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7. Find the maximum value of the expression
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2010}-x_{2011}\right)^{2}+\left(x_{2011}-x_{1}\right)^{2}
$$
for $x_{1}, \ldots, x_{2011} \in[0 ; 1]$.
#
|
# Answer: 2010.
Let's prove this using mathematical induction for $2n+1$ numbers $x_{1}, \ldots, x_{2n+1} \in [0; 1]$. Specifically, we will show that the maximum value of the expression
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2n}-x_{2n+1}\right)^{2}+\left(x_{2n+1}-x_{1}\right)^{2}
$$
for $x_{1}, \ldots, x_{2n+1} \in [0; 1]$ is $2n$.
First, note that for $x_{1}, \ldots, x_{2n+1} \in [0; 1]$, the following inequality always holds:
$$
\begin{aligned}
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2n}-x_{2n+1}\right)^{2}+\left(x_{2n+1}-x_{1}\right)^{2} \\
& \leqslant \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{1}\right|
\end{aligned}
$$
1) Let's check that for any numbers $x, y, z$ such that $x, y, z \in [0; 1]$, the inequality
$$
|x-y|+|y-z|+|z-x| \leqslant 2
$$
holds, and there exist numbers $x, y, z \in [0; 1]$ such that $|x-y|+|y-z|+|z-x|=2$. Indeed, without loss of generality, assume that $0 \leqslant x \leqslant y \leqslant z$. Then
$$
|x-y|+|y-z|+|z-x|=y-x+z-y+z-x=2(z-x) \leqslant 2
$$
Equality is achieved, for example, if $x=y=0, z=1$.
2) Assume that for $2n-1$ numbers, the maximum value of the expression
$$
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n-2}-x_{2n-1}\right|+\left|x_{2n-1}-x_{1}\right|
$$
is $2n-2$ for $x_{1}, \ldots, x_{2n-1} \in [0; 1]$.
3) Consider the expression for $2n+1$ numbers:
$$
\begin{gathered}
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n-2}-x_{2n-1}\right|+\left|x_{2n-1}-x_{2n}\right|+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{1}\right|= \\
=\underbrace{\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n-2}-x_{2n-1}\right|+\left|x_{2n-1}-x_{1}\right|}_{\leqslant 2n-2}-\left|x_{2n-1}-x_{1}\right|+\left|x_{2n-1}-x_{2n}\right|+ \\
+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{1}\right| \leqslant 2n-2-\left|x_{2n-1}-x_{1}\right|+\underbrace{\left|x_{2n-1}-x_{2n}\right|+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{2n-1}\right|}_{\leqslant 2}- \\
\quad-\left|x_{2n+1}-x_{2n-1}\right|+\left|x_{2n+1}-x_{1}\right|=2n+\left|x_{2n+1}-x_{1}\right|-\left|x_{2n-1}-x_{1}\right|-\left|x_{2n+1}-x_{2n-1}\right| \leqslant 2n
\end{gathered}
$$
since $\left|x_{2n+1}-x_{1}\right| \leqslant \left|x_{2n-1}-x_{1}\right|+\left|x_{2n+1}-x_{2n-1}\right|$.
The value $2n$ of the expression
$$
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\ldots+\left|x_{2n-2}-x_{2n-1}\right|+\left|x_{2n-1}-x_{2n}\right|+\left|x_{2n}-x_{2n+1}\right|+\left|x_{2n+1}-x_{1}\right|
$$
is achieved, for example, when $x_{1}=x_{3}=\ldots=x_{2n-1}=x_{2n+1}=1, x_{2}=x_{4}=\ldots=x_{2n}=0$.
In particular,
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{2n}-x_{2n+1}\right)^{2}+\left(x_{2n+1}-x_{1}\right)^{2}=2010
$$
when $x_{1}=x_{3}=\ldots=x_{2011}=x_{2n+1}=1, x_{2}=x_{4}=\ldots=x_{2010}=0$.
|
2010
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. There are 12 pencils of pairwise different lengths. In how many ways can they be placed in a box in 2 layers of 6 pencils each, so that in each layer the pencils are arranged in increasing order of length (from left to right), and each pencil in the upper layer lies strictly above a pencil in the lower layer and is shorter than it?
|
Answer: 132
Let $0 \leqslant m \leqslant n \leqslant 6$. Denote by $K(m, n)$ the set of all arrangements of $m+n$ pencils of different lengths with the conditions: 1) in the bottom row, there are $n$ pencils, starting from the right edge without gaps in decreasing order of length; 2) in the top row, there are also $m$ pencils; 3) each pencil in the top row is shorter than the pencil directly below it. Let $k(m, n)$ denote the number of such arrangements. In each arrangement from $K(m, n)$, the shortest pencil lies either as the leftmost in the bottom row (then $m<n$), or as the leftmost in the top row. Removing it, we get a correct arrangement from the set $K(m, n-1)$ in the first case, and a correct arrangement from the set $K(m-1, n)$ in the second case. Conversely, adding a short pencil to the bottom layer of any arrangement from $K(m, n-1)$ or to the top layer of any arrangement from $K(m-1, n)$ (only if $m-1<n$) will result in a correct arrangement from the set $K(m, n)$. From this, we get that $k(m, n)=k(m-1, n)+k(m, n-1)$, if $m<n$, and $k(m, n)=k(m-1, n)$, if $m=n$. In this case, $k(0, n)=1$ for all $n$.
Consider a grid board of size $7 \times 7$. Number the rows from bottom to top with numbers $0,1, \ldots, 6$, and the columns from left to right with numbers $0,1, \ldots, 6$. In the cell located in the row with number $m$ and in the column with number $n$, we will place the number $k(m, n)$. Using the above equality, fill in the table, moving from the bottom left corner to the top right. We get the table:
| | | | | | | 132 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | | | | | 42 | 132 |
| | | | | 14 | 42 | 90 |
| | | | 5 | 14 | 28 | 48 |
| | | 2 | 5 | 9 | 14 | 20 |
| | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
We get that $k(6,6)=132$.
|
132
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers?
|
Answer: 18.
Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces.
|
18
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The Slytherin faculty has 30 students. Some of them are friends (friendship is mutual), but there are no 3 people who are all friends with each other. On New Year's, everyone sent cards to all their friends. What is the maximum number of cards that could have been sent?
|
Answer: 450.
Solution: Let's find the person with the most friends. Suppose there are no fewer than 15, and denote their number as $15+a$. We will divide the students into two groups: the first group will consist of these $15+a$ students. According to the condition, they cannot be friends with each other, so each of them has no more than 15-a friends. The second group will consist of the remaining 15-a, each of whom has no more than $15+a$ friends. Thus, each group will send no more than 225- $a^{2}$ cards. Therefore, in total, no more than 450-2 $a^{2}$ cards will be sent, which does not exceed 450.
Note that this value is achievable. Divide the students into two groups of 15 people each, and let each representative of one group be friends with all representatives of the other group.
|
450
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given a sequence of natural numbers $a_{n}$, the terms of which satisfy the relations $a_{n+1}=k \cdot \frac{a_{n}}{a_{n-1}}$ (for $n \geq 2$). All terms of the sequence are integers.
It is known that $a_{1}=1$, and $a_{2018}=2020$. Find the smallest natural $k$ for which this is possible.
|
Answer: 2020
Solution: Let a2=x. Then all terms of the sequence will have the form $x^{m} k^{n}$.
The powers of $k$ will repeat with a period of 6: $0,0,1,2,2,1,0,0, \ldots$
The powers of $x$ will also repeat with a period of 6: $0,1,1,0,-1,-1,0,1, \ldots$
Since 2018 gives a remainder of 2 when divided by 6, then $a_{2018}=a_{2}=x=2020$. For all terms of the sequence to be integers, it is necessary for $k$ to be a multiple of $x$, the smallest such $k$ is 2020.
|
2020
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Points $A_{1}, \ldots, A_{12}$ are the vertices of a regular 12-gon. How many different 11-segment open broken lines without self-intersections with vertices at these points exist? Broken lines that can be transformed into each other by rotation are considered the same.
|
Answer: 1024.
Solution: The first vertex of the broken line can be chosen in 12 ways. Each subsequent vertex (except the last one) can be chosen in two ways - it must be adjacent to the already marked vertices to avoid self-intersections. The last vertex is chosen uniquely. We get $12^{*} 2^{10}$ ways. Considering 12 possible rotations, we get that each broken line will be counted 12 times, so this number must be divided by 12.
