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6. How many numbers from 1 to 999 are written in the Roman numeral system with the same number of symbols as in the decimal system?
(P. D. Mulyenko)
Reference. To write a number in Roman numerals, you need to break it down into place value addends, write each place value addend according to the table, and then write them sequentially from the largest to the smallest. For example, let's write the number 899. According to the table, $800 = \text{DCCC}$, $90 = \text{XC}$, $9 = \text{IX}$, we get DCCCXCIX.
| 1 I | $10 \text{X}$ | $100 \text{C}$ | $1000 \text{M}$ |
| :--- | :--- | :--- | :--- |
| $2 \text{II}$ | $20 \text{XX}$ | $200 \text{CC}$ | $2000 \text{MM}$ |
| 3 III | $30 \text{XXX}$ | $300 \text{CCC}$ | $3000 \text{MMM}$ |
| $4 \text{IV}$ | $40 \text{XL}$ | $400 \text{CD}$ | |
| 5 V | $50 \text{L}$ | $500 \text{D}$ | |
| 6 VI | $60 \text{LX}$ | $600 \text{DC}$ | |
| $7 \text{VII}$ | $70 \text{LXX}$ | $700 \text{DCC}$ | |
| 8 VIII | $80 \text{LXXX}$ | $800 \text{DCCC}$ | |
| $9 \text{IX}$ | $90 \text{XC}$ | $900 \text{CM}$ | |
| | | | | | Solution. Note that, regardless of the digit place and other digits of the number, a decimal digit a is written as:
- zero Roman numerals when $a=0$,
- one Roman numeral when $a=1$ and $a=5$,
- two Roman numerals when $a$ is $2,4,6,9$,
- three Roman numerals when $a=3$ and $a=7$,
- four Roman numerals when $a=8$.
Thus, for a number to meet the condition, it must:
1) either consist only of the digits 1 and 5,
2) or contain a pair of digits 0 and $x$, where $x$ is one of the digits $2,4,6$ or 9, and possibly one digit 1 or 5,
3) or consist of one digit 3 or 7 and two zeros.
There are a total of $2+2 \cdot 2+2 \cdot 2 \cdot 2=14$ numbers of the first type and 2 numbers of the third type (DCC and CCC). There are exactly 4 two-digit numbers of the second type, and 32 three-digit numbers $-2 \cdot 2 \cdot 4 \cdot 2=32$ (2 ways to place the digit "0", two ways to choose the position for the single-character digit, 4 ways to write the two-character digit, and 2 ways to write the single-character digit). In total, $14+$ $2+4+32=52$ numbers.
Criteria. Correct answer without justification - 2 points.
Each of the missing cases described above --2 points.
Arithmetic error in each of the cases --1 point. | 52 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Milla and Zhena came up with a number each and wrote down all the natural divisors of their numbers on the board. Milla wrote down 10 numbers, Zhena wrote down 9 numbers, and the largest number written on the board twice is 50. How many different numbers are written on the board? | Solution. Note that a number written twice is a common divisor of the original numbers; the largest such number is their GCD. Therefore, all numbers written twice are divisors of the number 50, that is, the numbers $1,2,5,10,25,50$. Thus, among the listed numbers, exactly 6 are repeated, and the number of different numbers is $10+9-6=13$.
Criteria.
An example of a 7-point solution: "The numbers 1,2,5,10,25,50 are written twice, and all other numbers are written once; in total, there are 13 different numbers." That is, we do not require justification that non-divisors of the number 50 cannot repeat, or even an explicit mention of this fact. We do not need to prove that the number 50 has no other divisors. Providing an example confirming the existence of such numbers is not required.
Errors. It is stated that the number 50 has six divisors, but it is not explained why there are six - minus one point.
One divisor of the number 50 is not accounted for - minus one point, two divisors are not accounted for - minus two points, more than two are not accounted for - the problem is considered unsolved. Similarly for extra divisors.
The person thinks that "different numbers" are only those that appear once (then the answer is 7) — minus one point.
Progress. It is stated that the divisors of the number 50 are repeated, but the divisors are not counted or there are more than two errors in the count - 2 points. If the word "divisors" is missing (for example, "1 and 50 are repeated, so there are 17 different numbers") — 0 points.
Only the answer $13-1$ point. | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. On graph paper, a polygon with a perimeter of 36 is drawn, with its sides running along the grid lines. What is the maximum area it can have? | Solution. Consider the extreme verticals and horizontals. Moving from them inward does not allow reducing the perimeter, but it decreases the area. Therefore, the rectangle has the largest area. If A and B are the lengths of its sides, then A + B = 18.
By trying different rectangles with a perimeter of 36, such as (1,17), (2,16), etc., we find that the maximum value of the area AB is achieved when A = B = 9, and this area is 81.
## Criteria.
A complete solution consists of two parts: (1) proving that rectangles are better than other shapes and (2) proving that the best rectangle is a square.
If (1) is stated but not proven, the problem is not solved (max. 2 points). If (2) is stated but not proven, the problem is solved with a deficiency (5 points). As proof, it is sufficient to mention what needs to be tried.
Only the answer (or the answer with the indication that it is obtained for a square) — 1 point. | 81 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The eldest received the most gold - 25 kg - and one eighth of all the silver. How much gold was in the treasure? | Solution. 1) The elder brother received 75 kg of silver, which is one-eighth of the total amount; therefore, the total mass of silver is 600 kg.
2) The others received more silver than the elder brother, i.e., each received more than 75 kg. If there are at least eight brothers, then in total they would receive more than 600 kg; therefore, there are no more than seven brothers.
3) But there must be at least seven of them, since the mass of the silver is 600 kg, which means the total mass of the treasure is more than 600 kg. Therefore, there are seven brothers.
4) The total mass of the treasure is 700 kg, so the mass of the gold is $700-600=100$ kg.
Criteria. The same as in problem 5.6. | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all the cars meet for the first time? | Solution. $A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 371st minute, car $D$ will be at the same point as the other three cars.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points. | 371 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The park is a $10 \times 10$ grid of cells. A lamp can be placed in any cell (but no more than one lamp per cell).
a) The park is called illuminated if, no matter which cell a visitor is in, there is a $3 \times 3$ square of 9 cells that contains both the visitor and at least one lamp. What is the minimum number of lamps in an illuminated park?
b) The park is called reliably illuminated if it remains illuminated even after the failure of any one lamp. What is the minimum number of lamps in a reliably illuminated park? | Solution. a) 4. Divide the park into 4 quarters (squares $5 \times 5$), then there must be at least one lamp in each quarter (to illuminate, for example, the corner cells). By placing one lamp in the center of each quarter, we get an example.
b) 10.
Estimate. In each corner square $3 \times 3$ there must be at least two lamps (to illuminate the corner cell). Temporarily leave only these 8 lamps. Each of them illuminates only within its own quarter, and if the lamp in the center of the quarter breaks (or if it is absent), then a five-cell strip inside this quarter, adjacent to another quarter, will definitely not be illuminated. Note that the union of two such strips for opposite quarters cannot be illuminated by one lamp in any case, so at least two more lamps are needed.
Example:
Criteria. In part a) 1 point is given for the estimate and 1 point for the example. In part b) 3 points are given for the estimate (1 point if it is proven that 8 lamps are insufficient), and 2 points for the example. | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Let's call the efficiency of a natural number $n$ the fraction of all natural numbers from 1 to $n$ inclusive that have a common divisor with $n$ greater than 1. For example, the efficiency of the number 6 is $\frac{2}{3}$.
a) Does there exist a number with an efficiency greater than $80\%$? If so, find the smallest such number.
b) Does there exist a number with the maximum efficiency (i.e., not less than the efficiency of any other number)? If so, find the smallest such number. | Solution. Let's move on to studying inefficiency (1 minus efficiency). From the formula for Euler's function, it follows that it is equal to $\frac{p_{1}-1}{p_{1}} \cdot \ldots \frac{p_{k}-1}{p_{k}}$, where $p_{1}, \ldots, p_{k}$ are all distinct prime divisors of $n$. Then, by adding a new prime factor, we can increase the efficiency, so the answer in part (b) is "no".
The smallest number with an efficiency greater than $80 \%$ is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13=30030$. Its efficiency is $1-\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \frac{10}{11} \cdot \frac{12}{13} =\frac{809}{1001}$. We will prove that it is more efficient than all smaller numbers. Indeed, the presence of a prime factor with a power higher than the first does not affect the efficiency, so the sought number has all factors in the first power. If the factors are not consecutive primes, then by replacing one of the prime numbers with a smaller one, the efficiency will increase. Therefore, "efficiency records" can only be set by numbers of the form "the product of the first few primes"; but the efficiency of the number $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ is too low.
Criteria. Part (a) is worth 5 points, part (b) - 2 points. | 30030 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Since a syllable consists of two different letters, identical letters can only appear at the junction of syllables.
First, let's find the number of combinations of two syllables with a matching letter at the junction. Such syllables (in terms of the arrangement of vowels and consonants) are either AMMO $(3 \cdot 8 \cdot 3$ variants) or MAAN $(8 \cdot 3 \cdot 8$ variants), totaling 264 variants.
From each such combination, a funny word can be formed in two ways - by adding an arbitrary syllable either at the beginning or at the end. Since the language has 48 syllables $(8 \ldots 3=24$ syllables of the form MA and another 24 syllables of the form AM), each of these methods yields $264 \cdot 48$ words.
However, some words are counted twice. These are words where the letters at the junction of the first syllable with the second and the letters at the junction of the second with the third syllable match. Clearly, all such words have the form AMMOON or MAANNO, and their number is $3 \cdot 8 \cdot 3 \cdot 8 + 8 \cdot 3 \cdot 8 \cdot 3 = 2 \cdot 24^{2}$. | Answer: $2 \cdot 264 \cdot 48-2 \cdot 24^{2}=24192$ funny words. | 24192 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Once Valera left home, walked to the cottage, painted 11 fence boards there, and returned home 2 hours after leaving. Another time, Valera and Olga went to the cottage together, painted 8 fence boards (without helping or hindering each other), and returned home together 3 hours after leaving. How many boards will Olga manage to paint alone if she needs to return home 1.5 hours after leaving? The physical abilities, diligence, and working conditions of Valera and Olga remain constant. (V. P. Fedorov) | Solution. This is a more complex version of problem 5 for 5th grade.
The strange result (that the characters accomplished less work in more time when working together) is explained by the different times spent walking, since the speed of "joint" walking is equal to the lower of the two walkers' speeds. The second time, Valera worked no more than $2 \cdot \frac{8}{11}$ hours, so they must have spent at least $3-\frac{16}{11}=\frac{17}{11}>1.5$ hours on the journey. This means that in one and a half hours, Olga would not have enough time to even walk to the cottage and back.
Answer: 0 boards.
Scoring. Only the answer - 0 points.
Evaluated progressions:
1) Understanding that two people walk at the speed of the slower one (this is stated explicitly or used in the solution) - 1 point.
2) With this in mind, it is indicated that Olga spends more than an hour on the journey (or at least an hour more than Valera, or something similar) - another +2 points.
3) It is calculated that in the second case, Valera would have completed the work alone in 16/11 hours (or that he would have completed it in less than 1.5 hours) - 2 points (added to progressions 1 and 2).
For example, the solution "In the second case, they spent an hour longer walking, so Olga spends an hour more on the journey than Valera, so in the third case, she will have a maximum of half an hour to work, and it's unclear what to do next" earns 3 points (progressions 1 and 2).
A solution like "In the second case, Valera would have worked alone for no more than $2 \cdot 8 / 11$ hours, and since it took them longer, Olga must have been in the way; so, alone, Olga would paint nothing (or a negative number of boards)" earns 2 points (only progression 3). | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a rectangular grid $20210 \times 1505$, two diagonals were drawn and all cells through which they passed were painted. How many cells were painted?
(O. A. Pyayve, A. A. Tessler) | Solution. This is a more complex version of problem 3 for 5th grade.
First, let's determine how many cells one diagonal crosses. Note that $20210=215$. $94,1505=215 \cdot 7$. Therefore, the diagonal passes through 215 rectangles of size $94 \times 7$. In each of these rectangles, it crosses 93 vertical and 6 horizontal grid lines (at non-coincident points), and 99 intersection points divide it into 100 segments, meaning it crosses 100 cells. In total, this results in 21500 cells.
Two diagonals should cross 43000 cells, but among these cells, there are overlapping ones (near the center of the rectangle) that are counted twice. Let's find out how many there are. The larger coordinate of the center is an integer, and the smaller one is a half-integer. The ratio of the sides of the rectangle is $94: 7$, so $x=\frac{1}{2} \cdot \frac{94}{7} \in (6 ; 7)$ (see the diagram). Therefore, the number of common cells is 14.
Answer: 42986.
Criteria. If the number of cells each diagonal crosses is correctly calculated but the overlap is not considered, -3 points. If, in addition to this, the "center" is considered but incorrectly, -4 points. If the GCD of the numbers is found without further progress, 1 point. | 42986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On a plane, an equilateral triangle and three circles with centers at its vertices are drawn, and the radius of each circle is less than the height of the triangle. A point on the plane is painted yellow if it lies inside exactly one of the circles; green if inside exactly two; and blue if inside all three. It turns out that the yellow area is 1000, the green area is 100, and the blue area is 1. Find the area of the triangle. (P. D. Mulyenko)
 | Solution. The sum of the areas of the three circles is $1000+2 \cdot 100+3 \cdot 1=1203$; the sum of the areas of the three "lenses" is $100+3 \cdot 1=103$ (a "lens" is the intersection of two circles).
The area of the triangle is $S_{1}-S_{2}+S_{3}$, where
$S_{1}=1203 / 6-$ the sum of the areas of the three 60-degree sectors,
$S_{2}=103 / 2$ - the sum of the areas of the halves of the three "lenses" lying inside the triangle;
$S_{3}=1-$ the area of the blue region.
Indeed, with this calculation, each yellow region inside the triangle is counted 1 time, each green region: $2-1=1$ time, the blue region: $3-3+1=1$ time.
In total, we get $\frac{1203}{6}-\frac{103}{2}+1=\frac{1203-309+6}{6}=\frac{900}{6}=150$.
Answer: 150.
Criteria. Up to 3 points will be deducted for errors in the inclusion-exclusion formula. | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. How many five-digit numbers are roots of the equation $x=[\sqrt{x}+1][\sqrt{x}]$? The symbol $[a]$ denotes the integer part of the number $a$, that is, the greatest integer not exceeding $a$.
(0. A. Pyayve) | Solution. Let $n=[\sqrt{x}]$, then $[\sqrt{x}+1]=[\sqrt{x}]+1=n+1$, which means $x=n(n+1)$.
