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7.5. A steamship from Gorky to Astrakhan takes 5 days, while from Astrakhan to Gorky it takes 7 days. How many days will it take for driftwood to float from Gorky to Astrakhan? | 7.5. Answer: 35. Let $v$ be the speed of the river current, $u$ be the speed of the steamship, and $l$ be the distance from Gorky to Astrakhan. According to the problem, $\frac{l}{u+v}=5$ and $\frac{l}{u-v}=7$; we need to find $l / v$. From the system of equations $l=5 u+5 v, l=7 u-7 v$, we find $l / v=35$. | 35 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,301 |
7.6. Two people $A$ and $B$ need to travel from point $M$ to point $N$, which is 15 km away from $M$. On foot, they can move at a speed of 6 km/h. In addition, they have a bicycle that can be ridden at a speed of 15 km/h. $A$ sets out on foot, while $B$ rides the bicycle until meeting pedestrian $C$, who is walking fro... | 7.6. a) For $A$ and $B$ to arrive at point $N$ simultaneously, they must walk the same distance $x$ km and cycle the same distance $15-x$. Then $C$ must also walk the distance $x$ before meeting $B$. Therefore, $C$ will cycle $15-2x$ before meeting $A$, and after that, $A$ will cycle $15-x$. Together, $A$ and $C$ will ... | \frac{3}{11}beforeABstarttheirjourneyfromM | Other | math-word-problem | Yes | Yes | olympiads | false | 38,302 |
7.7. Cities $A$ and $B$ are located on a river 10 km apart. Which will take the steamboat more time: to travel from $A$ to $B$ and back, or to travel 20 km on a lake? | 7.7. Let the speed of the steamboat be $v$, and the speed of the river current be $w$. If $v \leqslant w$, then the steamboat will not be able to move upstream at all. If, however, $v > w$, then the time required for the journey from $A$ to $B$ and back is
$$
\frac{10}{v+w}+\frac{10}{v-w}=\frac{20 v}{v^{2}-w^{2}}>\fra... | \frac{20v}{v^{2}-w^{2}}>\frac{20}{v} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,303 |
7.9. In a table of size $10 \times 10$, 100 numbers are written. In each row, the smallest number is selected, and among them, the largest number $A$ is chosen. In each column, the largest number is selected, and among them, the smallest number $B$ is chosen. Can it happen that $A>B$? | 7.9. Answer: no, it cannot. Let the number $x$ be at the intersection of the row containing the number $A$ and the column containing the number $B$. Then $A \leqslant x \leqslant B$. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,305 |
7.10. Seven mushroom pickers collected 100 mushrooms in total, and no two of them collected the same number of mushrooms. Prove that among them, there are three mushroom pickers who together collected at least 50 mushrooms. | 7.10. Let $a_{1}>a_{2}>\ldots>a_{7}$ be the number of mushrooms collected by the mushroom pickers. If $a_{3} \geqslant 16$, then $a_{2} \geqslant 17$ and $a_{1} \geqslant 18$, so $a_{1}+a_{2}+a_{3} \geqslant 51$. If $a_{3} \leqslant 15$, then $a_{4} \leqslant 14, a_{5} \leqslant 13, a_{6} \leqslant 12$ and $a_{7} \leqs... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,306 |
7.11. Students from one class went on two hiking trips. In each trip, the number of boys was less than $2 / 5$ of the total number of participants in the trip. Prove that in this class, boys make up less than $4 / 7$ of the total number of students, given that each student participated in at least one of the trips.
##... | 7.11. First solution. According to the condition, in each trip, the number of boys was less than $2 / 3$ of the girls participating in the trip. Therefore, in each trip, the number of boys was less than $2 / 3$ of all the girls in the class. Since every boy went on at least one trip, the number of boys in the class is ... | 7x<4(x+y) | Inequalities | proof | Yes | Yes | olympiads | false | 38,307 |
7.12. In the class, a test was conducted. It turned out that the average grade for boys is 4; for girls, it is 3.25; and for everyone together, it is 3.6. How many boys and how many girls took the test, if it is known that the class has more than 30 and fewer than 50 students | 7.12. Let there be $x$ boys and $y$ girls in the class. Then $\frac{4 x + 3.25 y}{x + y} = 3.6$, i.e., $8 x = 7 y$. Therefore, $x = 7 a$ and $y = 8 a$ for some integer $a$. Hence, the number $x + y = 15 a$ is divisible by 15. The only number divisible by 15 that is greater than 30 and less than 50 is 45. Thus, $x = 21$... | x=21,y=24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,308 |
7.13. In the report on the ski race, it is stated that $96 \%$ of its participants met the standard. It is known that this figure is given with an accuracy of $0.5 \%$. What is the smallest number of participants in this race? | 7.13. Let $n$ be the number of skiers in the race, and $k$ be the number of them who did not meet the standard. According to the condition, $3.5 \leqslant 100 k / n \leqslant 4.5$. Therefore, $k \geqslant 1$ and $n \geqslant \frac{100 k}{4.5}>22.2 k \geqslant 22.2$. Hence, $n \geqslant 23$. If there were 23 skiers in t... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,309 |
7.14. In a quarry, 120 granite slabs weighing 7 tons each and 80 slabs weighing 9 tons each are prepared. Up to 40 tons can be loaded onto a railway platform. What is the minimum number of platforms required to transport all the slabs? | 7.14. It is impossible to load 6 slabs, each weighing 7 tons, onto one platform. Therefore, at least $200 / 5=40$ platforms are needed. Forty platforms are sufficient: onto each platform, 3 slabs weighing 7 tons each and 2 slabs weighing 9 tons each can be loaded. | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,310 |
7.15. A cargo weighing 13.5 tons is packed into boxes such that the weight of each box does not exceed 350 kg. Prove that this cargo can be transported using 11 one-and-a-half-ton trucks. (The weight of the empty boxes can be neglected.) | 7.15. Let's allocate 8 trucks and load them sequentially, each time placing the box that can no longer be loaded next to the truck. The loaded boxes together with the boxes standing next to the trucks weigh more than $8 \cdot 1.5=12$ tons, so the remaining boxes weigh less than 1.5 tons; they can be transported on one ... | proof | Other | proof | Yes | Yes | olympiads | false | 38,311 |
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 | 38,312 |
7.17. At the consultation, there were 20 schoolchildren and 20 problems were discussed. It turned out that each of the schoolchildren solved two problems and each problem was solved by two schoolchildren. Prove that it is possible to organize the discussion of the problems in such a way that each schoolchild presents o... | 7.17. Let's start analyzing the tasks with a random student telling one of the problems he has solved. Suppose this student solved problems $a_{1}$ and $a_{2}$, and he told problem $a_{1}$. Then there is exactly one student who also solved problem $a_{2}$ (and another problem $a_{3}$). This student will tell problem $a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,313 |
7.18. In the library, two boards were set up. On one board, a reader records the number of readers they encountered upon entering the reading room, and on the other - how many readers remained when they left the library. Early in the morning and late in the evening, there were no readers in the library. Prove that over... | 7.18. It can be considered that the very last of the readers who came to the library always leaves last. Indeed, it does not matter who exactly writes down the number of readers remaining in the hall; one reader can delegate this to another - the result will not change. Then each reader will write down the same number ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 38,314 |
7.19. Consecutively, $n$ numbers are written down, among which there are positive and negative numbers. Each positive number is underlined, as well as each number whose sum with several immediately following numbers is positive. Prove that the sum of all underlined numbers is positive. | 7.19. Let the numbers $a_{1}, a_{2}, \ldots, a_{n}$ be written down. Underline the number $a_{i} k+1$ with lines, where $k$ is the smallest number for which $a_{i}+a_{i+1}+a_{i+2}+\ldots+a_{i+k}>0$ (a positive number $a_{1}$ is underlined with one line). Clearly, if the number $a_{i}$ is underlined with $k+1$ lines, th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,315 |
7.20. Prove that the number of all digits in the sequence $1,2,3, \ldots, 10^{k}$ is equal to the number of all zeros in the sequence $1,2,3, \ldots, 10^{k+1}$. | 7.20. We will match a digit from the first sequence to a zero from the second sequence as follows. We will write a zero after the given digit. As a result, we get a number from the second sequence with a marked zero. We match our digit precisely to this zero. Conversely, to a zero from the number of the second sequence... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,316 |
8.1. For which $n$ do there exist positive numbers $x_{1}, \ldots, x_{n}$, satisfying the system of equations
$$
x_{1}+x_{2}+\ldots+x_{n}=3, \quad \frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}=3 ?
