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5. In a right triangle $ABC$ ( $\angle C$ - right), a point $D$ is taken on side $BC$ such that $\angle CAD=30^{\circ}$. From point $D$, a perpendicular $DE$ is dropped to $AB$. Find the distance between the midpoints of segments $CE$ and $AD$, given that $AC=3 \sqrt{3}, DB=4$. | Answer: $\frac{9 \sqrt{57}}{38}$
## Solution.
Let point $G$ be the midpoint of segment $A D$, and point $F$ be the midpoint of segment $C E$. We need to find the length of segment $G F$.
S... | \frac{9\sqrt{57}}{38} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,445 |
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 | 16,446 |
4-7. Along a straight alley, 100 lamps are placed at equal intervals, numbered sequentially from 1 to 100. At the same time, from different ends of the alley, Petya and Vasya started walking towards each other at different constant speeds (Petya from the first lamp, Vasya from the hundredth). When Petya was at the 22nd... | Answer. At the 64th lamppost.
Solution. There are a total of 99 intervals between the lampposts. From the condition, it follows that while Petya walks 21 intervals, Vasya walks 12 intervals. This is exactly three times less than the length of the alley. Therefore, Petya should walk three times more to the meeting poin... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,450 |
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 | 16,451 |
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 l... | 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 ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,452 |
5-6. On a rectangular table of size $x$ cm $\times 80$ cm, identical sheets of paper of size 5 cm $\times 8$ cm are placed. 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 ... | 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 ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,453 |
5-7. On the faces of a die, the numbers $6,7,8,9,10,11$ are written. The die was rolled twice. The first time, the sum of the numbers on the four "vertical" (that is, excluding the bottom and top) faces was 33, and the second time - 35. What number can be written on the face opposite the face with the number 7? Find al... | Answer: 9 or 11.
Solution. The total sum of the numbers on the faces is $6+7+8+9+10+11=51$. Since the sum of the numbers on four faces the first time is 33, the sum of the numbers on the two remaining faces is $51-33=18$. Similarly, the sum of the numbers on two other opposite faces is $51-35=16$. Then, the sum on the... | 9or11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,454 |
6-3. The red segments in the figure have equal length. They overlap by equal segments of length $x$ cm. What is $x$ in centimeters?
 | Answer: 2.5.
Solution. Adding up the lengths of all the red segments, we get 98 cm. Why is this more than 83 cm - the distance from edge to edge? Because all overlapping parts of the red segments have been counted twice. There are 6 overlapping parts, each with a length of $x$. Therefore, the difference $98-83=15$ equ... | 2.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,456 |
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.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,459 |
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 ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,462 |
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... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,463 |
Task 1. Prove that the cube of any composite number can be represented as the difference of two squares of natural numbers in at least three different ways. | Solution. For any natural $n$, the following method is correct. $n^{3}=a^{2}-b^{2}=(a+b)(a-b)$. Let $a+b=n^{2}, a-b=n$, then by adding these two equations, we get that $2 a=n^{2}+n$, and by subtracting, we get that $2 b=n^{2}+n$. From this, $n^{3}={\frac{n^{2}+n}{2}}^{2}-{\frac{n^{2}-n}{2}}^{2}$. Notice that if $n$ is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,464 |
Problem 2. On the board, there are 2017 digits. From these, several numbers were formed, the sums of the digits of these numbers were calculated, and then the sum of all the numbers was subtracted by the sum of the sums of their digits. The resulting number was broken down into digits, and the above operation was repea... | Solution. Since the difference between a number and the sum of its digits is divisible by 9, the first operation will result in a number that is a multiple of 9. Moreover, if we take the sum of several numbers and subtract the sum of the digits of these numbers, the result will also be a multiple of 9. Continuing the c... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,465 |
Problem 5. In triangle $A B C$, the median $B M$ is drawn. Can the radius of the circle inscribed in triangle $B C M$ be twice as small as the radius of the circle inscribed in triangle $A B C$? | Solution. 1st method. From the formula $S=r p$ and the equality $S(B M C)=\frac{1}{2} S A B C$, it follows that the perimeters of triangles $B M C$ and $A B C$ are equal. But $\mathrm{P}(\mathrm{ABC})=$ $\mathrm{AB}+\mathrm{BC}+\mathrm{CA}=\mathrm{AB}+\mathrm{BC}+\mathrm{CM}+\mathrm{MA}=(\mathrm{AB}+\mathrm{AM})+\mathr... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,466 |
10.1. Do there exist eight natural numbers, among which exactly one is divisible by 8, exactly two are divisible by 7, exactly three are divisible by 6, ..., exactly seven are divisible by 2? | 10.1. Answer. They do not exist.
Suppose such eight numbers do exist. From the condition, it follows that exactly one of them is not divisible by 2 and exactly two of them are not divisible by 3. Therefore, among the considered numbers, at least five numbers are divisible by both 2 and 3, i.e., divisible by 6. However... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,467 |
10.2. Given numbers $a, b, c$. It is known that for any $x$ the inequality
$$
a x^{2}+b x+c \geqslant b x^{2}+c x+a \geqslant c x^{2}+a x+b
$$
holds. Prove that $a=b=c$. | 10.2. From the inequality, we obtain that the quadratic polynomials $\left(a x^{2}+b x+c\right)-\left(b x^{2}+c x+a\right)$ and $\left(b x^{2}+c x+a\right)-$ - $\left(c x^{2}+a x+b\right)$ take non-negative values for all values of $x$; hence, their leading coefficients $a-b$ and $b-c$ are non-negative, i.e., $a \geqsl... | =b= | Inequalities | proof | Yes | Yes | olympiads | false | 16,468 |
10.3. Given a triangle $A B C$. Through a point $X$ lying inside it, segments $c_{X}$, parallel to $A B$, with endpoints on sides $A C$ and $B C$, and segment $b_{X}$, parallel to $A C$, with endpoints on sides $A B$ and $C B$ are drawn. Prove that all points $X$ for which the lengths of segments $b_{X}$ and $c_{X}$ ar... | 10.3. Through the base $A_{1}$ of the bisector of angle $B A C$, draw lines parallel to $A C$ and $A B$, intersecting $A B$ and $A C$ at points $P$ and $Q$ respectively. By construction, $A P A_{1} Q$ is a parallelogram, and its diagonal $A A_{1}$ is the bisector, hence $A P A_{1} Q$ is a rhombus, which implies $A_{1} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,469 |
10.4. The number $\underbrace{111 \ldots 11}_{99 \text { ones }}$ is written on the board. Petya and Vasya play the following game, taking turns; Petya starts. On a turn, a player either writes a zero in place of one of the ones, except the first and last, or erases one of the zeros. The player loses if, after their mo... | 10.4. Answer. Petya.
