \mathbb{N}$, the numbers $x_{n}$ and $y_{n}$ are odd. Notice that
$$
2 x_{n+1}^{2}-y_{n+1}^{2}=2\left(3 x_{n}+2 y_{n}\right)^{2}-\left(4 x_{n}+3 y_{n}\right)^{2}=2 x_{n}^{2}-y_{n}^{2}
$$
Therefore, $2 x_{n}^{2}-y_{n}^{2}=\ldots=2 x_{1}^{2}-y_{1}^{2}=242-49=193$.
Let $p=1009$. ... | 1825 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. How many weeks can a year overlap? Assume that a year overlaps with a week if at least 6 and one day of this week falls within the given year. | Answer: On the 53rd 54th week. Solution. If there are 365 days in a year (a non-leap year), then since $365=52 \cdot 7+1$, there are no fewer than 53 weeks in a year. If it is a leap year, meaning there are 366 days in a year ( $366=52 \cdot 7+2$ ), there is a situation where the year starts with the last day of the we... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. How many numbers divisible by 4 and less than 1000 do not contain any of the digits $6,7,8,9$ or 0. | Answer: 31. Solution. According to the condition, these numbers consist only of the digits 1, 2, 3, 4, and 5. Out of such numbers, only one single-digit number is divisible by 4: 4. Among the two-digit numbers, the following are divisible by 4: $12, 24, 32, 44, 52$. If we prepend 1, 2, 3, 4, or 5 to all these two-digit... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In a right triangle $ABC$ with a right angle at $C$, the bisector $BD$ and the altitude $CH$ are drawn. A perpendicular $CK$ is dropped from vertex $C$ to the bisector $BD$. Find the angle $HCK$, if $BK: KD=3: 1$. | Answer: $30^{\circ}$.
Solution. Let $M$ be the midpoint of $B D$. Then $C M$ is the median of the right triangle $C B D$ and $C M=M B=M D$. In addition, $C K$ is the height and median of triangle $M C D$, so $M C=C D$ and triangle $C M D$ is equilateral. Then $\angle C D M=60^{\circ}$ and $\angle C B M=$ $\angle C B D... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Vanya thought of a two-digit number, then swapped its digits and multiplied the resulting number by itself. The result turned out to be four times larger than the number he thought of. What number did Vanya think of? | Answer: 81.
Solution. Let the intended number be $\overline{m n}=10 m+n$. Then $4 \overline{m n}=\overline{n m}^{2}$. Therefore, $\overline{n m}^{2}$ is divisible by 4, and $\overline{n m}$ is divisible by 2, so the digit $m$ is even (and not zero). Moreover, $\overline{m n}=\overline{n m}^{2}: 4=(\overline{n m}: 2)^{... | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the lateral sides $AB$ and $BC$ of an isosceles triangle $ABC$, points $M$ and $N$ are marked such that $AM = MN = NC$. On the side $AC$, points $P$ and $Q$ are chosen such that $MQ \parallel BC$ and $NP \parallel AB$. It is known that $PQ = BM$. Find the angle $MQB$. | Answer: $36^{\circ}$. The answer can also be given in the form $\arccos \left(\frac{\sqrt{5}+1}{4}\right)$, etc.
Solution. Let $\angle A=\angle C=\alpha$, then by the property of parallel lines $\angle N P C=$ $\angle A Q M=\alpha$. It is not difficult to prove that $M N \| A C$ (through the similarity of triangles, o... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.1. If the 200th day of some year is Sunday and the 100th day of the following year is also Sunday, then what day of the week was the 300th day of the previous year? Enter the number of this day of the week (if Monday, then 1, if Tuesday, then 2, etc.). | Answer. 1.
