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3.1. (12 points) The number
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}
$$
was written as an irreducible fraction with natural numerator and denominator. Find the last two digits of the numerator. | Answer: 99.
Solution. We have
$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\ldots+\left(\frac{1}{2017!}-\frac{1}{2018!}\right)=1-\frac{1}{2018!}=\frac{2018!-1}{2018!}$.
In the end, we obtained an irreducible fraction, and the las... | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. (12 points) The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $19 a, \frac{124 b}{13}, \frac{c}{13}$ is an arithmetic progression. Find the common ratio of the geometric progression. | Answer: 247.
Solution. Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{124 a q}{13}=19 a+\frac{a q^{2}}{13} \Leftrightarrow q^{2}-248 q+247=0$, from which $q=1$ or $q=247$. A decreasing geometric progression can only occur when $... | 247 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. (12 points) Yura has unusual clocks with several minute hands moving in different directions. Yura calculated that the minute hands coincided in pairs exactly 54 times in one hour. What is the maximum number of minute hands that can be on Yura's clocks?
# | # Answer: 28.
Solution. Let two arrows coincide, then after 30 seconds they will coincide again. Therefore, the arrows in each such pair will coincide exactly 2 times per minute. Thus, if $n$ arrows move in one direction, and $m$ arrows move in the other, then $2 m n=54, m n=27$. Therefore, $n$ can be $1,3,9$ or 27. T... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. (12 points) In how many ways can eight of the nine digits $1,2,3,4,5,6$, 7,8 and 9 be placed in a $4 \times 2$ table (4 rows, 2 columns) so that the sum of the digits in each row, starting from the second, is 1 more than in the previous one? | # Answer: 64.
Solution. The sum of all nine numbers is 45. Let $x$ be the sum of the two numbers in the first row, and let $a$ be the one number out of the nine that we do not place in the figure. Then $x+x+1+x+2+x+3=45-a$, from which $4 x+a=39$. Since $a$ is an integer from 1 to 9, we get 2 possible cases: either $x=... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.1. (12 points) For different natural numbers $k, l, m, n$, it is known that there exist such three natural numbers $a, b, c$ that each of the numbers $k, l, m, n$ is a root of either the equation $a x^{2}-b x+c=0$, or the equation $c x^{2}-16 b x+256 a=0$. Find $k^{2}+l^{2}+m^{2}+n^{2}$. | Answer: 325.
Solution. If $k, l$ are the roots of the first equation, then the roots of the second equation are the numbers $m=\frac{16}{k}$, $n=\frac{16}{l}$. Therefore, the numbers $k, l, m, n$ are divisors of the number 16. The divisors of 16 are the numbers 1, 2, 4, 8, and 16, but the number 4 does not fit, as by ... | 325 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Vanya thought of a two-digit number, then swapped its digits and multiplied the resulting number by itself. The result turned out to be four times larger than the number he thought of. What number did Vanya think of? | Answer: 81.
Solution. Let the intended number be $\overline{m n}=10 m+n$. Then $4 \overline{m n}=\overline{n m}^{2}$. Therefore, $\overline{n m}^{2}$ is divisible by 4, and $\overline{n m}$ is divisible by 2, so the digit $m$ is even (and not zero). Moreover, $\overline{m n}=\overline{n m}^{2}: 4=(\overline{n m}: 2)^{... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. The Martian traffic light consists of six identical bulbs arranged in two horizontal rows (one above the other) with three bulbs in each. A rover driver in the fog can distinguish the number and relative position of the lit bulbs on the traffic light (for example, if two bulbs are lit, whether they are in th... | Answer: 44.
Solution. If two traffic light signals differ only by the shift of the lit bulbs, then the driver cannot distinguish them (and vice versa). Therefore, any signal can either be transformed by a shift to the left and/or up into an indistinguishable signal, which has at least one bulb lit in the top horizonta... | 44 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.1. (2 points) The average age of employees in a company consisting of 13 people is 35 years. After a new employee is hired, the average age of the employees becomes 34 years. Find the age of the new employee. | Answer: 21.
Solution. The sum of the employees' ages before the new hire was $13 \cdot 35$, and after the new employee was hired, the sum of the ages became $14 \cdot 34$. Therefore, the age of the new employee is $14 \cdot 34 - 13 \cdot 35 = 35 - 14 = 21$. | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. (12 points) An eraser, 3 pens, and 2 markers cost 240 rubles. Two erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. (12 points) In an acute-angled triangle $A B C$, angle $A$ is $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of angle $B_{1}... | # Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ b... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.1. (12 points) The equation $x^{2}+5 x+1=0$ has roots $x_{1}$ and $x_{2}$. Find the value of the expression
$$
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}
$$ | Answer: 220.
Solution. Since $x_{1}^{2}=-5 x_{1}-1$, then $\left(1+x_{1}\right)^{2}=1+2 x_{1}-5 x_{1}-1=-3 x_{1}$. Therefore,
$$
\begin{gathered}
\left(\frac{x_{1} \sqrt{6}}{1+x_{2}}\right)^{2}+\left(\frac{x_{2} \sqrt{6}}{1+x_{1}}\right)^{2}=6\left(\frac{-5 x_{1}-1}{-3 x_{2}}+\frac{-5 x_{2}-1}{-3 x_{1}}\right)=\frac{... | 220 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cycl... | Answer: 40.
Solution. Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. (14 points) Mitya is 11 years older than Shura. When Mitya was as old as Shura is now, he was twice as old as she was. How old is Mitya? | Answer: 33.
Solution. 11 years ago, Shura was half as old as she is now. So, she is 22 years old, and Mitya is 33. | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. (14 points) Represent the number $\frac{201920192019}{191719171917}$ as an irreducible fraction. In the answer, write down the denominator of the resulting fraction. | Answer: 639.
