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# Problem № 6 (10 points)
Five identical balls are rolling towards each other on a smooth horizontal surface. The speeds of the first and second are \( v_{1} = v_{2} = 0.5 \) m/s, while the others are \( v_{3} = v_{4} = v_{5} = 0.1 \) m/s. The initial distances between the balls are the same, \( l = 2 \) m. All collis... | # Solution and Evaluation Criteria:
In the case of a perfectly elastic collision, identical balls "exchange" velocities.
Therefore, the situation can be considered as if the balls pass through each other with unchanged speeds. The first collision occurs between the second and third balls. The last collision will occu... | 10 | Other | olympiads | null | null | \(\boxed{10}\) |
# Problem No. 8 (10 points)
A water heater with a power of \( P = 500 \mathrm{W} \) is used to heat a certain amount of water. When the heater is turned on for \( t_{1} = 1 \) minute, the temperature of the water increases by \( \Delta T = 2^{\circ} \mathrm{C} \), and after the heater is turned off, the temperature de... | # Solution and evaluation criteria:
The law of conservation of energy during the heating of water: $P \cdot t_{1}=c m \Delta T+P_{\text {loss }} \cdot t_{1}$.
When the heating plate is turned off: $P_{\text {loss }}=\frac{c m \Delta T}{t_{2}}$.
As a result, we get: $m=\frac{P \cdot t_{1} \cdot t_{2}}{c \Delta T\left... | 2.38 | Other | olympiads | null | null | \(\boxed{2.38}\) |
2. (16 points) Aunt Masha decided to bake a cake. She kneaded the dough, which according to the recipe contains flour, butter, and sugar in the weight ratio of $3: 2: 1$, and mixed butter with sugar for the cream in the ratio of 2:3. Changing her mind about baking the cake, she combined both mixtures, added 200 g of fl... | Let the cake dough contain flour, butter, and sugar in the amounts of $3 x$, $2 x$, and $x$ respectively, and the cream contain butter and sugar in the amounts of $2 y$ and $3 y$ respectively. Using the ratios for the cookies, we get the system of equations $\left\{\begin{array}{l}\frac{3 x+200}{2 x+2 y}=\frac{5}{3}, \... | 480 | Algebra | olympiads | null | null | \(\boxed{480}\) |
2. (12 points) During the break, a fly flew into the math classroom and started crawling on a poster, on which the graph of a quadratic function $y=f(x)$ was depicted in the coordinate plane, with the leading coefficient equal to 1. Initially, the fly moved exactly along the parabola up to the point with an abscissa eq... | Let the quadratic function be of the form $y=x^{2}+b x+c$. The midpoint of the line segment has coordinates $\left(\frac{2+4}{2} ; \frac{f(2)+f(4)}{2}\right)$, on the other hand, $\left(3 ; 6069\right)$. Since $f(2)=4+2 b+c, f(4)=16+4 b+c$, then $20+6 b+2 c=12138$ or $3 b+c=6059$. Therefore, $f(3)=9+3 b+c=9+6059=6068$.... | 6068 | Algebra | olympiads | null | null | \(\boxed{6068}\) |
2. (12 points) During the break, a fly flew into the math classroom and started crawling on a poster, on which the graph of a quadratic function $y=f(x)$ was depicted in the coordinate plane, with the leading coefficient equal to -1. Initially, the fly moved exactly along the parabola up to the point with an abscissa o... | Let the quadratic function be of the form $y=-x^{2}+b x+c$. The midpoint of the line segment has coordinates $\left(\frac{2+4}{2} ; \frac{f(2)+f(4)}{2}\right)$, and on the other hand, $(3 ; 6069)$. Since $f(2)=-4+2 b+c, f(4)=-16+4 b+c$, then $-20+6 b+2 c=12138$ or $3 b+c=6079$. Therefore, $f(3)=-9+3 b+c=-9+6079=6070$.
... | 6070 | Algebra | olympiads | null | null | \(\boxed{6070}\) |
2. (12 points) Find a natural number $n$ such that the numbers $n+30$ and $n-17$ are squares of other numbers. | From the condition of the problem, it follows that $\left\{\begin{array}{l}n+30=k^{2} \\ n-17=m^{2}\end{array}\right.$. Subtracting the second equation from the first, we get $k^{2}-m^{2}=47(*)$ or $(k-m)(k+m)=47$. Since 47 is a prime number, the possible cases are $\left\{\begin{array}{l}k-m= \pm 1, \\ k+m= \pm 47\end... | 546 | Number Theory | olympiads | null | null | \(\boxed{546}\) |
2. (12 points) Find a natural number $n$ such that the numbers $n+15$ and $n-14$ are squares of other numbers. | From the condition of the problem, it follows that $\left\{\begin{array}{l}n+15=k^{2} \\ n-14=m^{2}\end{array}\right.$ Subtracting the second equation from the first, we get $k^{2}-m^{2}=29(*)$ or $(k-m)(k+m)=29$. Since 29 is a prime number, the possible cases are $\left\{\begin{array}{l}k-m= \pm 1 \\ k+m= \pm 29,\end{... | 210 | Number Theory | olympiads | null | null | \(\boxed{210}\) |
2. Fifteen numbers are arranged in a circle. The sum of any six consecutive numbers is 50. Petya covered one of the numbers with a card. The two numbers adjacent to the card are 7 and 10. What number is under the card? | Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 15$.) Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{4}=a_{10}=a_{1} .
