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78(1188). Check that the equality is correct:
a) $\sqrt{2 \frac{2}{3}}=2 \sqrt{\frac{2}{3}}$
b) $\sqrt{5 \frac{5}{24}}=5 \sqrt{\frac{5}{24}}$.
Specify the condition under which the observed pattern holds. Provide examples. | Solution. a) Method I. Transform the left side of the equation:
$$
\sqrt{2 \frac{2}{3}}=\sqrt{\frac{8}{3}}=\sqrt{\frac{4 \cdot 2}{3}}=2 \sqrt{\frac{2}{3}}
$$
Method II. Transform the right side of the equation:
$$
2 \sqrt{\frac{2}{3}}=\sqrt{\frac{8}{3}}=\sqrt{2 \frac{2}{3}}
$$
Method III. Square both sides of the e... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
101(1211). There are two arithmetic progressions $\left(a_{n}\right): 1 ; 5 ; 9 ; 13 ; 17 ; \ldots$ and $\left(b_{n}\right): 2 ; 5 ; 8 ; 11 ; 14 ; \ldots$. Prove that if you write down all the common terms of both progressions, the resulting sequence will also be an arithmetic progression. What is the common difference... | Solution. Method I. Let's find the $n$-th and $k$-th terms of the given arithmetic progressions. We have:
$$
\begin{aligned}
& a_{n}=a_{1}+d(n-1)=1+4(n-1)=4 n-3 \\
& b_{k}=b_{1}+d(k-1)=2+3(k-1)=3 k-1
\end{aligned}
$$
We need to find all such $n$ for which $a_{n}=b_{k}$, i.e., $4 n-3=3 k-1$, from which $k=\frac{4 n-2}... | 12 | Algebra | proof | Yes | Yes | olympiads | false |
109. In the equation $x^{2}-2 x+a=0$, the square of the difference of the roots is 20. Find $a$. | If $x_{1}$ and $x_{2}$ are the roots of the given quadratic equation, then we have the system of equations:
$$
\left\{\begin{array}{l}
\left(x_{1}-x_{2}\right)^{2}=20 \\
x_{1}+x_{2}=2
\end{array}\right.
$$
Squaring the second equation, we get:
$$
\left\{\begin{array}{c}
x_{1}^{2}-2 x_{1} x_{2}+x_{2}^{2}=20 \\
x_{1}^... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
140. Find the maximum value of the function
$$
y(x)=3 \sin x+4 \cos x
$$ | Solution. $y(x)=5\left(\frac{3}{5} \sin x+\frac{4}{5} \cos x\right)$. Since $\left(\frac{3}{5}\right)^{2}+$ $+\left(\frac{4}{5}\right)^{2}=1$, there exists an angle $x_{0}$ such that $\frac{3}{5}=\cos x_{0}, \frac{4}{5}=\sin x_{0}$. Then $y(x)=5 \sin \left(x+x_{0}\right)$. It is obvious that the maximum value of $y(x)$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
143. For which $n \in \boldsymbol{N}$ does the equality
$$
\sqrt[n]{17 \sqrt{5}+38}+\sqrt[n]{17 \sqrt{5}-38}=\sqrt{20} ?
$$ | S o l u t i o n. Denoting the first term by $a$, we will have $a+\frac{1}{a}=\sqrt{20}$, from which $a=\sqrt{5}+2$. Since $(\sqrt{5}+2)^{3}=17 \sqrt{5}+$ $+38=(\sqrt{5}+2)^{n}$, then $n=3$.
A n s w e r: $n=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
168. Two pedestrians set out towards each other at the same time: the first from point $A$, the second from point $B$. The first pedestrian walked 1 km more before the meeting than the second. The first pedestrian arrived at point $B$ 45 minutes after the meeting. The second pedestrian arrived at point $A$ 1 hour and 2... | S o l u t i o n. Let the speed of the first pedestrian who left point $A$ be $v_{1}$ kilometers per hour, and the speed of the second pedestrian be $v_{2}$ kilometers per hour. Since after meeting, the first pedestrian walked $\frac{3}{4} v_{1}$ kilometers, and the second $-\frac{4}{3} v_{2}$ kilometers, and the first ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A person has 12 pints of wine in a barrel (pint - an old French unit of volume, 1 pint ≈ 0.568 liters) and wants to give away half of the wine, but does not have a 6-pint container. However, there are two empty containers with capacities of 8 pints and 5 pints. How can one use them to measure out exactly 6 pints of ... | $\triangle$ The solution to the problem can be written as follows:
| Vessel with a capacity of 8 pints | 0 | 8 | 3 | 3 | 0 | 8 | 6 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Vessel with a capacity of 5 pints | 0 | 0 | 5 | 0 | 3 | 3 | 5 | 0 |
This should be understood as follows. Initiall... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. There is no less than 10 liters of milk in the bucket. How can you pour exactly 6 liters of milk from it using an empty nine-liter bucket and a five-liter bucket? | $\triangle$ Let's denote the initial amount of milk in the first bucket as $a$ liters. Let's think about how to use the fact that the number $a$ is not less than 10. The difference $a-10$ can be used, but the difference $a-11$ cannot. The solution is written as follows:
| Bucket with volume $a$ l | $a$ | $a-5$ | $a-5$... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
92. There are 9 coins, 8 of which are genuine and of the same weight, and one is counterfeit and heavier than the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the counterfeit coin? | $\triangle$ Let's number the coins with natural numbers from 1 to 9. Here, it is better to weigh the coins by placing three on each pan of the balance. Place coins numbered $1,2,3$ on the left pan, and coins numbered $4,5,6$ on the right pan. There are two possible outcomes, as in problem 90.
Suppose the balance is in... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
93. There are 10 coins, 9 of which are genuine and of the same weight, and one is counterfeit and lighter than the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the counterfeit coin? | $\triangle$ Let's number the coins. Place coins numbered $1,2,3$ on the left pan and coins numbered $4,5,6$ on the right pan. We can use Figure 30 for further steps.
If the scales balance, the counterfeit coin is among the remaining four. It takes two more weighings to determine the counterfeit coin from these four (s... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
97. There are 5 parts that are indistinguishable in appearance, 4 of which are standard and of the same mass, and one is defective, differing in mass from the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the defective part? | $\triangle$ Let's number the parts. Now try to figure out the weighing scheme presented below (Fig. 32).
As can be seen from this, it took three weighings to find the defective part.
Answer: in three. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
99. There are 6 identical-looking coins, but 4 are genuine, of the same weight, while 2 are counterfeit, lighter, and also weigh the same. What is the minimum number of weighings on a balance scale without weights that are needed to find both counterfeit coins? | $\triangle$ As usual, let's number the coins. During the first weighing, we will place coins numbered $1,2,3$ on the left pan, and all the others on the right (Fig. 33).