All numbers of the form $x=n(n+1)$ are suitable, since for them $n<\sqrt{x}<n+1$, meaning $[\sqrt{x}]$ is indeed equal to $n$.
It remains to count the five-digit numbers of this form. Note that 99$\cdot$100 $<10000<100 \cdot 101$, $315 \cdot 316<100000<316 \cdot 317$, so $x$ is a five-digit number for $n$ from 100 to 315 inclusive.
Answer: 216 numbers.
Criteria. No less than 3 points if the participant is looking for numbers of the form $n(n+1)$; no less than 4 points if he found the number 315 as the "upper bound".
5 points - the answer differs from the correct one by 1 in either direction.
6 points - the correct solution and answer, but with a vague explanation: for example, it is not proven (or even mentioned) that all numbers of the form $n(n+1)$ indeed fit. | 216 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Given a rectangle of size $2021 \times 4300$. Inside it, there is a billiard ball. It is launched in a straight line, forming a $45^{\circ}$ angle with the sides of the rectangle. Upon reaching a side, the ball reflects at a $45^{\circ}$ angle; if the ball hits a corner, it exits along the same line it entered. (An example of the beginning of the ball's path is shown in the figure.)
a) Is it true for any point that if a ball is launched from it according to these rules, it will return to that point again?
b) Suppose that starting from some point $A$, the ball returns to it after some time. What is the maximum number of bounces off the sides it can make before returning to point $A$ for the first time?
(O. A. Plyve)
 | # Solution.
We will replicate the rectangle multiple times by reflecting it relative to its sides. In adjacent columns (rows), copies of the rectangle will be oriented differently, but when shifted by an even number of columns and an even number of rows, the orientation will match the initial one. Now, we can consider that when the ball touches the edge, it transitions to the next copy of the rectangle but continues along the same straight line. We are interested in the moment when the ball reaches one of the copies of the initial point.
a) Let's introduce a coordinate system such that one of the corners of the board has coordinates \((0,0)\), the horizontal side is shorter than the vertical side, and at the beginning of the ball's movement
both coordinates increase.
If the coordinates of the initial point are \(\left(x_{0}, y_{0}\right)\), then any point on the path has coordinates \(\left(x_{0}+a, y_{0}+a\right)\). A shift by \(a=\operatorname{LCM}(2021 \cdot 2, 4300 \cdot 2)\) along each coordinate brings us to a copy of the initial point, since we have moved an even number of sides of the rectangle in each direction.
b) Note that \(2021 = 43 \cdot 47\) and \(4300 = 43 \cdot 100\), so \(\operatorname{LCM}(2021 \cdot 2, 4300 \cdot 2) = 2 \cdot 43 \cdot 47 \cdot 100\). A shift by \(a = 2 \cdot 43 \cdot 47 \cdot 100\) along each coordinate corresponds to a shift of \(2 \cdot 100\) sides of length 2021 to the right and \(2 \cdot 47\) sides of length 4300 upwards. This means that a total shift of 294 rectangles to the right and upwards will bring us to a copy of the initial point. Each transition to the next rectangle corresponds to one bounce of the ball, so the coincidence will occur after 294 bounces.
Note that a coincidence with a copy of the initial point in a rectangle oriented as the initial one cannot occur earlier, since \(a\) must divide both \(2 \cdot 2021\) and \(2 \cdot 4300\). However, the number of bounces before the first coincidence can be less for one of the following two reasons:
- the trajectory passes through a corner (then two bounces "merge" into one);
- the ball hits the initial point in one of the copies oriented differently than the initial rectangle.
Let's provide an example where this does not happen: let \(x_{0} = \sqrt{3}\) and \(y_{0} = \sqrt{2}\). The trajectory will not pass through a corner, since for this \(x - y\) must be an integer, while it is always \(\sqrt{3} - \sqrt{2}\). Let's study the question of copies of the point in "reflected" rectangles. The coordinates of copies of the initial points always take the form \(\left(2021 k \pm x_{0}, 4300 l \pm y_{0}\right)\), where the choice of signs depends on the parity of \(k\) and \(l\), and in "reflected" rectangles at least one of the signs is minus. Suppose, for example, \(x_{0} + a = 2021 k - x_{0}\), then \((a + 2 \sqrt{2})\) is an integer. Depending on the sign in the second coordinate, we get that \(a\) or \((a + 2 \sqrt{3})\) is an integer, but this is impossible.
Answer: a) yes b) 294 bounces.
Criteria. Part a - 2 points, part b - 5 points. Correct answer to part b - 1 point. | 294 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the country, there are 100 cities, and several non-stop air routes are in operation between them, such that one can travel from any city to any other, possibly with layovers. For each pair of cities, the minimum number of flights required to travel from one to the other was calculated. The transportation difficulty of the country is defined as the sum of the squares of these 4950 numbers. What is the maximum value that the transportation difficulty can take? The answer should be given as a number (in decimal notation). | Solution. First, we prove that the maximum transportation difficulty occurs when the cities are connected "in a chain."
Indeed, we can consider the graph as a tree (otherwise, we can remove some edges to leave a tree - the difficulty will increase). Choose the longest path in the tree. Suppose there are vertices not included in this path. Then among them, there is a pendant vertex $x$ (i.e., a vertex with only one edge leading to it). Let $y$ be the vertex on the longest path closest to $x$. Move the entire "branch" extending from $y$ and containing $x$ to the end of the longest path, the farthest from $y$. Note that as a result, the pairwise distances between the vertices within the "branch" do not change, nor do the distances between the vertices outside the branch. It is easy to see that for each vertex in the branch, the sum of the squares of the distances to all other vertices has increased. Indeed, if $k$ is the distance from some vertex $z$ in the branch to $y$, then the set of distances from $z$ to the vertices outside the branch (including $y$) was $k, k+1, k+1, k+2, k+2, \ldots, k+s, k+s, k+s+1, k+s+2, \ldots$, and it became $k, k+1, k+2, k+3, \ldots$
By such "re-hangings," we can bring the graph to the form of a chain, and the difficulty will always increase; hence, for the chain, it is maximal.
Now we calculate the difficulty for the "chain." The problem reduces to finding the sum $99 \cdot 1^{2} + 98 \cdot 2^{2} + 97 \cdot 3^{2} + \ldots + 1 \cdot 99^{2}$.
Let $S_{n}^{2} = 1^{2} + 2^{2} + \ldots + n^{2}$, $S_{n}^{3} = 1^{3} + 2^{3} + \ldots + n^{3}$. As is known, $S_{n}^{2} = \frac{n(n+1)(2n+1)}{6}$, $S_{n}^{3} = \left(\frac{n(n+1)}{2}\right)^{2}$ (this can be proven by induction).
Notice that the desired sum is equal to
$$
\begin{gathered}
100 \cdot \left(1^{2} + 2^{2} + \ldots + 99^{2}\right) - \left(1 \cdot 1^{2} + 2 \cdot 2^{2} + \ldots + 99 \cdot 99^{2}\right) = 100 S_{99}^{2} - S_{99}^{3} = \\
= 100 \cdot \frac{99 \cdot (99+1) \cdot (2 \cdot 99 + 1)}{6} - \left(\frac{99 \cdot (99+1)}{2}\right)^{2} = 32835000 - 24502500 = 8332500.
\end{gathered}
$$ | 8332500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Marina needs to buy a notebook, a pen, a ruler, a pencil, and an eraser to participate in the Olympiad. If she buys a notebook, a pencil, and an eraser, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. How much money will she need for the entire set, if the notebook costs 15 tugriks? | Solution. If Marina buys two sets from the condition, she will spend $47+58=105$ tugriks, but she will buy an extra notebook, so the full set of school supplies costs $105-15=90$ tugriks.
Criteria. Only the answer without explanation - 1 point. If in the solution they try to determine the cost of the pen and pencil (although it is not stated in the condition that the cost must necessarily be an integer) - 0 points. | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. An accident has occurred in the reactor of a research spacecraft, and toxic substances are leaking from it. All corridors between rooms are equipped with airtight doors, but there is no time to close individual doors. However, the captain can still give the command "Close $N$ doors," after which the ship's artificial intelligence will close $N$ random doors. What is the smallest $N$ so that the entire crew can be guaranteed to survive in the lounge? | Solution. There are a total of 23 corridors on the spaceship. If no more than 21 doors are closed, then the corridors between the reactor and the right engine, and between the right engine and the lounge, may remain open, which means the crew will be in danger. Therefore, at least 22 doors must be closed.
Criteria. Correct answer without justification - 0 points. Error in counting the number of corridors - 4 points.
 | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). He got the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 5? | Solution. The number must be divisible by 5, so the letter "A" is equal to 0 or 5. If it is equal to 0, then for the other letters ("G", "V", "T", "E", "M", "L") there are $A_{9}^{6}=9!/ 3$! options; if "A" is equal to 5, then for the other letters there are $8 \cdot A_{8}^{5}=8$!/3! options, since "G" cannot be zero. In total, $9!/ 6+8 \cdot 8!/ 6=114240$ ways.
Criteria. It is explicitly stated that for the letter "A" there are 2 options: 5 or $0-2$ points. If the two cases are further analyzed, that's another 5 points. If the answer is given as an expression not fully calculated to the end - do not deduct points. | 114240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Three cars $A, B$ and $C$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ drives counterclockwise. All cars move at constant (but pairwise distinct) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all three cars meet for the first time? | Solution. $A$ and $C$ meet every 7 minutes, and $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53=371$ minutes.
Criteria. If 53 minutes is replaced with 46 - 3 points. Solved by trial with lengths and speeds -1 point. Only the answer without explanation - 0 points. | 371 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. There is a rectangular sheet, white on one side and gray on the other. It was folded as shown in the picture. The perimeter of the first rectangle is 20 more than the perimeter of the second rectangle. And the perimeter of the second rectangle is 16 more than the perimeter of the third rectangle. Find the perimeter of the original sheet.
 | Solution. From the figure, it can be seen that when folding, the perimeter of the rectangle decreases by twice the short side, so the short side of rectangle-1 is $20 / 2=10$, the short side of rectangle-2 is $16 / 2=8$. Therefore, the long side of rectangle-1 is 18, and the long side of the original sheet is 28. Thus, the perimeter is: $(28+18) \cdot 2=92$.
Criteria. An answer without justification or found by trial and error - 0 points.
International Mathematical Olympiad
«Formula of Unity» / «The Third Millennium»
Year 2022/2023. Qualifying round
## Problems for grade R5
Each task is assessed at 7 points. Some problems have their own criteria (printed in gray). | 92 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Katya decided to calculate the sum of the cubes of all natural divisors of some natural number, and she got the result $M A T H$. But then she discovered that she had forgotten one of the divisors. Adding its cube, she got the correct result - MASS. Find the smallest possible value of the number $M A T H$. (MATH and $M A S S$ are four-digit numbers, where each digit is replaced by a letter, with the same digits replaced by the same letters, and different digits by different letters.) | Solution. Answer: 2017. The original natural number is $12 ; 12^{3}+6^{3}+4^{3}+2^{3}+1^{3}=$ 2017; if you add $3^{3}$, you get 2044.
We will prove that there are no smaller suitable numbers.
1) For any number less than 10, the sum of the cubes of all divisors, as is easily verified, is less than a thousand.
2) $10^{3}+5^{3}+2^{3}+1^{3}=1134$ and $11^{3}+1^{3}=1332$ cannot equal the number $M A S S$ (the last two digits are not equal).
3) For the number $12 M A S S=2044$; if you subtract a cube less than 27 from it, the result will be greater than 2017, and if a larger one, the result will already be less than 2000.
4) For $n \geqslant 13$ we get: $M A S S \geqslant n^{3} \geqslant 2197$, and therefore, $M A T H \geqslant 2100$. | 2017 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Several plants and zombies (no more than 20 creatures in total) attended the "Plants VS Zombies" gathering, and it turned out that all creatures were of different heights. Plants always tell the truth to those who are shorter than them and lie to those who are taller. Zombies, on the contrary, lie to shorter creatures and tell the truth to taller ones. Upon meeting, each participant approached every other and said either "I am taller than you" or "I am shorter." The phrase "I am shorter" was heard 20 times. When saying goodbye, each had to approach every other and say "I am taller and I am a plant." If a creature could not say this, it clapped. There were 18 claps. Determine how many creatures attended the gathering, and arrange them by height.
(P. D. Mulyenko) | Solution. Let the total number of beings be $n$, and exactly $z$ of them are zombies. When plants greet, they say to everyone “I am taller than you”, and zombies say to everyone “I am shorter”. Each zombie said this phrase to everyone except themselves, so we get $z(n-1)=20$. Given that $n-1<20$, the possible cases are: $z=2, n-1=10 ; \quad z=4, n-1=5 ; \quad z=5, n-1=4$ (larger $z$ are impossible since $z \leqslant n$).
Now let's examine the farewells. When a plant addresses someone who is shorter, it must tell the truth, and the phrase “I am taller and I am a plant” is true. When addressing someone who is taller, it must lie, and this phrase is a lie (the first part is false). For zombies, this phrase is always a lie, so they can say it to everyone who is shorter, but not to those who are taller. Therefore, all the slaps are performed by zombies to those who are taller. It is not hard to see that of the three cases, only $z=2, n=11$ fits (otherwise, there would be fewer than 18 slaps). Out of 11 beings, the zombies must be the shortest and the third from the bottom in height - only then do we get $10+8=18$ slaps (in other cases, it is either $10+9$ or no more than 17).
Answer: 11 beings; when ordered by increasing height - ZPZPPPPPPP.
Criteria. 3 points if the participant could determine that the number of zombies can be 2, 4, or 5.
Deduct 1 point if the arrangement of beings by height is not explicitly indicated in the solution, but everything else is explained. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. There are 28 students in the class. On March 8th, each boy gave each girl one flower - a tulip, a rose, or a daffodil. How many roses were given if it is known that there were 4 times as many roses as daffodils, but 10 times fewer than tulips? (A. A. Tesler) | Solution. This is a more complex version of problem 4 for 5th grade.
Let the number of narcissus be $x$, then the number of roses is $4x$, and the number of tulips is $40x$, so the total number of flowers is $45x$. The number of flowers is the product of the number of boys and the number of girls. If there are $m$ boys, then $m(28-m)$ is divisible by 45. Both factors cannot be divisible by 3, so one of them is divisible by 9, and the other by 5. The only option is $-18 \cdot 10$. Therefore, $45x = 180$, which means $x = 4$, and the required number of roses is $4x = 16$.
Answer: 16 roses.
