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Problem 10.3. Petya and Daniil are playing the following game. Petya has 36 candies. He lays out these candies in the cells of a $3 \times 3$ square (some cells may remain empty). After this, Daniil chooses four cells forming a $2 \times 2$ square and takes all the candies from there. What is the maximum number of candies that Daniil can guarantee to take?
|
# Answer: 9.
Solution. If Petya places 9 candies in each corner cell (and does not place any candies in the other cells), then in any $2 \times 2$ square there will be exactly 9 candies. After this, Daniil will be able to take only 9 candies.
Let's prove that Daniil can get at least 9 candies. Suppose the opposite: let him be able to take no more than 8 candies for any choice of a $2 \times 2$ square. Let him then sequentially take candies from all $x$ four $2 \times 2$ squares: first from one (no more than 8 candies), then the remaining candies from the next square (also no more than 8), then from the third square, and then from the fourth. At each step, he could take no more than 8 candies, so in total he will take no more than 32. But there should be 36. Contradiction.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.4. Roma thought of a natural number, the sum of the digits of which is divisible by 8. Then he added 2 to the thought number and again got a number, the sum of the digits of which is divisible by 8. Find the smallest number that Roma could have thought of.
|
Answer: 699.
Solution. If both numbers are divisible by 8, then their difference is also divisible by 8. If there was no carry-over when adding, the sum of the digits would differ by 2, which is not divisible by 8. If there was a carry-over but not into the hundreds place, the sum of the digits would differ by $9-2=7$, which is also not divisible by 8. If there was a carry-over into the hundreds place but not into the thousands place, the sum of the digits would change by $2 \cdot 9-2=16$, which is divisible by 8.
For a carry-over into the hundreds place when adding two, the last two digits must be 98 or 99. Now, let's find the smallest number in each case with a sum of digits divisible by 8. These numbers are 798 and 699; it is clear that 699 is smaller (and it obviously meets the condition).
It remains to note that if there was a carry-over into the thousands place or higher when adding 2, the original number would be at least 998, but the found number is already smaller.
|
699
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.5. On the side $AB$ of the rectangle $ABCD$, a circle $\omega$ is constructed with $AB$ as its diameter. Let $P$ be the second intersection point of the segment $AC$ and the circle $\omega$. The tangent to $\omega$ at point $P$ intersects the segment $BC$ at point $K$ and passes through point $D$. Find $AD$, given that $KD=36$.
|
Answer: 24.
Solution. Note that triangles $A D P$ and $C K P$ are similar (the equality $\angle D A P=\angle K C P$ and $\angle A D P=\angle C K P$ is ensured by the parallelism $A D \| B C$). Moreover, triangle $A D P$ is isosceles, as $A D$ and $D P$ are segments of tangents. Therefore, triangle $C K P$ is also isosceles, meaning $C K=K P$.
On the other hand, $K P=K B$, as these are also segments of tangents. We get $K C=K B$, so $K$ is the midpoint of side $B C$.
Then the similarity ratio of triangles $A D P$ and $C K P$ is the ratio $A D: K C$, which is 2. Therefore, $P K=\frac{1}{2} P D$, from which $A D=P D=\frac{2}{3} K D=24$.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.8. On the face $ABC$ of the tetrahedron $ABCD$, a point $P$ is marked. Points $A_1, B_1, C_1$ are the projections of point $P$ onto the faces $BCD, ACD, ABD$ respectively. It turns out that $PA_1 = PB_1 = PC_1$. Find $\angle BA_1C$, given that $\angle BC_1D = 136^\circ$ and $\angle CB_1D = 109^\circ$.
|
Answer: 115.
Solution. Consider triangles $D P A_{1}$ and $D P C_{1}$. They are equal by the leg $\left(P C_{1}=P A_{1}\right)$ and the hypotenuse (common $P D$). From this, we get that $D C_{1}=D A_{1}$.
Considering triangles $B P C_{1}$ and $B P A_{1}$, we similarly obtain $B C_{1}=B A_{1}$. Then triangles $B D C_{1}$ and $B A_{1} D$ are equal by three sides, which means $\angle B C_{1} D=\angle B A_{1} D$.
Similarly, we can obtain $\angle C B_{1} D=\angle C A_{1} D$. Now it is clear that the desired angle is calculated as
$$
\angle B A_{1} C=360^{\circ}-\angle B A_{1} D-\angle C A_{1} D=360^{\circ}-\angle B C_{1} D-\angle C B_{1} D=115^{\circ}
$$
Remark. If we draw a sphere with center at point $P$ and radius $P A_{1}$, then it will touch the faces $D A B, D B C, D C A$. Then the equalities of segments, for example, $D C_{1}=D A_{1}=D B_{1}$, will follow from the equality of the segments of tangents to the sphere.
|
115
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (7 points) Winnie-the-Pooh eats 3 cans of condensed milk and a jar of honey in 25 minutes, while Piglet takes 55 minutes. One can of condensed milk and 3 jars of honey, Pooh eats in 35 minutes, while Piglet takes 1 hour 25 minutes. How long will it take them to eat 6 cans of condensed milk together?
|
# Solution.
1st method. From the condition, it follows that Winnie-the-Pooh eats 4 cans of condensed milk and 4 jars of honey in 1 hour, while Piglet does so in 2 hours and 20 minutes. Therefore, one can of condensed milk and one jar of honey are eaten by Winnie-the-Pooh in 15 minutes, and by Piglet in 35 minutes.
Using the first condition, we get that 2 cans of condensed milk will be eaten by Winnie-the-Pooh in 10 minutes, and by Piglet in 20 minutes. Since Winnie-the-Pooh eats condensed milk twice as fast as Piglet, they will eat 6 cans of condensed milk in 20 minutes.
2nd method. Let Winnie-the-Pooh eat condensed milk at a speed of \( V_{1} \) cans per minute, and honey at a speed of \( V_{2} \) cans per minute. Then
\[
\left\{\begin{array}{l}
\frac{3}{V_{1}}+\frac{1}{V_{2}}=25, \\
\frac{1}{V_{1}}+\frac{3}{V_{2}}=35
\end{array} \Rightarrow V_{1}=\frac{1}{5}\right.
\]
Let Piglet eat condensed milk at a speed of \( U_{1} \) cans per minute, and honey at a speed of \( U_{2} \) cans per minute. Then
\[
\left\{\begin{array}{l}
\frac{3}{U_{1}}+\frac{1}{U_{2}}=55 \\
\frac{1}{U_{1}}+\frac{3}{U_{2}}=85
\end{array} \Rightarrow U_{1}=\frac{1}{10}\right.
\]
Thus, Winnie-the-Pooh and Piglet together will eat condensed milk at a speed of \( V_{1}+U_{1}=\frac{3}{10} \) cans per minute. Therefore, they will eat 6 cans of condensed milk in \( 6: \frac{3}{10}=20 \) minutes.
Answer. 20 minutes.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.6. Each of the 1000 gnomes has a hat, blue on the outside and red on the inside (or vice versa). If a gnome is wearing a red hat, he can only lie, and if it is blue, he can only tell the truth. During one day, each gnome said to each other, "You have a red hat!" (during the day, some gnomes turned their hats inside out). Find the minimum possible number of turnings. (I. Bogdanov)
|
9.6. Answer. 998 turnovers.
Let's call a gnome red or blue if he is wearing a cap of the corresponding color. Note that one gnome can say the required phrase to another if and only if these gnomes are of different colors: a blue gnome will tell the truth in this case, and a red one will lie. Now, if any three gnomes have not turned their caps, then two of them are of the same color, and they will not be able to say the required to each other, which is incorrect. Therefore, there are no more than two such gnomes, and the number of turnovers is not less than $1000-2=998$.
We will say that two gnomes have communicated if each of them has said the desired phrase to the other. Let's describe how it could happen that there were only 998 turnovers, for example, if at the beginning the gnome Vasya was blue, and the others were red. At the beginning of the day, each gnome communicated with Vasya. Then the red gnomes took turns turning their caps. In this case, after each turnover, all the red gnomes communicated with the one who changed color. When only one red gnome is left, any pair of gnomes will already have communicated with each other (at the moment when the first of them changed color), while 998 color changes have occurred.
Remark. An example with 998 color changes can be constructed starting from any situation where not all gnomes are of the same color.
|
998
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2.1. Points $A, B, C, D$ are marked on a line, in that exact order. Point $M$ is the midpoint of segment $A C$, and point $N$ is the midpoint of segment $B D$. Find the length of segment $M N$, given that $A D=68$ and $B C=20$.
## 68
|
Answer: 24.
Solution. Let $AC = x$, then $AM = \frac{x}{2}$. Now we will calculate the length of $ND$:
$$
ND = \frac{BD}{2} = \frac{20 + CD}{2} = \frac{20 + (68 - x)}{2} = 44 - \frac{x}{2}
$$
Now it is not difficult to calculate $MN$:
$$
MN = AD - AM - ND = 68 - \frac{x}{2} - \left(44 - \frac{x}{2}\right) = 24
$$
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.2. During a physical education class, 25 students from 5B class lined up. Each of the students is either an excellent student who always tells the truth, or a troublemaker who always lies.
Excellent student Vlad stood in the 13th place. Everyone except Vlad stated: "There are exactly 6 troublemakers between me and Vlad." How many troublemakers are there in the line?
|
Answer: 12.
Solution. Note that students in places $7-12$ are troublemakers, since there are fewer than 6 people between each of them and Vlad. Therefore, the student with number 6 is an excellent student. The same can be said about the student in the 5th place, then about the 4th, the 3rd, the 2nd, and the 1st.
Thus, the first six places are occupied by excellent students, the next six places are occupied by troublemakers, and Vlad is already at place 13. Similar reasoning can be applied to students in places $14-25$. Therefore, there are exactly $6+6=12$ troublemakers in the line.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 5.4. Masha drew two little people in her notebook. The area of each cell is 1.
Which of the little people has a larger area?
What is the difference? If the areas are the same, write "0" in the answer.
|
Answer: The area of the right human figure is 2 more than the area of the left one.
Solution. Let's calculate the areas of the human figures.
As can be seen from the figure on the right, the first human figure consists of
- 8 whole cells,
- 6 small triangles (each of which is equal to half a cell),
- 2 large triangles (each of which is equal to half a rectangle $1 \times 2$).
Every two small triangles "add up to one whole cell" in total, so their total area is 1. Every two large triangles "add up to one rectangle $1 \times 2$" in total, so their total area is 2.
Then the area of the first human figure is
$$
8 \cdot 1 + 3 \cdot 1 + 1 \cdot 2 = 13
$$
As can be seen from the figure on the right, the second human figure consists of
- 4 whole cells,
- 6 small triangles (each of which is equal to half a cell),
- 8 large triangles (each of which is equal to half a rectangle $1 \times 2$).
Then the area of the second human figure is
$$
4 \cdot 1 + 3 \cdot 1 + 4 \cdot 2 = 15
$$
Thus, the area of the second (right) human figure is 2 more than the area of the first (left) one.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.5. Denis has identical ten-ruble coins, identical two-ruble coins, and identical one-ruble coins (more than 20 coins of each type). In how many ways can Denis pay exactly 16 rubles for a pie without receiving change? It is not necessary to use coins of each type.
|
Answer: 13.
Solution. If Denis uses a ten-ruble coin, he will need to collect 6 rubles using two-ruble and one-ruble coins. There are 4 ways to do this: using from 0 to 3 two-ruble coins.
If Denis does not use the ten-ruble coin, he will need to collect 16 rubles using two-ruble and one-ruble coins. There are 9 ways to do this, using from 0 to 8 two-ruble coins.
In total, there are $4+9=13$ ways.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.7. In a class, there are 31 students. Three of them have exactly three friends each, the next three have six each, the next three have nine each, ..., and the next three have thirty each. How many friends does the 31st student have? (Friendship between people is mutual.)
Fig. 2: to the solution of problem 5.6
Fig. 3: to the solution of problem 5.6
#
|
# Answer: 15.
Solution. Let the 31st student have $x$ friends.
Consider three people, each of whom has 30 friends in the class. There are a total of 31 students in the class, so they are friends with all their classmates.
Let's expel them from the class. Then the number of friends for each person will decrease by 3:
- the first trio will have 0 friends each;
- the second trio will have 3 friends each;
- $\ldots$
- the ninth trio will have 24 friends each;
- the last student will have $x-3$ friends.
We will also expel the three students who have no friends. Then the configuration will look like this.
There are 25 students in the class. Three of them have exactly three friends each, the next three have six each, the next three have nine each, ..., the next three have 24. The last student has $x-3$ friends.
