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7.1. The numbers $p$ and $q$ are chosen such that the parabola $y=p x-x^{2}$ intersects the hyperbola $x y=q$ at three distinct points $A, B$, and $C$, and the sum of the squares of the sides of triangle $A B C$ is 324, while the centroid of the triangle is 2 units away from the origin. Find the product $p q$.
|
Answer: 42.
Solution. The system $y=p x-x^{2}, x y=q$ reduces to the cubic equation $x^{3}-p x^{2}+q=0$, which has, according to the problem, three distinct roots $x_{1}, x_{2}, x_{3}$, since the abscissas of any two different points on the curve $y=p x-x^{2}$ are distinct. By Vieta's theorem,
$$
x_{1}+x_{2}+x_{3}=p, \quad x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=0, \quad x_{1} x_{2} x_{3}=-q .
$$
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$. Note that if at least one of the numbers $x_{1}, x_{2}$, or $x_{3}$ is 0, then $q=0$, and the cubic equation will have only two roots, which contradicts the problem's condition. Then, from the system,
$$
y_{1}=\frac{q}{x_{1}}, \quad y_{2}=\frac{q}{x_{2}}, \quad y_{3}=\frac{q}{x_{3}}
$$
from which
$$
y_{1}+y_{2}+y_{3}=q \frac{x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}}{x_{1} x_{2} x_{3}}=0 ; \quad y_{1} y_{2}+y_{2} y_{3}+y_{1} y_{3}=q^{2} \frac{x_{1}+x_{2}+x_{3}}{x_{1} x_{2} x_{3}}=-p q
$$
The point $M$ of intersection of the medians of triangle $A B C$ has coordinates
$$
\left(\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right), \frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)\right)
$$
Taking into account Vieta's theorem, we have $M\left(\frac{p}{3}, 0\right)$. Now we will calculate the sum of the squares of the lengths of the sides of triangle $A B C$. It is equal to
$$
\begin{gathered}
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\left(x_{3}-x_{1}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}= \\
=2\left(x_{1}+x_{2}+x_{3}\right)^{2}-6\left(x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}\right)+2\left(y_{1}+y_{2}+y_{3}\right)^{2}-6\left(y_{1} y_{2}+y_{2} y_{3}+y_{1} y_{3}\right)=2 p^{2}+6 p q
\end{gathered}
$$
According to the problem's condition, $\left|\frac{p}{3}\right|=2,2 p^{2}+6 p q=324$, from which we find two pairs: $(p ; q)=(6 ; 7)$ or $(p ; q)=(-6 ;-7)$. In both cases, $p q=42$.
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. How many natural numbers $n \in [20182019 ; 20192018]$ are there for which the number $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}\right]$ is even? (Here, $[x]$ denotes the greatest integer not exceeding $x$.)
|
Answer: 4999.
Solution. Consider the sequences
$$
x_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}, \quad y_{n}=\left(\frac{1-\sqrt{5}}{2}\right)^{n}, \quad a_{n}=x_{n}+y_{n}
$$
Since $-1<\frac{1-\sqrt{5}}{2}<0$, we have $y_{n} \in(0 ; 1)$ for even $n$, and $y_{n} \in(-1 ; 0)$ for odd $n$. Then $\left[x_{n}\right]=\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}\right]$ for even $n$ equals $a_{n}-1$, and for odd $n$ it is simply $a_{n}$. Moreover, we have $a_{n+2}=a_{n+1}+a_{n}$ for all natural $n$. Since $a_{1}=1, a_{2}=3$, then $a_{3}=4$ and we get such a parity sequence for $a_{n}(n=1,2, \ldots)$: odd, odd, even, odd, odd, even, $\ldots$
Then $\left[x_{n}\right]$ is related to $a_{n}$ as follows:
| $n$ | $6 k$ | $6 k+1$ | $6 k+2$ | $6 k+3$ | $6 k+4$ | $6 k+5$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{n}$ | even | odd | odd | even | odd | odd |
| $\left[x_{n}\right]$ | $a_{n}-1$ | $a_{n}$ | $a_{n}-1$ | $a_{n}$ | $a_{n}-1$ | $a_{n}$ |
| $\left[x_{n}\right]$ | odd | odd | even | even | even | odd |
Since $x_{n}=a_{n}-y_{n}$, for numbers $n$ of the form $6 k, 6 k+1, 6 k+5$, the number $\left[x_{n}\right]$ is odd, and for numbers $n$ of the form $6 k+2, 6 k+3, 6 k+4$ it is even. Since $20182019=6 k_{1}+5, 20192018=6 k_{2}+2$, among the numbers $n=20182019, \ldots, 20192018$ the desired number is $3 \cdot \frac{20192018-20182022}{6}+1=4999$.
|
4999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The sum of positive integers is 11. In the first part of this equation, identical numbers are hidden behind cards with the same letters, and different numbers - behind cards with different letters.
Consider the equation: $\quad \mathbf{C}+\mathbf{y}+\mathbf{M}+\mathbf{M}+\mathbf{A}=11$.
Can you tell which number is hidden behind the letter M?
|
Answer: 1. Solution. If $M=1$, then the sum $C+y+A=9$, which is only possible with the set of numbers 2, 3, 4 (in any order). If $M=2$, then the sum $C+y+A=7$, which is impossible, as this sum is not less than $1+3+4=8$. If $M>2$, this is even more impossible.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Pete was given a new electric jigsaw on his birthday, with a feature to count the length of the cuts made. To try out the gift, Pete took a square piece of plywood with a side of 50 cm and cut it into squares with sides of 10 cm and 20 cm. How many squares in total were obtained, if the electric jigsaw shows a total cut length of $2 \mathbf{m} 80 \mathrm{~cm}$?
|
Answer: 16. Solution. The perimeter of the figures must be taken into account, in addition to the perimeter of the original square, from which we get that the total perimeter of the resulting squares is $280 \cdot 2 + 200 = 760$. Now, we can denote the number of squares through $x$ and $y$ respectively and solve the system $\left\{\begin{array}{c}10^{2} \cdot x + 20^{2} \cdot y = 50^{2}, \\ 4 \cdot 10 \cdot x + 4 \cdot 20 \cdot y = 760,\end{array}\right.$ or proceed with the following reasoning. If all the squares had a side of 10, there would be $6 \mathbf{~} 25$ of them, then the total perimeter would be $25 \cdot 40 = 1000$. If we replace four small squares $10 \times 10$ with one square $20 \times 20$, the perimeter would decrease by $160 - 80 = 80$ cm. To get 760 from 1000, this needs to be done 3 times. Therefore, there will be 3 squares $20 \times 20$ and 13 squares $10 \times 10$—a total of 16 squares.
## 2015/2016 Academic Year CRITERIA FOR DETERMINING WINNERS AND PRIZE WINNERS ${ }^{2}$ of the Lomonosov Mathematics Olympiad for school students 5-9 grades ELIMINATION STAGE
WINNER:
From 91 points inclusive and above.
PRIZE WINNER:
From 70 points to 90 points inclusive.
## FINAL STAGE
WINNER (Diploma I degree):
Not to be awarded.
PRIZE WINNER (Diploma II degree):
100 points inclusive.
PRIZE WINNER (Diploma III degree):
From 80 points to 99 points inclusive.
[^0]: ${ }^{2}$ Approved at the meeting of the jury of the Lomonosov Mathematics Olympiad for school students.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{224}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$ ANSWER:60.
|
Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$.
|
63
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then added: "We are all from the Ukh tribe." Another said: "There are no more than 17 of us in the hut," and then noted: "Some of us are from the Ah tribe." The third said: "There are five of us in the hut," and, looking around, added: "There are no fewer than three inhabitants from the Ukh tribe among us." How many inhabitants from the Ah tribe are in the hut?
|
Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 people in the hut. The third one is from tribe Ukh, as he said there are five of them in total. But since he said there are no fewer than three residents from tribe Ukh, there are no more than two. Two have already been found (the third and the first), so there are exactly two. Then the number of residents from tribe Ah is 15.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Photographs are archived in the order of their numbering in identical albums, with exactly 4 photographs per page. In this case, the 81st photograph in sequence landed on the 5th page of one of the albums, and the 171st photograph landed on the 3rd page of another. How many photographs can each album hold?
#
|
# Answer: 32.
Solution. Let $x, y$ be the album numbers in which the 81st and 171st photos are placed, respectively, and $n>4$ be the number of pages in the album. Then $4 n(x-1)+16<81 \leqslant 4 n(x-1)+20, 4 n(y-1)+8<171 \leqslant 4 n(y-1)+12$, i.e., $61 \leqslant 4 n(x-1)<65, 159 \leqslant 4 n(y-1)<163$. Therefore, $n(x-1)=16, n(y-1)=40$. From the first inequality, it follows that $n$ can be $1,2,4,8,16$, from the second $1,2,4,5,8,10,20$, 40. Thus, $n=8, 4 n=32$.
Answer to variant 2: 54.
Answer to variant 3: 40.
Answer to variant $4: 42$.
|
32
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Andrei likes all numbers that are not divisible by 3, and Tanya likes all numbers that do not contain digits divisible by 3.
a) How many four-digit numbers are liked by both Andrei and Tanya?
b) Find the total sum of the digits of all such four-digit numbers.
|
Answer: a) 810; b) 14580.
Solution. a) The required numbers must be composed of the digits $1,2,4,5,7,8$, and according to the divisibility rule by 3, the sum of the digits in each number should not be divisible by three. The digits 1, 4, and 7 (let's call them the digits of set $A$) give a remainder of 1 when divided by 3, while the digits $2,5,8$ (digits of set $B$) give a remainder of 2. Therefore, a number satisfying the condition must be composed in one of the following ways:
1) 4 digits from set $A$ - there are $3^{4}$ such numbers;
2) 4 digits from set $B$ - there are $3^{4}$ such numbers;
3) 3 digits from set $A$ and one digit from set $B$ - there are $4 \cdot 3^{4}$ such numbers;
4) 3 digits from set $B$ and one digit from set $A$ - there are $4 \cdot 3^{4}$ such numbers.
In total, there are $10 \cdot 3^{4}=810$ such numbers.
b) To find the total sum of the digits of all these numbers, we will divide them into pairs: the second number is obtained from the first by replacing all digits according to the principle $1 \leftrightarrow 8,2 \leftrightarrow 7,4 \leftrightarrow 5$. For example, the number 1545 has a pair 8454, the number 5271 has a pair 4728, and so on. The sum of the digits of any pair is $9 \cdot 4$, and the number of such pairs is $\frac{10 \cdot 3^{4}}{2}$. Therefore, the required sum of all digits is $\frac{9 \cdot 4 \cdot 10 \cdot 3^{4}}{2}=20 \cdot 3^{6}=14580$.
Answer to variant 2: a) 160; b) 2880.
Answer to variant 3: a) 810; b) 14580.
Answer to variant 4: a) 160; b) 2880.
|
810
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. All natural numbers are divided into "good" and "bad" according to the following rules:
a) From any bad number, you can subtract some natural number not exceeding its half so that the resulting difference becomes "good".
b) From a "good" number, you cannot subtract no more than half of it so that it remains "good".
It is known that the number 1 is "good". Find the nearest "good" number to 2015.
|
Answer: 2047. Solution. Considering the first few natural numbers, we notice that good numbers have the form $2^{n}-1$ (while $2^{n}, \ldots, 2 n+1-2$ are bad).
We will prove this by mathematical induction. For $n=1$, the statement is given in the condition. Suppose the statement is proven for $n-1$. Consider a number of the form $M=2^{n}+k$, where $k=0,1, \ldots, 2^{n}-2$. Then from such a number, we can subtract $k+1 \leqslant \frac{1}{2}\left(2^{n}+k\right)=\frac{M}{2}$.
On the other hand, consider a number of the form $N=2^{n}-1$. From it, we need to subtract at least $2^{n-1}$, which exceeds $\frac{N}{2}=2^{n-1}-\frac{1}{2}$.
Thus, the nearest good number to 2015 will be the number $2047=2^{11}-1$.
|
2047
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In a convex quadrilateral $A B C D$, the areas of triangles $A B D$ and $B C D$ are equal, and the area of $A C D$ is half the area of $A B D$. Find the length of the segment $C M$, where $M$ is the midpoint of side $A B$, if it is known that $A D=12$.
|
Answer: $C M=18$. Solution. From the equality of the areas of $A B C$ and $C B D$, it follows that $O$ - the intersection point of the diagonals $A C$ and $B D$ bisects $A C$ (see figure). And from the fact that $S(A C D) = \frac{1}{2} S(A B D)$, it follows that $S(A O D) = \frac{1}{3} S(A O B)$, therefore, $O D = \frac{1}{3} B O = G O$, where $G$ is the point of intersection of the medians of triangle $A B C$.
Thus, $A G C D$ is a parallelogram, so $A D = C G = \frac{2}{3} C M = 12$, from which $C M = 18$.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the smallest natural number that is greater than the sum of its digits by 1755 (the year of the founding of Moscow University).
|
Answer: 1770.
