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374. Can the numbers $10,11,12$ be members of the same geometric progression | Assumption. Suppose that the given numbers can be members of the same geometric progression, then
$$
11=10 \cdot q^{p}, 12=10 q^{n}
$$
where $p$ and $n$ are positive integers, or
$$
\left(\frac{11}{10}\right)^{n}=\left(\frac{12}{10}\right)^{p}
$$
But the last equality cannot hold, since
$$
10^{p-n} \cdot 11^{n} \n... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,539 |
377. Prove that if in a geometric progression each term is subtracted from the one following it, then the successive differences also form a geometric progression. | The general term of the obtained sequence will be $a_{1} q^{n-1}(q-1)$. By setting $n=1,2, \ldots, k$, we find the terms of this sequence:
$$
a_{1}(q-1), a_{1} q(q-1), a_{1} q^{2}(q-1), \ldots, a_{1} q^{k-1}(q-1)
$$
It forms a geometric progression with the first term $a_{1}(q-1)$ and the common ratio $q$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,540 |
378. Find the condition under which the product of two arbitrary terms of a geometric progression will be a term of the same progression. | Let there be three terms of a geometric progression: $a q^{n}, a q^{p}, a q^{k}$. According to the problem, it should be:
$$
a q^{n} a q^{p}=a q^{k}
$$
from which
$$
a=q^{k-n-p}
$$
Since the numbers $n, p, k$ are integers, the required condition is that the first term of the progression should be some integer power... | q^{k-n-p} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,541 |
379. Prove that if $A, B, C$ are respectively the $n$-th, $p$-th, and $k$-th terms of the same geometric progression, then $A^{p-k} \cdot B^{k-n} \cdot C^{n-p}=1$. | Given. For problem 373, under the conditions of this problem, we have:
$$
\left(\frac{A}{B}\right)^{n-k}=\left(\frac{A}{C}\right)^{n-p}
$$
from which
$$
A^{n-k} \cdot C^{n-p}=B^{n-k} \cdot A^{n-p}
$$
Dividing both sides of the equation by the expression on the right side, we get:
$$
A^{p-k} \cdot B^{k-n} \cdot C^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,542 |
381. The denominator of the geometric progression is $\frac{1+\sqrt{5}}{2}$. Prove that each term of it, starting from the second, is equal to the difference of two adjacent terms. | Instruction. If the denominator of the progression is $\frac{1+\sqrt{5}}{2}$, then its terms are:
$$
\begin{gathered}
a_{1}, a_{2}=a_{1}\left(\frac{1+\sqrt{5}}{2}\right), \ldots, a_{n-2}=a_{1}\left(\frac{1+\sqrt{5}}{2}\right)^{n-3} \\
a_{n-1}=a_{1}\left(\frac{1+\sqrt{5}}{2}\right)^{n-2} ; \quad a_{n}=a_{1}\left(\frac{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,544 |
382. Prove that if $a, b, c, d$ form a geometric progression, then
$$
\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}
$$ | Given the problem, we have:
$$
\begin{gathered}
\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)= \\
=\left(a^{2}+a^{2} q^{2}+a^{2} q^{4}\right)\left(a^{2} q^{2}+a^{2} q^{4}+a^{2} q^{6}\right)= \\
=a^{2}\left(1+q^{2}+q^{4}\right) a^{2} q^{2}\left(1+q^{2}+q^{4}\right)=a^{4} q^{2}\left(1+q^{2}+q^{4}\right)^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,545 |
383. Prove that if $a, b, c, d$ form a geometric progression, then
$$
(a-c)^{2}+(b-c)^{2}+(b-d)^{2}=(a-d)^{2}
$$ | Given the problem, $c^{2}=b d, b^{2}=a c, a d=b c$. Then
$$
\begin{gathered}
\quad(a-c)^{2}+(b-c)^{2}+(b-d)^{2}= \\
=a^{2}-2 a c+c^{2}+b^{2}-2 b c+c^{2}+b^{2}-2 b d+d^{2}= \\
=a^{2}-2 b c+d^{2}+2\left(c^{2}+b^{2}-a c-b d\right)=(a-d)^{2}
\end{gathered}
$$
since
$$
2 b c=2 a d, \quad c^{2}+b^{2}-a c-b d=0
$$
(by the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,546 |
384. Prove that if three numbers $a, b, c$ form a geometric progression, then
$$
(a+b+c)(a-b+c)=a^{2}+b^{2}+c^{2}
$$ | Instruction. Transform the left side of the equality to be proved:
$$
\begin{aligned}
(a+b+c)(a-b+c) & =(a+c)^{2}-b^{2}=a^{2}+2 a c+c^{2}-b^{2}= \\
& =a^{2}+b^{2}+c^{2}
\end{aligned}
$$
since
$$
2 a c=2 b^{2}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,547 |
385. Prove that if $a, b, c$ represent respectively the $n$-th, $2n$-th, and $4n$-th terms of a geometric progression, starting from the first, then
$$
b\left(b^{2}-a^{2}\right)=a^{2}(c-b)
$$ | Given the conditions of the problem,
$$
\frac{b}{a}=q^{n} ; \frac{c}{b}=q^{2 n} ;\left(\frac{b}{a}\right)^{2}=\frac{c}{b}
$$
Thus,
$$
\frac{b^{2}}{a^{2}}=\frac{c}{b} \text { or } \frac{b^{2}-a^{2}}{a^{2}}=\frac{c-b}{b},
$$
from which
$$
b\left(b^{2}-a^{2}\right)=a^{2}(c-b)
$$
## § 24. Sum of $n$ terms of a geomet... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,548 |
386. Derive the formula for the sum of $n$ terms of a geometric progression using its definition (the common ratio of the progression is not equal to 1). | Let
$$
S_{n}=a_{1}+a_{1} q+\ldots+a_{1} q^{n-1}=a_{1}\left(1+q+q^{2}+\ldots+q^{n-1}\right)
$$
The polynomial in parentheses represents the quotient of the division of the polynomial $q^{n}-1$ by $q-1$, i.e.,