The scales may balance. Then each of the triplets contains one counterfeit coin. To find them, we need two more weighings (see the solution to proble... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
111. How many five-digit natural numbers can be formed using the digits 1 and 0 if the digit 1 appears exactly three times in each number? | $\triangle$ We will search for the specified numbers by enumeration, ensuring that we do not miss any number. It is simpler to start by finding the positions for the two zeros, because if the positions for the zeros are determined, the three remaining positions will be filled with ones uniquely.
Let's fix one of the z... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
215. The sequence $\left(x_{n}\right)$ is defined by the recurrence relations
$$
x_{1}=1, x_{2}=2, x_{n}=\left|x_{n-1}-x_{n-2}\right|(n>2) .
$$
Find $x_{1994}$. | $\triangle$ First, let's compute the first few terms of the sequence: $x_{1}=1, x_{2}=2, x_{3}=|2-1|=1, x_{4}=|1-2|=1, x_{5}=|1-1|=0, x_{6}=$ $=|0-1|=1, x_{7}=1, x_{8}=0, x_{9}=1, x_{10}=1, x_{11}=0$.
Starting from the third term, the terms of this sequence are either 1 or 0. Only the terms with indices
$$
5,8,11, \l... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
234*. Once in a room, there were several inhabitants of an island where only truth-tellers and liars live. Three of them said the following.
- There are no more than three of us here. All of us are liars.
- There are no more than four of us here. Not all of us are liars.
- There are five of us. Three of us are liars.
... | $\triangle$ Let's consider the first of the three speakers. Suppose he is a truth-teller. Then both of his statements, including the second one, "All of us are liars," would be true, which means he is a liar. We have reached a contradiction. Therefore, he can only be a liar.
In this case, both statements of the first ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
299. There are 8 balls: 2 red, 2 blue, 2 white, and 2 black. Players A and B take turns attaching one ball to one of the vertices of a cube. Player A aims to achieve a vertex such that this vertex and its three adjacent vertices each have a ball of a different color, while Player B aims to prevent this. Who will win wi... | $\triangle$ The problem does not specify who starts the game - A or B. Therefore, let's consider two cases.
1) Suppose A starts the game.
If A, for example, attaches a red ball to one of the vertices of the cube, then B's best response is to attach a red ball to one of the adjacent vertices, and so on (Fig. 36). Cons... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
350. Among 18 coins, one is counterfeit. Genuine coins weigh the same, while the counterfeit coin differs in weight from the genuine ones. What is the minimum number of weighings on correct balance scales without weights needed to determine whether the counterfeit coin is lighter or heavier than the genuine ones? (Ther... | $\triangle$ Let's number the coins. Divide the set of coins into three piles, with 6 coins in each.
For the first weighing, place all the coins from the first pile on one scale pan, and all the coins from the second pile on the other. There are two possible cases.
1) Suppose the scales balance during this weighing. T... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
444. Eight hockey teams play against each other in a round-robin tournament to determine the final four. What is the minimum number of points that guarantees a team's advancement to the final four? | $\triangle$ All eight teams together will score $\frac{8 \cdot 7}{2} \cdot 2=56$ points. Therefore, 7, 8, and even 9 points do not guarantee a team's advancement to the final four.
What about 10 points? Imagine that there are five teams that played all their matches against each other to a draw, while the other three ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
454. Several identical boxes together weigh 10 tons, with each of them weighing no more than 1 ton. What is the minimum number of three-ton trucks needed to haul away all this cargo in one trip? | $\triangle$ Four trucks may not be enough. For example, if there are 13 identical boxes weighing $\frac{10}{13}$ tons each, then in one of the trucks, you cannot place more than three boxes, because with four boxes, the weight is $4 \cdot \frac{10}{13}$ tons, which is more than 3 tons. In this case, with four trucks, y... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 7.
The following very simple problem is one of many entertaining problems that have gained widespread popularity. In a dark room stands a wardrobe, in the drawer of which there are 24 red and 24 blue socks. How many socks should be taken from the drawer to ensure that at least one pair of socks of the same color ca... | 7. Usually, the question of the problem is given an incorrect answer: 25 socks. If the problem asked how many socks should be taken from the drawer to ensure that at least 2 socks of different colors are among them, then the correct answer would indeed be such: 25 socks. But in our problem, the question is about ensuri... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Alice returned the rattle to its rightful owner, but a few days later, another brother broke the rattle again. This time, the raven did not come to scare the brothers, and they began to beat each other with all their might. Alice grabbed the broken rattle and ran out of the forest.
After some time, Alice met the White... | 60. The chances are zero. Suppose the statement of the brother Alice met is true. Then the owner of the rattle should have been lying on the day of the meeting and, consequently, could not have been the brother Alice met. On the other hand, suppose the statement of the brother Alice met is false. Then the owner of the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 107.
On another island of knights, liars, and normal people, the king held opposite views and gave his daughter different paternal advice: “My dear, I don’t want you to marry any knight or liar. I would like your husband to be a solid, reputable normal person. You should not marry a knight because all knights are h... | 107. In both cases, a single statement would suffice. The King could be convinced by the true statement “I am not a knight” (such a statement could not belong to either a knight or a liar) and the false statement “I am a liar.” | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
132.
In one museum, I had the chance to see a pair of caskets adorned with the following inscriptions:
On the gold one
Both caskets in this set were made by members of the Cellini family
On the silver one
Neither of these caskets was made by either a son of Bellini or a son of Cellini
Whose work is each of the tw... | 132. The statement engraved on the lid of the golden casket cannot be true, for otherwise we would have arrived at a contradiction. Therefore, the golden casket was made by someone from the Cellini family. Since the inscription on the golden casket is false, both caskets could not have been made by members of the Celli... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 28. How Many Favorites?
- "Here's a different kind of puzzle," said the Gryphon. "Once, the Queen of Hearts held a reception for thirty guests. She needed to distribute one hundred gingerbread cookies among the guests. Instead of cutting the cookies into pieces, the Queen preferred to give four cookies to each of h... | 28. How many favorites? This problem, usually solved using algebra, is very simple if approached in the following way. First, let's distribute 3 pretzels to each of the 30 guests of the Queen. We will have 10 pretzels left. At this point, all non-favorites will have received all the pretzels they are entitled to, while... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 29. Pretzels and Pretzelkins
- Here's another problem,—began the Gryphon. One day, the Mock Turtle went to the store to buy pretzels for the next tea party.
- How much are your pretzels?—he asked the store owner.