$$
Now it is clear that for any $i$, $a_{i}... | 8 | Logic and Puzzles | olympiads | null | null | \(\boxed{8}\) |
2. Twenty numbers are arranged in a circle. It is known that the sum of any six consecutive numbers is 24. What is the number in the 12th position if the number in the 1st position is 1? | Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 20)$. Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{19}=a_{5}=a_{11}=a_{17}=a_{3}=a_{9}=a_{15}=a_{1}
$$
Now it is... | 7 | Logic and Puzzles | olympiads | null | null | \(\boxed{7}\) |
4. (13 points) Sixteen people are standing in a circle: each of them is either a truth-teller (he always tells the truth) or a liar (he always lies). Everyone said that both of their neighbors are liars. What is the maximum number of liars that can be in this circle? | A truthful person can only be next to liars. Three liars in a row cannot stand, so between any two nearest truthful persons, there is one or two liars. Then, if there are 5 or fewer truthful persons, in the intervals between them, there can be no more than 10 liars in total, so there are no more than 15 people in total... | 10 | Logic and Puzzles | olympiads | null | null | \(\boxed{10}\) |
4. (13 points) In a circle, there are 17 people: each of them is either a truth-teller (he always tells the truth) or a liar (he always lies). Everyone said that both of their neighbors are liars. What is the maximum number of liars that can be in this circle? | A truth-teller can only be surrounded by liars. Three liars in a row cannot stand, so between any two nearest truth-tellers, there is one or two liars. Then, if there are 5 or fewer truth-tellers, there can be no more than 10 liars in the gaps between them, making a total of no more than 15 people. This leads to a cont... | 11 | Logic and Puzzles | olympiads | null | null | \(\boxed{11}\) |
2. (17 points) Given a rectangle $A B C D$. On two sides of the rectangle, different points are chosen, five points on $A B$ and six on $B C$. How many different triangles exist with vertices at the chosen points? | To form a triangle, one needs to choose two points on one side and one point on another. We have: 5 ways to choose the first point on $AB$, 4 ways - the second, and since the triangle does not change with the permutation of its vertices, we divide $5 \cdot 4$ by 2. Thus, $\frac{5 \cdot 4}{2}=10$ ways to choose two poin... | 135 | Combinatorics | olympiads | null | null | \(\boxed{135}\) |
1. (16 points) Solve the equation $x-7=\frac{4 \cdot|x-3|}{x-3}$. If the equation has multiple roots, write their sum in the answer. | The equation has a restriction on the variable $x \neq 3$. We open the modulus: for $x>3, x-7=4, x=11$. For $x<3, \quad x-7=-4, x=3-$ extraneous root. | 11 | Algebra | olympiads | null | null | \(\boxed{11}\) |
2. (17 points) Given a rectangle $A B C D$. On two sides of the rectangle, different points are chosen, six points on $A B$ and seven - on $B C$. How many different triangles exist with vertices at the chosen points? | To form a triangle, one needs to choose two points on one side and one point on another. There are 6 ways to choose the first point on $AB$, 5 ways to choose the second, and since the triangle does not change with the permutation of its vertices, we divide $6 \cdot 5$ by 2. Thus, $\frac{6 \cdot 5}{2}=15$ ways to choose... | 231 | Combinatorics | olympiads | null | null | \(\boxed{231}\) |
1. (17 points) Masha's tablet, which she needed for a presentation at school, was completely drained. Using additional equipment, the tablet can be fully charged in 2 hours and 40 minutes, without it in 8 hours. Masha first put the discharged tablet on regular charging, and when she found the equipment, she switched to... | The tablet charges in 160 minutes on fast charging, and in 480 minutes on regular charging. Therefore, on fast charging, $\frac{1}{160}$ of the full charge is completed in 1 minute, and on regular charging, $\frac{1}{480}$ of the full charge is completed in 1 minute. Let $t-$ be the total charging time, then $\frac{t}{... | 288 | Algebra | olympiads | null | null | \(\boxed{288}\) |
2. (17 points) There are ten weights of different weights, each weighing an integer number of grams. It is known that the weight of the lightest and heaviest weight differs by 9 grams. One weight is lost. Find its weight if the total weight of the remaining weights is 2022 grams. | Let $x$ be the weight of the lightest weight. Denote the weight of the lost weight as $(x+y)$ $(0<y<9)$. Then $x+(x+1)+(x+2)+\cdots+(x+$ $9)-(x+y)=2022$. Combine like terms: $10 x+45-x-y=$ 2022 or $9 x=1977+y$. From this, $1977+y$ is divisible by 9. Considering the condition $0<y<9$, we get that $y=3$. Therefore, $x=19... | 223 | Number Theory | olympiads | null | null | \(\boxed{223}\) |
3. (16 points) In a garden plot, it was decided to create a rectangular flower bed. Due to a lack of space, the length of the flower bed was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the flower bed decreased by $12 \%$. However, this was not enough, so it was decided to red... | Let $x$ be the length of the flower bed, $y$ be the width of the flower bed. After the reduction: $0.9 x$ - length of the flower bed, $0.8 y$ - width of the flower bed, $2(0.9 x + 0.8 y)$ - perimeter. We get the equation: $2(0.9 x + 0.8 y) = 0.88 \cdot 2(x + y)$ or $x = 4 y$. The original perimeter: $10 y$. After the s... | 18 | Algebra | olympiads | null | null | \(\boxed{18}\) |
1. (17 points) Masha's tablet, which she needed for a presentation at school, was completely drained. Using additional equipment, the tablet can be fully charged in 3 hours, without it in 9 hours. Masha first put the discharged tablet on regular charging, and when she found the equipment, she switched to fast charging ... | The tablet charges in 180 minutes on fast charging, and in 540 minutes on regular charging. Therefore, on fast charging, $\frac{1}{180}$ of the full charge is completed in 1 minute, and on regular charging, $\frac{1}{540}$ of the full charge is completed in 1 minute. Let $t$ be the total charging time, then $\frac{t}{3... | 324 | Algebra | olympiads | null | null | \(\boxed{324}\) |
3. (16 points) In a garden plot, it was decided to create a rectangular flower bed. Due to a lack of space, the length of the flower bed was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the flower bed decreased by $12.5 \%$. However, this was not enough, so it was decided to r... | Let $x$ be the length of the flower bed, $y$ be the width of the flower bed. After the reduction: $0.9 x$ - length of the flower bed, $0.8 y$ - width of the flower bed, $2(0.9 x + 0.8 y)$ - perimeter. We get the equation: $\quad 2(0.9 x + 0.8 y) = 0.875 \cdot 2(x + y) \quad$ or $\quad x = 3 y$. The original perimeter: ... | 14 | Algebra | olympiads | null | null | \(\boxed{14}\) |
# Problem No. 6 (10 points)
A pot was filled with $2 \pi$ liters of water, taken at a temperature of $t=0{ }^{\circ} C$, and brought to a boil in 10 minutes. After that, without removing the pot from the stove, ice at a temperature of $t=0{ }^{\circ} \mathrm{C}$ was added. The water began to boil again only after 15 m... | # Solution and Evaluation Criteria:
Mass of the initial water: $m_{B}=\rho V=2$ kg
Power of the stove in the first case: $P=\frac{c_{B} m_{B} \Delta T}{t_{1}}$.
And in the second case: $P=\frac{\lambda m_{J}+c_{B} m_{\pi} \Delta T}{t_{2}}$.