$$ | 8.1. Using the inequality $x_{i}+1 / x_{i} \geqslant 2$, we get $3+3=\left(x_{1}+1 / x_{1}\right)+\ldots+\left(x_{n}+1 / x_{n}\right) \geqslant 2 n$, i.e., $n \leqslant 3$. For $n=3$, we obtain the solution $x_{1}=x_{2}=x_{3}=1$. For $n=2$, we obtain the solution $x_{1}=\frac{3+\sqrt{5}}{2}, x_{2}=\frac{3-\sqrt{5}}{2}$... | n=3,x_{1}=x_{2}=x_{3}=1;\,n=2,x_{1}=\frac{3+\sqrt{5}}{2},x_{2}=\frac{3-\sqrt{5}}{2};\,n=1, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,317 |
8.2. Prove that if the numbers $a_{1}, \ldots, a_{n}$ are positive and $a_{1} \ldots a_{n}=1$, then
$$
\left(1+a_{1}\right)\left(1+a_{2}\right) \ldots\left(1+a_{n}\right) \geqslant 2^{n}
$$ | 8.2. From the relation $a_{1} \ldots a_{n}=1$, it follows that
$$
\begin{aligned}
& \left(1+a_{1}\right)\left(1+a_{2}\right) \ldots\left(1+a_{n}\right)=\frac{\left(1+a_{1}\right)\left(1+a_{2}\right) \ldots\left(1+a_{n}\right)}{a_{1} \ldots a_{n}}=\left(1+\frac{1}{a_{1}}\right) \ldots\left(1+\frac{1}{a_{n}}\right)
\end... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,318 |
8.3. Prove that if $x^{5}-x^{3}+x=2$, then $3<x^{6}<4$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 8.3. From the equality $2+x^{3}=x^{5}+x$, it follows that $\frac{2}{x^{3}}+1=x^{2}+\frac{1}{x^{2}} \geqslant 2$. Therefore, $\frac{2}{x^{3}} \geqslant 1$, i.e., $x^{3} \leqslant 2$. But $x \neq 1$, so the inequality is strict.
Adding the equalities $x^{5}-x^{3}+x=2$ and $x^{7}-x^{5}+x^{3}=2 x^{2}$, we get $x^{7}+x=2+2... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 38,319 |
8.4. a) Prove that if $x_{1}, x_{2}, x_{3}$ are positive numbers, then
$$
\frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{1}}+\frac{x_{3}}{x_{1}+x_{2}} \geqslant \frac{3}{2}
$$
b) Prove that if $x_{1}, x_{2}, \ldots, x_{n}$ are positive numbers, and $n \geqslant 4$, then
$$
\frac{x_{1}}{x_{2}+x_{n}}+\frac{x_{2}}{x_{... | 8.4. a) Let $x_{2}+x_{3}=a, x_{3}+x_{1}=b, x_{1}+x_{2}=c$, i.e., $x_{1}=\frac{b+c-a}{2}$, $x_{2}=\frac{a+c-b}{2}, x_{3}=\frac{a+b-c}{2}$. Then the required inequality can be written as
$$
\frac{b+c-a}{2 a}+\frac{a+c-b}{2 b}+\frac{a+b-c}{2 c} \geqslant \frac{3}{2}
$$
i.e., $b / a+c / a+a / b+c / b+a / c+b / c \geqslan... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,320 |
8.5. Prove that positive numbers $a, b, c$ satisfy the triangle inequality if and only if $\left(a^{2}+b^{2}+c^{2}\right)^{2}>2\left(a^{4}+b^{4}+c^{4}\right)$. | 8.5. Positive numbers $a, b, c$ satisfy the triangle inequality if and only if
$$
(a+b-c)(b+c-a)(c+a-b)(a+b+c)>0
$$
Indeed, two expressions in parentheses cannot be negative at the same time. For example, inequalities $a+b< c$ and $b+c < a$ imply $b < 0$, which contradicts the positivity of $b$. Hence, at least one o... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,321 |
8.9. Prove the inequality
$$
\left(\sum_{i=1}^{k} a_{i} b_{i}\right)^{2} \leqslant\left(\sum_{i=1}^{k} a_{i}^{2}\right)\left(\sum_{i=1}^{k} b_{i}^{2}\right)
$$
The inequality in problem 8.9 is called the Cauchy inequality. Other proofs of it are given in problems 1.9 and 5.12. | 8.9. Let $A=\sqrt{\sum_{i=1}^{k} a_{i}^{2}}$ and $B=\sqrt{\sum_{i=1}^{k} b_{i}^{2}}$. Using the inequality $x y \leqslant\left(x^{2}+y^{2}\right) / 2$, we get
$$
\sum_{i=1}^{k} \frac{a_{i}}{A} \frac{b_{i}}{B} \leqslant \sum_{i=1}^{k} \frac{1}{2}\left(\frac{a_{i}^{2}}{A^{2}}+\frac{b_{i}^{2}}{B^{2}}\right)=\frac{1}{2 A^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,324 |
8.10. Prove the inequality
$$
a_{1}^{2}+\ldots+a_{n}^{2} \geqslant \frac{\left(a_{1}+\ldots+a_{n}\right)^{2}}{n}
$$ | 8.10. This inequality is a special case of Cauchy's inequality; it is sufficient to set $b_{1}=\ldots=b_{n}=1$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,325 |
8.11. Let \( x_{1}, \ldots, x_{n} \) be positive numbers. Prove the inequality
\[
\left(x_{1}+\ldots+x_{n}\right)\left(\frac{1}{x_{1}}+\ldots+\frac{1}{x_{n}}\right) \geqslant n^{2}
\]
98 | 8.11. This inequality obviously follows from the Cauchy inequality: it is sufficient to set \(a_{i}=\sqrt{x_{i}}\) and \(b_{i}=1 / \sqrt{x_{i}}\). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,326 |
8.12. Let $S=a_{1}+\ldots+a_{n}$, where $a_{1}, \ldots, a_{n}$ are positive numbers and $n \geqslant 2$. Prove that
$$
\frac{a_{1}}{S-a_{1}}+\ldots+\frac{a_{n}}{S-a_{n}} \geqslant \frac{n}{n-1}
$$
## 8.4. Inequality between the Arithmetic Mean and the Geometric Mean
| Let $a_{1}, \ldots, a_{n}$ be positive numbers. ... | 8.12. Let $b_{i}=S-a_{i}$. Then the required inequality can be written as
$\frac{S-b_{1}}{b_{1}}+\ldots+\frac{S-b_{n}}{b_{n}} \geqslant \frac{n}{n-1}$, i.e. $S\left(\frac{1}{b_{1}}+\ldots+\frac{1}{b_{n}}\right) \geqslant \frac{n}{n-1}+n=\frac{n^{2}}{n-1}$. But $b_{1}+\ldots+b_{n}=\left(S-a_{1}\right)+\ldots+\left(S-a_... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,327 |
8.13. Let \( a_{1}, \ldots, a_{n} \) be positive numbers. Prove the inequality
$$
\frac{a_{1}+\ldots+a_{n}}{n} \geqslant \sqrt[n]{a_{1} \ldots a_{n}}
$$
(If not all the given numbers are equal, the inequality is strict.) | 8.13. First solution. We apply induction on $n$. For $n=1$, the required statement is true. Suppose that the required inequality has been proven for any $n$ positive numbers. Consider the positive numbers $b_{1}=\sqrt[n+1]{a_{1}}, \ldots, b_{n+1}=\sqrt[n+1]{a_{n+1}}$. Clearly, $\left(b_{i}^{n}-b_{j}^{n}\right)\left(b_{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,328 |
8.14. Let $a, b>0$. Prove that $2 \sqrt{a}+3 \sqrt[3]{b} \geqslant 5 \sqrt[5]{a b}$. | 8.14. Let $a=x^{10}$ and $b=y^{15}$. According to the inequality between the arithmetic mean and the geometric mean
$$
\frac{x^{5}+x^{5}+y^{5}+y^{5}+y^{5}}{5} \geqslant \sqrt[5]{x^{10} y^{15}}
$$
i.e., $2 \sqrt{a}+3 \sqrt[3]{b} \geqslant 5 \sqrt[5]{a b}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,329 |
8.15. Prove that for any natural $n \geqslant 2$ the following inequalities hold:
$$
n(\sqrt[n]{n+1}-1)<1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}<n\left(1-\frac{1}{\sqrt[n]{n}}\right)+1
$$ | 8.15. Apply the inequality between the arithmetic mean and the geometric mean for the numbers $2, \frac{3}{2}, \frac{4}{3}, \ldots, \frac{n+1}{n}$. As a result, we get
$$
\frac{1}{n}\left(2+\frac{3}{2}+\frac{4}{3}+\ldots+\frac{n+1}{n}\right)>\sqrt[n]{n+1}
$$
i.e.,
$$
(1+1)+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,330 |
8.16. Let $a_{1}, \ldots, a_{n}, w_{1}, \ldots, w_{n}$ be positive numbers. Prove that
$$
\frac{1}{W}\left(w_{1} a_{1}+\ldots+w_{n} a_{n}\right) \geqslant\left(a_{1}^{w_{1}} \ldots a_{n}^{w_{n}}\right)^{1 / W}
$$
where $W=w_{1}+\ldots+w_{n}$.