- The price depends on the size: I can offer you small pretzelkins and large pretzels. One pretzel cost... | 29. Pretzels and pretzelkins. Since each pretzel costs as much as one pretzelkin, 7 pretzels cost as much as 21 pretzelkins, and 7 pretzels and 4 pretzelkins cost as much as 25 pretzelkins. On the other hand, 4 pretzels and 7 pretzelkins cost as much as 19 pretzelkins (since 4 pretzels cost as much as 12 pretzelkins). ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
33. How many people got lost in the mountains?
Alice and the Griffin had to wait several minutes before the Tortoise Quasi gathered his strength and could continue.
- You see,一began the Tortoise Quasi.
- I don't see anything!- the Griffin cut in.
The Tortoise Quasi did not respond, only grabbing his head with his fr... | 33. How many people got lost in the mountains? Let's call one portion the amount of supplies one person consumes in a day. Initially, 9 people had 45 portions (a 5-day food supply). On the second day, they had only 36 portions left. On the same day, they met a second group, and the 36 remaining portions were enough for... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
37. Will the cyclist make it to the train?
- Wasn't the previous problem sad? the Turtle Quasi asked. - Just think! The poor frog spent so many days in a dark well! And to get out of there, she had to undertake a climb like a real mountaineer!
- Nonsense! - the Griffin interrupted him. - The saddest part of the whole ... | 37. Will the cyclist make it to the train? The cyclist reasoned incorrectly: he averaged distances, not time. If he had traveled at 4 miles per hour, 8 miles per hour, and 12 miles per hour for the same amount of time, his average speed would indeed have been 8 miles per hour. However, he spent more time climbing the h... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
39. How far is it to school?
All the time while Alice and the Griffin were solving the previous problem, the Tortoise Quasi was weeping inconsolably.
- Can you tell me what is sad about this problem? - the Griffin barked at him angrily. - After all, the passenger caught up with the train. Or did I misunderstand somet... | 39. How far is it to school? The difference in time between being 5 minutes late and arriving 10 minutes before the start of the lesson is 15 minutes. Therefore, if the boy walks to school at a speed of 5 miles per hour, he will save 15 minutes (compared to how long it would take him to walk if he went at a speed of 4 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
32. Find the first 1963 digits after the decimal point in the decimal representation of the number $(\sqrt{26}+5)^{1963}$. | 63.32. Since the sum $(\sqrt{26}+5)^{1963}+(5-\sqrt{26})^{1963}$ is an integer (this is verified using the binomial theorem), and the inequality $-0.1 < 5 - \sqrt{26} < 0$ holds, then $-10^{-1963} < (5 - \sqrt{26})^{1963} < 0 \quad$ and, consequently, the first 1963 digits of the number $(\sqrt{26}+5)^{1963}$ after the... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
37. Solve the equation $x^{4}-2 y^{4}-4 z^{4}-8 t^{4}=$ $=0$ in integers. | 63.37. Answer. $x=y=z=t=0$. Hint. Consider an integer solution with the smallest absolute value of $x$ and try to find another solution with $x_{1}=x / 2$. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
40. For what values of $n$ is the expression $2^{n}+1$ a non-trivial power of a natural number? | 63.40. Only for $n=3$. Indeed, if $2^{n}+1=A^{p}$, then $2^{n}=A^{p}-1=(A-1)\left(A^{p-1}+A^{p-2}+\ldots+A+1\right)$. Then $A^{p-1}+A^{p-2}+\ldots+A+1$ is a power of two, not equal to 1, and since $A$ is odd, and the given sum is even, there is an even number of terms in it. Let $p=2 q$. Then $2^{n}=A^{2 q}-1=$ $=\left... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In the cells of a chessboard, natural numbers are placed such that each number is equal to the arithmetic mean of its neighbors. The sum of the numbers in the corners of the board is 16. Find the number standing on the field $e 2$. | 64.3. Consider the largest number standing in one of the cells. Obviously, all adjacent numbers to it are equal to it. Those adjacent to them are also equal to them, and so on. Therefore, all numbers on the board are equal. Hence, the number written in the field $e 2$ is 4. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
27. Find the maximum of the expression
$$
\begin{aligned}
& x_{1}+x_{2}+x_{3}+x_{4}-x_{1} x_{2}-x_{1} x_{3}-x_{1} x_{4}-x_{2} x_{3}-x_{2} x_{4}-x_{3} x_{4}+ \\
& +x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}-x_{1} x_{2} x_{3} x_{4}
\end{aligned}
$$ | 74.27. The given expression can be rewritten as follows:
$$
1-\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)\left(1-x_{4}\right)
$$
Since $\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)\left(1-x_{4}\right)$ is a non-negative number that can be equal to zero, the maximum value of the expres... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. In seven consecutive vertices of a regular 100-gon, chips of seven colors are placed. In one move, it is allowed to move any chip 10 fields clockwise to the $11-\mathrm{th}$, if it is free. It is required to collect the chips in the seven vertices following the initial ones. How many different arrangements of chips... | 75.18. Answer. There are seven possible arrangements. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
40. The angle $A$ at the vertex of the isosceles triangle $A B C$ is $100^{\circ}$. On the ray $A B$, a segment $A M$ is laid off, equal to the base $B C$. Find the measure of angle $B C M$. | 83.40. Answer. $10^{\circ}$. Indeed, let's lay off on the ray $A M$ a segment $B N$, equal in length to $B C$. Then, in the isosceles triangle $B N C$, the line $C M$ is the bisector of angle $C$. This follows from the fact that $B M / B N = C B / C N$. Since $\angle B C N = 20^{\circ}$, then $\angle B C M = 10^{\circ}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
58. Compute the integral
$$
\int_{-1}^{1} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}
$$ | 86.58. The main idea of the solution is to use the "non-evenness" of the function \( 1 /\left(1+x^{3}+\sqrt{1+x^{6}}\right) \), specifically:
\[
\begin{aligned}
& \text { of the function } 1 /\left(1+x^{3}+\sqrt{1+x^{6}}\right) \text {, specifically: } \\
& \int_{-1}^{1} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}=\int_{-1}^{0... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
6. Two players take turns placing crosses and noughts in the cells of a $9 \times 9$ square (the first player places crosses, and the opponent places noughts). At the end, the number of rows and columns where there are more crosses than noughts is counted - these are the points scored by the first player. The number of... | 87.6. Hint. The first one needs to use central symmetry. The maximum guaranteed number of his points is 10, i.e., he can guarantee to win two points over the second one. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
49*. On the board, there are 128 ones. In one move, you can replace a pair of numbers $a$ and $b$ with the number $a b + 1$. Let $A$ be the maximum number that can appear on the board after 127 such operations. What is its last digit?
## 11th grade | 92.49. Hint. Prove that to obtain the maximum number, you can act according to the following scheme: at each moment, perform the operation on the two smallest numbers.