Solution. Between the $200-\mathrm{th}$ day of the year and the $100-\mathrm{th}$ day of the next year, there are either $165+100=265$ or $166+100=266$ days. The number 266 is divisible by 7, while the number $265-$ is not. Therefore, the current year has 366 days, meaning it is a leap year. Thus, the previ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.1. How many terms will there be if we expand the expression $\left(4 x^{3}+x^{-3}+2\right)^{2016}$ and combine like terms? | Answer: 4033.
Solution. In the resulting sum, there will be monomials of the form $k_{n} x^{3 n}$ for all integers $n \in[-2016 ; 2016]$ with positive coefficients $k_{n}$, i.e., a total of $2 \cdot 2016+1=4033$ terms. | 4033 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. The function $f$ satisfies the equation $(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$ for each value of $x$, not equal to 0 and 1. Find $f\left(\frac{2016}{2017}\right)$. | Answer: 2017.
Solution. Substitute $\frac{1}{x}$ for $x$ in the equation. Together with the original equation, we get a system of two linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$:
$$
\left\{\begin{array}{l}
(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1} \\
\left(\frac{1}{x}-1\right) f\left(... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. From point $A$ to point $B$, which are 10 km apart, a car departed at 7:00. After traveling $2 / 3$ of the distance, the car passed point $C$, from which a cyclist immediately set off for point $A$. As soon as the car arrived at $B$, a bus immediately departed from $B$ in the opposite direction and arrived at $A$ ... | Answer: 6.
Solution. Let's plot the graphs of the car's movement (segment $K L$), the bus's movement (segment $L M$), and the cyclist's movement (segment $N P$) on the axes $(t ; s)$, where $t$ is time (in hours) and $s$ is the distance (in kilometers) from point $A$. Let $Q$ be the intersection point of $L M$ and $N ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. Find the sum of all integers \( x \in [-3 ; 13] \) that satisfy the inequality
$$
\left(1-\operatorname{ctg}^{2} \frac{\pi x}{12}\right)\left(1-3 \operatorname{ctg}^{2} \frac{\pi x}{12}\right)\left(1-\operatorname{tg} \frac{\pi x}{6} \cdot \operatorname{ctg} \frac{\pi x}{4}\right) \leqslant 16
$$ | Answer: 28.
Solution. Let $t=\frac{\pi x}{12}$, then the inequality takes the form
$$
\left(1-\operatorname{ctg}^{2} t\right)\left(1-3 \operatorname{ctg}^{2} t\right)(1-\operatorname{tg} 2 t \cdot \operatorname{ctg} 3 t) \leqslant 16
$$
Since
$$
\begin{gathered}
1-\operatorname{ctg}^{2} t=-\frac{\cos 2 t}{\sin ^{2}... | 28 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find the smallest natural number $n$ for which the decimal representation of $n$ together with $n^{2}$ uses all the digits from 1 to 9 exactly once. | Answer: 567.
Solution: If $n \geqslant 1000$, then $n^{2} \geqslant 10^{6}$, so the decimal representations of $n$ and $n^{2}$ together contain at least 11 digits. If, however, $n \leqslant 316$, then $n^{2} \leqslant 99856$, and the decimal representations of $n$ and $n^{2}$ together contain no more than 8 digits. Th... | 567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find a two-digit number, the digits of which are different and the square of which is equal to the cube of the sum of its digits. | Answer: 27.
Solution: Let's write the condition as $\overline{A B}^{2}=(A+B)^{3}$. Note that the sum of the digits $(A+B)$ does not exceed 17, i.e., $\overline{A B}^{2} \leq 17^{3}$. Moreover, this number $\overline{A B}^{2}=n^{6}$, where $n$ is some natural number that does not exceed 4. However, 1 and 2 do not work ... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On a circle, 100 points are marked, painted either red or blue. Some of the points are connected by segments, with each segment having one blue end and one red end. It is known that no two red points belong to the same number of segments. What is the maximum possible number of red points? | Answer: 50.