Solution. We have
$$
\frac{201920192019}{191719171917}=\frac{2019 \cdot 100010001}{1917 \cdot 100010001}=\frac{2019}{1917}=\frac{3 \cdot 673}{3 \cdot 639}=\frac{673}{639}
$$
Since $639=3^{2} \cdot 71$ and 673 is not divisible by 3 and 71, the resulting fraction is irreducible. | 639 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4.1. (14 points) How many six-digit numbers exist whose sum of digits equals 51? | Answer: 56.
Solution. Since the maximum sum of the digits of a six-digit number is $6 \cdot 9=54$, the sought numbers are $699999, 789999, 888999$, as well as all numbers obtained from them by rearranging the digits. The first of these numbers can form 6 six-digit numbers (the digit 6 can be in any of the 6 positions)... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. (14 points) Find the smallest natural number whose representation contains 5 zeros and 7 ones, and the sum of the digits in the even positions is equal to the sum of the digits in the odd positions. | Answer: 1000001111131.
Solution. The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Consider the case where the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. Thus, to find the smallest number, we choose... | 1000001111131 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 2.
B-1.
At the time when a lion cub, who was 5 minutes away, set off for the watering hole, the second cub, having already quenched its thirst, started heading back along the same road at 1.5 times the speed of the first. At the same time, a tortoise, which was half an hour away, set off for the watering hole ... | Answer: 28
Solution. Let's take the entire path of the turtle as 1 and let $x$ be the speed of the 1st lion cub. Then the speed of the 2nd lion cub is $1.5x$, and the speed of the turtle is $1/30$. The entire path to the watering hole for the 1st lion cub is $5x$. Therefore, the meeting with the 2nd lion cub occurred ... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
A square with a side of 75 mm was cut by two parallel cuts into three rectangles. It turned out that the perimeter of one of these rectangles is half the sum of the perimeters of the other two. What is this perimeter? Give your answer in centimeters. | # Answer: 20
Solution. For three rectangles, one of the sides is the same and equals 75 mm. The sum of the lengths of two such sides of one rectangle is half the sum of the lengths of all such sides of the other two. Therefore, the other side of the first rectangle is also half the sum of the other sides of the two ot... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
Find the smallest 12-digit natural number that is divisible by 36 and contains all 10 digits in its decimal representation. | Answer: 100023457896
Solution. The number is divisible by 4 and 9. Since the sum of ten digits is 45 (divisible by 9), two more digits should be added to these ten digits, the sum of which is 0, 9, or 18. Since we need the smallest number, we will add two digits 0 and place the number 10002345 at the beginning of the ... | 100023457896 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5.
B-1
Chicks hatch on the night from Sunday to Monday. For two weeks, the chick sits with its beak open, the third week it silently grows feathers, and on the fourth week, it flies away from the nest. Last week, 20 chicks were sitting with their beaks open, and 14 were growing feathers, while this week, 15... | Answer: 225
Solution. In fact, the chicks should be divided into three categories: one-week-old, two-week-old, and three-week-old. With each new week, each chick moves to the next category. So, if 11 are feathering this week, then last week there were 11 two-week-olds, and, accordingly, 9 one-week-olds ( $9+11=20$ ). ... | 225 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6.
In the alphabet of the inhabitants of the magical planet ABV2020, there are only three letters: A, B, and V, from which all words are formed. In any word, two identical letters cannot be adjacent, and each of the three letters must be present in any word. For example, the words ABV, VABAVAB, and BVBVAB ar... | # Answer: 1572858
Solution. The first letter can be any of the three, and for each subsequent letter, there are two options. This results in $3 \cdot 2^{19}$ words. However, we need to subtract from this the words that are made up of only two letters, not three. There are 6 such words. Therefore, we get $3 \cdot 2^{19... | 1572858 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 7.
Around a round table, 1001 people are sitting, each of whom is either a knight (always tells the truth) or a liar (always lies). It turned out that next to each knight sits exactly one liar, and next to each liar there is a knight. What is the minimum number of knights that can sit at the table? | Answer: 502
Solution. From the condition, it follows that a knight cannot sit between two knights or two liars, and a liar cannot sit between two liars. Thus, when moving around the table, knights will be encountered in pairs, while liars will be encountered singly or in pairs. From this, it follows that each liar can... | 502 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The numbers from 1 to 8 are arranged at the vertices of a cube such that the sum of the numbers in any three vertices lying on the same face is at least 10. What is the smallest possible sum of the numbers at the vertices of one face? | Answer: 16.
Solution. Each face has a vertex where a number not less than 6 is placed. Indeed, otherwise, one of the triples of the remaining largest numbers $2,3,4,5$ would give a sum less than 10 (specifically, the triple $2,3,4$ with a sum of 9).
Consider a face containing the vertex where the number 6 is placed. ... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. All natural numbers, the sum of the digits of each of which is equal to 5, were arranged in ascending order. What number is in the 125th place | Answer: 41000.
Solution. Let's calculate the number of $n$-digit numbers, the sum of the digits of each of which is equal to 5, for each natural $n$. Subtract 1 from the leading digit, we get a number (which can now start with zero), the sum of the digits of which is equal to 4. Represent the digits of this number as ... | 41000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In "Dragon Poker," the deck has four suits. An Ace brings 1 point, a Jack -2 points, a Two $-2^{2}$, a Three $-2^{3}$, ..., a Ten $-2^{10}=1024$ points. Kings and Queens are absent. You can choose any number of cards from the deck. In how many ways can you score 2018 points? | Answer: $C_{2021}^{3}=1373734330$.
Solution. In any suit, you can score any number of points from 0 to 2047, and this can be done in a unique way. This can be done as follows: write down this number in binary and select the cards corresponding to the binary positions containing 1 (i.e., if there is a 1 in the $k$-th p... | 1373734330 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8.
Inside a convex $n$-gon, 100 points are placed such that no three of these $n+100$ points lie on the same line. The polygon is divided into triangles, each of whose vertices are 3 of the given $n+100$ points. For what maximum value of $n$ can there not be more than 300 triangles? | Answer: 102
Solution. If each point is a vertex of some triangle, then the sum of the angles of all the obtained triangles is $180^{\circ} \cdot(n-2)+360^{\circ} \cdot 100=180^{\circ} \cdot(n-2+200)=180^{\circ} \cdot(n+198)$. Therefore, $\frac{180^{\circ} \cdot(n+198)}{180^{\circ}}=n+198$ triangles are obtained. If on... | 102 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.