Note that the number $A_{n}=1 \underbrace{101 \ldots 11}_{n \text { ones }}$ is not divisible by 11. Indeed, if $n$ is even, then $A_{n}=\underbrace{000 \ldots 00}_{n+1 \text { zeros }}+\underbrace{11 \ldots 11}_{n \text { ones }}$, and if $n$ is odd, then $A_{n}=\underbrace{900 \ldots 00}_{n \tex... | Petya | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,470 |
10.5. In a line, all natural numbers are written one after another without spaces in ascending order: $1234567891011 . .$. Which sequence of digits will appear in the string earlier: a sequence of exactly 2006 consecutive sixes (with non-sixes on both sides) or a sequence of exactly 2007 consecutive sevens (with non-se... | 10.5. Answer. A sequence of exactly 2007 sevens.
A sequence of exactly 2007 sevens will appear, for example, when listing the numbers \( N = \underbrace{777 \ldots 77}_{1004 \text{ sevens}} \) and \( N+1 = \underbrace{777 \ldots 7}_{1003 \text{ sevens}} 8 \), since the number \( N \) is preceded by the digit 6.
Suppo... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,471 |
10.6. Let $AB$ and $CD$ be two perpendicular chords of a circle with center $O$, intersecting at point $E$; let also $N$ and $T$ be the midpoints of segments $AC$ and $BD$ respectively. Prove that quadrilateral $ENOT$ is a parallelogram. | 10.6. Extend $T E$ to intersect line $A C$ at point $K$ (see Fig. 8). We will show that $E K$ is the altitude of triangle $C E A$. Let $\angle C A B=\alpha$. From the right triangle $A E C$, we get $\angle A C E=90^{\circ}-\alpha$. Further, $\angle C D B=\angle C A B=\alpha$ (as inscribed angles subtending the same arc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,472 |
10.7. On the board, natural numbers $1,2,3, \ldots, 10$ are written. It is allowed to write down the number $a^{2}$ if the number $a$ is already on the board, or to write down the least common multiple of numbers $a$ and $b$ if the numbers $a$ and $b$ are already written. Is it possible to obtain the number 1000000 usi... | 10.7. Answer. No.
Let $q(n)$ denote the greatest integer $k$ such that $n$ is divisible by $5^k$. For each $n$ in the initial set of numbers, $q(n)=0$ or $q(n)=1=2^0$. Note that $q\left(a^2\right)=2 q(a)$, and $q(\operatorname{LCM}(a, b))$ equals $q(a)$ or $q(b)$. From this,
. ${ }^{404} / 1616$ is exactly equal to $1 /... | \frac{403}{1617} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,475 |
7.2. At a round table, 7 aborigines are sitting: knights, liars, and bores (there is at least one representative of each tribe at the table). A knight always tells the truth, a liar always lies. A bore lies if at least one knight is sitting next to them, and in other cases, they can say anything.
How many knights coul... | Answer: 1 or 2.
Solution: Two knights cannot sit next to each other, so there are no more than three knights.
If there are 3 knights at the table, then there must be at least 4 bores, but in this case, there are no liars at the table. This is a contradiction.
Next, we will provide examples of seating arrangements wi... | 1or2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,476 |
7.3. It is known that both КРУГ and КУБ are cubes of some numbers. Find all possible values of the numbers КРУГ and КУБ. List them all and explain why there are no others. (Identical digits are replaced by the same letters, different digits - by different letters.)
Note: 8 and 1000 are cubes, since $8=2 \cdot 2 \cdot ... | Answer: 1728 and 125.
Solution: Notice that in the words КУБ (CUBE) and КРУГ (CIRCLE), two digits repeat - К and У.
Let's list all possible three-digit numbers that are cubes of some numbers:
$5^{3}=125, 6^{3}=216, 7^{3}=343, 8^{3}=512, 9^{3}=729$ (4^3=64 - two-digit; 10^3=1000 - four-digit). Notice that 343 does no... | 1728125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,477 |
7.4. A team of workers has to carve $n$ ice sculptures for an ice town over two days. On the first day, they carved 11 sculptures, with each worker completing an equal share of the work. On the second day, only 7 workers showed up because some were transferred to another project. As a result, each worker had to do thre... | Answer: 12, 14, 18, 22, or 32.
Solution: Let $k$ be the number of workers who started the job on the first day. The amount of work done by each worker on the first day is $\frac{11}{k}$, and on the second day, since each worker did three times more, it is $\frac{33}{k}$. On the other hand, on the second day, 7 workers... | 12,14,18,22,32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,478 |
7.5. The numbers from 1 to 20 are arranged in a circle. We will paint a number blue if it is divisible without a remainder by the number to its left. Otherwise, we will paint it red. What is the maximum number of blue numbers that could be in the circle? | # Solution.
Evaluation. It is obvious that numbers cannot be blue if the number to their left is greater than or equal to 11. That is, no more than 10 numbers can be blue.
Example. As an example, both any correct arrangement and a correct algorithm are counted. An example of a correct algorithm.
1) write down the nu... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,479 |
8.1. A dandelion blooms in the morning, remains yellow for this and the next day, turns white on the third morning, and by evening of the third day, it has lost its petals. Yesterday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and today there are 15 yellow and 11 white. How many yellow dandel... | Solution: All yellow dandelions the day before yesterday are the white dandelions of yesterday and the white dandelions of today. Therefore, the day before yesterday there were $14+11=25$ yellow dandelions.