Repeat a similar action 4 more times. Then only the last student will remain in the class, and he will have $x-15$ friends. Since at this moment there are no more people left in the class, $x=15$.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. 20 schoolchildren came to the mathematics Olympiad. Everyone who brought a pencil also brought a pen. 12 people forgot their pencils at home, and 2 schoolchildren forgot their pen. By how many fewer schoolchildren brought a pencil than those who brought a pen but forgot a pencil?
|
Answer: 2.
Solution: 8 students brought a pencil, which means they also brought a pen. 18 students brought a pen. Therefore, 10 students brought a pen without a pencil. Then 10-8=2.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.5. Four dolls and five robots cost 4100 rubles, while five dolls and four robots cost 4000. How much does one doll cost?
|
Answer: 400 rubles.
Solution. Let's briefly write down the condition of the problem:
$$
\begin{aligned}
& 4 \kappa + 5 p = 4100 \\
& 5 \kappa + 4 p = 4000 .
\end{aligned}
$$
Then $9 \kappa + 9 p = 8100$ and the cost of one doll together with one robot will be 900 rubles. Therefore, 4 dolls and 4 robots cost 3600 rubles, which means one doll costs 400 rubles.
|
400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. 1.1. An online electric scooter rental service charges a fixed amount for a ride and a certain amount for each minute of rental. Katya paid 78 rubles for a ride lasting 3 minutes. Lena paid 108 rubles for a ride lasting 8 minutes. It takes Kolya 5 minutes to get from home to work. How much will he pay for this ride?
|
Answer: 90.
## Solution.
1st method. If the rental of a scooter costs $x$ rubles, and a minute of use costs $y$ rubles, then $x+3y=78$, and $x+8y=108$. Solving this system, we get that $y=6, x=60$. Then Kolya will pay $60+30=90$ rubles.
2nd method. Notice that Lena's trip is 30 rubles more expensive than Katya's trip and lasts 5 minutes longer. It turns out that 1 minute, not counting the rental cost, is $30:5=6$ rubles. Then, since Katya paid 78 rubles for 3 minutes, $78-3 \times 6=60$ rubles. Therefore, Kolya paid $60+30=90$ rubles.
|
90
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. 2.1. Dealer Dima bought a "LADA Kalina" car from the manufacturer and increased its price by $A \%$. There was no demand, and Dima had to sell the car at a sale with a $20 \%$ discount. As a result, his profit was $20 \%$. Find $A$.
|
Answer: 50.
## Solution.
1st method. Let's assume Dima bought a car from the manufacturer for $X$ rubles. After raising the price, the car cost $X(1+A / 100)$ rubles. After reducing the price by $20 \%$, the car cost $0.8 X(1+A / 100)$. According to the condition, this is the same as $1.2 X$. Therefore, $1+A / 100=1.2 / 0.8$, which means $A=50$.
2nd method. After increasing by A\%, the price increased by ( $1+A / 100$ ) times. After reducing the price by $20 \%$, the price is multiplied by 0.8 and becomes 1.2 times the purchase price. Then, $(1+A / 100) \cdot 0.8=1.2$. Therefore, $1+A / 100=1.2: 0.8=1.5$, from which $=50$.
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. 3.1. Petya marked 5 points on a face of a cube, turned it and marked 6 points on an adjacent face, then turned it again and marked 7 points, and so on. He marked points on each face this way. What is the maximum number of points that can be on two opposite faces?
|
Answer: 18.
## Solution.
Since a cube has only 6 faces, the maximum number of points marked by Petya is 10. Then the number 9 will be on an adjacent face to the number 10 and cannot be on the opposite face. Therefore, the maximum sum will not exceed 18. The sum of 18 can be achieved, for example, as follows:
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. 5.1. An isosceles trapezoid \(ABCD\) is inscribed in a circle with diameter \(AD\) and center at point \(O\). A circle is inscribed in triangle \(BOC\) with center at point \(I\). Find the ratio of the areas of triangles \(AID\) and \(BIC\), given that \(AD=15, BC=5\).
|
Answer: 9.
## Solution.
From the fact that $A D$ is the diameter of the circle circumscribed around the trapezoid, it follows that $A O = B O = C O = D O$ as radii. Therefore, triangle $B O D$ is isosceles. Hence, $\angle O B D = \angle O D B = \angle D B C$ (the last equality is due to $A D \parallel B C$). Thus, $B D$ is the bisector of angle $O B C$. Similarly, $A C$ is the bisector of angle $O C B$. Since the center of the inscribed circle in triangle $B O C$ coincides with the intersection of the angle bisectors, we have: $I = A C \cap B D$. Further, due to the similarity of triangles $A I D$ and $B I C$, the areas are proportional to the square of the similarity coefficient: $\left(\frac{A D}{B C}\right)^{2} = \left(\frac{15}{5}\right)^{2} = 9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the largest natural number in which all digits are different and any two adjacent digits differ by 6 or 7.
|
Answer: 60718293.
## Solution.
We will map each digit from 0 to 9 to a vertex in a graph and connect the vertices with an edge if the corresponding digits differ by 6 or 7.
We see that the vertices corresponding to the digits 4 and 5 are not connected to anything, so the maximum number that can be obtained will be 8 digits long. The vertices corresponding to the digits 3 and 6 can only be the first or last in the number. Since we need to find the maximum number, it is uniquely determined by moving along the edges of the graph and will be equal to 60718293.
|
60718293
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Find the smallest positive integer $n$ such that $A_{n}=1+11+111+\ldots+1 \ldots 1$ (the last term contains $n$ ones) is divisible by 45.
|
Answer: 35.
## Solution.
For the sum $A_{n}$ to be divisible by 45, it must be divisible by 5 and by 9. By the divisibility rule for 5, we get that $n$ must be a multiple of 5. Since any natural number gives the same remainder when divided by 9 as the sum of its digits, $A_{n}$ gives the same remainder when divided by 9 as the sum $1+2+\ldots+n=\frac{n(n+1)}{2}$. Since 9 and 2 have no common divisors, $n(n+1)$ must be divisible by 9. Two consecutive natural numbers cannot both be divisible by 3, so either $n$ or $n+1$ must be divisible by 9. Next, consider the numbers $n$ that are multiples of 5, so that either $n$ or $n+1$ is divisible by 9 (it is obvious that $n \leq 45$).
| $n$ | 5 | 10 | 15 | 20 | 25 | 30 | 35 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $n+1$ | 6 | 11 | 16 | 21 | 26 | 31 | 36 |
From the table, we see that the smallest number is $n=35$.
|
35
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. 8.1. The vertices of a regular 22-sided polygon are numbered. In how many ways can four of its vertices be chosen to form a trapezoid? (A trapezoid is a quadrilateral with one pair of parallel sides and the other two sides not parallel).
|
Answer: 990.
## Solution.
Note that all trapezoids we obtain will be isosceles. We will count the number of inscribed quadrilaterals that have an axis of symmetry that does not pass through the vertices. Thus, we will count all trapezoids once (since each isosceles trapezoid has only one axis of symmetry), and count each rectangle twice (since a rectangle has two such axes of symmetry).
The axis of symmetry we consider can contain vertices of the polygon or not, see the figure.
Now let's count the number of the described quadrilaterals in a $2n$-gon. 1) Consider an axis of symmetry passing through the midpoint of a side of the $2n$-gon. There are exactly $n$ such axes. Choosing any two points on one side of the axis of symmetry uniquely determines our quadrilateral. The number of ways to choose two points from $n$ is $C_{n}^{2}=n(n-1) / 2$. Thus, in this case, we have counted $n^{2}(n-1) / 2$ quadrilaterals. 2) If the axis of symmetry passes through two vertices of the polygon, there are again $n$ such axes. Choosing any two points on one side of the axis of symmetry also uniquely determines our quadrilateral, but now there are $n-1$ points on each side of the axis. The number of ways to choose two points from $n-1$ is $C_{n-1}^{2}=(n-1)(n-2) / 2$. Thus, in this case, we have $n(n-1)(n-2) / 2$ quadrilaterals. We need to subtract twice the number of rectangles, and we will get the answer. Each rectangle is uniquely determined by two diameters of the circle. Therefore, their doubled number is $2 C_{n}^{2}=n(n-1)$. Thus, we finally get $n^{2}(n-1) / 2+n(n-1)(n-2) / 2-2 n(n-1) / 2=n(n-1)(n-2)$. In the case of a 22-gon, $n=11$ and the number of ways to choose a trapezoid will be $11 \cdot 10 \cdot 9=990$.
|
990
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. Petya writes on the board such different three-digit natural numbers that each of them is divisible by 3, and the first two digits differ by 2. What is the maximum number of such numbers he can write if they end in 6 or 7?
|
Answer: 9
Solution. A number is divisible by 3 if the sum of its digits is a multiple of 3. If the number ends in 6, then the sum of the other two digits leaves a remainder of 0 when divided by 3. Such numbers are: 2,4 and 4,2; 5,7 and 7,5. If the number ends in 7, then the sum of the other two digits leaves a remainder of 2 when divided by 3. Such numbers are: 2,0; 3,5 and 5,3; 6,8 and 8,6. In total, there are 9 numbers: 246, 426, 576, 756, 207, 357, 537, 687, 867.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. Every hour, between two adjacent nettle bushes in a row, two more of the same grow. How many bushes do you need to plant initially so that after three hours, the total number of bushes together is 190?
|
Answer: 8
Solution. If at the moment there are $\mathrm{n}$ bushes, then on the next move their number increases by $2(n-1)$. Thus, if after 3 hours the total number of bushes should be 190, then one hour before that, there should be $\frac{190-1}{3}+1=64$. One more hour back, $\frac{64-1}{3}+1=22$. And on the next move $\frac{22-1}{3}+1=8$.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. 40 people came into a room where there were 40 chairs, black and white, and sat on them. All of them said they were sitting on black chairs. Then they somehow resat, and exactly 16 claimed they were sitting on white chairs. Each of those sitting either lied both times or told the truth both times. How many of them lied the second time and were sitting on black chairs?
|
Answer: 8
Solution. Initially, everyone who told the truth sat on black chairs, and everyone who lied sat on white ones. After some of them switched chairs, 16 claimed they were sitting on white chairs. Obviously, this group includes those who told the truth and were sitting on white chairs, and those who switched with them - they lied and were sitting on black chairs. Since their numbers are equal, exactly half of them - 8 - lied and were sitting on black chairs.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. On a circular road, there are three cities: $A, B$ and $C$. It is known that the path from $A$ to $C$ is 3 times longer along the arc not containing $B$ than through $B$. The path from $B$ to $C$ is 4 times shorter along the arc not containing $A$ than through $A$. How many times shorter is the path from $A$ to $B$ along the arc not containing $C$ than through $C$?
|
Answer: 19
Solution. From the condition of the problem, the following relationship follows:
$$
\begin{aligned}
& x=3(y+z) \\
& 4 z=x+y \\
& 3 y+3 z=4 z-y \\
& 4 y=z \\
& x=15 y \\
& x+z=19 y
\end{aligned}
$$
|
19
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. On Fyodor's bookshelf, there are volumes of works by various authors. When a friend borrowed two volumes of Pushkin, Fyodor noticed that among the remaining books, he had read at least half of them in full. After the friend returned the two volumes of Pushkin and borrowed two volumes of Lermontov, Fyodor realized that among the remaining books, he had read less than one third of them in full. What is the greatest number of volumes that could have been on the shelf at Fyodor's?
|
Answer: 12
Solution. Let's number all of Fyodor's books from 1 to $\mathrm{n}$.
Let the first two volumes be Pushkin's, $s_{1}$ - the number of them read by Fyodor; the last two - Lermontov's, $s_{2}$ - the number of them read by Fyodor. $S$ - the number of read volumes among the books from 3 to $n-2$.
According to the condition,
$S+s_{2} \geqslant \frac{n-2}{2}, S+s_{1}<\frac{n-2}{3}$,
$3 S+3 s_{1}<n-2 \leqslant 2 S+2 s_{2}$,
$3 S+3 s_{1}<2 S+2 s_{2}$
$S<2 s_{2}-3 s_{1} \leqslant 4$
Thus, $S$ does not exceed 3. Then the maximum possible $S=3$. From this, $n \leqslant 12$. Example for 12 (1 read, 0 - not read): 001110000011.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.5. There are nuts in three boxes. In the first box, there are six fewer nuts than in the other two boxes combined, and in the second box, there are ten fewer nuts than in the other two boxes combined. How many nuts are in the third box? Justify your answer.
|
Solution: Let there be $x$ nuts in the first box, $y$ and $z$ in the second and third boxes, respectively. Then the condition of the problem is given by the equations $x+6=y+z$ and $x+z=y+10$. From the first equation, $x-y=z-6$, and from the second, $x-y=10-z$. Therefore, $z-6=10-z$, from which $z=8$.