We obtain:
$$
\frac{c_{B} m_{B} \Delta T}{t_{1}}=\frac{\lambda m_{J}+c_{B}... | 1.68 | Algebra | olympiads | null | null | \(\boxed{1.68}\) |
# Problem No. 6 (10 points)
A pot was filled with $3 \pi$ liters of water at a temperature of $t=0{ }^{\circ} C$, and it was brought to a boil in 12 minutes. After that, without removing the pot from the stove, ice at a temperature of $t=0{ }^{\circ} \mathrm{C}$ was added. The water began to boil again only after 15 m... | # Solution and evaluation criteria:
Mass of the initial water: $m_{B}=\rho V=3$ kg
Power of the stove in the first case: $P=\frac{c_{B} m_{B} \Delta T}{t_{1}}$.
And in the second case: $P=\frac{\lambda m_{J}+c_{B} m_{J} \Delta T}{t_{2}}$.
We get:
$\frac{c_{B} m_{B} \Delta T}{t_{1}}=\frac{\lambda m_{J}+c_{B} m_{J} ... | 2.1 | Algebra | olympiads | null | null | \(\boxed{2.1}\) |
# Problem No. 8 (15 points)
50 g of ice, taken at a temperature of $t_{\pi}=-10{ }^{\circ} \mathrm{C}$, was mixed with water taken at a temperature of $t_{B}=10{ }^{\circ} \mathrm{C}$. It is known that the final temperature in the container is $t=0{ }^{\circ} \mathrm{C}$. Determine the mass of the added water. The spe... | # Solution and Evaluation Criteria:
Two extreme situations possible in this problem:
First situation - only ice remains in the vessel at a temperature of $t=0{ }^{\circ} \mathrm{C}$
(2 points)
The heat balance equation in this case: $c_{L} m_{L} 10=c_{B} m_{B} 10+\lambda m_{B}$.
We get: $m_{B}=\frac{2100 \cdot 0.0... | 0.0028 | Other | olympiads | null | null | \(\boxed{0.0028}\) |
3. (16 points) Mitya, Anton, Gosha, and Boris bought a lottery ticket for 20 rubles. Mitya paid $24\%$ of the ticket's cost, Anton - 3 rubles 70 kopecks, Gosha - 0.21 of the ticket's cost, and Boris contributed the remaining amount. The boys agreed to divide the winnings in proportion to their contributions. The ticket... | The ticket costs 2000 kop. Mitya paid 480 kop, Anton - 370 kop, Gosha - 420 kop, therefore, Boris had to pay an additional 730 kop. Since the prize is 50 times the cost of the ticket, Boris is entitled to 365 rubles. | 365 | Algebra | olympiads | null | null | \(\boxed{365}\) |
3. (16 points) Mitya, Anton, Gosha, and Boris bought a lottery ticket for 20 rubles. Mitya paid $24\%$ of the ticket's cost, Anton - 3 rubles 70 kopecks, Gosha - $0.21$ of the ticket's cost, and Boris contributed the remaining amount. The boys agreed to divide the winnings in proportion to their contributions. The tick... | The ticket costs 2000 kop. Mitya paid 480 kop, Anton - 370 kop, Gosha - 420 kop, therefore, Boris had to pay an additional 730 kop. Since the prize is 40 times the cost of the ticket, Boris is entitled to 292 rubles. | 292 | Algebra | olympiads | null | null | \(\boxed{292}\) |
3. (17 points) Divide the number 90 into two parts such that $40\%$ of one part is 15 more than $30\%$ of the other part. Write the larger of the two parts in your answer. | Let one part of the number be $x$, then the other part will be $90-x$. We get the equation $0.4 \cdot x = 0.3 \cdot (90 - x) + 15$, solving it we get $x = 60$, and the other part of the number is 30. | 60 | Algebra | olympiads | null | null | \(\boxed{60}\) |
3. (17 points) Divide the number 80 into two parts such that $30\%$ of one part is 10 more than $20\%$ of the other part. Write the smaller of the two parts in your answer. | Let one part of the number be $x$, then the other part will be $80-x$. We get the equation $0.3 \cdot x = 0.2 \cdot (80 - x) + 10$, solving it we get $x = 52$, and the other part of the number is 28. | 28 | Algebra | olympiads | null | null | \(\boxed{28}\) |
2. (17 points) Find the smallest root of the equation
$$
\sin (\pi x)+\tan x=x+x^{3}
$$ | Obviously, 0 is a root of the equation (when $x=0$, both sides of the equation are equal to zero). If $x<0$, the right side of the equation is negative, while the left side of the equation is always non-negative. | 0 | Calculus | olympiads | null | null | \(\boxed{0}\) |
4. A circle is inscribed with 2019 numbers. For any two adjacent numbers $x$ and $y$, the inequalities $|x-y| \geqslant 2, x+y \geqslant 6$ are satisfied. Find the smallest possible sum of the recorded numbers. | Due to the odd number of total numbers, there will be three consecutive numbers $x, y$, and $z$ such that $x>y>z$. Adding the inequalities $y-z \geqslant 2$ and $y+z \geqslant 6$, we get $y \geqslant 4$. Then $x \geqslant y+2 \geqslant 6$. A number not less than 6 has been found. The remaining numbers can be divided in... | 6060 | Inequalities | olympiads | null | null | \(\boxed{6060}\) |
4. A circle is inscribed with 1001 numbers. For any two adjacent numbers $x$ and $y$, the inequalities $|x-y| \geqslant 4, x+y \geqslant 6$ are satisfied. Find the smallest possible sum of the recorded numbers. | Due to the odd number of total numbers, there will be three consecutive numbers $x, y$, and $z$ such that $x>y>z$. Adding the inequalities $y-z \geqslant 4$ and $y+z \geqslant 6$, we get $y \geqslant 5$. Then $x \geqslant y+4 \geqslant 9$. A number not less than 9 has been found. The remaining numbers can be divided in... | 3009 | Inequalities | olympiads | null | null | \(\boxed{3009}\) |
# Problem № 7 (10 points)
In the electrical circuit shown in the diagram, the resistances of the resistors are $R_{1}=10$ Ohms and $R_{2}=30$ Ohms. An ammeter is connected to points A and B in the circuit. When the polarity of the current source is reversed, the ammeter readings change by one and a half times. Determi... | # Solution and Evaluation Criteria:
When the positive terminal of the power supply is connected to point $A$, the current flows only through resistor $R_{2}$, and in this case: $I_{1}=\frac{\varepsilon}{R_{2}+r}$.