Solution. From the condition, it follows that the desired number cannot consist of three or fewer digits. We will look for the smallest such number in the form $\overline{a b c d}$, where $a, b, c, d$ are digits, and $a \neq 0$. We will form the equation:
$$
\begin{gathered}
\overline{a b c d}=a+b+c+d+1755 \Leftrightarrow 1000 a+100 b+10 c+d=a+b+c+d+1755 \Leftrightarrow \\
\Leftrightarrow 999 a+99 b+9 c=1755 \Leftrightarrow 111 a+11 b+c=195
\end{gathered}
$$
From the last equation, we conclude that the only possible value for the digit $a$ is 1. Next, we get the equation $11 b+c=84$, which means the digit $b$ can only be 7. Finally, we need to find the value of $c=84-11 \cdot 7=7$. Therefore, the desired number is 1770, as it is the smallest among all numbers of the form $\overline{177 d}$.
|
1770
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Having walked $2 / 5$ of the length of a narrow bridge, a pedestrian noticed that a car was approaching the bridge from behind. Then he walked back and met the car at the beginning of the bridge. If the pedestrian had continued walking forward, the car would have caught up with him at the end of the bridge. Find the ratio of the car's speed to the pedestrian's speed.
|
Answer: 5.
Solution. In the time $t$ that the pedestrian walked towards the car until they met at the beginning of the bridge, he covered $2 / 5$ of the bridge's length. Therefore, if the pedestrian continued walking forward, in time $t$ he would have covered another $2 / 5$ of the bridge's length, and he would have $1 / 5$ of the bridge's length left to walk. According to the problem, in time $t$, the car would have reached the beginning of the bridge and would have the entire bridge left to travel before meeting the pedestrian. Thus, the ratio of the car's speed to the pedestrian's speed is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. How many solutions does the equation
$$
\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}=\frac{2}{x^{2}} ?
$$
|
Answer: 1.
Solution. Let us consider the functions
$$
f(x)=\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}} \quad \text { and } \quad g(x)=\frac{2}{x^{2}}.
$$
For $x < 0$, both functions $f$ and $g$ are positive, and for $x \in (0 ; 1)$, the function $f$ is decreasing from $+\infty$ to 2 (not including), since $f^{\prime}(x) = -\frac{2}{(x-1)^{3}} - \frac{2}{(x-2)^{3}} < 0$ for $x \in (0 ; 1)$, while the function $g$ is decreasing from $+\infty$ to 2 (not including), since $g^{\prime}(x) = \frac{-4}{x^{3}} < 0$ for $x > 0$. Therefore, for $x \in (0 ; 1)$, the inequality $f(x) > g(x)$ holds, and for $x > 2$, the inequalities $0 < \frac{1}{(x-1)^{2}} < \frac{1}{x^{2}}$ and $\frac{1}{(x-2)^{2}} > \frac{1}{x^{2}}$ hold, so $f(x) > g(x)$, meaning that the equation has no solutions for $x > 2$.
## Second Solution.
The original equation, under the conditions $x \neq 0, x \neq 1, x \neq 2$, is equivalent to the equation $6 x^{3} - 21 x^{2} + 24 x - 8 = 0$. Consider the function $f(x) = 6 x^{3} - 21 x^{2} + 24 x - 8$. Since $f^{\prime}(x) = 18 x^{2} - 42 x + 24$, then $x = 1$ is a point of maximum, and $x = \frac{4}{3}$ is a point of minimum. The function $f$ is increasing on the domain $(-\infty, 1) \cup \left(\frac{4}{3} ; +\infty\right)$ and decreasing on the interval $\left(1 ; \frac{4}{3}\right)$. Since $f(0) = -8, f(1) = 1, f\left(\frac{4}{3}\right) = \frac{8}{9}$, the equation $f(x) = 0$ has a unique root, which lies in the interval $(0 ; 1)$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Given three points, the distances between which are 4, 6, and 7. How many pairwise distinct triangles exist for which each of these points is either a vertex or the midpoint of a side?
|
Answer: 11.
Solution. Let's list all the constructions of triangles that satisfy the condition of the problem, indicating the lengths of the sides.
| | Description of the Triangle | | Lengths of Sides |
| :---: | :---: | :---: | :---: |
| №1 | All points are vertices | | $4,6,7$ |
| | One point is a vertex, two are midpoints of sides | | |
| №2 | | The vertex is at the extension of both sides of lengths 4 and 6 | $8,12,14$ |
| №3 | | The extension of the side of length 4 | $8,14,2 \sqrt{94}$ |
| №4 | | The extension of the side of length 6 | $12,14,2 \sqrt{154}$ |
| №5 | | The vertex is at the extension of both sides of lengths 4 and 7 | $8,12,14$ |
| №6 | | The extension of the side of length 4 | $8,12,2 \sqrt{55}$ |
| №7 | | The extension of the side of length 7 | $12,14,2 \sqrt{154}$ |
| №8 | | The vertex is at the extension of both sides of lengths 6 and 7 | $8,12,14$ |
| №9 | | The extension of the side of length 6 | $8,12,2 \sqrt{55}$ |
| №10 | | The extension of the side of length 7 | $8,14,2 \sqrt{94}$ |
| | Two points are vertices, one is the midpoint of a side | | |
| №11 | | The vertices are on the side of length 4, the side of length 7 is extended | $4,14, \sqrt{154}$ |
| №12 | | The side of length 6 is extended | $4,12, \sqrt{154}$ |
| №13 | | The vertices are on the side of length 6, the side of length 4 is extended | $6,8, \sqrt{94}$ |
| №14 | | The side of length 7 is extended | $6,14, \sqrt{94}$ |
| №15 | | The vertices are on the side of length 7, the side of length 4 is extended | $7,8, \sqrt{55}$ |
| №16 | | The side of length 6 is extended | $7,12, \sqrt{55}$ |
| №17 | All points are midpoints of sides | | $8,12,14$ |
Thus, there are 17 ways of construction, leading to 11 different triangles.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. In a commercial football tournament, five teams participated. Each was supposed to play exactly one match against each other. Due to financial difficulties, the organizers canceled some games. In the end, it turned out that all teams had scored a different number of points, and no team had a zero in the points scored column. What is the minimum number of games that could have been played in the tournament, if three points were awarded for a win, one for a draw, and zero for a loss?
|
Answer: 6.
Solution. The minimum possible total score: $1+2+3+4+5=15$. In one game, a maximum of 3 points (in total) can be scored. Therefore, there were at least 5 games. However, if there were exactly 5 games, then all games would have to end with one of the teams winning, and then no team would have scored exactly 1 point.
Thus, there were at least 6 games. An example of the tournament results can be given by the table.
| $*$ | 1 | 2 | 3 | 4 | 5 | $\sum$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $*$ | $*$ | $*$ | 3 | 3 | 6 |
| 2 | $*$ | $*$ | $*$ | $*$ | 3 | 3 |
| 3 | $*$ | $*$ | $*$ | 1 | 3 | 4 |
| 4 | 0 | $*$ | 1 | $*$ | 1 | 2 |
| 5 | 0 | 0 | 0 | 1 | $*$ | 1 |
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. There were 21 consecutive natural numbers written on the board. When one of the numbers was erased, the sum of the remaining numbers became 2017. Which number was erased?
|
Answer: 104.
Solution: Let the numbers on the board be N-10, N-9,..,N, ..., N+10. Their sum is $21 \mathrm{~N}$. When one of these numbers - x - was erased, the sum became 2017, $21 \mathrm{~N}-$ $x=2017$. Therefore, $x=21 N-2017$, since this is one of these numbers, we get $N-10 \leq 21 N-2017 \leq N+10$. Solving the inequalities $\frac{2007}{20} \leq N \leq \frac{2027}{20}$, we get $N$ $=101$, therefore, $x=21 * 101-2017=104$.
|
104
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. 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
|
4. 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 $m+n$).
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
|
7. For the function $y=f(x)$, it is known that it is defined and continuous on the entire number line, odd, and periodic with a period of 5, and that $f(-1)=f(2)=-1$. What is the minimum number of roots that the equation $f(x)=0$ can have on the interval [1755; 2017]? Answer: 210.
|
Solution. Since the function $f$ is odd and defined at zero, we get $f(0)=-f(0)$ $\Rightarrow f(0)=0$. Due to the 5-periodicity, we then have $f(5)=f(0)=0$. Using the oddness again: $f(1)=-f(-1)=1$, and due to the 5-periodicity $f(3)=f(-2)=1, f(4)=$ $f(-1)=-1$. Thus, at points $1, 2, 3$, and $4$, the values of the function are $1, -1, 1$, and $-1$, respectively. Therefore, on each of the three intervals between these points, there is at least one zero of the function $f$. Hence, on the period $[0; 5)$, the function has at least 4 zeros (it is clear that this estimate is achievable: for example, one can take a piecewise-linear function, which will have exactly 4 zeros). On the interval $[1755; 2015)$, the period fits 52 times (on it, there are at least $52 \cdot 4=208$ zeros), plus the zero at the point 2015 and at least one on the interval (2016; 2017). In total, there are at least 210 zeros.
|
210
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Calculate $\sqrt{n}+\sqrt{n+524}$, given that this number is rational and that $n$ is a natural number.
|
Answer: 262.
Solution. Let the desired number be $a$. We have $\sqrt{n+524}=a-\sqrt{n}, n+524=a^{2}-2 a \sqrt{n}+n$. By the condition, $a$ is rational, so $\sqrt{n}$ is also rational. Therefore, $n=k^{2}, k \in \mathbb{N}$. Then the number $\sqrt{n+524}$ is also rational, so $n+524=m^{2}, m \in \mathbb{N}$. Thus, $m^{2}-k^{2}=524,(m-k)(m+k)=4 \cdot 131$. Note that the numbers $m-k$ and $m+k$ have the same parity, and the number 131 is prime. Therefore, $m-k=2$ and $m+k=2 \cdot 131$. Both equations are satisfied when $m=132, k=130$. Therefore, $a=m+k=262$.
|
262
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. We consider all possible sets consisting of 2017 different natural numbers, where in each set no number can be represented as the sum of two other numbers in the set. What is the smallest value that the largest number in such a set can take?
Answer: 4032.
|
Solution. Let for brevity $n=2017$. Consider the following set of $n$ numbers: $n-1, n, n+1, \ldots$, $2 n-3, 2 n-2$. Since $(n-1) + n = 2 n - 1 > 2 n - 2$, no number in this set can equal the sum of two others, meaning the given set satisfies the problem's condition.
Now, let there be an arbitrary set of $n$ natural numbers that satisfy the problem's condition. We will prove that the largest of these numbers $N$ is not less than $2 n - 2$. Suppose $N \leqslant 2 n - 3$. We will prove that in such a set of numbers, there is a pair of numbers less than $N$ whose sum equals $N$. Let's divide all natural numbers less than $N$ into pairs with a sum equal to $N: (1, N-1); (2, N-2), \ldots$. If there are $n-1$ or more such pairs, then the number of numbers less than $N$ will be no less than $2 n - 2$, which is impossible. Therefore, there are no more than $n-2$ such pairs, and there are $n$ numbers in the set. Thus, there will be at least one pair of numbers in the set whose sum equals $N$.
Thus, in any set of $n$ numbers that satisfy the problem's condition, the largest of these numbers is not less than $2 n - 2$ (and this estimate is achievable). In the case $n=2017$, this number is 4032.
Answer to variant 2: 4030.
|
4032
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In a full container, there are 150 watermelons and melons for a total of 24 thousand rubles, with all the watermelons together costing as much as all the melons. How much does one watermelon cost, given that the container can hold 120 melons (without watermelons) and 160 watermelons (without melons)?
|
Answer: 100 rubles. Solution. Let there be $x$ watermelons, then there will be $150-x$ melons. If a container can hold 120 melons, then one melon occupies $\frac{1}{120}$ of the container. Similarly, one watermelon occupies $\frac{1}{160}$ of the container. Therefore, if the container is fully loaded with 150 watermelons and melons (and the container is full), then $\frac{x}{160}+\frac{150-x}{120}=1$. From this, $x=120$. So, there were 120 watermelons and 30 melons. If a watermelon costs $a$ rubles, and a melon costs $b$ rubles, then (since the cost of watermelons and melons is the same): $120 a=30 b=12000$. Therefore, $a=100, b=400$.
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A $3 \times 3$ table needs to be filled with the numbers 2014, 2015, and 2016, such that the sum of the numbers in each row is the same. In how many different ways can this be done?
|
Answer: 831. Solution. Subtract 2015 from all numbers - the sums will remain the same, and the numbers in the table will take on values of 0 and $\pm 1$. Consider the 27 possible combinations of numbers in one row. Possible values of the sum of these numbers: $\pm 3$ - 1 combination each, $\pm 2$ - 3 combinations each, $\pm 1$ - 6 combinations each, $0-7$ combinations. Then the arrangement specified in the condition can be made in $2 \cdot 1^{3}+2 \cdot 3^{3}+2 \cdot 6^{3}+7^{3}=$ 831 ways.
|
831
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. At Andrei's birthday, Yana, who arrived last, gave him a ball, and Eduard, who arrived second to last, gave him a calculator. Testing the calculator, Andrei noticed that the product of the total number of his gifts and the number of gifts he had before Eduard arrived is exactly 16 more than the product of his age and the number of gifts he had before Yana arrived. How many gifts does Andrei have?
|
Answer: 18. Solution. If $n$ is the number of gifts, and $a$ is Andrei's age, then $n(n-2)=a(n-1)$. From this, $a=\frac{n^{2}-2 n-16}{n-1}=n-1-\frac{17}{n-1}$. Therefore, $n-1=17, a=16$.
|
18
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In an equilateral triangle $A B C$, points $A_{1}$ and $A_{2}$ are chosen on side $B C$ such that $B A_{1}=A_{1} A_{2}=A_{2} C$. On side $A C$, a point $B_{1}$ is chosen such that $A B_{1}: B_{1} C=1: 2$. Find the sum of the angles $\angle A A_{1} B_{1}+\angle A A_{2} B_{1}$.
|
Answer: $30^{\circ}$. Solution. Since $A_{1} B_{1} \| A B$, then $\angle B A A_{1}=\angle A A_{1} B_{1}$. From symmetry, $\angle B A A_{1}=\angle C A A_{2}$. It remains to note that $\angle C B_{1} A_{2}=$ $\angle B_{1} A A_{2}+\angle A A_{2} B_{1}$ as an exterior angle in $\triangle A A_{2} B_{1}$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the greatest possible value of $\gcd(x+2015 y, y+2015 x)$, given that $x$ and $y$ are coprime numbers.
|
Answer: $2015^{2}-1=4060224$. Solution. Note that the common divisor will also divide $(x+2015 y)-2015(y+2015 x)=\left(1-2015^{2}\right) x$. Similarly, it divides $\left(1-2015^{2}\right) y$, and since $(x, y)=1$, it divides $\left(1-2015^{2}\right)$. On the other hand, if we take $x=1, y=2015^{2}-2016$, then we get $\gcd(x+2015 y, y+2015 x)=2015^{2}-1$.
|
4060224
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A flea jumps along the number line, and the length of each jump cannot be less than $n$. It starts its movement from the origin and wants to visit all integer points belonging to the segment $[0,2013]$ (and only them!) exactly once. For what greatest value of $n$ will it be able to do this?
|
Answer: $n=1006$.