Solution: Take 50 red and 50 blue points. The first red point is not connected to any other, the second is connected to one blue, ..., the 50th is connected to 49 blues. Obviously, there cannot be more than 50 red points, because if there are 51 or more, then there are no more than 49 blues, hence the numb... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. How many diagonals in a regular 32-sided polygon are not parallel to any of its sides | Answer: 240.
Solution: In a 32-sided polygon, there are $32*(32-3)/2=464$ diagonals in total. We can divide the sides into 16 pairs of parallel sides. It is not hard to notice that if we fix a pair, i.e., 4 vertices, the remaining vertices can be connected in pairs by diagonals parallel to this pair. There will be a t... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. For natural numbers $m$ and $n$, it is known that $3 n^{3}=5 m^{2}$. Find the smallest possible value of $m+n$. | Answer: 60.
Solution: Obviously, if $m, n$ contain prime factors other than 3 or 5, then they can be canceled out (and reduce $\mathrm{m}+\mathrm{n}$).
Let $n=3^{a} \cdot 5^{b}, m=3^{c} \cdot 5^{d}$. Then the condition implies that $3 a+1=2 c, 3 b=2 d+1$. The smallest possible values are: $a=1, b=1, c=2, d=1$, from w... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, a tourist met five inhabitants of the island and asked them: "How many liars are there among you?" The first answered: "One," the second answered: "Two," the third answered: "Three," the fourth answered: "... | Answer: 4.
Solution: Among the islanders surveyed, exactly one is a knight, since they all gave different, and all possible answers. Then the remaining four are liars. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.1. On a grid paper, a square made up of several cells is shaded, with its sides lying on the grid lines. It is known that to get a larger square under the same condition, 47 more cells need to be shaded. Find the side length of the original square. | Answer: 23.
Solution. If the side of the original square was $n$, and the side of the obtained square became larger by $k$, then to obtain it, one needs to color $(n+k)^{2}-n^{2}=2 n k+k^{2}$ cells, i.e., $2 n k+k^{2}=47$. Therefore, $k$ is odd, and $k^{2}<47$, so $k \leqslant 5$. If $k=5$, then $10 n+25=47$, and $n$ ... | 23 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3.1. At a sumo wrestling tournament, 20 sumo-tori (sumo wrestlers) arrived. After weighing, it was found that the average weight of the sumo-tori is 125 kg. What is the maximum possible number of wrestlers who weigh more than 131 kg, given that according to sumo rules, people weighing less than 90 kg cannot participate... | Answer: 17.
Solution: Let $n$ be the number of sumo-tori that weigh more than 131 kg. Their total weight is more than $131 n$ kg, and the total weight of the remaining ones is no less than $90(20-n)$. Therefore, $\frac{131 n+90(20-n)}{20}<125$, from which $41 n<35 \cdot 20$, i.e., $n<\frac{700}{41}=17 \frac{3}{41}$. T... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five... | Answer: 7111765.
Solution. Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive... | 7111765 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.1. From one point on a circular track, a pedestrian and a cyclist started simultaneously in the same direction. The cyclist's speed is $55\%$ greater than the pedestrian's speed, and therefore the cyclist overtakes the pedestrian from time to time. At how many different points on the track will the overtakes occur? | Answer: 11.
Solution: Let's assume the length of the path is 55 (in some units), and the speeds of the pedestrian and the cyclist are $100 x$ and $155 x$. Then, the overtakes will occur every $1 / x$ units of time. During this time, the pedestrian covers 100 units, which means he will be 10 units away in the opposite ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. We will call a natural number interesting if all its digits, except the first and last, are less than the arithmetic mean of the two adjacent digits. Find the largest interesting number. | Answer: 96433469.