## (3 points)
When the polarity is reversed, the current flows only through resistance $R_{1}$, and:
$... | 30 | Other | olympiads | null | null | \(\boxed{30}\) |
8. A thin beam of light falls normally on a plane-parallel glass plate. Behind the plate, at some distance from it, stands an ideal mirror (its reflection coefficient is equal to one). The plane of the mirror is parallel to the plate. It is known that the intensity of the beam that has passed through this system is 16 ... | ## Solution.
Let $k$ be the reflection coefficient, then we get $I_{1}=I_{0}(1-k)$ (2 points). Similarly, $I_{3}=I_{2}=I_{1}(1-k)=I_{0}(1-k)^{2}$ (2 points). As a result: $I_{k}=I_{0}(1-k)^{4}$ (2 points). According to the condition $I_{0}=16 \cdot I_{k}=16 \cdot I_{0}(1-k)^{4}$ (2 points). In the end, we get $k=0.5$ ... | 0.5 | Other | olympiads | null | null | \(\boxed{0.5}\) |
4. 100 balls of the same mass move along a trough towards a metal wall with the same speed. After colliding with the wall, a ball bounces off it with the same speed. Upon collision of two balls, they scatter with the same speed. (The balls move only along the trough). Find the total number of collisions between the bal... | We will assume that each ball has a flag. Imagine that upon collision, the balls exchange flags. Then each flag flies to the wall at a constant speed, and after hitting the wall, it flies in the opposite direction. The number of collisions between the balls is equal to the number of flag exchanges. Any two flags will s... | 4950 | Combinatorics | olympiads | null | null | \(\boxed{4950}\) |
Problem 1. Grandfather Frost had 120 chocolate candies and 200 jelly candies. At the morning performance, he gave candies to the children: each child received one chocolate candy and one jelly candy. Counting the candies after the performance, Grandfather Frost found that there were three times as many jelly candies le... | Let the total number of children be $x$, then after the morning party, Grandfather Frost had $120-x$ chocolate candies and $200-x$ jelly candies left. Since there were three times as many jelly candies left as chocolate candies, we get the equation $3 \cdot(120-x)=200-x$. Solving this, we get $x=80$.
## Criteria
One ... | 80 | Algebra | olympiads | null | null | \(\boxed{80}\) |
Problem 7.1. Jerry has nine cards with digits from 1 to 9. He lays them out in a row, forming a nine-digit number. Tom writes down all 8 two-digit numbers formed by adjacent digits (for example, for the number 789456123, these numbers are $78, 89, 94, 45$, $56, 61, 12, 23$). For each two-digit number divisible by 9, To... | Note that among two-digit numbers, only 18, 27, 36, 45, and numbers obtained by swapping their digits are divisible by 9 (there are also 90 and 99, but we do not have the digit 0 and only one digit 9). Thus, only four pairs of digits from the available ones can form a number divisible by 9. To get an example, we need t... | 4 | Number Theory | olympiads | null | null | \(\boxed{4}\) |
Problem 2. Gosha entered a natural number into the calculator. Then he performed the following operation, consisting of two actions, three times: first, he extracted the square root, and then took the integer part of the obtained number. In the end, he got the number 1. What is the largest number that Gosha could have ... | Suppose he entered a number not less than 256. Then after the first operation, he would get a number not less than $[\sqrt{256}]=16$, after the second - not less than $[\sqrt{16}]=4$, after the third - not less than $[\sqrt{4}]=2$, which is a contradiction.
Let's assume Gosha entered the number 255. Then after the fir... | 255 | Number Theory | olympiads | null | null | \(\boxed{255}\) |
1. On an island, there live knights who always tell the truth and liars who always lie. The population of the island is 1000 people, distributed across 10 villages (with no fewer than two people in each village). One day, every islander claimed that all their fellow villagers are liars. How many liars live on the islan... | In one village, at least two knights cannot live, because otherwise the knights would lie. Also, in the village, they cannot all be liars, since then these liars would tell the truth. Therefore, in each village there is exactly one knight, and there are 10 knights in total, and 990 liars. | 990 | Logic and Puzzles | olympiads | null | null | \(\boxed{990}\) |
Problem 6.1. Jerry has nine cards with digits from 1 to 9. He lays them out in a row, forming a nine-digit number. Tom writes down all 8 two-digit numbers formed by adjacent digits (for example, for the number 789456123, these numbers are $78, 89, 94, 45$, $56, 61, 12, 23$). For each two-digit number divisible by 9, To... | Note that among two-digit numbers, only 18, 27, 36, 45, and numbers obtained by swapping their digits are divisible by 9 (there are also 90 and 99, but we do not have the digit 0 and only one digit 9). Thus, only four pairs of digits from the available ones can form a number divisible by 9. To get an example, we need t... | 4 | Number Theory | olympiads | null | null | \(\boxed{4}\) |
Problem 2. In a $3 \times 3$ table, natural numbers (not necessarily distinct) are placed such that the sums in all rows and columns are different. What is the minimum value that the sum of the numbers in the table can take? | Note that in each row and column, the sum of the numbers is no less than 3. Then, the doubled sum of all numbers in the table, which is equal to the sum of the sums of the numbers in the rows and columns, is no less than \(3+4+\ldots+8=33\), so the simple sum of the numbers in the table is no less than 17.
Example of ... | 17 | Combinatorics | olympiads | null | null | \(\boxed{17}\) |
1. Buses from Moscow to Oryol depart at the beginning of each hour (at 00 minutes). Buses from Oryol to Moscow depart in the middle of each hour (at 30 minutes). The journey between the cities takes 5 hours. How many buses from Oryol will the bus that left from Moscow meet on its way? | It is clear that all buses from Moscow will meet the same number of buses from Orel, and we can assume that a bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Orel at $7:30, 8:30, \ldots, 15:30, 16:30$ and only them. There are 10 such buses.