$$
S_{n}=\frac{a_{1} q^{n}-a_{1}}{q-1}
$$ | S_{n}=\frac{a_{1}q^{n}-a_{1}}{q-1} | Algebra | proof | Yes | Yes | olympiads | false | 42,549 |
388. The sum of $2 n$ terms of a geometric progression, where the first term is $a$ and the common ratio is $q$, is equal to the sum of $n$ terms of a geometric progression, where the first term is $b$ and the common ratio is $q^{2}$. Prove that $b$ equals the sum of the first two terms of the first progression. | If $S_{n}=S_{2 n}$, then
$$
\frac{b\left(q^{2 n}-1\right)}{q^{2}-1}=\frac{a\left(q^{2 n}-1\right)}{q-1}
$$
from which
$$
b=a(1+q)=a_{1}+a_{2}
$$ | b=(1+q)=a_{1}+a_{2} | Algebra | proof | Yes | Yes | olympiads | false | 42,551 |
391. Let $a_{n}$ and $S_{n}$ be the $n$-th term and the sum of the first $n$ terms of the geometric progression $a, a q, a q^{2}, \ldots, a q^{n-1}$;
$$
S_{n}^{\prime}=a+a q^{-1}+a q^{-2}+\ldots+a q^{-(n-1)}
$$
Prove that $a S_{n}=a_{n} S_{n}^{\prime}$. | According to the problem statement,
$$
\begin{gathered}
S_{n}^{\prime}=\frac{a\left(q^{-n}-1\right)}{q^{-1}-1}=\frac{a\left(1-q^{n}\right)}{q^{n-1}(1-q)}= \\
=\frac{a\left(q^{n}-1\right)}{q-1} \cdot \frac{1}{q^{n-1}}=S_{n} \cdot \frac{a}{a q^{n-1}}=S_{n} \cdot \frac{a}{a_{n}}
\end{gathered}
$$
from which
$$
a S_{n}=... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,554 |
392. Let $S_{n}$ and $P$ be the sum and product of $n$ consecutive terms of a given geometric progression, $S_{n}^{\prime}-$ the sum of the reciprocals of the terms of the given progression. Prove that
$$
P^{2}\left(S_{n}^{\prime}\right)^{n}=\left(S_{n}\right)^{n}
$$ | Instruction. By the condition of the problem,
$$
S_{n}^{\prime}=\frac{q^{n}-1}{(q-1) a q^{n-1}}, \quad P=a^{n} \cdot q^{\frac{n(n-1)}{2}}
$$
(see problem 369), from which
$$
P^{2}\left(S_{n}^{\prime}\right)^{n}=\frac{a^{2 n} q^{n(n-1)}\left(q^{n}-1\right)^{n}}{(q-1)^{n} a^{n} q^{n^{2}-n}}=\frac{a^{n}\left(q^{n}-1\ri... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,555 |
393. Prove that the sum of $n$ terms of a geometric progression, in which the $p$-th term is equal to $(-1)^{p} a^{4 p}$, for any value of $p$ is
$$
\frac{a^{4}}{a^{4}+1} \cdot\left[(-1)^{n} \cdot a^{4 n}-1\right]
$$ | Instruction. Using the condition of the problem, determine the first term of the progression $\left(-a^{4}\right)$, the common ratio of the progression $\left(-a^{4}\right)$, and the $n$-th term of the progression $(-1)^{n} a^{4 n}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,556 |
394. Prove that if $S_{n}, S_{2 n}, S_{3_{n}}$ are the sums of the first $n, 2 n$, and $3 n$ terms of the same geometric progression, then
$$
S_{n}\left(S_{3 n}-S_{2 n}\right)=\left(S_{2_{n}}-S_{n}\right)^{2}
$$ | Instruction. Find the expression for the left-hand side of the equation to be proved and transform it, keeping in mind that
\[
\begin{gathered}
q^{4 n}-2 q^{3 n}+q^{2 n}=\left(q^{2 n}-1\right)^{2}+\left(q^{n}-1\right)^{2}-2\left(q^{2 n}-1\right)\left(q^{n}-1\right)= \\
=\left(q^{2 n}-q^{n}\right)^{2}
\end{gathered}
\] | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,557 |
395. Prove that if $S$ is the sum of the terms of a decreasing infinite geometric progression: $1, a^{p}, a^{2 p}, \ldots ; S_{1}$ is the sum of the terms of a decreasing infinite geometric progression $1, a^{q}, a^{2 q}, \ldots$, then
$$
S^{q}\left(S_{1}-1\right)^{p}=\left(S_{1}\right)^{p}(S-1)^{q}
$$ | Instruction. According to the condition of the problem,
$$
S=\frac{1}{1-a^{p}} ; \quad S_{1}=\frac{1}{1-a^{q}},
$$
from which
$$
\begin{aligned}
a^{p} & =\frac{S-1}{S} \\
a^{q} & =\frac{S_{1}-1}{S_{1}}
\end{aligned}
$$
$8^{*}$
Raising the first equality to the power $q$, the second - to the power $p$, we get
$$
\... | S^{q}(S_{1}-1)^{p}=S_{1}^{p}(S-1)^{q} | Algebra | proof | Yes | Yes | olympiads | false | 42,558 |
396. Prove that if $x y, y^{2}, z^{2}$ are consecutive terms of an arithmetic progression, then $y, z$ and $(2 y-x)$ are consecutive terms of a geometric progression. | According to the problem,
$$
y^{2}-x y=z^{2}-y^{2}
$$
from which
$$
z=\sqrt{2 y^{2}-x y}=\sqrt{(2 y-x) \cdot y}
$$
i.e., $z$ is the geometric mean between $y$ and $(2 y-x)$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,559 |
397. Find the conditions under which the squares of three consecutive terms of an arithmetic progression are three consecutive terms of a geometric progression. | Let $a, a+d, a+2 d$ be three consecutive terms of an arithmetic progression. If the squares of these numbers form a geometric progression, then
$$
(a+d)^{2}=\sqrt{a^{2}(a+2 d)^{2}}
$$
or
$$
(a+d)^{2}= \pm a(a+2 d)
$$
If
$$
(a+d)^{2}=a(a+2 d)
$$
then $d=0$
If
$$
(a+d)^{2}=-a(a+2 d)
$$
then
$$
d=a(-2 \pm \sqrt{... | =0or=(-2\\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,560 |
398. Prove that if the $k$-th, $n$-th, and $p$-th terms of an arithmetic progression form three consecutive terms of a geometric progression, then its common ratio is $\frac{n-p}{k-n}$.