Answer: 25 dandelions.
Note: The number of yellow dandelions yesterday and today is not needed for solving the ... | 25 | Other | math-word-problem | Yes | Yes | olympiads | false | 16,480 |
8.2. A snail is crawling at a constant speed around the clock face along the circumference counterclockwise. It started at 12:00 from the 12 o'clock mark and completed a full circle exactly at 14:00 on the same day. What time did the clock show when the snail met the minute hand during its movement? Justify your answer... | Solution: In two hours of movement, the snail described a full circle, the minute hand - two full circles. Together they described three full circles. Therefore, one circle the snail and the hand together pass in $120: 3=40$ minutes - and this is the time between their consecutive meetings. Therefore, the times when th... | 12:4013:20 | Other | math-word-problem | Yes | Yes | olympiads | false | 16,481 |
8.3. Inside parallelogram $A B C D$, a point $E$ is taken such that $C E = C B$. Let $F$ and $G$ be the midpoints of segments $C D$ and $A E$ respectively. Prove that line $F G$ is perpendicular to line $B E$.
}=\frac{a b c}{a(1+b+b c)}=\frac{b c}{1+b+b c}
$$
Similarly, by replacing 1 with the number $a b c$, we have
$$
\frac{1}{1+c+c a}=\frac{a b}{1+a+a b}=\frac{a b^{2} c}{1+b+b c}=\frac{b}{1+b+b c} .
$$
Then
$$
\frac{1}{1+a+a b}+\frac{1}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,483 |
8.5. For breakfast, mom daily gives Seryozha either 1 sandwich and 3 candies, or 2 sandwiches and 4 candies, or 3 sandwiches and 5 candies. After several days, it turned out that Seryozha had eaten exactly 100 sandwiches. Could he have eaten exactly 166 candies during the same time? Justify your answer. | # Solution:
Method 1. Suppose this happened. Note that each day Petya eats 2 more candies than sandwiches. Therefore, the eating process took exactly $\frac{166-100}{2}=33$ days. But in 33 days, Petya could not have eaten more than $3 \cdot 33=99$ sandwiches. Contradiction.
Method 2. Let $x$ be the number of days Ser... | Hecouldnot | Other | math-word-problem | Yes | Yes | olympiads | false | 16,484 |
8.6. On a certain segment, its endpoints and three internal points were marked. It turned out that all pairwise distances between the five marked points are different and are expressed in whole centimeters. What is the smallest possible length of the segment? Justify your answer. | Solution: There are 5 points, so there are 10 pairwise distances. If all of them are expressed as positive whole numbers of centimeters and are distinct, at least one of them is not less than 10. Therefore, the length of the segment is not less than 10. Suppose the length of the segment is exactly 10. Then the pairwise... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,485 |
# 1.1. Condition:
On the International Space Station, there were electronic clocks displaying time in the format HH : MM. Due to an electromagnetic storm, the device started malfunctioning, and each digit on the display either increased by 1 or decreased by 1. What was the actual time when the storm occurred, if immed... | Answer: 11:18
Solution.
The first digit before the breakdown could have been 1 or 3, but since the electronic device does not display values above 24 hours, only 1 meets the condition. The second and third digits in sequence are zeros, so before the breakdown, they could only have been 1. The last digit is 9, which m... | 11:18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,486 |
# 1.2. Condition:
On the International Space Station, there were electronic clocks displaying time in the format HH : MM. Due to an electromagnetic storm, the device started malfunctioning, and each digit on the display either increased by 1 or decreased by 1. What was the actual time when the storm occurred, if immed... | Answer: at 11:48.
# | 11:48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,487 |
# 3.1. Condition:
The number 4597 is displayed on the computer screen. In one move, it is allowed to swap any two adjacent digits, but after this, 100 is subtracted from the resulting number. What is the largest number that can be obtained by making no more than two moves? | Answer: 8357
## Solution.
The first digit cannot exceed 8, since to obtain the other digits, you need to move the nine forward by two places and subtract one from it. Note that in any other example, the first digit will be less than 8, as we can only get an eight from a nine, and all other digits are less than 9. The... | 8357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,491 |
# 4.1. Condition:
In front of the elevator stand people weighing 50, 51, 55, 57, 58, 59, 60, 63, 75, and 140 kg. The elevator's load capacity is 180 kg. What is the minimum number of trips needed to get everyone up? | Answer: 4 (or 7)
## Solution.
In one trip, the elevator can move no more than three people, as the minimum possible weight of four people will be no less than $50+51+55+57=213>180$. Note that no one will be able to go up with the person weighing 140 kg, so a separate trip will be required for his ascent. For the rema... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,492 |
# 4.2. Condition:
In front of the elevator stand people weighing 150, 60, 70, 71, 72, 100, 101, 102, and 103 kg. The elevator's load capacity is 200 kg. What is the minimum number of trips needed to get everyone up? | Answer: 5 (or 9)
## Solution
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+70+71=201>200$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remaining... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,493 |
# 4.3. Condition:
In front of the elevator stand people weighing 150, 62, 63, 66, 70, 75, 79, 84, 95, 96, and 99 kg. The elevator's load capacity is 190 kg. What is the minimum number of trips needed to get everyone up? | # Answer: 6 (or 11)
## Solution.
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $62+63+66=191>190$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remai... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,494 |
# 4.4. Condition:
In front of the elevator stand people weighing $130,60,61,65,68,70,79,81,83,87,90,91$ and 95 kg. The elevator's load capacity is 175 kg. What is the minimum number of trips needed to get everyone up? | # Answer: 7 (or 13)
## Solution.
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+61+65=186>175$. Note that no one will be able to go up with the person weighing 135 kg, so a separate trip will be required for his ascent. Six trips wil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,495 |
# 5.1. Condition:
In the warehouse, there are 8 cabinets, each containing 4 boxes, each with 10 mobile phones. The warehouse, each cabinet, and each box are locked. The manager has been tasked with retrieving 52 mobile phones. What is the minimum number of keys the manager should take with them? | # Answer: 9
Solution. To retrieve 52 phones, at least 6 boxes need to be opened. To open 6 boxes, no fewer than 2 cabinets need to be opened. Additionally, 1 key to the warehouse is required. In total, $6+2+1=9$ keys need to be taken by the manager. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,496 |
# 6.1. Condition:
A large ladybug has 6 spots on its back, while a small one has 5 spots. Several ladybugs gathered together for a party on a large burdock leaf. In total, they had 43 spots. Indicate the number of ladybugs of both sizes that gathered for the party.