Remark. For $w_{1}=\ldots=w_{n}=1$, we obtain the inequality between the a... | 8.16. First solution. We apply induction on $n$ and use the inequality from problem 8.46. For convenience, let us denote $W_{n}=W$ and $W_{n-1}=w_{1}+\ldots+w_{n-1}$. It is clear that
$$
\begin{array}{r}
\left(a_{1}^{w_{1}} \ldots a_{n}^{w_{n}}\right)^{1 / W_{n}}=\left(a_{1}^{w_{1}} \ldots a_{n-1}^{w_{n-1}}\right)^{1 ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,331 |
8.17. Let $a_{1}, \ldots, a_{n}$ be positive numbers, and $\alpha_{1}, \ldots, \alpha_{n}$ be pairwise distinct positive numbers. For $x>0$, let $f(x)=\sum_{i=1}^{n} a_{i} x^{\alpha_{i}}$. Choose $a$ and $\alpha$ such that the graphs of the functions $y=f(x)$ and $y=a x^{\alpha}$ touch at $x=x_{0}$. Prove that $f(x) \g... | 8.17. Let $g(x)=a x^{\alpha}$. By the condition $f\left(x_{0}\right)=g\left(x_{0}\right)$ and $f^{\prime}\left(x_{0}\right)=g^{\prime}\left(x_{0}\right)$, i.e., $a x_{0}^{\alpha}=f\left(x_{0}\right)$ and $a \alpha x_{0}^{\alpha-1}=\sum_{i=1}^{n} a_{i} \alpha_{i} x_{0}^{\alpha_{i}-1}$. Therefore, $a=f\left(x_{0}\right) ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,332 |
8.18. Let $a, b, c$ be positive numbers. Prove that
$$
\sqrt{a^{2}-a b+b^{2}}+\sqrt{b^{2}-b c+c^{2}} \geqslant \sqrt{a^{2}+a c+c^{2}}
$$
with equality if and only if $1 / a+1 / c=1 / b$.
## 8.6. Cyclic Inequalities | 8.18. Consider the triangle $A O C$, where $A O=a, C O=c$ and the angle at vertex $O$ is $120^{\circ}$. Draw the angle bisector of angle $O$ and mark a segment $O B=b$ on it or its extension. Then, by the cosine rule, $A B=\sqrt{a^{2}-a b+b^{2}}, B C=\sqrt{b^{2}-b c+c^{2}}$ and $A C=\sqrt{a^{2}+a c+c^{2}}$. Therefore, ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,333 |
8.19. Let $a_{1}, \ldots, a_{n}$ be positive numbers. Prove that
$$
\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\ldots+\frac{a_{n}}{a_{1}} \geqslant n
$$ | 8.19. According to the inequality between the arithmetic mean and the geometric mean
$$
\frac{1}{n}\left(\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\ldots+\frac{a_{n}}{a_{1}}\right) \geqslant\left(\frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdot \ldots \cdot \frac{a_{n}}{a_{1}}\right)^{1 / n}=1
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,334 |
8.20. Prove that for any positive numbers $a$, $b, c$ the inequality
$$
a^{3} b+b^{3} c+c^{3} a \geqslant a^{2} b c+b^{2} c a+c^{2} a b
$$
holds. | 8.20. Let's complete the square:
$$
\begin{array}{r}
a^{3} b + b^{3} c + c^{3} a - a^{2} b c - b^{2} c a - c^{2} a b = \\
= \left(a^{3} b - 2 a^{2} b c + c^{2} a b\right) + \left(b^{3} c - 2 b^{2} c a + a^{2} b c\right) + \left(c^{3} a - 2 c^{2} a b + b^{2} c a\right) = \\
= a b(a - c)^{2} + b c(b - a)^{2} + c a(c - b... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,335 |
8.21. Let $0 \leqslant x_{1}, \ldots, x_{n} \leqslant 1$. Prove that for $n \geqslant 3$
$$
x_{1}+\ldots+x_{n}-x_{1} x_{2}-x_{2} x_{3}-\ldots-x_{n-1} x_{n}-x_{n} x_{1} \leqslant[n / 2]
$$ | 8.21. Fix all numbers $x_{1}, \ldots, x_{n}$ except one number $x_{i}=x$. Then the expression on the left side of the inequality has the form $a x+b$, where $a$ and $b$ are constant numbers. It is maximal when $x=0$ or $x=1$ (or is constant). Therefore, the expression on the left side of the inequality is maximal when ... | [n/2] | Inequalities | proof | Yes | Yes | olympiads | false | 38,336 |
8.22. Let $a, b, c$ be positive numbers. Prove that
$$
3+(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \geqslant \frac{3(a+1)(b+1)(c+1)}{a b c+1}
$$
## 8.7. Various Inequalities | 8.22. The required inequality can be easily transformed into the form $a^{2} b^{2} c^{2} \times$ $\times(a+b+c)-2 a b c(a b+b c+a c)-2 a b c(a+b+c)+\left(a^{2} c+b^{2} a+c^{2} b\right)+$ $+a b c\left(a^{2} c+b^{2} a+c^{2} b\right)+(a b+b c+a c) \geqslant 0$. But the left side of this inequality is equal to $a b(b+1)(a ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,337 |
8.23. Prove that if $a, b, c, d$ are positive numbers, and $a / b < c / d$, then $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$. | 8.23. The inequality $\frac{a}{b}<\frac{a+c}{b+d}$ is equivalent to the inequality $a b+a d<$ $<a b+b c$, i.e., $a d<b c$. And the inequality $\frac{a+c}{b+d}<\frac{c}{d}$ is equivalent to the inequality $a d+c d<b c+c d$, i.e., $a d<b c$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,338 |
8.24. Prove the inequality
$$
x^{2 n}+x^{2 n-1} y+x^{2 n-2} y^{2}+\ldots+y^{2 n} \geqslant 0
$$ | 8.24. If $x=y$, then the required inequality is obvious. If $x \neq y$, then the expression on the left side equals $\frac{x^{2 n+1}-y^{2 n+1}}{x-y}$. For
$x>y$ the numerator and denominator are positive, and for $x<y$ they are negative. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,339 |
8.25. Let $a>b>0$. Prove that $a^{a} b^{b}>a^{b} b^{a}$. | 8.25. Given that $a-b>0$ and $a / b>1$. Therefore, $(a / b)^{a-b}>1$, i.e., $a^{a-b}>b^{a-b}$. This inequality is equivalent to the required one. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,340 |
8.26. Suppose that $x_{1}, \ldots, x_{n}$ are real numbers, and it is known that all numbers $\sigma_{1}=\sum x_{i}, \sigma_{2}=$ $=\sum_{i<j} x_{i} x_{j}, \sigma_{3}=\sum_{i<j<k} x_{i} x_{j} x_{k}, \ldots, \sigma_{n}=x_{1} \ldots x_{n}$ are positive. Is it true that then the numbers $x_{1}, \ldots, x_{n}$ themselves a... | 8.26. A n s w e r: yes, it is correct. To prove this, consider the polynomial
$$
f(x)=x^{n}-\sigma_{1} x^{n-1}+\sigma_{2} x^{n-2}-\ldots+(-1)^{n} \sigma_{n}
$$
The numbers $x_{1}, \ldots, x_{n}$ are its roots. Therefore, it is sufficient to check that if $x>0$, then
$$
(-1)^{n} f(x)=(-x)^{n}+\sigma_{1}(-x)^{n-1}+\si... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,341 |
8.27. Prove that $x^{12}-x^{9}+x^{4}-x+1>0$ for all $x$. | 8.27. If $x \leqslant 0$, then $x^{12}, -x^{9}, x^{4}, -x \geqslant 0$. If $0 < x < 1$, then $x^{12} - x^{9} + x^{4} - x + 1 > 0$. If $x \geqslant 1$, then $x^{9}\left(x^{3}-1\right)+x\left(x^{3}-1\right)+1>0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,342 |
8.29. Let $n>1$ be a natural number. Prove that $x^{n}-n x+n-1 \geqslant 0$ for all $x>0$. | 8.29. 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 both expressions in parentheses are positive, and if $0<x<1$, they are negative. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,344 |
8.30. Prove that for all natural $n \geqslant 2$ the inequality
$$
\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2}
$$
holds. | 8.30. Let's add $n-1$ inequalities $\frac{1}{n+1}>\frac{1}{2 n}, \frac{1}{n+2}>\frac{1}{2 n}, \ldots, \frac{1}{2 n-1}>$ $>\frac{1}{2 n}$ and the equality $\frac{1}{2 n}=\frac{1}{2 n}$. As a result, we will obtain the required. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,345 |