116 | Let $a_{k}, a_{n}, a_{p}$ be the $k$-th, $n$-th, $p$-th terms of an arithmetic progression. If these same numbers are also consecutive terms of a geometric progression, then
$$
a_{n}=a_{k} q, \quad a_{p}=a_{n} q \quad \text { or } \quad a_{n}-a_{p}=q\left(a_{k}-a_{n}\right)
$$
hence
$$
q=\frac{a_{n}-a_{p}}{a_{k}-a_{... | \frac{n-p}{k-n} | Algebra | proof | Yes | Yes | olympiads | false | 42,561 |
399*. Can the numbers $2 ; \sqrt{6} ; 4.5$ be members of the same arithmetic or geometric progression? | If the given numbers were members of the same arithmetic progression, the following equality should hold:
$$
\frac{\sqrt{6}-2}{4.5-\sqrt{6}}=\frac{p}{k}
$$
where $p$ and $k$ are integers (see problem 336), or
$$
\begin{aligned}
& \sqrt{6}-2=\frac{p}{k}(4.5-\sqrt{6}) \\
& 6-4=\frac{p}{k}(2.5 \sqrt{6}+3)
\end{aligned}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,562 |
400**. Can the digits of a prime number not exceeding $10^{4}$ form an arithmetic or geometric progression? If they can, find these numbers. | Solution. 1) Suppose the digits of the number form an arithmetic progression.
a) For a three-digit number, we will have the digits: $x, x+d$, $x+2d$, where $d$ is the difference of the progression. Their sum $x+(x+d)+(x+2d)$ will be divisible by 3, i.e., the number formed by these digits is divisible by 3 and cannot b... | 4567,139,421 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,563 |
401. Prove that if three numbers, not equal to zero, form an arithmetic progression, then the numbers that are their reciprocals form a continuous harmonic proportion (harmonic progression). | Proof. Let $a, b, \quad c$ be consecutive terms of an arithmetic progression, i.e.,
$$
b-a=c-b
$$
where $a \neq 0, b \neq 0, c \neq 0$. Let also
$$
\frac{1}{a}=a_{1}, \quad \frac{1}{b}=b_{1}, \quad \frac{1}{c}=c_{1}
$$
or
$$
\frac{1}{a_{1}}=a, \frac{1}{b_{1}}=b, \frac{1}{c_{1}}=c
$$
Then
$$
\frac{1}{b_{1}}-\frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,564 |
402. If three numbers form a continuous harmonic proportion, then the numbers that are their reciprocals form an arithmetic progression. Prove. | Let $a, b, c$ be numbers forming a continuous harmonic proportion, i.e., $\frac{a-b}{b-c}=\frac{a}{c}$. Let $a_{1}, b_{1}$, $c_{1}$ be the numbers that are their reciprocals, i.e.,
$$
a=\frac{1}{a_{1}}, \quad b=\frac{1}{b_{1}}, \quad c=\frac{1}{c_{1}}
$$[^10]
Then
$$
\frac{\frac{1}{a_{1}}-\frac{1}{b_{1}}}{\frac{1}{b... | b_{1}-a_{1}=c_{1}-b_{1} | Algebra | proof | Yes | Yes | olympiads | false | 42,565 |
405*. If $x, y, z$ form a harmonic progression, then prove that
$$
\lg (x+z)+\lg (x-2 y+z)=2 \lg (x-z)
$$ | Given that if $x, y, z$ form a harmonic progression, then $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ should form an arithmetic progression, i.e.,
$$
\frac{2}{y}=\frac{1}{x}+\frac{1}{z}
$$
or
$$
2 x z=y(x+z)
$$
Consider the expression:
$$
\begin{gathered}
\lg (x+z)+\lg (x-2 y+z)=\lg (x+z)(x-2 y+z)= \\
=\lg \left[(x+z)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,566 |
406**. Prove that if the numbers $a_{1}, a_{2}, \ldots a_{n}$ form a harmonic progression, then
$$
a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{n-1} a_{n}=(n-1) a_{1} a_{n}
$$ | If $a_{1}, a_{2}, \ldots a_{n}$ are numbers forming a harmonic progression, then
$$
\frac{1}{a_{2}}-\frac{1}{a_{1}}=\frac{1}{a_{3}}-\frac{1}{a_{2}}=\ldots=\frac{1}{a_{n}}-\frac{1}{a_{n-1}}=d
$$
or
$$
\frac{a_{1}-a_{2}}{a_{1} a_{2}}=\frac{a_{2}-a_{3}}{a_{2} a_{3}}=\ldots=\frac{a_{n-1}-a_{n}}{a_{n-1} a_{n}}=d
$$
From... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,567 |
407. Prove, using the equality $a^{\log _{a} N}=N$, that the logarithm of the product of positive factors is equal to the sum of the logarithms of the factors. | Instruction. The task is to prove the validity of the formula
$$
\log _{a} x y=\log _{a} x+\log _{a} y
$$
By the definition of the logarithm:
$$
x=a^{\log _{a} x}, \quad y=a^{\log _{a} y}
$$
from which
$$
x y=a^{\log _{a} x+\log _{a} y}
$$
On the other hand, by definition as well:
$$
x y=a^{\log _{a} x y}
$$
th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,568 |
409. Prove that if the numbers $a, b, c$ form a geometric progression, then their logarithms form an arithmetic progression. | Instruction. Given the condition $b^{2}=a c$. Taking the logarithm of this equality to the base $k$, where $k$ is a positive number not equal to one, we get:
$$
\log _{k} b=\frac{\log _{k} a+\log _{k} c}{2}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,569 |
410. Prove that the logarithms of consecutive terms of a geometric progression form an arithmetic progression with the difference of the progression equal to the logarithm of the common ratio of the given geometric progression. | Let the members of the given geometric progression be: $a, a q, a q^{2} \ldots a q^{n}$. We form a sequence of logarithms of consecutive members of the given geometric progression to some base $k$, where $k>0, k \neq 1$:
$$
\log _{k} a, \log _{k} a+\log _{k} q, \log _{k} a+2 \log _{k} q, \ldots, \log _{k} a+n \log _{k... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,570 |
411. Prove that $\log _{a^{k}} a^{n}=\frac{n}{k}$. | By the definition of logarithm: $\left(a^{k}\right)^{\log _{a^{k^{a^{n}}}}}=a^{n}$, from which
$$
k \log _{a^{k}} a^{n}=n \text { and } \log _{a^{k}} a^{n}=\frac{n}{k}
$$ | \log_{^{k}}^{n}=\frac{n}{k} | Algebra | proof | Yes | Yes | olympiads | false | 42,571 |
412. Prove that the logarithm of $\sqrt[n]{a^{m}}$ with base $a^{\frac{n}{m}}$ is equal to $\frac{m^{2}}{n^{2}}$. | Instruction. Transform the given expression into the form
$$
\sqrt[n]{a^{m}}=a^{\frac{m}{n}}
$$
and use the conclusion from the previous problem. | \frac{^{2}}{n^{2}} | Algebra | proof | Yes | Yes | olympiads | false | 42,572 |
413. Prove that if $a^{2}+b^{2}=7 a b$, then
$$
\log _{k} \frac{a+b}{3}=\frac{1}{2}\left(\log _{k} a+\log _{k} b\right)
$$
where
$$
a>0, b>0, k>0, k \neq 1
$$ | Instruction. Add $2 a b$ to both sides of the given equality, we get:
$$
a^{2}+b^{2}+2 a b=9 a b \text { or }(a+b)^{2}=9 a b
$$
from which
$$
\log _{k} \frac{a+b}{3}=\frac{1}{2}\left(\log _{k} a+\log _{k} b\right)
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,573 |
414. Prove that $\lg \lg \lg 10^{10^{n p}}=\lg n p$. | Instruction. $\lg \lg \lg 10^{10^{n p}}=\lg \lg \lg \left(10^{10^{n p}}\right)=\lg \lg \left(10^{n p}\right)=\lg n p$, since $\lg 10=1$.