Criteria. An incomplete enumerative solution - no more than 4 points. | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The magical clock, in addition to the usual pair of hands, has a second pair that is symmetrical to the first at every moment relative to the vertical axis. It is impossible to determine which hands are real from a photograph of the clock. Furthermore, just like with ordinary clocks, it is impossible to distinguish morning from evening using the magical clock. Therefore, the same photograph of the clock can correspond to several different times (for example, 1:15, $10:45$, and $22:45$ look the same as shown on the right).
A robot takes several photographs of the clock over the course of a day (from 0:00 to 24:00). It remembers the order in which the photographs were taken but not the time they were taken. Sometimes, from such a series of photographs, it is possible to determine the exact time some of them were taken; such photographs will be called definite. If, however, there are several moments when a photograph could have been taken (even considering the other photographs in the series), it is called indefinite. For example, in the series of photographs shown on the right, photograph №2 is definite (it was taken at 9:00), while photograph №4 is indefinite (it could have been taken at either $16:00$ or 20:00).
Suppose there is a series of 100 photographs taken over the course of a day, none of which look the same, and none of which were taken at 0:00, $6:00$, $12:00$, $18:00$, or 24:00. What is the minimum number of indefinite photographs that can be among them?
(A. A. Tsel)
| Solution. See the solution to problem 6 for 10th grade.
Instead of the last paragraph of the solution to problem 6 for 10th grade, it is sufficient to provide an example with exactly three undefined photographs and prove that there are exactly three. For example, let the first three photos be taken at 5:50, 11:40, 17:30, and the rest in the interval from 20:00 to 23:00. Then, from the appearance of the photographs, it can be determined that the first one was taken no earlier than 5:50, which means the second one was taken no earlier than 11:40, and the third one no earlier than 17:30, and finally, the rest after 18:00, where only one option remains for them, meaning they are defined.
## Answer: 3.
Criteria. 1 point for providing an example with exactly 3 undefined photographs, +1 additional point if it is proven that the example is correct.
1 point for stating the proposition "each photograph corresponds to four possible times: $t, 12-t, 12+t, 24-t$." | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}=\frac{32}{y^{5}}+\frac{48}{y^{3}}+\frac{17}{y}-15 \\
\frac{1}{y}=\frac{32}{z^{5}}+\frac{48}{z^{3}}+\frac{17}{z}-15 \\
\frac{1}{z}=\frac{32}{x^{5}}+\frac{48}{x^{3}}+\frac{17}{x}-15
\end{array}\right.
$$
(A. B. Vladimirov) | Solution. Let $F(t)=32 t^{5}+48 t^{3}+17 t-15$. Then the system has the form $F\left(\frac{1}{y}\right)=\frac{1}{x}, F\left(\frac{1}{z}\right)=\frac{1}{y}$, $F\left(\frac{1}{x}\right)=\frac{1}{z}$. From this, it follows that $F\left(F\left(F\left(\frac{1}{x}\right)\right)\right)=\frac{1}{x}$. Note that $F(0.5)=0.5$ and the function $F(t)-t$ is strictly increasing. Therefore, if $t>0.5$, then $t < F(F(F(t)))$. But this means that $\frac{1}{x}=0.5$, and the original system has a unique solution $x=y=z=2$.
Criteria. 1 point is given for finding the solution. 3 points for the solution under the assumption $x=y=z$. If the uniqueness of the solution to $F(t)=t$ is not proven, 2 points are deducted. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Milla and Zhena came up with a number each and wrote down all the natural divisors of their numbers on the board. Milla wrote 10 numbers, Zhena - 9, and the number 6 was written twice. How many different numbers are on the board? | Solution. Since the number 6 is written twice, both original numbers (denote them as a and b) are divisible by 6.
If Vera's number has 10 divisors, then its factorization is either $p^{9}$ or $p^{1} \cdot q^{4}$ (where p and q are some prime numbers); the first is impossible since it is divisible by 6. Valya's number has 9 divisors, so its factorization is either $\mathrm{s}^{8}$ or $\mathrm{s}^{2} \mathrm{t}^{2}$; again, only the second is possible. In this case, the numbers p and q are 2 and 3 in some order, and the numbers $\mathrm{s}$ and $\mathrm{t}$ are also 2 and 3. It is easy to see that the GCD of such numbers is either $2 \cdot 3^{2}=18$ or $3 \cdot 2^{2}=12$, and in any case has $2 \cdot 3=6$ divisors. Therefore, among the listed numbers, exactly 6 are repeated, and the number of distinct numbers is $10+9-6=13$.
## Criteria.
A complete solution should include: (1) an indication of the form of the GCD factorization, with proof; (2) the calculation of the divisors of this GCD and the derivation of the answer, as in problem 7.1. Providing an example confirming the existence of such numbers is not required.
Errors. A person counts only the numbers that are written once (and gets the answer 7) — minus one point.
If the GCD factorization is correctly found but there are 1-2 errors in counting its divisors — deduct $1-2$ points.
Progress.
If the form of the GCD is correctly and justifiedly found, but then something goes wrong — 3 points can be given.
Reasonable attempts to determine the factorization of each of the original numbers — 1 point.
Incorrect but partially reasonable solutions.
"The numbers $1,2,3,6$ are written twice, so the number of different numbers is $10+9-4=15$ " — 1 point.
An example of suitable numbers is provided and the answer is found for it — 2 points. | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Three people want to travel from city A to city B, which is 45 kilometers away from A. They have two bicycles. The speed of a cyclist is 15 km/h, and the speed of a pedestrian is 5 km/h. What is the minimum time they can reach B, if the bicycle can be left unattended on the road? | Solution. Two people ride a bicycle for 10 kilometers, then one of them leaves the bicycle by the road and walks the next 10 kilometers, while the other continues for the next 10 kilometers and also leaves the bicycle (which the first one should pick up later), the third one walks the first 10 kilometers and then rides the bicycle picked up by the first one for the remaining distance.
In any case, they will collectively ride 60 km and walk 30 km. Therefore, there will be at least one of them who will walk no less than 10 kilometers and ride the rest. If he walks exactly 10 kilometers and rides 20 kilometers, it will take 3 hours and 20 minutes, and any increase in the 10 kilometers will only worsen the result.
Criteria. A complete solution consists of two parts: a description of how the cyclists should ride, and a proof that this result cannot be improved. If only one of these parts is present, the problem is not solved, and we give 2 points. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. On graph paper, a polygon with a perimeter of 2014 is drawn, with its sides running along the grid lines. What is the maximum area it can have? | Solution. First, consider the extreme verticals and horizontals. Moving from them inward does not allow reducing the perimeter, but it decreases the area. Therefore, the rectangle has the largest area. If A and B are the lengths of its sides, then $A + B = 1007$.
Now, among different rectangles with a perimeter of 2014, we need to find the rectangle with the largest value of the area $AB$. Since $4AB = (A + B)^2 - (A - B)^2 = 1007^2 - (A - B)^2$, to achieve the maximum value of the area $AB$, we need to choose A and B so that their difference is as small as possible. Since the sum of A and B is odd, they cannot be equal. Therefore, the smallest possible value of $A - B = 1$. Considering $A + B = 1007$, we find $A = 504, B = 503$, and $AB = \mathbf{253512}$.
## Criteria.
A complete solution consists of two parts: (1) proving that rectangles are better than other shapes and (2) proving that the best rectangle is a square.
If (1) is stated but not proven, the problem is not solved (max. 2 points). If (2) is stated but not proven, the problem is solved with a deficiency (5 points). As proof, the analysis of the quadratic function or a clear reference to the isoperimetric inequality is considered.
Only the answer (or the answer with the indication that it is obtained for a square) - 1 point. | 253512 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. It is known that on Monday, the painter painted twice as slowly as on Tuesday, Wednesday, and Thursday, and on Friday - twice as fast as on these three days, but worked 6 hours instead of 8. On Friday, he painted 300 meters more of the fence than on Monday. How many meters of the fence did the painter paint from Monday to Friday? | Solution. Let's take $100 \%$ of the fence length that the painter painted on Tuesday, Wednesday, and Thursday. Then Monday accounts for $50 \%$, and Friday for $-150 \%$. Therefore, 300 meters of the fence correspond to $150 \% - 50 \% = 100 \%$. For the entire week (from Monday to Friday), it is $500 \%$, i.e., 1500 meters.
Criteria. According to general rules. | 1500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of four-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3. | Solution. The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be 1, 4, 7; if the first is 4, the last can be 2, 5, 8; if the first is 6, the last can be 0, 3, (6 does not fit), 9; if the first is 8, the last can be 1, 4, 7. In total, there are 3 + 3 + 3 + 3 = 12 options for the first and last digits. For each of these options, there are 8 * 7 ways to choose the two middle digits. In total, 56 * 12 = 672 ways.
## Criteria.
Errors. Numbers of the form 6 ** 6 are incorrectly counted - two points are deducted.
Without explanation, it is stated that each first digit corresponds to three variants of the last - one point is deducted (this is not obvious, as it is only due to the impossibility of 6).
Numbers starting with zero are incorrectly counted - two points are deducted.
Arithmetic errors in the correct solution method - one point is deducted for each error.
Advancements. If it is written that the number of pairs (second digit, third digit) is not 7 * 8, but 9 * 9 or 8 * 8, etc., the problem is considered unsolved and no more than 2 points are given. | 672 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In the Olympionov family, it is a tradition to especially celebrate the day when a person turns as many years old as the sum of the digits of their birth year. Kolya Olympionov had such a celebration in 2013, and Tolya Olympionov had one in 2014. Who is older and by how many years? | Solution. Let's determine in which year a person could be born if adding the sum of the digits of their birth year results in 2013 or 2014. It is clear that the year of such a birth cannot be later than 2014. Since the sum of the digits of each number from 1 to 2014 does not exceed \(1+9+9+9=28\), the birth year cannot be earlier than 2013-28=1985. By checking all years from 1985 to 2013, we find that Kolya could have been born in 1992 or 2010, and Tolya in 1988 or 2006.
Therefore, the possible scenarios are: Tolya is 4 years older; Tolya is 22 years older; Kolya is 14 years older.
## Criteria.
Complete solution. The solution should include a check or at least indicate the range of the check. Justifications such as "It is sufficient to check only from 1985 to 2014 because the sum of the digits is no more than 28" are not required - it is enough for the range to include the years from 1988 to 2010.
Errors. If nothing is written about the check, and it is immediately stated, "Kolya could have been born in 1992 or 2010, and Tolya in 1988 or 2006, therefore... " and the correct answer follows, this is considered a solution with an error and is scored 5 points.
Progress. If one of the possible birth years for Kolya and one for Tolya is indicated - 1 point is given. If one variant is found for one brother and both for the other, 2 points are given.
If there are "extra" birth years given, the solution is scored no more than 2 points. A correct (complete) answer without indicating the years they could have been born - 2 points.
Note. Strictly speaking, we found the difference in birth years, but the difference in ages can lie in the range from 3 to 5 years or from 21 to 23 years or from -15 to -13 years. Mentioning this is not required but is not penalized. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Karlson bought several pancakes (25 rubles each) and several jars of honey (340 rubles each) at the buffet. When he told Little Man how much he spent at the buffet, Little Man was able to determine, based on this information alone, how many jars of honey and how many pancakes Karlson bought. Could this amount have exceeded 2000 rubles? | Solution. Could. For example, let's say Karlson spent $4 \cdot 340 + 25 \cdot 40 = 2360$ rubles. Suppose Karlson can make up this amount in some other way; for this, he should spend x rubles less on pancakes and x rubles more on honey (or vice versa). But then, for x rubles, he can buy both a whole number of pancakes and a whole number of pots of honey. This means that x is divisible by both 25 and 340. But the smallest such x is 1700; however, Karlson cannot spend 1700 rubles less on either honey or pancakes.
## Criteria.
Example of a correct solution (7 points). "Yes, it is possible, for example, if Karlson bought 4 jars of honey and 40 pancakes. The number of pancakes can only be changed by 68, and the number of jars of honey - only by 5, but neither is possible." In general, we do not require strict formulations - it is enough to demonstrate an intuitive understanding of what is happening.
Errors. An example is given of what Karlson could buy, but it is not explained why this quantity fits - we give 5 points.
Progress. The author tries to give an example and justify it in the correct way, but the example itself turns out to be unsuitable (for example, due to an arithmetic error) - we give 2 points.
Zero points: only for the answer "could"; for any attempts to prove that it could not. | 2360 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The eldest received the most gold - 30 kg - and one fifth of all the silver. How much gold was in the treasure | Solution. 1) The elder brother received 70 kg of silver, which is one fifth of the total amount; therefore, the total mass of silver is 350 kg.
2) The others received more silver than the elder brother, i.e., each received more than 70 kg. If there are at least five brothers, then in total they would receive more than 350 kg; therefore, there are no more than four brothers.
3) But there must be at least four of them, since the mass of silver exceeds 300 kg. Therefore, there are four brothers.
4) The total mass of the treasure is 400 kg, so the mass of gold is 400-350=50 kg.
## Criteria.
If it is not proven (but simply assumed out of the blue) that there are exactly four brothers, the problem is not solved (no more than 3 points).
Errors. In addition to the correct answer, the participant suggests the variant "five brothers, each received 30 kg of gold" - deduct 1 point.
Progress. "The total mass of silver is 350 kg" - 1 point.
"Brothers are no more than four" with reasoning close to the correct one - 2 points.
Only an example is provided (without proving that the answer cannot be different) — 1 point. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A fly is sitting at vertex $A$ of a triangular room $A B C$ ( $\angle B=60^{\circ}, \angle C=45^{\circ}, A C=5$ m). At some point, it flies out in a random direction, and each time it reaches a wall, it turns $60^{\circ}$ and continues flying in a straight line (see figure). Can it happen that after some time, the fly has flown more than 9.9 meters?
 | Solution. Let the fly take off at an angle of 60 degrees to the line $A C$. Consider the equilateral triangle $A K C$ with side $A C$. Note that its sides $A K$ and $K C$ can be divided into parts (into infinitely many parts) such that each part equals the next segment of the fly's trajectory. The sum of these parts is $A K + K C = 10$ m, so at some point, the fly will have flown more than 10 meters. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). After 12 games, Pasha is 2023 rubles richer than he was initially. How many of these games did he win?
(L. S. Korechkova, A. A. Tessler) | Answer: 9 (all except 4, 8, and 1024).
Solution. We need to place the signs in the equation $\pm 1 \pm 2 \pm 2^{2} \pm 2^{3} \pm \ldots \pm 2^{9} \pm 2^{10} \pm 2^{11}=2023$. If we choose all plus signs, the sum will be $2^{0}+\ldots+2^{11}=2^{12}-1=4095$, so we need to replace plus signs with minus signs before the numbers whose sum is $\frac{4095-2023}{2}=1036$. There is only one such set of numbers (due to the uniqueness of the binary representation of a number): $1036=1024+8+4$.