By now, carrying out the calculations in accordance with the derived algorithm, we get that the last digit of the maximum number is two. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. At the table, there are several boys and five girls, and on the table, on a plate, there are 30 buns. Each girl gave a bun (from the plate) to each boy she knew, and then each boy gave a bun (from the plate) to each girl he did not know. After this, it turned out that all the buns were given out. How many boys were ... | 93.3. Let the number of boys be $n$. Then there are $5 n$ different boy-girl pairs. Each such pair corresponds to
one passed bun: if the boy and girl are acquainted, the bun was passed by the girl, and if they are not acquainted - by the boy. Hence $5 n=30$. Answer. $n=6$. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. What is the greatest value that the expression
$$
a e k-a f h+b f g-b d k+c d h-c e g
$$
can take if each of the numbers $a, \ldots, k$ is equal to 1 or -1? | 93.11. Each of the six terms of the given sum is equal to 1 or -1, so this sum is even. It cannot take the value six, since in this case the terms $a e k, b f g, c d h$ must be equal to 1 and, therefore, their product is 1, and the terms $a f h, b d k, c e g$ must be equal to -1, and their product is -1. However, these... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 10.
Find all natural numbers $n$ such that the sum $S(n)$ of the digits in the decimal representation of the number $2^{n}$ is 5. | Solution. To find the solution, we will construct a table of values for the expression $2^{n}$:
| $n$ | $2^{n}$ | $S(n)$ | $n$ | $2^{n}$ | $S(n)$ |
| :---: | ---: | ---: | :---: | ---: | :---: |
| 1 | 2 | 2 | 16 | 65536 | 25 |
| 2 | 4 | 4 | 17 | 131072 | 14 |
| 3 | 8 | 8 | 18 | 262144 | 19 |
| 4 | 16 | 7 | 19 | 524288... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3 Task 1. For which natural $n(n \geqslant 2)$ is the equality
$$
-\sqrt[2]{17 \sqrt{5}+38}+\sqrt[2]{17 \sqrt{5}-38}=\sqrt{20} ?
$$ | Solution. Using a calculator, we find:
$$
\begin{gathered}
17 \sqrt{5}+38 \approx 38.013154+38 \approx 76.013154>1 \\
17 \sqrt{5}-38 \approx 0.0131541$, and the function $\varphi(n)=\sqrt[n]{17 \sqrt{5}-38}$ is increasing, and $\varphi(n)<1$. To discover some properties of the function $\psi(n)=f(n)+\varphi(n)$, we wi... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 17. The side of a regular triangle $ABC$ is 4. Point $D$ is the midpoint of side $BC$. A line passing through $B$ intersects side $AC$ at point $M$. Perpendiculars from points $D$ and $A$ to line $BM$ are $DH$ and $AK$. Calculate the length of segment $AM$, if
$$
AK^4 - DH^4 = 15
$$ | Instruction. Equation (1) has a unique solution. Let $\angle A B M=x$. Then $D H=2 \sin \left(60^{\circ}-x\right)$, $A K=4 \sin x$. Equation (1) takes the form:
$$
256 \sin ^{4} x-16 \sin ^{4}\left(60^{\circ}-x\right)=15
$$
Solving equation (2) on a microcalculator, we get $x=30^{\circ}$, and therefore, $A M=2$. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Find all values of $a$ for which the equation
$$
2 \lg (x+3)=\lg (a x)
$$
has a unique solution. | The first solution. Obviously, $x>-3, a x>0$, $x \neq 0$. From equation (1) we get:
$$
(x+3)^{2}=a x, a=\frac{(x+3)^{2}}{x}=\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right)^{2}, \text { if } x>0
$$
Obviously,
$$
\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right)=\left(\sqrt{x}-\frac{3}{\sqrt{x}}\right)^{2}+12
$$
From here it is clea... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. How many solutions does the equation
$$
\arcsin 2x + \arcsin x = \frac{\pi}{3} ?
$$ | Solution. The function $f(x)=\arcsin 2 x$ is defined on the interval $[-0.5 ; 0.5]$, and the function $\varphi(x)=\arcsin x$ is defined on the interval $[-1 ; 1]$. Therefore, the left side of equation (1) is defined on the interval $[-0.5 ; 0.5]$. The monotonic and continuous function $y=\arcsin 2 x+\arcsin x$ changes ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 13. Find the minimum value of the function
$$
\psi(x)=\sqrt{15-12 \cos x}+\sqrt{7-4 \sqrt{3} \sin x}
$$
on the interval $[0 ; 0.5 \pi]$. | Solution. Since $|\cos x| \leqslant 1$ and $|\sin x| \leqslant 1$, the function $\psi(x)$ is defined on the entire interval $[0 ; 0.5 \pi]$. To find the properties of the functions $f(x)=\sqrt{15-12 \cos x}$, $\varphi(x)=\sqrt{7-4 \sqrt{3} \sin x}, \quad \psi(x)=f(x)+\varphi(x) \quad$, we create a table of their values... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 16. Solve the equation
$$
\lg \left(x^{2}+9\right)-3 \cdot 2^{x}+5=0
$$ | Solution. The equation does not have negative roots because on $(-\infty ; 0)$ the function $P(x)=\lg \left(x^{2}+\right.$ $+9)+5$ is decreasing, the function $K(x)=3 \cdot 2^{x}$ is increasing, and $P(0)>K(0)$. To hypothesize about the number of roots of the given equation, we construct tables of the functions $P(x)$ ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3 a d a c h a 19. A circle with radius 1 and center at point $O$ is circumscribed around an equilateral triangle $A B C$. Prove that the sum of the squares of the distances from any point $M$ on the circle to the vertices of this triangle is equal to 6 (Fig. 9). | Solution. Obviously,
$\overline{C M}^{2}+\overline{M A}^{2}+\overline{M B}^{2}=(\overline{O M}-\overline{O C})^{2}+(\overline{O A}-\overline{O M})^{2}+$ $+(\overline{O B}-\overline{O M})^{2}=\overline{O M}^{2}+\overline{O C}^{2}-2 \overline{O M} \cdot \overline{O C}+\overline{O A}^{2}+$ $+\overline{O M}^{2}-2 \overlin... | 6 | Geometry | proof | Yes | Yes | olympiads | false |
Problem 5. $DABC$ is a triangular pyramid (Fig. 5). $\overline{AK}=\overline{KD}, \overline{BP}=\overline{PC}, \overline{DM}=0.4 \overline{DC}$. Find the area $S$ of the section of the pyramid by the plane $KMP$, if the vertex $A$ is at a distance $h=1$ from the plane $KMP$ and the volume of the pyramid $DABC$ is 5. | Solution. We construct the section of the pyramid $DABC$ using the method of parallel projections. We choose the line $KP$ as the direction of projection. In this case, the pyramid $\dot{D}ABC$ is represented by the parallelogram $ABDC$ (Fig. 6), and the cutting plane is represented by the line $KM$, which intersects t... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. The base of the pyramid $HABCD$ is a square $ABCD$, and the edge $HA$ is perpendicular to the base plane. $AB=3, HA=4$. Prove that a sphere can be inscribed in the pyramid, and find its radius.