## § 28. Different systems of logarithms | \lgnp | Algebra | proof | Yes | Yes | olympiads | false | 42,574 |
416. Prove that the logarithm of a number $N$ to a "new" base $b$ is equal to the logarithm of the same number to an "old" base $a$, divided by the logarithm of the new base to the old base. | By the definition of logarithm: $N=b^{\log _{b}{ }^{N}}$. Taking the logarithm of this equality to the base $a$, we get:
$$
\log _{a} N=\log _{b} N \cdot \log _{a} b
$$
or
$$
\log _{b} N=\frac{\log _{a} N}{\log _{a} b}
$$
The factor $\frac{1}{\log _{a} b}$, obtained in problem 416, is called the modulus of transiti... | \log_{b}N=\frac{\log_{}N}{\log_{}b} | Algebra | proof | Yes | Yes | olympiads | false | 42,575 |
417. Prove that the ratio $\frac{\log _{a} P}{\log _{a} K}$ does not depend on the base of the logarithms. | Instruction. For problem 416,
$$
\frac{\log _{a} P}{\log _{a} K}=\frac{\log _{b} P}{\log _{b} a}: \frac{\log _{b} K}{\log _{b} a}=\frac{\log _{b} P}{\log _{b} K}
$$
i.e., the ratio does not depend on the base of the logarithms $a$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,576 |
419. Prove that
$$
\frac{\log _{a} P}{\log _{a b} P}=1+\log _{a} b
$$
for
$$
\begin{gathered}
a>0, \quad b>0, a b \neq 1 \\
a \neq 1
\end{gathered}
$$ | Indication. By what has been proved earlier:
$$
\log _{a b} P=\frac{\log _{a} P}{\log _{a} a b}=\frac{\log _{a} P}{1+\log _{a} b}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,578 |
420. Prove that for $a>0, a \neq 1$
1) $\log _{a^{2}} P=\frac{\log _{a} P}{2}$
2) $\log _{\sqrt{a}} P=2 \log _{a} P$
3) $\log _{\frac{1}{a}} P=-\log _{a} P$. | Instruction. Use the dependence between logarithms of a number with different bases.
Convert the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Algebra | proof | Yes | Yes | olympiads | false | 42,579 | |
421. Prove that
$$
\log _{a^{k}} P=\frac{\log _{a} P}{k}
$$
for
$$
a>0, a \neq 1
$$ | Instruction. The validity of the given equality can be proved in two ways:
1) $\log _{a^{k}} P=\frac{\log _{a} P}{\log _{a} a^{k}}=\frac{\log _{a} P}{k}$.
2) Bearing in mind that, by definition,
$$
P=a^{\log _{a} P} \text { and } P=\left(a^{k}\right)^{\log _{a^{k}} P},
$$
from which
$$
\log _{a} P=k \log _{a^{k}} P... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,580 |
422. Prove that
1) $\log _{b n} a n=\frac{\log _{b} a+\log _{b} n}{1+\log _{b} n}$;
2) $\log _{b a^{n}} a^{n+1}=\frac{(n+1) \log _{b} a}{1+n \log _{b} a}$;
3) $\log _{b^{n+1}} a b^{n}=\frac{\log _{b} a+n}{1+n}$;
4) $\log _{b^{n}} a^{n}=\log _{b} a$. | Instruction. Use the dependence between the logarithms of a number with respect to the old and new bases, taking the "old" base as $b$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | Algebra | proof | Yes | Yes | olympiads | false | 42,581 | |
423. Prove that if $a, b, c$ are terms of a geometric progression, then
$$
\frac{\log _{a} P-\log _{b} P}{\log _{b} P-\log _{c} P}=\frac{\log _{a} P}{\log _{c} P}
$$
where $P>0, a>0, b>0, c>0, a \neq 1, b \neq 1, c \neq 1$. | Given the condition of the problem, $b=\sqrt{a c}$. According to problem 415,
$$
\log _{b} P=\frac{1}{\log _{P} b}=\frac{1}{\log _{P} \sqrt{a c}}=\frac{2}{\log _{P} a+\log _{P} c}=\frac{2 \log _{a} P \cdot \log _{c} P}{\log _{a} P+\log _{c} P}
$$
By replacing $\log _{b} P$ in the left part of the equation to be prove... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,582 |
424. Calculate $\log _{a b c \ldots k} x$, given
$$
\log _{a} x, \log _{b} x, \ldots, \log _{k} x, x \neq 1
$$ | Instruction. For problem 415,
\[
\begin{aligned}
\log _{a b c \ldots k} x= & \frac{1}{\log _{x}(a b c \ldots k)}=\frac{1}{\log _{x} a+\log _{x} b+\ldots+\log _{x} k}= \\
& =\frac{1}{\frac{1}{\log _{a} x}+\frac{1}{\log _{b} x}+\ldots+\frac{1}{\log _{k} x}}
\end{aligned}
\] | \frac{1}{\frac{1}{\log_{}x}+\frac{1}{\log_{b}x}+\ldots+\frac{1}{\log_{k}x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,583 |
425. Prove the identity:
\[
\begin{aligned}
& \log _{a} P \cdot \log _{b} P+\log _{b} P \cdot \log _{c} P+\log _{a} P \cdot \log _{c} P= \\
&= \frac{\log _{a} P \cdot \log _{b} P \cdot \log _{c} P}{\log _{a b c} P}
\end{aligned}
\] | Instruction. From problem 415 and the previous one, we have:
$$
\begin{gathered}
\log _{a b c} P=\frac{1}{\frac{1}{\log _{a} P}+\frac{1}{\log _{b} P}+\frac{1}{\log _{c} P}}= \\