Solution: For $n=1006$, a path can be constructed as follows:
$$
0 \rightarrow 1007 \rightarrow 1 \rightarrow 1008 \rightarrow \ldots \rightarrow 1005 \rightarrow 2012 \rightarrow 1006 \rightarrow 2013
$$
We will prove that $n$ cannot be greater than 1006. Indeed, suppose $n \geqslant 1007$. Then the point with coordinate 1007 can only be reached from the start of the segment (point 0). However, if the flea jumps from there to point 1007, it cannot jump back, so it must end its path at this point and will not visit other points on the segment.
|
1006
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In how many different ways can a chess king move from square $e 1$ to square $h 5$, if it is allowed to move only one square to the right, up, or diagonally to the right-up?
|
Answer: 129 ways.
Solution: Sequentially (starting from e1) find the number of ways to reach each cell. Each number (except 1) is obtained by summing the neighbors below, to the left, and diagonally left-below.
|
129
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. How many divisors of the number $2021^{2021}$ have a cube root that is a natural number?
|
Answer: 454276.
Solution. Since $2021=43 \cdot 47$, all divisors of the number $2021^{2021}$ have the form $43^{\alpha} \cdot 47^{\beta}$, where $\alpha, \beta \in[0 ; 2021]$. In this case, the exact cubes are numbers of the form $43^{3 n} \cdot 47^{3 k}$, where $3 n, 3 k \in[0 ; 2021]$, that is, $n, k \in[0 ; 673]$. There are $674^{2}=454276$ such numbers.
|
454276
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. How many triples of numbers $a, b, c$ exist, each of which is a root of the corresponding equation $a x^{2}+b x+c=0$?
|
Answer: 5.
Solution. If $a=0$ or $a=b=c$, then $a=b=c=0$. Otherwise: if $a=b \neq c$, then either $a=b=-1, c=0$, or $a=b=1, c=-2$; if $a \neq b=c$, then either $a=1, b=c=-0.5$; if $a=c \neq b$, then $a=c=c_{0}, b=1 / c_{0}$, where the number $c_{0}<0$ is the unique root of the equation $c^{3}+c=-1$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. (2 points) From the digits 1, 3, and 5, different three-digit numbers are formed, each of which has all distinct digits. Find the sum of all such three-digit numbers.
|
Answer: 1998.
Solution. All possible numbers: $135,153,315,351,513,531$.
|
1998
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. (12 points) The number
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}
$$
was written as an irreducible fraction with natural numerator and denominator. Find the last two digits of the numerator.
|
Answer: 99.
Solution. We have
$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\ldots+\left(\frac{1}{2017!}-\frac{1}{2018!}\right)=1-\frac{1}{2018!}=\frac{2018!-1}{2018!}$.
In the end, we obtained an irreducible fraction, and the last two digits of the number 2018! - 1 are two nines.
|
99
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. (12 points) The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $19 a, \frac{124 b}{13}, \frac{c}{13}$ is an arithmetic progression. Find the common ratio of the geometric progression.
|
Answer: 247.
Solution. Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{124 a q}{13}=19 a+\frac{a q^{2}}{13} \Leftrightarrow q^{2}-248 q+247=0$, from which $q=1$ or $q=247$. A decreasing geometric progression can only occur when $q=247$ (for example, if $a=-1$).
|
247
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. (12 points) Yura has unusual clocks with several minute hands moving in different directions. Yura calculated that the minute hands coincided in pairs exactly 54 times in one hour. What is the maximum number of minute hands that can be on Yura's clocks?
#
|
# Answer: 28.
Solution. Let two arrows coincide, then after 30 seconds they will coincide again. Therefore, the arrows in each such pair will coincide exactly 2 times per minute. Thus, if $n$ arrows move in one direction, and $m$ arrows move in the other, then $2 m n=54, m n=27$. Therefore, $n$ can be $1,3,9$ or 27. The largest sum $m+n$ is obtained when $n=1, m=27$ (or vice versa) and is equal to 28.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. (12 points) In how many ways can eight of the nine digits $1,2,3,4,5,6$, 7,8 and 9 be placed in a $4 \times 2$ table (4 rows, 2 columns) so that the sum of the digits in each row, starting from the second, is 1 more than in the previous one?
|
# Answer: 64.
Solution. The sum of all nine numbers is 45. Let $x$ be the sum of the two numbers in the first row, and let $a$ be the one number out of the nine that we do not place in the figure. Then $x+x+1+x+2+x+3=45-a$, from which $4 x+a=39$. Since $a$ is an integer from 1 to 9, we get 2 possible cases: either $x=9, a=3$, or $x=8, a=7$.
If $a=3$, then we need to place the numbers $1,2,4,5,6,7,8,9$, and the sum of the numbers in the rows should be 9, 10, 11, and 12, respectively. Possible variants for the numbers in the first row: $9=1+8=2+7=4+5$.
If the first row contains 1 and 8, then the second row should contain 4 and 6, the third row 2 and 9, and the last row 5 and 7.
If the first row contains 2 and 7, then the second row should contain 1 and 9 (with other variants, it is impossible to select numbers for the third row), the third row 5 and 6, and the last row 4 and 8.
If the first row contains 4 and 5, then the second row can have 1 and 9 or 2 and 8. But in both cases, it is impossible to find numbers for the third row.
If $a=7$, then we need to place the numbers $1,2,3,4,5,6,8,9$, and the sum of the numbers in the rows should be 8, 9, 10, and 11, respectively. The sum of the numbers in the first row is $8=5+3=6+2$.
If the first row contains 3 and 5, then the second row should contain 1 and 8, the third row 6 and 4, and the last row 2 and 9.
If the first row contains 6 and 2, then the second row can contain 1 and 8 or 4 and 5. If they contain 1 and 8, it is impossible to select numbers for the next row. Therefore, they contain 4 and 5, then the next row contains 1 and 9, and the last row 3 and 8.
Thus, we have 4 variants of number placement without considering the order of numbers in the rows. Since the numbers in each row can be swapped, we get $2^{4}=16$ variants for each placement. In total, we get 64 variants.
|
64
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. (12 points) For different natural numbers $k, l, m, n$, it is known that there exist such three natural numbers $a, b, c$ that each of the numbers $k, l, m, n$ is a root of either the equation $a x^{2}-b x+c=0$, or the equation $c x^{2}-16 b x+256 a=0$. Find $k^{2}+l^{2}+m^{2}+n^{2}$.
|
Answer: 325.
Solution. If $k, l$ are the roots of the first equation, then the roots of the second equation are the numbers $m=\frac{16}{k}$, $n=\frac{16}{l}$. Therefore, the numbers $k, l, m, n$ are divisors of the number 16. The divisors of 16 are the numbers 1, 2, 4, 8, and 16, but the number 4 does not fit, as by the condition all numbers $k, l, m, n$ are distinct. Thus, $k^{2}+l^{2}+m^{2}+n^{2}=1^{2}+2^{2}+8^{2}+16^{2}=325$.
|
325
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. Every time my brother tells the truth, our grandmother sneezes. One day, my brother said he got a "5" in math, but grandmother didn't sneeze. Then, slightly doubting his first words, he said he got a "4," and grandmother sneezed. Encouraged by grandmother's sneeze, he confirmed that he definitely got no less than 3, but grandmother didn't sneeze again. So, what grade did my brother actually get in math?
|
Answer: $« 2 »$.
Solution. If the grandmother did not sneeze, then the brother definitely lied, so he did not get a "5" and even more than that, less than 3. If the grandmother did sneeze, then it is not certain that he told the truth at that moment, as the condition does not prohibit the grandmother from sneezing when the brother is lying.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. 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
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. The Martian traffic light consists of six identical bulbs arranged in two horizontal rows (one above the other) with three bulbs in each. A rover driver in the fog can distinguish the number and relative position of the lit bulbs on the traffic light (for example, if two bulbs are lit, whether they are in the same horizontal row or in different rows, whether they are in the same vertical row, or in adjacent vertical rows, or in the two outer vertical rows). However, he cannot distinguish the unlit bulbs and the body of the traffic light. Therefore, if only one bulb is lit, it is impossible to determine which one of the six it is). How many signals of the Martian traffic light can the rover driver distinguish in the fog? If no bulb on the traffic light is lit, the driver cannot see it.
|
Answer: 44.
Solution. If two traffic light signals differ only by the shift of the lit bulbs, then the driver cannot distinguish them (and vice versa). Therefore, any signal can either be transformed by a shift to the left and/or up into an indistinguishable signal, which has at least one bulb lit in the top horizontal row and at least one in the left vertical row, or the signal already has this property. There are two cases.
|
44
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. (2 points) The average age of employees in a company consisting of 13 people is 35 years. After a new employee is hired, the average age of the employees becomes 34 years. Find the age of the new employee.
|
Answer: 21.
Solution. The sum of the employees' ages before the new hire was $13 \cdot 35$, and after the new employee was hired, the sum of the ages became $14 \cdot 34$. Therefore, the age of the new employee is $14 \cdot 34 - 13 \cdot 35 = 35 - 14 = 21$.
|
21
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. (12 points) An eraser, 3 pens, and 2 markers cost 240 rubles. Two erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers?
|
Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles.
|
520
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1. (12 points) In an acute-angled triangle $A B C$, angle $A$ is $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of angle $B_{1} K B_{2}$.
|
# Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ between points $B$ and $C_{2}$, and point $B_{1}$ lies on side $AC$ between points $C$ and $B_{2}$. Therefore, the point $K$ of intersection of lines $B_{1} C_{2}$ and $C_{1} B_{2}$ lies inside the triangle. Since $C_{1} B_{2}$ is the median of the right triangle $C C_{1} A$, triangle $C_{1} B_{2} A$ is isosceles. Therefore, $\angle A B_{1} C_{2}=35^{\circ}$. Similarly, we get $\angle A C_{1} B_{2}=35^{\circ}$, from which $\angle A B_{2} C_{1}=180^{\circ}-2 \cdot 35^{\circ}=110^{\circ}$. Then $\angle B_{1} K B_{2}=\angle A B_{2} K-\angle A B_{1} K=110^{\circ}-35^{\circ}=75^{\circ}$.
|
75
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$
|
Answer: 220.
Solution. Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\frac{-5 x_{2}-1}{-3 x_{1}}\right)=\frac{10\left(x_{1}^{2}+x_{2}^{2}\right)+2\left(x_{1}+x_{2}\right)}{x_{1} x_{2}}= \\
=\frac{10\left(x_{1}+x_{2}\right)^{2}-20 x_{1} x_{2}+2\left(x_{1}+x_{2}\right)}{x_{1} x_{2}}=\frac{10 \cdot 25-20-10}{1}=220
\end{gathered}
$$
|
220
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cyclist at point $D$, which is located $\frac{2}{3}$ km from point $C$. Find the speed of the bus (in km/h), if the distance between points $A$ and $B$ is 4 km, and the speeds of the bus and the cyclist are constant.
|
Answer: 40.
Solution. Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \frac{1}{6} v_{1}+s_{0}}{v_{2}}$ hours, where $s_{0}=\frac{1}{3}$ km. From this, we find $\frac{1}{4} v_{1}^{2}+v_{1}-24=0, v_{1}=8$ km/h (the second root is negative), $v_{2}=40 \text{km}/$ h.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. (2 points) In a nine-story building, there are 4 apartments on each floor. How many entrances are there in this building if there are a total of 180 apartments?
|
Answer: 5.
Solution. The number of entrances is $180:(4 \cdot 9)=5$.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. (14 points) Mitya is 11 years older than Shura. When Mitya was as old as Shura is now, he was twice as old as she was. How old is Mitya?
|
Answer: 33.
Solution. 11 years ago, Shura was half as old as she is now. So, she is 22 years old, and Mitya is 33.
|
33
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. (14 points) Represent the number $\frac{201920192019}{191719171917}$ as an irreducible fraction. In the answer, write down the denominator of the resulting fraction.
|
Answer: 639.