Remark: Here, it was implied in the condition that the broken line has a starting and ending point. If, however, we consider broken lines as geometric objects, i.e., without a distinguished "head" and "tail," this significantly complicates the problem.
|
1024
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square. Answer: 183
|
Solution: Let the required number be \(a\), then \(1000a + a + 1 = n^2\). We can write it as: \(1001a = (n - 1)(n + 1)\). Factorize \(1001 = 7 \times 11 \times 13\), so the product \((n - 1)(n + 1)\) must be divisible by 7, 11, and 13. Moreover, for the square to be a six-digit number, \(n\) must be in the interval \([317; 999]\).
Consider the following cases:
a) \(n - 1\) is divisible by 143, \(n + 1\) is divisible by 7, then we find \(n = 573\);
b) \(n - 1\) is divisible by 7, \(n + 1\) is divisible by 143, then \(n = 428\);
c) \(n - 1\) is divisible by 77, \(n + 1\) is divisible by 13, then \(n = 155\) - does not fit;
d) \(n - 1\) is divisible by 13, \(n + 1\) is divisible by 77, then \(n = 846\);
e) \(n - 1\) is divisible by 91, \(n + 1\) is divisible by 11, then \(n = 274\) - does not fit;
f) \(n - 1\) is divisible by 11, \(n + 1\) is divisible by 91, then \(n = 727\).
The smallest \(n = 428, n^2 = 428^2 = 183184\).
|
183
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
17. In a regular 1000-gon, all diagonals are drawn. What is the maximum number of diagonals that can be selected such that among any three of the selected diagonals, at least two have the same length?
|
Answer: 2000
Solution: For the condition of the problem to be met, it is necessary that the lengths of the diagonals take no more than two different values. The diagonals connecting diametrically opposite vertices are 500. Any other diagonal can be rotated to coincide with a diagonal of the corresponding length, i.e., there are 1000 of them. Therefore, 2000 can be chosen while satisfying the condition.
|
2000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. In how many ways can the number 1024 be factored into three natural factors such that the first factor is divisible by the second, and the second is divisible by the third?
|
Answer: 14
Solution: Note that the factors have the form $2^{a} \times 2^{b} \times 2^{c}$, where $\mathrm{a}+\mathrm{b}+\mathrm{c}=10$ and $a \geq b \geq c$. Obviously, $c$ is less than 4, otherwise the sum would be greater than 10. Let's consider the cases:
$c=0)$ Then $b=0, \ldots, 5, a=10-b-6$ options
$c=1)$ Then $b=1, . .4, a=9-b-4$ options
$c=2) b=2,3,4, a=8-b-3$ options
$c=3) b=3, a=4-1$ option.
In total, $6+4+3+1=14$ options.
In a trapezoid, the diagonals of which intersect at a right angle, it is known that the midline is 6.5 and one of the diagonals is 12. Find the second diagonal.
## Answer 5
Solution: Parallel translate one of the diagonals so that it forms a right triangle with the other. Then in this triangle, one leg is 12, and the hypotenuse is 13, so the remaining leg is 5.
|
14
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
19. Find the last two digits of the sum
$$
1^{2}+2^{2}+\ldots+50^{2}-51^{2}-\ldots-100^{2}+101^{2}+\ldots 150^{2}-151^{2}-\ldots 200^{2}+\ldots-2000^{2}+2001^{2}+\ldots+2017^{2}
$$
(i.e., 50 numbers with a plus sign, 50 with a minus sign, and so on.)
|
# Answer: 85
Solution: Note that the numbers $\mathrm{n}^{2}$ and $(\mathrm{n}+50)^{2}$ give the same remainder when divided by 100. Therefore, in each hundred, the sum will end in two zeros. The last digits of the squares from 2001 to 2017 are:
| 01 | 04 | 09 | 16 | 25 | 36 | 49 | 64 | 81 | 00 | 21 | 44 | 69 | 96 | 25 | 56 | 89 , which in sum |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
gives 685.
|
85
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20. Find the sum of the fourth powers of the real roots of the equation
$$
x^{4}-1000 x^{2}+2017=0
$$
|
# Answer: 1991932
Solution: The roots of the equation are of the form $\pm \sqrt{t_{1}}, \pm \sqrt{t_{2}}$, where $t_{1,2}$ are the roots of the equation $t^{2}-1000 t+2017=0$. Therefore, the sum of the fourth powers is $2\left(t_{1}^{2}+t_{2}^{2}\right)=2\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}=2 \cdot 1000^{2}-4 \cdot 2017=1991932$ - here we used Vieta's formulas.
## Variant 3-a
|
1991932
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. We will call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist?
|
Answer 36.
Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, otherwise the product will be at least 121. It is sufficient to check $p_{1}=2,3,5,7$. For 2, we get the second factor: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, or 47 - 14 options, for 3 we get: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 - 10 options, for 5: 3, 5, 7, 11, 13, 17, 19 - 7 options, and for 7: 3, 5, 7, 11, 13 - 5 options. In total, $14+10+7+5=36$ options.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Arrange the smallest square area using square tiles of sizes $1 \times 1$, $2 \times 2$, and $3 \times 3$, such that the number of tiles of each size is the same.
|
Answer:
Solution. Let $n$ be the number of squares of each type. Then $n + 4n + 9n = 14n$ must be a perfect square. The smallest $n$ for which this is possible is 14. The figure shows an example of how to construct a 14x14 square (other arrangements are possible, as long as there are 14 squares of each type).
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The company conducted a survey among its employees - which social networks they use: VKontakte or Odnoklassniki. Some employees said they use VKontakte, some - Odnoklassniki, some said they use both social networks, and 40 employees said they do not use social networks. Among all those who use social networks, 75% use VKontakte, and 65% use both networks. The proportion of employees who use Odnoklassniki from the total number of all employees is 5/6. How many employees work in the company
|
Answer: 540
Solution: Since 75% of social media users use VKontakte, it follows that only 25% use Odnoklassniki. Additionally, 65% use both networks, so in total, Odnoklassniki is used by $65+25=90\%$ of social media users. These $90\%$ constitute $5 / 6$ of the company's employees, so $100\%$ constitutes $10 / 9 * 5 / 6 = 50 / 54$ of all employees. Therefore, those who do not use social media constitute 1-50/54 = 4/54, and there are 40 such people. Thus, the total number of employees is 540.
|
540
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a regular 2017-gon, all diagonals are drawn. Petya randomly selects some $\mathrm{N}$ diagonals. What is the smallest $N$ such that among the selected diagonals, there are guaranteed to be two of the same length?
|
Answer: 1008.
Solution: Let's choose an arbitrary vertex and consider all the diagonals emanating from it. There are 2014 of them, and by length, they are divided into 1007 pairs. Clearly, by rotating the polygon, any of its diagonals can be aligned with one of these. Therefore, there are only 1007 different sizes of diagonals. Thus, by choosing 1008, Petya is guaranteed to get at least two of the same.
|
1008
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. How many zeros does the number
$$
(1 \underbrace{000 \ldots 001}_{2017 \text { zeros }})^{2017}-1 ?
$$
end with?
|
Answer 2018
Solution: Factorize:
$(1 \underbrace{000 \ldots 001}_{2017 \text { zeros }})^{2017}-1=(1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1-1) \times(1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1^{2016}+1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1^{2015}+$
$\ldots .+1 \underbrace{000 \ldots 00}_{2017 \text { zeros }} 1+1$ ). The first bracket ends with 2018 zeros, while the second is not divisible by 10.
|
2018
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. 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 is $9^{4}=6561$. Of these, 9 98x7x6=3024 consist of different digits. Therefore, 6561-3024=3537 passwords contain identical digits.
|
3537
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. 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 other computers via cables (possibly through intermediate computers). Initially, all computers formed one cluster. But one night, a malicious hacker cut several cables, resulting in 8 clusters. Find the maximum possible number of cables that were cut.
|
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 obtained by restoring 192 wires. Therefore, the hacker could have cut a maximum of 153 wires.
|
153
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In trapezoid $A B C D$ with bases $A D / / B C$
diagonals intersect at point $E$. It is known that
the areas $S(\triangle A D E)=12$ and $S(\triangle B C E)=3$. Find
the area of the trapezoid.
|
Answer: 27
Solution: Triangles ADE and CBE are similar, their areas are in the ratio of the square of the similarity coefficient. Therefore, this coefficient is 2. This means that point E divides the diagonals in the ratio 1:2. Therefore, the areas of triangles ABE and CDE are twice the area of BCE and are equal to 6.