For comparison, we will provide two solutions to this problem. | The first solution. First, we find the radius of the sphere. Then we prove its existence. For this, we apply the formula $3 v = S r$. Clearly, $V = \frac{1}{3} \cdot 3 \cdot 3 \cdot 4 = 12$; $H B = 5$; $S(ABCD) = 9$; $S(BAH) = S(HAD) = 0.5 \cdot 3 \cdot 4 = 6$; $S(HBC) = S(HDC) = 0.5 \cdot 3 \cdot 5 = 7.5$. Therefore, ... | 1 | Geometry | proof | Yes | Yes | olympiads | false |
17. Two thirds in addition (to the unknown), $\frac{1}{3}$ in subtraction, 10 remains (i.e., two thirds of the unknown are added to it, and one third is subtracted from the resulting sum. The remainder is 10). | 17. The equation will be:
$$
x+\frac{2}{3} x-\frac{1}{3}\left(x+\frac{2}{3} x\right)=10 ; x=9
$$
$\begin{array}{ll}\text { 18. } x=\frac{3}{10} ; x=\frac{15}{53} & \text { 19. } 4 \frac{1}{6} \text {. }\end{array}$ | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
28. In the Akhmim papyrus, the area of a circle, the circumference of which is the arithmetic mean of two given circumferences, is taken as the arithmetic mean of their areas. Show that this is incorrect, and find how large the error is in percentage, where the radii of the given circles \( r=5 \); \( R=10 \).
Greek mathematics reached its peak by the 3rd century BC (the Alexandrian period). From this time, a number of mathematical treatises (... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
92. Given a right triangle with sides 9, 12, 15. Find the diameter of the circle inscribed in this triangle.
## Problems of Marcus Junius Nipsus.
From the Erfurt Codex. | 92. Solution: $A B=A E+R ; A E=A D$,
$$
B C=F C+R ; F C=D C
$$
therefore,
$$
A B+B C=2 R+A C
$$
from which
$$
2 R=A B+B C-A C
$$
in this case
$$
2 R=9+12-15=6
$$
Fig. 33.
The Erfu... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
98. In a cage, there is an unknown number of pheasants and rabbits. It is only known that the entire cage contains 35 heads and 94 feet. The task is to determine the number of pheasants and the number of rabbits. | 98. If there were only pheasants in the cage, the number of legs would be 70, not 94. Therefore, the 24 extra legs belong to the rabbits, 2 per each. It is clear that there were 12 rabbits and, consequently, 23 pheasants. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
122. Find the number which, when multiplied by 3, then divided by 5, increased by 6, after which the square root is extracted, one is subtracted, and the result is squared, will give 4.
## Problem of Sridhara.
From the treatise "The Essence of Calculation." | 122. The method of inverse actions involves approaching the unknown by performing actions on numbers in the reverse order of those specified in the problem, and, obviously, all actions are replaced by their inverses. Sometimes this technique is called the "method of inversion."
$$
\sqrt{4}=2 ; 2+1=3 ; 3^{2}=9 ; 9-6=3 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
156. What number, when added to nine, gives its own square root?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note to the translator and should not be included in the translated text. Here is the final version:
156. What number, when added to nine, gives its own square root? | 156. $x^{2}+9=6 x ; x=3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
156. $x^{2}+9=6 x ; x=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
169. $x^{2}+10 x=39$.
169. $x^{2}+10 x=39$.
(Note: The equation is the same in both languages, so the translation is identical to the original text.) | 169. Al-Karaji is looking for a number which, when added to $x^{2}+10 x$, would make a complete square. Such a number is 25. Then
$x^{2}+10 x+25=(x+5)^{2}=39+25=64 ; x+5=8$ and $x=3$.
He calls this method the "method of solving in the manner of Diophantus". | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
181. Find the area of a rectangle where the base is twice the height, and the area is numerically equal to the perimeter. | 181. If the width is $x$, then the length is $2 x$. Thus, the area is $2 x^{2}$, and the perimeter is $6 x$. According to the condition, $2 x^{2}=6 x$, therefore, $x=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
192. If a number, when divided by 9, gives a remainder of 2 or 7, then the square of the number, when divided by 9, gives a remainder of 4. | 192. Let $N=9 k+2, \quad$ then $N^{2}=81 k^{2}+18 k+4=$ $=$ multiple of 9 +4 ; let $N_{1}^{2}=9 k+7$, then $N_{1}^{2}=81 k^{2}+$ $+126 k+49=9\left(9 k^{2}+14 k+5\right)+4$. | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
203. $\frac{1}{x^{2}}+2 \frac{1}{x}=1 \frac{1}{4}$.
## Problems of Begh-Edin.
From the treatise "The Essence of the Art of Calculation". | 203. Omar's Solution: $z=\frac{1}{x} ; \quad z^{2}+2 z=\frac{5}{4} ;$
$$
\begin{aligned}
& z^{2}+2 z+1=\frac{9}{4} ;(z+1)^{2}=\frac{9}{4} ; z+1=\frac{3}{2} ; z=\frac{1}{2} \\
& z^{2}=\frac{1}{4}, \text { therefore, } x^{2}=4 \text { and } x=2
\end{aligned}
$$
Bega Ed-Din (16th century), a Persian mathematician, autho... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
205. It is required to find a number which, when multiplied by itself, added to two, then doubled, added to three again, divided by 5, and finally multiplied by 10, results in 50. | 205. $\frac{50}{10}=5 ; 5 \cdot 5=25 ; 25-3=22 ; 11-2=9 ; \sqrt{9}=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
207. Zaimu is promised a reward in the form of the larger of two parts, which add up to 20, and the product of these parts is 96. How great is the reward? | 207. The ordinary solution: Let the larger part be $x$. Then the smaller one is $20-x$. According to the condition, $x(20-x)=96$, or $x^{2}-20 x+96=0 ; x=10 \pm \sqrt{100-96}=10 \pm 2 ; x_{1}=12 ; x_{2}=8$. But Al-Baghdadi's solution is simpler: put one number as $10+x$, then the other is $10-x$. Their product is $100-... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
252. $\frac{x}{10-x}+\frac{10-x}{x}=25$. | 252. $10 x=x^{2}+\frac{100}{27} ; x=5-\sqrt{21 \frac{8}{27}}$. But Regiomontanus also gives another solution, assuming $\frac{x}{10-x}=y$. Then $y+\frac{1}{y}=25 ; y=\frac{25}{2}-\sqrt{\frac{621}{4}}$. In both cases, Regiomontanus takes the roots with a minus sign. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
269. $3 x^{2}+12=12 x$ | 269. Two equal roots $(x=2)$.