=\frac{\log _{a} P \cdot \log _{b} P \cdot \log _{c} P}{\log _{b} P \cdot \log _{c} P+\log _{a} P \cdot \log _{c} P+\log _{a} P \cdot \log _{b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,584 |
426. Prove:
$$
a^{\frac{\log _{b}\left(\log _{b} a\right)}{\log _{b} a}}=\log _{b} a
$$
and determine the domain of valid values for the letters involved. 126 | Instruction. For the problem $415, \log _{a} b=\frac{1}{\log _{b} a}$, by the definition of logarithm $b^{\log _{b} a}=a$. We will transform the left part of the equality to be proven, keeping in mind these two remarks:
$$
\begin{gathered}
a^{\frac{\log _{b}\left(\log _{b} a\right)}{\log _{b} a}}=\left(a^{\left.\frac{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,585 |
427. Prove that if $a^{2}+b^{2}=c^{2}$, then
$$
\log _{c+b} a+\log _{c-b} a=2 \log _{c+b} a \cdot \log _{c-b} a
$$
given
$$
a>0, c>0, b>0, c>b
$$ | Given the condition of the problem, $a^{2}=(c+b)(c-b)$. By logarithmizing this equality first with base $c+b$, then with base $c-b$, we will have:
$$
\begin{aligned}
& 2 \log _{c+b} a=1+\log _{c+b}(c-b) \\
& 2 \log _{c-b} a=1+\log _{c-b}(c+b)
\end{aligned}
$$
Multiplying these equalities term by term, we get:
$$
\be... | \log_{+b}+\log_{-b}=2\log_{+b}\cdot\log_{-b} | Algebra | proof | Yes | Yes | olympiads | false | 42,586 |
428. Prove that if $(a c)^{\log _{a} b}=c^{2}$, then the numbers $\log _{a} P, \log _{b} P$, $\log _{c} P$ form an arithmetic progression, where
$$
a>0, b>0, c>0, P>0, a \neq 1, b \neq 1, c \neq 1
$$ | Instruction. Let's take the logarithm of the expression given in the problem to the base $c$:
$$
\log _{a} b\left(\log _{c} a+\log _{c} c\right)=2 \log _{c} c
$$
or
$$
\log _{a} b \cdot \log _{c} a+\log _{a} b=2
$$
We have
$$
\log _{a} b \cdot \log _{c} a \cdot \log _{b} P+\log _{a} b \cdot \log _{b} P=2 \log _{b}... | \log_{}P+\log_{}P=2\log_{b}P | Algebra | proof | Yes | Yes | olympiads | false | 42,587 |
429*. Prove that from the equality
$$
\frac{x(y+z-x)}{\log _{a} x}=\frac{y(x+z-y)}{\log _{a} y}=\frac{z(x+y-z)}{\log _{a} z}
$$
it follows that
$$
x^{y} \cdot y^{x}=z^{x} \cdot x^{z}=y^{z} \cdot z^{y}
$$ | Let each of the relations given in the condition of the problem be equal to $\frac{1}{n}$, then:
$$
\begin{gathered}
\log _{a} x=n x(y+z-x), \log _{a} y=n y(x+z-y) \\
\log _{a} z=n z(x+y-z)
\end{gathered}
$$
Transform these equalities by multiplying both sides by the same number: the first by $y$ and $z$, the second ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,588 |
430. Prove that the decimal logarithms of positive integers, not equal to an integer power of 10, are irrational numbers. | Let the number $N$ not be an integer power of 10, but suppose that it is a rational power of 10, i.e., $\lg N=\frac{n}{p}$, where $\frac{n}{p}$ is an irreducible rational fraction. Then $10^{\frac{n}{p}}=N$ or $10^{n}=N^{p}$, which contradicts the condition that $N$ is not an integer power of 10. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,589 |
431*. Prove that if $N$ is the number of integers whose logarithms have the characteristic $n$, and $M$ is the number of integers whose reciprocals have the characteristic $(-m)$, then
$$
\lg N - \lg M = n - m + 1
$$ | Instruction. All integers having the characteristic $n$, must lie between the numbers $10^{n}$ and $10^{n+1}$, their number will be determined as
$$
N=10^{n+1}-10^{n}=9 \cdot 10^{n}
$$
The characteristic (-m) is possessed by all fractions lying between the fractions: $\frac{1}{10^{m}}$ and $\frac{1}{10^{m-1}}$, and t... | \lgN-\lgM=n-+1 | Number Theory | proof | Yes | Yes | olympiads | false | 42,590 |
433*. Prove, without using logarithmic tables, that
1) $\frac{1}{\log _{2} \pi}+\frac{1}{\log _{5} \pi}>2$
2) $\frac{1}{\log _{2} \pi}+\frac{1}{\log _{\pi} 2}>2$ | Indication. 1) Let $\log _{2} \pi=a, \log _{5} \pi=b$. From the equalities $2^{a}=\pi, 5^{b}=\pi$ we obtain:
$$
\pi^{\frac{1}{a}}=2, \quad \pi^{\frac{1}{b}}=5, \quad \pi^{\frac{1}{a}} \cdot \pi^{\frac{1}{b}}=2 \cdot 5=10, \pi^{\frac{1}{a}+\frac{1}{b}}=10
$$
But $\pi^{2} \approx 3.14^{2}2$.
2) The second inequality i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,592 |
434. Prove that the sum of the first $n$ numbers of the natural series is equal to $\frac{n(n+1)}{2}$. | Proof. 1) For $n=1, S_{1}=\frac{1 \cdot 2}{2}=1-$ the formula is correct.