We get $S(ABCD)=12+3+6+6=27$.
|
27
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the smallest natural number ending in the digit 2 that doubles when this digit is moved to the beginning.
|
Answer: 105263157894736842
Solution: Let's write the number in the form ***...** 2 and gradually restore the "asterisks" by multiplying by 2:
$* * * . . . * * 2 \times 2=* * * . . * * 4$
$* * * . . . * 42 \times 2=* * * . . . * 84$
$* * * \ldots * 842 \times 2=* * * \ldots * 684$
$* * * . . * 6842 \times 2=* * * \ldots * 3684$
***...*36842 $2=* * * \ldots * 73684$
***... $* 736842 \times 2=* * * . . . * 473684$
***...*4736842 x $2={ }^{* * *} \ldots . . * 9473684$
...
$105263157894736842 \times 2=2105263157$
89473684
|
105263157894736842
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the smallest $n \geq 2017$, such that $1^{\mathrm{n}}+2^{\mathrm{n}}+3^{\mathrm{n}}+4^{\mathrm{n}}$ is not divisible by 10.
|
Answer: 2020.
Solution: Powers of 1 always end in 1. The last digit of powers of 2 changes with a period of 4:
$2,4,8,6$. Powers of 3 also follow the pattern $3,9,7,1$. Powers of 4 change with a period of 2: 4,6,4,6. Therefore, the last digit will repeat with a period of 4. Checking for $n=1,2,3$, we find that the last digit is 0, and for $n=4$ it is 4. Therefore, the smallest N>2016 will be 2020.
|
2020
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the Empire of Westeros, there were 1000 cities and 2017 roads (each road connected some two cities). From any city, you could travel to any other. One day, an evil wizard enchanted $N$ roads, making it impossible to travel on them. As a result, 7 kingdoms formed, such that within each kingdom, you could travel from any city to any other by roads, but you could not travel from one kingdom to another by roads. What is the largest $N$ for which this is possible?
|
Answer: 1024.
Solution: Suppose the evil wizard enchanted all 2017 roads. This would result in 1000 kingdoms (each consisting of one city). Now, imagine that the good wizard disenchants the roads so that there are 7 kingdoms. He must disenchant at least 993 roads, as each road can reduce the number of kingdoms by no more than 1. Therefore, the evil wizard could not have enchanted more than 2017-993=1024 roads.
For graders: Partial credit can be given for a correct answer obtained for some specific case - without a general proof.
|
1024
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Jack Sparrow needed to distribute 150 piastres into 10 purses. After placing a certain number of piastres in the first purse, he put more in each subsequent purse than in the previous one. As a result, it turned out that the number of piastres in the first purse was not less than half the number of piastres in the last purse. How many piastres are in the 6th purse?
|
Answer: 16
Solution: Let there be $\mathrm{x}$ piastres in the first purse. Then in the second there are no less than $\mathrm{x}+1$, in the third - no less than $\mathrm{x}+2$... in the 10th - no less than $\mathrm{x}+9$. Thus, on the one hand, $x+$ $x+1+\cdots+x+9=10 x+45 \leq 150$, from which $x \leq 10$. On the other hand, $x \geq(x+$ 9 )/2, from which $x \geq 9$. Therefore, in the first purse, there are 9 or 10 piastres. But 9 cannot be, because then in the last one there will be no more than 18 and the sum will not reach 150.
So, in the 1st - 10, in the 2nd - no less than 11, in the 3rd - no less than 12, in the 4th - no less than 13, in the 5th - no less than 14, and in the 6th - no less than 15. But with 15, the sum is less than 150. Therefore, 16. And it cannot be more, because then in the last purse there will be 21.
For graders: A partial score can be given for a correct example of distribution (trial and error) without proof of uniqueness.
Find the smallest natural $N$ such that the decimal representation of the number $N \times 999$ consists entirely of sevens (the " $x$ " symbol denotes multiplication of numbers).
Answer 778556334111889667445223
Solution: $N \times 999=77 \ldots 7$, then $N$ is divisible by 7, denote $n=N / 7$. We get $999 n=$ $1000 n-n=11 \ldots 1$, so $1000 \mathrm{n}-111 \ldots 1=$ n. Write it as a subtraction in a column and repeat the found digits of $N$ with a shift of 3 to the left
********000
$* * * * * * 889$
$* * * * 889000$
1111111111
$* * * 777889$
777889000
111111111
6667778889
Notice that the digits repeat every 3, so we get $n=111222333444555666777889$.
Therefore, $N=7 n=778556334111889667445223$.
For graders: An alternative approach is to take the number 111... 1 and start dividing it by 999 in a column until it divides evenly. I think the score should not be reduced if at the very end the number $\mathrm{n}$ was incorrectly multiplied by 7.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. 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, which means no more than 49 participants could have won 2 games (someone must lose 2 games to them).
We will show that there could have been 49. Let's say β3 won against β1 and β2, β5 - against β3 and β4, ... β99 - against β97 and β98, and β100 won against β99. Then all participants with odd numbers (except the first) won exactly 2 games.
|
49
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. We will call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist?
|
Answer 36.
Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, because otherwise the product will be at least 121. It is sufficient
to check $p_{1}=2,3,5,7$. For 2, we get the second factor: $3,5,7,11,13,17,19$, $23,29,31,37,41,43$ or 47 - 14 options, for 3 we get: 3, 5, 7, 11, 13, 17, 19, 23, 29, $31-10$ options, for $5-: 3, \quad 5, \quad 7,11,13,17,19,-7$ options and for $7: 3,5, \quad 7,11,13,-5$ options. In total $14+10+7+5=36$ options.
|
36
|
Number Theory
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math-word-problem
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Yes
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Yes
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olympiads
| false
|
4. 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 be inclined, i.e., lie on the segment BC. If we fix some diagonal segment, the remaining vertex is uniquely determined. That is, we need to choose 2 points out of 8, which can be done in $7 \times 8 \backslash 2=28$ ways.
|
28
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of distinct digits Π, B, and Π, for which the equality $\overline{\Pi B \Gamma}=(Π+B+\Gamma) \times(Π+B+\Gamma+1)$ holds.
|
Answer: 156.
Solution: Note that $P+B+\Gamma \geq 3$ and $\leq 24$ (since the digits are different). Moreover, the numbers $\overline{P B \Gamma}$ and $(P + B + \Gamma)$ should give the same remainder when divided by 9. This is only possible when $P+B+\Gamma$ is a multiple of 3. Note that $P+B+\Gamma=9$ does not work, since $(P + B + \Gamma) \times (P+B+\Gamma+1)=90$ is a two-digit number. By trying $12,15,18,21,24$, we get $\overline{P B \Gamma}=$ $156=12 \times 13$.
|
156
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. We will call a point "equidistant" if there exist two vertices of the cube for which this point is the midpoint of the segment. How many "equidistant" points are there in the cube
|
Answer: 19.
Solution: Midpoints of 12 edges, centers of 6 faces, and the center of the cube.
|
19
|
Geometry
|
math-word-problem
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Yes
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Yes
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olympiads
| false
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8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-3=0$. Find $\left(x_{1}^{5}-20\right) \cdot\left(3 x_{2}^{4}-2 x_{2}-35\right)$.
|
Answer: -1063.
Solution: If $x$ is one of the roots of the equation, then $x^{2}=x+3$. Squaring, we get $x^{4}=$ $x^{2}+6 x+9=7 x+12$. Multiplying by $x$, we get $x^{5}=7 x^{2}+12 x=19 x+21$. Substituting the roots of the equation for $x$, we get: $\left(x_{1}^{5}-20\right) \cdot\left(3 x_{2}^{4}-2 x_{2}-35\right)=\left(19 x_{1}+21-20\right) \cdot\left(21 x_{2}+36-\right.$ $\left.2 x_{2}-35\right)=\left(19 x_{1}+1\right)\left(19 x_{2}+1\right)=361 x_{1} x_{2}+19\left(x_{1}+x_{2}\right)+1$. By Vieta's theorem $x_{1} x_{2}=$ $-3, x_{1}+x_{2}=1$. Substituting, we get $361 *(-3)+19+1=-1063$.