Solution. We have
$$
\frac{201920192019}{191719171917}=\frac{2019 \cdot 100010001}{1917 \cdot 100010001}=\frac{2019}{1917}=\frac{3 \cdot 673}{3 \cdot 639}=\frac{673}{639}
$$
Since $639=3^{2} \cdot 71$ and 673 is not divisible by 3 and 71, the resulting fraction is irreducible.
|
639
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1. (14 points) How many six-digit numbers exist whose sum of digits equals 51?
|
Answer: 56.
Solution. Since the maximum sum of the digits of a six-digit number is $6 \cdot 9=54$, the sought numbers are $699999, 789999, 888999$, as well as all numbers obtained from them by rearranging the digits. The first of these numbers can form 6 six-digit numbers (the digit 6 can be in any of the 6 positions), the second can form $6 \cdot 5=30$ numbers (6 options for placing one of the digits 7, 8, and another 5 for the other), and the third can form $\frac{6 \cdot 5 \cdot 4}{1 \cdot 2 \cdot 3}=20$ numbers (6 options for one digit 8, another 5 options for the second, and 4 options for the third, but since these digits are the same, we have counted each number $1 \cdot 2 \cdot 3=6$ times). In total, $6+30+20=56$ numbers.
|
56
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. (14 points) The old man was pulling the turnip, and one by one, the old woman, the granddaughter, the dog, and the cat joined him. They pulled and pulled, but couldn't pull out the turnip! The cat called the mouse. They pulled and pulled, and finally pulled out the turnip! It is known that each subsequent participant pulls with a quarter of the strength of the previous one. How many men from the village, pulling with the same strength as the old man, should have been called to help the old man and the cat pull out the turnip?
|
Answer: 2.
Solution. The old woman, the granddaughter, Zhuchka, and the mouse pull with a combined force of
$$
\frac{3}{4}+\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4}\right)^{3}+\left(\frac{3}{4}\right)^{5}=\frac{2019}{1024}=1 \frac{995}{1024}
$$
of the force of the old man. Therefore, two men need to be called.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. (14 points) Find the smallest natural number whose representation contains 5 zeros and 7 ones, and the sum of the digits in the even positions is equal to the sum of the digits in the odd positions.
|
Answer: 1000001111131.
Solution. The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Consider the case where the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. Thus, to find the smallest number, we choose it so that this digit is as far to the right in the number as possible. Let's try to find the desired number in the form 100000111111x. Equating the sums of the digits in the even and odd positions, we get the equation $4+x=3$, from which $x=1$, which is impossible. Let's check the number in the form $a=10000011111 x 1$. Calculating the sums of the digits in the even and odd positions, we get $5=2+x \Rightarrow x=3$. Therefore, the desired number is 1000001111131.
|
1000001111131
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. (14 points) Misha drew a triangle with a perimeter of 11 and cut it into parts with three straight cuts parallel to the sides, as shown in the figure. The perimeters of the three shaded figures (trapezoids) turned out to be 5, 7, and 9. Find the perimeter of the small triangle that resulted from the cutting.
#
|
# Answer: 10.
Solution. Since opposite sides of parallelograms are equal, the lateral sides of each resulting trapezoid are equal to the corresponding segments on the sides of the original triangle. Therefore, the sum of the perimeters of the three trapezoids is equal to the sum of the perimeter of the original triangle (the sum of the larger bases and lateral sides of the three trapezoids) and the perimeter of the new triangle (the sum of the smaller bases of the trapezoids). Thus, the perimeter of the new triangle is equal to the sum of the perimeters of the trapezoids minus the perimeter of the original triangle: \(a + b + c - P\).
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. (14 points) Two people spend time playing a game: they take turns naming prime numbers not exceeding 100 such that the last digit of the number named by one player is equal to the first digit of the number named by the next player (except for the first prime number named in the game). Repeating numbers that have already been named is not allowed. The player who cannot name the next prime number according to these rules loses. Prove that one of the players can act in such a way as to guarantee a win, and find the smallest possible number of prime numbers that will be used by both players in such a game.
#
|
# Answer: 3.
Solution. We will describe a winning strategy for the first player. First, the first player names a prime number ending in 9 and different from 79 (for example, 19). Since among the numbers $90, \ldots, 99$ the only prime is 97, the second player must name this number on their next move. Then, on the third move, the first player names 79, and the second player loses. The first player cannot win in fewer moves, as for any digit from 1 to 9, there exists a prime number in the first hundred that starts with this digit.
|
3
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
# Task 1.
Mom told Dima that he should eat 13 spoons of porridge. Dima told his friend that he had eaten 26 spoons of porridge. Each child, telling about Dima's feat, increased the amount of porridge Dima had eaten by 2 or 3 times. As a result, one of the children told Dima's mom about 33696 spoons of porridge. How many times in total did the children, including Dima, tell about the feat?
|
Answer: 9
Solution. $33696=2^{5} \cdot 3^{4} \cdot 13$. From this, it follows that the story of the feat was retold 5 times with the amount of porridge doubling and 4 times with it tripling, totaling 9 times.
## B-2
Dad persuaded Natasha to eat 11 spoons of porridge. Natasha told her friend that she had eaten 22 spoons of porridge. Then each child, retelling the story of Natasha's feat, increased the amount of porridge she had eaten by 2 or 3 times. As a result, one of the children told Natasha's dad about 21384 spoons of porridge. How many times in total did the children, including Natasha, retell the story?
## Answer: 8
## B-3
Mom told Grisha that he had to eat 3 spoons of porridge. Grisha told his friend that he had eaten 9 spoons of porridge. Then each child, retelling the story of Grisha's feat, increased the amount of porridge he had eaten by 3 or 5 times. As a result, one of the children told Grisha's mom about 18225 spoons of porridge. How many times in total did the children, including Grisha, retell the story?
Answer: 7
## B-4
Dad persuaded Tanya to eat 7 spoons of porridge. Tanya told her friend that she had eaten 21 spoons of porridge. Then each child, retelling the story of Tanya's feat, increased the amount of porridge she had eaten by 2 or 3 times. As a result, one of the children told Tanya's dad about 27216 spoons of porridge. How many times in total did the children, including Tanya, retell the story?
Answer: 9
## Lomonosov Olympiad for Schoolchildren in Mathematics
Preliminary Stage 2020/21 academic year for 5-6 grades
#
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 2.
B-1.
At the time when a lion cub, who was 5 minutes away, set off for the watering hole, the second cub, having already quenched its thirst, started heading back along the same road at 1.5 times the speed of the first. At the same time, a tortoise, which was half an hour away, set off for the watering hole along the same road. After some time, all three met at one point and then continued on their way. How many minutes after the meeting did the tortoise reach the watering hole, given that all three moved at constant speeds?
|
Answer: 28
Solution. Let's take the entire path of the turtle as 1 and let $x$ be the speed of the 1st lion cub. Then the speed of the 2nd lion cub is $1.5x$, and the speed of the turtle is $1/30$. The entire path to the watering hole for the 1st lion cub is $5x$. Therefore, the meeting with the 2nd lion cub occurred after $5x/(x+1.5x)=2$ minutes from the start of the movement. The turtle also moved for the same amount of time before the meeting. The remaining distance was covered by the turtle in $30-2=28$ minutes.
## B-2
At the time when one lion cub, located 7.5 minutes away from the watering hole, set off for it, the second, having quenched his thirst, headed back along the same path at twice the speed of the first. At the same time, a turtle, located half an hour away from the watering hole, set off for it along the same path. After some time, all three met at one point and continued on their way. How many minutes after the meeting did the turtle reach the watering hole, given that all three moved at constant speeds?
Answer: 27.5
## B-3
At the time when one lion cub, located 7 minutes away from the watering hole, set off for it, the second, having quenched his thirst, headed back along the same path at 2.5 times the speed of the first. At the same time, a turtle, located half an hour away from the watering hole, set off for it along the same path. After some time, all three met at one point and continued on their way. How many minutes after the meeting did the turtle reach the watering hole, given that all three moved at constant speeds?
Answer: 28
## B-4
At the time when one lion cub, located 6 minutes away from the watering hole, set off for it, the second, having quenched his thirst, headed back along the same path at three times the speed of the first. At the same time, a turtle, located half an hour away from the watering hole, set off for it along the same path. After some time, all three met at one point and continued on their way. How many minutes after the meeting did the turtle reach the watering hole, given that all three moved at constant speeds?
Answer: 38.5
## Lomonosov School Olympiad in Mathematics
Preliminary stage $2020 / 21$ academic year for $5-6$ grades
## B-1
#
|
28
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 3.
A square with a side of 75 mm was cut by two parallel cuts into three rectangles. It turned out that the perimeter of one of these rectangles is half the sum of the perimeters of the other two. What is this perimeter? Give your answer in centimeters.
|
# Answer: 20
Solution. For three rectangles, one of the sides is the same and equals 75 mm. The sum of the lengths of two such sides of one rectangle is half the sum of the lengths of all such sides of the other two. Therefore, the other side of the first rectangle is also half the sum of the other sides of the two other rectangles. Thus, it equals $75: 3=25$ mm, from which the perimeter of the first rectangle is $2 \cdot(25+75)=200$ mm $=20 \mathrm{cm}$.
## B-2
A square with a side of 105 mm was cut by two parallel cuts into three rectangles. It turned out that the perimeter of one of these rectangles is half the sum of the perimeters of the other two. What is this perimeter? Give your answer in centimeters.
Answer: 28
## B-3
A square with a side of 135 mm was cut by two parallel cuts into three rectangles. It turned out that the perimeter of one of these rectangles is half the sum of the perimeters of the other two. What is this perimeter? Give your answer in centimeters.
Answer: 36
## B-4
A square with a side of 165 mm was cut by two parallel cuts into three rectangles. It turned out that the perimeter of one of these rectangles is half the sum of the perimeters of the other two. What is this perimeter? Give your answer in centimeters.
Answer: 44
## Lomonosov School Olympiad in Mathematics
Preliminary stage $2020 / 21$ academic year for $5-6$ grades
## B-1
#
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4.
Find the smallest 12-digit natural number that is divisible by 36 and contains all 10 digits in its decimal representation.
|
Answer: 100023457896
Solution. The number is divisible by 4 and 9. Since the sum of ten digits is 45 (divisible by 9), two more digits should be added to these ten digits, the sum of which is 0, 9, or 18. Since we need the smallest number, we will add two digits 0 and place the number 10002345 at the beginning of the required number (this is the smallest possible "start" of the number). The remaining digits $6,7,8$ and 9 need to be arranged so that the number is divisible by 4. This means the last two digits should be 68, 76, or 96. The minimum is 7896.
## B-2
Find the smallest 13-digit natural number that is divisible by 75 and contains all 10 digits in its notation.
Answer: 1000023468975
Solution. The number is divisible by 3 and 25. The sum of ten digits is 45 (divisible by three), so the sum of the three additional digits in the number's notation should be $0,3,6, \ldots, 27$. The older the digit place, the smaller the digit it is better to put in it, so we will take three zeros and place the number 10000 at the beginning of the required number. To make the number divisible by 25, the last two digits should be 25, 50, or 75. The first two options use small digits, which are better placed in older places, so we choose 75 and get the answer: 1000023468975.
## B-3
Find the largest 12-digit natural number that is divisible by 36 and contains all 10 digits in its decimal notation.
Answer: 999876543120
The number is divisible by 4 and 9. Since the sum of ten digits is 45 (divisible by 9), two more digits should be added to these ten digits, the sum of which is 0, 9, or 18. The older the digit place, the larger the digit it is better to put in it, so we will add two nines to the ten digits and start the number as: 999.... For divisibility by 4, the last two digits should form a number divisible by 4: $00,04,08,12,16,20,24, \ldots, 96$. Of these options, 20 is the best because it leaves the larger digits for the older places (we have only one zero, so 00 is out). In the end, we get the number 999876543120.
## B-4
Find the largest 13-digit natural number that is divisible by 75 and contains all 10 digits in its notation.
Answer: 9999876432150
Solution. The number is divisible by 3 and 25. The sum of ten digits is 45 (divisible by three), so the sum of the three additional digits in the number's notation should be $0,3,6, \ldots, 27$. The older the digit place, the larger the digit it is better to put in it, so we will take three nines and start the number as: $9999 \ldots$. To make the number divisible by 25, the last two digits should be 25, 50, or 75. The five is used in any case, and it is best to choose 50 to leave the larger digits 7 and 2 for the older places. In the end, we get the number 9999876432150.
## Lomonosov Olympiad for Schoolchildren in Mathematics
Preliminary Stage 2020/21 academic year for 5-6 grades
#
|
100023457896
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5.
B-1
Chicks hatch on the night from Sunday to Monday. For two weeks, the chick sits with its beak open, the third week it silently grows feathers, and on the fourth week, it flies away from the nest. Last week, 20 chicks were sitting with their beaks open, and 14 were growing feathers, while this week, 15 were sitting with their beaks open and 11 were growing feathers. a) How many chicks were sitting with their beaks open two weeks ago? b) How many chicks will be growing feathers next week? In the answer, write the product of these numbers.
|
Answer: 225
Solution. In fact, the chicks should be divided into three categories: one-week-old, two-week-old, and three-week-old. With each new week, each chick moves to the next category. So, if 11 are feathering this week, then last week there were 11 two-week-olds, and, accordingly, 9 one-week-olds ( $9+11=20$ ). This means that this week there are 9 two-week-olds, and they will be feathering next week. Last week, there were 11 two-week-olds and 14 three-week-olds - which means that the week before last, there were 11 one-week-olds and 14 two-week-olds, and it was they who were opening their beaks. The answer is $9(11+14)$.