## Number of large ladybugs: 3
Number of small lady... | Solution. Note that if there were 8 small ladybugs on the burdock leaf, the number of spots would be 40. If a small ladybug is replaced with a large one, the total number of spots on their backs would increase by 1. Let's make such a replacement three times and we will get a total of 43 spots.
# | Number\of\large\ladybugs:\3,\Number\of\small\ladybugs:\5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,497 |
# 8.1. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 3 meters away from me."
- The rightmost in the row said: "My fellow ... | # Answer:
The second islander -2 m
The third islander -1 m; 3 m; 4 m.
## Solution.
Let's number the islanders from left to right. Suppose the first one is a knight. Then, from his statement, it follows that the fourth one must also be a knight. The fourth one said that his fellow islander stands 2 meters away, and ... | notfound | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 16,502 |
# 8.2. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 2 meters away from me."
- The rightmost in the row said: "My fellow ... | # Answer:
The second islander -1 m
The third islander -1 m.
## Solution.
Let's number the islanders from left to right. Suppose the first one is a knight. Then from his statement, it follows that the third one is also a knight; by the principle of exclusion, the second and fourth must be liars. The fourth said that... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,503 |
# 8.3. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 1 meter away from me."
- The second from the left said: "My fellow t... | # Answer:
The third islander -1 m; 3 m; 4 m.
The fourth islander -2 m.
## Solution.
Let's number the islanders from right to left. Suppose the first one is a knight. Then, from his statement, it follows that the second one is also a knight. However, the second one said that his fellow tribesman is two meters away f... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,504 |
# 8.4. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The second from the left said: "My fellow tribesman in this line stands 1 meter away from me."
- The third from the left said: "My fellow ... | # Answer:
The first islander - 1 m; 2 m; 4 m.
The fourth islander - 1 m; 2 m; 4 m.
## Solution.
Number the islanders from right to left. Suppose the second is a liar. Then from his statement, it follows that both the first and the third must be knights, because if either of them lied, the second would have told the... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,505 |
9-1. Postman Pechkin calculated that he walked half the distance (at a speed of 5 km/h) and only a third of the time he was cycling (at a speed of 12 km/h). Did he make a mistake in his calculations? | Answer. Mistaken.
Solution. Let's denote the entire distance Pechkin traveled as $2 S$ km. Then, on foot, he covered a distance of $S$ km and spent $S / 5 = 0.2 S$ (hours) on it. According to the problem, this constituted $2 / 3$ of the total time spent, meaning the entire journey took $0.2 S : 2 / 3 = 0.3 S$ (hours),... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,506 |
9-2. The school volleyball team played several matches. After they won another match, the share of victories increased by $1 / 6$. To increase the share of victories by another 1/6, the volleyball players had to win two more consecutive matches. What is the minimum number of victories the team needs to achieve to incre... | Answer: 6.
Solution: Let the team initially play $n$ matches, of which $k$ were won. Then, after the next win, the share of victories increased by $\frac{k+1}{n+1}-\frac{k}{n}=\frac{1}{6}$. Similarly, after two more wins, the increase was $\frac{k+3}{n+3}-\frac{k+1}{n+1}=\frac{1}{6}$. Simplifying each equation, we get... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,507 |
9-3. Can we find at least one cube of an integer among the numbers $2^{2^{n}}+1, n=0,1,2, \ldots$? | Answer: No.
Solution. Suppose there exist natural numbers $k$ and $n$ such that $2^{2^{n}}+1=k^{3}$. Then $k$ is odd, and $2^{2^{n}}=k^{3}-1=(k-1)\left(k^{2}+k+1\right)$. Therefore, $k-1=2^{s}$ and $k^{2}+k+1=2^{t}$, where $s$ and $t$ are some natural numbers. Now we have $2^{2 s}=(k-1)^{2}=k^{2}-2 k+1$ and $2^{t}-2^{... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,508 |
9-4. In an isosceles right triangle \(ABC\), angle \(A\) is \(90^\circ\), and point \(M\) is the midpoint of \(AB\). A line passing through point \(A\) and perpendicular to \(CM\) intersects side \(BC\) at point \(P\). Prove that \(\angle AMC = \angle BMP\). | Solution. Let's complete the isosceles right triangle to form a square \(ABKC\) (see figure). Let \(N\) be the intersection point of \(AP\) and \(BK\). The lines \(CM\) and \(AN\) are perpendicular to each other, so \(\angle AMC = \angle BNA\). From this, it follows that the right triangles \(MAC\) and \(BNA\) are equa... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,509 |
9-5. Two three-digit numbers are written on the board in a multiplication example. If the multiplication sign is replaced with 0, a seven-digit number is obtained, which is an integer multiple of the product. By what factor exactly | Answer: 73.
Solution. Let the original numbers be denoted by $a$ and $b$. Then the specified seven-digit number will have the form $10000a + b$. According to the condition, $10000a + b = nab$, from which we get $b = \frac{10000}{na - 1}$.
Note that the numbers $a$ and $a-1$ do not have common divisors, so $na - 1 = p... | 73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,510 |
9.5. From the digits $1,2,3,4,5,6,7,8,9$, nine (not necessarily distinct) nine-digit numbers are formed; each digit is used exactly once in each number. What is the maximum number of zeros that the sum of these nine numbers can end with?
(N. Agakhanov) | Answer: Up to 8 zeros.