## Variant 1-b
|
-1063
|
Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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4. On graph paper, a stepped right-angled triangle with legs equal to 6 cells (see fig.) was drawn. Then all the grid lines inside the triangle were outlined. What is the maximum number of rectangles that can be found in this drawing?
|
Answer: 126
Solution: For each cell, find the number of rectangles in which this cell is the top right corner. This is not difficult to do, you can simply multiply the cell number horizontally and vertically (if starting from the lower left corner and numbering from one).
| | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | | | | | | |
| 5 | 10 | | | | | |
| 4 | 8 | 12 | | | | |
| 3 | 6 | 9 | 12 | | | |
| 2 | 4 | 6 | 8 | 10 | | |
| 1 | 2 | 3 | 4 | 5 | 6 | |
| | | | | | | |
| | | | | | | |
Summing the numbers by columns: $1+\ldots+6+2(1+\ldots 5)+3(1+\ldots 4)+4(1+\ldots 3)+5(1+2)+6=21+$ $30+30+24+15+6=126$.
|
126
|
Combinatorics
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math-word-problem
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Yes
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Yes
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olympiads
| false
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8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-4=0$. Find $\left(x_{1}^{5}-20 x_{1}\right) \cdot\left(x_{2}^{4}+16\right)$.
|
Answer 76
Solution: If $x$ is one of the roots of the equation, then $x^{2}=x+4$. Squaring, we get $x^{4}=$ $x^{2}+8 x+16=9 x+20$. Multiplying by $x$, we get $x^{5}=9 x^{2}+20 x=29 x+36$. Substituting $x$ with the roots of the equation, we get: $\left(x_{1}^{5}-20 x_{1}\right) \cdot\left(x_{2}^{4}+16\right) .=\left(29 x_{1}+36-20 x_{1}\right) \cdot\left(9 x_{2}+16\right)=$ $\left(9 x_{1}+16\right)\left(9 x_{2}+16\right)=81 x_{1} x_{2}+144\left(x_{1}+x_{2}\right)+256$. By Vieta's theorem $x_{1} x_{2}=-4, x_{1}+$ $x_{2}=1$. Substituting, we get $81 *(-4)+144+256=76$
All problems were scored out of 15 points.
|
76
|
Algebra
|
math-word-problem
|
Yes
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Yes
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olympiads
| false
|
8. Find $\sqrt{\frac{x}{63}-32} \times \sqrt{\frac{y}{63}-32}$, given that $\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}$.
ANSWER: 32.
|
Solution: Let's make the substitution: $a=x / 63, b=y / 63$. Then we can rewrite the condition as: find $\sqrt{(a-32) \cdot(b-32)}$, given that $\frac{1}{a}+\frac{1}{b}=\frac{1}{32}$, i.e., $a b=32(a+b)$. Transforming: $\sqrt{(a-32) \cdot(b-32)}=\sqrt{a b-32(a+b)+1024}=32$.
|
32
|
Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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3. Find the smallest $n>2016$, such that $1^{n}+2^{n}+3^{n}+4^{n}$ is not divisible by 10.
|
Answer: 2020.
Solution: Powers of 1 always end in 1. The last digit of powers of 2 changes with a period of 4: 2,4,8,6. Powers of 3 also change with a period of 4: 3,9,7,1. Powers of 4 change with a period of 2: 4,6,4,6. That is, the last digit will repeat with a period of 4. Checking for $n=1,2,3$ we get that the last digit is 0, and for $n=4-4$. Therefore, the smallest N>2016 will be 2020.
|
2020
|
Number Theory
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math-word-problem
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Yes
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Yes
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olympiads
| false
|
4. Masha has 2 kg of "Swallow" candies, 3 kg of "Truffle" candies, 4 kg of "Bird's Milk" candies, and 5 kg of "Citron" candies. What is the maximum number of New Year's gifts she can make if each gift must contain 3 different types of candies, 100 grams of each?
|
Answer: 45
Solution: Even if Masha puts "Citron" in all the gifts, she will still have 2+3+4 = 9 kg of candies left, and in each gift, she must put at least 200g (and if she doesn't put Citron in all of them, then more than 200g). This means there can be no more than 45 gifts. 45 gifts can be made if she creates 5 gifts of the type Swallow+Truffle+Citron, 15 gifts of the type Swallow+Bird's Milk + Citron, and
25 gifts of the type Truffle+Bird's Milk + Citron.
Comment for graders: Here, half a point can be given for the correct selection without proving that it is optimal.
|
45
|
Combinatorics
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math-word-problem
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Yes
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Yes
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olympiads
| false
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6. In the Empire of Westeros, there were 1000 cities and 2017 roads (each road connected some two cities). From any city, you could travel to any other. One day, an evil wizard enchanted $N$ roads, making it impossible to travel on them. As a result, 7 kingdoms formed, such that within each kingdom, you could travel from any city to any other by roads, but you could not travel from one kingdom to another by roads. What is the largest $N$ for which this is possible?
|
# Answer 1024.
Solution: Suppose the evil wizard enchanted all 2017 roads. This would result in 1000 kingdoms (each consisting of one city). Now, imagine that the good wizard disenchants the roads so that there are 7 kingdoms. He must disenchant at least 993 roads, as each road can reduce the number of kingdoms by no more than 1. Therefore, the evil wizard could not have enchanted more than 2017-993=1024 roads.
For graders: Partial credit can be given for a correct answer obtained for a specific case - without a general proof.
## Variant 3-b
|
1024
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In a regular $1000-$gon, all diagonals are drawn. What is the maximum number of diagonals that can be selected such that among any three of the selected diagonals, at least two have the same length?
|
Answer: 2000
Solution: For the condition of the problem to be met, it is necessary that the lengths of the diagonals take no more than two different values. The diagonals connecting diametrically opposite vertices are 500. Any other diagonal can be rotated to coincide with a diagonal of the corresponding length, i.e., there are 1000 of them. Therefore, 2000 can be chosen while meeting the condition.
|
2000
|
Combinatorics
|
math-word-problem
|
Yes
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Yes
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olympiads
| false
|
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
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Yes
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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 night, a malicious hacker cut several cables, resulting in 8 clusters. Find the maximum possible number of cables that were cut.
|
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 obtained by restoring 192 wires. Therefore, the hacker could have cut a maximum of 153 wires.
## Variant 1-a
|
153
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
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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 β3 won against β1 and β2, β5 - against β3 and β4, ... β99 - against β97 and β98, and β100 won against β99. Then all participants with odd numbers (except the first) won exactly 2 games.
|
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 determined. That is, we need to choose 2 points out of 8, which can be done in $7 \times 8 \backslash 2=28$ ways.
|
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 Π+Π $\overline{\Pi Π \Gamma}$ and (Π + Π + Π) should give the same remainder when divided by 9. This is only possible when $Π+Π+\Gamma$ is a multiple of 3. Note that $Π+Π+\Gamma=9$ - does not work, because (Π + Π + $\Gamma) \times(Π+B+\Gamma+1)=90$-two-digit. By trying $12,15,18,21,24$, we get $\overline{\Pi Π \Gamma}=$ $156=12 \times 13$.
## Variant 1-b
|
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 22 of them, so the remaining 900-22 $=878$ are not squares.
|
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 β4 win against β1,2,3; β7 - against β4,5,6,... β199 - against β196,197,198, and β200 win against β199. Then all participants with numbers giving a remainder of 1 when divided by 3 (except the first) won exactly 3 games.
|
66
|
Combinatorics
|
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
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Yes
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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 $n=-2015-2016+2016 \cdot 2015 p$. The smallest natural $n$ is $2016 \cdot 2015-2015-2016=2015^{2}-2016=4058209$.
Lomonosov Moscow State University
## School Olympiad "Conquer Sparrow Hills" in Mathematics
Final stage tasks for the 2015/2016 academic year for 9th grade
|
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
|
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
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math-word-problem
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Yes
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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 total, if it is known that recently he bought an additional album for them, which can hold 150 stamps, as such an old album was no longer sufficient
OTBET: 208.
|
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
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math-word-problem
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Yes
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Yes
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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
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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
one triangle (for example, AOB and BOA).
ANSWER: 20.
|
Solution: First, consider the triangles that do not contain point O. These are two equilateral triangles and 6 triangles with an angle of $120^{\circ}$. With vertex $O$, there are 6 equilateral triangles and 6 triangles with an angle of $120^{\circ}$.
|
20
|
Combinatorics
|
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 the same (for example, AOB and BOA).
ANSWER: 20.
|
Solution: First, consider the triangles that do not contain point O. These are two equilateral triangles and 6 triangles with an angle of $120^{\circ}$. With vertex $O$, there are 6 equilateral triangles and 6 triangles with an angle of $120^{\circ}$.
|
20
|
Combinatorics
|
math-word-problem
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Yes
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Yes
|
olympiads
| false
|
16. Given 2024 sets, each consisting of 44 elements. The union of any two of these sets contains 87 elements. How many elements does the union of all 2024 sets contain?
|
Answer: 87033.
II. Find the number of natural numbers $n$, not exceeding 500, for which the equation $x^{[x]}=n$ has a solution. Here $[x]$ is the greatest integer not exceeding $x$.