2) Assume that the formula is valid for $n=k$, i.e., $S_{k}=\frac{k(k+1)}{2}$, and we will prove that it is also valid for $n=k+1$, i.e.,
$$
S_{k+1}=\frac{(k+1)(k+2)}{2}
$$
3) Indeed, $S_{k+1}=S_{k}+(k+1)$, but $S_{k}=\frac{k(k+1)}{2}$ by assu... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,593 |
435. Prove that the sum of the first $n$ odd numbers is equal to the square of the number of them. | Proof. 1) For $n=1$ we have $S_{1}=1, n^{2}=1$, so the formula is correct.
2) Suppose the formula is true for $n=k$, i.e., $S_{k}=k^{2}$, and we will prove that it is also true for $n=k+1$, i.e., $S_{k+1}=(k+1)^{2}$.
$9^{*}$
3) Indeed, $S_{k+1}=S_{k}+(2 k+1)$, but $S_{k}=k^{2}$ by assumption, so
$$
\mathrm{S}_{k+1}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,594 |
436. Prove that the sum of the squares of the first $n$ natural numbers is equal to $\frac{n(n+1)(2 n+1)}{6}$. | $$
\begin{gathered}
S_{k+1}=\mathrm{S}_{k}+(k+1)^{2}=\frac{k(k+1)(2 k+1)}{6}+(k+1)^{2}= \\
=\frac{(k+1)(2 k+3)(k+2)}{6}=\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}
\end{gathered}
$$
Indication.
$$
\begin{gathered}
S_{k+1}=\mathrm{S}_{k}+(k+1)^{2}=\frac{k(k+1)(2 k+1)}{6}+(k+1)^{2}= \\
=\frac{(k+1)(2 k+3)(k+2)}{6}=\frac{(k+1)[(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,595 |
437. Prove that
$$
1^{2}+3^{2}+\ldots+(2 n-1)^{2}=\frac{n(2 n-1)(2 n+1)}{3}
$$
where $n$ is any natural number. | $$
\begin{gathered}
S_{k+1}=S_{k}+(2 k+1)^{2}=\frac{k(2 k-1)(2 k+1)}{3}+(2 k+1)^{2}= \\
=\frac{(2 k+1)\left(2 k^{2}+5 k+3\right)}{3}=\frac{(2 k+1)(k+1)(2 k+3)}{3}= \\
=\frac{(k+1)[2(k+1)-1][2(k+1)+1]}{3}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,596 |
438. Prove that
$$
1-2^{2}+3^{2}-4^{2}+\ldots+(-1)^{n-1} n^{2}=(-1)^{n-1} \frac{n(n+1)}{2}
$$
where $n$ is any natural number. | $$
\begin{gathered}
S_{k+1}=S_{k}+(-1)^{k}(k+1)^{2}=(-1)^{k-1} \frac{k(k+1)}{2}+ \\
+(-1)^{k}(k+1)^{2}=(-1)^{k} \cdot \frac{(k+1)(k+2)}{2}= \\
=(-1)^{k} \cdot \frac{(k+1)[(k+1)+1]}{2}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,597 |
439. Prove that
$$
1+2+2^{2}+\ldots+2^{n-1}=2^{n}-1
$$
where $n$ is any natural number.
132 | Indication. $S_{k+1}=S_{k}+2^{k}=2^{k}-1+2^{k}=2^{k+1}-1$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,598 |
440. Prove the identity:
$$
S_{n}=1+2^{3}+3^{3}+\ldots+n^{3}=\frac{n^{2}(n+1)^{2}}{4}
$$
for any natural $n$. | $$
\begin{aligned}
& S_{k+1}=S_{k}+(k+1)^{3}=\frac{k^{2}(k+1)^{2}}{4}+(k+1)^{3}= \\
& \quad=\frac{(k+1)^{2}(k+2)^{2}}{4}=\frac{[k+1]^{2}[(k+1)+1]^{2}}{4}
\end{aligned}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,599 |
441. Prove that for any natural numbers $n$ and $p$, the following equality holds:
\[
\begin{gathered}
S_{n}=1 \cdot 2 \ldots p+2 \cdot 3 \ldots p(p+1)+\ldots+n(n+1) \ldots \\
\ldots(n+p-1)=\frac{n(n+1)(n+2) \ldots(n+p)}{p+1}
\end{gathered}
\] | Indication.
$$
\begin{aligned}
& S_{k+1}=S_{k}+(k+1)(k+2) \ldots(k+p)=\frac{k(k+1) \ldots(k+p)}{p+1}+ \\
& +(k+1)(k+2) \ldots(k+p)=\frac{(k+1)(k+2) \ldots(k+p+1)}{p+1}
\end{aligned}
$$
By setting $p=1, p=2, p=3$ in this equality, we get:
1) $1+2+3+\ldots+n=\frac{n(n+1)}{2}$.
2) $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,600 |
442. Prove the identity
$$
P_{n}=(n+1)(n+2) \ldots(n+n)=2^{n} \cdot 1 \cdot 3 \ldots(2 n-1)
$$
where $n$ is any natural number. | $$
\begin{gathered}
P_{k+1}=\frac{P_{k}(2 k+1)(2 k+2)}{k+1}=\frac{2^{k} \cdot 1 \cdot 3 \cdot 5 \ldots(2 k-1)(2 k+1)(2 k+2)}{k+1}= \\
=2^{k+1} \cdot 1 \cdot 3 \cdot 5 \ldots(2 k+1)
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,601 |
443. Prove that $S_{n}=1 \cdot 1!+2 \cdot 2!+\ldots+n \cdot n!=(n+1)!-1$, where $n-$ is any natural number. | ## Instruction.
$$
\begin{gathered}
S_{k+1}=S_{k}+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!= \\
=(k+2)!-1=[(k+1)+1]!-1
\end{gathered}
$$
Note. The "instructions" we provide cannot be considered complete proofs; they are only auxiliary considerations for conducting the induction step in a proof by mathematical induction. Stude... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,602 |
444*. Prove that the sum of the members of each horizontal row of the given table is equal to the square of the number of numbers in it:
$$
\begin{aligned}
& 1 \\
& 2,3,4 \\
& 3,4,5,6,7 \\
& 4,5,6,7,8,9,10
\end{aligned}
$$ | Instruction. General form of the rows of the table: $n, n+1, n+2, \ldots$, $3 n-2$; a total of $2 n-1$ numbers.