Given a polynomial $P(x)$ of degree 10 with the leading coefficient 1. The graph of $y=P(x)$ lies entirely above the $O x$ axis. The polynomial $-P(x)$ is factored into irreducible factors (i.e., polynomials that cannot be expressed as the product of two non-constant polynomials). It is known that at $x=2... | Answer: 243
Solution. Note that since the graph of $y=P(x)$ is entirely above the $O x$ axis, the polynomial $P(x)$ has no real roots. This means that all irreducible polynomials in the factorization have degree 2, and their number is 5. Since the polynomial $-P(x)$ will have the same number of irreducible polynomials... | 243 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 10.
Set $A$ on the plane $O x y$ is defined by the equation $x^{2}+y^{2}=2 x+2 y+23$. Set $B$ on the same plane is defined by the equation $|x-1|+|y-1|=5$. Set $C$ is the intersection of sets $A$ and $B$. What is the maximum value that the product of the lengths of $n$ segments $X Y_{1} \cdot X Y_{2} \cdot X... | Answer: 1250
Solution. The set $A$ is $(x-1)^{2}+(y-1)^{2}=25$, which is a circle with center $(1 ; 1)$ and radius 5. The set $B$ is a square with vertices $(-4 ; 1),(1 ; 6),(6 ; 1),(1 ;-4)$. The center of the square is the point $(1 ; 1)$, the diagonals of the square are 10, and the sides are $5 \sqrt{2}$. The set $C... | 1250 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. After treating the garden with a caterpillar control agent, the gardener noticed that from 12 blackcurrant bushes he was getting the same harvest as before from 15 bushes. By what percentage did the blackcurrant yield in the garden increase? | Answer: $25 \%$.
Solution. The yield from 12 bushes increased by $15 / 12=1.25$ times, which means the yield of currants increased by $25 \%$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. What is the degree measure of angle $\angle A$, if its bisector forms an angle with one of its sides that is three times smaller than the angle adjacent to $\angle A$? | Answer: $72^{\circ}$.
Solution. Let $x$ be the degree measure of the angle formed by the angle bisector of $\angle A$ with one of its sides. Then the degree measure of the angle $\angle A$ is $2 x$, and the degree measure of the adjacent angle is $3 x$. Therefore, $2 x+3 x=180$, from which $x=36$ and $\angle A=72^{\ci... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. In "Dragon Poker," the deck has four suits. An Ace brings 1 point, a Jack -2 points, a Two $-2^{2}$, a Three $-2^{3}, \ldots$, a Ten $-2^{10}=1024$ points. Kings and Queens are absent. You can choose any number of cards from the deck. In how many ways can you score 2018 points? | Answer: $C_{2021}^{3}=1373734330$.
Solution. In any suit, you can score any number of points from 0 to 2047, and this can be done in a unique way. This can be done as follows: write down this number in binary and select the cards corresponding to the binary positions containing 1 (i.e., if there is a 1 in the $k$-th p... | 1373734330 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.1. Calculate: $\frac{1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+2013 \cdot 2014}{(1+2+3+\ldots+2014) \cdot \frac{1}{5}}$. If necessary, round the answer to the nearest hundredths. | Answer: 6710. Solution. Let's calculate the sum in the numerator: $\sum_{k=1}^{n} k(k+1)=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$ $=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$. Here, the well-known formula for the sum of squares of natural numbers is used, which can be proven in various ways (these pr... | 6710 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find the sum of all integer values of $a$ belonging to the interval $[-10 ; 10]$, for each of which the double inequality $5 \leq x \leq 10$ implies the inequality $a x+3 a^{2}-12 a+12>a^{2} \sqrt{x-1}$. | Answer: -47. Solution. Let $t=\sqrt{x-1}$, then the original problem reduces to finding such values of $a$ for which the inequality $f(t) \equiv a t^{2}-a^{2} t+3 a^{2}-11 a+12>0$ holds for all $t \in[2 ; 3]$. This means that the minimum of the function $f(t)$ on the interval $2 \leq t \leq 3$ is positive.
If $a=0$, t... | -47 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3.1. (14 points) Svetlana, Katya, Olya, Masha, and Tanya attend a math club, in which more than $60 \%$ of the students are boys. What is the smallest number of schoolchildren that can be in this club? | Answer: 13.
Solution. Let $M$ be the number of boys, $D$ be the number of girls in the club. Then $\frac{M}{M+D}>\frac{3}{5}$, hence $M>\frac{3}{2} D \geqslant \frac{15}{2}$. The minimum possible value of $M$ is 8, and the minimum possible value of $D$ is 5, making a total of 13 children. | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4.1. (14 points) Find all integer values that the fraction $\frac{8 n+157}{4 n+7}$ can take for natural $n$. In your answer, write the sum of the found values. | Answer: 18.
Solution. We have $\frac{8 n+157}{4 n+7}=2+\frac{143}{4 n+7}$. Since the divisors of the number 143 are only $1, 11, 13$, and 143, integer values of the fraction are obtained only when $n=1$ and $n=34$, which are 15 and 3, respectively, and their sum is 18. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. (14 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offer... | Answer: 199.
Solution. If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If th... | 199 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. (14 points) We will call a natural number a snail if its representation consists of the representations of three consecutive natural numbers, concatenated in some order: for example, 312 or 121413. Snail numbers can sometimes be squares of natural numbers: for example, $324=18^{2}$ or $576=24^{2}$. Find a four-dig... | Answer: 1089.
Solution. Note that a snail number can only be a four-digit number if the three numbers from which its record is formed are $8, 9, 10$ (numbers $7, 8, 9$ and smaller still form a three-digit number, while $9, 10, 11$ and larger - already form a number of no less than five digits). It remains to figure ou... | 1089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. The sequence $\left\{x_{n}\right\}$ is defined by the conditions $x_{1}=20, x_{2}=17, x_{n+1}=x_{n}-x_{n-1}(n \geqslant 2)$. Find $x_{2018}$. | Answer: 17.
