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1. Misha, Petya, Kolya, and Vasya played "podkidnoy durak" (a card game), a total of 16 games were played. Each of them was left "in the fool" at least once. It is known that Misha was left the most, and Petya and Kolya together were left 9 times. How many times was Vasya left "in the fool"?
ANSWER: 1. | Solution: If Petya or Kolya was left behind no less than 5 times, it means Misha was left behind no less than 6 times, therefore Vasya was left behind once (he could not have been left behind 0 times according to the condition). | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,997 |
2. Find the smallest natural N such that N+2 is divisible (without remainder) by 2, N+3 by 3, ..., N+10 by 10.
ANSWER: 2520. | Solution: Note that $N$ must be divisible by $2,3,4, \ldots, 10$, therefore, $N=$ LCM $(2,3,4, . ., 10)=2^{3} \times 3^{2} \times 5 \times 7=2520$. | 2520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,998 |
3. Five runners ran a relay. If the first runner ran twice as fast, they would have spent $5 \%$ less time. If the second runner ran twice as fast, they would have spent $10 \%$ less time. If the third runner ran twice as fast, they would have spent $12 \%$ less time. If the fourth runner ran twice as fast, they would ... | Solution: If each ran twice as fast, they would run 50% faster. This means that if the 5th ran faster, the time would decrease by $50-5-10-12-15=8 \%$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,999 |
4. Can the numbers 2016, 2017, ..., 2024 be placed in the specified positions (see fig.), so that the sum of the numbers on each side of the triangle is the same?
ANSWER: Yes, it is possible... | Solution: First, place the average value 2020 at all points. Then arrange $0, \pm1, \pm2, \pm3, \pm4$ in such a way that the sum on each side is equal to 0 (see the figure). Other arrangements are also possible.
 How many different solutions does the equation
$$
10 \sqrt{6} \cos x \cdot \operatorname{ctg} x - 2 \sqrt{6} \operatorname{ctg} x + 5 \cos x - 1 = 0
$$
have on the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$? Find these solutions. | 2. 2 solutions, $\pm \arccos \frac{1}{5}$. | \\arccos\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,001 |
3. (4 points) Positive numbers $b_{1}, b_{2}, b_{3}, b_{4}, b_{5}$ form a geometric progression. The sum of the logarithms base 2 of these numbers is 15. Find these numbers if $\log _{2} b_{1} \cdot \log _{2} b_{5}=8$. | 3. $4, 4 \sqrt{2}, 8, 8 \sqrt{2}, 16$. | 4,4\sqrt{2},8,8\sqrt{2},16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,002 |
5. The continuation of the bisector $C D$ of triangle $A B C$ intersects the circumcircle of this triangle at point $E$. The circumcircle of triangle $A D E$ intersects the line $A C$ at point $F$, different from $A$. Find $B C$, if $A C=b, A F=a$. | 5. If $A$ is between $C$ and $F$, then $B C=a+b$, if $F$ is between $A$ and $C$, then $B C=b-a$. | BC=bifAisbetweenCF,BC=ifFisbetweenAC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,006 |
1. Determine which of the numbers is greater
$$
\operatorname{arctg}(2+\sqrt{5})+\operatorname{arcctg}(2-\sqrt{5}) \quad \text { or } \quad \frac{5 \sqrt{7}}{4} .
$$ | # Problem 1.
Answer: the second number is greater.
Solution. If $a>0$ and $\varphi=\operatorname{arctg} a$, then $0<\varphi<\frac{\pi}{2}$ and $\operatorname{arctg} \frac{1}{a}=\frac{\pi}{2}-\varphi$. Therefore, $\operatorname{arctg} a + \operatorname{arctg} \frac{1}{a} = \frac{\pi}{2}$. If $a<0$, then $-\frac{1}{a}>... | \frac{5\sqrt{7}}{4}>\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,007 |
2. Find all values of $a$, for each of which the system
$$
\left\{\begin{array}{l}
(x-2)^{2}+(|y-1|-1)^{2}=4 \\
y=b|x-1|+a
\end{array}\right.
$$
has solutions for any value of $b$ | # Problem 2.
Answer: $-\sqrt{3} \leqslant a \leqslant 2+\sqrt{3}$.
Solution. Let's represent the solution of the system on the coordinate plane. The first equation of the system defines the union of two arcs of circles: $(x+1)^{2}+(y-2)^{2}=4, y \geqslant 1$ and $(x+1)^{2}+y^{2}=4, y<1$. The second equation of the sy... | -\sqrt{3}\leqslant\leqslant2+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,008 |
3. In an isosceles triangle $ABC$, the angle at the base is $\arcsin \frac{2}{3}$. A circle with its center at the midpoint of the base $BC$ intersects the line $AC$ at points $K$ and $L$, and the line $AB$ at points $M$ and $N$, such that segments $KM$ and $LN$ intersect. Find its radius if the radius of the circle pa... | # Problem 3.
Answer: $\frac{4 \sqrt{5}}{3}$.
Solution. Let point $O$ be the midpoint of the base $B C$. Without loss of generality, we assume that points $L$ and $M$ are closer to vertex $A$ than points $K$ and $N$ respectively. Due to symmetry with respect to the line $A O$, segment $M L$ is parallel to the base $B ... | \frac{4\sqrt{5}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,009 |
4. Solve the equation
$$
\log _{3}(x+2) \cdot \log _{3}(2 x+1) \cdot\left(3-\log _{3}\left(2 x^{2}+5 x+2\right)\right)=1
$$ | Problem 4.
Answer: $x=1$.
Solution. For admissible values $x>-\frac{1}{2}$ we have: $\log _{3}(x+2)>0$.
If $\log _{3}(2 x+1)0$ and the left side of the equation is negative.
Thus, all three factors on the left side of the equation are positive, and the roots should be sought only among those $x$ for which
$$
\left... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,010 |
5. Two equal cones are arranged so that the axis of each is the generatrix of the other. The angles at the vertices in the axial sections of these cones are each $90^{\circ}$. Find the angle between the two generatrices along which these cones intersect. | Problem 5.
Answer: $2 \arccos \sqrt{\frac{2}{2+\sqrt{2}}}$.
Solution. Let $S$ be the common vertex of the considered cones, $S A_{1}$ and $S A_{2}$ be their axes. Denote by $S B$ and $S C$ their common generators and by $\alpha$ the desired angle $\angle B S C$. The configuration described in the problem has two plan... | 2\arccos\sqrt{\frac{2}{2+\sqrt{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,011 |
3. Petrov lists the odd numbers: $1,3,5, \ldots, 2013$, while Vasechkin lists the even numbers: $2,4, \ldots, 2012$. Each of them calculated the sum of all digits of all their numbers and told the excellent student Masha. Masha subtracted Vasechkin's result from Petrov's result. What did she get? | Answer: 1007.
Solution: Let's break down the numbers of Petrov and Vasechkin into pairs as follows: $(2,3),(4,5), \ldots,(98,99),(100,101), \ldots$ (2012,2013), with 1 left unpaired for Petrov. Notice that in each pair, the sum of the digits of the second number is 1 greater than that of the first (since they differ o... | 1007 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,014 |
4. From point $P$, located on the hypotenuse $A B$ of an isosceles right triangle $\triangle A B C$, perpendiculars are dropped to the legs. These perpendiculars divide $\triangle A B C$ into three parts - two triangles and a rectangle. Can the area of each of these parts be less than $\frac{4}{9}$ of the area of the o... | Answer: No.