Solution. If $[x]=0$, then the solution is: $0 \leqslant x < 1$, and $n=1$.
If $[x]=1$, then the solution is: $1 \leqslant x < 2$, and $n=1$.
If $[x]=2$, then the solution is: $2 \leqslant x < 3$, and $n=4, 5, 6, 7, 8$.
If $[x]=3$, then the solution is: $3 \leqslant x < 4$, and $n=27, 28, \ldots, 63$.
If $[x]=4$, then the solution is: $4 \leqslant x < 5$, and $n=256, 257, \ldots, 500$.
Therefore, we get: $n=1,4,5,6,7,8,27,28, \ldots, 62,63,256,257, \ldots, 500$. In total, there are 288 numbers.
Answer: 288.
|
288
|
Combinatorics
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math-word-problem
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Yes
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Yes
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olympiads
| false
|
4. Between the pairs of solutions $(x, y)$ 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 inequality. Perform this transformation enough times so that instead of $(x, y)$ we get the pair $(1,1)$, and instead of $f(x, y)$ we get the pair $(5,7)$. But then it would follow that there is another solution to the equation between the pairs $(1,1)$ and $(5,7)$, which is a contradiction.
Conclusion. The recursive sequence of pairs
$$
\left(x_{1}, y_{1}\right)=(1,1), \ldots,\left(x_{k}, y_{k}\right),\left(x_{k+1}, y_{k+1}\right)=\left(3 x_{k}+2 y_{k}, 4 x_{k}+3 y_{k}\right), \ldots
$$
exhausts the set of all natural solutions to the equation.
We find the first few pairs of this sequence:
$$
\begin{aligned}
& x_{2}=5, y_{2}=7, \quad x_{3}=29, y_{3}=41, \quad x_{4}=169, y_{4}=239, \\
& x_{5}=985, y_{5}=1393, \quad x_{6}=5741, y_{6}=8119, \ldots
\end{aligned}
$$
It is clear that only the pair $x_{5}=985, y_{5}=1393$ (and the corresponding values $k=492, n=696$) satisfy the conditions of the problem.
|
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, inclusive. All rooms in the hotel are numbered consecutively from 1 to 1593, and Know-it-all lives in a room with a number greater than 200. Determine the room number where Don't-know-it lives.
If there are several room numbers, write their sum in the answer. If there are no such room numbers, write the number 0.
Answer: 492.
V-3. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 25-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, inclusive. All rooms in the hotel are numbered consecutively from 1 to 4591, and Know-it-all lives in a room with a number greater than 700. Determine the room number where Know-it-all lives.
If there are several room numbers, write their sum in the answer. If there are no such room numbers, write the number 0.
Answer: 4059.
V-4. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 25-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, inclusive. All rooms in the hotel are numbered consecutively from 1 to 4599, and Know-it-all lives in a room with a number greater than 700. Determine the room number where Don't-know-it lives.
If there are several room numbers, write their sum in the answer. If there are no such room numbers, write the number 0.
Answer: 2870.
V-5. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 17-story hotel by the sea. Know-it-all noticed that the sum of all room numbers from the first to his own, inclusive, is three times the sum of all room numbers from the first to the one where Don't-know-it stayed, inclusive. All rooms in the hotel are numbered consecutively from 1 to 1799, and Know-it-all lives in a room with a number greater than 200. Determine the room number where Know-it-all lives.
If there are several room numbers, write their sum in the answer. If there are no such room numbers, write the number 0.
Solution. According to the problem, the room number of Don't-know-it $k$ and the room number of Know-it-all $n$ satisfy the relation $n^{2}+n=3\left(k^{2}+k\right)$. After the substitution $x=2 k+1, y=2 n+1$, this relation reduces to the form
$$
3 x^{2}-y^{2}=2
$$
The set of all natural solutions to the equation is exhausted by the following recursive sequence of pairs
$$
\left(x_{1}, y_{1}\right)=(1,1), \ldots,\left(x_{k}, y_{k}\right),\left(x_{k+1}, y_{k+1}\right)=\left(2 x_{k}+y_{k}, 3 x_{k}+2 y_{k}\right), \ldots
$$
We find the first few pairs of this sequence:
$$
\begin{aligned}
& x_{2}=3, y_{2}=5, \quad x_{3}=11, y_{3}=19, \quad x_{4}=41, y_{4}=71 \\
& x_{5}=153, y_{5}=265, \quad x_{6}=571, y_{6}=989, \quad x_{7}=2131, y_{7}=3691, \ldots
\end{aligned}
$$
Obviously, only the pair $x_{6}=571, y_{6}=989$ (and the corresponding values $k=285, n=494$) satisfy the conditions of the problem.
Answer: 494.
V-6. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 17-story hotel by the sea. Know-it-all noticed that the sum of all room numbers from the first to his own, inclusive, is three times the sum of all room numbers from the first to the one where Don't-know-it stayed, inclusive. All rooms in the hotel are numbered consecutively from 1 to 1781, and Know-it-all lives in a room with a number greater than 200. Determine the room number where Don't-know-it lives.
If there are several room numbers, write their sum in the answer. If there are no such room numbers, write the number 0.
Answer: 285.
V-7. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 17-story hotel by the sea. Know-it-all noticed that the sum of all room numbers from the first to his own, inclusive, is three times the sum of all room numbers from the first to the one where Don't-know-it stayed, inclusive. All rooms in the hotel are numbered consecutively from 1 to 3791, and Know-it-all lives in a room with a number greater than 600. Determine the room number where Know-it-all lives.
If there are several room numbers, write their sum in the answer. If there are no such room numbers, write the number 0.
Answer: 1845.
V-8. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 17-story hotel by the sea. Know-it-all noticed that the sum of all room numbers from the first to his own, inclusive, is three times the sum of all room numbers from the first to the one where Don't-know-it stayed, inclusive. All rooms in the hotel are numbered consecutively from 1 to 3799, and Know-it-all lives in a room with a number greater than 600. Determine the room number where Don't-know-it lives.
If there are several room numbers, write their sum in the answer. If there are no such room numbers, write the number 0.
Answer: 1065.
V-9. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 17-story hotel by the sea. Know-it-all noticed that the sum of all room numbers from the first to his own, inclusive, is one and a half times the sum of all room numbers from the first to the one where Don't-know-it stayed, inclusive. All rooms in the hotel are numbered consecutively from 1 to 1799, and Know-it-all lives in a room with a number greater than 200. Determine the room number where Know-it-all lives.
If there are several room numbers, write their sum in the answer. If there are no such room numbers, write the number 0.
Solution. According to the problem, the room number of Don't-know-it $k$ and the room number of Know-it-all $n$ satisfy the relation $2\left(n^{2}+n\right)=3\left(k^{2}+k\right)$. After the substitution $x=2 k+1, y=2 n+1$, this relation reduces to the form
$$
3 x^{2}-2 y^{2}=1
$$
The set of all natural solutions to the equation is exhausted by the following recursive sequence of pairs
$$
\left(x_{1}, y_{1}\right)=(1,1), \ldots,\left(x_{k}, y_{k}\right),\left(x_{k+1}, y_{k+1}\right)=\left(5 x_{k}+4 y_{k}, 6 x_{k}+5 y_{k}\right), \ldots
$$
We find the first few pairs of this sequence:
$$
\begin{aligned}
& x_{2}=9, y_{2}=11, \quad x_{3}=89, y_{3}=109, \quad x_{4}=881, y_{4}=1079 \\
& x_{5}=8721, y_{5}=10681, \ldots
\end{aligned}
$$
Obviously, only the pair $x_{4}=881, y_{4}=1079$ (and the corresponding values $k=440, n=539$
|
539
|
Number Theory
|
proof
|
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 2012, we get a total of 402 such numbers.
|
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
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5 \\
x^{2}+z^{2}=13 \\
x^{2}+y^{2}=10
\end{array}\right.
$$
Fig. 6:
From here, we find $x=3, y=1, z=2$. Let the lengths of the edges $S A, S B$, and $S C$ be $a, b$, and $c$ respectively. Since points $A, B, C$, and $D$ lie in the same plane, the relation $\frac{3}{a}+\frac{1}{b}+\frac{2}{c}=1$ holds.