Solution: From the condition, it follows that $x_{n+3}=x_{n+2}-x_{n+1}=x_{n+1}-x_{n}-x_{n+1}=-x_{n}$, therefore $x_{n+6}=$ $-x_{n+3}=x_{n}$, i.e., the sequence is periodic with a period of 6. Since $2018=6 \cdot 336+2$, we get $x_{2018}=x_{2}=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Philatelist Andrey decided to distribute all his stamps equally into 2 envelopes, but it turned out that one stamp was left over. When he distributed them equally into 3 envelopes, one stamp was again left over; when he distributed them equally into 5 envelopes, 3 stamps were left over; finally, when he tried to d... | Answer: 223.
Solution. If the desired number is $x$, then from the first sentence it follows that $x$ is odd, and from the rest it follows that the number $x+2$ must be divisible by 3, 5, and 9, i.e., has the form $5 \cdot 9 \cdot p$. Therefore, $x=45(2 k-1)-2=90 k-47$. According to the condition $150<x \leqslant 300$... | 223 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{11 \ldots 11}_{2017} \underbrace{22 \ldots .22}_{2018} 5}$. | Answer: 6056 (the number from the condition of the problem is $\underbrace{33 \ldots 33}_{2017} 5$).
Solution. Since
$\underbrace{11 \ldots 11}_{2017} \underbrace{22 \ldots 22}_{2018} 5=\frac{10^{2017}-1}{9} \cdot 10^{2019}+\frac{10^{2018}-1}{9} \cdot 20+5=\frac{10^{4036}+10^{2019}+25}{9}=\left(\frac{10^{2018}+5}{3}\... | 6056 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.2. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{11 \ldots 11}_{2018} \underbrace{55 \ldots 55}_{2017} 6}$. | Answer: 6055 (the number from the condition of the problem is $\underbrace{33 \ldots 33}_{2017} 4$ ). | 6055 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.3. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{44.44}_{2017} \underbrace{22 \ldots 22}_{2018}} 5$. | Answer: 12107 (the number from the condition of the problem is $\underbrace{66 \ldots 66}_{2017} 5$ ). | 12107 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.4. Find the sum of the digits in the decimal representation of the integer part of the number $\sqrt{\underbrace{44.44}_{2018} \underbrace{88 \ldots 88}_{2017} 9}$. | Answer: 12109 (the number from the condition of the problem is equal to $\underbrace{66 \ldots 66}_{2017} 7$ ). | 12109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. Find all integer solutions of the equation $x \ln 27 \log _{13} e=27 \log _{13} y$. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 70. | Answer: 117.
Solution: The original equation is equivalent to the equation $\frac{x \ln 27}{\ln 13}=\frac{27 \ln y}{\ln 13} \Leftrightarrow x \ln 27=27 \ln y \Leftrightarrow$ $\ln 27^{x}=\ln y^{27} \Leftrightarrow 27^{x}=y^{27} \Leftrightarrow 3^{x}=y^{9}$, with $y \geqslant 1$, and thus $x \geqslant 0$. Since 3 is a ... | 117 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.1. Find the smallest 12-digit natural number that is divisible by 36 and contains each of the 10 digits at least once. | Answer: 100023457896.
Solution. The number must be divisible by 4 and by 9. Since the sum of ten different digits is 45, the sum of the two remaining digits must be 0, 9, or 18. We need the smallest number, so we add two digits 0 to the digits $0,1, \ldots, 9$ and place the digits 10002345 at the beginning of the desi... | 100023457896 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1. The sine of the dihedral angle at the lateral edge of a regular quadrilateral pyramid is $\frac{15}{17}$. Find the area of the lateral surface of the pyramid if the area of its diagonal section is $3 \sqrt{34}$. | Answer: 68.
Solution. Let the linear angle of the dihedral angle given in the problem be denoted as $\alpha$. This angle is always obtuse, so $\cos \alpha=-\frac{8}{17}$.
The projection of the lateral face onto the diagonal section is a triangle, the area of which is half the area of this section. Since the dihedral ... | 68 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.1. A batch of tablets from four different brands was delivered to a computer store. Among them, Lenovo, Samsung, and Huawei tablets made up less than a third, with Samsung tablets being 6 more than Lenovo. All other tablets were Apple iPads, and there were three times as many of them as Huawei. If the number of Lenov... | Answer: 94.
Solution. Let $n$ be the total number of tablets we are looking for, with $x$ being the number of Lenovo brand tablets, and $y$ being the number of Huawei brand tablets. Then the number of Samsung tablets is $x+6$, and the number of Apple iPad tablets is $n-2x-y-6=3y$. From the problem statement, we also h... | 94 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. Solve the equation $x^{x+y}=y^{y-x}$ in natural numbers. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 1500. | Answer: 2744.
Solution. Let $y = t x$, where $t \in \mathbb{Q}$ for $x, y \in \mathbb{N}$. Then, substituting this into the equation, we find
$$
x = t^{\frac{t-1}{2}}, \quad y = t^{\frac{t+1}{2}}
$$
We will show that the number $t$ can only be an integer. Suppose the contrary: let $t$ be a rational number different ... | 2744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1.
In-1 Find $f(1)+f(2)+f(3)+\ldots+f(13)$, if $f(n)=4 n^{3}-6 n^{2}+4 n+13$. | Answer: 28743.
Solution. Since $f(n)=n^{4}-(n-1)^{4}+14$, then
$f(1)+f(2)+f(3)+\ldots+f(13)=1^{4}+2^{4}+\ldots+13^{4}-\left(0^{4}+1^{4}+\ldots+12^{4}\right)+14 \cdot 13=13^{4}+14 \cdot 13=28743$.
V-2 Find $f(1)+f(2)+f(3)+\ldots+f(11)$, if $f(n)=4 n^{3}+6 n^{2}+4 n+9$.