$$
\begin{gathered}
S_{k+1}=S_{k}-k+(3 k-1)+3 k+(3 k+1)=(2 k-1)^{2}-k+ \\
+(3 k-1)+3 k+(3 k+1)=(2 k-1)^{2}+8 k=(2 k+1)^{2}= \\
=[2(k+1)-1]^{2}
\end{gathered}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,603 |
445. Prove that
$$
\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}
$$ | $$
\begin{gathered}
S_{k+1}=\mathrm{S}_{k}+\frac{1}{(2 k+1)(2 k+3)}=\frac{k}{2 k+1}+\frac{1}{(2 k+1)(2 k+3)}= \\
=\frac{(k+1)(2 k+1)}{(2 k+1)(2 k+3)}=\frac{k+1}{2(k+1)+1}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,604 |
446. Prove that
$$
\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}
$$ | $$
\begin{gathered}
S_{k+1}=S_{k}+\frac{1}{(3 k+1)(3 k+4)}=\frac{k}{3 k+1}+\frac{1}{(3 k+1)(3 k+4)}= \\
=\frac{k+1}{3 k+4}=\frac{k+1}{3(k+1)+1}
\end{gathered}
$$
134 | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,605 |
447. Prove that
$$
\frac{1}{1 \cdot 5}+\frac{1}{5 \cdot 9}+\frac{1}{9 \cdot 13}+\ldots+\frac{1}{(4 n-3)(4 n+1)}=\frac{n}{4 n+1}
$$ | $$
\begin{gathered}
S_{k+1}=S_{k}+\frac{1}{(4 k+1)(4 k+5)}=\frac{k}{4 k+1}+\frac{1}{(4 k+1)(4 k+5)}= \\
=\frac{k+1}{4 k+5}=\frac{k+1}{4(k+1)+1}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,606 |
448. Prove that
$$
\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\ldots+\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{(n+1)(n+2)}\right]
$$ | $$
\begin{gathered}
S_{k+1}=S_{k}+\frac{1}{(k+1)(k+2)(k+3)}=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{(k+1)(k+2)}\right]+ \\
+\frac{1}{(k+1)(k+2)(k+3)}=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{(k+2)(k+3)}\right]= \\
=\frac{1}{2}\left\{\frac{1}{2}-\frac{1}{[(k+1)+1] \cdot[(k+1)+2]}\right\}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,607 |
449. Prove that
$$
\begin{gathered}
\frac{1}{1 \cdot 3 \cdot 5}+\frac{2}{3 \cdot 5 \cdot 7}+\ldots+\frac{n}{(2 n-1)(2 n+1)(2 n+3)}= \\
=\frac{n(n+1)}{2(2 n+1)(2 n+3)}
\end{gathered}
$$ | $$
\begin{gathered}
S_{k+1}=S_{k}+\frac{k+1}{(2 k+1)(2 k+3)(2 k+5)}=\frac{k(k+1)(2 k+5)+2(k+1)}{2(2 k+1)(2 k+3)(2 k+5)}= \\
=\frac{(k+1)(k+2)}{2(2 k+3)(2 k+5)}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,608 |
450. Prove the identity
$$
\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\ldots+\frac{1}{(a+n-1)(a+n)}=\frac{n}{a(a+n)}
$$ | $$
\begin{gathered}
S_{k+1}=S_{k}+\frac{1}{(a+k)(a+k+1)}=\frac{k}{a(a+k)}+\frac{1}{(a+k)(a+k+1)}= \\
=\frac{k+1}{a(a+k+1)}=\frac{k+1}{a[a+(k+1)]}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,609 |
451. Prove that
$$
x+2 x^{2}+3 x^{3}+\ldots+p x^{p}=\frac{x-(p+1) x^{p+1}+p x^{p+2}}{(1-x)^{2}} \text { for } x \neq 1
$$ | $$
\begin{aligned}
& S_{k+1}=S_{k}+(k+1) x^{k+1}=\frac{x-(k+1) x^{k+1}+k x^{k+2}}{(1-x)^{2}}+ \\
& \quad+(k+1) x^{k+1}=\frac{x-(k+2) x^{k+2}+(k+1) x^{k+3}}{(1-x)^{2}}
\end{aligned}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,610 |
452. Prove the identity:
$$
\frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\ldots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{1-x}+\frac{2^{n+1}}{1-x^{2^{n+1}}}
$$
for $x \neq 1$. | $$
\begin{gathered}
S_{k+1}=S_{k}+\frac{2^{k+1}}{1+x^{2^{k+1}}}=\frac{1}{1-x}+\frac{2^{k+1}}{1-x^{2^{k+1}}}+\frac{2^{k+1}}{1+x^{2^{k+1}}}= \\
=\frac{1}{1-x}+\frac{2^{k+2}}{1-x^{2^{k+2}}}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,611 |
453. Prove that
$$
\frac{x}{1-x^{2}}+\frac{x^{2}}{1-x^{4}}+\frac{x^{4}}{1-x^{8}}+\ldots+\frac{x^{2^{n-1}}}{1-x^{2^{n}}}=\frac{1}{1-x} \cdot \frac{x-x^{2 n}}{1-x^{2 n}}
$$
for $|x| \neq 1$. | Instruction.
$$
\begin{aligned}
& S_{k+1}=S_{k}+\frac{x^{2^{k}}}{1-x^{2^{k+1}}}=\frac{1}{1-x} \cdot \frac{x-x^{2^{k}}}{1-x^{2^{k}}}+\frac{x^{2^{k}}}{1-x^{2^{k+1}}}= \\
= & \frac{1}{1-x}\left[\frac{\left(x-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)+x^{2^{k}}(1-x)}{1-x^{2^{k+1}}}\right]=\frac{1}{1-x} \cdot \frac{x-x^{2 k+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,612 |
454. Prove that the sum of the cubes of three consecutive natural numbers is divisible by 9.
136 | Indication.