Using the inequality between the arithmetic mean and the geometric mean for three variables, we get:
$$
\begin{aligned}
& \frac{\frac{3}{a}+\frac{1}{b}+\frac{2}{c}}{3} \geqslant \sqrt[3]{\frac{3}{a} \cdot \frac{1}{b} \cdot \frac{2}{c}}=\sqrt[3]{\frac{6}{a b c}} \Longleftrightarrow \\
& \Longleftrightarrow 1=\left(\frac{3}{a}+\frac{1}{b}+\frac{2}{c}\right)^{3} \geqslant \frac{6 \cdot 27}{a b c} \Longleftrightarrow a b c \geqslant 6 \cdot 27
\end{aligned}
$$
with equality holding when $\frac{3}{a}=\frac{1}{b}=\frac{2}{c}=\frac{1}{3}$. The volume of the pyramid $V=\frac{a b c}{6}$, so $V \geqslant 27$. Equality holds when $a=9, b=3, c=6$.
|
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:
$$
\left\{\begin{array} { l }
{ 5 x = 2 ( y + z ) , } \\
{ 7 z = 3 ( x + y ) . }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ 5 x - 2 y = 2 z , } \\
{ 3 x + 3 y = 7 z }
\end{array} \Rightarrow \left\{\begin{array}{l}
21 x=20 z \\
21 y=29 z
\end{array}\right.\right.\right.
$$
This means that the minimum values can only be: $x=20, y=29$, $z=21$.
|
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 $a \in\left(0 ; \frac{\pi}{12}\right]$, then $2^{\frac{1}{\sin ^{2}(2 a)}}>16$, and $2^{-4 \operatorname{tg}(3 a)}<1$. Therefore, the minimum value of the function $f(x)=\left|x-2^{\frac{1}{\sin ^{2}(2 a)}}\right|+\left|x-2^{-4 \operatorname{tg}(3 a)}\right|$ is not less than 15. On the other hand, the absolute value of the expression $g(a)=a\left(a+\frac{\pi}{12}\right)^{2}\left(a-\frac{\pi}{12}\right)$ on the half-interval $\left(0 ; \frac{\pi}{12}\right]$ is certainly no more than $1:|g(a)|<a\left(a+\frac{\pi}{12}\right)^{3}<1$.
Therefore, when $a \in\left(0 ; \frac{\pi}{12}\right]$, there are no solutions.
|
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,
$$
g(x)=(x-2019)(x-2020)(x-2021)(x-2022)(x-2023),
$$
and
$$
f(x)=(x-2019)(x-2020)(x-2021)(x-2022)(x-2023)-(x-2021)^{2}+4
$$
Thus,
$$
f(2018)=(-1)(-2)(-3)(-4)(-5)-(-3)^{2}+4=-120-9+4=-125
$$
|
Answer: -125. Answer to option: 7-2: -115. 7-3: 115. 7-4: 125.
|
-125
|
Algebra
|
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 $18+b=27$, from which $b=9$, which contradicts the condition $a \neq b$.
Now consider $a=8$. Then the number $81 a=648=2^{3} \cdot 3^{4}$ must be divisible by $16+b$ without a remainder, which is possible only when $b=2$ and $b=8$ (but the latter contradicts the condition $a \neq b$). Therefore, $a=8, b=2$.
|
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 PRIZE WINNERS ${ }^{1}$
## of the school students' competition "CONQUER THE SPARROW MOUNTAINS!" IN MATHEMATICS
ELIMINATION STAGE
WINNER:
From 95 points inclusive and above.
PRIZE WINNER:
From
91
points to
94
points inclusive.
FINAL STAGE
WINNER (Diploma I degree):
From 90 points inclusive and above.
PRIZE WINNER (Diploma II degree):
From 75 points to 89 points inclusive.
PRIZE WINNER (Diploma III degree):
from 60 points to 74 points inclusive.[^0]
[^0]: ${ }^{1}$ Approved at the meeting of the jury of the school students' competition "Conquer the Sparrow Mountains!" in mathematics
|
80
|
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, $x \leq 25$.
If each of the shareholders, except M, owns $\frac{75}{99}=\frac{25}{33} \%$ of the shares, then any 66 of them own exactly $50 \%$, and M owns exactly $25 \%$ of the shares.
Answer: $25 \%$ of the shares.
|
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, a grandmother-pensioner, and four schoolchild children, should pay for visiting the museum?
|
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 | "child" 60 | 840 |
| free | "family ticket" 650 | | | | "child" 60 | "child" 60 | 770 |
The minimum amount that a family, including a father, a mother, a grandmother-pensioner, and four school-age children, should pay for visiting the museum is 770 rubles.
Answer: 770 rubles
|
770
|
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
|
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 remove this extreme point, the number of painted points will decrease by at least two. Next, we remove another extreme point. Then the number of painted points will decrease by at least two more. And so on until only three points remain. We get that the number of painted points is not less than \(3 + (n-3) \cdot 2 = 3 + 2n - 6 = 2n - 3\), where \(n\) is the number of marked points. In the case where there are 3025 points on the line, the number of midpoints marked is not less than \(3 \cdot 3025 - 3 = 6047\). This is possible if all marked points are located at equal distances from their neighbors. Suppose that the number of points of each color is no more than 2015. Then there should be no more than \(3 \cdot 2015 = 6045\) points, while the number of painted points is not less than 6047. Therefore, the number of points painted in one of the colors on the line is not less than 2016.
Criteria: Full solution - 7 points. Proved that the number of painted points is not less than 6047 - 4 points. Provided an example for 6047 painted points - 2 points. Proved that out of 6047 painted points, at least 2016 are painted in one color - 1 point.
|
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
|
8.6. 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 these tires to ensure that the "Gnu Antelope" travels the maximum possible distance? What is this distance?
|
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, respectively. Thus, the total distance that can be traveled is no more than
$$
x+25000\left(1-\frac{x}{15000}\right)=25000-\frac{2}{3} x
$$
and no more than
$$
x+15000\left(1-\frac{x}{25000}\right)=15000+\frac{2}{5} x
$$
The maximum distance can be traveled if these expressions are equal (otherwise, either the first or the second set of tires will wear out earlier, since when the first expression increases, the second decreases and vice versa). Thus,
$$
25000-\frac{2}{3} x=15000+\frac{2}{5} x
$$
from which $10000=\frac{16}{15} x$, or $x=9375$.
Answer: The tires should be swapped after 9375 km, then a total of 18750 km can be traveled.
|
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: $9 + 2 \cdot 90 + 3 \cdot (x - 99) = 1392$, the solution to which is $x = 500$.
Answer: 500.
|
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, $20 \cdot 2 + 10 \cdot 3 = 70$ cuts are needed.
Answer. 70.
|
70
|
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 $\sqrt{a}+\sqrt{b}=7(\sqrt{a}-\sqrt{b})$. Therefore, $3\sqrt{a}=4\sqrt{b}$. Thus, $a:b=16:9$. Considering that $a$ and $b$ are two-digit numbers, the sum will be the largest for $a=16 \cdot 6=96, b=9 \cdot 6=54$. Their arithmetic mean is $\frac{96+54}{2}=75$.
|
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 $1 * * 7$ - there are also 200 such numbers. In addition, there are 2007 and 2017 - two more 7s.
|
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 $-1, 1, -3$, or 3, respectively, i.e., it will change by 2 or 6. It is clear that if we add two sums, the remainder when divided by 4 does not change. Therefore, we can obtain the numbers $10, 6, 2, -2$, $-6, -10$. The number -14, obviously, cannot be obtained, since this would require making all numbers equal to -1. For the other values, it is easy to construct corresponding examples.
|
-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 form $4n+2$ cannot be obtained as the difference of squares. There is exactly one such number (of the form $4n+2$) in every set of four consecutive numbers, so the total number of such numbers from 1 to 1000 will be $1000 / 4 = 250$.
|
250
|
Number Theory
|
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 $1 * * 7$ - there are also 200. In addition, there are 2007 and 2017 - two more 7s.
|
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 the form $4 n+2$ cannot be obtained as the difference of squares. There is exactly one such number (of the form $4 n+2$) in every set of four consecutive numbers, so the total number of such numbers from 1 to 1000 will be $1000 / 4=250$.
|
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 write the answer as $x y=43 \cdot 61=2623$.
|
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 different. It is known that if he had taken one more shirt, he could have conducted 18 more lessons; if he had taken one more pair of pants, he could have conducted 63 more lessons; if he had taken one more pair of shoes, he could have conducted 42 more lessons. What is the maximum number of lessons he could conduct under these conditions?
|
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 has one more pair of trousers, the number of lessons will increase by $3 x z$. If he has one more pair of shoes, the number of lessons will increase by $3 y z$. Thus, we get a system of three equations: $3 y z=18, 3 x z=63, 3 x y=42$. Therefore, $y z=6, x z=21, x y=14$, and thus $(x y z)^{2}=6 \cdot 21 \cdot 14, x y z=42$ (although it is not necessary, but we can calculate that $x=7, y=2, z=3$). The desired value: $3 x y z=126$.