## B-2
In a workshop, wooden horses are made. A figurine is painted for two days, on the third day it is lacquered, and on the fourth day, it is shipped to the store. Today, 12 figurines are being painted, and 7 are being lacquered. Yesterday, 11 figurines were painted, and 10 were lacquered. a) How many figurines were painted the day before yesterday? b) How many figurines will be lacquered tomorrow? Write the product of these numbers as the answer.
Answer: 68
## B-3
In an enclosed area of the Universe, stars flare up. Each star shines brightly for two billion years, then glows dimly for a billion years, and then goes out completely. Today, in that area of the Universe, there are 7 bright and 12 dim stars, a billion years ago there were 16 bright and 8 dim. a) How many dim stars will you count there a billion years from now? b) How many bright stars were there two billion years ago? Write the product of the obtained numbers as the answer.
Answer: 80
## B-4
Every year, new models of phones appear on the market. In the first year after release, a model is considered the cutting edge of progress, in the second and third years, the model is downgraded to just "decent," and after that, it hopelessly becomes outdated. This year, there are 8 cutting-edge models and 11 decent ones. Last year, there were 4 cutting-edge models and 17 decent ones. a) How many models were at the cutting edge of progress by the standards of the year before last? b) How many decent models will there be next year? Write the product of the obtained numbers as the answer.
Answer: 84
## Lomonosov Olympiad for Schoolchildren in Mathematics
Preliminary stage $2020 / 21$ academic year for $5-6$ grades
## B-1
#
|
225
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 6.
In the alphabet of the inhabitants of the magical planet ABV2020, there are only three letters: A, B, and V, from which all words are formed. In any word, two identical letters cannot be adjacent, and each of the three letters must be present in any word. For example, the words ABV, VABAVAB, and BVBVAB are permissible, while the words VAV, ABAAVA, and AVABBB are not. How many 20-letter words are there in the dictionary of this planet?
|
# Answer: 1572858
Solution. The first letter can be any of the three, and for each subsequent letter, there are two options. This results in $3 \cdot 2^{19}$ words. However, we need to subtract from this the words that are made up of only two letters, not three. There are 6 such words. Therefore, we get $3 \cdot 2^{19}-6=1572858$ different words.
## B-2
In the alphabet of the inhabitants of the magical planet ABV2020, there are only three letters: A, B, and V, from which all words are formed. In any word, two identical letters cannot be adjacent, and each of the three letters must be present in any word. For example, the words AVB, VAVAVA, BVVVBVA are valid, while the words VAV, ABAAVA, AVABB are not. How many 18-letter words are there in the dictionary of this planet?
Answer: 393210
## B-3
In the alphabet of the inhabitants of the magical planet ABV2020, there are only three letters: A, B, and V, from which all words are formed. In any word, two identical letters cannot be adjacent, and each of the three letters must be present in any word. For example, the words AVB, VAVAVA, BVVVBVA are valid, while the words VAV, ABAAVA, AVABB are not. How many 19-letter words are there in the dictionary of this planet?
Answer: 786426
## B-4
In the alphabet of the inhabitants of the magical planet ABV2020, there are only three letters: A, B, and V, from which all words are formed. In any word, two identical letters cannot be adjacent, and each of the three letters must be present in any word. For example, the words AVB, VAVAVA, BVVVBVA are valid, while the words VAV, ABAAVA, AVABB are not. How many 17-letter words are there in the dictionary of this planet?
Answer: 196602
## Lomonosov Olympiad for Schoolchildren in Mathematics
Preliminary stage $2020 / 21$ academic year for $5-6$ grades
## B-1
#
|
1572858
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 7.
Around a round table, 1001 people are sitting, each of whom is either a knight (always tells the truth) or a liar (always lies). It turned out that next to each knight sits exactly one liar, and next to each liar there is a knight. What is the minimum number of knights that can sit at the table?
|
Answer: 502
Solution. From the condition, it follows that a knight cannot sit between two knights or two liars, and a liar cannot sit between two liars. Thus, when moving around the table, knights will be encountered in pairs, while liars will be encountered singly or in pairs. From this, it follows that each liar can be paired with a knight (and there will still be knights who have not been paired). Therefore, the number of knights at the table is more than half of the total number of people and is also an even number, i.e., at least 502.
An example seating arrangement of 502 knights and 499 liars that satisfies the condition: first, seat 248 groups of "knight, knight, liar, liar," then three more groups of "knight, knight, liar."
## B-2
At a round table, 2001 people are sitting, each of whom is either a knight (always tells the truth) or a liar (always lies). It turns out that next to each knight sits exactly one liar, and next to each liar, there is a knight. What is the maximum number of liars that can sit at the table?
Answer: 999
## B-3
At a round table, 3001 people are sitting, each of whom is either a knight (always tells the truth) or a liar (always lies). It turns out that next to each knight sits exactly one liar, and next to each liar, there is a knight. What is the minimum number of knights that can sit at the table?
Answer: 1502
## B-4
At a round table, 4001 people are sitting, each of whom is either a knight (always tells the truth) or a liar (always lies). It turns out that next to each knight sits exactly one liar, and next to each liar, there is a knight. What is the maximum number of liars that can sit at the table?
Answer: 1999
|
502
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The numbers from 1 to 8 are arranged at the vertices of a cube such that the sum of the numbers in any three vertices lying on the same face is at least 10. What is the smallest possible sum of the numbers at the vertices of one face?
|
Answer: 16.
Solution. Each face has a vertex where a number not less than 6 is placed. Indeed, otherwise, one of the triples of the remaining largest numbers $2,3,4,5$ would give a sum less than 10 (specifically, the triple $2,3,4$ with a sum of 9).
Consider a face containing the vertex where the number 6 is placed. Since the sum of the numbers in the other three vertices is not less than 10, the sum of all numbers in the vertices of this face is not less than 16.
An example of an arrangement where the smallest sum of numbers in the vertices of one face is 16 is shown in the figure: the sum of the numbers in the front face is $2+3+5+6=16$.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On graph paper (the side of a cell is 1 cm) a rectangle is drawn, the sides of which lie on the grid lines, with one side being 7 cm shorter than the other. It turned out that it can be cut into several pieces along the grid lines and form a square from them. What can the side of this square be? Find all possible values.
|
Answer: 6 cm.
Solution. Let the larger side of the rectangle be $k$ cm, where $k>7$, and the side of the resulting square be $n$ cm. Then, from the equality of the areas, we get $k(k-7)=n^{2}$. Since $n<k$, let $n=k-m$, where $m \geqslant 1$. Then $k^{2}-7 k=(k-m)^{2}, k^{2}-7 k=k^{2}-2 m k+m^{2}$, from which $m(2 k-m)=7 k$. Therefore, one of the numbers $m$ or $2 k-m$ is divisible by 7.
If $m=7 s$, then we get $s(2 k-7 s)=k$, from which
$$
k=\frac{7 s^{2}}{2 s-1}=7\left(\frac{s}{2}+\frac{1}{4}+\frac{1}{4(2 s-1)}\right) \quad \Rightarrow \quad 4 k=14 s+7+\frac{7}{2 s-1}
$$
This means that 7 is divisible by $2 s-1$, which is possible only when $s=1$ or $s=4$. In the first case, $m=7, k=7$ and $n=0$, which is impossible. In the second case, we get $m=28, k=16$ and $n=12$.
The case where $2 k-m$ is divisible by 7 reduces to the previous one by substituting $m_{1}=2 k-m, m=$ $2 k-m_{1}$.
Thus, the only possible value of $n$ is 12, with $k=16$. The rectangle $16 \times 9$ can be cut along the grid lines and assembled into a square $12 \times 12$ in many ways.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. All natural numbers, the sum of the digits of each of which is equal to 5, were arranged in ascending order. What number is in the 125th place
|
Answer: 41000.
Solution. Let's calculate the number of $n$-digit numbers, the sum of the digits of each of which is equal to 5, for each natural $n$. Subtract 1 from the leading digit, we get a number (which can now start with zero), the sum of the digits of which is equal to 4. Represent the digits of this number as cells, in each of which there is a number of balls equal to the digit in the corresponding place. Distributing 4 balls into $n$ cells is the same as placing $n-1$ partitions between 4 balls (there may be no balls between some partitions). This can be done in $C_{n+3}^{4}=\frac{(n+3)(n+2)(n+1) n}{24}$ ways, which is also the number of the desired $n$-digit numbers.
For $n=1,2,3,4,5$, we get $C_{4}^{4}=1, C_{5}^{4}=5, C_{6}^{4}=15, C_{7}^{4}=35, C_{8}^{4}=70$, totaling 126 numbers. The 126th number is the largest five-digit such number, i.e., 50000. Therefore, the 125th number is the previous one - 41000.
|
41000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. In "Dragon Poker," the deck has four suits. An Ace brings 1 point, a Jack -2 points, a Two $-2^{2}$, a Three $-2^{3}$, ..., a Ten $-2^{10}=1024$ points. Kings and Queens are absent. You can choose any number of cards from the deck. In how many ways can you score 2018 points?
|
Answer: $C_{2021}^{3}=1373734330$.
Solution. In any suit, you can score any number of points from 0 to 2047, and this can be done in a unique way. This can be done as follows: write down this number in binary and select the cards corresponding to the binary positions containing 1 (i.e., if there is a 1 in the $k$-th position, then you should take the card that brings $2^{k}$ points).
Thus, the problem reduces to partitioning the number 2018 into 4 non-negative integer addends. Reasoning similarly to the previous problem, we get the desired number of ways: $C_{2021}^{3}=1373734330$.
|
1373734330
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 8.
Inside a convex $n$-gon, 100 points are placed such that no three of these $n+100$ points lie on the same line. The polygon is divided into triangles, each of whose vertices are 3 of the given $n+100$ points. For what maximum value of $n$ can there not be more than 300 triangles?
|
Answer: 102
Solution. If each point is a vertex of some triangle, then the sum of the angles of all the obtained triangles is $180^{\circ} \cdot(n-2)+360^{\circ} \cdot 100=180^{\circ} \cdot(n-2+200)=180^{\circ} \cdot(n+198)$. Therefore, $\frac{180^{\circ} \cdot(n+198)}{180^{\circ}}=n+198$ triangles are obtained. If one of the points does not participate in the cutting, there will be fewer triangles. Therefore, $n+198 \leqslant 300, n \leqslant 102$.
## B-2
Inside a convex 200-gon, $n$ points are placed such that no three of these $n+200$ points lie on the same line. The polygon is cut into triangles, the vertices of each of which are 3 of the given $n+200$ points. For what maximum value of $n$ can there not be more than 400 triangles?
Answer: 101
## B-3
Inside a convex $n$-gon, 200 points are placed such that no three of these $n+200$ points lie on the same line. The polygon is cut into triangles, the vertices of each of which are 3 of the given $n+200$ points. For what values of $n$ can there not be more than 600 triangles?
Answer: 202
## B-4
Inside a convex 100-gon, $n$ points are placed such that no three of these $n+100$ points lie on the same line. The polygon is cut into triangles, the vertices of each of which are 3 of the given $n+100$ points. For what values of $n$ can there not be more than 500 triangles?
Answer: 201
## Lomonosov Olympiad for Schoolchildren in Mathematics
Preliminary Stage 2020/21 academic year for $10-11$ grades
## B-1
|
102
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 9.
Given a polynomial $P(x)$ of degree 10 with the leading coefficient 1. The graph of $y=P(x)$ lies entirely above the $O x$ axis. The polynomial $-P(x)$ is factored into irreducible factors (i.e., polynomials that cannot be expressed as the product of two non-constant polynomials). It is known that at $x=2020$, all the obtained irreducible polynomials take the value -3. Find $P(2020)$.
|
Answer: 243
Solution. Note that since the graph of $y=P(x)$ is entirely above the $O x$ axis, the polynomial $P(x)$ has no real roots. This means that all irreducible polynomials in the factorization have degree 2, and their number is 5. Since the polynomial $-P(x)$ will have the same number of irreducible polynomials in its factorization, and all of them take the value -3 at the point $x=2020$, then $-P(2020)=(-3)^{5}=-243$. Hence, $P(2020)=243$.
## B-2
Given a polynomial $P(x)$ of degree 12 with the leading coefficient 1. The graph of $y=P(x)$ is entirely above the $O x$ axis. The polynomial $P(x)$ is factored into irreducible factors (i.e., polynomials that cannot be represented as a product of two non-constant polynomials). It is known that at $x=2020$ all the obtained irreducible polynomials take the value -3. Find $P(2020)$.
Answer: 729
## B-3
Given a polynomial $P(x)$ of degree 20 with the leading coefficient 1. The graph of $y=P(x)$ is entirely above the $O x$ axis. This polynomial is factored into irreducible factors (i.e., polynomials that cannot be represented as a product of two non-constant polynomials). It is known that at $x=2020$ all the obtained irreducible polynomials take the value 2. Find $P(2020)$.
Answer: 1024
## B-4
Given a polynomial $P(x)$ of degree 18 with the leading coefficient -1. The graph of $y=P(x)$ is entirely below the $O x$ axis. The polynomial $P(x)$ is factored into irreducible factors (i.e., polynomials that cannot be represented as a product of two non-constant polynomials). It is known that at $x=2020$ all the obtained irreducible polynomials take the value -2. Find $P(2020)$.