Solution: Without loss of generality, we can assume $S(\triangle A B C)=1$. Suppose $A P=p \times A B$, where $p$ varies from 0 to 1. Denoting the feet of the perpendiculars as $A_{1}$ and $B_{1}$, we can express the areas of the parts in terms of $p$: $S\left(\triangle A A_{1} P\right)=p^{2}, S\left(\tria... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,015 |
6. There are no fewer than 150 boys studying at the school, and there are $15 \%$ more girls than boys. When the boys went on a trip, 6 buses were needed, and each bus had the same number of students. How many people in total study at the school, given that the total number of students is no more than 400? | Answer: 387.
Solution: The number of boys is a multiple of 6, let's denote it as $6n$, obviously, $n \geqslant 25$. Then the number of girls is $6n \times 1.15 = 6.9n$. The total number of students is $12.9n \leqslant 400$, so $n \leqslant 31$. Considering that $6.9n$ must be an integer, and therefore $n$ must be a mu... | 387 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,017 |
7. In the school Spartakiad, teams from classes $8^{\mathrm{A}}$, $8^{\text {Б }}$, and $8^{\mathrm{B}}$ participated. In each of the competitions, one of these teams took 1st place, another took 2nd place, and another took 3rd place. After the Spartakiad, points were tallied: $x$ points were awarded for 1st place, $y$... | Answer: 5 competitions, $8^{\text {B }}$.
Solution: Let $n \geqslant 2$ be the number of competitions in the Spartakiad, then the total number of points scored by all teams is $n(x+y+z)=22+9+9=40$. But $z \geqslant 1, y \geqslant 2, x \geqslant 3$, so $x+y+z \geqslant 6$. Consider the possible cases: $x+y+z=8, n=5$; $... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,018 |
8. Famous skater Tony Hawk is riding a skateboard (segment $A B$) in a ramp, which is a semicircle with diameter $P Q$. Point $M$ is the midpoint of the skateboard, $C$ is the foot of the perpendicular dropped from point $A$ to the diameter $P Q$. What values can the angle $\angle A C M$ take if it is known that the an... | Answer: $12^{\circ}$.
Solution: Extend the line $A C$ to intersect the circle at point $D$ (see figure). The chord $A D$ is perpendicular to the diameter $P Q$, therefore, it is bisected by it. Thus, $C M$ is the midline of triangle $A B D$, so $C M \| B D$ and, therefore, $\angle A C M=\angle A D B$. The angle $\angl... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,019 |
9. Find the number of natural numbers from 1 to 100 that have exactly four natural divisors, at least three of which do not exceed 10. | Answer: 8.
Solution: A number has exactly 4 natural divisors if it is either the cube of a prime number or the product of two prime numbers. The cubes of prime numbers (satisfying the conditions) are: 8 and 27. Prime numbers not greater than 10 are - 2, 3, 5, and 7. All their pairwise products satisfy the conditions, ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,020 |
1. Karlson filled a conical glass with lemonade and drank half of it by height (measuring from the surface of the liquid to the apex of the cone), and Little Man drank the second half. How many times more lemonade did Karlson drink compared to Little Man? | # Answer: 7 times.
Solution. Let $r$ and $h$ be the radius of the base and the height of the conical glass, respectively. Then the volume of lemonade in the entire glass is $V_{\text {glass}}=\frac{1}{3} \pi r^{2} h$. The volume of lemonade drunk by Little One is $V_{\text {Little}}=\frac{1}{3} \pi(r / 2)^{2}(h / 2)=\... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,022 |
2. Find all integer solutions of the equation
$$
x+\frac{1}{y+\frac{1}{z}}=\frac{7}{3}
$$ | Answer: $(2 ; 2 ; 1) ;(2 ; 4 ;-1) ;(3 ;-1 ;-2) ;(3 ;-2 ; 2) ;(1 ; 1 ;-4)$.
Solution. First, consider the case when $\left|y+\frac{1}{z}\right| \geqslant 1 \Leftrightarrow \frac{1}{\left|y+\frac{1}{z}\right|} \leqslant 1$.
Possible scenarios: 1) $x=2 \Rightarrow \frac{1}{y+\frac{1}{z}}=\frac{1}{3} \Rightarrow y+\frac{... | (2;2;1);(2;4;-1);(3;-1;-2);(3;-2;2);(1;1;-4) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,023 |
3. Solve the equation
$$
\sin \left(x+3^{0} \cdot \frac{2 \pi}{7}\right)+\sin \left(x+3^{1} \cdot \frac{2 \pi}{7}\right)+\ldots+\sin \left(x+3^{5} \cdot \frac{2 \pi}{7}\right)=1
$$ | Answer: $x=-\frac{\pi}{2}+2 \pi n, \quad n \in \mathbb{Z}$.
Solution. First, note that
$$
\sin x+\sin \left(x+\frac{2 \pi}{7}\right)+\ldots+\sin \left(x+6 \cdot \frac{2 \pi}{7}\right)=0
$$
Indeed, seven points on the unit circle of the form $x+i \cdot \frac{2 \pi}{7}(i=0,1, \ldots, 6)$ form the vertices of a regular... | -\frac{\pi}{2}+2\pin,\quadn\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,024 |
4. What values can the angle at the vertex of an isosceles triangle take if it is known that exactly three lines can be drawn that divide both the area and the perimeter of this triangle in half. | Answer: $2 \arcsin (\sqrt{2}-1)\sqrt{2}-1$, one - if $\sin \frac{\alpha}{2}=\sqrt{2}-1$ and none if $\sin \frac{\alpha}{2}<\frac{1}{2}$, then the roots of equation (1)
$$
x_{1,2}=\frac{2 b(1+t) \pm 2 b \sqrt{t^{2}+2 t-1}}{4 b}=\frac{1+t \pm \sqrt{t^{2}+2 t-1}}{2}
$$
have the property:
$x_{2}=\frac{1+t+\sqrt{t^{2}+2 ... | 2\arcsin(\sqrt{2}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,025 |
5. Malvina and Buratino play according to the following rules: Malvina writes six different numbers in a row on the board, and Buratino comes up with his own four numbers $x_{1}, x_{2}, x_{3}, x_{4}$ and writes under each of Malvina's numbers one of the sums $x_{1}+x_{2}, x_{1}+x_{3}, x_{1}+x_{4}$, $x_{2}+x_{3}, x_{2}+... | # Answer: 14.