Criteria. 1 point - it is stated that the total sum of money is 4095;
another 1 point - it is found that Pasha won 3059;
2 points - the correct answer with an example;
-2 points for the lack of proof of the uniqueness of the representation of 1036. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Given a right triangle $ABC$ with a right angle at $A$. On the leg $AC$, a point $D$ is marked such that $AD: DC = 1: 3$. Circles $\Gamma_{1}$ and $\Gamma_{2}$ are then constructed with centers at $A$ and $C$ respectively, passing through point $D$. $\Gamma_{2}$ intersects the hypotenuse at point $E$. Circle $\Gamma_{3}$ with center $B$ and radius $BE$ intersects $\Gamma_{1}$ inside the triangle at a point $F$ such that angle $AFB$ is a right angle. Find $BC$ if $AB = 5$.
(P. D. Mulyenko) | Answer: 13.
Solution. Let $AC = x$. Then $AD = x / 4, DC = CE = 3x / 4, BE = BC - CE = \sqrt{x^2 + 25} - 3x / 4$. According to the problem, $\angle AFB = 90^\circ$, so $AF^2 + FB^2 = AB^2$, which means $AD^2 + BE^2 = 25$. Expressing everything in terms of $x$ and simplifying, we get $13x = 12 \sqrt{x^2 + 25}$. Squaring both sides, we obtain $x^2 = 144$, so $x = 12$ and
$BC = 13$. | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, points $A(0,0)$ and $B(1000,0)$ were marked, as well as points $C_{1}(1,1)$, $C_{2}(2,1), \ldots, C_{999}(999,1)$. Then all possible lines $A C_{i}$ and $B C_{i}(1 \leqslant i \leqslant 999)$ were drawn. How many integer-coordinate intersection points do all these lines have? (An integer-coordinate point is one where both coordinates are integers.)
(O. A. Pyayve) | Solution. Let $a_{n}$ and $b_{n}$ denote the lines passing through $A$ and $B$ respectively, as well as through a point on $l$ with an abscissa that is $n$ greater than the abscissa of $A$ (where $1 \leqslant n \leqslant 999$). The lines $a_{n}$ and $a_{m}$ for $n \neq m$ intersect at a non-integer point (between $A B$ and $l$). Finally, the lines $a_{n}$ and $b_{m}$ for $n \leqslant m$ intersect at a point at a distance $k$ from $A B$ such that $1000(k-1)=k(m-n)$ (it is an integer if and only if $k$ is an integer). Therefore, $k$ is a divisor of 1000, and conversely, for each divisor, the corresponding $m-n$ is an integer. For each of them, there are $\frac{1000}{k}-1$ suitable pairs $(n, m)$, so the answer is $1+2+4+8+5+10+20+40+25+50+100+200+125+$ $250+500+1000-16+2=2326$.
Criteria. 2 points - for the equation $1000(k-1)=k(m-n)$;
5 points - for the statement about $\frac{1000}{k}-1$ suitable pairs $(n, m)$ or something equivalent to it;
-1 point for a calculation error in the final step. | 2326 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of five-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3. | Solution. The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be $1, 4, 7$; if the first is 4, the last can be $2, 5, 8$; if the first is 6, the last can be $0, 3, (6$ does not fit), 9; if the first is 8, the last can be $1, 4, 7$. In total, there are $3+3+3+3=13$ options for the first and last digits. For each of these options, there are $8 \cdot 7 \cdot 6=336$ ways to choose the two middle digits. In total, $336 \cdot 12=4032$ ways.
Criteria. The same as in 5.4. | 4032 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of all roots of the equation:
$$
\begin{gathered}
\sqrt{2 x^{2}-2024 x+1023131}+\sqrt{3 x^{2}-2025 x+1023132}+\sqrt{4 x^{2}-2026 x+1023133}= \\
=\sqrt{x^{2}-x+1}+\sqrt{2 x^{2}-2 x+2}+\sqrt{3 x^{2}-3 x+3}
\end{gathered}
$$
(L. S. Korechkova) | Solution. Note that the radicands in the left part are obtained from the corresponding radicands in the right part by adding $x^{2}-2023 x+1023130=$ $(x-1010)(x-1013)$. Since all radicands are positive (it is sufficient to check for $x^{2}-x+1$ and for $\left.2 x^{2}-2024 x+1023131=2(x-506)^{2}+511059\right)$, the left part is less than the right part when $1010<x<1013$ and greater when $x \notin[1010,1013]$. Equality is achieved only at $x=1010$ and $x=1013$, so the answer is -2023.
Criteria. If the answer is found but it is not proven that there are no other roots of the equation, no more than 2 points. | -2023 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). At the beginning of the game, Pasha had a single-digit amount of money, and Igor had a four-digit amount. By the end, Igor had a two-digit amount, and Pasha had a three-digit amount. What is the minimum number of games Pasha could have won? The players cannot go into debt.
(L. S. Korechkova, A. A. Tessler) | Solution. Let $n$ be the amount of money Pasha has become richer (and Igor poorer). Note that Pasha won the last game (otherwise, he would have lost more money than he gained in all previous stages). Therefore, the sequence of games can be divided into series, in each of which Pasha won the last game and lost all the others (a series can consist of just one game). If a series started with game number $k$ and ended with game number $m$, then Pasha won $-2^{k}-2^{k+1}-\ldots-2^{m-2}+2^{m-1}=2^{k}$ rubles for it. Thus, the binary representation of the number $n$ uniquely describes the set of games Pasha won (except for the number of the last game): the term $2^{k}$ means that the next series started with game number $k+1$, i.e., Pasha won game number $k$.
According to the problem, $901 \leqslant n \leqslant 998$. But all numbers from 901 to 998 contain $2^{7}+2^{8}+2^{9}$ in their binary representation, so Pasha won the seventh, eighth, and ninth games, as well as the last one (its number is greater than 9, otherwise there would be no term $2^{9}$) - already a minimum of 4 games.
In addition, Pasha must have won at least 3 times in the first 6 games:
1) at least one out of the first four games, since $9-1-2-4-8<0$;
2) at least one out of the next two, since $9 \pm 1 \pm 2 \pm 4 \pm 8-16-32<0$;
3) if only one out of the first four was won, then after them the sum is no more than 10, and the fifth and sixth must definitely be won.
Thus, Pasha won at least 7 games. Here is an example for 7 games: initially, Pasha had 9 rubles, and Igor had 1000 rubles, a total of 10 games were played. Then $n=985=2^{0}+2^{3}+$ $2^{4}+2^{6}+2^{7}+2^{8}+2^{9}=\left(-2^{0}-2^{1}+2^{2}\right)+\left(2^{3}\right)+\left(2^{4}\right)+\left(-2^{5}+2^{6}\right)+\left(2^{7}\right)+\left(2^{8}\right)+\left(2^{9}\right)$, i.e., Pasha won games with numbers $3,4,6,7,8,9,10$, and Igor - games $1,2,5$. In the end, Pasha will have 994 rubles, and Igor - 15 rubles.
Answer: 7 games.
Criteria. 5 points are given for the estimate (1 point for estimating that there were at least 10 games in total), 2 points for the example. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. How many solutions in natural numbers does the equation $(a+1)(b+1)(c+1)=2 a b c$ have? | Solution. Rewrite the equation as $(1+1 / a)(1+1 / b)(1+1 / c)=2$. Due to symmetry, it is sufficient to find all solutions with $a \leqslant b \leqslant c$. Then $(1+1 / a)^{3} \geqslant 2$, which means $a \leqslant(\sqrt[3]{2}-1)^{-1}<4$ and $a \in\{1,2,3\}$. In the case $a=1$, the inequality $2(1+1 / b)^{2} \geqslant 2$ holds, so there are no solutions. If $a=2$, then $\frac{3}{2}(1+1 / b)^{2} \geqslant 2$, which means $2 \leqslant b \leqslant\left(\frac{2 \sqrt{3}}{3}-1\right)^{-1}<7$. In this case, there are 3 solutions $(a, b, c)=(2,4,15),(2,5,9),(2,6,7)$ (for $b=2$ and $b=3$, the equation for $c$ has no solutions in natural numbers). Finally, if $a=3$, then $\frac{4}{3}(1+1 / b)^{2} \geqslant 2$, which means $3 \leqslant b \leqslant\left(\sqrt{\frac{3}{2}}-1\right)^{-1}<5$. This gives 2 more solutions $(a, b, c)=(3,3,8),(3,4,5)$. Taking into account permutations, there are a total of 27 solutions.
Criteria. If only part of the solutions is found, no more than 2 points are given. | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. $f(x)$ is a linear function, and the equation $f(f(x))=x+1$ has no solutions. Find all possible values of the quantity $f(f(f(f(f(2022)))))-f(f(f(2022)))-f(f(2022))$. | Solution. Let $f(x)=k x+b$, then $f(f(x))=k(k x+b)+b=k^{2} x+k b+b$. The equation can have no solutions only when $k^{2}=1$, that is, for functions $x+b$ or $-x+b$, so the answer is either $(2022+5 b)-(2022+3 b)-(2022+2 b)=-2022$, or $(-2022+b)-(-2022+b)-2022=-2022$.
Answer: -2022.
Criteria. Up to 2 points will be deducted for gaps in the proof that $k= \pm 1$. | -2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Let's call the efficiency of a natural number $n$ the fraction of all natural numbers from 1 to $n$ inclusive that have a common divisor with $n$ greater than 1. For example, the efficiency of the number 6 is $\frac{2}{3}$.
a) Does there exist a number with an efficiency greater than $80\%$? If so, find the smallest such number.
b) Does there exist a number with the maximum efficiency (i.e., not less than the efficiency of any other number)? If so, find the smallest such number. | Solution. Let's move on to studying inefficiency (1 minus efficiency). From the formula for Euler's function, it follows that it is equal to $\frac{p_{1}-1}{p_{1}} \cdot \ldots \frac{p_{k}-1}{p_{k}}$, where $p_{1}, \ldots, p_{k}$ are all distinct prime divisors of $n$. Then, by adding a new prime factor, we can increase the efficiency, so the answer in part (b) is "no".
The smallest number with an efficiency greater than $80\%$ is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 = 30030$. Its efficiency is $1 - \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \frac{10}{11} \cdot \frac{12}{13} = \frac{809}{1001}$. We will prove that it is more efficient than all smaller numbers. Indeed, the presence of a prime factor with a power higher than one does not affect the efficiency, so the sought number has all factors in the first power. If the factors are not consecutive primes, then replacing one of the primes with a smaller one will increase the efficiency. Therefore, "efficiency records" can only be set by numbers of the form "the product of the first few primes"; but the efficiency of the number $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ is too low.
Criteria. Part (a) is scored 5 points, part (b) - 2 points.
International Mathematical Olympiad
«Formula of Unity» / «The Third Millennium»
Year 2022/2023. Qualifying round
## Problems for grade R10
Each task is assessed at 7 points. Some problems have their own criteria (printed in gray). | 30030 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the area of the set of points on the coordinate plane that satisfy the inequality $(y+\sqrt{x})\left(y-x^{2}\right) \sqrt{1-x} \leqslant 0$. | Solution. The left side makes sense only for $0 \leqslant x \leqslant 1$. In this case, it is required that $y+\sqrt{x}$ and $y-x^{2}$ have different signs (or one of them equals zero), or that $x$ equals 1. If we exclude the case $x=1$ (which gives a zero area), we are left with a part of the plane bounded by the segment of the line $x=1$ and parts of the parabolas $y=-\sqrt{x}$ and $y=x^{2}$. By cutting this figure into two parts along the x-axis and placing the upper part below the lower one, we get a square of area 1.
Answer: the area is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. A natural number $n$ is called cubish if $n^{3}+13 n-273$ is a cube of a natural number. Find the sum of all cubish numbers. | Solution. If $0<13n-273<13\cdot21$, so it remains to check all other numbers.
If $n=21$, then $13n-273=0$, so 21 is cubic. For $n-3n^{2}+3n-1$, the number $n$ will not be cubic (i.e., for $8<n<21$).
If $n=8$, then $13n-273=-169=-3\cdot8^{2}+3\cdot8-1$, so it is cubic. For $n \leqslant 5$, the expression $n^{3}+13n-273$ will be negative, so they are definitely not cubic. The numbers 6 and 7 are not cubic, which can be verified directly. In total, the answer is $8+21=29$. | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Pasha draws dots at the intersections of the lines on graph paper.
He likes it when four dots form a figure resembling a "kite," as shown on the right (the kite must be of this exact shape and size, but can be rotated). For example, the 10 dots shown in the second image form only two kites. Is it possible to draw a certain number of dots so that the number of kites is greater than the number of dots themselves?
 | Solution. For example, like this. Here there are 21 points and 24 snakes (6 snakes in each direction).
 | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Flint has five sailors and 60 gold coins. He wants to distribute them into wallets, and then give the wallets to the sailors so that each gets an equal number of coins. But he doesn't know how many sailors will be alive by the time of the distribution. Therefore, he wants to distribute the coins in such a way that they can be evenly divided among two, three, four, or five sailors. What is the minimum number of wallets he will need? Don't forget to prove that the number you found is the smallest. | Solution. Answer: 9 wallets. Example: $12,12,8,7,6,5,4,3,3$.
We will prove that 8 wallets are insufficient.
1) Note that each wallet should contain no more than 12 coins. Therefore, 15 coins must be made up of at least two wallets. This means that when we divide 8 wallets among four pirates, each should receive two wallets. Thus, they form four pairs with a sum of 15.
2) When dividing among five pirates, at least two pirates will receive one wallet each. Therefore, there are two wallets with 12 coins.
3) From point 1, it follows that these two wallets must be paired with two wallets containing 3 coins.
4) If there are only two wallets with 12 coins, the rest must form pairs with a sum of 12. Then, in addition to each wallet with 3 coins, there must be a wallet with 9 coins. And then each of them needs a pair with 6 coins (to make a sum of 15). Thus, the set of wallets is: $12,12,3,3,9,9,6,6$. Clearly, 20 coins cannot be obtained with these (20 is not divisible by 3).
5) If there are more than two wallets with 12 coins, i.e., at least three, then there must be at least three wallets with 3 coins (which complement them to 15 coins, see point 1). However, from such a set of wallets, it is impossible to form even one portion of 20 coins; to be able to get three such portions, at least three more wallets need to be added, then there will be no fewer than nine. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In a certain triangle, the sum of the tangents of the angles turned out to be 2016. Estimate (at least to the nearest degree) the magnitude of the largest of its angles. | Solution. One of the tangents must exceed 600. This is only possible for an angle very close to $90^{\circ}$. We will prove that it exceeds $89.5^{\circ}$. This is equivalent to the statement that $\operatorname{tg} 0.5^{\circ}>$ $1 / 600$.