Answer: -512
## B-5
Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-210 x+11000}{\sqrt[2021]{x^{2021}-420 x^{2020}}}>1
$$
Answer: 421
Solution. We will solve the problem in general, specifying the constraints on the parameters $a, b, c$: Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-(a+b) x+a b}{\sqrt[2021]{x^{2021}-c x^{2020}}}>1
$$
We assume that $a, b, c$ are natural numbers, $1 \leqslant a < b$, and $c \geqslant a+b$. For $x \leqslant a$, the numerator is non-positive, and the denominator is positive, so the left side of the inequality is non-positive. For $x \geqslant b$, the numerator is non-negative, and the denominator is positive, so the left side of the inequality is non-negative. For $a < x < b$, the numerator is positive, and the denominator is positive, so the left side of the inequality is positive. We need to find the smallest $x$ such that the left side of the inequality is greater than 1.
For $x = a$, the numerator is 0, and the denominator is positive, so the left side of the inequality is 0. For $x = b$, the numerator is 0, and the denominator is positive, so the left side of the inequality is 0. For $x = c$, the numerator is positive, and the denominator is positive, so the left side of the inequality is positive. We need to find the smallest $x$ such that the left side of the inequality is greater than 1.
For $x = c+1$, the numerator is positive, and the denominator is positive, so the left side of the inequality is positive. We need to check if the left side of the inequality is greater than 1.
The numerator is
$$
x^{2}-(a+b) x+a b = (x-a)(x-b)
$$
The denominator is
$$
\sqrt[2021]{x^{2021}-c x^{2020}} = x \sqrt[2021]{1 - \frac{c}{x}}
$$
For $x = c+1$, the denominator is
$$
(c+1) \sqrt[2021]{1 - \frac{c}{c+1}} = (c+1) \sqrt[2021]{\frac{1}{c+1}} = (c+1) \frac{1}{(c+1)^{1/2021}} = (c+1)^{1 - 1/2021}
$$
The left side of the inequality is
$$
\frac{(c+1-a)(c+1-b)}{(c+1)^{1 - 1/2021}}
$$
Since $c \geqslant a+b$, we have $c+1 > a+b$, so $(c+1-a)(c+1-b) > 1$. Since $(c+1)^{1 - 1/2021} < c+1$, we have
$$
\frac{(c+1-a)(c+1-b)}{(c+1)^{1 - 1/2021}} > 1
$$
Therefore, the smallest integer satisfying the inequality is $c+1$.
Answer: $c+1$
## B-6
Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-171 x+7290}{\sqrt[2021]{x^{2021}-215 x^{2020}}}>1
$$
Answer: 216
## B-7
Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-251 x+15730}{\sqrt[2021]{x^{2021}-374 x^{2020}}}>1
$$
Answer: 375
## B-8
Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-294 x+21600}{\sqrt[2021]{x^{2021}-517 x^{2020}}}>1
$$
Answer: 518
## B-9
Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-210 x+11000}{\sqrt[2021]{x^{2021}-425 x^{2020}}}>1
$$
Answer: 426
## B-10
Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-171 x+2290}{\sqrt[2021]{x^{2021}-219 x^{2020}}}>1
$$
Answer: 220
## B-11
Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-251 x+15730}{\sqrt[2021]{x^{2021}-379 x^{2020}}}>1
$$
Answer: 380
## B-12
Find the smallest integer satisfying the inequality
$$
\frac{x^{2}-294 x+21600}{\sqrt[2021]{x^{2021}-519 x^{2020}}}>1
$$
Answer: 520
## Lomonosov Olympiad for Schoolchildren
Preliminary Round 2020/21 academic year for $10-11$ grades
## B-1
#
|
243
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 10.
Set $A$ on the plane $O x y$ is defined by the equation $x^{2}+y^{2}=2 x+2 y+23$. Set $B$ on the same plane is defined by the equation $|x-1|+|y-1|=5$. Set $C$ is the intersection of sets $A$ and $B$. What is the maximum value that the product of the lengths of $n$ segments $X Y_{1} \cdot X Y_{2} \cdot X Y_{3} \cdot \ldots \cdot X Y_{n}$ can take, where point $X$ is an arbitrarily chosen point from set $A$, and points $Y_{1}, Y_{2}, Y_{3}, \ldots, Y_{n}$ are all elements of set $C$?
|
Answer: 1250
Solution. The set $A$ is $(x-1)^{2}+(y-1)^{2}=25$, which is a circle with center $(1 ; 1)$ and radius 5. The set $B$ is a square with vertices $(-4 ; 1),(1 ; 6),(6 ; 1),(1 ;-4)$. The center of the square is the point $(1 ; 1)$, the diagonals of the square are 10, and the sides are $5 \sqrt{2}$. The set $C$ consists of 4 points on the circle: $(-4 ; 1),(1 ; 6),(6 ; 1),(1 ;-4)$, which are also the vertices of the inscribed square. Let's denote them as $Y_{1}, Y_{2}, Y_{3}, Y_{4}$. If we now take a point $X$ on the arc $Y_{2} Y_{3}$ (for definiteness), denote the distance from $X$ to the line $Y_{1} Y_{2}$ as $a$, and note that $\angle Y_{2} X Y_{1}=\frac{\angle Y_{2} O Y_{1}}{2}=45^{\circ}$, then we get: $X Y_{1} \cdot X Y_{2}=\frac{X Y_{1} \cdot X Y_{2} \cdot 2 \sin 45^{\circ}}{2 \sin 45^{\circ}}=S_{X Y_{1} Y_{2}} \cdot \frac{2}{\sin 45^{\circ}}=\frac{Y_{1} Y_{2} \cdot a}{2} \cdot \frac{2 \cdot 2}{\sqrt{2}}=\frac{5 \sqrt{2} \cdot a}{2} \cdot \frac{4}{\sqrt{2}}=10 a$. Similarly, $X Y_{3} \cdot X Y_{4}=10(5 \sqrt{2}-a)$. Then $X Y_{1} \cdot X Y_{2} \cdot X Y_{3} \cdot X Y_{4}=100 a(5 \sqrt{2}-a)$. The maximum of this quadratic function is achieved at $a=\frac{5 \sqrt{2}}{2}$ and is equal to 1250. Note that this means that the point $X$ is at the midpoint of the arc $Y_{2} Y_{3}$ (or any of the other three arcs).
## B-2
The set $A$ on the plane $O x y$ is defined by the equation $x^{2}+y^{2}=2 x+2 y+14$. The set $B$ on the same plane is defined by the equation $|x-1|+|y-1|=4$. The set $C$ is the intersection of sets $A$ and $B$. What is the maximum value that the product of the lengths of $n$ segments $X Y_{1} \cdot X Y_{2} \cdot X Y_{3} \cdot \ldots \cdot X Y_{n}$ can take, where the point $X$ is an arbitrarily chosen point from set $A$, and the points $Y_{1}, Y_{2}, Y_{3}, \ldots, Y_{n}$ are all elements of set $C$.
Answer: 512
## B-3
The set $A$ on the plane $O x y$ is defined by the equation $x^{2}+y^{2}=2 x+2 y+7$. The set $B$ on the same plane is defined by the equation $|x-1|+|y-1|=3$. The set $C$ is the intersection of sets $A$ and $B$. What is the maximum value that the product of the lengths of $n$ segments $X Y_{1} \cdot X Y_{2} \cdot X Y_{3} \cdot \ldots \cdot X Y_{n}$ can take, where the point $X$ is an arbitrarily chosen point from set $A$, and the points $Y_{1}, Y_{2}, Y_{3}, \ldots, Y_{n}$ are all elements of set $C$.
Answer: 162
## B-4
The set $A$ on the plane $O x y$ is defined by the equation $x^{2}+y^{2}=2 x-2 y+34$. The set $B$ on the same plane is defined by the equation $|x-1|+|y+1|=6$. The set $C$ is the intersection of sets $A$ and $B$. What is the maximum value that the product of the lengths of $n$ segments $X Y_{1} \cdot X Y_{2} \cdot X Y_{3} \cdot \ldots \cdot X Y_{n}$ can take, where the point $X$ is an arbitrarily chosen point from set $A$, and the points $Y_{1}, Y_{2}, Y_{3}, \ldots, Y_{n}$ are all elements of set $C$.
Answer: 2592
|
1250
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. After treating the garden with a caterpillar control agent, the gardener noticed that from 12 blackcurrant bushes he was getting the same harvest as before from 15 bushes. By what percentage did the blackcurrant yield in the garden increase?
|
Answer: $25 \%$.
Solution. The yield from 12 bushes increased by $15 / 12=1.25$ times, which means the yield of currants increased by $25 \%$.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. What is the degree measure of angle $\angle A$, if its bisector forms an angle with one of its sides that is three times smaller than the angle adjacent to $\angle A$?
|
Answer: $72^{\circ}$.
Solution. Let $x$ be the degree measure of the angle formed by the angle bisector of $\angle A$ with one of its sides. Then the degree measure of the angle $\angle A$ is $2 x$, and the degree measure of the adjacent angle is $3 x$. Therefore, $2 x+3 x=180$, from which $x=36$ and $\angle A=72^{\circ}$.
|
72
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On a vast meadow, a lamb is grazing, tied with two ropes to two stakes (each stake with its own rope).
a) What shape will the area on the meadow be, where the lamb can eat the grass?
b) In the middle between the stakes, a rose grows, and the distance between the stakes is 20 meters. What should the lengths of the ropes be so that the lamb cannot eat the rose?
|
Answer: b) Exactly one of the ropes is shorter than 10 m.
Solution. a) The desired figure is the intersection of two circles with centers at the stakes and radii equal to the lengths $l_{1}$ and $l_{2}$ of the first and second ropes, respectively. Therefore, it can be either the figure shaded in Fig. 3, or a complete circle. The latter case occurs if $\left|l_{1}-l_{2}\right| \geqslant r$, where $r$ is the distance between the stakes, since then one of the circles is completely inside the other, and the desired figure coincides with the circle of the smaller radius.
b) For the rose not to fall within the area on the meadow where the lamb can eat grass, it must lie outside one of the circles bounding this figure. The distance from it to each of the stakes is $r / 2=20 / 2=10 \mathrm{m}$, so the length of one of the ropes must be less than $10 \mathrm{m}$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the number of numbers from 1 to 3400 that are divisible by 34 and have exactly 2 odd natural divisors. For example, the number 34 has divisors $1,2,17$ and 34, exactly two of which are odd.
|
Answer: 7. Solution. It is obvious that if a number is divisible by 34, then its divisors will always include 1 and 17. According to the condition, there should be no other odd divisors, so these must be numbers of the form $17 \cdot 2^{k}$, $k \geqslant 1$ (and only they). These numbers fall within the specified range for $k=1,2, \ldots, 7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the sum
$$
\begin{aligned}
& \frac{1}{(\sqrt[4]{1}+\sqrt[4]{2})(\sqrt{1}+\sqrt{2})}+\frac{1}{(\sqrt[4]{2}+\sqrt[4]{3})(\sqrt{2}+\sqrt{3})}+ \\
& +\ldots+\frac{1}{(\sqrt[4]{9999}+\sqrt[4]{10000})(\sqrt{9999}+\sqrt{10000})}
\end{aligned}
$$
|
Answer: 9. Solution. If we multiply the numerator and denominator of the fraction $\frac{1}{(\sqrt[4]{n}+\sqrt[4]{n+1})(\sqrt{n}+\sqrt{n+1})}$ by $\sqrt[4]{n}-\sqrt[4]{n+1}$, we get $\sqrt[4]{n+1}-\sqrt[4]{n}$. The specified sum transforms to $(\sqrt[4]{2}-\sqrt[4]{1})+(\sqrt[4]{3}-\sqrt[4]{2})+\ldots+(\sqrt[4]{10000}-\sqrt[4]{9999})=$ $\sqrt[4]{10000}-\sqrt[4]{1}=9$
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. In "Dragon Poker," the deck has four suits. An Ace brings 1 point, a Jack -2 points, a Two $-2^{2}$, a Three $-2^{3}, \ldots$, a Ten $-2^{10}=1024$ points. Kings and Queens are absent. You can choose any number of cards from the deck. In how many ways can you score 2018 points?
|
Answer: $C_{2021}^{3}=1373734330$.
Solution. In any suit, you can score any number of points from 0 to 2047, and this can be done in a unique way. This can be done as follows: write down this number in binary and select the cards corresponding to the binary positions containing 1 (i.e., if there is a 1 in the $k$-th position, then you should take the card that brings $2^{k}$ points).
Thus, the problem reduces to partitioning the number 2018 into 4 non-negative integer addends. Reasoning similarly to the previous problem, we get the desired number of ways: $C_{2021}^{3}=1373734330$.
|
1373734330
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. Trapezoid $A B C D$ with base $A D=6$ is inscribed in a circle. The tangent to the circle at point $A$ intersects lines $B D$ and $C D$ at points $M$ and $N$ respectively. Find $A N$, if $A B \perp M D$ and $A M=3$.
|
Answer: 12. Solution. The condition $A B \perp M D$ means that $\angle A B D=90^{\circ}$, that is, the base $A D$ is the diameter of the circle. Since the trapezoid is inscribed, it is isosceles. $\triangle D N A$ and $\triangle D A C$ are similar as right triangles with a common angle. From the equality of angles $\angle C A D$ and $\angle B D A$, the similarity of $\triangle D A C$ and $\triangle M D A$ follows. Therefore, $\triangle D N A \square \triangle M D A$, that is, $\frac{A N}{A D}=\frac{A D}{A M}=2$, hence $A N=12$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. The parabola $y=x^{2}$ intersects with the line $y=25$. A circle is constructed on the segment between the intersection points of the parabola and the line as its diameter. Find the area of the convex polygon whose vertices are the intersection points of the given circle and the parabola. Provide the nearest integer to the value of this area.
|
Answer: 10. Solution. The parabola $y=x^{2}$ intersects the line $y=a$ at points with coordinates $( \pm \sqrt{a} ; a)$. Each of the intersection points of the circle and the parabola mentioned in the problem has coordinates $\left(x ; x^{2}\right)$. For such a point, the distance to the line $y=a$ is $\left|a-x^{2}\right|$, the distance to the y-axis is $|x|$, and the distance to the intersection point of these lines (the center of the circle) is $\sqrt{a}$. By the Pythagorean theorem, $x^{2}+\left(a-x^{2}\right)^{2}=a$, which simplifies to $\left(a-x^{2}\right)\left(a-x^{2}-1\right)=0$. Besides the two points with coordinates $( \pm \sqrt{a} ; a)$, the circle and the parabola also intersect at two points $( \pm \sqrt{a-1} ; a-1)$, which lie on the line $y=a-1$.