Remark. There is a possible interpretation of the problem condition in which it is required to find the maximum number of lessons given that the teacher will take with him one more shirt, one more pair of trousers, and one more pair of shoes. The corresponding answer $3(x+1)(y+1)(z+1)=288$ is also considered correct.
|
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 different. It is known that if he had taken one more shirt, he could have conducted 36 more lessons; if he had taken one more pair of pants, he could have conducted 72 more lessons; if he had taken one more pair of shoes, he could have conducted 54 more lessons. What is the maximum number of lessons he could conduct under these conditions?
|
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
$$
2^{k}=2 \cdot(1+1)^{k-1}=2 \sum_{i=0}^{k-1} C_{k-1}^{i}=2 \sum_{i=0}^{k-1} \frac{(k-1)(k-2) \ldots(k-i)}{i!}
$$
Note that the sum on the right side of this equation is a polynomial in $k$ of degree $k-1$, and if we add terms corresponding to $i=k, k+1, \ldots, n$, it will not change, since $(k-1)(k-2) \ldots(k-i)=0$ for $i \geqslant k$ (each such product contains a factor $(k-k)=0$). Therefore,
$$
2^{k}=2 \sum_{i=0}^{k-1} \frac{(k-1)(k-2) \ldots(k-i)}{i!}=2 \sum_{i=0}^{n} \frac{(k-1)(k-2) \ldots(k-i)}{i!} \quad \text { for } k \in \mathbb{N}, \quad n \geqslant k-1
$$
Consider the polynomial
$$
Q(x)=2 \sum_{i=0}^{n} \frac{(x-1)(x-2) \ldots(x-i)}{i!}
$$
This is a polynomial of degree $n$, and by the proven result, $Q(k)=2^{k}, k=1,2,3, \ldots, n+1$. Therefore, $P(x)=Q(x)$ as polynomials of degree $n$ that coincide at $n+1$ points.
Substituting $x=n+m+2(m \geqslant 0)$, we find
$$
P(n+m+2)=2 \sum_{i=0}^{n} C_{n+m+1}^{i}=2 \sum_{i=0}^{n+m+1} C_{n+m+1}^{i}-2 \sum_{i=n+1}^{n+m+1} C_{n+m+1}^{i}=2^{n+m+2}-2 \sum_{i=0}^{m} C_{n+m+1}^{i}
$$
(in the last equality, we again used the binomial theorem and the equality of binomial coefficients $\left.C_{n+m+1}^{i}=C_{n+m+1}^{n+m+1-i}\right)$. For $m=1$ we get
$$
P(n+3)=2^{n+3}-2-2(n+2)=2^{n+3}-2 n-6
$$
In particular, for the polynomial in the problem statement $n=9, P(12)=2^{12}-2 \cdot 9-6=4072$.
Another way to solve. A polynomial of degree $n$ is uniquely determined by its values at $n+1$ points $k=1,2,3, \ldots, n+1$. By direct substitution of these points, we verify that the condition of the problem is satisfied by the polynomial (Lagrange interpolation polynomial)
$$
\begin{aligned}
P(x)= & 2 \frac{(x-2)(x-3)(x-4) \ldots(x-(n+1))}{(1-2)(1-3)(1-4) \ldots(1-(n+1))}+2^{2} \frac{(x-1)(x-3)(x-4) \ldots(x-(n+1))}{(2-1)(2-3)(2-4) \ldots(2-(n+1))}+ \\
& +2^{3} \frac{(x-1)(x-2)(x-4) \ldots(x-(n+1))}{(3-1)(3-2)(3-4) \ldots(3-(n+1))}+\ldots+ \\
& +2^{n+1} \frac{(x-1)(x-2)(x-3) \ldots(x-n)}{(n+1-1)(n+1-2)(n+1-3) \ldots(n+1-n)}= \\
= & 2(-1)^{n} \frac{(x-2) \ldots(x-(n+1))}{n!}-2^{1} \frac{(x-1)(x-3) \ldots(x-(n+1))}{1!(n-1)!}+ \\
& \left.+2^{2} \frac{(x-1)(x-2)(x-4) \ldots(x-(n+1))}{2!(n-2)!}-\ldots+(-1)^{n} 2^{n} \frac{(x-1) \ldots(x-n)}{n!}\right) .
\end{aligned}
$$
Its value at the point $x=n+3$ is
$$
\begin{aligned}
P(n+3)= & 2(-1)^{n}\left(\frac{(n+1) n(n-1) \ldots 3 \cdot 2}{n!}-2^{1} \frac{(n+2) n(n-1) \ldots \cdot 3 \cdot 2}{1!(n-1)!}+\right. \\
& \left.+2^{2} \frac{(n+2)(n+1)(n-1) \ldots \cdot 3 \cdot 2}{2!(n-2)!}-\ldots+(-1)^{n} 2^{n} \frac{(n+2)(n+1) \ldots \cdot 4 \cdot 3}{n!}\right)= \\
= & \left.2(-1)^{n}(n+2)(n+1)\left(\frac{1}{n+2}-2^{1} C_{n}^{1} \cdot \frac{1}{n+1}+2^{2} C_{n}^{2} \cdot \frac{1}{n}+\ldots+(-1)^{n} 2^{n} C_{n}^{n} \cdot \frac{1}{2}\right)\right)= \\
= & 2(-1)^{n}(n+2)(n+1) \sum_{k=0}^{n}(-2)^{k} \frac{C_{n}^{k}}{n-k+2}=2(-1)^{n}(n+2)(n+1) S(1)
\end{aligned}
$$
where
$$
S(x)=\sum_{k=0}^{n} C_{n}^{k} \frac{(-2)^{k} x^{n-k+2}}{n-k+2}
$$
Since
$$
S^{\prime}(x)=\left(\sum_{k=0}^{n} C_{n}^{k} \frac{(-2)^{k} x^{n-k+2}}{n-k+2}\right)^{\prime}=\sum_{k=0}^{n} C_{n}^{k}(-2)^{k} x^{n-k+1}=x(x-2)^{n}=(x-2)^{n+1}+2(x-2)^{n}
$$
with the condition $S(0)=0$ we find $S(x)=\frac{(x-2)^{n+2}-(-2)^{n+2}}{n+2}+2 \frac{(x-2)^{n+1}-(-2)^{n+1}}{n+1}$. Therefore,
$$
\begin{aligned}
P(n+3) & =2(-1)^{n}(n+2)(n+1)\left(\frac{(-1)^{n+2}-(-2)^{n+2}}{n+2}+2 \frac{(-1)^{n+1}-(-2)^{n+1}}{n+1}\right)= \\
& =-2(n+1)(n+2)\left(\frac{2^{n+2}-1}{n+2}-\frac{2^{n+2}-2}{n+1}\right)=2\left(2^{n+2}-1\right)-2(n+2)=2^{n+3}-2 n-6
\end{aligned}
$$
|
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 $p=1009$. This number is odd and prime, so the number $\left(y_{n}^{2}-2 x_{n}^{2}\right)^{p}=(-193)^{p}$ gives the same remainder when divided by $2 p$, as $y_{n}^{2 p}-2^{p} x_{n}^{2 p}$, since all binomial coefficients $C_{p}^{1}, \ldots, C_{p}^{p-1}$ are divisible by $p$. Moreover, by Fermat's little theorem, $2^{p}$ gives a remainder of 2 when divided by $2 p=2018$ (and therefore the remainders of the numbers $y_{n}^{2 p}-2^{p} x_{n}^{2 p}$ and $y_{n}^{2 p}-2 x_{n}^{2 p}$ are the same), and $(-193)^{p}$ gives a remainder of $2018-193=1825$ when divided by $2 p=2018$.
|
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 week, followed by 52 full weeks, and then the first day of the following week - a total of 54 different weeks. If there are 55 (or more) weeks, then there must be no fewer than 53 full weeks, which is impossible, as $53 \cdot 7=371>366$.
|
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 numbers, they will also be divisible by 4. There will be no other three-digit numbers divisible by 4 (the rule for divisibility by 4). Therefore, the total number of such numbers is: $1 + 5 \cdot 6 = 31$.
|
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 $\angle C B M=$ $\angle C B D=90^{\circ}-60^{\circ}=30^{\circ}$, therefore, $\angle C B A=2 \angle C B D=60^{\circ}$. Hence $\angle B C H=30^{\circ}$, $\angle H C K=90^{\circ}-30^{\circ}-30^{\circ}=30^{\circ}$.