8.31. Let \( x \) and \( y \) be arbitrary real numbers. Prove that
\[
\frac{x+y}{2} \cdot \frac{x^{2}+y^{2}}{2} \cdot \frac{x^{3}+y^{3}}{2} \leqslant \frac{x^{6}+y^{6}}{2}
\] | 8.31. Let's first prove that if $m$ and $n$ are natural numbers of the same parity, then
$$
\frac{x^{m}+y^{m}}{2} \cdot \frac{x^{n}+y^{n}}{2} \leqslant \frac{x^{m+n}+y^{m+n}}{2}
$$
This inequality can be easily transformed into the form $\left(x^{m}-y^{m}\right)\left(x^{n}-y^{n}\right) \geqslant 0$. If the numbers $m... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,346 |
8.32. Prove that for all natural $n$ the inequality holds
$$
\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{2 n-1}{2 n}<\frac{1}{\sqrt{2 n+1}}
$$ | 8.32. It is clear that $\left(\frac{1}{2} \cdot \frac{3}{4} \cdot \ldots \cdot \frac{2 n-1}{2 n}\right)^{2}=\frac{1 \cdot 3}{2^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdot \ldots \cdot \frac{(2 n-1)(2 n+1)}{(2 n)^{2}} \times$ $\times \frac{1}{2 n+1}<\frac{1}{2 n+1}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,347 |
8.33. Prove that for all natural $n>1$
$$
\frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{2 n}}<\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{2 n-1}{2 n}<\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2 n}}
$$ | 8.33. Let $P_{n}=\frac{1 \cdot 3 \cdot \ldots \cdot(2 n-1)}{2 \cdot 4 \cdot \ldots \cdot 2 n}$. Then
$$
P_{n}^{2}=\frac{1^{2}}{2^{2}} \cdot \frac{3^{2}}{4^{2}} \cdot \frac{5^{2}}{6^{2}} \cdot \ldots \cdot \frac{(2 n-1)^{2}}{(2 n)^{2}}=\frac{1}{2} \cdot \frac{3^{2}}{2 \cdot 4} \cdot \frac{5^{2}}{4 \cdot 6} \cdot \ldots... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,348 |
8.35. The numbers $1,2, \ldots, n$ are arranged in a circle in some arbitrary order. Prove that the sum of the absolute values of the differences of adjacent numbers is not less than $2 n-2$.
保留源文本的换行和格式，直接输出翻译结果。 | 8.35. The points where the numbers 1 and $n$ are located divide the circle into two arcs. On each arc, the sum of the differences of adjacent numbers is $n-1$ (if we go from $n$ to 1), and the sum of the absolute values of the numbers is not less than their sum. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,350 |
8.37. Let $0<a \leqslant x_{i} \leqslant b$. Prove that
$$
\left(x_{1}+\ldots+x_{n}\right)\left(\frac{1}{x_{1}}+\ldots+\frac{1}{x_{n}}\right) \leqslant \frac{(a+b)^{2}}{4 a b} n^{2}
$$ | 8.37. Given that $\left(x_{i}-a\right)\left(x_{i}-b\right) \leqslant 0$ and $x_{i}>0$, therefore $\frac{a b}{x_{i}}+x_{i} \leqslant$ $\leqslant a+b$. Adding such inequalities, we get $a b \sum \frac{1}{x_{i}}+\sum x_{i} \leqslant$ $\leqslant n(a+b)$. Let $x^{*}=\frac{1}{n} \sum x_{i}$ and $y^{*}=\frac{1}{n} \sum \frac{... | \frac{(+b)^{2}}{4} | Inequalities | proof | Yes | Yes | olympiads | false | 38,352 |
8.38. Prove that if $-1 \leqslant x_{1}, \ldots, x_{n} \leqslant 1$ and $x_{1}^{3}+\ldots+x_{n}^{3}=$ $=0$, then $x_{1}+\ldots+x_{n} \leqslant n / 3$. | 8.38. Consider the polynomial $P(x)=4(x+1)(x-1 / 2)^{2}=(x+1) \times$ $\times\left(4 x^{2}-4 x+1\right)=4 x^{3}-3 x+1$. It is clear that $P\left(x_{i}\right) \geqslant 0$. By adding the inequalities $P\left(x_{1}\right) \geqslant 0, \ldots, P\left(x_{n}\right) \geqslant 0$, we get $-3\left(x_{1}+\ldots+x_{n}\right)+n \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,353 |
8.39. Prove that if $x y=4$ and $z^{2}+4 w^{2}=4$, then $(x-z)^{2}+$ $+(y-w)^{2} \geqslant 1.6$. | 8.39. Geometrically, this inequality means that the distance between the ellipse $x^{2}+4 y^{2}=4$ and the hyperbola $x y=4$ is not less than $\sqrt{1.6}$. For the proof, it is convenient to use the geometric formulation. We will prove the following statement:
The tangent to the ellipse $x^{2}+4 y^{2}=4$ at the point ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,354 |
8.40. Let \( A(x) = A_{1}(x) + \ldots + A_{5}(x) \), where \( x = (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) \) and \( A_{i}(x) = \prod_{j \neq i} (x_{i} - x_{j}) \). Prove that \( A(x) \geqslant 0 \) for all \( x \). | 8.40. The value of $A(x)$ does not change under any permutation of the variables, so we can assume that $x_{1} \geqslant x_{2} \geqslant x_{3} \geqslant x_{4} \geqslant x_{5}$. In this case,
\[
\begin{aligned}
A_{1}(x)+A_{2}(x)=\left(x_{1}-x_{2}\right)\left[\left(x_{1}-x_{3}\right)\right. & \left(x_{1}-x_{4}\right)\le... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,355 |
8.41. One hundred positive numbers $x_{1}, x_{2}, \ldots, x_{100}$ satisfy the conditions
$$
\begin{aligned}
& x_{1}^{2}+x_{2}^{2}+\ldots+x_{100}^{2}>10000 \\
& x_{1}+x_{2}+\ldots+x_{100}<300
\end{aligned}
$$
Prove that among them, there are three numbers whose sum is greater than 100.
See also problem 18.4.
## 8.8... | 8.41. We can assume that $x_{1} \geqslant x_{2} \geqslant \ldots \geqslant x_{100}>0$. If $x_{1} \geqslant 100$, then $x_{1}+x_{2}+x_{3}>100$. Therefore, we will assume that $x_{1}<100$, $100-x_{2}>0$, $x_{1}-x_{3} \geqslant 0$ and $x_{2}-x_{3} \geqslant 0$, hence
$$
\begin{gathered}
100\left(x_{1}+x_{2}+x_{3}\right) ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,356 |
8.43. Compare the numbers $\sqrt[3]{60}$ and $2+\sqrt[3]{7}$.
## 8.9. Hölder and Minkowski Inequalities | 8.43. Apply the inequality from problem 8.42 for \( n=3, a=\sqrt[3]{8} \) and \( b=\sqrt[3]{7} \). As a result, we get \(\frac{\sqrt[3]{8}+\sqrt[3]{7}}{2}<\sqrt[3]{\frac{7+8}{2}}\), i.e., \(2+\sqrt[3]{7}<\sqrt[3]{60}\). | 2+\sqrt[3]{7}<\sqrt[3]{60} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,358 |
8.44. a) Let $a$ be a rational number, such that $a>1$ or $a<0$. For all $x>0, x \neq 1$, the inequality $x^{a}-a x+a-1>0$ holds.
b) Let $a$ be a rational number, such that $0<a<1$. For all $x>0, x \neq 1$, the inequality $x^{a}-a x+a-1<0$ holds. | 8.44. Let $n$ be a natural number. The identity
$$
\frac{y^{n+1}-1}{n+1}-\frac{y^{n}-1}{n}=\frac{y-1}{n(n+1)}\left(n y^{n}-y^{n-1}-\ldots-y-1\right)
$$
shows that $\frac{y^{n+1}-1}{n+1}-\frac{y^{n}-1}{n}>0$ for $y>0, y \neq 1$. Indeed, if $y>1$, then $y-1>0$ and $n y^{n}>y^{n-1}+\ldots+y+1$, and if $0<y<1$, then $y-1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,359 |
8.46. Let $A$ and $B$ be positive numbers, and $p$ and $q$ be rational numbers related by the equation $1 / p + 1 / q = 1$. Prove that if $p > 1$, then $A^{1 / p} B^{1 / q} \leqslant A / p + B / q$, and if $p < 1$, then $A^{1 / p} B^{1 / q} \geqslant A / p + B / q$. (If $A \neq B$, then both inequalities are strict.) | 8.46. Let $x=A / B, a=1 / p$ and $1-a=1 / q$. Then
$$
B\left(x^{a}-a x+a-1\right)=A^{1 / p} B^{1 / q}-\frac{A}{p}-\frac{B}{q} .
$$
It remains to use the result of problem 8.44.