Answer: 20823.
V-3 Find $f(1)+f(2)+f(3)+\ldots... | 28743 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
In-1 The numbers $a, b, c$ are such that each of the two equations $x^{2}+b x+a=0$ and $x^{2}+c x+a=1$ has two integer roots, and all these roots are less than (-1). Find the smallest value of $a$. | Answer: 15.
Solution. By Vieta's theorem, the product of the roots of the first equation is $a$, and the product of the roots of the second equation is $a-1$. Since the roots are integers and less than -1, their product is greater than 1, so each of the two consecutive numbers $a-1$ and $a$ is the product of two diffe... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. At Andrei's birthday, Yana was the last to arrive, giving him a ball, and Eduard was the second to last, giving him a calculator. Testing the calculator, Andrei noticed that the product of the total number of his gifts and the number of gifts he had before Eduard arrived is exactly 16 more than the product of his ag... | Answer: 18. Solution. If $n$ is the number of gifts, and $a$ is Andrei's age, then $n(n-2)=a(n-1)$. From this, $a=\frac{n^{2}-2 n-16}{n-1}=n-1-\frac{17}{n-1}$. Therefore, $n-1=17, a=16$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In an equilateral triangle $ABC$, points $A_{1}$ and $A_{2}$ are chosen on side $BC$ such that $B A_{1}=A_{1} A_{2}=A_{2} C$. On side $AC$, a point $B_{1}$ is chosen such that $A B_{1}: B_{1}C=1: 2$. Find the sum of the angles $\angle A A_{1} B_{1}+\angle A A_{2} B_{1}$. | Answer: $30^{\circ}$. Solution. Since $A_{1} B_{1} \| A B$, then $\angle B A A_{1}=\angle A A_{1} B_{1}$. From symmetry, $\angle B A A_{1}=\angle C A A_{2}$. It remains to note that $\angle C B_{1} A_{2}=$ $\angle B_{1} A A_{2}+\angle A A_{2} B_{1}$ as an exterior angle in $\triangle A A_{2} B_{1}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. All natural numbers are divided into "good" and "bad" according to the following rules:
a) From any bad number, you can subtract some natural number not exceeding its half so that the resulting difference becomes "good".
b) From a "good" number, you cannot subtract no more than half of it so that it remains "good"... | Answer: 2047. Solution. Note that good numbers have the form $2^{n}-1$. Indeed, let a number have the form $M=2^{n}+k$, where $k=0,1, \ldots, 2^{n}-2$. Then from such a number, you can subtract $k+1 \leqslant \frac{1}{2}\left(2^{n}+k\right)=\frac{M}{2}$. On the other hand, from a number of the form $N=2^{n}-1$, you nee... | 2047 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Calculate
$$
\frac{2 a b\left(a^{3}-b^{3}\right)}{a^{2}+a b+b^{2}}-\frac{(a-b)\left(a^{4}-b^{4}\right)}{a^{2}-b^{2}} \quad \text { for } \quad a=-1, \underbrace{5 \ldots 5}_{2010} 6, \quad b=5, \underbrace{4 \ldots 44}_{2011}
$$
Answer: 343. | Solution. Given $a$ and $b$, we get
$$
\begin{gathered}
\frac{2 a b\left(a^{3}-b^{3}\right)}{a^{2}+a b+b^{2}}-\frac{(a-b)\left(a^{4}-b^{4}\right)}{a^{2}-b^{2}}=2 a b(a-b)-(a-b)\left(a^{2}+b^{2}\right)= \\
\quad=-(a-b)\left(a^{2}-2 a b+b^{2}\right)=-(a-b)^{3}=-(-7)^{3}=343
\end{gathered}
$$ | 343 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the smallest natural number that is greater than the sum of its digits by 1755 (the year of the founding of Moscow University). | Answer: 1770.
Solution. From the condition, it follows that the desired number cannot consist of three or fewer digits. We will look for the smallest such number in the form $\overline{a b c d}$, where $a, b, c, d$ are digits, and $a \neq 0$. We will form the equation:
$$
\begin{gathered}
\overline{a b c d}=a+b+c+d+1... | 1770 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.
Let $A(n)$ denote the greatest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9, A(64)=1$. Find the sum $A(111)+A(112)+\ldots+A(218)+A(219)$. | Answer: 12045
Solution. The largest odd divisors of any two of the given numbers cannot coincide, as numbers with the same largest odd divisors are either equal or differ by at least a factor of 2. Therefore, the largest odd divisors of the numbers $111, 112, \ldots, 218, 219$ are distinct. This means that the largest... | 12045 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Two circles touch each other internally at point K. A chord $AB$ of the larger circle touches the smaller circle at point $L$, and $AL=10$. Find $BL$, if $AK: BK=2: 5$. | Answer: $B L=25$.
Solution. Draw the chord $P Q$ and the common tangent $M N$ (see fig.), then
$$
\angle P Q K=\frac{1}{2} \smile P K=\angle P K M=\frac{1}{2} \smile A K=\angle A B K
$$
th... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. What is the smallest (same) number of pencils that need to be placed in each of 6 boxes so that in any 4 boxes there are pencils of any of 26 predefined colors (there are enough pencils available)? | Answer: 13.
Solution. On the one hand, pencils of each color should appear in at least three out of six boxes, since if pencils of a certain color are in no more than two boxes, then in the remaining boxes, which are at least four, there are no pencils of this color. Therefore, the total number of pencils should be no... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
Masha tightly packed 165 identical balls into the shape of a regular triangular pyramid. How many balls are at the base? | Answer: 45
Solution. We will solve the problem in general for $\frac{1}{6} n(n+1)(n+2)$ balls. At the base of a triangular pyramid of height $n$ rows lies the $n$-th triangular number of balls $1+2+3+\ldots+n=\frac{1}{2} n(n+1)$. Then the total number of balls in the pyramid is $\sum_{k=1}^{n} \frac{k(k+1)}{2}=\frac{1... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6.