Let's start with the equality $\sin 30^{\circ}=1 / 2$. Note that $\sin 2 x=2 \sin x \cos x$, so $\sin x=\frac{\sin 2 x}{2 \cos x}>\frac{\sin 2 x}{2}$ for acute angles. Therefore:
$\sin 32^{\circ}>1 / 2 ; \quad \sin 16^{\circ}>1 / 4 ; \quad \sin 8^{\circ}>1 / 8$
$\sin 4^{\circ}>1 / 16 ; \quad \sin 2^{\circ}>1 / 32 ; \quad \sin 1^{\circ}>1 / 64$
$\sin 0.5^{\circ}>1 / 128>1 / 600 ; \quad \operatorname{tg} 0.5^{\circ}>1 / 600$.
Thus, one of the angles of the triangle is in the interval from 89.5 to 90 degrees. Note that it may not be the largest; but in this case, the largest angle is less than $90.5^{\circ}$. Therefore, to the nearest degree, the largest angle is $90^{\circ}$ in any case. | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural numbers $n$ for which $2^{n}+n^{2016}$ is a prime number. | Solution. Let's consider three cases.
- If $n$ is even, then the given number is also even (and greater than two for $n>0$).
- If $n$ is odd and not divisible by 3, then $2^{n}$ gives a remainder of 2 when divided by 3, and $n^{2016}=\left(n^{504}\right)^{4}$ gives a remainder of 1 when divided by 3, so the sum is divisible by 3 (and greater than three for $n>1$). For $n=1$, the result is 3, which is a prime number.
- Finally, let $n$ be divisible by 3 (and odd, which is not used). Then the number in question is the sum of cubes: if $n=3k$, then
$2^{n}+n^{2016}=\left(2^{k}\right)^{3}+\left(n^{672}\right)^{3}=\left(2^{k}+n^{672}\right) \cdot\left(2^{2k}-2^{k} \cdot n^{672}+n^{2 \cdot 672}\right)-$ a composite number (it is obvious that $1<2^{k}+n^{672}<2^{n}+n^{2016}$ for $n \geqslant 3$).
Answer: $n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In the game "set," all possible four-digit numbers consisting of the digits $1,2,3$ (each digit appearing exactly once) are used. It is said that a triplet of numbers forms a set if, in each digit place, either all three numbers contain the same digit or all three numbers contain different digits.
The complexity of a set is defined as the number of digit places where all three digits are different.
For example, the numbers 1232, 2213, 3221 form a set of complexity 3 (in the first place, all three digits are different, in the second place, only the digit 2 appears, in the third place, all three digits are different, and in the fourth place, all three digits are different); the numbers $1231,1232,1233$ form a set of complexity 1 (in the first three places, the digits are the same, and only in the fourth place are all digits different). The numbers 1123, 2231, 3311 do not form a set at all (in the last place, two ones and a three appear).
Which sets have the highest complexity in the game, and why? | Solution. Note that for any two numbers, there exists exactly one set in which they occur. Indeed, the third number of this set is constructed as follows: in the positions where the first two numbers coincide, the third number has the same digit; in the position where the first two numbers differ, the third number gets the remaining digit. For example, for the numbers 1231 and 1223, the third in the set will be 1212.
Let's call an "ordered set" a set of three four-digit numbers with their order taken into account. Note that each unordered set $\{a, b, c\}$ corresponds to six ordered sets: $(a, b, c)$, $(a, c, b)$, $(b, a, c)$, $(b, c, a)$, $(c, a, b)$, $(c, b, a)$. Therefore, instead of the number of unordered sets, we can compare the number of ordered sets (which is six times more). Each ordered set $(a, b, c)$ is uniquely determined by an ordered pair of numbers $(a, b)$.
Let's count the number of ordered sets of complexity $k>0$. Each such set can start with any number $a$ (81 options). In this number, we need to choose $k$ positions ($C_{4}^{k}$ ways to choose) and replace the digit in each of them with one of the two different ones ($2^{k}$ ways). As a result, we get a number $b$, which uniquely determines the set. In total, we get $81 \cdot C_{4}^{k} \cdot 2^{k}$ ordered sets of complexity $k$.
Comparing the numbers $f(k)=C_{4}^{k} \cdot 2^{k}$ for different $k$, we get: $f(1)=8$, $f(2)=24$, $f(3)=32$, $f(4)=16$. As we can see, the largest number of sets has complexity $k=3$.
Answer: the most sets have complexity 3. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Each cell of a $100 \times 100$ board is painted blue or white. We will call a cell balanced if among its neighbors there are an equal number of blue and white cells. What is the maximum number of balanced cells that can be on the board? (Cells are considered neighbors if they share a side.) | Solution. Cells lying on the border of the board but not in the corner cannot be equilibrium cells, since they have an odd number of neighbors (three). There are $4 \cdot 98=392$ such cells.
All other cells can be made equilibrium, for example, with a striped coloring (the first row is blue, the second row is white, the third row is blue, and so on). The number of these cells is $10000-392=9608$.
Answer: 9608. | 9608 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On a sheet of notebook paper, two rectangles are outlined. The first rectangle has a vertical side shorter than the horizontal side, while the second has the opposite. Find the maximum possible area of their common part, if each rectangle contains more than 2010 but less than 2020 cells. | Solution. The common part of these two rectangles (if it is non-empty) is a rectangle. Let's define the maximum possible lengths of its sides.
The vertical side of the first rectangle is shorter than the horizontal one, so it is less than $\sqrt{2020}$, which is less than 45; the same is true for the horizontal side of the second rectangle. At the same time, the sides of the intersection cannot exceed the corresponding sides of the original rectangles (but they can obviously be equal to them).
Let's find the greatest possible length of each of these sides. For this, we need to find the greatest number less than 45 that is a divisor of any number from 2011 to 2019 inclusive. It can be noted that none of the numbers $2011, \ldots, 2019$ are divisible by 44 (since $2024=45^{2}-1^{2}=$ $44 \cdot 46$ is divisible by 44), nor by 43 (since $2021=45^{2}-2^{2}=43 \cdot 48$ is divisible by 43). However, the number 2016 is divisible by 42 (since $2024=45^{2}-3^{2}=42 \cdot 48$), so the maximum possible length of each of the sides of the intersection is 42. Therefore, the maximum area of the intersection is $42^{2}=1764$. This value is achieved for rectangles $48 \times 42$ and $42 \times 48$, aligned, for example, by their top-left cells.
Answer: 1764. | 1764 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the game "set," all possible four-digit numbers consisting of the digits $1,2,3$ (each digit appearing exactly once) are used. It is said that a triplet of numbers forms a set if, in each digit place, either all three numbers contain the same digit, or all three numbers contain different digits.
For example, the numbers 1232, 2213, 3221 form a set (in the first place, all three digits appear, in the second place, only the digit two appears, in the third place, all three digits appear, and in the fourth place, all three digits appear). The numbers $1123,2231,3311$ do not form a set (in the last place, two ones and a three appear).
How many sets exist in the game?
(Permuting the numbers does not create a new set: $1232,2213,3221$ and $2213,1232,3221$ are the same set.) | Solution. Note that for any two numbers, there exists exactly one set in which they occur. Indeed, the third number of this set is constructed as follows: in the positions where the first two numbers coincide, the third number has the same digit; in the position where the first two numbers differ, the third number gets the remaining digit. For example, for the numbers 1231 and 1223, the third in the set will be 1212.
Let's call an "ordered set" a set of three four-digit numbers with their order taken into account. In such a sequence, the first place can be occupied by any of the 81 numbers, the second by any of the 80 remaining; the third is exactly one (according to the principle described above). In total, there are $81 \cdot 80 \cdot 1$ ordered sets. Note that each unordered set $\{a, b, c\}$ corresponds to six ordered sets: $(a, b, c),(a, c, b),(b, a, c),(b, c, a),(c, a, b),(c, b, a)$. Therefore, there are six times fewer ordered sets.
Answer: $81 \cdot 80 / 6=1080$. | 1080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On a sheet of lined paper, two rectangles are outlined. The first rectangle has a vertical side shorter than the horizontal side, while the second has the opposite. Find the maximum possible area of their common part, if the first rectangle contains 2015 cells, and the second - 2016. | Solution. The common part of these two rectangles (if it is non-empty) is a rectangle. Let's determine the maximum possible lengths of its sides.
The vertical side of the first rectangle is shorter than the horizontal one, so it is less than $\sqrt{2015}$, which is less than 45; the same is true for the horizontal side of the second rectangle. At the same time, the sides of the intersection cannot exceed the corresponding sides of the original rectangles (but they can obviously be equal to them).
Note that $2015=5 \cdot 403=5 \cdot 13 \cdot 31 ; 2016=4 \cdot 504=4 \cdot 4 \cdot 126=4 \cdot 4 \cdot 2 \cdot 7 \cdot 9=2^{5} \cdot 3^{2} \cdot 7$.
Taking this into account, the vertical side is no more than 31, and the horizontal side is no more than 42 (these are the largest divisors of the numbers 2015 and 2016 that are less than 45). Therefore, the sides of the intersection do not exceed 31 and 42, and the area is no more than $31 \cdot 42=1302$. The case of equality is obviously achieved for rectangles $65 \times 31$ and $42 \times 48$, aligned, for example, by the top-left cells.
Answer: 1302. | 1302 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A motorcyclist set out from point $A$ with an initial speed of 90 km/h, uniformly increasing his speed (that is, over equal time intervals, his speed increases by the same amount). After three hours, the motorcyclist arrived at point $B$, passing through $C$ along the way. After that, he turned around and, still uniformly increasing his speed, headed back. Two hours later, he passed point $C$ again at a speed of 110 km/h and continued on to $A$. Find the distance between points $A$ and $C$. | Solution. In 5 hours, the speed changed from 90 km/h to 110 km/h, so the acceleration is 4 km/h$^2$. From $A$ to $B$ the distance is
$$
90 \cdot 3+\frac{4}{2} \cdot 3^{2}=270+18=288(\text{km})
$$
from $B$ to $C-$
$$
110 \cdot 2-\frac{4}{2} \cdot 2^{2}=220-8=212(\text{km})
$$
And the required distance is 76 km. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A natural number $n$ is called cubish if $n^{3}+13 n-273$ is a cube of a natural number. Find the sum of all cubish numbers. | Solution. If $021$, so it remains to check all other numbers.
If $n=21$, then $13 n-273=0$, so 21 is cubic. For $n-3 n^{2}+3 n-1$, the number $n$ will not be cubic (i.e., for $8<n<21)$
If $n=8$, then $13 n-273=-169=-3 \cdot 8^{2}+3 \cdot 8-1$, so it is cubic. For $n \leqslant 5$, the expression $n^{3}+13 n-273$ will be negative, so they are definitely not cubic. The numbers 6 and 7 are not cubic, which can be verified directly. In total, the answer is $8+21=29$. | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Let $a$ and $n$ be natural numbers, and it is known that $a^{n}-2014$ is a number with $2014$ digits. Find the smallest natural number $k$ such that $a$ cannot be a $k$-digit number. | Solution. Let $a$ be a $k$-digit number, then $10^{k-1} \leq a2013$, i.e., $k>45$.
We will check for $k$ starting from 46, whether the condition of the absence of an integer in the specified interval is met. We will find that
(for $k=46$) $2013 / 462013$ );
(for $k=47$) $2013 / 472013$ );
(for $k=48$) $2013 / 482013$ );
but (for $k=49$) $2013 / 49 > 41$ (since $\quad 41 \cdot 49 = 45^2 - 1^2 = 2025 - 16 < 2013$).
Thus, the interval $[2013 / 49, 2013 / 48)$ does not contain any integers, and for $k=49$, there are no suitable $n$.
## Answer: $\mathbf{k}=49$.
## Criteria.
The problem is complex, so don't be stingy with points for reasonable thoughts. Any answer in the vicinity of $\sqrt{2014}$, which is preceded by something reasonable, is worth points. A point is deducted for the absence of what is in small print. | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a certain language, there are 3 vowels and 5 consonants. A syllable can consist of any vowel and any consonant in any order, and a word can consist of any two syllables. How many words are there in this language? | 1. The language has $3 \cdot 5=15$ syllables of the form "consonant+vowel" and the same number of syllables of the form "vowel+consonant", making a total of 30 different syllables. The total number of two-syllable words is $30 \cdot 30=900$ | 900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Masha and Lena left home and went to the store for ice cream. Masha walked faster and got to the store in 12 minutes. Spending 2 minutes buying the ice cream, she headed back. After another 2 minutes, she met Lena. Walking a bit more, Masha finished her ice cream and, deciding to buy another one, turned around and went back to the store. As a result, Masha arrived at the store with Lena. How many minutes did Lena walk to the store? | 3. Masha covers $1 / 6$ of the entire distance in 2 minutes, which means Lena covered $5 / 6$ of the distance by the time they met, and it took her 16 minutes. Therefore, $1 / 6$ of the distance takes Lena $16: 5=3 \frac{1}{5}$ minutes, i.e., 3 minutes and 12 seconds. For the entire distance, she would need 19 minutes and 12 seconds. | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Petya prints five digits on the computer screen, none of which are zeros. Every second, the computer removes the initial digit and appends to the end the last digit of the sum of the remaining four digits. (For example, if Petya enters 12345, after one second it will become 23454, then 34546, and so on. However, he can enter any five digits, not necessarily 12345.) At some point, Petya stops the process. What is the minimum possible sum of the five digits that could be on the screen at that moment?
(A. A. Tesler) | Answer: 2.
Solution. The record 00000 cannot appear on the screen, as it can only result from 00000. A record with four zeros and one also cannot appear, since in that case, the last digit would not equal the remainder of the division of the sum of the first four by 10.
However, a sum of digits equal to 2 is possible. For example, by working backwards, we can find an example of obtaining the record 00011 (or 10001):
$00011 \leftarrow 10001 \leftarrow 91000 \leftarrow 09100 \leftarrow 00910 \leftarrow 20091 \leftarrow 72009 \leftarrow 17200 \leftarrow 01720 \leftarrow 40172 \leftarrow$ $24017 \leftarrow 52401 \leftarrow 95240 \leftarrow 89524$.
Criteria. 5 points for the example, 2 points for the proof of the impossibility of smaller sums. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a certain language, there are 5 vowels and 7 consonants. A syllable can consist of any vowel and any consonant in any order, and a word can consist of any two syllables. How many words are there in this language? | 1. The language has $5 \cdot 7=35$ syllables of the form "consonant+vowel" and the same number of syllables of the form "vowel+consonant", making a total of 70 different syllables. The total number of two-syllable words is $70 \cdot 70=4900$. | 4900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given an isosceles triangle $A B C$, where $\angle A=30^{\circ}, A B=A C$. Point $D$ is the midpoint of $B C$. On segment $A D$, point $P$ is chosen, and on side $A B$, point $Q$ is chosen such that $P B=P Q$. What is the measure of angle $P Q C ?$ (S. S. Korechkova) | Answer: $15^{\circ}$.