Answer: 8 nuts.
| in progress | points |
| :--- | :--- |
| correct and fully justified answer | 7 points |
| an incorrect answer due solely to arithmetic errors | 6 points |
| the system of equations describing the problem condition is correctly set up but not solved | 3 points |
| a specific example (examples) of the distribution of nuts in the boxes (and, therefore, the correct answer) is provided | 1 point |
| only the correct answer is given without justification (or with incorrect justification) | 0 points |
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. Find $f(2021)$ if for any real $x$ and $y$ the equality $f(x+f(y))=x+y$ holds.
|
Answer: 2021.
Solution. Let $f(0)=b$, then for $y=0$ we get $f(x+b)=x$, from which $f(x)=x-b$. Thus, $f(x+f(y))=f(x+(y-b))=f(x+y-b)=x+y-b-b=x+y-2b$.
Since $f(x+f(y))=x+y$ for any real $x$ and $y$, then $b=0$, so $f(x)=x$, and $f(2021)=2021$.
Comment. Only the example of the function $f(x)=x-1$ gets a point.
|
2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4. There is a grid board $2015 \times 2015$. Dima places a detector in $k$ cells. Then Kolya places a grid ship in the shape of a square $1500 \times 1500$ on the board. The detector in a cell reports to Dima whether this cell is covered by the ship or not. For what smallest $k$ can Dima place the detectors in such a way as to guarantee the reconstruction of the ship's position?
(O. Dmitriev, R. Zhendarov)
|
Answer. $k=2(2015-1500)=1030$.
Solution. We will show that 1030 detectors are enough for Dima. Let him place 515 detectors in the 515 leftmost cells of the middle row of the square, and the remaining 515 detectors in the 515 top cells of the middle column. Note that for any position of the ship, its left column lies in one of the 516 leftmost columns of the board. If this column is one of the 515 farthest left, then the ship will cover a detector from this column, lying in the middle row; otherwise, it will not cover any detector from this row. Therefore, the columns in which the ship lies can be determined from the readings of the detectors in this row. Similarly, the rows in which it is located can be determined from the readings of the detectors in the middle column.
Now consider an arbitrary arrangement of $k$ detectors that meet the requirements. Consider two positions of the ship, differing by a horizontal shift of 1. The readings of some detector will differ for them only if this detector lies in the far left column of the left ship or in the far right column of the right ship. This means that in any two vertical rectangles $1500 \times 1$, differing by a horizontal shift of 1500, there is at least one detector. Similarly, in any two horizontal rectangles $1 \times 1500$, differing by a vertical shift of 1500, there is at least one detector. Let's call such pairs of rectangles vertical and horizontal, respectively.
Highlight all vertical pairs lying in the lower 1500 and upper 1500 rows of the board (there are $2 \cdot 515=1030$ such pairs). Similarly, highlight all 1030 horizontal pairs lying in the left 1500 and right 1500 columns. Divide the board into 9 rectangular areas as shown in Fig. 3. The highlighted pairs do not cover cells in $E$; each cell in the other areas is covered by two highlighted pairs (in $D$ and $F$ - by two vertical pairs, in $B$ and $H$ - by two horizontal pairs, and in areas $A, C, G$, and $I$ - by one horizontal and one vertical pair). Thus, each detector lies in no more than two highlighted pairs; therefore, to have at least one detector in each highlighted pair, at least $2 \cdot 1030 / 2=1030$ detectors are required.
Remark. There are many other examples of placing 1030 detectors that meet the requirements.
Comment. An example of placing 1030 detectors that meet the requirements is given - 1 point.
Only proved that $k \geqslant 1030-$ 5 points.
Fig. 5
If, in an otherwise correct reasoning, there is one or more errors of one unit in the count of detectors in the example or in the estimate (for example, it is claimed that the number of highlighted vertical pairs in the lower rows is 516, not 515) - 1 point is deducted.
|
1030
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
|
Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm.
|
120
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4-5. Given a figure consisting of 33 circles. You need to choose three circles that are consecutive in one of the directions. In how many ways can this be done? The image shows three of the desired ways.
|
Answer: 57.
Solution. The number of options along the long side is $1+2+3+4+5+6=21$. Along each of the other two directions, it is $-4+4+4+3+2+1=18$. The total number of options is $21+18 \cdot 2=57$
|
57
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
|
Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$
|
500
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th lamp, Boris was at the 321st lamp. At which lamp will their meeting occur? If the meeting occurs between two lamps, indicate the smaller number of these two.
|
Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to the meeting point than to the 55th lamppost, i.e., $54 \cdot 3=162$ intervals. And she will be at the 163rd lamppost.
|
163
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right corner. What is the length $x$ in centimeters?
|
Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such a rectangle the current one. Notice that for each new current rectangle, both the width and the height are 1 cm larger than the previous one. Initially, when there was only one sheet of paper, the width of the large rectangle was 8 cm, and at the end, it was 80 cm. Thus, a total of $(80-8): 1=72$ sheets of paper were added. The height of the current rectangle also increased by $72 \cdot 1$ cm, initially it was 5 cm, so $x=5+72=77$.
Solution II. As in the first solution, let's look at the length and width of the current rectangles. Again, notice that for each new current rectangle, both the length and the width are 1 cm larger than the previous one. However, we will draw a different conclusion: specifically, the difference between the width and the height of the current rectangle is always the same! (Such a value that does not change during a certain process is called an invariant.) Since initially the width was 3 cm greater than the height, i.e., $8-5=3$ cm, at the end it should also be 3 cm greater, so the answer is $x=80-3=77$ cm.
|
77
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
|
Answer: 65.
Solution. The area of the white part is $8 \cdot 10-37=43$, so the area of the gray part is $12 \cdot 9-43=65$
|
65
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-2. In square $A B C D$, a segment $C E$ is drawn such that the angles shown in the diagram are $7 \alpha$ and $8 \alpha$. Find the value of angle $\alpha$ in degrees. If necessary, round the answer to 0.01 or write the answer as a common fraction.
|
Answer: $9^{\circ}$.
Solution. In triangle $D F E$, the angles are $7 \alpha, 8 \alpha$ and $45^{\circ}$.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha + 8 \alpha + 45^{\circ} = 180^{\circ}$, from which $\alpha = \frac{1}{15} \cdot 135^{\circ} = 9^{\circ}$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-1. Segment $P Q$ is divided into several smaller segments. On each of them, a square is constructed (see figure).
What is the length of the path along the arrows if the length of segment $P Q$ is 73? If necessary, round the answer to 0.01 or write the answer as a common fraction.
|
Answer: 219.
Solution. Note that in each square, instead of going along one side, we go along three sides. Therefore, the length of the path along the arrows is 3 times the length of the path along the segment, hence the answer $73 \cdot 3=219$.
|
219
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
|
Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then $\angle A C O=\angle O A C=90^{\circ}-67^{\circ}=23^{\circ}$ (here we used the fact that triangle $A C D$ is a right triangle: angle $A C D$, which subtends the diameter, is a right angle).
Thus, $x=32^{\circ}-23^{\circ}=9^{\circ}$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
|
Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they are radii of the circle, so $\angle OCA = \angle ACO = 48^{\circ}$ and $x = \angle OBC = \angle OCB = 48^{\circ} + 10^{\circ} = 58^{\circ}$.
|
58
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.6. A square $100 \times 100$ is divided into squares $2 \times 2$. Then it is divided into dominoes (rectangles $1 \times 2$ and $2 \times 1$). What is the smallest number of dominoes that could end up inside the squares of the division?
(C. Berlov)
#
|
# Answer. 100.
Solution. Example. We will divide the top and bottom horizontals into horizontal dominoes - they will end up in $2 \times 2$ squares. The remaining rectangle $98 \times 100$ will be divided into vertical dominoes - they will not end up in $2 \times 2$ squares.
Estimate. Consider the squares $A_{1}, A_{3}, \ldots, A_{99}$ of sizes $1 \times 1, 3 \times 3, \ldots, 99 \times 99$, whose lower left corner coincides with the lower left corner of the original $100 \times 100$ square. For each of the squares $A_{i} (i=1,3,5, \ldots, 99)$, there is a domino $X_{i}$ that intersects its side (since squares of odd area cannot be divided into dominoes). It is easy to see that $X_{i}$ lies within a $2 \times 2$ square from the division.
Similarly, considering the squares $B_{1}, B_{3}, \ldots, B_{99}$ of sizes $1 \times 1, 3 \times 3, \ldots, 99 \times 99$, whose upper right corner coincides with the upper right corner of the original $100 \times 100$ square, we find another 50 dominoes $Y_{j} (j=1,3,5, \ldots, 99)$ that we need.
This completes the solution (it is obvious that all dominoes $X_{1}, X_{3}, \ldots, X_{99}, Y_{1}, Y_{3}, \ldots, Y_{99}$ are distinct).
Remark. We will present a slightly different proof of the estimate.
Suppose that no more than 99 dominoes ended up inside the $2 \times 2$ squares.
Draw 50 vertical grid lines $v_{1}, v_{3}, v_{5}, \ldots, v_{99}$ such that $v_{i}$ separates $i$ columns on the left. It is easy to see that any domino intersected by one of the lines $v_{1}, v_{3}, v_{5}, \ldots, v_{99}$ is suitable for us. Each vertical line intersects an even number of dominoes, since there is an even number of cells to the left of this line. Therefore, among the lines $v_{1}, v_{3}, v_{5}, \ldots, v_{99}$, there is a line $v_{i}$ that does not intersect any dominoes, otherwise we would have already found at least $2 \cdot 50 = 100$ dominoes that we need. We will conduct a similar reasoning for 50 horizontal grid lines $h_{1}, h_{3}, h_{5}, \ldots, h_{99}$ and find among them a line $h_{j}$ that does not intersect any dominoes. But $v_{i}$ and $h_{j}$ divide the board into regions with an odd number of cells, so at least one of these two lines must intersect a domino. Contradiction.
Comment. Only the example -1 point.
Proved only that the number of dominoes is even - 0 points.
Criteria for a solution with paths of good (lying in $2 \times 2$ squares) dominoes are as follows.
Only the idea of paths -0 points.
Proved that next to each good domino there is at least one more good one, and there is an idea of constructing a path from these dominoes -1 point.
For a complete solution of the problem, it remains to solve the problem of looping and crossing paths of dominoes, and there is a division into regions by shifting the grid and constructing a graph with even degrees of vertices, and there is an example - a total of 3 points.
For a complete solution of the problem, it remains to solve the problem of looping and crossing paths of dominoes, and there is a division into regions by shifting the grid and constructing a graph with even degrees of vertices, and there is an example - a total of 5 points.
An attempt to close the problem of looping by the fact that in this case an area of odd area is obtained, WITHOUT a complete and correct proof -0 points.
The problem is solved, but there are minor errors in the proof of the intersection of paths -6 points.
Criteria for a solution with a division into regions, each of which contains a good (lying in a $2 \times 2$ square) domino, are as follows.
There is a division into regions, in each of which there is indeed at least one good domino (but this is not proved) -1 point.
There is a division into regions and a proof that the problem is solved if we prove that in each region there is at least one good domino; but this statement is not proved or contains significant errors - a total of 3 points.
There is a division into regions and a proof that the problem is solved if we prove that in each region there is at least one good domino, but this statement is proved with minor errors - 5 or 6 points.
Attempts to count the number of boundaries of $2 \times 2$ squares (internal or external), not leading to a correct solution - 0 points.
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the smallest 10-digit number, the sum of whose digits is not less than that of any smaller number.
|
Answer: 1899999999.
Solution. Among 9-digit numbers, the largest sum of digits is for the number 999999999, which is 81. Since the desired 10-digit number is greater than 999999999, we need to find the smallest number with a sum of digits no less than 81. The sum of the last eight digits of the desired number is no more than $9 \cdot 8=72$, so the sum of the first two digits is no less than $81-72=9$. Since the desired number is the smallest, its first digit is 1, and the second digit is 8. Then the sum of the remaining eight digits is no less than 72, which means the remaining digits are nines.
Criteria. Correct answer without explanation - 1 point. Proven that the sum of the digits of the desired number is no less than $81-3$ points. Complete solution - 7 points.
|
1899999999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In a chess tournament, everyone played against each other once. The winner won half of the games and drew the other half. It turned out that he scored 13 times fewer points than all the others. (1 point for a win, 0.5 for a draw, 0 for a loss.) How many chess players were there in the tournament?
|
Answer: 21 chess players.