270. $x=5 \pm \sqrt{21}$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
309. 26 persons together spent 88 coins, with each man spending 6, each woman 4, and each girl 2 coins. How many were men, women, and girls? | 309. 10 solutions; from $x+y+z=26$
and
$$
6 x+4 y+2 z=88
$$
we get:
$$
2 x+y=18
$$
from which
$$
y=18-2 x
$$
therefore,
$$
2 x \leqslant 18
$$
H
$$
x \leqslant 9
$$
T. e
$$
\begin{aligned}
& x=0, \quad 1, \quad 2, \quad 3, \quad 4, \quad 5, \quad 6, \quad 7, \quad 8, \quad 9 \\
& y=18,16,14,12,10, \quad 8, ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
314. Divide $\sqrt[3]{216}$ by $\sqrt[4]{16}$. | 314. Rudolf's Solution:
$$
\begin{aligned}
& =\sqrt[12]{531441}=\sqrt[6]{729}=\sqrt[3]{27}=3
\end{aligned}
$$
Indeed: $\quad \sqrt[3]{216}: \sqrt[4]{16}=6: 2=3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
340. Find the positive root of the equation by construction:
$$
x^{2}+6 x=91
$$ | 340. Cardano's Solution: Let the square $F D$ (Fig. 74) be $x^{2}$, hence its side $F H=x$. $D G=D B=3$ (half the coefficient of $x$). Construct the square $A F E C$. The rectangle $A D$ equals the rectangle $D E$, i.e., equals $3 x$. The sum of the square $F D$ and the two rectangles equals $x^{2}+6 x$, which, by the ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
351. Show that
$$
\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}=4
$$ | 351. $\sqrt[3]{2+v \overline{-121}}=\sqrt[3]{2+11 i}=\sqrt[3]{8+12 i-6-i}=$
$$
\begin{aligned}
& =\sqrt[3]{2^{3}+3 \cdot 2^{2} i+3 \cdot 2 \cdot i^{2}+i^{3}}=\sqrt[3]{(2+i)^{3}}=2+i \\
& \sqrt[3]{2-\sqrt{-121}}=\sqrt[3]{2-11 i}=\sqrt[3]{8-12 i-6+i}= \\
& =\sqrt[3]{2^{3}-3 \cdot 2^{2} \cdot i+3 \cdot 2 i^{2}-i^{3}}=\sq... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
382. What is the smallest number of weights and of what weight can be used to weigh any whole number of pounds from 1 to 40 on a balance scale, given that during weighing, weights can be placed on both pans of the scale. | 382. The solution to the problem is based on the properties of the ternary system of numeration, which can express any numbers; since the number $40=27+9+3+1=3^{3}+3^{2}+3+1$, it is possible with four weights of $27, 9, 3$ and 1 pound, as is easy to verify by checking, to weigh any load up to forty pounds. It is suffic... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
393. Several people had dinner together and had to pay 175 shillings for the bill. It turned out that two of them had no money, so each of the rest had to pay 10 shillings more than their share. How many people had dinner? | 393. Let $x$ be the number of people who had lunch. It is easy to form the equation
$$
\frac{175}{x-2}-\frac{175}{x}=10
$$
from which
$$
x^{2}-2 x=35
$$
Maclaurin would solve it as follows:
$$
\begin{gathered}
x^{2}-2 x+1=36 \\
(x-1)^{2}=36 ; x-1= \pm 6 ; x_{1}=7 \text { or } x_{2}=-5
\end{gathered}
$$
The second... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
409. Several merchants contributed to a common fund 100 times as many rubles as there were merchants. They sent a trusted person to Venice, who received a number of rubles from each hundred rubles, twice the number of merchants. The question is: how many merchants were there, if the trusted person received 2662 rubles? | 409. Let the number of merchants be $x$. Each contributed $100 x$, and the total capital is $100 x^{2}$. The profit on the capital is $2 x^{3}=2662$, from which $x^{3}=1331$, hence $x=11$. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
426. A lazy worker was told that he would receive 24 su for each day he worked, with the condition of deducting 6 su for each day he skipped. After 30 days, it turned out that he did not receive anything. The question is: how many days did he work? | 426. We form the equation $24 x=6(30-x)$, from which $x=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
433. Several people have to pay 800 francs for the production of a lawsuit. But three of them have no money, so the rest have to add 60 francs to their share. How many participants are there in the payment of legal costs? | 433. We form the equation $\frac{80}{x+3}=\frac{800}{x}-60$, where $x-$ is the number of participants in the payment. $x^{2}+3 x=40$, from which $x=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
471. Someone has twelve pints of wine and wants to give away half of it, but he does not have a six-pint container. He has two containers, one holds 8 pints and the other 5 pints; the question is: how can he pour six pints into the eight-pint container?
## Lebesgue's Problem. | 471. Regarding this problem, Arago tells that it decided the fate of Poisson, as, by becoming interested in it, he thereby discovered his vocation and devoted his whole life to mathematics. We have already seen a similar problem (336). This one also has 2 solutions:
| 12 | 8 | 5 |
| ---: | ---: | ---: |
| 12 | 0 | 0 |... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
472. Solve the system in integers:
$$
\begin{aligned}
& 2 x+3 y+7 z=131 \\
& 2 x+3 y+8 z=140
\end{aligned}
$$
60
## Schläfli Problems. | 472. Immediately find $z=9$ and $2 x+3 y=68$, from which in the usual way: $y=0,2,4, \ldots, 22$. In total, there are 12 solutions.