Solution. Since $D$ is the midpoint of the base of the isosceles triangle, $A D$ is the median, bisector, and altitude of the triangle. Draw the segment $P C$. Since $\triangle P D B = \triangle P D C$ (by two sides and the right angle between them), $P C = P B = P Q$, meaning that all three triangles $\triangle P B C$, $\triangle P B Q$, and $\triangle P Q C$ are isosceles.
Notice that $\angle A B D = \angle A C B = (180^{\circ} - 30^{\circ}) : 2 = 75^{\circ}$. Let $\angle P B D = \angle P C D = \alpha$, then $\angle P B Q = \angle P Q B = 75^{\circ} - \alpha$, and $\angle P Q C = \angle P C Q = \beta$. The sum of the angles in triangle $\triangle Q C B$ is $2 \alpha + 2 \cdot (75^{\circ} - \alpha) + 2 \beta = 180^{\circ}$, from which $\beta = 15^{\circ}$.
Criteria. Correct answer without justification - 0 points.
Proven that $P B = P C = P Q - 3$ points.
 | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A few years ago, in the computer game "Minecraft," there were 9 different pictures (see the figure): one horizontal picture measuring $2 \times 1$ and $4 \times 2$, one square picture measuring $2 \times 2$, and two each of pictures measuring $1 \times 1$, $4 \times 3$ (horizontal), and $4 \times 4$. In how many ways can all 9 pictures be placed on a rectangular wall measuring 12 blocks in length and 6 in height? The pictures should not overlap; they cannot be rotated.
 | Answer: 896.
Solution. We will say that two paintings are in different columns if no block of the first painting is in the same column as any block of the second. It is clear that the $4 \times 4$ paintings are in different columns from each other and from the $4 \times 3$ paintings regardless of their placement. Thus, the $4 \times 3$ paintings will necessarily be strictly one below the other. The $4 \times 4$ paintings will be pressed against the floor or ceiling, as there must still be $4 \times 2$ and $2 \times 2$ paintings on the wall.
Consider the case where the $4 \times 2$ is strictly above or below one of the $4 \times 4$ paintings. It is clear that there are $3 \cdot 2^{2} \cdot 2^{3}=96$ ways to place all the paintings of width 4 (if we consider the small paintings as one $4 \times 2$ painting, we get 3 ways
| $4 \times 3$ | $4 \times 2$ | |
| :---: | :---: | :---: |
| | | |
| $4 \times 3$ | $4 \times 4$ | $4 \times 4$ |
to choose the column for the $4 \times 3$ paintings, $2^{2}$ ways to choose the attachment to the floor/ceiling in each of the other columns, and $2^{3}$ ways to swap paintings of the same size). The remaining paintings can be placed in $2^{3}=8$ ways in the remaining $4 \times 2$ wall space, as the $2 \times 2$ and $2 \times 1$ paintings must be in different columns.
Now let the $4 \times 2$ painting be above or below both $4 \times 4$ paintings. Since it and the $2 \times 2$ and $2 \times 1$ paintings must be in different columns, there are 16 ways to place all the paintings of width 4 (2 ways to choose the column for the $4 \times 3$, 2
| $4 \times 2$ | | | $4 \times 3$ |
| :---: | :---: | :---: | :---: | :---: |
| $4 \times 4$ | $4 \times 4$ | $4 \times 3$ | |
ways to choose "ceiling/floor" and $2^{2}=4$ ways to swap paintings of the same size). The remaining two $2 \times 2$ areas can be filled in 8 ways.
In total, $8 \cdot(96+16)=896$ options.
Criteria. It is shown that the three smallest paintings (two $1 \times 1$ and one $1 \times 2$) can only be placed together as a $2 \times 2$ square - 1 point.
Correct reasoning about the placement of the $4 \times 4$ and $4 \times 3$ paintings - another 2 points.
Any significant error in calculating the ways to place the paintings of width 4 (for example: not considering that the number of options in cases where the $4 \times 2$ painting is strictly below or above a $4 \times 4$ painting or below or above two of them, are different; or not considering that in cases where the $4 \times 4$ paintings are in adjacent columns, the number of remaining cases is different depending on their position (both above, both below, or one above and one below)) -2 points. Arithmetic error in counting cases - - -1 point. | 896 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In each cell of a $100 \times 100$ table, a natural number was written. It turned out that each number is either greater than all its neighbors or less than all its neighbors. (Two numbers are called neighbors if they are in cells sharing a common side.) What is the smallest value that the sum of all the numbers can take? | Solution. Let's divide the board into dominoes. In each domino, the numbers are different, so their sum is at least $1+2=3$. Then the total sum of the numbers on the board is at least 15000. This estimate is achievable if we alternate ones and twos in a checkerboard pattern. | 15000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The pond has a square shape. On the first frosty day, the part of the pond that is no more than 10 meters away from the nearest point on the shore froze. On the second day, the part no more than 20 meters away froze, on the third day, the part no more than 30 meters away, and so on. On the first day, the area of open water decreased by $35 \%$. On which day will the pond be completely frozen? | Solution. Note that the larger the side of the pond, the smaller the percentage that will freeze on the first day. If the side of the pond is 100 m, then 36% will freeze on the first day, and if the side is 120 m, then on the first day, $1-\frac{100^{2}}{120^{2}}=11 / 36<1 / 3$ of the pond's area will freeze. Therefore, the side of the pond is between 100 and 120 meters. Therefore, it will freeze completely on the sixth day. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find all such numbers $k$ for which
$$
(k / 2)!(k / 4)=2016+k^{2}
$$
The symbol $n!$ denotes the factorial of the number $n$, which is the product of all integers from 1 to $n$ inclusive (defined only for non-negative integers; $0!=1$). | Solution. Note that the left side makes sense only for even values of $k$. We directly verify that $k=2,4,6,8,10$ do not work, while $k=12$ gives a correct equality.
With each further increase of $k$ by 2, the expression $(k / 2)!$ increases by at least 7 times, i.e., the left side grows by more than 7 times. At the same time, the right side increases by less than double: $(k+1)^{2}-k^{2}<12$. Therefore, for $k>12$, the left side is greater than the right side.
Answer: $k=12$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the coordinate plane, an isosceles triangle $A B C$ was drawn: $A B=2016, B C=$ $A C=1533$, with vertices $A$ and $B$ lying on nodes on the same horizontal line. Determine how many nodes lie within the triangle $A B C$ (including nodes lying on the sides). A node is a point on the coordinate plane where both coordinates are integers. | Solution. Note that $1533^{2}-1008^{2}=(1533-1008)(1533+1008)=525 \cdot 2541=21 \cdot 25 \cdot 7 \cdot 363=$ $7 \cdot 3 \cdot 5^{2} \cdot 7 \cdot 3 \cdot 11^{2}=(7 \cdot 3 \cdot 5 \cdot 11)^{2}=1155^{2}$. Therefore, the height of the triangle is 1155.
We see that the GCD of 1155 and 1008 is 21. This means that there are 22 nodes (including the vertices) on the side, which divide it into 21 equal parts.
The further part of the solution is based on the fact that the triangle gradually transforms into a rectangle $1008 \times 1155$, the number of nodes in which is easy to calculate. This process is schematically shown in the figure. Let's denote the required number of nodes in the triangle by $T$.
1) Cut the triangle along the axis of symmetry into two halves (two right triangles) and consider each of them as a separate figure. The sum of the number of nodes in these halves is 1156 more than in the original triangle, that is, $T+1156$, since the nodes that were on the axis of symmetry have been duplicated.
2) Combine these two halves into a rectangle $1008 \times 1155$. The inner parts of the halves completely cover the rectangle, while 22 nodes, located in these halves, coincide. Therefore, the number of nodes in the rectangle is 22 less than in the two halves, that is, $T+1156-22$. At the same time, this number is obviously equal to $1009 \cdot 1156$. Thus, $T+1156-22=1009 \cdot 1156$; $T=1008 \cdot 1156+22=1156000+9248+22=1165270$.
Answer: 1165270 nodes. | 1165270 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Let's call a rectangular parallelepiped typical if all its dimensions (length, width, and height) are different. What is the smallest number of typical parallelepipeds into which a cube can be cut? Don't forget to prove that this is indeed the smallest number. | Solution. A cube can be cut into four typical parallelepipeds. For example, a cube $5 \times 5 \times 5$ can be cut into parallelepipeds $5 \times 3 \times 1, 5 \times 3 \times 4, 5 \times 2 \times 1, 5 \times 2 \times 4$.
It is impossible to cut the cube into a smaller number of typical parallelepipeds. Indeed, a cube has 8 vertices; if it is cut into three or fewer parallelepipeds, then at least one of them contains at least three vertices of the cube.
If all three vertices are located on the same face of the cube (for example, on the top face), then the parallelepiped contains the entire top face of the cube; thus, it has two identical dimensions. If two vertices are located on the top face of the cube and one on the bottom, then the parallelepiped contains at least one edge of the top face, and its height is equal to the height of the cube. Again, we have two identical dimensions.
Answer: into 4 typical parallelepipeds. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Lydia likes a five-digit number if none of the digits in its notation is divisible by 3. Find the total sum of the digits of all five-digit numbers that Lydia likes. | Solution. Lidia likes five-digit numbers composed only of the digits $1,2,4,5,7,8$. Note that all such numbers can be paired in the following way: each digit $a$ is replaced by $9-a$, that is, 1 is replaced by 8, 2 by 7, 4 by 5, and vice versa. For example, the number 42718 is paired with the number 57281. This partition is good because the sum of the digits in each pair is 45 (since the sum of the two digits in each place is 9).
It remains to determine how many such pairs there are. To do this, we find the total number of numbers. In the first place, any of the six digits can stand, in the second place - also any of the six, and so on. The total number of numbers is $6^{5}$. The number of pairs is $6^{5} / 2$, and the sum of the digits in each of them is 45. The total sum of the digits is $45 \cdot 6^{5} / 2 = 45 \cdot 7776 / 2 = 45 \cdot 3888 = 174960$.
Another solution. We find the total sum of the digits in the highest place of all numbers. There are $6^{4}$ numbers that start with one, and the sum of their first digits is $1 \cdot 6^{4}$. There are also $6^{4}$ numbers that start with two, and the sum of the first digits in them is $2 \cdot 6^{4}$. We proceed similarly with numbers starting with $3,4,7,8$. We get the total sum of the digits in the highest place $(1+2+4+5+7+8) \cdot 6^{4} = 27 \cdot 6^{4}$. In each of the other places, the sum of the digits is the same, so we have $5 \cdot 27 \cdot 6^{4} = 174960$.
Remark. The digit 0 is divisible by $3 (0: 3=0)$, so it cannot be included in the specified numbers.
Answer: 174960. | 174960 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petya's favorite TV game is called "Sofa Lottery." During the game, viewers can send SMS messages with three-digit numbers containing only the digits $1, 2, 3$, and 4. At the end of the game, the host announces a three-digit number, also consisting only of these digits. An SMS is considered a winning one if the number in it differs from the host's number by no more than one digit (for example, if the host announced the number 423, then the messages 443 and 123 are winning, but 243 and 224 are not).
Petya wants to send as few messages as possible to ensure that at least one is definitely a winner. How many SMS messages will he have to send? (L. S. Korechkova) | Answer: 8.
Solution. An example of eight suitable SMS messages: 111, 122, 212, 221, 333, 344, 434, 443. Indeed, no matter what number the host names, it contains either at least two digits from the set $\{1,2\}$, or at least two from the set $\{3,4\}$. If the third digit is from the other set, we replace it with a digit from the same set as the other two, so that the sum of the digits is odd - this will definitely result in one of the specified options.
Now suppose there are fewer than 8 messages. Then some digit (for example, the digit 1) begins at most one message. Without loss of generality, let this message be 111 (if it exists at all). Consider the cases when the host names the numbers 122, 123, $124,132,133,134,142,143,144$. There are 9 of them, and for each of them, a separate message is needed (since the first digit is not 1, the second and third must be guessed).
Remark. The problem can be reformulated as: "What is the minimum number of rooks needed to cover a $4 \times 4 \times 4$ chessboard?". The general solution (for a board $n \times n \times n$) is provided, for example, in the comment at http://math.hashcode.ru/questions/52123.
Criteria. 4 points are given for the estimate, 3 points for the example. 1 point is given for the assertion that at least 7 SMS messages are necessary. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. What is the maximum number of numbers that can be chosen from the set $\{1,2, \ldots, 12\}$ so that the product of no three chosen numbers is a perfect cube? | Solution. 9: all except $4,9,12$. Note that to remove the cubes, we need to remove at least one element from each of the sets $\{1,2,4\},\{3,6,12\},\{2,4,8\},\{1,3,9\},\{2,9,12\}$, $\{3,8,9\},\{4,6,9\}$. Note that all numbers, except 9, are in no more than three of these seven triples. Therefore, if we remove two numbers different from 9, at least one triple will remain. If, however, we remove the number 9, then among the remaining triples there will be two non-intersecting $(\{1,2,4\},\{3,6,12\})$, so no matter which second number we remove, one of the triples will remain. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest possible value of the expression
$$
\left(\frac{x y}{z}+\frac{z x}{y}+\frac{y z}{x}\right)\left(\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}\right)
$$
where $x, y, z$ are non-zero real numbers. | Solution. Note that the signs of all six numbers $\frac{x y}{z}, \frac{z x}{y}$, etc., are the same. If all of them are negative, then replace the numbers $x, y, z$ with their absolute values. As a result, each term ($\frac{x y}{z}$, etc.) will change its sign. The modulus of each bracket will remain the same, but the sign will change, so the product of the two brackets will remain the same. Therefore, any value taken by the expression is also taken when $x, y, z$ are positive.
For positive values of $x, y, z$, we use the inequality of the arithmetic mean and the geometric mean. We get:
$\left(\frac{x y}{z}+\frac{z x}{y}+\frac{y z}{x}\right)\left(\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}\right) \geqslant 3 \sqrt{\frac{x y}{z} \cdot \frac{z x}{y} \cdot \frac{y z}{x}} \cdot 3 \sqrt{\frac{x}{y z} \cdot \frac{y}{z x} \cdot \frac{z}{x y}}=9 \sqrt{\frac{(x y z)^{2}}{x y z} \cdot \frac{x y z}{(x y z)^{2}}}=9$.