$\pm$ Correct reasoning with an ... | 10 | Logic and Puzzles | olympiads | null | null | \(\boxed{10}\) |
2. Three generous friends, each of whom has candies, redistribute them as follows: Vasya gives some of his candies to Petya and Kolya, doubling the number of candies they have. After this, Petya gives some of his candies to Kolya and Vasya, doubling the number of candies they have as well. Finally, Kolya gives some of ... | Let's track the number of candies Kolya has. After the first redistribution, he has 72, and after the second - 144. Therefore, he gave away $144-36=108$ candies, and during this time, the number of candies Vasya and Petya had doubled. So, the total number of candies Vasya, Petya, and Kolya have together is $2 \cdot 108... | 252 | Logic and Puzzles | olympiads | null | null | \(\boxed{252}\) |
1. Nезнayka, Doctor Pilulkin, Knopochka, Vintik, and Znayka participated in a math competition. Each problem in the competition was solved by exactly four of them. Znayka solved strictly more than each of the others - 10 problems, while Nезнayka solved strictly fewer than each of the others - 6 problems. How many probl... | Each of Dr. Pill, Knopochka, and Vintik, according to the condition, solved from 7 to 9 problems. Therefore, the total number of solved problems ranges from $10+6+3 \cdot 7=37$ to $10+6+3 \cdot 9=43$. Note that this number should be equal to four times the number of problems. Among the numbers from 37 to 43, only one i... | 10 | Logic and Puzzles | olympiads | null | null | \(\boxed{10}\) |
Problem 10.5. In each cell of a square table of size $200 \times 200$, a real number not exceeding 1 in absolute value was written. It turned out that the sum of all the numbers is zero. For what smallest $S$ can we assert that in some row or some column, the sum of the numbers will definitely not exceed $S$ in absolut... | First, we show that $S<40000$.
$$
This means that one of the numbers $A$ or $D$ in absolute value exceeds 10000. However, each of the corresponding squares contains only 10000 cells, and the numbers in them do not exceed 1 in absolute value. Contradiction.
## Criteria
Any correct solution to the problem is worth 7 p... | 100 | Combinatorics | olympiads | null | null | \(\boxed{100}\) |
1. Several numbers were written on the board, their arithmetic mean was equal to $M$. They added the number 15, after which the arithmetic mean increased to $M+2$. After that, they added the number 1, and the arithmetic mean decreased to $M+1$. How many numbers were on the board initially? (Find all options and prove t... | Let there be $k$ numbers in the original list with a sum of $S$. Then, by the condition,
$$
\frac{S+15}{k+1}-\frac{S}{k}=2, \quad \frac{S+15}{k+1}-\frac{S+16}{k+2}=1.
$$
By bringing to a common denominator and transforming in an obvious way, we get that these equations are equivalent to the following two:
$$
15 k-S=... | 4 | Algebra | olympiads | null | null | \(\boxed{4}\) |
4. The teacher gave Vasya and Petya two identical cardboard $n$-gons. Vasya cut his polygon into 33-gons along non-intersecting diagonals, while Petya cut his polygon into 67-gons along non-intersecting diagonals. Find the smallest possible value of $n$. | The sum of the angles of an $n$-sided polygon is $(n-2) \cdot 180^{\circ}$. If it is cut into $k$ 33-sided polygons, then $(n-2) \cdot 180^{\circ}=k \cdot (33-2) \cdot 180^{\circ}$, hence $n-2 \vdots 31$. Similarly, from the second condition, it follows that $n-2 \vdots 65$. Since 31 and 65 are coprime, the smallest po... | 2017 | Combinatorics | olympiads | null | null | \(\boxed{2017}\) |
# Task 2. (20 points)
Find the maximum possible value of the ratio of a three-digit number to the sum of its digits.
# | # Solution.
Let $N=\overline{a b c}$, where $a, b, c$ are the digits of the number. Clearly, for "round" numbers $N=$ $100, 200, \ldots, 900$, we have $\frac{N}{a+b+c}=100$. Furthermore, if the number $N$ is not round, then $b+c>0$ and $a+b+c \geq a+1$. Since the leading digit of the number $N$ is $a$, we have $N<(a+1... | 100 | Number Theory | olympiads | null | null | \(\boxed{100}\) |
4. Consider the game "Battleship" on a $5 \times 5$ square grid. What is the minimum number of shots needed to guarantee hitting a ship of size $1 \times 4$ cells? | Obviously, less than five shots are not enough, since at least one shot must be made in each vertical and each horizontal. We will show that five shots are not enough. Indeed, a shot made in the first vertical leaves four consecutive cells on the given horizontal, which means at least one more shot must be made in this... | 6 | Logic and Puzzles | olympiads | null | null | \(\boxed{6}\) |
1. Senya thought of two numbers, then subtracted the smaller from the larger, added both numbers and the difference, and got 68. What was the larger of the numbers Senya thought of? | The subtrahend plus the difference equals the minuend. Therefore, the doubled minuend equals 68.
Comment. Correct solution - 20 points. The result is obtained based on examples and the pattern is noticed but not explained - 15 points. The result is obtained based on one example - 10 points. The solution is started, bu... | 34 | Algebra | olympiads | null | null | \(\boxed{34}\) |
1. There are 28 students in the class. 17 have a cat at home, and 10 have a dog. 5 students have neither a cat nor a dog. How many students have both a cat and a dog? | The number of students with cats or dogs is $28-5=23$. If we add $17+10=27$, it exceeds 23 due to those who have both a cat and a dog, who were counted twice. The number of such students is $27-23=4$.
Comment: Correct solution - 20 points. Solution started, with some progress - 5 points. Solution started, but progress... | 4 | Combinatorics | olympiads | null | null | \(\boxed{4}\) |
2. The little squirrel has several bags of nuts. In two bags, there are 2 nuts each, in three bags, there are 3 nuts each, in four bags, there are 4 nuts each, and in five bags, there are 5 nuts each. Help the little squirrel arrange the bags on two shelves so that there are an equal number of bags and nuts on each she... | For example, $5+5+5+4+4+2+2=27$ nuts in 7 bags - the first shelf, $5+5+4+4+3+3+3=27$ nuts in 7 bags - the second shelf.
Comment. Correct solution - 20 points. Solution started, some progress made - 5 points. Solution started, but progress insignificant - 1 point. Solution incorrect or absent - 0 points. | 27 | Logic and Puzzles | olympiads | null | null | \(\boxed{27}\) |
5. Sixth-graders were discussing how old their principal is. Anya said: "He is older than 38 years." Borya said: "He is younger than 35 years." Vova: "He is younger than 40 years." Galya: "He is older than 40 years." Dima: "Borya and Vova are right." Sasha: "You are all wrong." It turned out that the boys and girls mad... | Note that Anya and Vova cannot be wrong at the same time, so Sasha is wrong. Also, at least one of the pair "Anya-Borya" and at least one of the pair "Vova-Galya" is wrong. Thus, there are no fewer than three wrong answers, and due to the evenness, there are exactly four: two boys and two girls. This means that no more... | 39 | Logic and Puzzles | olympiads | null | null | \(\boxed{39}\) |
2. On 8 balls, numbers are written: $2,3,4,5,6,7,8,9$. In how many ways can the balls be placed into three boxes so that no box contains a number and its divisor? | The numbers 5 and 7 can be placed in any box, the number of ways is $3 \cdot 3=9$. The numbers $2, 4, 8$ must be in different boxes, the number of ways is $3 \cdot 2 \cdot 1=6$. Thus, the numbers $2, 4, 5, 7, 8$ can be arranged in 54 ways. Suppose the numbers 2 and 3 are placed in the same box (this box is already chos... | 432 | Combinatorics | olympiads | null | null | \(\boxed{432}\) |
3. For a children's party, pastries were prepared: 10 eclairs, 20 mini tarts, 30 chocolate brownies, 40 cream puffs. What is the maximum number of children who can each take three different pastries? | From eclairs, baskets, and brownies, at least 2 pastries must be taken, and there are 60 of them in total, meaning no more than 30 children can take three different pastries. They can do this as follows: 10 children will take an eclair, a brownie, and a roll, and 20 children will take a basket, a brownie, and a roll.