Remark. For other proofs, see problems 26.24 (case $p>1$) and 28.44. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,361 |
8.47. Let $x_{i}$ and $y_{i}$ be positive numbers, and $p$ and $q$ be rational numbers related by the equation $1 / p + 1 / q = 1$. Prove that if $p > 1$, then
$$
x_{1} y_{1} + \ldots + x_{n} y_{n} \leqslant \left(x_{1}^{p} + \ldots + x_{n}^{p}\right)^{1 / p} \left(y_{1}^{q} + \ldots + y_{n}^{q}\right)^{1 / q}
$$
and... | 8.47. Let $X=x_{1}^{p}+\ldots+x_{n}^{p}, Y=y_{1}^{q}+\ldots+y_{n}^{q}, A=x_{i}^{p} / X$ and $B=y_{i}^{q} / Y$. If $p>1$, then according to problem $8.46 \frac{x_{i}}{X^{1 / p}} \frac{y_{i}}{Y^{1 / q}} \leqslant \frac{x_{i}^{p}}{p X}+\frac{y_{i}^{q}}{q Y}$. Adding such inequalities for $i=1, \ldots, n$, we get
$$
\frac... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,362 |
8.48. Let $x_{i}$ and $y_{i}$ be positive numbers, and $p>1$ a rational number. Prove that
$$
\left(\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)^{p}\right)^{1 / p} \leqslant\left(\sum_{i=1}^{n} x_{i}^{p}\right)^{1 / p}+\left(\sum_{i=1}^{n} y_{i}^{p}\right)^{1 / p}
$$
If $p<1$, the inequality is reversed (Minkowski's inequa... | 8.48. Let $p>1$. It is clear that
$$
\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)^{p}=\sum_{i=1}^{n} x_{i}\left(x_{i}+y_{i}\right)^{p-1}+\sum_{i=1}^{n} y_{i}\left(x_{i}+y_{i}\right)^{p-1}
$$
According to Hölder's inequality (problem 8.47)
$$
\sum_{i=1}^{n} x_{i}\left(x_{i}+y_{i}\right)^{p-1} \leqslant\left(\sum_{i=1}^{n} ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,363 |
9.1. Let $a_{1}, \ldots, a_{n}$ be an arithmetic progression. Prove that
$$
\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\ldots+\frac{1}{a_{n-1} a_{n}}=\frac{n-1}{a_{1} a_{n}}
$$ | 9.1. It is clear that $\frac{1}{a_{k}}-\frac{1}{a_{k+1}}=\frac{a_{k+1}-a_{k}}{a_{k} a_{k+1}}=\frac{d}{a_{k} a_{k+1}}$, where $d$ is the common difference of the arithmetic progression. Therefore, the considered sum is equal to
$$
\begin{aligned}
\frac{1}{d}\left(\frac{1}{a_{1}}-\frac{1}{a_{2}}+\frac{1}{a_{2}}-\frac{1}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,364 |
9.2. Let $a_{1}, \ldots, a_{n}$ be an arithmetic progression with positive terms. Prove that
$$
\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}
$$
- A decimal fraction $\alpha=0, a_{1} a_{2} a_{3} \ldots$ is cal... | 9.2. Note that
$$
\begin{aligned}
\frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_{k}}}{\sqrt{a_{k+1}}-\sqrt{a_{k}}} \cdot \frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}} & = \\
& =\frac{\sqrt{a_{k+1}}-\sqrt{a_{k}}}{a_{k+1}-a_{k}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_{k}}}{d}
\end{aligned}
$$
where \( d \) is the co... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,365 |
9.3. Prove that a periodic decimal fraction is a rational number.
For the converse statement, see problem 17.1. | 9.3. It is clear that
$$
\begin{aligned}
& 0, a_{1} a_{2} \ldots a_{N-1}\left(a_{N} a_{N+1} \ldots a_{N+n-1}\right)=a_{1} a_{2} \ldots a_{N-1} 10^{-N+1}+ \\
& \quad+10^{-N} a_{N} q+10^{-N-1} a_{N+1} q+\ldots+10^{-N-n+1} a_{N+n-1} q
\end{aligned}
$$
where \( q=1+10^{-n}+10^{-2 n}+10^{-3 n}+\ldots=\frac{1}{1-10^{-n}} \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,366 |
9.4. Is the number $0.1234567891011 \ldots$ (all natural numbers written in sequence) rational? | 9.4. Suppose this number is rational. Then, according to problem 17.1, it is written as a periodic decimal fraction $0, a_{1} a_{2} a_{3} \ldots$, where $a_{k+n}=a_{k}$ for all $k \geqslant N$ ($n$ is the length of the period). We choose $m$ to be sufficiently large so that the number $10^{m+2 n}$ is written after $a_{... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,367 |
9.5. Prove that any number of the form $n^{k}$, where $n$ and $k$ are natural numbers different from 1, can be represented as the sum of $n$ consecutive odd numbers. | 9.5. For the number $a+(a+2)+\ldots+(a+2 n-2)=n(a+n-1)$ to equal $n^{k}$, we need to set $a+n-1=n^{k-1}$, i.e., $a=n^{k-1}-n+1$. It is clear that the number $a$ is odd in this case. | n^{k-1}-n+1 | Number Theory | proof | Yes | Yes | olympiads | false | 38,368 |
9.6. Find the sum $1+2 x+3 x^{2}+\ldots+(n+1) x^{n}$. | 9.6. To obtain the required sum, we need to add $1+x+$ $+x^{2}+\ldots+x^{n}=\frac{x^{n+1}-1}{x-1}, x+x^{2}+\ldots+x^{n}=\frac{x^{n+1}-x}{x-1}, \ldots, x^{n}=\frac{x^{n+1}-x^{n}}{x-1}$. As a result, we get
$$
\begin{aligned}
& \frac{(n+1) x^{n+1}-\left(1+x+\ldots+x^{n}\right)}{x-1}=\frac{(n+1) x^{n+1}-\frac{x^{n+1}-1}{... | \frac{(n+1)x^{n+2}-(n+2)x^{n+1}+1}{(x-1)^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,369 |
9.7. Let $a, a+d, a+2d, \ldots$ be an infinite arithmetic progression, where $a, d > 0$. Prove that it is possible to select an infinite subsequence from it that forms a geometric progression if and only if the number $a / d$ is rational. | 9.7. Let $a / d=m / n$, where $m$ and $n$ are natural numbers. For all natural $k$, the number $(1+n)^{k}-1$ is divisible by $n$, so the number $b_{k}=\frac{a(1+n)^{k}-a}{d}=\frac{m}{n}\left((1+n)^{k}-1\right)$ is an integer. Therefore, all numbers $a(1+n)^{k}=a+b_{k} d$ belong to the given arithmetic progression.
Let... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,370 |
9.8. There are $4 n$ positive numbers such that any four pairwise distinct numbers can form a geometric progression. Prove that among these numbers, there are $n$ identical ones. | 9.8. We will show that among the given numbers, there cannot be more than four pairwise distinct numbers. Let's combine equal numbers into groups, select one number from each group, and arrange the selected numbers in descending order: $a>b>c>d>e>\ldots$ The numbers $a, b, c, d$ form a geometric progression by the cond... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,371 |
9.9. A geometric progression is given, the denominator of which is an integer (not equal to 0 and -1). Prove that the sum of any number of arbitrarily chosen terms of this progression cannot equal any term of this progression.