Solution: We will show that the sum cannot end with 9 zeros. Each of the numbers formed is divisible by 9, since the sum of its digits is divisible by 9. Therefore, their sum is also divisible by 9. The smallest natural number divisible by 9 and ending with nine zeros is $9 \cdot 10^{9}$, so the... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,511 |
9.6. A square is divided into $n^{2} \geqslant 4$ rectangles by $2(n-1)$ lines, of which $n-1$ are parallel to one side of the square, and the remaining $n-1$ are parallel to the other side. Prove that one can choose $2 n$ rectangles from the partition such that for any two chosen rectangles, one can be placed inside t... | Solution. Let's call a pair of rectangles embeddable if one of them can be embedded into the other.
Suppose the horizontal side of the square is divided into segments of lengths $a_{1}, \ldots, a_{n}$ (from left to right), and the vertical side is divided into segments of lengths $b_{1}, \ldots, b_{n}$ (from top to bo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,512 |
9.7. The circle $\omega$ is inscribed in triangle $ABC$, where $AB < AC$. The excircle of this triangle touches side $BC$ at point $A'$. Point $X$ is chosen on segment $A'A$ such that segment $A'X$ does not intersect $\omega$. The tangents drawn from $X$ to $\omega$ intersect segment $BC$ at points $Y$ and $Z$. Prove t... | Solution. We will assume that point $Y$ is closer to point $B$ than $Z$; moreover, we assume that side $BC$ is horizontal, and $A$ lies above it (see Fig. 2).
Let $\omega_{A}$ be the excircle of triangle $ABC$ touching side $BC$, and let $\omega^{\prime}$ be the excircle of triangle $XYZ$ touching side $XZ$. Let $\ome... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,513 |
9.8. The sum of positive numbers $a, b, c$ and $d$ is 3. Prove the inequality
$$
\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}} \leqslant \frac{1}{a^{2} b^{2} c^{2} d^{2}}
$$ | Solution. Multiplying the inequality to be proved by $a^{2} b^{2} c^{2} d^{2}$, we get
$$
a^{2} b^{2} c^{2}+a^{2} b^{2} d^{2}+a^{2} c^{2} d^{2}+b^{2} c^{2} d^{2} \leqslant 1
$$
Since the inequality is symmetric, we can assume that $a \geqslant b \geqslant c \geqslant d$. By the inequality of means for the numbers $a,... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,514 |
1.1. Polina has two closed boxes - a square one and a round one. She was told that the round one contains 4 white and 6 black balls, while the square one contains 10 black balls. In one move, Polina can take a ball from any box without looking and either throw it away or move it to the other box. Polina wants to make t... | Answer: 15
Solution. Suppose that in the end, both boxes still contain both white and black balls. Then the last action Polina took was to draw a ball of a certain color from one of the boxes, and there were still both black and white balls left. But she could not have done this with certainty, meaning she could have ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,515 |
1.2. In Olya's black box, there are 5 apples and 7 pears, and in the white box, there are 12 pears. In one move, Olya can blindly take a fruit from any box and either eat it or move it to the other box. Olya wants the contents of the boxes to be the same. What is the minimum number of moves Olya can guarantee to achiev... | # Answer: 18
Solution. Suppose that in the end both boxes still contain both apples and pears. Then the last action Olya took was to take some fruit from a box, and there were still both apples and pears left. Because of this, she could not guarantee that she would take the needed fruit, as she might have picked the w... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,516 |
1.3. Inna has two closed boxes - a square one and a round one. She was told that the round box contains 3 white and 10 black balls, while the square box contains 8 black balls. In one move, Inna can, without looking, take a ball from any box and either throw it away or move it to the other box. Inna wants to make the c... | Answer: 17
Solution. Suppose that in both boxes, there are both white and black balls left in the end. Then the last action Inna took was to draw a ball of a certain color from one of the boxes, and there were still both black and white balls left. However, she could not have done this with certainty, i.e., she might ... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,517 |
1.4. In Zhenya's black box, there are 8 bananas and 10 mangoes, and in the white box - 12 mangoes. In one move, Zhenya, without looking, can take a fruit from any box and either eat it or move it to the other box. Zhenya wants the contents of the boxes to be the same. What is the minimum number of moves Zhenya can guar... | # Answer: 24
Solution. Suppose that in the end, both boxes still contain both bananas and mangoes. Then the last action was Zhenya taking some fruit from a box, and both types of fruit remained. Because of this, she could not guarantee picking the right fruit, as she might have picked the wrong one.