... | 30 | Combinatorics | olympiads | null | null | \(\boxed{30}\) |
3. (7-8 grade) Vasya thought of 5 natural numbers and told Petya all their pairwise sums (in some order): $122,124,126,127,128,129,130,131,132,135$. Help Petya determine the smallest of the numbers Vasya thought of. Answer: 60. | By adding all 10 pairwise sums, we get the sum of all 5 numbers, multiplied by 4, since each number participates in it exactly 4 times. $122+124+126+127+128+129+130+131+132+135=1284=4 S$, therefore, $S=321$. The sum of the two smallest numbers is 122, and the sum of the two largest is 135, so the third largest number i... | 60 | Algebra | olympiads | null | null | \(\boxed{60}\) |
5. (7-8 grade) Maria Ivanovna is a strict algebra teacher. She only puts twos, threes, and fours in the grade book, and she never gives the same student two twos in a row. It is known that she gave Vovochka 6 grades for the quarter. In how many different ways could she have done this? Answer: 448 ways. | Let $a_{n}$ be the number of ways to assign $n$ grades. It is easy to notice that $a_{1}=3$, $a_{2}=8$. Note that 3 or 4 can be placed after any grade, while 2 can only be placed if a 3 or 4 preceded it. Thus, a sequence of length $n$ can be obtained by appending 3 or 4 to a sequence of length $n-1$ or by appending 32 ... | 448 | Combinatorics | olympiads | null | null | \(\boxed{448}\) |
7. 7-8 grade Excellent student Kolya found the sum of the digits of all numbers from 0 to 2012 and added them all together. What number did he get? Answer: 28077. | Note that the numbers 0 and 999, 1 and 998, ..., 499 and 500 complement each other to 999, i.e., the sum of their digit sums is 27. Adding their digit sums, we get $27 \times 500 = 13500$. The sum of the digits of the numbers from 1000 to 1999, by similar reasoning, is 14500 (here we account for the fact that each numb... | 28077 | Number Theory | olympiads | null | null | \(\boxed{28077}\) |
11. (9th grade) In how many ways can the numbers $1,2,3,4,5,6$ be written in a row so that for any three consecutive numbers $a, b, c$, the quantity $a c-b^{2}$ is divisible by 7? Answer: 12. | Each subsequent member will be obtained from the previous one by multiplying by a certain fixed value (modulo 7). This value can only be 3 or 5. And the first term can be any, so there are 12 options in total. | 12 | Combinatorics | olympiads | null | null | \(\boxed{12}\) |
7. Olga Ivanovna, the class teacher of 5B, is staging a "Mathematical Ballet." She wants to arrange the boys and girls so that exactly 2 boys are 5 meters away from each girl. What is the maximum number of girls that can participate in the ballet if it is known that 5 boys are participating?
Answer: 20 girls. | Let's choose and fix two arbitrary boys - $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. Suppose they are 5 meters away from some girl - G. Then $\mathrm{M}_{1}, \mathrm{M}_{2}$, and G form an isosceles triangle with the legs being 5 meters. Given the fixed positions of $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$, there are no more ... | 20 | Combinatorics | olympiads | null | null | \(\boxed{20}\) |
. An apple, a pear, an orange, and a banana were placed in four boxes (one fruit per box). Inscriptions were made on the boxes:
On the 1st: Here lies an orange.
On the 2nd: Here lies a pear.
On the 3rd: If in the first box lies a banana, then here lies an apple or a pear.
On the 4th: Here lies an apple.
It is know... | The inscription on the 3rd box is incorrect, so in the first box lies a banana, and in the third - not an apple and not a pear, therefore, an orange. From the inscription on the 4th box, it follows that there is no apple there, so since the banana is in the 1st, and the orange is in the 2nd, then in the 4th lies a pear... | 2431 | Logic and Puzzles | olympiads | null | null | \(\boxed{2431}\) |
Problem 2. Beginner millionaire Bill buys a bouquet of 7 roses for $20 for the entire bouquet. Then he can sell a bouquet of 5 roses for $20 per bouquet. How many bouquets does he need to buy to earn a difference of $1000? | Let's call "operation" the purchase of 5 bouquets (= 35 roses) and the subsequent sale of 7 bouquets (= 35 roses). The purchase cost is $5 \cdot 20=\$ 100$, and the selling price is $7 \cdot 20=\$ 140$. The profit from one operation is $\$ 40$.
Since $\frac{1000}{40}=25$, 25 such operations are needed. Therefore, Bill... | 125 | Algebra | olympiads | null | null | \(\boxed{125}\) |
. Find a natural number $N(N>1)$, if the numbers 1743, 2019, and 3008 give the same remainder when divided by $N$. | From the condition, it follows that the numbers $2019-1743=276$ and $3008-2019=989$ are divisible by $N$. Since $276=2^{2} \cdot 3 \cdot 23$, and $989=23 \cdot 43$, then $N=23$. | 23 | Number Theory | olympiads | null | null | \(\boxed{23}\) |
Problem 6. Master Li Si Qing makes fans. Each fan consists of 6 sectors, painted on both sides in red and blue (see fig.). Moreover, if one side of a sector is painted red, the opposite side is painted blue and vice versa. Any two fans made by the master differ in coloring (if one coloring can be transformed into anoth... | ## Solution:
The coloring of one side can be chosen in $2^{6}=64$ ways. It uniquely determines the coloring of the opposite side. However, some fans - those that transform into each other when flipped, we have counted twice. To find their number, let's see how many fans transform into themselves when flipped. There ar... | 36 | Combinatorics | olympiads | null | null | \(\boxed{36}\) |
4.