Solution. Let $a_{n}$ be the n-th digit of the desired number. By the condition, $a_{n}9$. Similarly, there cannot be four negative differences. Therefore, the maximum number of such differences can be 7, and the desired number is an eight-digit number. To make it the largest, the first digit should ... | 96433469 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. Find the smallest natural solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}>2^{2017}$. Answer. 17. | Solution. The value $x=17$ is clearly a solution. If $x \leqslant 16$, then we have
$$
2^{x}+2^{x+1}+\ldots+2^{x+2000} \leqslant 2^{16}+2^{17}+\ldots+2^{2015}+2^{2016}<2^{2017}
$$
since this inequality reduces sequentially to the following: $2^{16}+2^{17}+\ldots+2^{2015}<$ $2^{2017}-2^{2016}=2^{2016}, 2^{16}+2^{17}+\... | 17 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. First-grader Petya was laying out a contour of an equilateral triangle with the tokens he had, so that each of its sides, including the vertices, contained the same number of tokens. Then, with the same tokens, he managed to lay out the contour of a square in the same way. How many tokens does Petya have, if each si... | Answer: 24.
Solution. Let the side of the triangle contain $x$ chips, and the side of the square - $y$ chips. The total number of chips, counted in two ways, is $3 x-3=4 y-4$ (we account for the corner chips being counted twice). From the problem statement, it follows that $y=x-2$. Therefore, we get the equation $3(x-... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petrov and Vasechkin were solving the same arithmetic problem. A certain number had to be divided by 2, multiplied by 7, and 1001 subtracted. Petrov performed all the operations correctly, while Vasechkin got everything mixed up: he divided by 8, squared the result, and also subtracted 1001. It is known that Petrov ... | Answer: 295.
Solution. Note that the number 1001 is divisible by 7 without a remainder. This means that the number Petrov obtained must be divisible by 7. But it is a prime number, so Petrov got 7. Let's reverse the operations Petrov performed and find the original number: $\frac{7+1001}{7} \cdot 2=288$. Now, let's re... | 295 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The numbers from 1 to 8 are arranged at the vertices of a cube such that the sum of the numbers in any three vertices lying on the same face is at least 10. What is the smallest possible sum of the numbers at the vertices of one face? | Answer: 16.
Solution. Each face has a vertex where a number not less than 6 is placed. Indeed, otherwise, one of the triples of the remaining largest numbers $2,3,4,5$ would give a sum less than 10 (specifically, the triple $2,3,4$ with a sum of 9).
Consider a face containing the vertex where the number 6 is placed. ... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. On graph paper (the side of a cell is 1 cm) a rectangle is drawn, the sides of which lie on the grid lines, with one side being 5 cm shorter than the other. It turned out that it can be cut into several pieces along the grid lines and form a square from them. What can the side of this square be? Find all possible va... | # Answer: 6 cm.
Solution. Let the larger side of the rectangle be $k$ cm, $k>5$, and the side of the obtained square be $n$ cm. Then, from the equality of areas, we get $k(k-5)=n^{2}$. Note that $k^{2}-5k=k^{2}-6k+9=(k-3)^{2}$ for $k>9$. Therefore, $6 \leqslant k \leqslant 9$. For $k=6,7,8,9$, we get that $k(k-5)$ equ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. All natural numbers, the sum of the digits of each of which is equal to 5, were arranged in ascending order. What number is in the 125th place?
Otvet: 41000. | Solution. Let's calculate the number of $n$-digit numbers, the sum of the digits of each of which is equal to 5, for each natural $n$. Subtract 1 from the leading digit, we get a number (which can now start with zero), the sum of the digits of which is equal to 4. Represent the digits of this number as cells, in each o... | 41000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1. Find the integer part of the number $a+\frac{9}{b}$, where $a$ and $b-$ are respectively the integer and fractional part of the number $\sqrt{76-42 \sqrt{3}}$. | Answer: 12.