If removing mango... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,518 |
2.1. At the beginning of January 2020, Borya and Vitya started saving money for a new phone. On the 20th of each month, Borya saved 200 rubles, and Vitya saved 300 rubles. One day, Borya got tired of saving money and stopped (but he did not spend the money he had already saved). Vitya continued to save. On May 11, 2021... | Answer: May, 2020
Solution. By May 1, 2021, Vitya will have been saving for $12+4=16$ months, by which time he will have accumulated $16 \cdot 300=4800$ rubles. If at this point his savings are 6 times more than Borya's, then Borya will have saved $4800: 6=800$ rubles. He will save such an amount over $800: 200=4$ mon... | May,2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,519 |
2.2. At the beginning of January 2019, Borya and Vitya started saving money for a new phone. On the 20th of each month, Borya saved 150 rubles, and Vitya saved 100 rubles. One day, Vitya got tired of saving money and stopped saving (but he did not spend the money he had already saved). Borya continued to save. On May 1... | Answer: July, 2019
Solution. By May 2021, Borya will have been saving for 28 months, that is, he will have set aside $28 \cdot 150=$ 4200 rubles. This means Vitya saved $4200: 7=600$ rubles, that is, he saved for a total of 6 months. This means that in June he was still saving money, but in July he was not. | July,2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,520 |
2.3. On January 11, 2011, Baba Yaga and Koschei the Deathless decided to compete in who was kinder. Each month, Baba Yaga did good deeds for 77 children, while Koschei the Deathless did so for 12. But one day, Koschei got tired of being kind. On January 11, 2021, Baba Yaga calculated that the number of her good deeds h... | Answer: March, 2012
Solution. Baba Yaga did good for 10 years, which is 120 months in total. Therefore, Baba Yaga did 7 $\cdot$ 120 $=840$ good deeds, and Koschei did $840: 5=168$ deeds. $168: 12=14$, so Koschei seemed good for 14 months, and stopped being good on the fifteenth month. | March,2012 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,521 |
2.4. At the beginning of January 2018, the mother and father started saving for a new apartment. Their salaries were received on the 15th of each month, and after each salary, the mother saved 12 thousand rubles, and the father saved 22 thousand rubles. One day, the father's income decreased, and the family decided to ... | Answer: October, 2018
Solution. By October 2020, the years 2018 and 2019 had fully passed, and 9 months of 2020 had also passed, totaling 33 months. The mother's savings from her salary amounted to $33 \cdot 12 = 396$ thousand rubles. Therefore, the father's savings amounted to (396:2):22 = 9 months. This means the in... | October,2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,522 |
3.1. A chocolate bar weighed 250 g and cost 50 rubles. Recently, to save money, the manufacturer reduced the weight of the bar to 200 g, and increased its price to 52 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 30
Solution. We will calculate by what percentage the cost of one kilogram has increased. Before the increase, 1 kg cost 200 rubles, and after the increase, it costs $52 \cdot 5=260$ rubles. The increase is 60 rubles per kg, which originally cost 200 rubles. This is $\frac{60}{200} \cdot 100 \% = 30 \%$ | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,523 |
3.2. A chocolate bar weighed 400 g and cost 150 rubles. Recently, to save money, the manufacturer reduced the weight of the bar to 300 g, and increased its price to 180 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 60
Solution. We will calculate by what percentage the cost of 1 kg 200 g has increased. Before the increase, this amount cost 450 rubles, and after the increase, it costs $180 \cdot 4=720$ rubles. The increase is 270 rubles on 1.2 kg, which originally cost 450 rubles. This is $\frac{270}{450} \cdot 100 \%=$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,524 |
3.3. A package of milk with a volume of 1 liter cost 80 rubles. Recently, in order to save money, the manufacturer reduced the volume of the package to 0.9 liters, and increased the price to 99 rubles. By what percentage did the manufacturer's revenue increase? | Answer: 37.5
Solution. We will calculate by what percentage the cost of 9 liters of milk has increased. Before the increase, 9 liters cost $9 \cdot 80=720$ rubles, and after the increase - $99 \cdot 10=990$ rubles. The increase amounts to 270 rubles for 9 liters. This amount previously cost 720 rubles. This is $\frac{... | 37.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,525 |
3.4. A package of milk with a volume of 1 liter cost 60 rubles. Recently, in order to save money, the manufacturer reduced the volume of the package to 0.9 liters, and increased its price to 81 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 50
Solution. We will calculate by what percentage the cost of 9 liters of milk has increased. Before the increase, 9 liters cost $9 \cdot 60=540$ rubles, and after the increase, $-81 \cdot 10=810$ rubles. The increase amounts to 270 rubles for 9 liters. This amount previously cost 540 rubles. This is $\frac{... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,526 |
4.1. Galia thought of a number, multiplied it by N, then added N to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 2021 less than the originally thought number. What is N? | Answer: 2022
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numb... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,527 |
4.2. Galia thought of a number, multiplied it by $\mathrm{N}$, then added $\mathrm{N}$ to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 7729 less than the originally thought number. What is N? | Answer: 7730
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numb... | 7730 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,528 |
4.3. Galia thought of a number, multiplied it by N, then added N to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 100 less than the originally thought number. What is N? | Answer: 101
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numbe... | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,529 |
5.1. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 14 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 13 girls in thi... | # Answer: 15
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 7 pizzas. Then $m: 13=7: x$, from which $m x=91$. The number 91 has only one divisor greater than 13, which is 91. Therefore, $m=91, x=1$,... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,530 |
5.2. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 10 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 11 girls in thi... | # Answer: 11
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 5 pizzas. Then $m: 11 = 5: x$, from which we get $m x = 55$. The number 55 has only one divisor greater than 11, which is 55. Therefore, $... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,531 |
5.3. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 22 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 13 girls in thi... | # Answer: 23
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 11 pizzas. Then $m: 13 = 11: x$, from which $m x = 143$. The number 143 has only one divisor greater than 13, which is 143. Therefore, $m ... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,532 |
5.4. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 10 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 17 girls in thi... | Answer: 11
Solution. Let the number of boys be $m$, and the number of pizzas that the girls got be $x$. If each boy had eaten as much as each girl, the boys would have eaten 5 pizzas. Then $m: 17 = 5: x$, from which $m x = 85$. The number 85 has only one divisor greater than 17, which is 85. Therefore, $m=85, x=1$, th... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,533 |
6.1. How many natural numbers $\mathrm{N}$ greater than 900 exist such that among the numbers $3 \mathrm{~N}, \mathrm{~N}-$ $900, N+15,2 N$ exactly two are four-digit numbers? | Answer: 5069
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+15$, and if we write the numbers in ascending order, we get $\mathrm{N}-900, \mathrm{~N}+15,2 \mathrm{~N}, 3 \mathrm{~N}$. Four-digit numbers can only be two consecutive ones. If the four-digit numbers are $\mathrm{N}-900$ and $\mathrm{N}+15$, then $2 \mathrm{... | 5069 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,534 |