## 9.2. Change of Summation Order | 9.9. Each member of a geometric progression is represented in the form $a q^{n}, n \geqslant 0$. The case when $q=1$ is obvious, so we will assume that $q \neq 1$. Suppose there exist different non-negative integers $k_{1}, k_{2}, \ldots, k_{m+1}$ ( $m \geqslant 2$ ), for which
$$
a q^{k_{1}}+a q^{k_{2}}+\ldots+a q^{k... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,372 |
9.10. Prove that if $n=p+q-1$, then
\[
\begin{aligned}
& \left(a_{1}+a_{2}+\ldots+a_{p}\right)+\left(a_{2}+\ldots+a_{p+1}\right)+\ldots+\left(a_{n-p+1}+\ldots+a_{n}\right)= \\
& \quad=\left(a_{1}+a_{2}+\ldots+a_{q}\right)+\left(a_{2}+\ldots+a_{q+1}\right)+\ldots+\left(a_{n-q+1}+\ldots+a_{n}\right)
\end{aligned}
\] | 9.10. Consider the table
$$
\begin{array}{cccc}
a_{1} & a_{2} & \ldots & a_{p} \\
a_{2} & a_{3} & \ldots & a_{p+1} \\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots & \ldots \ldots \\
a_{n-p+1} & a_{n-p+2} & \ldots & a_{n}
\end{array}
$$
By the condition $n-p+1=q$ and $p=n-q+1$, so on the left side stands the sum o... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,373 |
9.11. The numbers $a_{1}, a_{2}, \ldots, a_{n}$ are such that the sum of any seven consecutive numbers is negative, and the sum of any eleven consecutive numbers is positive. For what largest $n$ is this possible?
118 Chapter 9. Computation of Sums and Products
$$
\text { 9.3. Sums } S_{k}(n)=1^{k}+2^{k}+\ldots+n^{k}... | 9.11. According to problem $9.10\left(a_{1}+a_{2}+\ldots+a_{7}\right)+\left(a_{2}+\ldots+a_{8}\right)+\ldots$ $\ldots+\left(a_{11}+\ldots+a_{17}\right)=\left(a_{1}+a_{2}+\ldots+a_{11}\right)+\left(a_{2}+\ldots+a_{12}\right)+\ldots+$ $+\left(a_{7}+\ldots+a_{17}\right)$. Therefore, $n<17$.
For $n=16$, such a sequence ex... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,374 |
9.12. Calculate in the same way the sums $S_{2}(n)=1^{2}+$ $+2^{2}+\ldots+n^{2}$ and $S_{3}(n)=1^{3}+2^{3}+\ldots+n^{3}$. | 9.12. By summing the equalities \((k+1)^{3}=k^{3}+3 k^{2}+3 k+1\) for \(k=1,2, \ldots, \ldots, n\), we get \((n+1)^{3}=1+3 S_{2}(n)+3 S_{1}(n)+n\). Therefore,
\[
3 S_{2}(n)=(n+1)^{3}-\frac{3}{2} n(n+1)-(n+1)=\frac{n(n+1)(2 n+1)}{2}
\]
By summing the equalities \((k+1)^{4}=k^{4}+4 k^{3}+6 k^{2}+4 k+1\) for \(k=1,2, \l... | S_{2}(n)=\frac{n(n+1)(2n+1)}{6},\quadS_{3}(n)=(\frac{n(n+1)}{2})^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,375 |
9.13. a) Prove that
$$
\begin{aligned}
& C_{k+1}^{k} S_{k}(n)+C_{k+1}^{k-1} S_{k-1}(n)+\ldots+C_{k+1}^{1} S_{1}(n)+S_{0}(n)= \\
& \quad=(n+1)^{k+1}-1
\end{aligned}
$$
b) Prove that for a fixed $k$, the sum $S_{k}(n)$ is a polynomial of degree $k+1$ in $n$ with the leading coefficient $\frac{n^{k+1}}{k+1}$. | 9.13. a) On one hand, $\sum_{j=1}^{n}\left((j+1)^{k+1}-j^{k+1}\right)=(n+1)^{k+1}-1$. On the other hand, $\sum_{j=1}^{n}\left((j+1)^{k+1}-j^{k+1}\right)=\sum_{j=1}^{n}\left(C_{k+1}^{k} j^{k}+C_{k+1}^{k-1} j^{k-1}+\ldots\right.$ $\left.\ldots+C_{k+1}^{1} j+1\right)=C_{k+1}^{k} S_{k}(n)+C_{k+1}^{k-1} S_{k-1}(n)+\ldots+C_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,376 |
9.14. a) Let
$$
S=C_{k}^{1} S_{2 k-1}(n)+C_{k}^{3} S_{2 k-3}(n)+C_{k}^{5} S_{2 k-5}(n)+\ldots
$$
where the last term in this sum is $C_{k}^{k} S_{k}(n)$ for odd $k$ and $C_{k}^{k-1} S_{k+1}(n)$ for even $k$. Prove that $S=\frac{n^{k}(n+1)^{k}}{2}$.
b) Prove that $S_{2 k-1}(n)$ is a polynomial of degree $k$ in $\frac... | 9.14. a) On one hand, $\sum_{j=1}^{n}\left(j^{k}(j+1)^{k}-j^{k}(j-1)^{k}\right)=n^{k}(n+1)^{k}$. On the other hand, this sum is equal to the sum of expressions of the form
$$
\begin{array}{r}
j^{k}+C_{k}^{1} j^{2 k-1}+C_{k}^{2} j^{2 k-2}+\ldots+C_{k}^{k-1} j^{k-1}+C_{k}^{k} j^{k} \\
-j^{k}+C_{k}^{1} j^{2 k-1}-C_{k}^{2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,377 |
9.15. Let
$$
S=C_{k+1}^{1} S_{k}(n)+C_{k+1}^{3} S_{k-2}(n)+\ldots
$$
where the last term in this sum is $C_{k+1}^{k+1} S_{0}(n)$ for even $k$ and $C_{k+1}^{k} S_{1}(n)$ for odd $k$. Prove that $S=\frac{(n+1)^{k+1}+n^{k+1}-1}{2}$. | 9.15. On the one hand, $\sum_{j=1}^{n}\left((j+1)^{k+1}-(j-1)^{k+1}\right)=(n+1)^{k+1}+$ $+n^{k+1}-1$. On the other hand, this sum is equal to $2 S$. | \frac{(n+1)^{k+1}+n^{k+1}-1}{2} | Combinatorics | proof | Yes | Yes | olympiads | false | 38,378 |
9.16. Find the sum
$$
1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}
$$
## 9.4. Pairing
## | 9.16. According to problem $9.121^{3}+2^{3}+3^{3}+\ldots+m^{3}=\left(\frac{m(m+1)}{2}\right)^{2}$. Therefore, $1^{3}+2^{3}+3^{3}+\ldots+(2 n-1)^{3}+(2 n)^{3}=\left(\frac{2 n(2 n+1)}{2}\right)^{2}$, i.e.
$1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}+2^{3}\left(1^{3}+2^{3}+\ldots+n^{3}\right)=\left(\frac{2 n(2 n+1)}{2}\right)^{2... | n^{2}(2n^{2}-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,379 |
9.17. a) Prove that for any prime number $p>2$ the numerator of the fraction
$$
\frac{m}{n}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p-1}
$$
is divisible by $p$.
b) Prove that for any prime number $p>3$ the numerator of the fraction
$$
\frac{m}{n}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p-1}
$$
is divisible by ... | 9.17. a) The considered sum can be broken down into addends $\frac{1}{k}+\frac{1}{p-k}, k=1,2, \ldots, \frac{p-1}{2}$. In this case, $\frac{1}{k}+\frac{1}{p-k}=\frac{p}{k(p-k)}$. After bringing such fractions to a common denominator, we get $\frac{m}{n}=\frac{p q}{1 \cdot 2 \cdot 3 \cdot \ldots \cdot(p-1)}$. It remains... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,380 |
9.18. Let $\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{4 k-1}-$ be an irreducible fraction, and let the number $6 k-1$ be prime. Prove that $p$ is divisible by $6 k-1$. | 9.18. First, note that
$$
\begin{aligned}
& 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{4 k-1}= \\
& =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{4 k-1}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{4 k-2}\right)= \\
& =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{4 k-1}-1-\fr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,381 |
9.19. Let $p>2$ be a prime number, $a_{k}$ be the remainder of the division of $k^{p}$ by $p^{2}$. Prove that $a_{1}+a_{2}+\ldots+a_{p-1}=\frac{p^{3}-p^{2}}{2}$.
## 9.5. Calculation of a sum in two ways | 9.19. Let's show that $k^{p}+(p-k)^{p}$ is divisible by $p^{2}$. Indeed, $(x-y)^{p}=-y^{p}+p x y^{p-1}+\ldots$, where the ellipsis denotes terms divisible by $x^{2}$. In our case, $(p-k)^{p} \equiv -k^{p} + p k^{p-1} p \equiv -k^{p} \pmod{p^{2}}$, so $k^{p}+(p-k)^{p}$ is divisible by $p^{2}$.