Schlömilch, Oscar (1823--1901), a well-known German mathematician, whose name is associated with the expression for the remainder term of a Taylor series. Author of the very useful course... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
475. Prove that
$$
\sqrt[3]{20+1+\sqrt{2}}+\sqrt[3]{20-14 \sqrt{ } 2}=4
$$ | 475. Cubing:
$$
\begin{gathered}
20+14 \sqrt{2}+20-14 \sqrt{2}+ \\
+3 \sqrt[3]{20+14 \sqrt{2}} \sqrt[3]{20-14 \sqrt{2}} \cdot 4=64 \\
40+12 \sqrt[3]{40-392}=40+24
\end{gathered}
$$
Another method:
$$
\begin{gathered}
\sqrt[3]{20+14 \sqrt{2}}=\sqrt[3]{8+12+12 \sqrt{2}+2 \sqrt{2}}= \\
=\sqrt[3]{2^{3}+3 \cdot 2} \cdot(... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
2.8. Let $n>1$ be a natural number. Find all positive solutions of the equation $x^{n}-n x+n-1=0$.
## 2.3. Equations with Radicals
In problems $2.9-2.15$, it is assumed that the values of square roots are non-negative. We are interested only in the real roots of the equations. | 2.8. A n s w e r: $x=1$. It is clear that
$$
x^{n}-n x+n-1=\left(1+x+\ldots+x^{n-1}-n\right)(x-1)
$$
If $x>1$, then $1+x+\ldots+x^{n-1}-n>0$, and if $0<x<1$, then $1+x+\ldots$ $\cdots+x^{n-1}-n<0$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.9. Solve the equation $\sqrt{2 x-6}+\sqrt{x+4}=5$.
### 2.10. Solve the equation
$$
\sqrt[m]{(1+x)^{2}}-\sqrt[m]{(1-x)^{2}}=\sqrt[m]{1-x^{2}}
$$ | 2.9. Let $y=\sqrt{x+4}$. Then $x=y^{2}-4$, so $2 x-6=2 y^{2}-$ -14. Therefore, we get the equation $\sqrt{2 y^{2}-14}+y=5$. Move $y$ to the right side and square both sides. As a result, we get the equation $y^{2}+10 y-39=0$. Its roots are 3 and -13. But $y \geqslant 0$, so only the root $y=3$ remains, which correspond... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.11. Solve the equation
$$
3 \sqrt{x^{2}-9}+4 \sqrt{x^{2}-16}+5 \sqrt{x^{2}-25}=\frac{120}{x}
$$ | 2.11. The expression on the left side increases as $x$ increases, while the expression on the right side decreases. Therefore, the equation has no more than one solution. It is easy to verify that $x=5$ is a solution. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.12. Solve the equation $x+\sqrt{3+\sqrt{x}}=3$. | 2.12. If $x>1$, then $x+\sqrt{3+\sqrt{x}}>3$, and if $x<1$, then $x+$ $+\sqrt{3+\sqrt{x}}<3$. Only the root $x=1$ remains. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.13. $\left\{\begin{array}{l}\left(x_{3}+x_{4}+x_{5}\right)^{5}=3 x_{1}, \\ \left(x_{4}+x_{5}+x_{1}\right)^{5}=3 x_{2}, \\ \left(x_{5}+x_{1}+x_{2}\right)^{5}=3 x_{3}, \\ \left(x_{1}+x_{2}+x_{3}\right)^{5}=3 x_{4}, \\ \left(x_{2}+x_{3}+x_{4}\right)^{5}=3 x_{5} .\end{array}\right.$
$. The function $f(x)=x^{5}$ is monotonically increasing, so $3 x_{2}=\left(x_{4}+x_{5}+x_{1}\right)^{5} \geqslant\left(x_{3}+x_{4}+x_{5}\right)^{5}=3 x_{1}$. Therefore, $x_{1}=x_{2}$ and $x_{3}=x_{1}$. Moreover, $3 x_{4... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.48. Find the remainder when the number
$$
10^{10}+10^{\left(10^{2}\right)}+10^{\left(10^{3}\right)}+\ldots+10^{\left(10^{10}\right)}
$$
is divided by 7. | 4.48. Answer: 5. Note that $10^{6} \equiv 1(\bmod 7)$, since $10^{3}+1$ is divisible by 7, and $10^{k} \equiv 4(\bmod 6)$ for $k \geqslant 1$, since the number $99 \ldots 96$ is even and divisible by 3. Therefore, $10^{10^{k}} \equiv 10^{4}(\bmod 7)$ for $k \geqslant 1$. Hence, the required remainder is the remainder o... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.12. Prove that the number $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ is rational. | 6.12. The identity $(2 \pm \sqrt{2})^{3}=20 \pm 14 \sqrt{2}$ shows that $\sqrt[3]{20 \pm 14 \sqrt{2}}=2 \pm \sqrt{2}$. Therefore, the considered number is 4. | 4 | Algebra | proof | Yes | Yes | olympiads | false |
6.23. Find the first digit after the decimal point of the number $(2+\sqrt{3})^{1000}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6.23. Answer: 9. Let $(2+\sqrt{3})^{n}=A_{n}+B_{n} \sqrt{3}$, where $A_{n}$ and $B_{n}$ are natural numbers. According to problem 6.22, $(2-\sqrt{3})^{n}=A_{n}-B_{n} \sqrt{3}$. Therefore, $(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}$ is a natural number. But $2-\sqrt{3} \approx$ $\approx 0.2679 < 0.3$, so $(2-\sqrt{3})^{1000} \a... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7.2. A pedestrian left point $A$ for point $B$, which are 5 km apart. At the same time, a cyclist left point $B$ towards the pedestrian, with a speed twice that of the pedestrian. Upon meeting the pedestrian, he turned around and rode back to $B$. Upon reaching $B$, the cyclist turned again and rode towards the pedestr... | 7.2. A n s w e r: 10 km. We can assume that the cyclist is always moving in one direction (the length of the path does not change from this).
His speed is twice the speed of the pedestrian, so in the time it takes the pedestrian to walk 5 km, the cyclist will travel 10 km. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.16. Ten workers need to assemble 50 items from parts. First, the parts of each item need to be painted; it takes one worker 10 minutes to do this. After painting, the parts dry for 5 minutes. Assembling an item takes one worker 20 minutes. How many workers should be assigned as painters and how many as assemblers to ... | 7.16. Answer: 3 painters and 6 assemblers (one worker is extra, i.e., you can assign 4 painters and 6 assemblers or 3 painters and 7 assemblers). It is easy to verify that 3 painters and 6 assemblers can complete the work in 195 minutes. Indeed, after 15 minutes from the start of the painters' work, 3 items are ready f... | 3 | Other | math-word-problem | Yes | Yes | olympiads | false |
10.9. When dividing the polynomial $x^{1951}-1$ by $x^{4}+x^{3}+2 x^{2}+$ $+x+1$, a quotient and a remainder are obtained. Find the coefficient of $x^{14}$ in the quotient.