6.2. How many natural numbers $\mathrm{N}$ greater than 300 exist such that among the numbers $4 \mathrm{~N}, \mathrm{~N}-$ $300, N+45,2 N$ exactly two are four-digit numbers? | Answer: 5410
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+15$, and if we write the numbers in ascending order, we get $\mathrm{N}-300, \mathrm{~N}+45,2 \mathrm{~N}, 4 \mathrm{~N}$. Only two consecutive numbers can be four-digit. If the four-digit numbers are $\mathrm{N}-300$ and $\mathrm{N}+15$, then $2 \mathrm{~N}$ ... | 5410 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,535 |
6.3. How many natural numbers $\mathrm{N}$ greater than 700 exist such that among the numbers $3 \mathrm{~N}, \mathrm{~N}-$ $700, N+35,2 N$ exactly two are four-digit numbers? | Answer: 5229
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+35$, and if we write the numbers in ascending order, we get $\mathrm{N}-700, \mathrm{~N}+35,2 \mathrm{~N}, 3 \mathrm{~N}$. Only two consecutive numbers can be four-digit. If the four-digit numbers are $\mathrm{N}-700$ and $\mathrm{N}+35$, then $2 \mathrm{~N}$ ... | 5229 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,536 |
8.1. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved two seats to the right, Galia had moved one seat to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. Wh... | # Answer: 2
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 2, Galia's decreased by 1, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have decrease... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,538 |
8.2. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved three seats to the right, Galia had moved one seat to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. ... | Answer: 3
Solution. Let's see how the seat number changed for everyone except Anya. Varya's seat number increased by 3, Galia's decreased by 1, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have decreased by ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,539 |
8.3. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved one seat to the right, Galia had moved three seats to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. ... | # Answer: 3
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 1, Galia's decreased by 3, and the sum of Diana's and El's seat numbers did not change. At the same time, the total sum of the seats did not change, so Anya's seat number must have increased by 2. ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,540 |
8.4. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved one seat to the right, Galia had moved two seats to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. Wh... | # Answer: 4
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 1, Galia's decreased by 2, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have increase... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,541 |
1. A box and 100 identical aluminum balls inside it weigh 510 g. The same box and 100 identical plastic balls inside it weigh 490 g. How much will the box and 20 aluminum balls and 80 plastic balls inside it weigh? | Solution. Method 1. Let's take 4 boxes with plastic balls and one with aluminum ones. Their total weight is 4*490+510=2970 (g). Redistribute the balls so that each box contains 20 aluminum and 80 plastic balls. Then the weight of all boxes is the same, and the weight of one box with balls is $2970: 5=494(g)$.
Method 2... | 494() | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,542 |
3. Misha calculated the products $1 \times 2, 2 \times 3$, $3 \times 4, \ldots, 2017 \times 2018$. For how many of them is the last digit zero? | Solution. The last digit of the product depends on the last digits of the factors. In the sequence of natural numbers, the last digits repeat every ten. In each ten, in the sequence of products, four products end in zero: ... $4 \times \ldots 5, \ldots 5 \times \ldots 6, \ldots 9 \times \ldots 0, \ldots 0 \times \ldots... | 806 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,544 |
4. There are 17 button-lamps arranged in a circle. Initially, all the lamps are on. With one press of a button, the state of the pressed button-lamp and its neighbors changes (from on to off and vice versa). Is it possible to turn off all the button-lamps with such operations? | Solution. After seventeen single presses on each button-lamp, they will all change their state three times, which means they will not be lit. Answer: possible.
Criteria. Any correct solution: 7 points.
The correct pressing algorithm is indicated, but it is not justified why it leads to the goal: 6 points.
It is stat... | possible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,545 |
5. In the pantry, Winnie-the-Pooh keeps 11 pots, seven of which contain jam, and four contain honey. All the pots are lined up, and Winnie remembers that the pots with honey are standing together. What is the minimum number of pots Winnie-the-Pooh needs to check to find a pot with honey? | Answer: one.
Solution. Let's number the pots from 1 to 11 in the order of their arrangement in a row. Exactly one of the pots numbered 4 and 8 contains honey. Therefore, it is sufficient to check one of them. It is impossible to identify the pot with honey without checking, as any pot may contain either honey or jam.
... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,546 |
114. Do there exist distinct natural numbers $x, y, z$ satisfying the equation $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=3$? a) distinct natural numbers $x, y, z$? b) distinct integers $x, y, z$? | Answer a) do not exist, b) exist. Hint a) From the inequality of means for three positive numbers, it follows that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3 \sqrt{\frac{x}{y} \frac{y}{z} \frac{z}{x}}=3$, and equality is achieved only in the case of equality of the numbers $\frac{x}{y}, \frac{y}{z}, \frac{z}{x}$, i.e.... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,547 |
2. Gena encrypted a sequence of five integers: B, AN, AX, NO, FF. Here, different digits are denoted by different letters, and the same digits by the same letters; a comma separates adjacent numbers (and in “NO” the second character is the letter “O”). Gena forgot the encrypted numbers but remembers that the difference... | Answer: $5,12,19,26,33$.
Solution. All these numbers can be determined if we know the first number and the difference $d$ between two adjacent numbers. The first digits of the second and third numbers coincide, meaning they are in the same decade, and their difference, equal to $d$, does not exceed 9. Therefore, by ad... | 5,12,19,26,33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,549 |
3. Kolya says that two chocolate bars are more expensive than five gums, Sasha - that three chocolate bars are more expensive than eight gums. When this was checked, only one of them was right. Is it true that 7 chocolate bars are more expensive than 19 gums? Don't forget to justify your answer. | Solution. Let the price of a chocolate bar be $c$, and the price of a gum be $g$. Kolya says that $2 c>5 g$ or $6 c>15 g$, while Sasha says that $3 c>8 g$ or $6 c>16 g$. If Sasha were right, then Kolya would also be right, which contradicts the condition. Therefore, Kolya is right, but not Sasha, and in fact $3 c \leqs... | false | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,550 |
5. There is a $6 \times 6$ square, all cells of which are white. In one move, it is allowed to change the color of both cells in any domino (a rectangle of two cells) to the opposite. What is the minimum number of moves required to obtain a square with a checkerboard pattern? Don't forget to explain why a smaller numbe... | Answer: 18 moves.
Solution. Note that in the chessboard coloring of a $6 \times 6$ square, there are 18 black cells. At the same time, no two of them can be turned black in one move, because they are not covered by one domino. Therefore, at least 18 moves are required (at least one move per cell). This can be done in ... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,552 |
9.1. Given positive numbers $p$ and $r$. Let $f(x)$ and $g(x)$ be linear functions with roots $p$ and $r$. Find all roots of the equation $f(x) g(x)=f(0) g(0)$. | Answer. $x_{1}=0, x_{2}=p+r$.