Obviously, the value 9 is achieved, for example, when $x=y=z=1$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In each cell of a $10 \times 10$ table, a natural number was written. Then, each cell was shaded if the number written in it was less than one of its neighbors but greater than another neighbor. (Two numbers are called neighbors if they are in cells sharing a common side.) As a result, only two cells remained unshaded, and neither of them is in a corner. What is the minimum possible sum of the numbers in these two cells? | Solution. Answer: 20. The estimate matches the estimate in problem 7.5. One of the possible examples is shown below.
| 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 8 | 10 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 |
| 7 | 8 | 10 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| 6 | 7 | 8 | 10 | 12 | 13 | 14 | 15 | 16 | 17 |
| 5 | 6 | 7 | 8 | 10 | 12 | 13 | 14 | 15 | 16 |
| 4 | 5 | 6 | 7 | 8 | 10 | 12 | 13 | 14 | 15 |
| 3 | 4 | 5 | 6 | 7 | 8 | 10 | 12 | 13 | 14 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 10 | 12 | 13 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 10 | 12 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The edge of a regular tetrahedron $A B C D$ is 1. Through a point $M$, lying on the face $A B C$ (but not on the edge), planes parallel to the other three faces are drawn. These planes divide the tetrahedron into parts. Find the sum of the lengths of the edges of the part that contains point $D$. | 5. See the figure at the end of the file.
Note that the part of the tetrahedron we are interested in is bounded by three of its faces containing point $D$, and by three planes parallel to the faces. Therefore, this is a parallelepiped.
Consider the faces $A B C$ and $A D C$. Align these faces so that vertex $B$ coincides with $D$, and edge $A C$ remains in place. Note that the lines of intersection of these faces with the plane passing through $M$ parallel to $(B C D)$ will coincide, since they are at the same distance from $C D$ (from $B C$). The same is true for the plane parallel to $(A B D)$. As a result, one of the faces of our parallelepiped transforms into a parallelogram with diagonal $BM$.
Reasoning similarly for the other two faces, we see that three faces of our parallelepiped are equal to three parallelograms with vertex $M$ lying in the face $A B C$. If we denote the edges of the parallelepiped as $a, b$, and $c$, then the result shown in Figure 2 is obtained. Since all triangles in this figure are equilateral, $a+b+c=1$. Therefore, the sum of the lengths of the edges of the parallelepiped is $4 a+4 b+4 c=4$.
Answer: 4. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.2. Two circles with radii 1 and 2 have a common center $O$. The area of the shaded region is three times smaller than the area of the larger circle. Find the angle $\angle A O B$.
 | Answer: $60^{\circ}$
Solution: Let $\angle A O B=\alpha$. Then the area of the shaded part is $\frac{360^{\circ}-\alpha}{360^{\circ}} \pi+$ $4 \frac{\alpha}{360^{\circ}} \pi-\frac{\alpha}{360^{\circ}} \pi=\left(\frac{360^{\circ}-\alpha}{360^{\circ}}+3 \frac{\alpha}{360^{\circ}}\right) \pi$. The area of the large circle is $4 \pi$. According to the condition, $4 \pi=3\left(\frac{360^{\circ}-\alpha}{360^{\circ}}+3 \frac{\alpha}{360^{\circ}}\right) \pi, 4 \times 360^{\circ}=3\left(360^{\circ}-\alpha\right)+9 \alpha, 1440^{\circ}=1080^{\circ}-6 \alpha, 6 \alpha=$ $360^{\circ}, \alpha=60^{\circ}$.
Criteria: Only the correct answer - 1 point. The formula for calculating the area of a sector is written - 1 point. If it is proven that for an angle of $60^{\circ}$, the given ratio of areas holds - 4 points. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3. Prove that $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$ is a rational number. | Solution: Let $x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$. Then $x^{3}=(\sqrt{5}+2)-(\sqrt{5}-2)-$ $3 \sqrt[3]{(\sqrt{5}+2)(\sqrt{5}-2)} \times(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2})=4-3 x ; x^{3}+3 x-4=0.0=x^{3}+$ $3 x-4=(x-1)\left(x^{2}+x+4\right), x-1=0$ or $x^{2}+x+4=0$. The second equation has no roots because its discriminant is negative. Therefore, $x=1$ is a rational number.
Criteria: It is written that the value of the given sum is equal to 1 - 1 point. The root of the obtained equation is found and it is not proven that there are no other roots - 5 points. | 1 | Algebra | proof | Yes | Yes | olympiads | false |
11.4. In a row, $n$ integers are written such that the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative. For what largest $n$ is this possible? | Answer: 16
Solution: Let's provide an example for $n=16$:
$80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80$. We will prove that for $n \geq 17$, it will not be possible to write down a sequence of numbers that satisfy the conditions of the problem. Let's construct a table for the first 17 numbers in this sequence:
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ |
| $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ |
| $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ |
| $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ | $a_{17}$ |
According to the condition: the sum of the numbers in each row must be negative, and in each column positive. Therefore, the sum of all the numbers in the table should be positive on one side and negative on the other. This is a contradiction.
Criteria: Only the correct answer - 1 point. Correct answer with an example - 2 points. | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. Given a function $f(x)$, satisfying the condition
$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$
What is $f(2017)$, if it is known that $f(0)=1$? | Answer: 2018
Solution: $f(0 \cdot 0+1)=f(0) f(0)-f(0)-0+2, f(1)=1-1+2=2$,
$$
f(2017 \cdot 0+1)=f(2017) f(0)-f(0)-2017+2
$$
$$
\begin{gathered}
f(1)=f(2017)-1-2017+2=2 \\
f(2017)=2018
\end{gathered}
$$
Criteria: Only for the correct answer - 1 point. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation
$$
x^{2018}+\frac{1}{x^{2018}}=1+x^{2019}
$$ | Answer: $x=1$.
Solution: $x^{2018}+\frac{1}{x^{2018}} \geq 2$ for $x \neq 0$, because $x^{4036}-2 x^{2018}+1=\left(x^{2018}-1\right)^{2} \geq 0 . x^{2019}=x^{2018}+\frac{1}{x^{2018}}-1 \geq 2-1=1 . x \geq 1$. If $\quad x>1, \quad x^{2019}+1>$ $x^{2018}+\frac{1}{x^{2018}}$, because $x^{2019}=x * x^{2018}>x^{2018}$ and $1>\frac{1}{x^{2018}}$. Contradiction, so $x=1, x<=1$, so $x=1$. $x=1$ is indeed a solution to the equation. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. When one of two integers was increased 1996 times, and the other was reduced 96 times, their sum did not change. What can their quotient be?
Solution. Let the first number be $a$, and the second $b$. Then the equation $1996 a+\frac{b}{96}=a+b$ must hold, from which we find that $2016 a=b$. Therefore, their quotient is 2016 or $\frac{1}{2016}$. | Answer: 2016 or $\frac{1}{2016}$.
Criteria: Full solution - 14 points; correct answer without solution - 2 points. | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A black and white chocolate bar consists of individual pieces forming $n$ horizontal and $m$ vertical rows, arranged in a checkerboard pattern. Yan ate all the black pieces, and Maxim ate all the white pieces. What is $m+n$, if it is known that Yan ate $8 \frac{1}{3} \%$ more pieces than Maxim. | Solution. The number of black and white segments can only differ by 1. Therefore, Yan ate 1 segment more than Maksim. If 1 segment is 8 $\frac{1}{3} \%$, then Maksim ate 12 segments, Yan ate 13 segments, and together they ate 25 segments. This means the chocolate bar was $5 \times 5$.
Answer: 10.
Criteria: 14 points for a justified correct solution. 12 points for a correct solution with an arithmetic error. 6 points for stating that Yan ate 1 segment more than Maksim. 2 points for the correct answer only. (Do not deduct points for the answer $26-25+1$.) | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. There are books of three colors: white, blue, and green. To make the shelf look beautiful, the boy first arranged the white books, and then placed blue books in each gap between them. Finally, he placed green books in each gap between the standing books. In the end, there were 41 books on the shelf. How many white books did the boy place on the shelf? | Answer: 11
9. The sum of two natural numbers is 2017. If you append 9 to the end of the first number and remove the digit 8 from the end of the second number, the numbers will be equal. Find the largest of these numbers.
Answer: 1998 | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the sum of the digits of all numbers in the sequence $1,2,3, \ldots, 199,200$.
untranslated text remains the same as requested. | 2. Find the sum of the digits of all numbers in the sequence $1,2,3, \ldots, 199,200$.
Answer: 1902 | 1902 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. A parallelepiped is composed of white and black unit cubes in a checkerboard pattern. It is known that the number of black cubes is $1 \frac{12}{13} \%$ more than the number of white cubes. Find the surface area of the parallelepiped, given that each side of the parallelepiped is greater than 1. | Solution. The number of black and white cubes can differ by only 1. Therefore, 1 cube is $1 \frac{12}{13} \%$ of the quantity of white cubes. Thus, there are 52 white cubes, 53 black cubes, and a total of 205 cubes. That is, our parallelepiped is $3 \times 5 \times 7$. The surface area is 142.
Answer: 142.
Criteria: 14 points for a justified correct solution. 12 points for a correct solution with an arithmetic error. 6 points for stating that there is 1 more black cube than white cubes. 2 points for the correct answer only. | 142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. How many three-digit numbers exist where all digits are odd numbers, and all two-digit numbers that can be obtained by erasing one of these digits are not divisible by 5? | 2. How many three-digit numbers exist where all digits are odd numbers, and all two-digit numbers that can be obtained by erasing one of these digits are not divisible by 5?
Answer: 80 | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.2. Let $f(n)$ be equal to the product of the even digits of the natural number $\mathrm{n}$ or be zero if there are no even digits. Find the sum $f(1)+f(2)+\cdots+f(100)$. | Answer: 620
Solution: for a single-digit number $n$, $f(n)$ will be equal to $n$ itself if it is even and 0 if it is odd. For single-digit $\mathrm{n}$, we get the sum $2+4+6+8=20$. If $\mathrm{n}$ is a two-digit number, let's consider the cases:
1) If both digits are even. Then the first digit can be $2,4,6$ or 8, and the second digit can be $0,2,4,6$ or 8. The total sum will be $2 * 0+2 * 2+\ldots+2 * 8+4 * 0+\ldots+8 * 8=2 *(0+2+4+6+8)+\ldots+8 *(0+\ldots+8)=(2+4+6+8) *(0+2+4+6+8)=20$ $* 20=400$.
2) If the first digit is even and the second is odd. Then for each of the even first digits ( $2,4,6$ or 8 ), there will be 5 odd second digits, so the total sum is $2 * 5+4 * 5+6 * 5+8 * 5=(2+4+6+8) * 5=20 * 5=100$.
3) If the first digit is odd and the second is even. Then for each of the even last digits ( $0,2,4,6$ or 8 ), there will be 5 odd first digits, so the total sum is $0 * 5+2 * 5+4 * 5+6 * 5+8 * 5=(0+2+4+6+8) * 5=20 * 5=100$.
4) If both digits are odd, the value of the function is 0.
Summing up the obtained values: $20+400+100+100+0=620$.
Criteria: Only for the correct answer - 1 point. | 620 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. In a row, $n$ integers are written such that the sum of any three consecutive numbers is positive, while the sum of any five consecutive numbers is negative. For what largest $n$ is this possible | Answer: 6.
Solution: Let's provide an example for $n=6: 3,-5,3,3,-5,3$. We will prove that for $n \geq 7$ it will not be possible to write down a sequence of numbers that satisfy the condition of the problem. We will construct a table for the first 7 numbers in this sequence
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ |
| :--- | :--- | :--- | :--- | :--- |
| $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ |
| $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ |
By the condition: the sum of the numbers in each row must be negative, and in each column positive. Therefore, the sum of all the numbers in the table should be positive on one side and negative on the other. Contradiction.
Criteria: Only for the correct answer - 1 point. Correct answer with an example - 2 points. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students? | Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students?
| Solution | Criteria |
| :---: | :---: |
| Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm{a}_{4}, \mathrm{a}_{5}$ be the number of problems solved by the students. Note that no more than one student could have solved more than 4 problems, otherwise there would be two students who solved more than 9 problems in total. Without loss of generality, we can assume that $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}$ are no more than 4. Then $\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\mathrm{a}_{4}+\mathrm{a}_{5} \leq 4+4+4+\left(\mathrm{a}_{4}+\mathrm{a}_{5}\right) \leq 4+4+4+9=21$. Let's provide an example: $4,4,4,4,5$. In this example, all conditions of the problem are met and the sum of the numbers is 21. |  | | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Find the value of the expression $\sqrt[3]{7+5 \sqrt{2}}-\sqrt[3]{5 \sqrt{2}-7}$ | Answer: 2
Task 2.B The bases $AD$ and $BC$ of an isosceles trapezoid $ABCD$ are $16 \sqrt{3}$ and $8 \sqrt{3}$, respectively, and the acute angle at the base is $30^{\circ}$. What is the length of the lateral side of the trapezoid?
Answer: 8 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 4. Solve the system of equations $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=10 \\ \sqrt[4]{x}+\sqrt[4]{y}=4\end{array}\right.$ and find the value of the product $x y$. | Answer: 81
Problem 5. From the vertex of the right angle $K$ of triangle $MNK$, a perpendicular $KL$ is drawn to the plane of the triangle, equal to 280. Find the distance from point $L$ to the line $MN$, given that the height of the triangle dropped from vertex $K$ is 96.
Answer: 296
Problem 6. The bank allocated a certain amount of money for loans to two companies for a period of one year. Company A received a loan amounting to $60\%$ of the allocated sum at an annual interest rate of $30\%$, and Company B received the remaining amount. After one year, when the loans were repaid, it turned out that the bank had earned a profit of $24\%$. At what interest rate was the loan given to Company B?
Answer: 15
Problem 7. In triangle $ABC$, sides $AC$ and $AB$ are equal to $6\sqrt{2}$ and $\sqrt{39}$, respectively, and the height $BH$ is equal to $\sqrt{7}$. Find the length of the median $BM$ of triangle $ABC$.
Answer: 3 | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 12. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side $3 \sqrt{2}$. Find the volume of a regular tetrahedron, one vertex of which coincides with point $A$, and the other three vertices belong to the plane $C M A_{1} N$, where $M$ and $N$ are the midpoints of edges $D D_{1}$ and $B B_{1}$. | Answer: 9
Problem 13. Solve the equations for all natural $n$:
$$
\cos ^{4} x+\sin x(\sin x+1)\left(\cos ^{2} x+\sin x-1\right)=n
$$
In the answer, write the number of roots in the interval $[0,2 \pi]$
Answer: 3
Problem 14. $M$ is the midpoint of the lateral side $A B$ of trapezoid $A B C D$, and $E$ is the intersection point of segments $M D$ and $A C$. Find the area of triangle $A D E$, given that $B C: A D=1: 3$, and the area of trapezoid $A B C D$ is 196.