Solution. If the number of participants is $n$, then each played $n-1$ games. The winner won half of the games and scored $\frac{1}{2}(n-1)$ points. The winner drew half of the games and scored another $\frac{1}{4}(n-1)$ points. In total, the winner scored $\frac{1}{2}(n-1)+\frac{1}{4}(n-1)=\frac{3}{4}(n-1)$ points.
In the tournament, $\frac{1}{2} n(n-1)$ games were played. In each game, the players played for one point, so they scored a total of $\frac{1}{2} n(n-1)$ points. Therefore,
$$
13 \cdot \frac{3(n-1)}{4}=\frac{n(n-1)}{2}-\frac{3(n-1)}{4}
$$
from which $n=21$. Thus, in a tournament of 21 chess players, the winner played 20 games and scored $10+5=15$ points, and the rest scored $\frac{1}{2} \cdot 21(21-1)-15=195$ points, which is 13 times more than the winner.
Criteria. Only the answer - 0 points. Proven that for $n=21$ the tournament meets the condition, but not justified that there are no other solutions - 2 points. The equation for the number of participants is correctly set up - 5 points. Complete solution - 7 points.
|
21
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The length of the diagonal of a rectangular parallelepiped is 3. What is the maximum possible value of the surface area of such a parallelepiped?
|
Answer: 18.
Solution. Let the edges of the rectangular parallelepiped be $-a, b$, and $c$. According to the problem, its diagonal is $\sqrt{a^{2}+b^{2}+c^{2}}=3$, and thus, $a^{2}+b^{2}+c^{2}=9$. The surface area of the parallelepiped is $f=2(a b+b c+c a)$. We will prove the inequality $a b+b c+c a \leqslant a^{2}+b^{2}+c^{2}$. Indeed, multiply both sides by 2 and write it in an equivalent form:
$$
2 a b+2 b c+2 c a \leqslant 2 a^{2}+2 b^{2}+2 c^{2} \quad \Longleftrightarrow \quad(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \leqslant 0
$$
From the proven inequality, it follows that the maximum value of the function $f$ does not exceed $2 \cdot 9=18$, and the equality sign is possible only for a rectangular parallelepiped where all edges are equal, $a=b=c=\sqrt{3}$, i.e., for a cube with edge $\sqrt{3}$.
Criteria. Correct answer without explanation - 0 points. Correctly formulated function $f-1$ point. Noted (without justification) that the maximum value is achieved for a cube and correctly calculated the length of its edge - another 1 point. Proof of the estimate without specifying an example - 5 points. Complete solution - 7 points.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. All values of the quadratic trinomial $f(x)=a x^{2}+b x+c$ on the interval $[0 ; 2]$ do not exceed 1 in absolute value. What is the greatest value that the quantity $|a|+|b|+|c|$ can have under these conditions? For which function $f(x)$ is this value achieved?
|
Answer: the maximum value is 7; for example, it is achieved for $f(x)=2 x^{2}-4 x+1$.
Solution. By the condition, the values of $f(x)=a x^{2}+b x+c$ on the interval $[0 ; 2]$ do not exceed one in absolute value. In particular, $|f(0)| \leqslant 1,|f(1)| \leqslant 1,|f(2)| \leqslant 1$, which is equivalent to the system
$$
\left\{\begin{array}{r}
|c| \leqslant 1 \\
|a+b+c| \leqslant 1 \\
|4 a+2 b+c| \leqslant 1
\end{array}\right.
$$
Since the absolute value of the sum of numbers does not exceed the sum of their absolute values, we obtain the inequalities
$$
\begin{aligned}
& |2 a|=|(4 a+2 b+c)-2(a+b+c)+c| \leqslant|4 a+2 b+c|+2|(a+b+c)|+|c| \leqslant 4 \\
& |2 b|=|4(a+b+c)-(4 a+2 b+c)-3 c| \leqslant 4|a+b+c|+|4 a+2 b+c|+3|c| \leqslant 8
\end{aligned}
$$
from which $|a| \leqslant 2,|b| \leqslant 4$, and thus, $|a|+|b|+|c| \leqslant 7$.
For the function $f(x)=2 x^{2}-4 x+1$, the quantity $|a|+|b|+|c|$ is exactly 7. Moreover, the quadratic trinomial $f(x)=2 x^{2}-4 x+1=2(x-1)^{2}-1$ satisfies the condition of the problem, since for $x \in[0 ; 2]$
$$
-1=f(1) \leqslant f(x) \leqslant f(0)=f(2)=1
$$
Thus, the maximum value of the quantity $|a|+|b|+|c|$ under the given conditions is 7.
Criteria. Correct answer -0 points. The inequality $|c| \leqslant 1$ is proven -1 point. Only one of the inequalities $|a| \leqslant 2$ or $|b| \leqslant 4$ is proven - another 2 points. It is proven that the sum of the absolute values does not exceed $7,-5$ points (not cumulative with previous points). Example of a function without justification of the estimate 2 points. Full solution -7 points.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. It is known that the quadratic equations in $x$, $2017 x^{2} + p x + q = 0$ and $u p x^{2} + q x + 2017 = 0$ (where $p$ and $q$ are given real numbers) have one common root. Find all possible values of this common root and prove that there are no others.
|
Solution: Let $x_{0}$ be the common root of these equations, that is, the equalities $2017 x_{0}^{2} + p x_{0} + q = 0$ and $p x_{0}^{2} + q x_{0} + 2017 = 0$ are satisfied. Multiply both sides of the first equation by the number $x_{0}$ and subtract the second equation. We get the consequential equation: $2017 x_{0}^{3} - 2017 = 0$, from which $x_{0} = 1$. Substituting this root into any of the original equations, we get the equality $p + q + 2017 = 0$. This means that when $p + q = -2017$, both equations indeed have a common root equal to 1, and if this sum equals another number, there are simply no common roots.
Answer: 1.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| Correctly written roots of both quadratic equations and at least one case of coinciding roots is correctly investigated | 4 points |
| Noted that if $1$ is a root, then this root is common, but not proven that there can be no other common roots | 1 point |
| Correct answer without justification and/or calculations from which the solution process is not visible | 0 points |
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.2. On a plane, a semicircle with diameter $A B=36$ cm was constructed; inside it, a semicircle with diameter $O B=18$ cm was constructed ($O-$ the center of the larger semicircle). Then, a circle was constructed that touches both semicircles and the segment АО. Find the radius of this circle. Justify your answer.
|
Solution: Let the desired radius be $x$. Let points $C$ and $Q$ be the centers of the smaller semicircle and the inscribed circle, respectively, and $D$ be the point of tangency of the inscribed circle with the diameter $AB$. Due to the tangency of the circle and the larger semicircle, the ray $OQ$ passes through the point of tangency, so $OQ = 18 - x$. Due to the tangency of the circle and the smaller semicircle, the segment $CQ$ passes through the point of tangency, so $CQ = 9 + x$. Due to the tangency of the circle and the segment $AB$, the segment $QD$ equals $x$ and is the height of triangle $QOC$. In triangle $QOC$, all three sides and the height are expressed through $x$. We will find the area of the triangle in two ways. The semiperimeter of the triangle is $\frac{9 + (18 - x) + (9 + x)}{2} = 18$ and by Heron's formula, its area is $\sqrt{18 \cdot (18 - 9) \cdot (18 - (18 - x)) \cdot (18 - (9 + x))} = \sqrt{18 \cdot 9 \cdot x \cdot (9 - x)}$. This area is also equal to half the product of the height and the side to which the height is drawn, that is, it is equal to $\frac{9 \cdot x}{2}$. From the equation $\sqrt{18 \cdot 9 \cdot x \cdot (9 - x)} = \frac{9 \cdot x}{2}$, we find $x = 8$.
Answer: 8 cm
Recommendations for checking:
| present in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| The correct approach to the solution resulted in an incorrect answer due to errors in calculations (algebraic or arithmetic) | 6 points |
| The elements of triangle $QOC$ are correctly written (in terms of the unknown), fully defining the triangle, but the solution is not completed to the answer | 5 points |
| Triangle $QOC$ is considered, but not all elements defining it are found (expressed) | 3 points |
| The correct application of the coordinate method, which is not completed to the answer (depending on the degree of progress) | 1 - 3 points |
| Correct answer without justification and/or any calculations from which the solution process is not clear | 0 points |
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. We will call a natural number semi-prime if it is greater than 25 and is the sum of two distinct prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime? Justify your answer.
|
Solution: There are many sets of five consecutive semiprime numbers, for example $30(=13+17), 31(=2+29), 32(=3+29), 33(=2+31), 34$ $(=5+29)$ or $102(=5+97), 103(=2+101), 104(=31+73), 105(=2+103)$, $106(=47+59)$. We will show that there are no six consecutive semiprime numbers.
Indeed, among any six consecutive numbers, exactly three are odd, and these odd numbers increase by a step of 2. If an odd number is represented as the sum of two natural numbers, then one of the addends is even, and the other is odd. If both addends are prime, then the even one must be 2. Therefore, these odd semiprime numbers have the form $2+p_{1}, 2+p_{2}$, and $2+p_{3}$. In this case, the numbers $p_{1}, p_{2}$, $p_{3}$ are consecutive odd numbers. Among any three consecutive odd numbers, one is divisible by 3. Since it is prime, it must be 3. Then one of the numbers is 5, and it is not semiprime because it is less than 25. Contradiction.
Answer: 5 numbers.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| If the proof that there are no more than 5 numbers is correct, but the example of 5 consecutive semiprime numbers is missing | 4 points |
| Only an example (or examples) of 5 consecutive semiprime numbers is provided | 2 points |
| Correct answer without justification and/or examples of consecutive semiprime numbers in a quantity less than 5, and/or a more crude estimate of their quantity. | 0 points |
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.6. Ali-baba came to a cave where there was gold and diamonds. Ali-baba had one large bag with him. It is known that a full bag of gold weighs 200 kg, and if the entire bag is filled with diamonds, it will weigh 40 kg (the empty bag weighs nothing). A kilogram of gold costs 20 dinars, and a kilogram of diamonds costs 60 dinars. What is the maximum amount of money Ali-baba can get if he can carry no more than 100 kg in this bag? Justify your answer.
|
# Solution:
Method 1. Let Ali-Baba have placed a set of gold and diamonds that gives him the maximum possible revenue - we will call such a bag optimal. Then either the bag is filled to the brim, or it weighs exactly 100 kg: otherwise, more diamonds could be added to it and the revenue would increase. Consider the first situation, when the bag is full. It is clear that it contains diamonds, since the full weight of a bag of gold is more than 100 kg. Note that a bag filled with only gold is worth $200 \cdot 20=40000$ dinars, and one with only diamonds is $-40 \cdot 60=24000$ dinars, so an equal volume of gold yields more profit than the same volume of diamonds. If the weight of the optimal bag is less than 100 kg, we can replace some of the diamonds with an equal volume of gold (so that the weight of the bag increases but does not exceed 100 kg). In this case, the value of the bag will increase - a contradiction to the optimality of the bag. Therefore, the weight of the optimal bag is exactly 100 kg. Suppose the optimal bag is not full. Then it contains gold, and Ali-Baba can replace part of it with diamonds of the same weight (while ensuring that the volume of the bag does not become too large). Since a kilogram of diamonds is more expensive than a kilogram of gold, Ali-Baba's profit will increase - again a contradiction to the optimality.
Thus, the bag is full. Let there be $x$ kg of diamonds in the bag. Then there is $100-x$ kg of gold in it. The diamonds occupy $x / 40$ of the bag, and the gold - $(100-x) / 200$. Since the bag is full, we have the equation $\frac{x}{40}+\frac{100-x}{200}=1$, from which $x=25$. The value of the optimal bag is now trivial to find. It turns out to be 3000 dinars.
Method 2. Let Ali-Baba put $x$ kg of gold and $y$ kg of diamonds in the bag. Then he will earn $20 x+60 y$ dinars. Ali-Baba has two constraints: first, he cannot carry more than 100 kg, so $x+y \leqslant 100$. Second, such an amount of gold and diamonds must fit in the bag. Since one kilogram of gold occupies $1 / 200$ of the bag, and one kilogram of diamonds $-1 / 40$, another inequality arises: $\frac{x}{200}+\frac{y}{40} \leqslant 1$, or $x+5 y \leqslant 200$. In addition, it is clear that the variables $x$ and $y$ are non-negative. We have the problem: find the maximum value of the expression $20 x+60 y$ on the set of all pairs of non-negative numbers $(x ; y)$ that satisfy the system of inequalities $\left\{\begin{aligned} x+y & \leqslant 100 \\ x+5 y & \leqslant 200\end{aligned}\right.$. This is a classic problem in mathematics with many solution methods. We will solve it graphically. The set of points on the plane, the coordinates of which satisfy our conditions, is the shaded quadrilateral shown in the figure.