## Solution.
Solution. Let Malvina write the numbers $a_{1}>a_{2}>a_{3}>a_{4}>a_{5}>a_{6}$. If Buratino comes up with the numbers $x_{1}=\left(a_{1}+a_{2}-a_{3}\right) / 2, x_{2}=\left(a_{1}+a_{3}-a_{2}\right) / 2, x_{3}=\left(a_{2}+a_{3}-a_{1}\right) / 2, x_{4}=a_{4}-x_{3}$, then by writing $x_{1}+x_{2... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,026 |
6. Find all four-digit numbers $\overline{a b c d}$ (where $a, b, c, d$ are the digits of the decimal representation), each of which is a divisor of at least one of the three four-digit numbers $\overline{b c d a}, \overline{c d a b}, \overline{d a b c}$ formed from it. | Answer: All numbers of the form $\overline{a b a b}$, where $a$ and $b$ are any digits except zero (there are 81 such numbers).
## Solution.
From the problem statement, it follows that there exists $k \in \mathbb{N}$ such that at least one of the following equalities holds:
(I) $k \cdot \overline{a b c d}=\overline{b... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,027 |
7. For what values of $a$ does the equation
$$
[x]^{2}+2012 x+a=0
$$
(where $[x]$ is the integer part of $x$, i.e., the greatest integer not exceeding $x$) have the maximum number of solutions? What is this number? | Answer: 89 solutions at $1006^{2}-20120
\end{array}\right.
$
has the maximum number of integer solutions.
The solution to the first inequality is the interval $\left[-1006-\sqrt{1006^{2}-a} ;-1006+\sqrt{1006^{2}-a}\right]$ under the condition $a \leqslant 1006^{2}$.
The solution to the second inequality for $a>1006^... | 89 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,028 |
8. In coordinate space, find the length of the shortest path between the points $(0 ; 1 ; 2)$ and $(22 ; 4 ; 2)$ along the surface of a rectangular parallelepiped bounded by the planes $x=22$, $y=5$, $z=4$ and the three coordinate planes. | Answer: $\sqrt{657}$.
Solution. Due to the symmetry of the configuration relative to the plane $\pi(x ; y ; 2)$ (passing through points $K$ and $L$ parallel to the base $A B C D$), we can assume that the shortest path does not go above the plane $\pi$ and thus does not have any common points with the upper base $A^{\p... | \sqrt{657} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,029 |
9. In a class consisting of 21 students, any three students have done homework together exactly once, either in mathematics or in Russian language. Can we assert that in this class there exists a quartet of students, any three of whom have done homework together in the same subject? | # Answer: Possible
Solution. Let's separate one person from a group of 21 (assign this person the number 0), and for the remaining 20 people, draw a graph where each person corresponds to a vertex and is assigned a number from 1 to 20. We will connect pairs of vertices with edges if and only if the trio, consisting of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 5,030 |
Task 1. Determine how many zeros the number $N!$ ends with! | Solution. Let $N=2014$. Among the first 2014 natural numbers, 402 numbers are divisible by 5, of which 80 numbers are divisible by 25. Among these 80 numbers, 16 numbers are divisible by 125, of which 3 numbers are divisible by 625. There are more even numbers among the first 2014 natural numbers than those divisible b... | 501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,031 |
2. By how much is the sum of the squares of the first hundred even numbers greater than the sum of the squares of the first hundred odd numbers | Answer: 20100.
Solution: Group the terms as $\left(2^{2}-1^{2}\right)+\left(4^{2}-3^{2}\right)+\cdots+\left(200^{2}-199^{2}\right)=$ $(2-1) \cdot(2+1)+(4-3) \cdot(4+3)+\ldots+(200-199) \cdot(200+199)=1+2+\cdots+$ $199+200$. Divide the terms into pairs that give the same sum: $1+200=2+199=\ldots=100+101=201$. There wil... | 20100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,033 |
3. Petrov and Vasechkin were repairing a fence. Each had to nail a certain number of boards (the same amount). Petrov nailed two nails into some boards, and three nails into the rest. Vasechkin nailed three nails into some boards, and five nails into the rest. Find out how many boards each of them nailed, if it is know... | Answer: 30.
Solution: If Petrov had nailed 2 nails into each board, he would have nailed 43 boards and had one extra nail. If he had nailed 3 nails into each board, he would have nailed 29 boards. Therefore, the desired number lies between 29 and 43 (inclusive). Similarly, if Vasechkin had nailed 3 nails into each boa... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,034 |
4. Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers? | Answer: 18.
Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces.
Thus, the sum cannot be less than 18. We will show that 18 can be achieved - place the number 1 on... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,035 |
5. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square. | Answer: 183
Solution: Let the required number be a, then $1000a + a + 1 = n^2$. We can rewrite this as: $1001a = (n - 1)(n + 1)$. Factorize $1001 = 7 \times 11 \times 13$, meaning the product (n-1)(n+1) must be divisible by 7, 11, and 13. Additionally, for the square to be a six-digit number, $n$ must be in the interv... | 183 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,036 |
6. Petya formed all possible natural numbers that can be formed from the digits $2,0,1$, 8 (each digit can be used no more than once). Find their sum. Answer: 78331 | Solution: First, consider the units place. Each of the digits 1, 2, 8 appears once in this place for single-digit numbers, twice for two-digit numbers, four times for three-digit numbers, and four times for four-digit numbers - a total of 11 times.
In the tens place, each of them appears 3 times for two-digit numbers,... | 78331 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,037 |
1. (5-7,8,9) There are 2014 boxes on the table, some of which contain candies, while the others are empty.
On the first box, it is written: “All boxes are empty.”
On the second - “At least 2013 boxes are empty.”
On the third - “At least 2012 boxes are empty.”
...
On the 2014th - “At least one box is empty.”
It is... | Answer: 1007.
Solution: Suppose that $N$ boxes are empty, then $2014-N$ boxes contain candies. Note that on the box with number $k$, it is written that there are at least $2015-k$ empty boxes. Therefore, the inscriptions on the boxes with numbers $1,2, \ldots, 2014-N$ are false. Consequently, $N=2014-N$, from which $N... | 1007 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,038 |
2. (5-7) Sergei collects toy trains. He has several sets, each with a different number of cars. If all the sets are combined into one train, there will be 112 cars. If you take the three smallest sets, there will be 25 cars in them, and in the three largest - 50 cars. How many sets does Sergei have? How many cars are i... | Answer: 9 sets. 18 or 19 cars.
Solution: Let $a_{1}, a_{2}, \ldots, a_{n}$ be the number of cars in the sets, ordered in ascending order. Note that $a_{3} \geqslant 9$, otherwise the total length of the three smallest sets would be less than $a_{1}+a_{2}+a_{3} \leqslant 7+8+9=2450$.
Thus, the remaining $112-50-25=37$... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,039 |
3. $(5-7,8)$ Nезнayka and Ponchik have the same amounts of money, composed of coins worth $1, 3, 5$, and 7 ferthings.
Nезнayka has as many 1-ferthing coins as Ponchik has 3-ferthing coins;
3-ferthing coins - as many as Ponchik has 5-ferthing coins; 5-ferthing coins - as many as Ponchik has 7-ferthing coins; and 7-fer... | Answer: 5 coins.