First solution. Let the given functions be of the form: $f(x)=a x+b$ and $g(x)=c x+d$. Then the equation becomes $(a x+b)(c x+d)-b d=0$, that is, $x(a c x+a d+b c)=0$. One root of this equation is $x_{1}=0$, and the second is $x_{2}=\frac{-a d-b c}{a c}$, which means $x_{2}=-\frac{d}{c}-\... | x_{1}=0,x_{2}=p+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,553 |
9.2. During breaks, schoolchildren played table tennis. Any two schoolchildren played no more than one game with each other. By the end of the week, it turned out that Petya played half, Kolya - a third, and Vasya - a fifth of the total number of games played during the week. How many games could have been played durin... | Answer: 30.
Solution: From the condition, it follows that half, a third, and a fifth of the total number of games played are integers. Since the least common multiple of the denominators - the numbers 2, 3, 5 - is 30, the total number of games played is also a multiple of 30. Let this number be $30p$. Then, Pete playe... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,554 |
9.3. Tangents $A B$ and $A C$ are drawn from point $A$ to a circle with center $O$ (points $B$ and $C$ are the points of tangency). Let $M$ be the midpoint of segment $A O$. Prove that the circle circumscribed around triangle $A B M$ is tangent to the line $A C$. | Solution. The statement of the problem is equivalent to the angle $OAC$ being equal to the inscribed angle $ABM$ (see Fig. 4). But the radius $OB$ is perpendicular to the tangent $AB$, so $BM$ is the median of the right triangle $OBA$. Therefore, $BM = \frac{AO}{2} = AM$, and thus $\angle ABM = \angle BAM = \angle BAO$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,555 |
9.4. To the number $A$, consisting of eight non-zero digits, a seven-digit number, consisting of identical digits, was added, and the eight-digit number $B$ was obtained. It turned out that the number $B$ can be obtained from the number $A$ by rearranging some of the digits. What digit can the number $A$ start with if ... | Answer: 5.
Solution: Since the numbers $A$ and $B$ have the same sum of digits, their difference is divisible by 9. Therefore, the added seven-digit number with identical digits is divisible by 9. This means it consists of nines. That is, we can consider that $10^7$ was added to the number $A$ and 1 was subtracted. Th... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,556 |
9.5. On a checkerboard of size $8 \times 8$, 8 checkerboard ships of size $1 \times 3$ are placed such that no two cells occupied by different ships share any points. One shot is allowed to pierce all 8 cells of one row or one column. What is the minimum number of shots needed to guarantee hitting at least one ship? | Answer: 2 shots.
Solution. We will make 2 shots as shown in Fig. 5. Suppose we did not hit any ship. Then there are no ships in area 1. In each of areas 2 and 3, there is no more than 1 ship. Therefore, there are at least 6 ships in area 4.
Area 4 is a $5 \times 5$ square. Then in this area, horizontally placed ships... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,557 |
1. Prove that for any natural numbers $x$ and $y$, the number $2022 x^{2}+349 x+72 x y+12 y+2$ is composite. | Solution.
$2022 x^{2}+349 x+72 x y+12 y+2=\underline{6 \cdot 337 x^{2}}+337 x+\underline{12 x}+\underline{6 \cdot 12 x y}+12 y+2=6 x \cdot(337 x+$ $12 y+2)+337 x+12 y+2=(6 x+1)(337 x+12 y+2)$. Since $x$ and $y$ are natural numbers, both brackets are greater than 1. Therefore, the number is composite. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,558 |
2. If you take three different digits, form all six possible two-digit numbers using two different digits, and add these numbers, the result is 462. Find these digits. Provide all variants and prove that there are no others. | Solution. Let these three digits be denoted by $a, b, c$. We obtain six numbers $10 a+b, 10 b+ a, 10 c+b, 10 b+c, 10 a+c, 10 c+a$. The sum of these numbers is $22(a+b+c)$. According to the condition, $a+b+c=462: 22=21$. By trial, we find the sets of digits $(6,7,8),(4,8,9),(5,7,9)$. These sets can be obtained, for exam... | (6,7,8),(4,8,9),(5,7,9) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,559 |
3. Petya says that on the way from home to school, he walked half the distance at a speed of 4 km/h, and half the time at a speed of 5 km/h. Could Petya's story be true? Provide an example or prove that it cannot be. | Solution: Let $S$ be the distance that Petya has traveled, and $T$ be the total time it took him to travel the entire distance $S$. Consider Petya's first statement. From the condition, it follows that Petya walked at a speed of 4 km/h for no more than half of the time. The time it took Petya to walk half the distance ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,560 |
4. A tea set consists of six identical cups and six identical saucers. Six different sets were packed into two boxes, all saucers in one box, and all cups in the other. All items are wrapped in opaque paper and are indistinguishable by touch. Find the minimum number of items that need to be taken out of these boxes to ... | Solution. Explanation by example. 18 cups and 12 saucers may not be enough. Let's number the sets. By randomly picking 18 cups, we might get all the cups from sets №1, 2, 3.
By randomly picking 12 saucers, we might get all the saucers from sets №5, 6. It is impossible to form a cup-saucer pair. Adding one more saucer ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,561 |
5. In triangle $ABC$, $AM$ and $CN$ are angle bisectors. Point $X$ is chosen on the angle bisector $AM$ such that the distance $BX$ is the smallest possible. Point $Y$ is chosen on the angle bisector $CN$ such that the distance $BY$ is the smallest possible. Prove that lines $AC$ and $XY$ are parallel. | Solution. From the condition, we immediately obtain that $B X$ and $B Y$ are perpendiculars drawn to the angle bisectors. Extend these perpendiculars to intersect the line $A C$ at points $F$ and $E$ respectively. In triangle $A B F$, segment $A X$ is both a bisector and an altitude, so triangle $A B F$ is isosceles, w... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,562 |
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