The number $p$ is prime, ... | \frac{p^{3}-p^{2}}{2} | Number Theory | proof | Yes | Yes | olympiads | false | 38,382 |
9.20. Around a circle, $a$ plus signs and $b$ minus signs are arranged in an arbitrary order. Let $p$ be the number of pairs of adjacent plus signs, and $q$ be the number of pairs of adjacent minus signs. Prove that $a-b=p-q$. | 9.20. Between adjacent plus signs, write the number +2, between adjacent minus signs, write -2, and between adjacent plus and minus signs, write 0. On one hand, the sum of all the written numbers is equal to $2(a-b)$. On the other hand, it is equal to $2(p-q)$. | p-q | Combinatorics | proof | Yes | Yes | olympiads | false | 38,383 |
10.1. Represent the polynomial $(x+1)(x+2)(x+3)(x+4)+1$ as a perfect square.
## 10.2. Roots of Polynomials | 10.1. The equalities $(x+1)(x+2)(x+3)(x+4)+1=x^{4}+10 x^{3}+\ldots+25$ and $\left(x^{2}+a x+b\right)^{2}=x^{4}+2 a x^{3}+\ldots+b^{2}$ show that suitable candidates are the quadratic trinomials $x^{2}+5 x \pm 5$.
Next, note that $\left(x^{2}+5 x+5\right)^{2}-1=\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)$ and $x^{... | (x^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,384 |
10.2. a) Prove that the remainder of the division of a polynomial $f(x)$ by $x-a$ is $f(a)$ (Bezout).
b) Let $x_{0}$ be a root of the polynomial $f(x)$. Prove that $f(x)$ is divisible by $x-x_{0}$. | 10.2. a) Let's divide $f(x)$ by $x-a$ with a remainder. As a result, we get $f(x)=(x-a) g(x)+r$, where $r$ is some number. Let $x=a$. As a result, we get $f(a)=r$.
b) This follows directly from a), since $f\left(x_{0}\right)=0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,385 |
10.3. Let $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$ be a polynomial with integer coefficients. Suppose it has a rational root $x_{0}=p / q$, where $p / q$ is an irreducible fraction. Prove that $a_{n}$ is divisible by $q$, and $a_{0}$ is divisible by $p$. | 10.3. The equality $a_{n} x_{0}^{n}+a_{n-1} x_{0}^{n-1}+\ldots+a_{1} x_{0}+a_{0}=0$ after multiplying by $q^{n}$ can be written as $a_{n} p^{n}+a_{n-1} p^{n-1} q+\ldots+a_{1} p q^{n-1}+a_{0} q^{n}=0$. The number $a_{n} p^{n}=-\left(a_{n-1} p^{n-1} q+\ldots+a_{1} p q^{n-1}+a_{0} q^{n}\right)$ is divisible by $q$, and si... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,386 |
10.4. Find a polynomial with integer coefficients for which the number $\sqrt{2}+\sqrt{3}$ is a root. | 10.4. Let $x=\sqrt{2}+\sqrt{3}$. Then $x^{2}=5+2 \sqrt{6}$ and $\left(x^{2}-5\right)^{2}=24$, i.e., $x^{4}-10 x^{2}+1=0$. | x^{4}-10x^{2}+1=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,387 |
10.5. Find a polynomial with integer coefficients for which the number $\sqrt[3]{2}+\sqrt[3]{3}$ is a root. | 10.5. Simple calculations show that
\[
\begin{gathered}
(\sqrt[3]{2}+\sqrt[3]{3})^{3}=5+3 \alpha \\
(\sqrt[3]{2}+\sqrt[3]{3})^{6}=133+30 \alpha+9 \beta \\
(\sqrt[3]{2}+\sqrt[3]{3})^{9}=2555+711 \alpha+135 \beta
\end{gathered}
\]
where \(\alpha=\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})\) and \(\beta=\sqrt[3]{36}(\sqrt[3]{4... | x^{9}-15x^{6}-87x^{3}-125=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,388 |
10.6. Find all polynomials of the form $x^{n} \pm x^{n-1} \pm x^{n-2} \pm \ldots$ $\ldots \pm x \pm 1$, which have all real roots.
## 10.3. Coefficients of a Polynomial | 10.6. Let $x_{1}, \ldots, x_{n}$ be the roots of the polynomial $x^{n} + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \ldots$, where $a_{i} = \pm 1$. According to Vieta's theorem, $x_{1} + \ldots + x_{n} = -a_{n-1}$ and $\sum_{1 \leqslant i < j \leqslant n} x_{i} x_{j} = a_{n-2}$, so $x_{1}^{2} + \ldots + x_{n}^{2} = a_{n-1}^{2... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,389 |
10.7. Determine the coefficients that will stand by \(x^{17}\) and \(x^{18}\) after expanding the brackets and combining like terms in the expression
\[
\left(1+x^{5}+x^{7}\right)^{20}
\] | 10.7. The number 18 cannot be represented as the sum of several numbers 5 and 7, so the coefficient of $x^{18}$ will be zero.
The number 17 can be represented as the sum of several numbers 5 and 7 as follows: $17=7+5+5$; this representation is unique up to the permutation of the addends. In one of the 20 expressions $... | 3420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,390 |
10.8. Let $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$ be a polynomial with real coefficients, and $a_{n} \geqslant 1$. Prove that if the number $m$ is greater than any of the numbers $\left|a_{n-1}\right|+1, \ldots$ $\ldots,\left|a_{0}\right|+1$, then $Q(x)=P(x+m)$ is a polynomial with positive coefficients. | 10.8. Let's write the polynomial $P(x)$ in the form $P(x)=\left(a_{n}-1\right) x^{n}+$ $+\left(x+a_{n-1}-1\right) x^{n-1}+\left(x+a_{n-2}-1\right) x^{n-2}+\ldots+\left(x+a_{0}-1\right)+1$. This representation shows that $Q(x)=\left(a_{n}-1\right)(x+m)^{n}+\left(x+\left(m+a_{n-1}-1\right)\right) x^{n-1}+\ldots$ $\ldots+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,391 |
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
$$
\begin{aligned}
& x^{4}+x^{3}+2 x^{2}+x+1=\frac{x^{12}-1}{(x-1)\left(x^{3}+1\ri... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,392 |
10.10. Let $x_{1}, \ldots, x_{n}$ be the roots of the polynomial
$$
x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\ldots+a_{0}
$$
Prove that $x_{1}+x_{2}+\ldots+x_{n}=-a_{n-1}$,
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}=a_{n-2}, \quad \sum_{1 \leqslant i<j<k \leqslant n} x_{i} x_{j} x_{k}=-a_{n-3}, \quad \ldots
$$
... | 10.10. The considered polynomial has the form
\[
\begin{aligned}
\left(x-x_{1}\right) \ldots\left(x-x_{n}\right)=x^{n} & -\left(\sum_{1 \leqslant i \leqslant n} x_{i}\right) x^{n-1}+\left(\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\right) x^{n-2}- \\
& -\left(\sum_{1 \leqslant i<j<k \leqslant n} x_{i} x_{j} x_{k}\r... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,393 |
10.11. What condition must the coefficients of the polynomial $x^{3}+a x^{2}+b x+c$ satisfy so that its three roots form an arithmetic progression? | 10.11. The numbers $x_{1}, x_{2}, x_{3}$ form an arithmetic progression if and only if $x_{1}+x_{3}=2 x_{2}$ (possibly after renumbering), i.e., $x_{1}+x_{2}+x_{3}=3 x_{2}$. According to Vieta's theorem, $x_{1}+x_{2}+x_{3}=-a$. Therefore, the number $x_{2}=-a / 3$ must be a root of the given polynomial. An equivalent c... | \frac{2}{27}^{3}-\frac{}{3}+=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,394 |
10.12. a) Let $\alpha, \beta$ and $\gamma$ be the roots of the polynomial $x^{3}-9 x+9$. Prove that $\alpha^{2}+\alpha-6=\beta$ or $\gamma$.
b) Let $\alpha, \beta$ and $\gamma$ be the roots of the polynomial $x^{3}-21 x+35$. Prove that $\alpha^{2}+2 \alpha-14=\beta$ or $\gamma$. | 10.12. a) Let $y=\alpha^{2}+\alpha-6$. It is required to prove that $(y-\beta) \times$ $\times(y-\gamma)=0$. By Vieta's theorem, $(y-\beta)(y-\gamma)=y^{2}-(\beta+\gamma) y+\beta \gamma=$ $=y^{2}+\alpha y-\frac{9}{\alpha}$. We will compute $y^{2}+\alpha y-\frac{9}{\alpha}$, substituting $y=\alpha^{2}+\alpha-6$ and usin... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,395 |
10.13. The polynomial $f(x)=x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x+1$ with non-negative coefficients has $n$ real roots. Prove that $f(2) \geqslant 3^{n}$.