Solution. The given number is $\sqrt{76-42 \sqrt{3}}=\sqrt{(7-3 \sqrt{3})^{2}}=7-3 \sqrt{3}=1+(6-3 \sqrt{3})$, where $6-3 \sqrt{3} \in(0 ; 1)$. Therefore, $a=1, b=6-3 \sqrt{3}$. Thus, $a+\frac{9}{b}=1+\frac{9}{6-3 \sqrt{3}}=1+\frac{3}{2-\sqrt{3}}=1+3(2+\sqrt{3})=7+3 \sqrt{3}$. Since $12<7+3 \sqrt{3}<13$, t... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. For the sequence $\left\{a_{n}\right\}$, it is known that $a_{1}=1.5$ and $a_{n}=\frac{1}{n^{2}-1}$ for $n \in \mathbb{N}, n>1$. Are there such values of $n$ for which the sum of the first $n$ terms of this sequence differs from 2.25 by less than 0.01? If yes, find the smallest one. | Answer: yes, $n=100$.
Solution. The general formula for the terms of the sequence (except the first) can be written as $(n \geqslant 2)$:
$$
a_{n}=\frac{1}{n^{2}-1}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)
$$
As a result, the sum of the first $n$ terms of the sequence, except the first, takes the form:
$... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. How many values of the parameter $a$ exist for which the equation
$$
4 a^{2}+3 x \lg x+3 \lg ^{2} x=13 a \lg x+a x
$$
has a unique solution | Answer: 2.
Solution. The given equation is equivalent to the equation $(3 \lg x-a)(x+\lg x-4 a)=0$, from which we have the system
$$
\left[\begin{array}{l}
\lg x=\frac{a}{3} \\
x+\lg x=4 a
\end{array}\right.
$$
Each equation in this system has a unique positive solution for any $a$, since the continuous functions $f... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Can non-negative integers be placed on the faces of two cubes so that when they are randomly thrown, the sum of the points showing can be equal to any divisor of the number 36? If this is possible, indicate the sum of all 12 numbers on the faces; if it is impossible - indicate 0. | Answer: The sum is 111. Solution. The numbers $1,2,3,4,5$ and 6 can be placed on the faces of one die, and $0,6,12,18,24,30$ on the faces of the other. Then the sum of the points rolled can be equal to any of the numbers from 1 to 36. The sum of the numbers on all faces is 111. This sum does not depend on how the dice ... | 111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find all three-digit numbers $\overline{L O M}$, composed of distinct digits $L, O$ and $M$, for which the following equality holds:
$$
\overline{L O M}=(L+O+M)^{2}+L+O+M
$$ | Answer: 156. Instructions. Let $x=L+O+M$. Then $\overline{L O M}=x(x+1)$. In this case, $x \geq 10$ (otherwise $x(x+1)<100)$ and $x \leq 24$ (the sum of digits does not exceed $9+8+7=24$). Therefore, $x \in[10 ; 24]$. From the relation $100 \cdot L+10 \cdot O+M=x^{2}+L+O+M$ it follows that $x^{2}=99 \cdot L+9 \cdot O$,... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Pete was given a new electric jigsaw for his birthday, with a function to count the length of the cuts made. To try out the gift, Pete took a square piece of plywood with a side of $50 \mathrm{cm}$ and cut it into squares with sides of 10 cm and squares with sides of 20 cm. How many squares in total were obtained, i... | Answer: 16. Solution. Each protruding edge is part of the perimeter of two figures, in addition, the perimeter of the original square must be taken into account, from which we get that the total perimeter of the resulting small squares is $280 \cdot 2+200=760$. Now we can denote the number of squares through $x$ and $y... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. How many four-digit numbers exist that have the following properties: all digits of the number are even; the number is divisible by four, if the last digit is erased, the resulting three-digit number is not divisible by four? | Answer: 120. Solution. The last digit of the number must be divisible by 4 (that is, it can be 0, 4 or 8), and the second to last - not (2 or 6). In addition, the first digit is not zero. Therefore, we get $4 \cdot 5 \cdot 2 \cdot 3=120$ variants. | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.