Solution: Let $x, y, z, t$ be the number of 1, 3, 5, and 7-ferting coins that Nезнайка has. It is known that $x+y+z+t=20$ and $x+$ $3 y+5 z+7 t=3 x+5 y+7 z+t$. From the last equation, it follows that $6 t=2(x+y+z)$. Substituting $x+y+z=20-t$, we get the equation $6 t=2(20-t)$, the solution of which is... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,040 |
4. $(5-7,8)$ The phone PIN code consists of 4 digits (and can start with zero, for example, 0951). Petya calls "lucky" those PIN codes where the sum of the outer digits equals the sum of the middle digits, for example $1357: 1+7=3+5$. In his phone, he uses only "lucky" PIN codes. Petya says that even if he forgets one ... | Answer: a) $10 ;$ b) 670 .
Solution: a) Obviously, in the worst case, Petya tries all possible values for one digit and reconstructs the PIN code from them. Thus, no more than 10 combinations need to be tried. Using the example of the PIN code 0099, we can see that if you forget the two middle digits, you need to try ... | 670 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,041 |
5. (5-7,8) There are 10 segments, the length of each of which is expressed as an integer not exceeding some $N$. a) Let $N=100$. Provide an example of a set of 10 segments such that no three of them can form a triangle. b) Find the maximum $N$ for which it can be guaranteed that there will be three segments that can fo... | Answer: a) $1,1,2,3,5,8,13,21,34,55 ;$ b) $N=54$.
Solution: a) If each new segment is chosen to be equal to the sum of the two largest of the remaining ones, then it is impossible to form a triangle with its participation.
b) From the previous point, it is clear that for $N=55$ such a sequence can be constructed.
We... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,042 |
6. (5-7,8) Can we find 100 consecutive natural numbers, the first of which is divisible by 3, the second by 5, the third by 7, ..., the 100th by 201? | Answer: Yes.
Solution: Start with $n=\frac{3+3 \cdot 5 \cdot 7 \cdot \ldots \cdot 201}{2}$ - this number is an integer, since the numerator is even and divisible by 3, because the numerator is divisible by 3. Similarly, $n+$ $1=\frac{5+3 \cdot 5 \cdot 7 \cdot \ldots \cdot 201}{2}$ is divisible by $5, n+2=\frac{7+3 \cd... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,043 |
7. $(5-7,8,9)$ a) In a $3 \times 4$ table, numbers from 1 to 12 need to be arranged such that the difference between any two numbers in the same row is divisible by 3, and the difference between any two numbers in the same column is divisible by 4. An example of such an arrangement is:
| 1 | 4 | 7 | 10 |
| :---: | :--... | Answer: a) $144=3!\cdot 4!$; b) No.
Solution: a) Each column can be associated with a remainder of division by 4, and each row with a remainder of division by 3. Considering all possible permutations of remainders of division by 3 and by 4, we get $3!\cdot 4!=144$. b) Suppose such a table exists. We associate rows wit... | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,044 |
8. $(8,9)$ It is known that the number $\frac{(2+\sqrt{3})^{4}+(2-\sqrt{3})^{4}}{(2+\sqrt{3})^{6}+(2-\sqrt{3})^{6}}$ is rational. Write this number as an irreducible fraction. | Answer: $\frac{97}{1351}$.
Solution: Let $a=2+\sqrt{3}, b=2-\sqrt{3}$. Notice that $a+b=4$, $a b=1$, therefore, $a^{2}+b^{2}=(a+b)^{2}-2 a b=14$. Similarly, we find $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2(a b)^{2}=194$ and $a^{6}+b^{6}=\left(a^{2}+\right.$ $\left.b^{2}\right) \cdot\left(a^{4}-a^{2} b^{2}+b^{4}\rig... | \frac{97}{1351} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,045 |
9. $(8,9)$ What is the maximum possible area of quadrilateral $A B C D$, the sides of which are $A B=1, B C=8, C D=7$ and $D A=4$? | Answer: 18.
Solution: Note that $1^{2}+8^{2}=7^{2}+4^{2}=65$. With fixed lengths of $A B$ and $B C$, the area of $\triangle A B C$ will be maximized if $\angle A B C=$ $90^{\circ}$. In this case, $A C=\sqrt{65}$, and thus $\angle B C D=$ $90^{\circ}$ as well, and the area of $\triangle B C D$ is also maximized. | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,046 |
10. $(8,9)$ Find the smallest possible value of $\left|2015 m^{5}-2014 n^{4}\right|$, where $m$ and $n$ are natural numbers. | Answer: 0.
Solution: Consider numbers of the form $m=2014^{a} \cdot 2015^{b}$ and $n=2014^{c} \cdot 2015^{d}$. Then $\left|2015 m^{5}-2014 n^{4}\right|=\left|2014^{5 a} \cdot 2015^{5 b+1}-2014^{4 c+1} \cdot 2015^{4 d}\right|$. This value equals 0 in the case $5 a=4 c+1,5 b+1=4 d$. It is not difficult to find such numb... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,047 |
11. (9) Integers $a, b$ and $c$ are such that $a \cdot\left(1-\frac{\sqrt{7}}{2}\right)^{2}+b \cdot\left(1-\frac{\sqrt{7}}{2}\right)+c=5$. What is the minimum value of $|a+b+c|$ under this condition? | Answer: 2.
Solution: If $a=0$, then $b=0$ and $c=5$, hence $|a+b+c|=5$. If $a \neq 0$, then consider the quadratic function $f(x)=a x^{2}+b x+c$. Notice that $f(x)-5$ has roots $1 \pm \frac{\sqrt{7}}{2}$. Therefore, $f(x)=$ $5+k \cdot\left(4 x^{2}-8 x-3\right), k \neq 0$, i.e., $|a+b+c|=|5-7 k|$, The minimum value of ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,048 |
12. (9) On a plane, there are 9 points arranged in a $3 \times 3$ grid, as shown in the figure.
a) Through all possible pairs of points, lines are drawn. How many different lines are obtain... | Answer: a) 20 ; b) 76.
Solution: a) The number of pairs of points will be $C_{9}^{2}=36$, but lines coincide when three points lie on the same line. There are $8=3$ horizontals +3 verticals +2 diagonals such cases. Therefore, subtract $36-8 \times 2=20$.
b) The number of triplets of points is $C_{9}^{3}=84$, but not ... | )20;b)76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,049 |
13. (9) Solve the inequality $\sqrt{x^{2}-1}+\sqrt{x^{2}-4 x+3}+\sqrt{2 x+3-x^{2}} \geqslant 2$. | Answer: $x \in\{-1,1,3\}$.
Solution: Let's find the domain of the given inequality. It consists of 3 points: $\{-1,1,3\}$. By direct substitution, we can verify that all of them are valid. | x\in{-1,1,3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,050 |
14. (9) In parallelogram $A B C D$, angle $\angle B A C$ is twice the angle $\angle B D C$. Find the area of the parallelogram, given that $A B=$ $A C=2$. | Answer: $2 \sqrt{3}$.
Solution: The specified parallelogram is a rhombus with an angle of 60 degrees. This can be proven by constructing a circle centered at point $A$ with a radius of 2. Then points $B, C, D$ will lie on this circle. The side of the rhombus is 2, so its area $S(A B C D)=A B \cdot A D$.
$\sin 60^{\ci... | 2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,051 |
15. (9) Find $q$, for which $x^{2}+x+q=0$ has two distinct real roots satisfying the relation $x_{1}^{4}+2 x_{1} x_{2}^{2}-x_{2}=19$. | Answer: $q=-3$.