The equation $20 x+60 y=C$ for different $C$ defines a family of parallel lines; the higher the line on the plane - the greater the value of $C$. (One such line is shown in the figure for $C=1200$.) The highest line parallel to it passes through the vertex $A$ of the quadrilateral shown. The coordinates of the vertex are found from the system $\left\{\begin{aligned} x+y & =100 \\ x+5 y & =200\end{aligned}\right.$; they are $x=75, y=25$. Thus, the maximum possible value of $C$ is 3000.
Answer: 3000 dinars.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| The correct solution path leads to an incorrect answer due to arithmetic errors | 6 points |
| Both facts (see criteria for 2 points) are proven | 4 points |
| The problem is correctly reduced to the analysis of a system or there is a correct graphical interpretation of the problem, but the solution is not completed | 3 points |
| One of the two facts is proven: 1) the optimal bag weighs exactly 100 kg; 2) the optimal bag is full; | 2 points |
| The correct answer is obtained under the assumption that the optimal bag is full and weighs 100 kg, but neither point of the assumption is proven | 1 point |
| Correct answer without justification | |
|
3000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Inside a convex pentagon, a point is marked and connected to all vertices. What is the maximum number of the ten segments drawn (five sides and five segments connecting the marked point to the vertices of the pentagon) that can have a length of 1? (A. Kuznetsov)
|
Answer: 9 segments.
Solution. First, we prove that all 10 segments cannot have a length of 1. Assume the opposite. Let $A B C D E$ be a pentagon, $O$ be a point inside it, and all 10 drawn segments have a length of 1 (see Fig. 6). Then triangles $O A B, O B C$, $O C D, O D E$, and $O E A$ are equilateral, so $\angle A O B = \angle B O C = \angle C O D = \angle D O E = \angle A O E = 60^{\circ}$. The sum of these angles should be $360^{\circ}$, but $5 \cdot 60^{\circ} = 300^{\circ}$, which is a contradiction.
Fig. 6
Fig. 7
It remains to provide an example where 9 segments have a length of 1 (see Fig. 7). Mark points $A$ and $O$ on the plane at a distance of 1. Choose points $B, C, D$, and $E$ sequentially so that triangles $A O B, B O C, C O D$, and $D O E$ are equilateral. Then point $O$ lies inside the pentagon $A B C D E$, and out of the 10 drawn segments, all except $A E$ have a length of 1.
Remark. In the given example, points $A, B, C, D$, and $E$ are five vertices of a regular hexagon with side length 1.
Comment. Only the proof that there are no more than 9 segments of unit length - 3 points.
Only the example showing that 9 segments of unit length can be found - 3 points.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. Let's call a trapezoid with bases 1 and 3 a "boat," which is obtained by gluing two triangular half-cells to the opposite sides of a unit square. In a $100 \times 100$ square, an invisible boat is placed (it can be rotated, it does not go beyond the boundaries of the square, its middle cell is entirely on one of the cells of the square). With one shot, you can cover any triangular half-cell. If the shot intersects the interior of the boat (i.e., the intersection of the triangular shot with the boat has a non-zero area), the boat is considered sunk. What is the minimum number of shots sufficient to definitely sink the boat?
(S. Berlov, N. Vlasova)
|
Answer: 4000 shots.
First solution. We will call a boat horizontal or vertical depending on whether its parallel sides are horizontal or vertical.
First, we will show that 4000 shots are enough. We divide the $100 \times 100$ square into 400 squares of size $5 \times 5$, and in each square, we make 10 shots as shown in Fig. 9. It is not hard to see that in each row and each column, a boat cannot be placed between adjacent shots; therefore, one of the shots will definitely sink the boat.
It remains to show that it is impossible to guarantee sinking the boat with fewer than 4000 shots. This time, we divide the board into 2000 horizontal rectangles of size $1 \times 5$ and show that at least two shots must be made in each such rectangle. Indeed, at least one shot must be made in the left three cells of the rectangle, otherwise an unsunk boat could be placed there; the same is true for the right three cells. Thus, if only one shot was made in the rectangle, it could only be in the central cell of the rectangle. Without loss of generality, this shot was made in the lower left triangle of this cell; but then a boat placed as in Fig. 10 would not be sunk.
Fig. 9
Fig. 10
Second solution. We will provide another proof that at least 4000 shots are required. We divide the square into 100 horizontal strips of size $1 \times 100$ and show that even to guarantee the sinking of a horizontal boat, at least 40 shots are required in each strip.
Let's count the number of different ways to place a horizontal boat in a strip. The central cell of such a boat can be in any cell of the strip except the edge ones. For each of these 98 cells, there are two possible placements of the horizontal boat with this central cell. Therefore, the total number of ways is $98 \times 2 = 196$.
On the other hand, a shot in any triangle in the strip can sink at most five of these possible boats. Indeed, if this triangle, say, is the upper left one in its cell $c$, then it will sink any of the two boats with the central cell $c$, any of the two boats with the central cell immediately to the left of $c$, and one of the two boats with the central cell immediately to the right of $c$. (Note that some of these boats may extend beyond the edges of the board.)
Thus, if fewer than 40 shots are made in the strip, they can sink at most $39 \times 5 = 195$ possible placements of the boat, meaning there will be a placement that will not be sunk. Therefore, at least 40 shots must be made in each strip, totaling at least $40 \times 100 = 4000$ shots.
Remark. A similar argument can be applied to any rectangle of size $1 \times 5$: there are 6 ways to place the boat in it, so at least two shots must be made in it.
Comment. An example is given showing that the boat can be guaranteed to be sunk with 4000 shots - 2 points.
It is only proven that it is impossible to guarantee sinking the boat with fewer than 4000 shots - 5 points.
|
4000
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.1. Masha lives in apartment No. 290, which is located in the 4th entrance of a 17-story building. On which floor does Masha live? (The number of apartments is the same in all entrances of the building on all 17 floors; apartment numbers start from 1.$)$
|
# Answer: 7.
Solution. Let $x$ be the number of apartments per floor, then there are $17 x$ apartments in each entrance. Thus, in the first three entrances, there are $51 x$ apartments, and in the first four $68 x$.
If $x \geqslant 6$, then in the first three entrances there are at least 306 apartments, so apartment No. 290 cannot be located in the fourth entrance. And if $x \leqslant 4$, then in the first four entrances there will be no more than 272 apartments, which again means that apartment No. 290 cannot be located in the fourth entrance. The only remaining option is when $x=5$.
Then in the first three entrances, there are 255 apartments, and apartment No. 290 is the 35th in the fourth entrance, i.e., it is located on the 7th floor.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 11.2. On the table, there are 30 coins: 23 ten-ruble coins and 7 five-ruble coins, with 20 of these coins lying heads up and the remaining 10 tails up. What is the smallest $k$ such that among any $k$ randomly selected coins, there will definitely be a ten-ruble coin lying heads up?
|
Answer: 18.
Solution. If you choose 18 coins, then among them there will be no more than 10 lying heads down, so at least 8 coins will be lying heads up. Among these coins, no more than 7 will be five-ruble coins, so at least one will be a ten-ruble coin, which is what we need.
On the other hand, if initially on the table there are 7 five-ruble coins heads up, 10 ten-ruble coins heads down, and 13 ten-ruble coins heads up, then among 17 coins there might be only coins of the first two types, so 17 coins (or fewer) might not be enough.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.3. The product of positive numbers $a$ and $b$ is 1. It is known that
$$
(3 a+2 b)(3 b+2 a)=295
$$
Find $a+b$.
|
Answer: 7.
Solution. Expanding the brackets, we get
$$
295=6 a^{2}+6 b^{2}+13 a b=6\left(a^{2}+b^{2}\right)+13
$$
from which $a^{2}+b^{2}=47$. Then
$$
(a+b)^{2}=a^{2}+b^{2}+2 a b=47+2=49=7^{2}
$$
which gives $a+b=7$ (note that $a+b>0$, since $a>0$ and $b>0$).
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.4. A convex pentagon $A B C D E$ is such that $\angle A B C=128^{\circ}, \angle C D E=9^{\circ}$ $104^{\circ}, A B=B C, A E=E D$. How many degrees does the angle $A D B$ measure?
|
Answer: 26.
Fig. 14: to the solution of problem 11.4
Solution. The angles at the base of the isosceles triangle $ABC$ are equal to $\frac{1}{2}\left(180^{\circ}-128^{\circ}\right)=26^{\circ}$, and in the isosceles triangle $AED$ they are equal to $\frac{1}{2}\left(180^{\circ}-104^{\circ}\right)=38^{\circ}$.
Notice that $\angle ADC=90^{\circ}-38^{\circ}=52^{\circ}$, and the quadrilateral $ABCD$ is cyclic (Fig. 14), since $\angle ABC + \angle ADC = 128^{\circ} + 52^{\circ} = 180^{\circ}$. Therefore, $\angle ADB = \angle ACB = 26^{\circ}$.
|
26
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.5. For what least natural $n$ can the numbers from 1 to $n$ be arranged in a circle so that each number is either greater than all 40 following it in the clockwise direction, or less than all 30 following it in the clockwise direction?
|
Answer: 70.
Solution. If $n \leqslant 39$, then for the number $n$ the condition cannot be satisfied: it cannot be greater than the next 40 numbers (since it is not greater than itself), nor can it be less than the next 30 numbers (since it is the largest).
If $40 \leqslant n \leqslant 69$, then for the number 40 the condition cannot be satisfied: there are neither 40 numbers less than it, nor 30 numbers greater than it.
However, if $n=70$, the numbers can be arranged. For this, they can be placed in the following order clockwise: $1,2,3, \ldots, 40,70,69,68, \ldots, 41$ (Fig. 15). Then the numbers from 1 to 40 will be less than the next 30 following them, and the numbers from 70 to 41 will be greater than the next 40 following them.
Fig. 15: to the solution of problem 11.5
|
70
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.6. The polynomial $P(x)$ has all coefficients as non-negative integers. It is known that $P(1)=4$ and $P(5)=152$. What is $P(11) ?$
|
Answer: 1454.
Solution. Suppose the degree of the polynomial $P$ is not less than 4, then its leading coefficient is not less than 1. Since all other coefficients of $P(x)$ are non-negative, then $P(5) \geqslant 5^{4}=625$, which contradicts the condition $P(5)=152$.
Therefore, the degree of the polynomial $P$ is not more than 3, and $P(x)=a x^{3}+b x^{2}+c x+d$ for some non-negative integers $a, b, c, d$. Then
$$
a+b+c+d=P(1)=4, \quad 125 a+25 b+5 c+d=P(5)=152
$$
Suppose $a \geqslant 2$. Then $P(5) \geqslant 125 \cdot 2=250>152$, a contradiction.
Suppose $a=0$. Then $P(5)=25 b+5 c+d \leqslant 25(b+c+d)=25 \cdot 427$, a contradiction.
Suppose $b=0$. Then $27=5 c+d \leqslant 5(c+d)=5 \cdot 3<27$, a contradiction.
Thus, $b=1$. Then
$$
c+d=2, \quad 5 c+d=2
$$
From this, it easily follows that $c=0$ and $d=2$, i.e., $P(x)=x^{3}+x^{2}+2$. Therefore, $P(11)=11^{3}+11^{2}+2=1454$.
Another solution. One can notice that $P(1)=4$ is in any case the sum of the coefficients of the polynomial $P(x)=a_{n} x^{n}+\ldots+a_{1} x+a_{0}$, from which each of them does not exceed 4. Then these coefficients can be considered as digits in the base-5 number system, and the value of the polynomial at the point 5 is the number written by these digits:
$$
152=P(5)=5^{n} a_{n}+\ldots+5 a_{1}+a_{0}
$$
Since $152_{10}=1102_{5}$, and the representations of numbers in the base-5 system are unique, then $P(x)=1 x^{3}+1 x^{2}+0 x+2$.
|
1454
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.8. In a $28 \times 35$ table, some $k$ cells are painted red, another $r-$ in pink, and the remaining $s-$ in blue. It is known that
- $k \geqslant r \geqslant s$
- each boundary cell has at least 2 neighbors of the same color;
- each non-boundary cell has at least 3 neighbors of the same color.
What is the smallest value that the quantity $k-s$ can take?
(A cell is called boundary if it is adjacent to the edge of the table. Neighbors are cells that share a common side.)
|
Answer: 28.
Solution. From the condition, it is easy to understand that each cell can have no more than one neighbor of a different color.