Thus, the polygon mentioned in the problem is a trapezoid with height 1 and bases $2 \sqrt{a}$ and $2 \sqrt{a-1}$. Its area is $\sqrt{a}+\sqrt{a-1}$. For $a=25$, we get $5+\sqrt{24} \approx 9.898979 \ldots$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1. Solve the equation $\left(x^{2}-2 x+4\right)^{x^{2}-2 x+3}=625$. In the answer, specify the sum of the squares of all its roots. If there are no roots, put 0.
|
Answer: 6. Solution. Since $x^{2}-2 x+3>0$, then $x^{2}-2 x+4>1$. The function $f(z)=z^{z-1}$ is increasing for $z>1$ (if $1<z_{1}<z_{2}$, then $f\left(z_{1}\right)=z_{1}^{z_{1}-1}<z_{1}^{z_{2}-1}<z_{2}^{z_{2}-1}=f\left(z_{2}\right)$. Therefore, the original equation, which is of the form $f\left(x^{2}-2 x+4\right)=f(5)$, is equivalent to the equation $x^{2}-2 x+4=5$. From this, $x=1 \pm \sqrt{2}$. The sum of the squares of the roots of the equation is $(1+\sqrt{2})^{2}+(1-\sqrt{2})^{2}=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. Calculate: $\frac{1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+2013 \cdot 2014}{(1+2+3+\ldots+2014) \cdot \frac{1}{5}}$. If necessary, round the answer to the nearest hundredths.
|
Answer: 6710. Solution. Let's calculate the sum in the numerator: $\sum_{k=1}^{n} k(k+1)=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$ $=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$. Here, the well-known formula for the sum of squares of natural numbers is used, which can be proven in various ways (these proofs can easily be found in literature).
Thus, the fraction specified in the condition is equal to $\frac{2013 \cdot 2014 \cdot 2015 \cdot 2 \cdot 5}{3 \cdot 2014 \cdot 2015}=\frac{2013 \cdot 10}{3}=6710$.
|
6710
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. A team of athletes, one third of whom are snowboarders, descended from the mountain. Some of them got into a cable car carriage, which can accommodate no more than 10 people, while all the others descended on their own, and their number turned out to be more than $45 \%$, but less than $50 \%$ of the total number. Determine the number of snowboarders (if it is determined from the condition of the problem ambiguously, then write in the answer the sum of all possible its values).
|
Answer: 5. Solution. If there were $x$ snowboarders and $y$ people descended on the cable car, then there were a total of $3 x$ athletes and $\left\{\begin{array}{l}\frac{9}{20}=\frac{45}{100}<\frac{3 x-y}{3 x}<\frac{50}{100}=\frac{10}{20}, \\ y \leq 10\end{array} \Rightarrow\right.$ $\left\{\begin{array}{l}x<\frac{20}{30} y \leq \frac{20}{3}<7, \\ y \in\left(\frac{30}{20} x ; \frac{33}{20} x\right) .\end{array}\right.$
By trying the values $x=1,2, \ldots, 6$, we find that only when $x=5$ the obtained interval $y \in\left(\frac{30}{20} x ; \frac{33}{20} x\right)$ contains an integer $(y=8)$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Find the number of roots of the equation
$$
\operatorname{arctg}\left(\operatorname{tg}\left(\sqrt{13 \pi^{2}+12 \pi x-12 x^{2}}\right)\right)=\arcsin \left(\sin \sqrt{\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}}\right)
$$
|
Answer: 9. Solution. Let's make the substitution $t=\sqrt{\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}}$. Since $\frac{13 \pi^{2}}{4}+3 \pi x-3 x^{2}=4 \pi^{2}-3\left(x-\frac{\pi}{2}\right)^{2}$, then $0 \leq t \leq 2 \pi$. The original equation, after the specified substitution, transforms into the equation $\operatorname{arctg}(\operatorname{tg} 2 t)=\arcsin (\sin t)$.
The graphs of the functions $f(t)=\operatorname{arctg}(\operatorname{tg} 2 t)$ and $g(t)=\arcsin (\sin t)$ on the interval $[0 ; 2 \pi]$ have five points of intersection: $t_{1}=0, t_{2} \in\left(\frac{\pi}{2} ; \pi\right), t_{3}=\pi, t_{4} \in\left(\pi ; \frac{3 \pi}{2}\right)$, and $t_{5}=2 \pi$.
The value of $t_{2}$ is found from the system: $\left\{\begin{array}{c}y=\pi-t, \\ y=2\left(t-\frac{\pi}{2}\right)\end{array} \Leftrightarrow t_{2}=\frac{2 \pi}{3}\right.$.
Similarly, we find $t_{4}:\left\{\begin{array}{c}y=\pi-t, \\ y=2\left(t-\frac{3 \pi}{2}\right)\end{array} \Leftrightarrow t_{4}=\frac{4 \pi}{3}\right.$.
The equations $\sqrt{4 \pi^{2}-3\left(x-\frac{\pi}{2}\right)^{2}}=t_{k}$ have two distinct solutions for $k=1,2,3,4$. The equation $\sqrt{4 \pi^{2}-3\left(x-\frac{\pi}{2}\right)^{2}}=t_{6}$ has a unique solution. In total, we get 9 solutions.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Find the sum of all integer values of $a$ belonging to the interval $[-10 ; 10]$, for each of which the double inequality $5 \leq x \leq 10$ implies the inequality $a x+3 a^{2}-12 a+12>a^{2} \sqrt{x-1}$.
|
Answer: -47. Solution. Let $t=\sqrt{x-1}$, then the original problem reduces to finding such values of $a$ for which the inequality $f(t) \equiv a t^{2}-a^{2} t+3 a^{2}-11 a+12>0$ holds for all $t \in[2 ; 3]$. This means that the minimum of the function $f(t)$ on the interval $2 \leq t \leq 3$ is positive.
If $a=0$, then $f(t) \equiv 12$. If $a \neq 0$, then there are two cases:
a) $\left\{\begin{array}{l}f\left(\frac{a}{2}\right)=-\frac{1}{4}(a-2)(a-4)(a-6)>0, \\ 2 \leq \frac{a}{2} \leq 3\end{array} \Leftrightarrow 40, \\ f(3)=2(6-a)>0\end{array} \Leftrightarrow a<3\right.$.
In the end, $a \in(-\infty ; 3) \cup(4 ; 6)$, and the sum of the sought values of $a$ on the interval $[-10 ; 10]$ is: $-10-9-8-\ldots-1+0+1+2+5=(-10-9-\ldots-6)-4-3=-47$.
|
-47
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. (2 points) In a large family, one of the children has 3 brothers and 6 sisters, while another has 4 brothers and 5 sisters. How many boys are there in this family
|
Answer: 4.
Solution. In the family, boys have fewer brothers than girls have, and girls have fewer sisters than boys have. Therefore, there are 4 boys and 6 girls in the family.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. (14 points) Svetlana, Katya, Olya, Masha, and Tanya attend a math club, in which more than $60 \%$ of the students are boys. What is the smallest number of schoolchildren that can be in this club?
|
Answer: 13.
Solution. Let $M$ be the number of boys, $D$ be the number of girls in the club. Then $\frac{M}{M+D}>\frac{3}{5}$, hence $M>\frac{3}{2} D \geqslant \frac{15}{2}$. The minimum possible value of $M$ is 8, and the minimum possible value of $D$ is 5, making a total of 13 children.
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1. (14 points) Find all integer values that the fraction $\frac{8 n+157}{4 n+7}$ can take for natural $n$. In your answer, write the sum of the found values.
|
Answer: 18.
Solution. We have $\frac{8 n+157}{4 n+7}=2+\frac{143}{4 n+7}$. Since the divisors of the number 143 are only $1, 11, 13$, and 143, integer values of the fraction are obtained only when $n=1$ and $n=34$, which are 15 and 3, respectively, and their sum is 18.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. (14 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offered him to name a sum. What is the largest number of coins the guard can ask for so that the stranger, guided solely by his own profit, would definitely pay the guard?
|
Answer: 199.
Solution. If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If the guard demands 200, the outsider might refuse, as there is no difference in profit. If the guard demands more, it is more profitable for the outsider to refuse. The guard can ask for less, but the problem requires finding the largest amount. Thus, the answer is 199 coins.
|
199
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. (14 points) We will call a natural number a snail if its representation consists of the representations of three consecutive natural numbers, concatenated in some order: for example, 312 or 121413. Snail numbers can sometimes be squares of natural numbers: for example, $324=18^{2}$ or $576=24^{2}$. Find a four-digit snail number that is the square of some natural number.
|
Answer: 1089.
Solution. Note that a snail number can only be a four-digit number if the three numbers from which its record is formed are $8, 9, 10$ (numbers $7, 8, 9$ and smaller still form a three-digit number, while $9, 10, 11$ and larger - already form a number of no less than five digits). It remains to figure out the order of the selected numbers. The square of a natural number cannot end in 10 (there should be an even number of zeros or none at all). Also, a square cannot end in the digit 8. Therefore, the last digit must be 9. The remaining options are 8109 and 1089. Since $8109=90^{2}+9$ and $1089=33^{2}$, the only four-digit snail number that is a square is 1089.
|
1089
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. The sequence $\left\{x_{n}\right\}$ is defined by the conditions $x_{1}=20, x_{2}=17, x_{n+1}=x_{n}-x_{n-1}(n \geqslant 2)$. Find $x_{2018}$.
|
Answer: 17.
Solution: From the condition, it follows that $x_{n+3}=x_{n+2}-x_{n+1}=x_{n+1}-x_{n}-x_{n+1}=-x_{n}$, therefore $x_{n+6}=$ $-x_{n+3}=x_{n}$, i.e., the sequence is periodic with a period of 6. Since $2018=6 \cdot 336+2$, we get $x_{2018}=x_{2}=17$.
|
17
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. Philatelist Andrey decided to distribute all his stamps equally into 2 envelopes, but it turned out that one stamp was left over. When he distributed them equally into 3 envelopes, one stamp was again left over; when he distributed them equally into 5 envelopes, 3 stamps were left over; finally, when he tried to distribute them equally into 9 envelopes, 7 stamps were left over. How many stamps does Andrey have in total, if recently, to accommodate them all, he had to buy a second album for 150 stamps, as one such album was no longer sufficient?
|
Answer: 223.
Solution. If the desired number is $x$, then from the first sentence it follows that $x$ is odd, and from the rest it follows that the number $x+2$ must be divisible by 3, 5, and 9, i.e., has the form $5 \cdot 9 \cdot p$. Therefore, $x=45(2 k-1)-2=90 k-47$. According to the condition $150<x \leqslant 300$, so $k=3$. Therefore, $x=223$.
|
223
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{11 \ldots 11}_{2017} \underbrace{22 \ldots .22}_{2018} 5}$.
|
Answer: 6056 (the number from the condition of the problem is $\underbrace{33 \ldots 33}_{2017} 5$).
Solution. Since
$\underbrace{11 \ldots 11}_{2017} \underbrace{22 \ldots 22}_{2018} 5=\frac{10^{2017}-1}{9} \cdot 10^{2019}+\frac{10^{2018}-1}{9} \cdot 20+5=\frac{10^{4036}+10^{2019}+25}{9}=\left(\frac{10^{2018}+5}{3}\right)^{2}$,
the number from the condition of the problem is $\frac{10^{2018}+5}{3}=\overbrace{\frac{99 \ldots 99}{2018}+6}^{3}=\underbrace{33 \ldots 33}_{2018}+2=\underbrace{33 \ldots 3}_{2017} 5$, it is an integer, and the sum of its digits is $2017 \cdot 3+5=6056$.
|
6056
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.2. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{11 \ldots 11}_{2018} \underbrace{55 \ldots 55}_{2017} 6}$.
|
Answer: 6055 (the number from the condition of the problem is $\underbrace{33 \ldots 33}_{2017} 4$ ).
|
6055
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.3. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{44.44}_{2017} \underbrace{22 \ldots 22}_{2018}} 5$.
|
Answer: 12107 (the number from the condition of the problem is $\underbrace{66 \ldots 66}_{2017} 5$ ).
|
12107
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.4. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{44.44}_{2018} \underbrace{88 \ldots 88}_{2017} 9}$.
|
Answer: 12109 (the number from the condition of the problem is equal to $\underbrace{66 \ldots 66}_{2017} 7$ ).
|
12109
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. Find all integer solutions of the equation $x \ln 27 \log _{13} e=27 \log _{13} y$. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 70.
|
Answer: 117.