Solution: If $x$ is a root, then $x^{2}=-x-q$, hence $x^{4}=x^{2}+$ $2 q x+q^{2}=(2 q-1) x+q^{2}-q$. Then $x_{1}^{4}+2 x_{1} x_{2}^{2}-x_{2}=(2 q-1) x_{1}+q^{2}-q+$ $2 q x_{2}-x_{2}=(2 q-1)\left(x_{1}+x_{2}\right)+q^{2}-q$. Using Vieta's theorem
$x_{1}+x_{2}=-1$, we get $q^{2}-3 q+1=19$. Solving, we ob... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,052 |
5. Rewrite the equation as $\left(n^{2}-1\right)\left(3^{a-1}+3^{1-a}\right)=\frac{16 n}{3}$. The values $n= \pm 1$ and $n=0$ are not solutions. Since $3^{a-1}+3^{1-a} \geqslant 2$, it is necessary that $\frac{16 n}{3 n^{2}-3} \geqslant 2$ $\Leftrightarrow 3 n^{2}-8 n-3 \leqslant 0$ (considering that $|n|>1$). $n \in\l... | Answer: $a=1, a=1+\log _{3} \frac{16 \pm 5 \sqrt{7}}{9}$.
Solution of the second option:
Rewrite the equation as $\left(1-n^{2}\right)\left(2^{a}+2^{-a}\right)=\frac{16 n}{3}$. The values $n= \pm 1$ and $n=0$ are not solutions. Since $2^{a}+2^{-a} \geqslant 2$, it is necessary that $\frac{16 n}{3-3 n^{2}} \geqslant 2... | =1,=1+\log_{3}\frac{16\5\sqrt{7}}{9},=0,=\log_{2}\frac{16\5\sqrt{7}}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,057 |
1. From the condition about 15 kg, it follows that the container can contain either only items weighing 5 and 10 kg, or only items weighing 2, 3, and 10 kg. Let $x, y, z, u$ denote the number of items weighing $2, 3, 5$, and 10 kg, respectively.
In the first case, the conditions of the problem give the system $\left\{... | Answer: 2 items weighing 2 and 3 kg, 9 items weighing 10 kg.
Answer to variant 4-2: 2 weighing 3 and 5 kg, 12 weighing 7 kg. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,058 |
5. Let $a=\sin x, b=\cos y$. We obtain $\left\{\begin{array}{l}a^{2} b^{2}-2 a+b^{2}=0, \\ 2 a^{2}-4 a+b^{3}+3=0 .\end{array}\right.$ Consider the equations in the system as quadratic in terms of $a$. In the first equation, the discriminant is $4-4 b^{2}$, in the second equation, it is $-8-8 b^{3}$. The discriminants m... | Answer: $(\pi / 2 + 2 \pi n ; \pi + 2 \pi k), n, k \in Z$.
Answer to the option: $4-2: (2 \pi n ; 3 \pi / 2 + 2 \pi k), n, k \in Z$.
## Lomonosov Moscow State University
## Olympiad "Conquer Sparrow Hills" Variant 5-1 | (\pi/2+2\pin;\pi+2\pik),n,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,062 |
4. Let the carrying capacity of the truck be $a$ tons, and for transporting 60 tons of sand, $n$ trips were required. Then there are two possible situations:
a) $n-1$ full trucks and no more than half a truck: then to transport 120 tons, $2(n-1)+1=2n-1$ trips will be needed. According to the condition, $2n-1-n=5$, so $... | Answer: $\left[10 \frac{10}{11} ; 13 \frac{1}{3}\right)$ tons.
Answer to the variant: $5-2:\left[9 \frac{1}{11} ; 11 \frac{1}{9}\right)$ tons. | [10\frac{10}{11};13\frac{1}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,065 |
5. Rewrite the equation as
$$
5^{(x-1)^{2}} \log _{7}\left((x-1)^{2}+2\right)=5^{2|x-a|} \log _{7}(2|x-a|+2)
$$
Let $f(t)=5^{t} \log _{7}(t+2)$. This function is increasing. Therefore,
$$
f\left((x-1)^{2}\right)=f(2|x-a|) \Longleftrightarrow(x-1)^{2}=2|x-a|
$$
There will be three solutions in the case of tangency f... | Answer: $a=1 / 2 ; 1 ; 3 / 2$.
Answer to the option: $6-2: a=-3 / 2 ;-1 ;-1 / 2$.
## Lomonosov Moscow State University
## Olympiad "Conquer Sparrow Hills" Variant 7-1 | 1/2;1;3/2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,071 |
1. The oldest can be either Boris or Viktor. But Viktor could not take second place, as he lost to both Anton and Grigory. Therefore, Boris is second. This means Grigory is first, Anton is third, and Viktor is fourth. | Answer: Grigory, Boris, Anton, Viktor.
Answer to option: 7-2: Andrey, Valery, Gennady, Boris. | Grigory,Boris,Anton,Viktor | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,072 |
4. First, we solve the equation:
$$
\begin{aligned}
& \cos 4 x-6 \cos ^{2} 2 x+8 \cos ^{2} x=0 \Longleftrightarrow \\
& 2 \cos ^{2} 2 x-1-6 \cos ^{2} 2 x+4(1+\cos 2 x)=0 \Longleftrightarrow \\
& 4 \cos ^{2} 2 x-4 \cos 2 x-3=0 \Longleftrightarrow \\
& \cos 2 x=-1 / 2 \Longleftrightarrow \\
& x= \pm \frac{\pi}{3}+\pi n,... | Answer: $\pi / 3 ; 2 \pi / 3 ; 4 \pi / 3$. Answer to option: $7-2: \pi / 3 ; 2 \pi / 3 ; 4 \pi / 3$. | \pi/3;2\pi/3;4\pi/3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,075 |
5. Solution. It can be regrouped as follows:
$$
\begin{aligned}
(|x-1|+|x+1|+|x-2|+|x+2|+\cdots & +|x-2015|+|x+2015|-4030 x)+ \\
& +(x-a)^{2}+(x-4030+a)^{2}=0
\end{aligned}
$$
The first expression in parentheses is non-negative, and it equals zero if and only if $x \geqslant 2015$. Therefore, the original equation is... | Answer: If $a=2015$, then $x=2015$; if $a \neq 2015$, then there are no solutions. Answer to variant: 7-2: If $c=2014$, then $x=2014$; if $c \neq 2014$, then there are no solutions, | If\=2015,\then\2015;\if\\\neq\2015,\then\there\\no\solutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,076 |
1. The robotics club accepts only those who know mathematics, physics, or programming. It is known that 8 members of the club know physics, 7 know mathematics, and 11 know programming. Additionally, it is known that at least two know both physics and mathematics, at least three know both mathematics and programming, an... | Answer 19.
2. From the sequence of natural numbers $1,2,3, \ldots$, all perfect squares (squares of integers) have been removed. What number will be in the 2018th position among the remaining numbers?
Answer: 2063. | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,077 |
5. We will call a number $\mathrm{X}$ "25-supporting" if for any 25 real numbers $a_{1}, \ldots, a_{25}$, the sum of which is an integer, there exists at least one for which $\left|a_{i}-\frac{1}{2}\right| \geq X$.