We will prove that the coloring of the table must be "striped," meaning that either each row or each column is completely colored in one color. To do this, it is sufficient to show that either all pairs of adjacent cells of different colors are horizontal neighbors, or all such pairs are vertical neighbors.
Fig. 17: Solution to problem 11.8
Consider any pair of adjacent cells of different colors - if there are cells of different colors in the table, such a pair will exist. The other neighbors of these cells match their colors, so the boundary separating the colors will continue in both directions (Fig. 17), and further, until it reaches the edges of the table.
This means that any boundary between different colors must run from edge to edge of the table, and on each side of it will be a strip of cells of the same color. But this means that vertical and horizontal boundaries cannot coexist simultaneously. Therefore, either all boundaries are horizontal, or all boundaries are vertical, and the coloring is "striped" in any case. (The width of each strip should be at least 2 cells - however, this does not matter for the solution.)
This means that either the number of cells of each color is divisible by the height of the table, or by its width, as is the difference $k-s$. On the other hand, this difference cannot be equal to 0, because in this case $k=r=s$ and the total number of cells is $3k$, but the number $28 \cdot 35$ is not divisible by 3. Therefore, $k-s \geqslant 28$ or $k-s \geqslant 35$.
The equality $k-s=28$ is possible if the red cells occupy 12 columns (28 cells each), the pink cells occupy the same number, and the blue cells occupy 11 columns. If the columns of one color are placed together, all the conditions of the problem will be satisfied.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. At a round table, 30 people are sitting, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each person was asked: "How many knights are among your neighbors?" (Two people are called neighbors of each other if there is no one else sitting between them.) 10 people answered "one," 10 - "two," and 10 - "none." What is the maximum number of knights that can be at the table?
|
Answer: 22 knights.
Solution. From the condition of the problem, it follows that the number of knights who have two knight-neighbors does not exceed 10; the same can be said about the number of knights who have one knight-neighbor. Therefore, the number of pairs of knight-neighbors does not exceed $(2 \cdot 10 + 1 \cdot 10) : 2 = 15$. Since there are 30 pairs of neighbors at the table, at least $30 - 15 = 15$ pairs include people who are not knights; the number of such people is no less than $\left[\frac{15}{2}\right] + 1 = 8$.
Let's provide an example showing that there can be exactly $30 - 8 = 22$ knights at the table. Suppose those sitting are numbered from 1 to 30 clockwise, and the knights are all those whose numbers are different from $1, 3, 5, 7, 11, 15, 19,$ and 23. Then all the necessary conditions will be met if No. 1 answers "two," and Nos. $3, 5, 7, 11, 15, 19,$ and 23 answer "none."
Criteria. Only the answer - 0 points. Proven that the number of knights is no more than 22 (estimation) - 3 points. Provided an example with 22 knights - 3 points. Complete solution - 7 points.
|
22
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In triangle $A B C$, angle $A$ is $75^{\circ}$, and angle $C$ is $60^{\circ}$. On the extension of side $A C$ beyond point $C$, segment $C D$ is laid off, equal to half of side $A C$. Find angle $B D C$.
|
Answer: $\angle B D C=45^{\circ}$.
Solution. In this triangle, angle $A B C$ is obviously equal to $180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ}$. Draw the altitude $A H$ and note that triangle $A H B$ is isosceles, since $\angle B A H=90^{\circ}-45^{\circ}=\angle A B H$ (Fig. 1). Therefore, $A H=B H$. In the right triangle $A H C$, angle $H A C$ is equal to $75^{\circ}-45^{\circ}=30^{\circ}$, so the leg $H C$ is half the hypotenuse $A C$, which means it is equal to $C D$. Thus, triangle $H C D$ is isosceles, and $\angle H C D=120^{\circ}$, so angles $C H D$ and $C D H$ are equal to $\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=30^{\circ}$. Since $\angle H A D=\angle A D H=30^{\circ}$, triangle $A H D$ is also isosceles and $B H=A H=D H$, meaning triangle $B H D$ is isosceles as well. By the property of the exterior angle, $\angle H B D=\angle H D B=\frac{1}{2} \angle C H D=15^{\circ}$. Therefore, $\angle B D C=\angle B D H+\angle H D C=45^{\circ}$.
Fig. 1
Criteria. Only the answer - 0 points. Isosceles property of triangle $A H B$ - 1 point, triangle $A H D$ - 4 points, triangle $B H D$ - 5 points. Full solution - 7 points.
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 10.5
The field is a $41 \times 41$ grid, in one of the cells of which a tank is hidden. The fighter aircraft shoots one cell per shot. If a hit occurs, the tank moves to an adjacent cell along a side; if not, it remains in place. After each shot, the pilot does not know whether a hit occurred. To destroy the tank, it must be hit twice. What is the minimum number of shots required to guarantee that the tank is destroyed?
## Points 7
#
|
# Answer:
2521 shots
## Solution
## Example.
Let's color the cells in a checkerboard pattern so that the corner cells are black. Suppose the pilot first shoots at all the white cells, then at all the black ones, and then again at all the white ones. If the tank was on a white cell, the pilot will destroy it in the first and second series; if it was on a black cell, then in the second and third series. In this case, the pilot will make no more than $41^{2}+$ $1 / 2\left(41^{2}-1\right)=1 / 2\left(3 \cdot 41^{2}-1\right)=2521$ shots.
## Estimation.
Suppose the pilot has a sequence of shots after which the tank will be guaranteed to be destroyed. Clearly, he must shoot at least once at any cell (otherwise, the tank in that cell would not be destroyed). Assume there are two adjacent cells $A$ and $B$, at which he shot exactly once, and the shot at $B$ occurred later. Then, if the tank was initially in $B$, it could have moved to $A$ after the shot at $B$, and the second hit would not have occurred. Therefore, such pairs of cells do not exist. We can divide the entire board into $1 / 2\left(41^{2}-1\right)$ dominoes $1 \times 2$ and one cell. The pilot must make at least three shots into each domino and at least one shot into the remaining cell. In total, he made no fewer than $1 / 23 \cdot\left(41^{2}-1\right)+1=1 / 2\left(3 \cdot 41^{2}-1\right)$ shots.
|
2521
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Real numbers $a$ and $b$ are such that $a^{5}+b^{5}=3, a^{15}+b^{15}=9$. Find the value of the expression $a^{10}+b^{10}$.
|
Answer: 5.
Solution. Let $x=a^{5}, y=b^{5}$. Then $x+y=3, x^{3}+y^{3}=9 ;(x+y)\left(x^{2}-x y+y^{2}\right)=9$; $3\left(x^{2}-x y+y^{2}\right)=9 ; x^{2}-x y+y^{2}=3 ; x^{2}+2 x y+y^{2}=9 ; 3 x y=6 ; x y=2 ; x^{2}+y^{2}=9-2 x y=9-$ $2 \cdot 2=5 ; a^{10}+b^{10}=x^{2}+y^{2}=5$
Criteria. Simplification using substitution was done, but the solution is not complete: 1 point.
The value $x y=2$ was found, but the solution is not complete: 3 points.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a company of 8 people, each person is acquainted with exactly 6 others. In how many ways can four people be chosen such that any two of them are acquainted? (We assume that if A is acquainted with B, then B is also acquainted with A, and that a person is not acquainted with themselves, as the concept of acquaintance applies to two different people.)
|
Answer: 16.
Solution. Each person in the company does not know $8-1-6=1$ person in this company. This means the company can be divided into four pairs of strangers. From each pair, we can only take one person to form a company of four people who are all acquainted with each other. The selection of such a company consists of four steps, each with two choices (from each pair), and by the multiplication rule, the company can be chosen in $2^{4}=16$ ways.
Criteria. It is shown that each person is not acquainted with exactly one of the others and only: 1 point.
It is indicated that exactly one person must be chosen from each pair of strangers, but the calculations are incorrect: 3 points.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. In each cell of a $6 \times 6$ table, numbers are written. All the numbers in the top row and all the numbers in the left column are the same. Each of the other numbers in the table is equal to the sum of the numbers written in the two adjacent cells - the cell to the left and the cell above. What number can be written in the top-left corner cell if the number in the bottom-right corner cell is 2016?
|
Answer: 8. Let $a$ be the number written in the corner cell at the top left. Then the table is filled in sequentially. For the bottom right corner, we get the relation $252 a = 2016$, from which it follows that $a = 8$.
Comment. An answer with a chain of calculations - 7 points. An answer obtained using combinatorial considerations - Pascal's triangle - 7 points.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4. A set of composite numbers from the set $\{1,2,3,4, \ldots, 2016\}$ is called good if any two numbers in this set do not have common divisors (other than 1). What is the maximum number of numbers that a good set can have?
|
Answer: 14 numbers.
Example. $2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}, 19^{2}, 23^{2}, 29^{2}, 31^{2}, 37^{2}, 41^{2}, 43^{2}$. All the presented numbers are composite, as they are squares of natural numbers greater than 1. It is also clear that different prime numbers do not have common divisors greater than 1.
Estimate. Suppose a set of composite numbers $a_{1}, a_{2}, \ldots, a_{n}$ satisfies the condition of the problem and $a_{1}<a_{2}<\ldots<a_{n}$. Note then that the inequality $p_{n}^{2} \leq a_{n}$ holds, where $p_{n}$ is the $n$-th prime number in ascending order. Let $q_{1}, q_{2}, \ldots, q_{n}$ be the smallest prime numbers in the factorization of the composite numbers $a_{1}, a_{2}, \ldots, a_{n}$. If for some index $k, 1 \leq k<n$, the inequality $p_{n}<q_{k}$ holds, then $p_{n}^{2}<q_{k}^{2} \leq a_{k}<a_{n}$. If there is no such index, then the prime $q_{n}$ is greater than some $n-1$ prime number. This means that $p_{n}^{2} \leq q_{n}^{2} \leq a_{n}$. Therefore, if there is a set of composite numbers $a_{1}, a_{2}, \ldots, a_{15}$ satisfying the condition of the problem $a_{1}<a_{2}<\ldots<a_{15}$, then $2209=47^{2}=p_{15}^{2} \leq a_{15}$. Contradiction.
Comment. Answer - 1 point, example - 2 points, estimate - 3 points. Points are summed for progress.
|
14
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. Write the number 2013 several times in a row so that the resulting number is divisible by 9. Explain your answer.
|
Answer: for example, 201320132013.
Solution. We will provide several ways to justify this.
First method. The sum of the digits of the written number is $(2+0+1+3) \cdot 3=18$, so it is divisible by 9 (by the divisibility rule for 9).
Second method. The sum of the digits of the number 2013 is divisible by 3, so if we write 2013 three times in a row, the sum of the digits of the resulting number will be divisible by $3 \cdot 3=9$, meaning the number itself will be divisible by 9 (by the divisibility rule for 9).
Third method. Dividing the written number by 9, we get: $201320132013: 9=22368903557$.
Fourth method. Dividing the written number by 3, we get: $201320132013: 3=67106710671$. The resulting number is divisible by 3, as the sum of its digits is 42. Therefore, the original number is divisible by 9.
There are other examples: we can write the number 2013 six times in a row, nine times in a row, and so on (any number of times that is a multiple of three). The justifications are similar to those provided above.
# Grading criteria:
+ one or more correct answers are provided and explained correctly
$\pm$ along with the correct answer, an incorrect one is also provided, but the correct answer is explained
干 the correct answer is provided without explanation or with an incorrect explanation
- only an incorrect answer is provided
|
201320132013
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. In the sum $+1+3+9+27+81+243+729$, any addends can be crossed out and the signs before some of the remaining numbers can be changed from “+” to “-”. Masha wants to use this method to first obtain an expression whose value is 1, then, starting over, obtain an expression whose value is 2, then (starting over again) obtain 3, and so on. Up to what largest integer will she be able to do this without skipping any?
|
Answer: up to the number 1093 (inclusive).
Solution. The number 1 is obtained by erasing all addends except the first. Then Masha can get the numbers $2=-1+3, 3=+3$, and $4=+1+3$.
We will show that by adding the addend 9, one can obtain any integer from $5=9-4$ to $13=9+4$. Indeed, $5=-1-3+9; 6=-3+9; 7=+1-3+9; 8=-1+9; 9=+9; 10=+1+9; 11=-1+3+9; 12=+3+9; 13=+1+3+9$.
By adding 27 and acting similarly, one can obtain any integer from $14=-1-3-9+27$ to $40=+1+3+9+27$, then get all numbers from $41=-1-3-9-27+81$ to $121=1+3+9+27+81$, and so on. Therefore, Masha will gradually be able to obtain all integers from 1 to $1+3+9+27+81+243+729=1093$.