See also problem 28.28.
## 10.5. Divisibility | 10.13. All coefficients of the polynomial $f(x)$ are non-negative, so it has no positive roots. Let $-x_{1}, \ldots,-x_{n}$ be the roots of the polynomial $f(x)$. Then by Vieta's formulas, $a_{k}=\sum x_{i_{1}} \ldots x_{i_{k}}$, and the numbers $x_{1}, \ldots, x_{n}$ are positive and their product is 1. The arithmetic... | 3^{n} | Algebra | proof | Yes | Yes | olympiads | false | 38,396 |
10.14. Let $P(x)$ be a polynomial with integer coefficients, and let $a$ and $b$ be integers such that $a > b$. Prove that the number $P(a) - P(b)$ is divisible by $a - b$. | 10.14. Let $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$. Then $P(a)-$ $-P(b)=a_{n}\left(a^{n}-b^{n}\right)+a_{n-1}\left(a^{n-1}+b^{n-1}\right)+\ldots+a_{1}(a-b)$. Therefore, it is sufficient to check that for any natural number $k$ the number $a^{k}-b^{k}$ is divisible by $a-b$. For this, one can use the ide... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,397 |
10.15. Does there exist a polynomial $P(x)$ with integer coefficients for which $P(7)=11$ and $P(11)=13$? | 10.15. A n s w e r: no, it does not exist. According to problem 10.14, the difference $P(11)-P(7)=2$ should be divisible by $11-7=4$. | no,itdoesnotexist | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,398 |
10.16. Let $P(x)$ be a polynomial with integer coefficients, and for some integer $n$ the numbers $P(n)$, $P(n+1)$, and $P(n+2)$ are divisible by 3. Prove that then $P(m)$ is divisible by 3 for any integer $m$. | 10.16. If $a$ and $b$ are integers, and $a > b$, then the number $P(a) - P(b)$ is divisible by $a - b$ (Problem 10.14). For any integer $m$, one of the numbers $m - n, m - (n + 1), m - (n + 2)$ is divisible by 3. Therefore, one of the numbers $P(m) - P(n), P(m) - P(n + 1), P(m) - P(n + 2)$ is divisible by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,399 |
10.17. Prove that any polynomial with integer coefficients, different from a constant, takes a value that is a composite number for some natural value of the argument. | 10.17. Let $P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}$, where the numbers $a_{0}, a_{1}, \ldots, \ldots, a_{n}$ are integers. Suppose that $P(k)=p$ is a prime number (here $k$ is some natural number). The polynomial $P(x)$ cannot take the same value at more than $n$ different points, so one can choose a natural numbe... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,400 |
10.18. Provide an example of a polynomial $P(x)$ that is divisible by $x^{2}+1$, and at the same time $P(x)-1$ is divisible by $x^{3}+1$. | 10.18. Answer: $P(x)=-\frac{1}{2}\left(x^{2}+1\right)\left(x^{2}+x-1\right)$.
Let $P(x)=U(x)\left(x^{2}+1\right)$ and $P(x)=V(x)\left(x^{3}+1\right)+1$. Then $U(x)\left(x^{2}+1\right)-V(x)\left(x^{3}+1\right)=1$. To find such polynomials $U$ and $V$, we need to apply the Euclidean algorithm to $a=x^{3}+1$ and $b=x^{2}... | P(x)=-\frac{1}{2}(x^{2}+1)(x^{2}+x-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,401 |
10.19. Let \( a_{0}=0, a_{n}=P\left(a_{n-1}\right) \) for \( n=1,2, \ldots \), where \( P(x) \) is a polynomial with integer coefficients, and \( P(x)>0 \) for \( x \geqslant 0 \). Prove that if \( m, n>0 \), then \(\gcd \left(a_{m}, a_{n}\right)=a_{d}\), where \( d=\gcd (m, n) \). | 10.19. Let $P_{1}(x)=P(x), P_{n}(x)=P\left(P_{n-1}(x)\right)$ for $n \geqslant 2$. Then $a_{n}=P_{n}\left(a_{0}\right)$ and generally $a_{n}=P_{n-k}\left(a_{k}\right)$ for $0 \leqslant k \leqslant n \geqslant 1$, so GCD ( $a_{m}, a_{n}$ ) $=$ $=\operatorname{GCD}\left(P_{m-n}\left(a_{n}\right), a_{n}\right)=\operatorna... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,402 |
10.20. Prove that the polynomial $x^{15}-1$ has divisors of all degrees from 1 to 14, i.e., for any natural number $k \leqslant 14$ there exists a polynomial of degree $k$ with integer coefficients that divides $x^{15}-1$. | 10.20. The polynomial $x^{15}-1$ can be factored into factors of degrees $1, 2, 4$, and $8$. Indeed, $x^{15}-1=\left(x^{5}-1\right)\left(x^{10}+\right.$ $+x^{5}+1\right), x^{5}-1=(x-1)\left(x^{4}+x^{3}+x^{2}+x+1\right)$, and the polynomial $x^{10}+x^{5}+1$ can be factored into factors of degrees 2 and 8 (Problem 5.6). ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,403 |
10.21. Prove that the polynomial $x^{2 n}+x^{n}+1$ is divisible by $x^{2}+x+1$ if and only if $n$ is not divisible by 3. | 10.21. For $n=0$ and 1, the statement is obvious. For $n=2$, we can divide the polynomial $x^{4}+x^{2}+1$ by $x^{2}+x+1$; the result is $x^{2}-x+1$. The difference of the polynomials $x^{2(n+3)}+x^{n+3}+1$ and $x^{2 n}+x^{n}+1$ is divisible by $x^{2}+x+1$, since the polynomials $x^{2 n+6}-x^{2 n}=x^{2 n}\left(x^{6}-1\r... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,404 |
10.22. a) It is known that $a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are given integers, is divisible by 5 for any integer $x$. Prove that all numbers $a, b, c, d$ are divisible by 5.
b) It is known that $a x^{4}+b x^{3}+c x^{2}+d x+e$, where $a, b, c, d, e$ are given integers, is divisible by 7 for any integer $x$.... | 10.22. a) Substituting $x=0$, we get that $d$ is divisible by 5. Considering this and substituting $x= \pm 1$, we get that $a+b+c$ and $-a+b-c$ are divisible by 5. Therefore, $2 b$ and $2 a+2 c$ are divisible by 5, which means $b$ and $a+c$ are divisible by 5. Substituting $x=2$, we get that $a+5 a+2(a+c)+4 b+d$ is div... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,405 |
10.23. Prove that if $p / q$ is an irreducible rational fraction that is a root of the polynomial
$$
f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}
$$
with integer coefficients, then $p-k q$ is a divisor of the number $f(k)$ for any integer $k$. | 10.23. The polynomial $f(x)$ is divisible by $x-\frac{p}{q}$, so $f(x)=g(x) \times \left(x-\frac{p}{q}\right)$. Let $g(x)=b_{0} x^{n-1}+b_{2} x^{n-2}+\ldots+b_{n-1}$. Then $a_{0}=b_{0}$, $a_{1}=b_{1}-b_{0} \frac{p}{q}, a_{2}=b_{2}-b_{1} \frac{p}{q}, \ldots, a_{n-1}=b_{n-1}-b_{n-2} \frac{p}{q}, a_{n}=-b_{n-1} \frac{p}{q... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,406 |
10.24. Prove that for no integer $A$ the polynomial $3 x^{2 n}+A x^{n}+2$ is divisible by the polynomial $2 x^{2 m}+A x^{m}+3$.
## 10.6. Inequalities for Roots | 10.24. Suppose the polynomial $3 x^{2 n}+A x^{n}+2$ is divisible by the polynomial $2 x^{2 m}+A x^{m}+3$. Then any root of the polynomial $2 x^{2 m}+A x^{m}+3$ is also a root of the polynomial $3 x^{2 n}+A x^{n}+2$. If $x_{i}$ is a root of the polynomial $2 x^{2 m}+A x^{m}+3$, then $x_{i}^{m}=\frac{-A \pm \sqrt{A^{2}-2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,407 |
10.25. Prove that the positive root of the equation $x(x+1) \ldots(x+n)=1$ is less than $1 / n!$.
## 10.7. The number of real roots of a polynomial | 10.25. If $x>0$ and $x(x+1) \ldots(x+n)=1$, then
$$
x=\frac{1}{(x+1) \ldots(x+n)}<\frac{1}{1 \cdot 2 \cdot \ldots \cdot n}=\frac{1}{n!}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,408 |
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