B-1
Find the minimum value of the expression $4 x+9 y+\frac{1}{x-4}+\frac{1}{y-5}$ given that $x>4$ and $y>5$. | Answer: 71
Solution. $4 x+9 y+\frac{1}{x-4}+\frac{1}{y-5}=4 x-16+\frac{4}{4 x-16}+9 y-45+\frac{9}{9 y-45}+61 \geq 2 \sqrt{4}+2 \sqrt{9}+61=$ 71 (inequality of means). Equality is achieved when $4 x-16=2,9 y-45=3$, that is, when $x=\frac{9}{2}, y=\frac{16}{3}$.
B-2
Find the minimum value of the expression $9 x+4 y+\f... | 71 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of the digits of the number $A$, if $A=2^{63} \cdot 4^{25} \cdot 5^{106}-2^{22} \cdot 4^{44} \cdot 5^{105}-1$. Answer: 959. | Solution. Since $2^{63} \cdot 4^{25} \cdot 5^{106}=2^{113} \cdot 5^{106}=2^{7} \cdot 10^{106}$ and $2^{22} \cdot 4^{44} \cdot 5^{105}=$ $2^{110} \cdot 5^{105}=2^{5} \cdot 10^{105}$, then $A+1=2^{7} \cdot 10^{106}-2^{5} \cdot 10^{105}=\left(2^{7} \cdot 10-2^{5}\right) \cdot 10^{105}=$ $1248 \cdot 10^{105}$. Therefore, t... | 959 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Each Kinder Surprise contains exactly 3 different gnomes, and there are 12 different types of gnomes in total. In the box, there are enough Kinder Surprises, and in any two of them, the triplets of gnomes are not the same. What is the minimum number of Kinder Surprises that need to be bought to ensure that after the... | Answer: 166.
Solution. From 11 different gnomes, a maximum of $C_{11}^{3}=\frac{11 \cdot 10 \cdot 9}{3 \cdot 2 \cdot 1}=165$ different Kinder can be formed, so if you take one more Kinder, there cannot be fewer than 12 different gnomes among them. | 166 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that $m, n, k$ are distinct natural numbers greater than 1, the number $\log _{m} n$ is rational, and, moreover,
$$
k^{\sqrt{\log _{m} n}}=m^{\sqrt{\log _{n} k}}
$$
Find the minimum of the possible values of the sum $k+5 m+n$. | Answer: 278.
Solution. Transform the original equation, taking into account that the numbers $m, n, k$ are greater than 1 and the corresponding logarithms are positive:
$$
\begin{aligned}
& k \sqrt{\log _{m} n}=m \sqrt{\log _{n} k} \Leftrightarrow \log _{m} k \cdot \sqrt{\log _{m} n}=\sqrt{\log _{n} k} \Leftrightarro... | 278 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. To guard the object, the client agreed with the guards on the following: all of them will indicate the time intervals of their proposed shifts with the only condition that their union should form a predetermined time interval set by the client, and he will choose any set of these intervals that also satisfies the sa... | 6. $\frac{90000}{2 \cdot 300}=150 \text{h}$. | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.1. Find the sum of all integer values of $a$ from the interval $[-2012 ; 2013]$, for which the equation $(a-3) x^{2}+2(3-a) x+\frac{a-7}{a+2}=0$ has at least one solution. | Answer: 2011. Solution. When $a=3$ there are no solutions. For other $a$, it should be $\frac{D}{4}=(a-3)^{2}-\frac{(a-3)(a-7)}{a+2} \geq 0$. The last inequality (plus the condition $a \neq 3$) has the solution $a \in(-\infty ;-2) \bigcup\{1\} \cup(3 ;+\infty)$. The desired sum is: $-2012-2011-\ldots-5-4-3+1+4+5+\ldots... | 2011 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. Among the numbers exceeding 2013, find the smallest even number $N$ for which the fraction $\frac{15 N-7}{22 N-5}$ is reducible. | Answer: 2144. Solution: The presence of a common factor in the numbers $15 N-7$ and $22 N-5$ implies that the same factor is present in the number $(22 N-5)-(15 N-7)=7 N+2$, and subsequently in the numbers $(15 N-7)-2 \cdot(7 N+2)=N-11,(7 N+2)-7 \cdot(N-11)=79$. Since 79 is a prime number, the fraction is reducible by ... | 2144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. Find the area of the figure defined on the coordinate plane by the inequality $\sqrt{\arcsin \frac{x}{3}} \leq \sqrt{\arccos \frac{y}{3}} \cdot$ In your answer, specify the integer closest to the found value of the area. | Answer: 16. Solution. The domain of admissible values of the inequality is determined by the system $\left\{\begin{array}{l}\arcsin \frac{x}{3} \geq 0, \\ \arccos \frac{y}{3} \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}0 \leq x \leq 3, \\ -3 \leq y \leq 3 .\end{array}\right.\right.$ If $-3 \leq y \leq 0$, t... | 16 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7.1. Given two different geometric progressions, the first terms of which are equal to 1, and the sum of the denominators is 3. Find the sum of the fifth terms of these progressions, if the sum of the sixth terms is 573. If the answer to the question is not unique, specify the sum of all possible values of the desired ... | Answer: 161. Solution. Let the denominators of the progressions be $p$ and $q$. According to the condition, we get: $\left\{\begin{array}{c}p+q=3, \\ p^{5}+q^{5}=573\end{array}\right.$. We need to find $p^{4}+q^{4}$. Denoting $p+q=a, p q=b$, we express:
$$
\begin{gathered}
p^{4}+q^{4}=\left((p+q)^{2}-2 p q\right)^{2}-... | 161 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Chords $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$ of a sphere intersect at a common point $S$. Find the sum $S A^{\prime}+S B^{\prime}+S C^{\prime}$, if $A S=6, B S=3, C S=2$, and the volumes of pyramids $S A B C$ and $S A^{\prime} B^{\prime} C^{\prime}$ are in the ratio $2: 9$. If the answer is not an integ... | Answer: 18. Solution. If $a, b, c$ are the given edges, $x, y, z$ are the corresponding extensions of these edges, and $k$ is the given ratio, then by the theorem of intersecting chords and the lemma on the ratio of volumes of pyramids with equal (vertical) trihedral angles at the vertex, we have: $\left\{\begin{array}... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find the sum of all such integers $a \in[0 ; 400]$, for each of which the equation $x^{4}-6 x^{2}+4=\sin \frac{\pi a}{200}-2\left[x^{2}\right]$ has exactly six roots. Here the standard notation is used: $[t]$ - the integer part of the number $t$ (the greatest integer not exceeding $t$). | Answer: 60100. Solution. Let $b=\sin \frac{\pi a}{200}, t=x^{2}$. Using the equality $[t]=t-\{t\}$ (here $\{t\}$ is the fractional part of $t$), we get that the original equation is equivalent to the equation $t^{2}-6 t+4=b-2 t+2\{t\}$. Note that any positive $t$ corresponds to two different $x$, two different positive... | 60100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In a box, there are a hundred colorful balls: 28 red, 20 green, 13 yellow, 19 blue, 11 white, and 9 black. What is the smallest number of balls that need to be pulled out without looking into the box to ensure that there are at least 15 balls of the same color among them? | Answer: 76. Solution. Worst-case scenario: 14 red, 14 green, 13 yellow, 14 blue, 11 white, and 9 black balls will be drawn - a total of 75 balls. The next ball will definitely be the 15th ball of one of the colors: either red, green, or blue.