In your answer, specify the largest 25-supporting X, rounded to the hundredths according to standard mat... | Answer $\frac{1}{50}=\mathbf{0 . 0 2}$. | 0.02 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,080 |
6. The sequence $a_{n}$ is defined as follows:
$$
a_{1}=1, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}, \text { for } n \geq 1 . \text { Find } a_{200}
$$ | Answer: 20100.
## Variant 3a (Chelyabinsk) | 20100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,081 |
6. The sequence $a_{n}$ is defined as follows:
$a_{1}=1, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{999}$. | Answer: 499500.
## School Olympiad "Conquer Sparrow Hills"
## Tasks for 7-8 Grades
## Variant 1b (Zheleznovodsk) | 499500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,083 |
3. Thirteen millionaires arrived at the economic forum and settled in the "Super Luxury+" hotel. The hotel has rooms of 3 different types: 6-star, 7-star, and 8-star. The millionaires need to be accommodated in such a way that all three types of rooms are used (i.e., at least one person must be placed in a 6-star room,... | Answer: $C_{12}^{2}=66$. | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,084 |
6. The sequence $a_{n}$ is defined as follows:
$a_{1}=2, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{100}$. | Answer: 10100.
## Variant 2b (Saratov) | 10100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,085 |
3. A school coach decided to reward 12 students who ran the distance in the best time. Each of them should be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used (at least once), and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
... | Answer: $C_{11}^{2}=55$. | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,087 |
5. Find all values of $a$ for each of which the equation
$$
x^{3}+(a-1) x^{2}+\left(a^{3}+3 a^{2}-4 a-8\right) x+(a+2)^{3}=0
$$
has three distinct roots that form a geometric progression (specify these roots). | 5. For $a=2$
$$
\frac{3-\sqrt{5}}{2},-1, \frac{3+\sqrt{5}}{2}; \text { for } a=4 \quad \frac{3-\sqrt{5}}{2}, 1, \frac{3+\sqrt{5}}{2} \text {. }
$$
# Answers to Moscow variant 3-2 | 2,\frac{3-\sqrt{5}}{2},-1,\frac{3+\sqrt{5}}{2}; | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,091 |
12. (9) On a plane, there are 9 points arranged in a $3 \times 3$ grid, as shown in the figure.
a) Through all possible pairs of points, lines are drawn. How many different lines are obtain... | Answer: a) 20 ; b) 76.
Solution: a) The number of pairs of points will be $C_{9}^{2}=36$, but lines coincide when three points lie on the same line. There are $8=3$ horizontals +3 verticals +2 diagonals such cases. Therefore, subtract $36-8 \times 2=20$.
b) The number of triplets of points is $C_{9}^{3}=84$, but not ... | )20;b)76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,092 |
3. Petrov lists the odd numbers: $1,3,5, \ldots, 2013$, while Vasechkin lists the even numbers $2,4, \ldots, 2012$. Each of them calculated the sum of all digits of all their numbers and told the result to the excellent student Masha. Masha subtracted the result of Vasechkin from the result of Petrov. What did she get? | Answer: 1007.
Solution: Let's break down the numbers of Petrov and Vasechkin into pairs as follows: $(2,3),(4,5), \ldots,(98,99),(100,101), \ldots$ (2012,2013), with 1 left unpaired for Petrov. Notice that in each pair, the sum of the digits of the second number is 1 greater than that of the first (since they differ o... | 1007 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,094 |
4. Lenochka decided to bake a pie for a birthday. She rolled out the dough into an isosceles right triangle \(\triangle ABC\). Then she thought that the dough would be enough for two pies and made two straight cuts parallel to the legs of the triangle (see figure).
Two triangles and one rectangle were formed. Lenochka... | Answer: No.
Solution: Let $p=\frac{A P}{A B}$, then the triangles cut off (which went into the cabbage pie) are similar to the original with coefficients $p$ and $1-p$. If the area of the original triangle is denoted by $S$, then the total area of the cabbage pie is $S_{\text {cabbage }}=$ $p^{2} S+(1-p)^{2} S=\left(2... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,095 |
5. The numbers $1,2, \ldots, 9$ are arranged in a $3 \times 3$ square. We will call such arrangements "feng shui" if, when any three cells are chosen, located in different columns and different rows, the sum of the numbers in the chosen cells is equal to 15. An example of a "feng shui" arrangement is shown in the figur... | Answer: a), b)
| 9 | 7 | 8 | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 3 | 1 | 2 | |
| 6 | 7 | 8 | |
| 6 | 4 | 5 | |
| 6 | 4 | 5 | |
| | 3 | 1 | 2 |
Solution: Notice that the number 6 can only be placed in the first column, as it cannot be in a different column from 9 (since the sum would exceed 15). It c... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,096 |
6. Dima went to school in the morning, but after walking exactly half the distance, he realized he had forgotten his mobile phone at home. Dima estimated (he had an A in mental arithmetic) that if he continued walking at the same speed, he would arrive at school 3 minutes before the first bell, but if he ran home for t... | Answer: 2.
Solution: Let $x$ be the time it takes for Dima to walk from home to school, $y$ be the time it takes for Dima to run from home to school, and $T$ be the remaining time until the bell rings (at the moment Dima noticed the loss). Then the conditions of the problem can be written as $\left\{\begin{array}{l}\f... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,097 |
7. Famous skater Tony Hawk is riding a skateboard (segment $A B$) in a ramp, which is a semicircle with diameter $P Q$. Point $M$ is the midpoint of the skateboard, $C$ is the foot of the perpendicular dropped from point $A$ to the diameter $P Q$. What values can the angle $\angle A C M$ take if it is known that the an... | Answer: $12^{\circ}$.
Solution: Extend the line $A C$ to intersect the circle at point $D$ (see figure). The chord $A D$ is perpendicular to the diameter $P Q$, therefore, it is bisected by it. Thus, $C M$ is the midline of triangle $A B D$, so $C M \| B D$ and, therefore, $\angle A C M=\angle A D B$. The angle $\angl... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,098 |
8. Can the number $n^{2}+2 n+2014$ be divisible (evenly) by 121 for some integer $n$? | Answer: It cannot.
Solution: Let's write this number in the form $n^{2}+2 n+2014=(n+1)^{2}+$ 11. 183. If it is divisible by 121, then it must also be divisible by 11, hence $n+1$ is also divisible by 11. Then $(n+1)^{2}$ is a multiple of 121, but $11 \cdot 183$ is not, therefore, their sum is not divisible by 121. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,099 |
2. Vasya and Petya simultaneously ran out from the starting point of a circular running track and ran in opposite directions at constant speeds. At some point, they met. Vasya ran a full circle and, continuing to run in the same direction, reached the place of their first meeting at the moment when Petya ran a full cir... | 2. $\frac{1+\sqrt{5}}{2}$ | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,100 |
5. Consider the plane sections of a regular pyramid $S A B C D$ parallel to the lateral edge $S B$ and the diagonal of the base $A C$, into which a circle can be inscribed. What values can the radii of these circles take if $A C=1, \cos \angle S B D=\frac{2}{3} ?$ | 5. $0<r \leqslant \frac{1}{6}, r=\frac{1}{3}$. | 0<r\leqslant\frac{1}{6},r=\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,101 |
1. Place the numbers $1,2,3,4,5,6,7,8$ and 9 in the nine cells of the figure shown in the diagram, so that the sum of the numbers in each column, starting from the second, is 1 more than in the previous one. It is sufficient to find at least one such arrangement. In your answer, indicate the number in the first column.... | Answer: 7.