Solution: The original equation is equivalent to the equation $\frac{x \ln 27}{\ln 13}=\frac{27 \ln y}{\ln 13} \Leftrightarrow x \ln 27=27 \ln y \Leftrightarrow$ $\ln 27^{x}=\ln y^{27} \Leftrightarrow 27^{x}=y^{27} \Leftrightarrow 3^{x}=y^{9}$, with $y \geqslant 1$, and thus $x \geqslant 0$. Since 3 is a prime number, either $y=1$ (then $x=0$), or $y$ is divisible by 3 and has no other prime divisors. Therefore, $y=3^{n}$, where $n \geqslant 0$. Hence, $x=9n$. Since $y>70$, then $n \geqslant 4$, the solution sought is: $(x, y)=(36,81)$. The answer is $x+y=117$.
|
117
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. Find the smallest 12-digit natural number that is divisible by 36 and contains each of the 10 digits at least once.
|
Answer: 100023457896.
Solution. The number must be divisible by 4 and by 9. Since the sum of ten different digits is 45, the sum of the two remaining digits must be 0, 9, or 18. We need the smallest number, so we add two digits 0 to the digits $0,1, \ldots, 9$ and place the digits 10002345 at the beginning of the desired number (this is the smallest possible "start" of the number). The remaining digits $6,7,8,9$ must ensure divisibility by 4. Therefore, the last two digits can be 76, 96, or 68. The smallest option: 7896.
|
100023457896
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. The sine of the dihedral angle at the lateral edge of a regular quadrilateral pyramid is $\frac{15}{17}$. Find the area of the lateral surface of the pyramid if the area of its diagonal section is $3 \sqrt{34}$.
|
Answer: 68.
Solution. Let the linear angle of the dihedral angle given in the problem be denoted as $\alpha$. This angle is always obtuse, so $\cos \alpha=-\frac{8}{17}$.
The projection of the lateral face onto the diagonal section is a triangle, the area of which is half the area of this section. Since the dihedral angle between the lateral face and the diagonal section is $\frac{\alpha}{2}$, we have $S_{\text {face }} \cdot \cos \frac{\alpha}{2}=\frac{1}{2} S_{\text {section }}$. Therefore, $S_{\text {lateral }}=4 \cdot \frac{S_{\text {section }}}{2 \cos \frac{\alpha}{2}}=\frac{2 \sqrt{2} S_{\text {section }}}{\sqrt{1+\cos \alpha}}=68$.
|
68
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1. A batch of tablets from four different brands was delivered to a computer store. Among them, Lenovo, Samsung, and Huawei tablets made up less than a third, with Samsung tablets being 6 more than Lenovo. All other tablets were Apple iPads, and there were three times as many of them as Huawei. If the number of Lenovo tablets were tripled, and the number of Samsung and Huawei tablets remained the same (with the total number of all tablets unchanged), then there would be 59 Apple iPads. How many tablets were delivered to the store in total?
|
Answer: 94.
Solution. Let $n$ be the total number of tablets we are looking for, with $x$ being the number of Lenovo brand tablets, and $y$ being the number of Huawei brand tablets. Then the number of Samsung tablets is $x+6$, and the number of Apple iPad tablets is $n-2x-y-6=3y$. From the problem statement, we also have the equation $4x+y+6=n-59$. From these two equations, we find $x=\frac{3n-254}{14}$ and $y=\frac{n+53}{7}$. Since $x \geqslant 1$, we get $3n-254 \geqslant 14$, which leads to $n \geqslant \frac{268}{3}=89 \frac{1}{3}$. On the other hand, from the problem statement, we also have the inequality $2x+y+6<\frac{n}{3}$, which gives $\frac{33-254}{7}+\frac{n+53}{7}+6<\frac{n}{3}$, or $n<\frac{477}{5}=95 \frac{2}{5}$. Additionally, since $n+53$ is divisible by 7, the number $n$ gives a remainder of 3 when divided by 7. Among the integers from 90 to 95, there is only one such number, namely $n=94$.
|
94
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. Let $S(n)$ be the sum of the digits in the decimal representation of the number $n$. Find $S\left(S\left(S\left(S\left(2017^{2017}\right)\right)\right)\right)$. Answer. 1.
|
Solution. Since $2017^{2017}<10000^{2017}$, the number of digits in the representation of $2017^{2017}$ does not exceed $4 \cdot 2017=$ 8068, and their sum $S\left(2017^{2017}\right)$ does not exceed $9 \cdot 8068=72612$. Then we sequentially obtain $S\left(S\left(2017^{2017}\right)\right) \leqslant 6+9 \cdot 4=42, S\left(S\left(S\left(2017^{2017}\right)\right)\right) \leqslant 3+9=12, S\left(S\left(S\left(S\left(2017^{2017}\right)\right)\right)\right) \leqslant 9$. Note also that the sum of the digits of a number gives the same remainder when divided by 9 as the number itself. Since 2016 is divisible by 9, the number $2017^{2017}=(2016+1)^{2017}$ gives a remainder of 1 when divided by 9. Therefore, $\left.S\left(S\left(S\left(2017^{2017}\right)\right)\right)\right)=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Solve the equation $x^{x+y}=y^{y-x}$ in natural numbers. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 1500.
|
Answer: 2744.
Solution. Let $y = t x$, where $t \in \mathbb{Q}$ for $x, y \in \mathbb{N}$. Then, substituting this into the equation, we find
$$
x = t^{\frac{t-1}{2}}, \quad y = t^{\frac{t+1}{2}}
$$
We will show that the number $t$ can only be an integer. Suppose the contrary: let $t$ be a rational number different from an integer, i.e., $t = \frac{p}{q}$, where the numbers $p$ and $q$ are coprime, $q \neq 1$. Then $y = \frac{p x}{q}$, and the equation becomes
$$
x^{x + \frac{p x}{q}} = \left(\frac{p x}{q}\right)^{\frac{p x}{q} - x} \Leftrightarrow x^{x q + p x} = \left(\frac{p x}{q}\right)^{p x - q x} \Leftrightarrow x^{x q} = p^{p x - x q} \cdot x^{-q x} \cdot q^{x(q - p)} \Leftrightarrow x^{2 q x} = \left(\frac{p}{q}\right)^{x(p - q)}
$$
In the last equation, the left side is always an integer, while the right side is not an integer. Contradiction.
Thus, $t \in \mathbb{N}$, so the numbers $x$ and $y$ will be natural only in the following two cases:
1) the number $t$ is odd: $t = 2k - 1$, then $x = (2k - 1)^{k - 1}, y = (2k - 1)^{k}, k \in \mathbb{N}$;
2) the number $t$ is a perfect square: $t = k^2$, then $x = k^{k^2 - 1}, y = k^{k^2 + 1}, k \in \mathbb{N}$.
We will summarize the obtained solutions in a table:
| $t$ | 1 | 3 | 4 | 5 | 7 | 9 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | 1 | 3 | 8 | 25 | 343 | 6561 | $\ldots$ |
| $y$ | 1 | 9 | 32 | 125 | 2401 | 59049 | $\ldots$ |
The solution $(x, y)$, in which $y$ is the smallest, exceeding 1500, is $x = 343, y = 2401$, in which case $x + y = 2744$.
|
2744
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 1.
In-1 Find $f(1)+f(2)+f(3)+\ldots+f(13)$, if $f(n)=4 n^{3}-6 n^{2}+4 n+13$.
|
Answer: 28743.
Solution. Since $f(n)=n^{4}-(n-1)^{4}+14$, then
$f(1)+f(2)+f(3)+\ldots+f(13)=1^{4}+2^{4}+\ldots+13^{4}-\left(0^{4}+1^{4}+\ldots+12^{4}\right)+14 \cdot 13=13^{4}+14 \cdot 13=28743$.
V-2 Find $f(1)+f(2)+f(3)+\ldots+f(11)$, if $f(n)=4 n^{3}+6 n^{2}+4 n+9$.
Answer: 20823.
V-3 Find $f(1)+f(2)+f(3)+\ldots+f(12)$, if $f(n)=4 n^{3}-6 n^{2}+4 n+7$.
Answer: 20832.
V-4 Find $f(1)+f(2)+f(3)+\ldots+f(14)$, if $f(n)=4 n^{3}+6 n^{2}+4 n+5$.
Answer: 50680.
V-5 Let $f(x)=x^{2}+10 x+20$. Solve the equation
$$
f(f(f(f(f(x)))))=0
$$
Answer: $-5 \pm \sqrt[32]{5}$.
Solution. Note that $f(x)=(x+5)^{2}-5$. Therefore, $f(f(x))=(x+5)^{4}-5, f(f(f(x)))=(x+5)^{8}-5, \ldots, f(f(f(f(f(x)))))=(x+5)^{32}-5=0$, from which $x=-5 \pm \sqrt[32]{5}$.
V-6 Let $f(x)=x^{2}+6 x+6$. Solve the equation
$$
f(f(f(f(f(x)))))=0
$$
Answer: $-3 \pm \sqrt[32]{3}$.
V-7 Let $f(x)=x^{2}+14 x+42$. Solve the equation
$$
f(f(f(f(x))))=0
$$
Answer: $-7 \pm \sqrt[16]{7}$.
V-8 Let $f(x)=x^{2}+18 x+72$. Solve the equation
$$
f(f(f(f(x))))=0
$$
Answer: $-9 \pm \sqrt[16]{9}=-9 \pm \sqrt[8]{3}$.
## Lomonosov Olympiad for Schoolchildren in Mathematics
Final Stage 2020/21 Academic Year for 10-11 Grades
#
|
28743
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 3.
In-1 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=0$ and $x^{2}+c x+a=1$ has two integer roots, and all these roots are less than (-1). Find the smallest value of $a$.
|
Answer: 15.
Solution. By Vieta's theorem, the product of the roots of the first equation is $a$, and the product of the roots of the second equation is $a-1$. Since the roots are integers and less than -1, their product is greater than 1, so each of the two consecutive numbers $a-1$ and $a$ is the product of two different integers greater than 1. Since the first odd number that is neither prime nor the square of a prime is 15, we get $a-1=14, a=15$. Then the roots of the first equation are -3 and -5 (in which case $b=8$), and the roots of the second equation are -2 and -7 (in which case $c=9$).
V-2 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=0$ and $x^{2}+c x+a=-1$ has two integer roots, and all these roots are greater than 1. Find the smallest value of $a$.
Answer: 14.
V-3 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=1$ and $x^{2}+c x+a=0$ has two integer roots, and all these roots are less than -1. Find the smallest value of $a$.
Answer: 15.
V-4 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=-1$ and $x^{2}+c x+a=0$ has two integer roots, and all these roots are greater than 1. Find the smallest value of $a$.
Answer: 14.
V-5 How many different polynomials of the form $P(x)=x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+E$, where $A, B, C, D, E$ are positive integers, exist for which $P(-1)=11, P(1)=21$?
Answer: 315.
Solution. Since $P(-1)=11$, we have $-1+A-B+C-D+E=11$; since $P(1)=21$, we have $1+A+B+C+D+E=21$. Therefore, $A+C+E=16, B+D=4$. The second of these two equations has 3 pairs of solutions in natural numbers, and the first equation has $C_{15}^{2}=\frac{15 \cdot 14}{2}=105$ solutions (arrange 16 "balls" in a row and divide them into three groups using two "partitions"). This corresponds to $3 \cdot 105=315$ different polynomials.
V-6 How many different polynomials of the form $P(x)=x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+E$, where $A, B, C, D, E$ are positive integers, exist for which $P(-1)=8, P(1)=22$?
Answer: 455.
V-7 How many different polynomials of the form $P(x)=x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+E$, where $A, B, C, D, E$ are positive integers, exist for which $P(-1)=9, P(1)=21$?
Answer: 364.
V-8 How many different polynomials of the form $P(x)=x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+E$, where $A, B, C, D, E$ are positive integers, exist for which $P(-1)=10, P(1)=22$?
Answer: 420.
## Lomonosov Olympiad for Schoolchildren in Mathematics
Final Stage 2020/21 Academic Year for 10-11 Grades
#
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. At Andrei's birthday, Yana was the last to arrive, giving him a ball, and Eduard was the second to last, giving him a calculator. Testing the calculator, Andrei noticed that the product of the total number of his gifts and the number of gifts he had before Eduard arrived is exactly 16 more than the product of his age and the number of gifts he had before Yana arrived. How many gifts does Andrei have?
|
Answer: 18. Solution. If $n$ is the number of gifts, and $a$ is Andrei's age, then $n(n-2)=a(n-1)$. From this, $a=\frac{n^{2}-2 n-16}{n-1}=n-1-\frac{17}{n-1}$. Therefore, $n-1=17, a=16$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In an equilateral triangle $ABC$, points $A_{1}$ and $A_{2}$ are chosen on side $BC$ such that $B A_{1}=A_{1} A_{2}=A_{2} C$. On side $AC$, a point $B_{1}$ is chosen such that $A B_{1}: B_{1}C=1: 2$. Find the sum of the angles $\angle A A_{1} B_{1}+\angle A A_{2} B_{1}$.
|
Answer: $30^{\circ}$. Solution. Since $A_{1} B_{1} \| A B$, then $\angle B A A_{1}=\angle A A_{1} B_{1}$. From symmetry, $\angle B A A_{1}=\angle C A A_{2}$. It remains to note that $\angle C B_{1} A_{2}=$ $\angle B_{1} A A_{2}+\angle A A_{2} B_{1}$ as an exterior angle in $\triangle A A_{2} B_{1}$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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