Answer to variant 2: $30^{\circ}$.
|
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, $\overline{m n}=\overline{n m}^{2}: 4=(\overline{n m}: 2)^{2}$, which means $\overline{m n}$ is a square of a natural number, starting with an even digit. Therefore, $\overline{m n}$ can be $25, 49, 64$ or 81. Checking shows that only the last one satisfies the condition.
|
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, or through the equality of distances from points $M, N$ to the line $A C$, or through the calculation of angles), therefore $\angle B M N=$ $\angle Q M N=\angle B N M=\angle M N P=\alpha$ and $A M N P$ is a rhombus, from which $A P=A M=M N$. (Seventh-graders can prove this equality through the equality $\triangle A M N=\Delta A P N$.)
First case. Suppose point $P$ lies on segment $A Q$. Then, taking into account the above, $\triangle A M Q=\triangle A P B$ by two sides and the angle between them, so $\angle A B P=\angle A Q M=\alpha$; similarly, $\angle Q B C=\angle C P N=\alpha$. $\angle B P Q=2 \alpha$ as an external angle for $\triangle A P B$, hence $\angle B P N=2 \alpha-\alpha=$ $\alpha$; similarly, $\angle M Q B=\alpha$. Since $\triangle A B Q$ is isosceles, $\angle A B Q=\angle A Q B=2 \alpha$, from which $\angle P B Q=\alpha$. By the theorem on the sum of angles in a triangle for $\triangle A B C$ we get $5 \alpha=180^{\circ}$, hence $\angle M Q B=\alpha=36^{\circ}$.
Second case. Suppose point $P$ lies on segment $C Q$. Then $M N=A P>P Q=B M$. In $\triangle B M N$ the larger angle lies opposite the larger side, so $\angle M B N=180^{\circ}-2 \alpha>\angle B N M=\alpha$, from which $\alphaA Q=A P-Q P=A M-Q P$, this means that $\angle A=\alpha>\angle A M Q=180^{\circ}-2 \alpha$, from which $\alpha>60^{\circ}$. The obtained contradiction shows that the second case is impossible.
|
36
|
Geometry
|
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 $f\left(\frac{1}{x}\right)$:
$$
\left\{\begin{array}{l}
(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1} \\
\left(\frac{1}{x}-1\right) f\left(\frac{1}{x}\right)+f(x)=\frac{x}{1-x}
\end{array}\right.
$$
Subtracting the first equation, multiplied by $\frac{1-x}{x}$, from the second equation, we get $\left(1+\frac{(1-x)^{2}}{x}\right) f(x)=\frac{x}{1-x}+\frac{1}{x}$, which simplifies to $\frac{x^{2}-x+1}{x} f(x)=\frac{x^{2}-x+1}{x(1-x)}$, or $f(x)=\frac{1}{1-x}$. Therefore, $f\left(\frac{2016}{2017}\right)=2017$.
|
2017
|
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
$$
\begin{gathered}
1-\operatorname{ctg}^{2} t=-\frac{\cos 2 t}{\sin ^{2} t}, \quad 1-3 \operatorname{ctg}^{2} t=\frac{\sin ^{2} t-3 \cos ^{2} t}{\sin ^{2} t}=\frac{4 \sin ^{2} t-3}{\sin ^{2} t}=-\frac{\sin 3 t}{\sin ^{3} t} \\
1-\operatorname{tg} 2 t \cdot \operatorname{ctg} 3 t=1-\frac{\sin 2 t \cdot \cos 3 t}{\cos 2 t \cdot \sin 3 t}=\frac{\sin t}{\cos 2 t \cdot \sin 3 t}
\end{gathered}
$$
the inequality reduces to $\frac{1}{\sin ^{4} t} \leqslant 16$, that is, $|\sin t| \geqslant \frac{1}{2}$. Therefore, $t \in\left[\frac{\pi}{6}+\pi n ; \frac{5 \pi}{6}+\pi n\right]$, $n \in \mathbb{Z}$. It is also necessary to take into account the conditions $\sin t \neq 0, \cos 2 t \neq 0, \sin 3 t \neq 0$. Thus, the points $\frac{\pi}{4}+\frac{\pi k}{2}$ and $\pm \frac{\pi}{3}+\pi k, k \in \mathbb{Z}$ need to be excluded from the obtained solution. Therefore, $x \in$ $[2+12 n ; 10+12 n], x \neq 3+6 n, x \neq \pm 4+12 n$. In the interval $[-3 ; 13]$, the numbers $-2,2,5,6,7$, 10 fall. Their sum is 28.
|
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. Therefore, $317 \leqslant n \leqslant 999$.
The sum of all digits from 1 to 9 is 45 and is divisible by 9, so $n+n^{2}$ is also divisible by 9. Since $n+n^{2}=n(n+1)$ is the product of two consecutive numbers, it is divisible by 9 only in two cases: 1) $n$ is divisible by 9; 2) $n+1$ is divisible by 9, i.e., $n$ has a remainder of 8 when divided by 9.
The first such numbers greater than 316 are 323 and 324. Therefore, the desired numbers lie in the arithmetic progressions $323+9 k$ and $324+9 k, k \in \mathbb{N}$. Moreover, the decimal representation of $n$ should not contain identical digits or the digit 0, and it should not end in 1, 5, or 6 (otherwise, its square would end in the same digit). Additionally, if $n$ contains the digit 4 and ends in 2 or 8, it is also not suitable, as then $n^{2}$ would end in 4. Taking this into account, we will sequentially list the remaining possible values of $n \geqslant 317$ in a table:
| $n$ | $n^{2}$ |
| :---: | :---: |
| 324 | 104976 |
| 359 | 128881 |
| 368 | 135424 |
| 369 | 136161 |
| $n$ | $n^{2}$ |
| :---: | :---: |
| 378 | 142884 |
| 387 | 149769 |
| 413 | 170569 |
| 423 | 178929 |
| $n$ | $n^{2}$ |
| :---: | :---: |
| 459 | 210681 |
| 467 | 218089 |
| 512 | 262144 |
| 513 | 263169 |
| $n$ | $n^{2}$ |
| :---: | :---: |
| 539 | 290521 |
| 549 | 301401 |
| 567 | 321489 |
| $\cdots$ | $\ldots$ |
Among the checked numbers, the first one that satisfies the condition of the problem is the number 567. Similarly, by checking possible values of $n \leqslant 999$ "from the end," the largest such number can be found, which is 854 (in this case, the search can be shortened by noting that for $n \geqslant 953$, the decimal representation of $n^{2} \geqslant 908209$ also contains the digit 9).
|
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 because their cubes are single-digit numbers. Only 3 and 4 remain, and direct verification shows that $27^{2}=(2+7)^{3}=729$
|
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 number of connection options is no more than 50, i.e., (by the pigeonhole principle) some two red points will belong to the same number of segments.
|
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 total of (32-4)/2 = 14 such diagonals. Therefore, the number of diagonals parallel to any side is $14 * 16=224$. The number of diagonals not parallel to any side is $464-224=240$.
|
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 which $n=15, m=45$.
|
60
|
Number Theory
|
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 $n$ is not an integer. If $k=3$, then $k=3$, then $6 n+9=47$, and again we get a non-integer $n$. If, however, $k=1$, then $2 n+1=47$, from which $n=23$.
|
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 in the competition?
|
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., $n<\frac{700}{41}=17 \frac{3}{41}$. The largest integer $n$ satisfying this condition is 17. If 17 sumo-tori weigh $131 \frac{3}{17}$ kg each, and the remaining 3 weigh 90 kg each, the conditions of the problem are met.
|
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 digits formed a palindrome, - noted Anton.
- Seven-digit numbers are not memorized as a whole; they are broken down into three groups: first three digits, and then two groups of two digits each. I think the three-digit number obtained in this way was divisible by 9 - remarked Nikita.
- That's right, - supported Mitya, - and there were three consecutive ones in the phone number.
- Only one of the two-digit numbers obtained by Nikita's method was prime, - added Sasha.
Help the guys restore Misha's phone number.
|
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 natural numbers, there will be no three consecutive ones in the number. Therefore, the second, third, and fourth digits are ones, then the first and fifth are sevens. This means that the first two-digit number in the Nikitin partition is 17 (a prime number). The last three digits will be consecutive natural numbers if the second two-digit number is 89 or 65. But since 89 is a prime number, the only option left is 65. Therefore, the desired number is 7111765.
|
7111765
|
Logic and Puzzles
|
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 be 9, and the first differences between the digits should be as small as possible. If the differences are sequentially $3,2,1,0,-1,-2,-3$, then we get the answer: 96433469.
|
96433469
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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