In essence, this solution uses inductive reasoning. Initially, several first numbers are obtained, and then it is shown that, being able to obtain all integers from 1 to $k$ and adding the next addend, equal to $2k+1$, Masha will be able to obtain all integers from $k+1=(2k+1)-k$ to $3k+1=(2k+1)+k$.
Note that the addends in the given sum are consecutive powers of the number 3, so in the ternary system, this sum is written as 11111113. The proposed problem, on a specific example, illustrates the fact that any integer can be written as a sum of powers of three with coefficients 0, 1, and -1 instead of the usual 0, 1, and 2. (It can be proven that such a representation is also unique.)
## Evaluation Criteria:
+ correct answer and solution with full justification provided
$\pm$ correct reasoning with full justification provided, but an arithmetic error was made in the answer
Ғ correct answer provided and the idea of its justification is present, but the justification itself is missing
- only the answer is provided
|
1093
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On January 1, 2013, a little boy was given a bag of chocolate candies, containing 300 candies. Each day, the little boy ate one candy. On Sundays, Karlson would fly over, and the little boy would treat him to a couple of candies. How many candies did Karlson eat? (January 1, 2013, was a Tuesday).
2. Petya can swap any two digits of different parity in the number 1974835. What is the largest number he can obtain this way?
|
1. Answer: 66.
Solution. In a full week, Little One eats 7 candies (one each day), and Karlson eats 2 (on Sunday), so together they eat 9 candies. Dividing 300 by 9 with a remainder, we get the quotient 33 and the remainder 3 (33$\cdot$9+3=300). In the last incomplete week, Little One will eat 3 candies: on Wednesday, Thursday, and Friday. Therefore, in 33 full weeks, Karlson will eat 33$\cdot$2=66 candies.
Grading criteria.
Correct answer without justification: 1 point.
|
66
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A $5 \times 5$ square is to be cut into two types of rectangles: $1 \times 4$ and $1 \times 3$. How many rectangles can result from the cutting? Justify your answer.
|
5. Answer: 7.
Solution. Example:
If there are no more than 6 rectangles, then they occupy no more than $6 \cdot 4 = 24$ cells, while there are 25 cells. Therefore, there must be at least 7 rectangles.
A rectangle that is completely located in a column will be called vertical, and one that is completely located in a row will be called horizontal. Suppose we managed to cut the square into such rectangles. Consider the rectangle that contains the central cell of the $5 \times 5$ square. If necessary, rotate the $5 \times 5$ square so that this rectangle is horizontal. The cells of this rectangle are located in at least three columns, and there are no vertical rectangles in these columns. There are a total of 10 possible rectangles: 5 vertical and 5 horizontal. But 3 vertical rectangles are impossible, so there can be no more than $10 - 3 = 7$ rectangles.
The number of rectangles can only be 7.
## Grading Criteria.
Only the example: 2 points.
Proved that there are at least 7 rectangles: 1 point.
Shown that there are no more than 9 rectangles, as 9 would occupy at least 27 cells: 1 point.
Attempted to enumerate to prove that there cannot be 8 rectangles: 1 point.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. 50 students from fifth to ninth grade published a total of 60 photos on Instagram, each not less than one. All students of the same grade (same parallel) published an equal number of photos, while students of different grades (different parallels) published a different number. How many students published only one photo?
|
Solution. Let each of the students publish one photo first, during which 50 photos out of 60 will be published. There are 10 photos left to be published, which we will call additional photos. We have a total of five different classes (parallels), and these 10 additional photos must be published by different students from different classes, and in different quantities. It remains to note that $1+2+3+4=10$, i.e., one additional photo can be published by only one student from some class, two additional photos by another student from another class, and so on, a total of four students from four different classes. Therefore, 46 students from some class (parallel) published one photo each, and four students from the other four classes published the remaining 14 photos, one published two, one published three, one published four, and one published five.
Answer. 46 students published one photo each.
|
46
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find the value of the expression
$$
2^{2}+4^{2}+6^{2}+\ldots+2018^{2}+2020^{2}-1^{2}-3^{2}-5^{2}-\ldots-2017^{2}-2019^{2}
$$
|
# Solution.
$$
\begin{aligned}
& 2020^{2}-2019^{2}+2018^{2}-2017^{2}+\ldots+6^{2}-5^{2}+4^{2}-3^{2}+2^{2}-1^{2}= \\
& =(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+\ldots+ \\
& +(6-5)(6+5)+(4-3)(4+3)+(2-1)(2+1)= \\
& =2020+2019+2018+2017+\ldots+6+5+4+3+2+1= \\
& =\frac{2020+1}{2} \cdot 2020=2041210
\end{aligned}
$$
Answer. 2041210.
|
2041210
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Given the natural numbers $1,2,3 \ldots, 10,11,12$. Divide them into two groups such that the quotient of the product of all numbers in the first group divided by the product of all numbers in the second group is an integer, and takes the smallest possible value. What is this quotient?
|
Solution. Let's factorize the given numbers into prime factors and find their product.
$$
1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 10 \cdot 11 \cdot 12=2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11
$$
The factors 7 and 11 do not have pairs. One of the factors 3 does not have a pair. Therefore, to make the quotient an integer, these prime factors without pairs must be placed in the numerator. The quotient cannot be less than what these three irreducible factors provide. The remaining threes, fives, and twos should be paired in the numerator and denominator so that the pairs cancel out. One way to do this is written below
$$
\frac{7 \cdot 11 \cdot 1 \cdot 2 \cdot 3 \cdot 8 \cdot 9 \cdot 10}{4 \cdot 5 \cdot 6 \cdot 12}=231
$$
Answer. 231.
|
231
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. For what value of the parameter m is the sum of the squares of the roots of the equation
$$
x^{2}-(m+1) x+m-1=0
$$
the smallest?
|
3. $\mathrm{m}=3$.
Note that the discriminant $\mathrm{D}$ of the equation is equal to
$$
\mathrm{D}=(\mathrm{m}+1)^{2}-4(\mathrm{~m}-1)=(\mathrm{m}-1)^{2}+4>0
$$
therefore, the equation has two roots for any $\mathrm{m}$. Let's represent the sum of the squares of the roots $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ as
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}
$$
From Vieta's theorem, we get
$$
x_{1}^{2}+x_{2}^{2}=\frac{(m+1)^{2}}{4}-2(m-1)
$$
The minimum value of $x_{1}^{2}+x_{2}^{2}$ is achieved at the minimum point of the function $\mathrm{f}(\mathrm{m})=(\mathrm{m}+1)^{2}-8(\mathrm{~m}-1)$, which is $\mathrm{m}_{\min }=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. Arina wrote down a number and the number of the month of her birthday, multiplied them and got 248. In which month was Arina born? Write the number of the month in your answer.
|
Answer: 8
Solution. $248=2 * 2 * 2 * 31$. Obviously, 31 is the number, then 8 is the month number.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. A number was multiplied by its first digit and the result was 494, by its second digit - 988, by its third digit - 1729. Find this number.
|
Answer: 247
Solution. It can be noticed that all three obtained numbers are divisible by 247, from which we get the answer.
|
247
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.1. It is known that in the past chess tournament, in each round all players were paired, the loser was eliminated (there were no draws). It is known that the winner played 6 games. How many participants in the tournament won at least 2 games more than they lost?
|
Answer: 8
Solution. Since the winner played 6 games, there were a total of $2^{6}=64$ players. The win-loss ratio for those eliminated in the 1st round is -1; for those who lost in the 2nd round, it is 0; for those who lost in the 3rd round, it is +1; for those who lost in the 4th round and beyond, it is at least +2. 8 participants reached the $4th$ round.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.1. In a basketball team, there is a certain number of players. The coach added up all their heights and divided by the number of players (let's call this the average height), getting 190 cm. After the first game, the coach removed Nikolai, who is 197 cm tall, and replaced him with Petr, who is 181 cm tall, after which the average height of the players in the team was 188 cm. How many people are in the team?
|
Answer: 8
Solution. The change in the total height of the players by $197-181=16$ cm leads to a change in the average height of the team by $190-188=2$ cm, so the team has $\frac{16}{2}=8$ people.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. On the counter lie 10 weights with masses $n, n+1, \ldots, n+9$. The seller took one of them, after which the total weight of all the remaining ones was 1457. Which weight did the seller take?
|
# Answer: 158
Solution. The total weight of all 10 weights is $10 n+45$. Suppose the weight taken has a weight of $n+x$, where $x<10$. Then
$10 n+45-n-x=1457$.
$9 n+45-x=1457$.
We can see that 1457, when divided by 9, gives a remainder of 8, so $x=1$. Then
$9 n+45-1=1457$
$9 n=1413$.
$n=157$.
$n+x=158$.
|
158
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. In a circle, 58 balls of two colors - red and blue - are arranged. It is known that the number of triples of consecutive balls, among which there are more red ones, is the same as the number of triples with a majority of blue ones. What is the smallest number of red balls that could be present?
|
Answer: 20
Evaluation
There are a total of 58 threes, so the number of threes with dominance of each of the two colors is 29. Therefore, the number of red balls should be no less than $\frac{2}{3} * 29$, which is no less than 20.
Example
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In a $7 \times 7$ table (see figure), the elements of each row and each column form arithmetic progressions. What is the number $x$ in the central cell?
|
Answer: 111.
Solution. The middle term of an arithmetic progression is equal to the half-sum of the extreme terms, so in the middle cell of the first row, the number is $(3+143) / 2=73$, and in the middle cell of the last row, the number is $(82+216) / 2=149$.
The elements of the middle column also form an arithmetic progression, so $x=(73+$ $149) / 2=111$.
Comment. The correct answer obtained by sequentially filling in the table - 7 points. For arithmetic errors with correct reasoning, deduct $2-3$ points.
|
111
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In a school checkers tournament, each participant played against every other participant once. 2 points were awarded for a win, 1 point for a draw, and 0 points for a loss. There were 9 times fewer girls than boys, and the boys together scored 4 times more points than the girls together. How many points did the girls score?
|
Answer: 18 points.
Solution. Let $x$ be the number of girls participating in the tournament. Then there were $10 x$ participants in total, and they scored $10 x(10 x-1)$ points. According to the problem, the ratio of the number of points scored by girls to the number of points scored by boys is 1:4. Therefore, the girls scored one-fifth of the points, which is $2 x(10 x-1)$ points, meaning each girl won all $10 x-1$ games she played. If there were two girls among the participants, they would have had to both win their game against each other, which is impossible. Therefore, only one girl participated in the tournament; she scored 18 points.
Comment. The expression for the total number of points is written - 1 point, the expression for the number of points scored by girls is written - 2 points, the conclusion about the number of wins by girls is made - 2 points, the conclusion about the number of girls and the answer is obtained - 2 points. Points are summed.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find all natural $\boldsymbol{n}$ such that the value of the expression $\frac{2 n^{2}-2}{n^{3}-n}$ is a natural number.
|
Solution. $\frac{2 n^{2}-2}{n^{3}-n}=\frac{2(n-1)(n+1)}{n(n-1)(n+1)}$. For $n \neq \pm 1$, this expression is identically equal to $\frac{2}{n}$. Since $n \neq 1$, the value of the last expression is a natural number only when $n=2$.
Answer: $n=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) The number 20 is written on the board. In one move, it is allowed to either double the number or erase its last digit. Is it possible to get the number 25 in several moves?
|
Answer. Possible.
Solution. The number 25 can be obtained by erasing the last digit of the number 256, which is a power of two. Thus, the required chain of transformations can look like this:
$$
20 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 32 \rightarrow 64 \rightarrow 128 \rightarrow 256 \rightarrow 25
$$
There are other solutions as well.
## Grading Criteria.
- Any complete correct solution -7 points.
- Incomplete solution (for example, it is stated that 25 can be obtained from the number 256, but it is not specified how to get 256) -3 points.
- Only the answer is provided -0 points.
|
25
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
|
Answer: 52.
Solution. All sides of a square are equal, and its perimeter is 24, so each side is $24: 4=6$. The perimeter of the rectangle is 16, and its two largest sides are each 6, so the two smallest sides are each $(16-6 \cdot 2): 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimeter is $2 \cdot(8+18)=$ 52.
|
52
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How much does the triangular weight weigh?
|
Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ grams.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
#
|
# Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. Then the emeralds are in the two remaining boxes, and there are a total of $5+7=12$.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
|
Answer: 22.
Solution. Note that there are a total of 91 coins, so after all moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will be transferred from the box with 20 coins. In the end, these two boxes should have a total of 26 coins, so at least \(25 + 7 - 26 = 6\) coins need to be moved from these boxes. Next, consider the boxes with 17 and 18 coins, from which at least 4 and 5 coins need to be moved, respectively. In total, there should be at least \(7 + 6 + 4 + 5 = 22\) moves.