Solution: For now, we will not pay attention to the order of numbers in one column.
The sum of the given numbers is 45. Let $x$ be the number in the bottom-left cell. Then $5x + 10 = 45$, from which $x = 7$. Therefore, the sum of the numbers in the second column is $8 = 5 + 3 = 6 + 2$. If the second column... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,102 |
2. The villages "Upper Vasyuki" and "Lower Vasyuki" are located on the bank of a river. A steamboat takes one hour to travel from Upper Vasyuki to Lower Vasyuki, while a motorboat takes 45 minutes. It is known that the speed of the motorboat in still water is twice the speed of the steamboat (also in still water). Dete... | Answer: 90 minutes.
Solution: Let's take the distance between the villages as a unit of length (unit). Then the speed of the steamboat downstream is 1 unit/h, and the speed of the motorboat is $4 / 3$ units/h. Therefore, the own speed of the steamboat is $1 / 3$ units/h, from which the speed of the current is $2 / 3$ ... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,103 |
4. The board for playing "Battleship" is $10 \times 10$. What is the maximum number of ships of size $1 \times 4$ that can be placed on it? | Answer: 24.
Solution: It is easy to show by trial that 24 ships can be placed. Let's color the board in 4 colors as shown in the figure. Note that each ship of size $1 \times 4$ contains one cell of each color, and there will be 24 yellow cells.
, 1 \leqslant x, y \leqslant 1000$, such that $x^{2}+y^{2}$ is divisible by 5. | Answer: 360000.
Solution: We have 200 numbers for each of the remainders $0,1,2,3,5$ when divided by 5. There are two cases: a) the numbers $x, y$ are both divisible by 5; b) or one number gives a remainder of 1 or 4, and the other gives a remainder of 2 or 3 when divided by 5. In the first case, we get $200 \times 20... | 360000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,105 |
6. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Answer: 999.
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$
also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the s... | 999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,106 |
1. Solution. From the formulas for the sum of a geometric progression, we know
$$
\begin{aligned}
& \frac{b_{1}}{1-q}=60 \\
& \frac{b_{1}^{2}}{1-q^{2}}=1200
\end{aligned}
$$
By dividing the second equation by the first, we get $\frac{b_{1}}{1+q}=20$, which is the answer.
Remark. We could also find $b_{1}=30, q=1 / 2... | Answer: 20 (option $1-2:-30)$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task is to restate it as a translation, here it is:
... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,107 |
1. Solution. The inequality is equivalent to
$$
\begin{aligned}
& \sqrt{9-x} \cdot\left(\left|x^{2}-10 x+13\right|-\left|x^{2}-1\right|\right) \geqslant 0 \Longleftrightarrow \\
& \sqrt{9-x} \cdot\left(x^{2}-10 x+13-x^{2}+1\right)\left(x^{2}-10 x+13+x^{2}-1\right) \geqslant 0 \Longleftrightarrow \\
& \sqrt{9-x} \cdot(... | Answer: $(-\infty ; 7 / 5] \cup[2 ; 3] \cup\{9\}$ (variant $2-2:(-\infty ; 15 / 7] \cup[3 ; 4] \cup\{7\})$. | (-\infty;7/5]\cup[2;3]\cup{9} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,111 |
5. Solution. The second equation can be transformed into the form $(x-3)^{2}+y^{2}=1$. Therefore, if the pair of numbers $\left(3+t_{0} ; y_{0}\right)$ is a solution to the system, then the pair $\left(3-t_{0} ; y_{0}\right)$ will also be a solution. This means that the only possible solutions are $(3 ; 1)$ or $(3 ;-1)... | Answer: 1 and 4 (the answer is the same for option $2-2$).
Answer: $x=\frac{2 \pi k}{7}, x= \pm \frac{\arccos ((\sqrt{3}-1) / 2)}{7}+\frac{2 \pi l}{7}, k, l \in \mathbb{Z}$.
## Lomonosov Moscow State University
Olympiad "Conquer Sparrow Hills"
Variant 3-1 (Voronezh) | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,112 |
1. Solution. The original equation is equivalent to $(3 x-1)(y+1)=1007$. It remains to note that $1007=19 \cdot 53$, and by checking the integer divisors solve the equation. | Answer: $x=18, y=18$ (in the variant $3-2: x=12, y=18$ ). | 18,18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,113 |
5. Solution. The original equation is equivalent to
$$
(\sin 7 x+\sin 3 x)^{2}+(\cos 3 x+\cos x)^{2}+\sin ^{2} \pi a=0
$$ | Answer: For $a \in \mathbb{Z}, x=\pi / 4+\pi k / 2, k \in \mathbb{Z}$, there are no solutions for other $a$.
## Lomonosov Moscow State University
Olympiad "Conquer Sparrow Hills" Variant 4-1 (Cheboksary) | \in\mathbb{Z},\pi/4+\pik/2,k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,116 |
1. Solution. The equation is equivalent to $\sin (\pi / 4-11 x)=\sin (17 x+\pi / 4)$. From this, either $\pi / 4-11 x=17 x+\pi / 4+2 \pi k, k \in \mathbb{Z}$, or $\pi-(\pi / 4-11 x)=$ $17 x+\pi / 4+2 \pi n, n \in \mathbb{Z}$. | Answer: $\pi k / 14, \pi / 12+\pi n / 3, k, n \in \mathbb{Z}$ (in the variant $5-2: \pi k / 17,-\pi / 12+\pi n / 3$, $k, n \in \mathbb{Z})$. | \pik/14,\pi/12+\pin/3,k,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,121 |
5. Solution. Let's take the side of the tetrahedron as 12. We will find the angle through the cosine, which is equal to the ratio of the area \( S_{1} \) of the triangle \( K_{1} L_{1} M_{1} \) - the projection of the triangle \( K L M \) onto the base plane, to the area \( S_{2} \) of the triangle \( K L M \) - the se... | Answer: $\cos \gamma=\frac{5 \sqrt{3}}{\sqrt{131}}$.