## 10.4. Vieta's Theorem | 10.9. Answer: -1. The equalities $x^{4}+x^{3}+2 x^{2}+x+1=\left(x^{2}+1\right) \times$ $\times\left(x^{2}+x+1\right)$ and $x^{12}-1=(x-1)\left(x^{2}+x+1\right)\left(x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)$ show that
$$
\begin{aligned}
& x^{4}+x^{3}+2 x^{2}+x+1=\frac{x^{12}-1}{(x-1)\left(x^{3}+1\ri... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.31. What remainder does $x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}$ give when divided by $(x-1) ?$ | 10.31. Answer: 6. Let $x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}=P(x) \times$ $\times(x-1)+r$. By setting $x=1$, we get $r=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12.6. Solve the equation $x+y=x y$ in natural numbers. 12.7. Solve the equation $2 x y+3 x+y=0$ in integers. | 12.6. It is clear that $(x-1)(y-1)=x y-x-y+1=1$, therefore $x-1=y-1=1$, i.e., $x=y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12.15. Solve the equation in integers
$$
x^{3}-2 y^{3}-4 z^{3}=0
$$ | 12.15. Answer: $x=y=z=0$.
Let $x^{3}-2 y^{3}-4 z^{3}=0$, where $x, y, z$ are integers. Then the number $x$ is even. After substituting $x=2 x_{1}$, we get the equation $8 x_{1}^{3}-2 y^{3}-4 z^{3}=0$. Dividing by 2: $4 x_{1}^{3}-y^{3}-2 z^{3}=0$. Therefore, the number $y$ is even. After substituting $y=2 y_{1}$, we ge... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14.4. How many necklaces can be made from five white beads and two black ones? | 14.4. Answer: 3. Black beads divide the white beads into two groups (one of these groups may contain 0 beads). The appearance of the necklace is completely determined by the number of beads in the smaller group. This number can be 0, 1, or 2. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14.48. Which sum of the numbers rolled is more likely when throwing two dice: 9 or $10?$ | 14.48. A n s w e r: 9. The numbers 9 and 10 can be obtained in two different ways: $9=3+6=4+5$ and $10=4+6=5+5$. However, we need to consider the order in which the numbers fall on the dice. Therefore, the number 9 can be obtained in four different ways, while the number 10 can only be obtained in three: $9=3+6=6+3=4+5... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14.51. In a drawer, there are red and black socks. If two socks are randomly pulled out of the drawer, the probability that both are red is $1 / 2$.
a) What is the smallest number of socks that can be in the drawer?
b) What is the smallest number of socks that can be in the drawer if it is known that the number of bl... | 14.51. a) Answer: 4. Let there be $m$ red socks and $n$ black socks in the box. The probability that the first selected sock is red is $\frac{m}{n+m}$. Given that the first selected sock is red, the probability that the second selected sock is also red is $\frac{m-1}{n+m-1}$. Therefore, the probability that both socks ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
16.13. a) There is a piece of chain consisting of 60 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, ..., 60 g can be formed (an unbuckled link also weighs 1 g)?
b) The same question for a chain consisting of 150 l... | 16.13. a) Answer: 3 links. Let's determine the largest $n$ for which it is sufficient to break $k$ links of an $n$-link chain so that all integer weights from 1 to $n$ can be formed from the resulting parts. If $k$ links are broken, any number of links from 1 to $k$ can be formed from them. But we cannot form $k+1$ lin... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
17.21. We took three numbers $x, y, z$ and computed the absolute values of their pairwise differences $x_{1}=|x-y|, y_{1}=|y-z|$, $z_{1}=|z-x|$. In the same way, from the numbers $x_{1}, y_{1}, z_{1}$ we constructed the numbers $x_{2}, y_{2}, z_{2}$ and so on. It turned out that for some $n$ we got $x_{n}=x, y_{n}=y, z... | 17.21. Answer: $y=z=0$. The numbers $x_{n}, y_{n}, z_{n}$ are non-negative, so the numbers $x, y, z$ are also non-negative. If all the numbers $x$, $y, z$ were positive, then the largest of the numbers $x_{1}, y_{1}, z_{1}$ would be strictly less than the largest of the numbers $x, y, z$, and then the largest of the nu... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20.16. Some of the 20 metal cubes, identical in size and appearance, are aluminum, the rest* are duralumin (heavier). How can you determine the number of duralumin cubes using 11 weighings on a balance with two pans and no weights? | 20.16. Let's put one cube on each pan of the balance. There are two possible cases.
Case 1. One of the cubes turned out to be heavier in the first weighing.
In this case, one of the selected cubes is aluminum, and the other is duralumin. Place the selected cubes on one pan and compare them with the remaining cubes. S... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
21.3. The numbers $2^{n}$ and $5^{n}$ start with the digit $a$. What is $a$? | 21.3. Answer: 3. By the condition $a \cdot 10^{p}<2^{n}<(a+1) 10^{p}$ and $a \cdot 10^{q}<5^{n}<$ $<(a+1) 10^{q}$. Therefore, $a^{2} 10^{p+q}<10^{n}<(a+1)^{2} 10^{p+q}$, i.e., $a^{2}<10^{n-p-q}<$ $<(a+1)^{2}$. At the same time, $(a+1)^{2} \leqslant 100$. Thus, $a^{2}<10<(a+1)^{2}$, i.e., $a=3$. The number 3 is the lead... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
21.16. All integers are written in a row, starting from one. What digit stands at the 206,788th place?
## 21.6. Periods of Decimal Fractions and Repunits
| Let $p$ be a prime number different from 2 and 5. The length of the period of the number $p$ is the number of digits in the period of the decimal representation o... | 21.16. Answer: the digit 7. There are exactly 9 single-digit numbers, 99-9=90 two-digit numbers, 999-99-9=900 three-digit numbers, 9000 four-digit numbers, and so on. Single-digit numbers occupy the first 9 positions in the written sequence, two-digit numbers occupy $90 \cdot 2=180$ positions, three-digit numbers occup... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
21.33. If we consider digits in different positions as different, then in a $d$-ary numeral system, $n d$ digits allow us to write $d^{n}$ numbers (from 0 to $d^{n}-1$). Which numeral system is the most economical in this respect, i.e., allows recording the largest number of numbers using a given number of digits? (Whe... | 21.33. Answer: system with base 3. Using $m=d n$ digits, we can write $d^{m / d}$ numbers. Therefore, we need to prove that $3^{m / 3} \geqslant d^{m / d}$, i.e., $3^{d} \geqslant d^{3}$ for any natural $d$. This inequality is proven in the solution to problem 13.11. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
23.21. For which natural numbers $n$ is the expression $a^{n}(b-c)+$ $+b^{n}(c-a)+c^{n}(a-b)$ divisible by $a^{2}+b^{2}+c^{2}+a b+b c+c a$? | 23.21. Answer: only for $n=4$. It is easy to verify that for $n=4$ the result of the division is $(a-b)(b-c)(a-c)$. We will show that for all other natural $n$ the expression $a^{n}(b-c)+b^{n}(c-a)+c^{n}(a-b)$ does not divide $a^{2}+b^{2}+c^{2}+ab+bc+ca$. It is sufficient to check that the first expression does not div... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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