Let's provide an example of how to equalize all the boxes in exactly 22 moves. From the box with 20 coins, we move 3 coins to the box with 10 (which then becomes 13), and 4 coins to the box with 15 (which then becomes 19). From the box that now has 19 coins, we move 6 coins to the box with 5 coins (which then becomes 11). From the box with 17 coins, we move 2 coins to the box with 11, and 2 coins to the box with 6 (which then becomes 8). Finally, we move 5 coins from the box with 18 coins to the box with 8 coins. In the end, each box has exactly 13 coins. It is easy to verify that exactly 22 moves were made.
|
22
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
|
Answer: 7.
Solution. Since $A B C D$ is a square, then $A B=B C=C D=A D$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle A B L$ to $90^{\circ}$. Then the right triangles $A B K$ and $C B L$ are equal by the acute angle and the leg $A B=B C$ (Fig. 1). Therefore, $A K=C L=6$. Then
$$
L D=C D-C L=A D-C L=(K D-A K)-C L=K D-2 \cdot C L=19-2 \cdot 6=7
$$
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued faces, the same number of dots is marked. All dots were erased from all faces except nine, as shown in the diagram. What is the total number of dots that were initially marked on the surface of the figure?
|
Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, the dots were not erased, while on the other, they were erased. It is not difficult to understand that originally there were the same number of dots on these faces.
Let's calculate the total number of dots that were originally on all 9 pairs of faces. We get
$$
2 \cdot(1+1+6+2+2+5+3+3+4)=54
$$
There are 6 faces $1 \times 1$ left, about which we still know nothing. However, it can be noticed that they are divided into pairs according to the following principle: in one pair, there will be faces from one $1 \times 1 \times 1$ cube. In each such pair, the sum of the numbers was 7. Then we get that the number of dots on the surface of the figure originally was
$$
54+3 \cdot 7=75
$$
|
75
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
|
Answer: 83.
Solution. Mark a point $K$ on the ray $AB$ such that $AK = AC$. Then the triangle $KAC$ is equilateral; in particular, $\angle AKC = 60^{\circ}$ and $KC = AC$. At the same time, $BK = AK - AB = AC - AB = AD$. This means that triangles $BKC$ and $DAC$ are equal by two sides and the angle $60^{\circ}$ between them (Fig. 2).
It remains to note that the angle $\angle ABC$ - the exterior angle of triangle $BKC$ - is equal to the exterior angle at vertex $D$ of triangle $DAC$, which is calculated as the sum of two interior angles: $60^{\circ} + 23^{\circ} = 83^{\circ}$.
|
83
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.5. On the base $AC$ of isosceles triangle $ABC (AB = BC)$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
|
Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angle at vertex \(B\) is \(30^\circ\), so \(HM = \frac{1}{2} BM = 1.5\). Then \(AC = 2AH = 2(AM + MH) = 2 \cdot (7 + 1.5) = 17\).
Fig. 4: to the solution of problem 9.5
Another solution. Mark a point \(K\) on \(MC\) such that \(\angle BKM = 60^\circ\) (Fig. 4; such a point lies exactly on the segment \(MC\) because \(\angle BCM = \angle BAM = \angle BMC - \angle ABM < 60^\circ\)). Note that in triangle \(BMK\), two angles are \(60^\circ\), so it is equilateral and \(BK = MK = BM = 3\). Also note that triangles \(ABM\) and \(CBK\) are equal because \(BC = AB\), \(\angle A = \angle C\), \(\angle AMB = \angle CKB = 120^\circ\) (therefore, \(\angle ABM = \angle CBK\)). Then \(CK = AM = 7\) and \(AC = AM + MK + KC = 7 + 3 + 7 = 17\).
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
|
Answer: 18.
Solution. Let the lines $B M$ and $A D$ intersect at point $K$ (Fig. 5). Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 5: to the solution of problem 9.8
In the isosceles triangle $A D H$, draw the height and median $D S$. Then, in triangle $A H K$, the segment $D S$ passes through the midpoint of side $A H$ and is parallel to $H K$. Therefore, $D S$ is the midline of this triangle, and $A D = D K = 18$.
Remark. If in a right triangle a point on the hypotenuse is equidistant from two vertices of the triangle, then it is equidistant from all three vertices of the triangle.
## 10th Grade
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.3. On the side $AD$ of rectangle $ABCD$, a point $E$ is marked. On the segment $EC$, there is a point $M$ such that $AB = BM, AE = EM$. Find the length of side $BC$, given that $ED = 16, CD = 12$.
|
Answer: 20.
Solution. Note that triangles $A B E$ and $M B E$ are equal to each other by three sides. Then $\angle B M E=\angle B A E=90^{\circ}$.
Fig. 6: to the solution of problem 10.3
From the parallelism of $A D$ and $B C$, it follows that $\angle B C M=\angle C E D$ (Fig. 6). Therefore, right triangles $B C M$ and $C E D$ are equal by an acute angle and the leg $B M=A B=C D$. Using the Pythagorean theorem for triangle $C D E$, we conclude
$$
B C=C E=\sqrt{C D^{2}+E D^{2}}=\sqrt{12^{2}+16^{2}}=20
$$
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure?
|
Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D=$ $180^{\circ}-\angle C A D-\angle A D C=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}$.
Fig. 7: to the solution of problem 10.6
Draw segment $C M$ (Fig. 7). Since in a right triangle, the median to the hypotenuse is equal to half of it, we have $C M=D M=A M=B M$. Triangle $M C D$ is isosceles with an angle of $60^{\circ}$ at the base, so it is equilateral, and $\angle C M D=60^{\circ}$. Then $\angle B M C=180^{\circ}-\angle A M B-\angle C M D=180^{\circ}-70^{\circ}-60^{\circ}=50^{\circ}$. Since triangle $B M C$ is isosceles with vertex $M$, we have $\angle C B M=$ $\frac{1}{2}\left(180^{\circ}-\angle B M C\right)=\frac{1}{2}\left(180^{\circ}-50^{\circ}\right)=65^{\circ}$. Finally,
$$
\angle B C A=180^{\circ}-\angle C B M-\angle B O C=180^{\circ}-65^{\circ}-80^{\circ}=35^{\circ} \text {. }
$$
Remark. There are other solutions that use the fact that $A B C D$ is a cyclic quadrilateral with the center of the circumscribed circle $M$.
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
|
Answer: 58.
Solution. As is known, in a circle, inscribed angles subtended by equal chords are either equal or supplementary to $180^{\circ}$. Since $B C=C D$ and $\angle B A D<180^{\circ}$, we get that $\angle B A C=\angle D A C$.
Fig. 9: to the solution of problem 11.5
Since $\angle B D A=\angle B C A=64^{\circ}$, we get that $\angle B D O=\angle A D O=32^{\circ}$. Therefore, point $O$ lies on the two angle bisectors of triangle $A B D$, i.e., it is the point of intersection of the angle bisectors (Fig. 9). Then
$$
\angle B O C=\angle B A O+\angle A B O=\frac{\angle B A D+\angle A B D}{2}=\frac{180^{\circ}-\angle B D A}{2}=\frac{180^{\circ}-64^{\circ}}{2}=58^{\circ}
$$
Remark. The solution could also have been completed differently: since $O$ is the point of intersection of the angle bisectors of triangle $A B D$, by the trident lemma $C D=C B=C O$, from which $\angle C O B$ is easily found.
|
58
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4 Around a circle, 100 integers are written, the sum of which is 1. A chain is defined as several numbers (possibly one) standing consecutively. Find the number of chains, the sum of the numbers in which is positive.
#
|
# Answer: 4951
Solution. Let's divide all chains (except the chain consisting of all numbers) into pairs that complement each other. If the sum of the numbers in one chain of the pair is $s$, and in the second is $t$, then $s+t=1$. Since $s$ and $t$ are integers, exactly one of them is positive. Therefore, exactly half of all chains have a positive sum. The total number of chains is $100 \cdot 99$ (two complementary chains are determined by the choice of two positions between them). This gives us 4950 chains with a positive sum. Adding to them the chain of all numbers.
|
4951
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5 The square of a natural number a, when divided by a natural number n, gives a remainder of 8. The cube of the number a, when divided by n, gives a remainder of 25. Find n.
|
Answer: $n=113$
Solution. Note that the number $x=a^{6}-8^{3}=(a^{2})^{3}-8^{3}=(a^{2}-8)(a^{4}+8a^{2}+64)$ is divisible by $n$. Also note that the number $y=a^{6}-25^{2}=(a^{3})^{2}-25^{2}=(a^{3}-25)(a^{3}+25)$ is divisible by $n$. Then the difference $x-y=(a^{6}-8^{3})-(a^{6}-25^{2})=625-512=113$ is divisible by $n$. The number 113 is a prime number, it has only two divisors: 1 and 113. Since by the condition $n>25$, then $n=113$.
|
113
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The number 1232 is divisible by the sum of its digits $1+2+3+2=8$. What is the next number that is divisible by the sum of its digits?
|
Answer: 1233.
Solution: One can notice that the next number fits, as 1233 is divisible by the sum of its digits $1+2+3+3=9$. Indeed, $1233: 9=137$.
Comment: One could have avoided performing the division and instead used the divisibility rule for 9.
|
1233
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A farmer wants to start growing watermelons. He wants to sell at least 10,000 watermelons every year. Watermelons are grown from seeds (one seed grows into one watermelon). Each watermelon can produce 250 good seeds that can be planted the following year, but then this watermelon cannot be sold. What is the minimum number of watermelon seeds the farmer should buy to start growing watermelons, and never have to buy seeds again?
|
Answer: 10041.
Solution: Let the smallest number of seeds the farmer needs to buy be $10000+x$. Then he will be able to sell 10000 watermelons, and the remaining $x$ watermelons can be used for seeds. For the seeds to be sufficient for the next season, the inequality $250 x \geqslant 10000+x$ must hold. From this, $x \geqslant \frac{10000}{249}=40 \frac{40}{249}$. Therefore, $x=41$.
|
10041
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. For the function $f(x)$, it is known that it is odd, i.e., $f(-x) = -f(x)$ for every real $x$. Additionally, it is known that for every $x$, $f(x+5) = f(x)$, and also $f(1 / 3) = 2022$, and $f(1 / 2) = 17$. What is the value of
$$
f(-7) + f(12) + f\left(\frac{16}{3}\right) + f\left(\frac{9}{2}\right) ?
$$
|
Answer: 2005.
Solution. Let's find the values of $f(-7)+f(12)$, $f\left(\frac{16}{3}\right)$, and $f\left(\frac{9}{2}\right)$ separately.
- $f(-7)+f(12)=f(-7+5)+f(12-2 \cdot 5)=f(-2)+f(2)=0$;
- $f\left(\frac{16}{3}\right)=f\left(5+\frac{1}{3}\right)=f\left(\frac{1}{3}\right)=2022$.
- $f\left(\frac{9}{2}\right)=f\left(5-\frac{1}{2}\right)=f\left(-\frac{1}{2}\right)=-f\left(\frac{1}{2}\right)=-17$
The answer is $0+2022-17=2005$.
|
2005
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. For the numbers $1000^{2}, 1001^{2}, 1002^{2}, \ldots$, the last three digits are discarded. How many of the first terms of the resulting sequence form an arithmetic progression?
|
Answer: 32.
Solution. $(1000+k)^{2}=1000000+2000 k+k^{2}$. As long as $k^{2}<1000$, after discarding the last three digits, the number $1000+2 k$ will remain, i.e., each term of the sequence will be 2 more than the previous one. When $k=31: k^{2}=961<1000$, and when $k=32: k^{2}=1024$. Therefore, the element obtained from $1032^{2}$ will differ from the previous one by 3. Thus, the length of the progression is the number of numbers from 1000 to 1031, i.e., 32.
|
32
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. While tidying up the children's room before the guests arrived, mom found 9 socks. Among any four socks, at least two belong to the same owner. And among any five socks, no more than three belong to the same owner. How many children scattered the socks, and how many socks does each child own?
Answer. There are three children, and each owns three socks.
|
Solution. No child owned more than three socks, as otherwise the condition "among any five socks, no more than three belonged to one owner" would not be met. There are a total of 9 socks, so there are no fewer than three children. On the other hand, among any four socks, there are two socks belonging to one child, so there are fewer than four children. Therefore, there are three children in the family, and each scattered no more than three socks, with a total of 9 socks. This means that each child owns 3 of the socks found by the mother.
Grading Criteria.
- Complete answer with correct explanation - 7 points.
- Justified that there are three children - 5 points.
- Correct reasoning, but the solution is not fully completed - 1-2 points.
- Answer without justification - 0 points.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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