Answer: 63
Problem 15. Solve the equation
$$
x+P(x)+P(P(x))+P(P(P(x)))=2014
$$
where $P(x)$ is the sum of the digits of the natural number $x$. In the answer, write the sum of the roots of this equation.
Answer: 3986 | 3986 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2.
 | Problem 4. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2.
Answer: 14 | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the table, there are 10 stacks of playing cards (the number of cards in the stacks can be different, there should be no empty stacks). The total number of cards on the table is 2015. If a stack has an even number of cards, remove half of the cards. If the number of remaining cards in the stack is still even, remove half again, and so on, until the number of cards in the stack becomes odd. Do this for each stack. Explain:
a) What is the maximum possible number of cards remaining on the table?
b) What is the minimum possible number of cards remaining on the table?
In each case, provide an example of how the playing cards can be distributed among the stacks. | # Solution
a) Since 2014 is an odd number, in any distribution of cards into 10 piles, there will be at least one pile with an even number of cards. Therefore, the number of cards in this pile, and thus the total number of cards, will definitely decrease by at least one card (if there are 2 cards in this pile). This means the final number of cards on the table will always be less than 2015. With the distribution $\underbrace{1,1, \ldots, 1}_{8}, 2,2005$, it is easy to verify that the final number of cards remaining on the table is 2014.
b) The smallest number of cards remaining in a pile is 1. Therefore, the final number of cards remaining on the table cannot be less than 10. In the following example, the final number of cards remaining on the table is 10:
1024, 512, 256, 128, 64, 16, 8, 4, 2, 1. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 6. In an acute-angled triangle $\mathrm{ABC}$, angle $\mathrm{B}$ is $30^{\circ}, \mathrm{BC}=12$. The altitude CD of triangle $\mathrm{ABC}$ and the altitude DE of triangle BDC are drawn. Find BE. | Answer: 9
Task 7.3a 2016 The number of books in the school library fund increased by $0.4 \%$, and in 2017 - by $0.8 \%$, remaining less than 50 thousand. By how many books did the library fund increase in 2017?
Answer: 251
Task 8. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2?
Answer: 14 | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 7. On a circle, there are 25 non-overlapping arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the arc following it in a clockwise direction. What can the sum of all the numbers be? | Problem 7. On a circle, there are 25 non-intersecting arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the next arc clockwise. What can the sum of all the numbers be?
| Solution | Criteria |
| :---: | :---: |
| Let the numbers written be $\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right), \ldots,\left(a_{25}, b_{25}\right)$. Note that if $a_{i} b_{i} \geq a_{i}+b_{i}$, where $a_{i}$ and $b_{i}$ are prime numbers, since prime numbers are at least 2. Now, let's write down the condition of the problem: $a_{1}+b_{1} \geq a_{2} b_{2} \geq a_{2}+b_{2} \geq a_{3} b_{3} \geq$ $a_{3}+b_{3} \geq \cdots \geq a_{1}+b_{1}$. That is, the inequality holds only in the case of equality. Thus, $a_{i} b_{i}=a_{i}+b_{i}$ for all $i$. This equality is only satisfied if $a_{i}=b_{i}=2$. Therefore, all the numbers written on the circle are equal to 2. Consequently, the sum of the numbers is $2 \times 2 \times 25=100$. Answer: 100. | Only the correct answer - 1 point. Noted that the product of two prime numbers is not less than their sum - 3 points. | | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a convex quadrilateral $A B C D: A B=A C=A D=B D$ and $\angle B A C=\angle C B D$. Find $\angle A C D$. | Answer: $70^{\circ}$.
Solution: Triangle $A B D$ is equilateral, so the angles $\angle A B D=\angle B D A=$ $\angle D A B=60^{\circ}$. Let $\angle B A C=\angle C B D=\alpha$, then $\angle A B C=60^{\circ}+\alpha . A B=A C$, thus $\angle A C B=\angle A B C=60^{\circ}+\alpha$. The sum of the angles in triangle $A B C$ is $\alpha+\left(60^{\circ}+\right.$ $\alpha)+\left(60^{\circ}+\alpha\right)=180^{\circ}, 3 \alpha=60^{\circ}, \alpha=20^{\circ}$. Angle $\quad \angle C A D=\angle B A D-\angle B A C=60^{\circ}-$ $20^{\circ}=40^{\circ}$. $A C=A D$, so $\angle A C D=\angle A D C=\beta$, the sum of the angles in triangle $A C D 40^{\circ}+2 \beta=180^{\circ}, 2 \beta=140^{\circ}, \angle A C D=\beta=70^{\circ}$.
Criteria: The angle $\angle B A C=\angle C B D=20^{\circ}$ is found - 4 points. | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation in integers $x^{2}+y^{2}=3 x y$.
---
Note: The translation maintains the original format and line breaks as requested. | Answer: $x=y=0$.
Solution: If both numbers are not equal to 0, divide the numbers $x$ and $y$ by their greatest common divisor, resulting in coprime numbers $a$ and $b$. The right side of the equation is divisible by 3, so the left side must also be. The square of an integer can give a remainder of 0 or 1 when divided by 3, so $a^{2}$ and $b^{2}$ are divisible by 3, which means $a$ and $b$ are divisible by 3. This is a contradiction because $a$ and $b$ are coprime and have a common divisor 3, which is greater than 1.
Criteria: Only for the correct answer - 1 point. Noted that $x$ and $y$ must be divisible by $3-3$ points. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In how many ways can a pile of 100 stones be divided into heaps so that the number of stones in any two heaps differs by no more than one? | Answer: 99.
Solution: We will prove that for any $k$ from 2 to 100, the pile can be divided into $\mathrm{k}$ such piles in a unique way. $100 = q k + r$, where $q$ and $r$ are the quotient and remainder of 100 when divided by $k$, respectively. Suppose there is a pile with no more than $q-1$ stones, then in the remaining $k-1$ piles there are no fewer than $101 - q = q(k-1) + (r+1)$ stones. Then in some pile there are no fewer than $q+1$ stones, otherwise the number of stones in $k-1$ piles would not exceed $q(k-1)$, which is less than $q(k-1) + (r+1) = 101 - q$. Two piles have been found that do not meet the condition (one with no fewer than $q+1$ and one with no more than $q-1$). Contradiction. Suppose there is a pile with no fewer than $q+2$ stones. Then in the remaining $k-1$ piles there are no more than $99 - q$ stones. Then in some pile there are no more than $q$ stones, otherwise the number of stones in $k-1$ piles would be no less than $(k-1)(q+1) = (k q + k) - 1 - q > (k q + r) - 1 - q > 100 - 1 - q = 99 - q$. Two piles have been found where the number of stones differs by at least 2. Then in all piles there are $q$ or $q+1$ stones. Let the number of piles with $q$ stones be $t$, then the number of piles with $q+1$ stones is $k-t$, the total number of stones is $100 = q t + (q+1)(k-t) = q t + q k - q t + k - t = (q k + r) - r + k - t = 100 - r + k - t, t = k - r$, i.e., the method is unique.
Criteria: 1 point for the correct answer "99". 1 point for the answer "100" as well. | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 6. Find the value of the expression $\left(\sqrt[3]{x^{2}} \cdot x^{-0.5}\right):\left(\left(\sqrt[6]{x^{2}}\right)^{2} \cdot \sqrt{x}\right)$ at $x=\frac{1}{2}$ | Answer: 2
Problem 7. The diagonals of trapezoid $ABCD (BC \| AD)$ are perpendicular to each other, and $CD = \sqrt{129}$. Find the length of the midline of the trapezoid, given that $BO = \sqrt{13}, CO = 2\sqrt{3}$, where $O$ is the point of intersection of the diagonals of the trapezoid.
Answer: 10
Problem 8. Four numbers are given, the first three of which form a geometric progression, and the last three form an arithmetic progression with a difference of 999. It is known that the first and last numbers are the same. Find these numbers and give their sum in the answer.
Answer: 2331
Problem 9. The function $f(x)$ is defined for all $x$ except 1 and satisfies the equation: $(x+2) f\left(\frac{x+1}{x-1}\right) = x - f(x)$. Find $f(-1)$.
Answer: 1
Problem 10. Find the number of integer solutions to the inequality $\frac{\left(6 x-5-x^{2}\right)\left(2-x-x^{2}\right)(1+x)}{\left(x^{2}-3 x-10\right)(1-x)} \geq 0$
Answer: 2.
Problem 11. A tourist left the camp and walked along a country road to a mountain, climbed to the top of the mountain, and then returned to the camp along the same route, spending a total of 5 hours on the entire journey. What distance did the tourist travel if his speed along the country road was 4 km/h, he climbed the mountain at a speed of 3 km/h, and descended from the mountain at a speed of 6 km/h?
Answer: 20 km
Problem 12. Find the sum of all solutions to the equation
$$
(x-2)^{2}-\frac{24}{4 x-x^{2}}=18
$$
Answer: 8
Problem 13. In trapezoid $ABCD$, the bases $BC$ and $AD$ are in the ratio $1:3$, $AC = 12 \sqrt[4]{5}$, $AD = 18 \sqrt[4]{5}$, $\cos \angle CAD = \frac{1}{9}$. Find the area of the trapezoid.
Answer: 320
Problem 14. Solve the equation in integers for all natural $n$:
$$
x^{4} + n = y(2-y) + 1
$$
In the answer, write the number of solutions.
Answer: 5
Problem 15. Two cars are participating in a race. The first car moves at a speed of 120 km/h, the second at 180 km/h. There are gas stations along the route. The first car spends $a$ minutes at each gas station, and the second car spends twice as long. They started and finished at the same time. The length of the route is 300 km. How many gas stations could there be on the route if $a$ is an integer greater than 5? In the answer, indicate the number of solutions.
Answer: 3 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Find the remainder when the number 20172017... 2017 (2017 written 100 times) is divided by 8. | Answer: 1
8. MN and $\mathrm{PQ}$ are two parallel chords located on opposite sides of the center of a circle with radius $10 . \mathrm{MN}=12, \mathrm{PQ}=16$. Find the distance between the chords.
Answer: 14 | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one? | Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one?
## Solutions and Evaluation Criteria
| Task Number | Solution | Criteria |
| :---: | :---: | :---: |
| 1 | Answer: No. Suppose this is possible. No digit is 0, because then the product on one side is 0, while there is no such equality on the other side. Therefore, all 8 letters are digits from 1 to 9, meaning they include all digits except one. Among them, there is a 5 or 7. But then one of the products is divisible by 5 (7), while the other is not. Contradiction. | 7 points for the correct solution. 1 point for proving that none of the digits is 0. 2 points for proving that there is no 0 and 5-2 points. |
| 2 | Answer: No. Color the vertices in white and black so that the colors alternate. Then the difference between the sum of the numbers in the white vertices and the sum of the numbers in the black vertices is initially 1, and this difference does not change with any move (1 is added to both sums). Suppose the opposite. Then both sums are divisible by 3, because all numbers are divisible by 3. Therefore, their difference is also divisible by 3, but 1 is not divisible by 3. Contradiction. | 7 points for the correct solution, otherwise 0 points. |
| 3 | Answer: $70^{\circ}$. Solution: Triangle $ABD$ is equilateral, so the angles $\angle ABD = \angle BDA = \angle DAB = 60^{\circ}$. Let $\angle BAC = \angle CBD = \alpha$, then $\angle ABC = 60^{\circ} + \alpha$. Since $AB = AC$, $\angle ACB = \angle ABC = 60^{\circ} + \alpha$. The sum of the angles in triangle $ABC$ is $\alpha + (60^{\circ} + \alpha) + (60^{\circ} + \alpha) = 180^{\circ}$, so $3\alpha = 60^{\circ}$, $\alpha = 20^{\circ}$. The angle $\angle CAD = \angle BAD - \angle BAC = 60^{\circ} - 20^{\circ} = 40^{\circ}$. Since $AC = AD$, $\angle ACD = \angle ADC = \beta$, the sum of the angles in triangle $ACD$ is $40^{\circ} + 2\beta = 180^{\circ}$, so $2\beta = 140^{\circ}$, $\angle ACD = \beta = 70^{\circ}$. | 7 points for the correct solution. 4 points for finding the angle $\angle BAC = \angle CBD = 20^{\circ}$. |
| :---: | :---: | :---: |
| 4 | Answer: $x = y = 0$. Solution: If both numbers are not 0, divide the numbers $x$ and $y$ by their greatest common divisor, resulting in coprime numbers $a$ and $b$. The right side of the equation is divisible by 3, so the left side must also be. The square of an integer can give a remainder of 0 or 1 when divided by 3, so $a^2$ and $b^2$ are divisible by 3, meaning $a$ and $b$ are divisible by 3. Contradiction, because $a$ and $b$ are coprime and have a common divisor 3, which is greater than 1. | 7 points for the correct solution. 1 point for the correct answer only. 3 points for noticing that $x$ and $y$ are divisible by 3. |
| 5 | Answer: 99. Solution: We will prove that for any $k$ from 2 to 100, the pile can be divided into $k$ such piles in a unique way. $100 = qk + r$, where $q$ and $r$ are the quotient and remainder of 100 when divided by $k$, respectively. Suppose there is a pile with no more than $q-1$ stones, then in the remaining $k-1$ piles, there are at least $101 - q = q(k-1) + (r+1)$ stones. Then in some pile, there are at least $q+1$ stones, otherwise, the number of stones in $k-1$ piles would not exceed $q(k-1)$, which is less than $q(k-1) + (r+1) = 101 - q$. Two piles have been found that do not meet the condition (one with at least $q+1$ and the other with no more than $q-1$). Contradiction. Suppose there is a pile with at least $q+2$ stones. Then in the remaining $k-1$ piles, there are no more than $99 - q$ stones. Then in some pile, there are no more than $q$ stones, otherwise, the number of stones in $k-1$ piles would be at least $(k-1)(q+1) = (kq + k) - 1 - | For the correct solution - 7 points. For the correct answer "99" - 1 point. For the answer "100" - also 1 point. |
| | $q > (kq + r) - 1 - q > 100 - 1 - q = 99 - q$. Two piles have been found where the number of stones differs by at least 2. Then in all piles, there are $q$ or $q+1$ stones. Let the number of piles with $q$ stones be $t$, then the number of piles with $q+1$ stones is $k-t$, the total number of stones is $100 = qt + (q+1)(k-t) = qt + qk - qt + k - t = (qk + r) - r + k - t = 100 - r + k - t$, so $t = k - r$, i.e., the way is unique. |
| :--- | :--- | :--- | | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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