Answer of variant 152: 109.
Answer of variant 153: 83.
Answer of variant 1... | 76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Find the largest four-digit number in which all digits are different, and moreover, no two of them can be swapped to form a smaller number. | Answer: 7089.
Solution. If the digit 0 is not used, then the digits in each such number must be in ascending order, and the largest of such numbers is 6789. Since a number cannot start with zero, using 0, one can obtain additional suitable numbers where 0 is in the hundreds place, and the other digits are in ascending... | 7089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 3. On the board, all such natural numbers from 3 to 223 inclusive are written, which when divided by 4 leave a remainder of 3. Every minute, Borya erases any two of the written numbers and instead writes their sum, decreased by 2. In the end, only one number remains on the board. What can it be? | Answer: 6218.
Solution. Initially, the total number of written numbers is $224: 4=56$, and their sum is $(3+223) \cdot 56: 2=6328$. Every minute, the number of written numbers decreases by 1, and their sum decreases by 2. Therefore, one number will remain on the board after 55 minutes and will be equal to $6328-55 \cd... | 6218 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Masha has seven different dolls, which she arranges in six different doll houses so that each house has at least one doll. In how many ways can Masha do this? It is important which doll ends up in which house. It does not matter how the dolls are seated in the house where there are two of them. | Answer: 15120.
Solution. In each such seating arrangement, in one of the cabins there will be two dolls, and in the other cabins - one each. First, we choose which two dolls will sit in one cabin. This can be done in $7 \cdot 6: 2=21$ ways. Then we choose which cabin to seat these two dolls in. This can be done in 6 w... | 15120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{1}{32}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$
ANSWER: 60. | Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$. | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. There are 5 identical buckets, each with a maximum capacity of some integer number of liters, and a 30-liter barrel containing an integer number of liters of water. The water from the barrel was distributed among the buckets, with the first bucket being half full, the second one-third full, the third one-quarter ful... | The solution $-\frac{x}{2}+\frac{x}{3}+\frac{x}{4}+\frac{x}{5}+\frac{x}{6}=\frac{87 x}{60}=\frac{29 x}{20}$ will be an integer only when $x$ is a multiple of 20. But $x>20$ cannot be taken, as the sum will exceed 30 liters. | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Two oligarchs, Alejandro and Maximilian, looted their country in 2012. It is known that Alejandro's wealth at the end of 2012 equals twice Maximilian's wealth at the end of 2011. And Maximilian's wealth at the end of 2012 is less than Alejandro's wealth at the end of 2011. Which is greater: Maximilian's wealth or th... | Answer: Maximilian's state is greater.
Solution: Consider the table where $z, x-$ are the states of Alejandro and Maximilian in 2011:
| | 2011 | 2012 |
| :--- | :--- | :--- |
| $\mathrm{A}$ | $z$ | $2 x$ |
| $\mathrm{M}$ | $x$ | $y$ |
Then $N=(2 x+y)-(x+z)-$ are the national riches of the country. Subtracting $x$ f... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Several boys and girls gathered around a round table. It is known that exactly for 7 girls, girls are sitting to their left, and for 12 - boys. It is also known that for $75\%$ of the boys, girls are sitting to their right. How many people are sitting at the table? | Answer: 35 people.
Solution: From the condition, it is clear that there are exactly 19 girls.
Note that the number of girls who have boys sitting to their left is equal to the number of boys who have girls sitting to their right. Thus, $75\%$ of the boys equals 12, i.e., there are 16 boys in total sitting at the tabl... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Answer: 18108.
Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+... | 18108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. A trip in Moscow using the "Troyka" card in 2016 costs 32 rubles for one subway ride and 31 rubles for one trip on surface transport. What is the minimum total number of trips that can be made under these rates, spending exactly 5000 rubles? | Answer: 157.
Solution: If $x$ is the number of trips by surface transport and $y$ is the number of trips by subway, then we have $31x + 32y = 5000$, from which $32(x + y) = 5000 + x \geqslant 5000$. Therefore, $x + y \geqslant 5000 / 32 = 156 \frac{1}{4}$. The smallest integer value of $x + y$ is 157, which is achieve... | 157 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. Determine the number of natural divisors of the number $11!=1 \cdot 2 \cdot \ldots \cdot 10 \cdot 11$ that are multiples of three. Answer. 432. | Solution. The prime factorization of the given number is $11!=2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11$. All multiples of three that are divisors of this number have the form $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta} \cdot 11^{\varphi}$, where $\alpha \in[0 ; 8], \beta \in[1 ; 4], \gamma \in[0... | 432 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,4,6,9,10,11,15,16,20,22$. Determine which numbers are written on the board. In your answer, write their product. | Answer: -4914 (numbers on the board: $-3,2,7,9,13$).