In the test, there are 4 sections, each containing the same number of questions. Andrey answered 20 questions correctly. The percentage of his correct answers was more than 60 but less than 70. How many questions were in the test? | According to the condition $\frac{60}{100}<\frac{20}{x}<\frac{70}{100}$, hence $28 \frac{4}{7}=\frac{200}{7}<x<\frac{100}{3}=33 \frac{1}{3}$, that is $29 \leq x \leq 33$. From the first condition of the problem, it follows that the number of questions must be divisible by 4. | 32 | Logic and Puzzles | olympiads | null | null | \(\boxed{32}\) |
5. Find the largest natural number that cannot be represented as the sum of two composite numbers.
ANSWER: 11 | Even numbers greater than 8 can be represented as the sum of two even numbers greater than 2. Odd numbers greater than 12 can be represented as the sum of 9 and an even composite number. By direct verification, we can see that 11 cannot be represented in such a way.
Lomonosov Moscow State University
## School Olympia... | 11 | Number Theory | olympiads | null | null | \(\boxed{11}\) |
5. Philatelist Andrey decided to distribute all his stamps equally into 3 envelopes, but it turned out that one stamp was extra. When he distributed them equally into 5 envelopes, 3 stamps were extra; finally, when he distributed them equally into 7 envelopes, 5 stamps remained. How many stamps does Andrey have in tota... | If the desired number is $x$, then the number $x+2$ must be divisible by 3, 5, and 7, i.e., it has the form $3 \cdot 5 \cdot 7 \cdot p$. Therefore, $x=105p-2$. Since according to the condition $150 < x < 300$ for mathematics
Final stage tasks for the 2015/2016 academic year for 5-6 grades | 208 | Number Theory | olympiads | null | null | \(\boxed{208}\) |
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. | 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 | olympiads | null | null | \(\boxed{1}\) |
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 ... | 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 | olympiads | null | null | \(\boxed{8}\) |
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? | 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 only in the last digit). T... | 1007 | Number Theory | olympiads | null | null | \(\boxed{1007}\) |
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? | 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 multiple of 10, we get tha... | 387 | Algebra | olympiads | null | null | \(\boxed{387}\) |
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. | 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, and their number is 6. | 8 | Number Theory | olympiads | null | null | \(\boxed{8}\) |
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 | 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 will be exactly 100 such pair... | 20100 | Number Theory | olympiads | null | null | \(\boxed{20100}\) |
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... | 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 board, he would have naile... | 30 | Algebra | olympiads | null | null | \(\boxed{30}\) |
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? | 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 the top and bottom fac... | 18 | Logic and Puzzles | olympiads | null | null | \(\boxed{18}\) |
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 | 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, 4 times f... | 78331 | Combinatorics | olympiads | null | null | \(\boxed{78331}\) |
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... | 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 $N=1007$. | 1007 | Logic and Puzzles | olympiads | null | null | \(\boxed{1007}\) |
10. $(8,9)$ Find the smallest possible value of $\left|2015 m^{5}-2014 n^{4}\right|$, where $m$ and $n$ are natural numbers. | 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 numbers, for example, $a=c... | 0 | Number Theory | olympiads | null | null | \(\boxed{0}\) |
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? | 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 is achieved when $k=... | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) |
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? | 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 only in the last digit). T... | 1007 | Number Theory | olympiads | null | null | \(\boxed{1007}\) |
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... | 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}\frac{x}{2}=T-3 \\ \frac... | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) |
5. Find the number of pairs of natural numbers $(x, y), 1 \leqslant x, y \leqslant 1000$, such that $x^{2}+y^{2}$ is divisible by 5. | 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 $200 \times 200=40000$ options, in the se... | 360000 | Number Theory | olympiads | null | null | \(\boxed{360000}\) |
6. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | 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 sum of the remaining digi... | 999 | Number Theory | olympiads | null | null | \(\boxed{999}\) |
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... | 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 board, he would have naile... | 30 | Number Theory | olympiads | null | null | \(\boxed{30}\) |
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? | 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 the top and bottom fac... | 18 | Logic and Puzzles | olympiads | null | null | \(\boxed{18}\) |
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... | Let's find the person with the most friends. Suppose there are no fewer than 15, and denote their number as $15+a$. We will divide the students into two groups: the first group will consist of these $15+a$ students. According to the condition, they cannot be friends with each other, so each of them has no more than 15-... | 450 | Combinatorics | olympiads | null | null | \(\boxed{450}\) |
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. | The first vertex of the broken line can be chosen in 12 ways. Each subsequent vertex (except the last one) can be chosen in two ways - it must be adjacent to the already marked vertices to avoid self-intersections. The last vertex is chosen uniquely. We get $12 * 2^{10}$ ways. Considering 12 possible rotations, we get ... | 1024 | Combinatorics | olympiads | null | null | \(\boxed{1024}\) |
6. Pete came up with all the numbers that can be formed using the digits 2, 0, 1, 8 (each digit can be used no more than once). Find their sum. | 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, 4 times f... | 78331 | Combinatorics | olympiads | null | null | \(\boxed{78331}\) |
7. Given a sequence of natural numbers $a_{n}$, the terms of which satisfy
the relations $a_{n+1}=k \cdot \frac{a_{n}}{a_{n-1}}$ (for $n \geq 2$). All terms of the sequence are integers. It is known that $a_{1}=1$, and $a_{2018}=2020$. Find the smallest natural $k$ for which this is possible. | Let $a_2=x$. Then all terms of the sequence will have the form $x^{m} k^{n}$.
The powers of $k$ will repeat with a period of 6: $0,0,1,2,2,1,0,0, \ldots$
The powers of $x$ will also repeat with a period of 6: 0,1,1,0,-1,-1,0,1,...
Since 2018 gives a remainder of 2 when divided by 6, then $a_{2018}=a_{2}=x=2020$. For... | 2020 | Algebra | olympiads | null | null | \(\boxed{2020}\) |
7. A natural number $N$ ends in ...70, and it has exactly 72 natural divisors (including 1 and itself). How many natural divisors does the number 80N have? ANSWER: 324. | Each divisor of the number $\mathrm{N}$ can be represented in the form $2^{a} \cdot 5^{b} \cdot q$, where q is not divisible by 2 and 5, and the numbers $a, b$ are 0 or 1. Therefore, there are only 4 different combinations for the pair a and b, which means there are $72 / 4=18$ possible values for q. These correspond t... | 324 | Number Theory | olympiads | null | null | \(\boxed{324}\) |
4. The number 2015 can be represented as the sum of consecutive integers in various ways, for example, $2015=1007+1008$ or $2015=$ $401+402+403+404+405$. In how many ways can this be done? | The sum of $k$ numbers starting from $n$ is $S(k, n)=\frac{1}{2}(2 n+k-1) \cdot k$. That is, we need to solve the equation $(2 n+k-1) \cdot k=4030$ in integers. Obviously, $k$ can be any divisor of $4030=2 \times 5 \times 13 \times 31$. Note that each of the prime factors $(2,5,13$ and 31$)$ can be in the power of 0 or... | 16 | Number Theory | olympiads | null | null | \(\boxed{16}\) |
1. Natural numbers, for which the sum of the digits equals 5, were arranged in ascending order. What number is in the $122-$nd position? | Let's calculate the number of such numbers for different digit counts.