## Lomonosov Moscow State University
Olympiad "Conquer Sparrow Hills"
Variant $6-1$ (Irkutsk) | \cos\gamma=\frac{5\sqrt{3}}{\sqrt{131}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,124 |
1. Answer: 1 (in all options). | Solution. (Option 6-1). Let's represent $k$ as $k=10 a+b$, where $a, b$ are integers and $1 \leqslant b \leqslant 9$. Then $k^{2}+6 k=10\left(10 a^{2}+2 a b+6 a\right)+b^{2}+6 b$. The expression $b^{2}+6 b$ ends in 6 only if $b=2$. But in this case $k^{2}+6 k=$ $100\left(a^{2}+a\right)+16$, which means the second-to-la... | 1 | Other | MCQ | Yes | Yes | olympiads | false | 5,125 |
1. In the periodic decimal fraction $0.242424 \ldots$, the first digit after the decimal point is replaced by 4. How many times greater is the resulting number than the original? | Solution. The original fraction is found as the sum of an infinite decreasing geometric progression: $\frac{24}{100\left(1-\frac{1}{100}\right)}=\frac{24}{99}$ (another way: let the original fraction be $x$, then $100 x=24+x$, so $\left.x=\frac{24}{99}\right)$. After replacing the digit, the number became $\frac{24}{99... | \frac{73}{40} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,129 |
2. Find all values of $y$, for each of which no value of $x$ satisfying the inequality $\log _{2}(|x|+|y|) \leqslant 2, \quad$ satisfies the inequality $\log _{\frac{1}{2}}(|x|+|y+4|) \geqslant-2$ | Solution. The solutions of the first inequality on the Oxy plane form a square with vertices at points $(0 ; 4),(4 ; 0),(0 ;-4),(-4 ; 0)$, excluding the point with coordinates $(0 ; 0)$. The solution to the first inequality exists only if $y \in[-4 ; 4]$.
The solutions of the second inequality on the Oxy plane form a ... | (-\infty;-4]\cup[0;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,130 |
3. A circle of radius $a$ passes through vertices $A$ and $B$ of triangle $ABC$ and intersects sides $AC$ and $BC$ at points $M$ and $K$ respectively. Find the area of triangle $ABC$, if $AB = a \sqrt{3}$, $MK = a$, and the center of the circle is inside triangle $ABC$ at a distance $b$ from point $C$. In different ver... | Solution. Let $C M=x, C K=y$. Triangles $A B C$ and $K M C$ are similar, since $\angle C M K=\pi-\angle A M K=\angle A B C$. Then $A C=y \sqrt{3}, B C=x \sqrt{3}$ and the area of $\triangle A B C$ is $\frac{3}{2} x y \sin \angle C$. From the fact that $A C \cdot M C$ equals the square of the length of the tangent from ... | \frac{\sqrt{3}}{4}(b^{2}-^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,131 |
4. Determine the minimum value of the quantity $|x+y|$ given that the numbers $x$ and $y$ satisfy the relation $5 \cos (x+4 y)-3 \cos (x-4 y)-4 \sin (x-4 y)=10$. | Solution. Let $\varphi=\arccos \frac{3}{5}$ and we get $\cos (x+4 y)-\cos (x-4 y-\varphi)=2 \Leftrightarrow$ $\left\{\begin{array}{c}\cos (x+4 y)=1, \\ \cos (x-4 y-\varphi)=-1 .\end{array}\right.$ Therefore, $\left\{\begin{array}{c}x=\frac{\varphi+\pi}{2}+\pi(n+k), \\ y=-\frac{\varphi+\pi}{8}+\frac{\pi(n-k)}{4}\end{arr... | \frac{\pi}{8}-\frac{3}{8}\arccos\frac{3}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,132 |
5. One base of a regular $n$-sided prism ( $n \geqslant 3$ ) has $n$ common points with a sphere of radius 3; the other base has one common point with this sphere. What values can the volume of the prism take? | Solution. If the base of the prism has $n$ common points with the sphere, then this base is inscribed in the corresponding circle. Then the second base touches the sphere at the center of the base. We will describe a cylinder around the prism (its volume is greater than the volume of the prism) and look for possible va... | (0;\frac{32}{27}\piR^{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,133 |
6. Let's call "sawing" the following operation on a polygon:
a) Divide each side of the polygon into three equal parts.
b) Choose the middle part as the base of an equilateral triangle located outside the polygon.
c) Remove the base and add the other two sides.
Let $M_{0}$ be an equilateral triangle, $M_{1}$ be the... | Answer: 4.8.
## Variant 2a (Ufa) | 4.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,136 |
6. Let's call "sawing" the following operation on a polygon:
a) Divide each side of the polygon into three equal parts.
b) Choose the middle part as the base of an equilateral triangle located outside the polygon.
c) Remove the base and add the other two sides.
Let $M_{0}$ be an equilateral triangle, $M_{1}$ be the... | Answer: 3.2.
## Variant 3a (Chelyabinsk) | 3.2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,138 |
2. Thirteen millionaires arrived at the economic forum and settled in the "Super Luxury+" hotel. The hotel has rooms of 3 different types: 6-star, 7-star, and 8-star. The millionaires need to be accommodated in such a way that all three types of rooms are used (i.e., at least one person must be placed in a 6-star room,... | Answer: $C_{12}^{2}=66$. | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,141 |
4. We will call a number $\mathrm{X}$ "25-supporting" if for any 25 real numbers $a_{1}, \ldots, a_{25}$, the sum of which is an integer, there exists at least one for which $\left|a_{i}-\frac{1}{2}\right| \geq X$.
In your answer, specify the largest 25-supporting $X$, rounded to the hundredths according to standard m... | Answer $\frac{1}{50}=\mathbf{0 . 0 2}$. | 0.02 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,144 |
2. A school coach decided to reward 12 students who ran the distance in the best time. Each of them must be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used, and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
In how many ways c... | Answer: $C_{11}^{2}=55$. | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,146 |
1. Petrov and Vasechkin were repairing a fence. Each had to nail a certain number of boards (the same amount). Petrov nailed two nails into some boards and three nails into others. Vasechkin nailed three nails into some boards and five nails into the rest. Find out how many boards each of them nailed, given that Petrov... | Answer: 30.
Solution: If Petrov had nailed 2 nails into each board, he would have nailed 43 boards and had one extra nail. If he had nailed 3 nails into each board, he would have nailed 29 boards. Therefore, the desired number lies between 29 and 43 (inclusive). Similarly, if Vasechkin had nailed 3 nails into each boa... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,147 |
2. Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers? | Answer: 18.
Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces.
Thus, the sum cannot be less than 18. We will show that 18 can be achieved - place the number 1 on... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,148 |
3. In the Slytherin faculty at Hogwarts, there are 30 students. Some are friends (friendship is mutual, i.e., if A is friends with B, then B is also friends with A), but there are no 3 people who are pairwise friends with each other. On New Year's, each student sent cards to all their friends. What is the maximum numbe... | Answer: 450.
Solution: Let's find the person with the most friends. Suppose there are no fewer than 15, and denote their number as $15+a$. We will divide the students into two groups: the first group will consist of these $15+a$ students. According to the condition, they cannot be friends with each other, so each of t... | 450 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,149 |
4. Points $A_{1}, \ldots, A_{12}$ are the vertices of a regular 12-gon. How many different 11-segment open broken lines without self-intersections with vertices at these points exist? Broken lines that can be transformed into each other by rotation are considered the same. | Answer: 1024.
Solution: The first vertex of the broken line can be chosen in 12 ways. Each subsequent vertex (except the last one) can be chosen in two ways - it must be adjacent to the already marked vertices to avoid self-intersections. The last vertex is chosen uniquely. We get $12 * 2^{10}$ ways. Considering 12 po... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,150 |
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