Solution. The sum of the numbers in the obtained set is 112. Each of the original five numbers appears 4 times in this sum. Therefore, the sum of the required numbers is $112: 4=28$. The sum of the two smallest numbers is -1, and the sum of the two largest numbers i... | -4914 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.2. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $3,8,9,16,17,17,18,22,23,31$. Determine which numbers are written on the board. In your answer, write their product. | Answer: 3360 (numbers on the board: $1,2,7,15,16$). | 3360 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.3. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $2,6,10,10,12,14,16,18,20,24$. Determine which numbers are written on the board. In your answer, write their product | Answer: -3003 (numbers on the board: $-1,3,7,11,13$ ). | -3003 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.4. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $6,9,10,13,13,14,17,17,20,21$. Determine which numbers are written on the board. In your answer, write their product. | Answer: 4320 (numbers on the board: $1,5,8,9,12$ ). | 4320 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.5. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,5,8,9,11,12,14,18,20,24$. Determine which numbers are written on the board. In your answer, write their product. | Answer: -2002 (numbers on the board: $-2,1,7,11,13$ ). | -2002 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.6. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $5,8,9,13,14,14,15,17,18,23$. Determine which numbers are written on the board. In your answer, write their product. | Answer: 4752 (numbers on the board: $2,3,6,11,12$). | 4752 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.7. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,2,6,7,8,11,13,14,16,20$. Determine which numbers are written on the board. In your answer, write their product. | Answer: -2970 (numbers on the board: $-3,2,5,9,11$). | -2970 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.8. On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $5,9,10,11,12,16,16,17,21,23$. Determine which numbers are written on the board. In your answer, write their product. | Answer: 5292 (numbers on the board: $2,3,7,9,14$). | 5292 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. In the expansion of the function $f(x)=\left(1+x-x^{2}\right)^{20}$ in powers of $x$, find the coefficient of $x^{3 n}$, where $n$ is equal to the sum of all coefficients of the expansion. | Answer: 760.
Solution. The expansion in powers of $x$ has the form $f(x)=\sum_{k=0}^{40} a_{k} x^{k}$. The sum of all coefficients is $\sum_{k=0}^{40} a_{k}=f(1)=1$. Therefore, we need to find the coefficient of $x^{3}$. We have:
$$
\begin{aligned}
f(x)=\left(\left(1-x^{2}\right)+x\right)^{20}=(1- & \left.x^{2}\right... | 760 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find the smallest four-digit number that is not a multiple of 10 and has the following property: if the digits are reversed, the resulting number is a divisor of the original number, and the quotient is different from one.
Answer. 8712. | Solution. Let's find all four-digit numbers with the property specified in the problem. Let $\overline{a b c d}$ be the desired number, then $\overline{c b a d}$ is the number obtained from it by reversing the digits, $a, d \neq 0$. According to the condition, $\overline{a b c d}=k \cdot \overline{c b a d}$, where $k$ ... | 8712 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{224}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$ ANSWER:60. | Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$. | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Construct on the plane the set of points whose coordinates satisfy the inequality $|3 x+4|+|4 y-3| \leq 12$. In your answer, specify the area of the resulting figure.
ANSWER: 24. | Solution: We will shift the coordinate axes by $-\frac{4}{3}$ along the Ox axis, and by $\frac{3}{4}$ along the Oy axis. In the new coordinates, the obtained figure is described by the inequality $|3 x|+|4 y| \leq 12$. It is not difficult to show that this will be a rhombus with diagonals of 6 and 8, therefore, its are... | 24 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. Find the largest ten-digit number of the form $\overline{a_{9} a_{8} a_{7} a_{6} a_{5} a_{4} a_{3} a_{2} a_{1} a_{0}}$, possessing the following property: the digit equal to $\mathrm{a}_{\mathrm{i}}$ appears in its representation exactly $\mathrm{a}_{9-\mathrm{i}}$ times (for example, the digit equal to $\mathrm{a}_... | Solution: We will divide all digits into pairs $a_{0}-a_{9}, \ldots, a_{4}-a_{5}$. We will prove that the digit 9 cannot be present in the given number. Suppose $a_{i}=9$, then the paired digit $a_{9-i}$ appears 9 times. If $a_{9-i} \neq 9$, then 9 appears only once, so we get a number consisting of one nine and nine o... | 8888228888 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.1. (2 points) On two bookshelves, there are mathematics books, with an equal number on each. If 5 books are moved from the first shelf to the second, then the second shelf will have twice as many books as the first. How many books are there in total on both shelves? | Answer: 30.
Solution. If $x$ is the number of books on both shelves, then $\frac{x}{3}+5=\frac{2 x}{3}-5$, from which $x=30$. | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. (14 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the share (in percent) of masters in t... | Answer: 76.
Solution. Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from which we find $19x = 6y$. Therefore, the proportion of masters is $\frac{y}{x+y} = \frac{19y}{19x + 19y} = \frac{19y}{25y} = 0.76$, i.e., $76\%$. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. (14 points) Calculate $x^{3}+\frac{1}{x^{3}}$, given that $x+\frac{1}{x}=3$. | Answer: 18.
Solution. If $x+\frac{1}{x}=a$, then $a^{3}=\left(x+\frac{1}{x}\right)^{3}=x^{3}+3\left(x+\frac{1}{x}\right)+\frac{1}{x^{3}}$, from which $x^{3}+\frac{1}{x^{3}}=a^{3}-3 a=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. (14 points) An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. (14 points) In an acute-angled triangle $A B C$, angle $A$ is equal to $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of ang... | Answer: 75.
Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ be... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.1. (14 points) In a hut, several residents of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The residents of the Ah tribe always tell the truth, while the residents of the Ukh tribe always lie. One of the residents said: "There are no more than 16 of us in the hut," and then added: "We... | Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 peop... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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