Let $n$ be the number of digits. Subtract 1 from the most significant digit, we get a number (which can now start with zero), the sum of whose digits is 4. Represent this as follows - there are 4 balls, between which $n-1$ partitions are placed. Th... | 40001 | Number Theory | olympiads | null | null | \(\boxed{40001}\) |
3. On a line, there are 16 points $A_{1}, \ldots A_{16}$, spaced $1 \mathrm{~cm}$ apart. Misha constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{16}$.
c) None of these... | We can represent such a system of circles as a tree with 16 leaves.
In such a tree, there cannot be more than 31 nodes. It is easy to construct a binary tree in which there are exactly 31 nodes. | 31 | Combinatorics | olympiads | null | null | \(\boxed{31}\) |
5. 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 $a_{999}$. | Consider the beginning of the sequence: $a_{1}=2, a_{2}=6, a_{3}=12, a_{4}=20, \ldots$
We can notice a pattern - the difference between consecutive terms forms an arithmetic progression: $4,6,8, \ldots$.
From this, we get the formula for the $\mathrm{n}$-th term $a_{n}=n \cdot(n+1)$, which can be proven by induction.... | 999000 | Algebra | olympiads | null | null | \(\boxed{999000}\) |
3. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | 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 sum of the remaining digi... | 999 | Number Theory | olympiads | null | null | \(\boxed{999}\) |
9. Calculate (without using a calculator) $\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}$, given that this number is an integer. | Note that $2<\sqrt[3]{9+4 \sqrt{5}}<3$ and $0<\sqrt[3]{9-4 \sqrt{5}}<1$. Therefore, $2<\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}<4$, hence this number is 3. | 3 | Algebra | olympiads | null | null | \(\boxed{3}\) |
10. The company conducted a survey among its employees - which social networks they use: VKontakte or Odnoklassniki. Some employees said they use VKontakte, some - Odnoklassniki, some said they use both social networks, and 40 employees said they do not use social networks. Among all those who use social networks, 75% ... | Since 75% of social media users use VKontakte, it follows that only 25% use Odnoklassniki. Additionally, 65% use both networks, so the total percentage of users of Odnoklassniki is $65 + 25 = 90\%$ of social media users. These $90\%$ constitute $5 / 6$ of the company's employees, so $100\%$ constitutes $10 / 9 * 5 / 6 ... | 540 | Algebra | olympiads | null | null | \(\boxed{540}\) |
11. In a regular 2017-gon, all diagonals are drawn. Petya randomly selects some $\mathrm{N}$ diagonals. What is the smallest $N$ such that among the selected diagonals, there are guaranteed to be two of the same length? | Let's choose an arbitrary vertex and consider all the diagonals emanating from it. There are 2014 of them, and by length, they are divided into 1007 pairs. Clearly, by rotating the polygon, any of its diagonals can be aligned with one of these. Therefore, there are only 1007 different sizes of diagonals. Thus, by choos... | 1008 | Combinatorics | olympiads | null | null | \(\boxed{1008}\) |
2. Let's call an integer "extraordinary" if it has exactly one even divisor other than 2. How many extraordinary numbers exist in the interval $[1 ; 75]$? | This number should be equal to a prime multiplied by 2. There are 12 such numbers:
$\begin{array}{llllllllllll}2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37\end{array}$ | 12 | Number Theory | olympiads | null | null | \(\boxed{12}\) |
5. In a regular 1000-gon, all diagonals are drawn. What is the maximum number of diagonals that can be selected such that among any three of the selected diagonals, at least two have the same length? | For the condition of the problem to be met, it is necessary that the lengths of the diagonals take no more than two different values. The diagonals connecting diametrically opposite vertices are 500. Any other diagonal can be rotated to coincide with a diagonal of the corresponding length, i.e., there are 1000 of them.... | 2000 | Combinatorics | olympiads | null | null | \(\boxed{2000}\) |
6. In how many ways can the number 1024 be factored into three natural factors such that the first factor is divisible by the second, and the second is divisible by the third? | Note that the factors have the form $2^{a} \times 2^{b} \times 2^{c}$, where $\mathrm{a}+\mathrm{b}+\mathrm{c}=10$ and $a \geq b \geq c$. Obviously, c is less than 4, otherwise the sum would be greater than 10. Let's consider the cases:
$c=0)$ Then $b=0, \ldots, 5, a=10-b-6$ options
c=1) Then $b=1, . .4, a=9-b-4$ opt... | 14 | Number Theory | olympiads | null | null | \(\boxed{14}\) |
4. Vasya is coming up with a 4-digit password for a code lock. He doesn't like the digit 2, so he doesn't use it. Moreover, he doesn't like when two identical digits are next to each other. Additionally, he wants the first digit to match the last digit. How many options need to be tried to guarantee guessing Vasya's pa... | The password must have the form ABCA, where A, B, C are different digits (not equal to 2). They can be chosen in $9 * 8 * 7=504$ ways.
Comment for graders: Half a point can be given to those who consider that the first digit cannot be zero. | 504 | Combinatorics | olympiads | null | null | \(\boxed{504}\) |
5. In the Empire of Westeros, there were 1000 cities and 2017 roads (each road connected some two cities). From any city, you could travel to any other. One day, an evil wizard enchanted $N$ roads, making it impossible to travel on them. As a result, 7 kingdoms formed, such that within each kingdom, you could travel fr... | Suppose the evil wizard enchanted all 2017 roads. This would result in 1000 kingdoms (each consisting of one city). Now, imagine that the good wizard disenchants the roads so that there are 7 kingdoms. He must disenchant at least 993 roads, as each road can reduce the number of kingdoms by no more than 1. Therefore, th... | 1024 | Combinatorics | olympiads | null | null | \(\boxed{1024}\) |
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