$S_{k+1}=(k+1)^{3}+(k+2)^{3}+(k+3)^{3}=k^{3}+(k+1)^{3}+(k+2)^{3}+$
$+9\left(k^{2}+3 k+3\right)=S_{k}+9\left(k^{2}+3 k+3\right)$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,613 |
455. Prove that $n^{3}+5 n$ is divisible by 6, where $n$ is any natural number. | $$
\begin{gathered}
A_{k+1}=(k+1)^{3}+5(k+1)=k^{3}+5 k+3 k(k+1)+6= \\
=A_{k}+3 k(k+1)+6
\end{gathered}
$$
Indication.
$$
\begin{gathered}
A_{k+1}=(k+1)^{3}+5(k+1)=k^{3}+5 k+3 k(k+1)+6= \\
=A_{k}+3 k(k+1)+6
\end{gathered}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,614 |
456. Prove that the expression $m^{3}+11 m$ is divisible by 6, where $m$ is any natural number. | Indication.
$A_{k+1}=(k+1)^{3}+11(k+1)=k^{3}+11 k+3 k(k+1)+12=$ $=A_{k}+3 k(k+1)+12$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,615 |
457. Prove that if $m$ is an even number, then $m^{3}+20 m$ is divisible by 48. | $$
\begin{aligned}
& A_{2 k+2}=(2 k+2)^{3}+20(2 k+2)=(2 k)^{3}+20(2 k)+ \\
& \quad+24(k+1) k+48=A_{k}+24(k+1) k+48
\end{aligned}
$$
The translation is as follows:
$$
\begin{aligned}
& A_{2 k+2}=(2 k+2)^{3}+20(2 k+2)=(2 k)^{3}+20(2 k)+ \\
& \quad+24(k+1) k+48=A_{k}+24(k+1) k+48
\end{aligned}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,616 |
458*. Prove that for any integer $a$ the expression $a^{7}-a$ is divisible by 42. | Instruction. The task is to prove that the given expression is divisible by 6 and by 7. To prove its divisibility by 6, the given expression should be transformed as follows:
$$
\begin{aligned}
& a^{7}-a=a\left(a^{6}-1\right)=a\left(a^{3}-1\right)\left(a^{3}+1\right)= \\
& =a(a-1)(a+1)\left(a^{2}+a+1\right)\left(a^{2}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,617 |
459*. Prove that if $p$ is a prime number, then for any integer $a$ the difference $a^{p}-a$ is divisible by $p$ (Fermat's little theorem). | Proof. 1) When $a=1$, the statement is obvious, since $a^{p}-a=0$.
2) Suppose the statement is true for $a=k$, i.e., that $k^{p}-k$ is divisible by $p$, and we will prove that then $(k+1)^{p}-(k+1)$ is divisible by $p$.
3) We will use the binomial formula:
$$
\begin{gathered}
(k+1)^{p}-(k+1)=k^{p}+C_{p}^{1} k^{p-1}+\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,618 |
460. Prove that for a natural number $n \geqslant 5$ the inequality is valid:
$$
2^{n}>n^{2}
$$ | Instruction. The condition of the problem can be expressed differently: prove that $2^{n+4}>(n+4)^{2}$, where $n$ is any natural number. Then
$$
\begin{gathered}
2^{(k+1)+4}=2^{k+5}=2 \cdot 2^{k+4}>2(k+4)^{2}= \\
=2 k^{2}+16 k+32>\underline{(k+5)^{2}}
\end{gathered}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,619 |
463*. Prove that
$$
\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{13}{24}
$$
where $n$ is a natural number greater than one. | Indication.
$$
S_{k+1}=S_{k}-\frac{1}{k+1}+\frac{1}{2 k+1}+\frac{1}{2 k+2}
$$
or
$$
S_{k+1}-S_{k}=\frac{1}{2 k+1}+\frac{1}{2 k+2}-\frac{1}{k+1}=\frac{1}{2(2 k+1)(k+1)}>0
$$
i.e.
$$
S_{k+1}>S_{k}
$$
but
$$
S_{k}>\frac{13}{24}
$$
thus,
$$
S_{k+1}>\frac{13}{24}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,622 |
466*. Prove that for $x>0$ and natural $n$, greater than 1, the inequality $(1+x)^{n}>1+n x$ (Bernoulli's inequality) holds. | Proof. 1) For $n=2$ we have
$$
(1+x)^{2}=1+2 x+x^{2}>1+2 x
$$
since
$$
x>0
$$
thus, the inequality holds.
2) Assume that
$$
(1+x)^{k}>1+k x
$$
and we will prove that then
$$
(1+x)^{k+1}>1+(1+k) x
$$
3) To prove this, multiply both sides of inequality (1) by $1+x>0$, we get:
$$
(1+x)^{k+1}>(1+k x)(1+x)
$$
or
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,624 |
467*. Prove that for any positive integers
$$
\begin{gathered}
a_{1}, a_{2}, \ldots a_{n} \\
\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\ldots+\frac{a_{n}}{a_{1}} \geqslant n
\end{gathered}
$$ | Proof. 1) For $n=2$ the inequality holds (see problem 285 of the chapter "Inequalities").
2) Suppose the inequality holds for $n=k-1$,
i.e.
$$
\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\ldots+\frac{a_{k-1}}{a_{1}} \geqslant k-1
$$
and we will prove its validity for $n=k$, i.e., for the numbers
$$
a_{1}, a_{2}, \ldot... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,625 |
486. Prove that the product of two even or two odd functions is an even function. | Let $F(x)=f(x) \cdot \varphi(x)$, where $f(x)$ and $\varphi(x)$ are even functions. Then:
$$
F(-x)=f(-x) \cdot \varphi(-x)=f(x) \cdot \varphi(x)=F(x)
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,627 |
488. Prove that any function $f(x)$ defined on some domain symmetric with respect to the origin can be represented (and in a unique way) as the sum of an even and an odd function on this domain. | Proof. Let $f(x)$ be a function satisfying the condition of the problem. We construct the function:
$$
\varphi(x)=f(x)+f(-x)
$$
The function $\varphi(x)$ will be even in this domain. Indeed,
$$
\varphi(-x)=f(-x)+f(x)=\varphi(x)
$$
Consider another function: $\psi(x)=f(x)-f(-x)$. This function will be odd, because:
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,628 |
491. Show that if $l$ is a period of a function, then $2l$ and $3l$, generally $nl$, where $n$ is a natural number, are also periods of this function.
Note. In this and subsequent problems, it is assumed that all values of the argument belong to the domain of permissible values. | Instruction. $f(x+2l)=f[(x+l)+l]=f(x+l)=f(x)$ and so on. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,629 |
493. Show that the function $f(x)=x-\mathrm{E}(x)$ over the interval $(-\infty,+\infty)$, where $\mathrm{E}(x)$ is the greatest integer not exceeding $x$ (for example, $\mathrm{E}(3.5)=3 ; \mathrm{E}(2)=2 ; \mathrm{E}(-0.5)=-1$), is periodic with a period of 1. Sketch the graph of this function. | Indication.
$f(x+1)=(x+1)-\mathrm{E}(x+1)=x+1-[\mathrm{E}(x)+1]=x-\mathrm{E}(x)=f(x)$ and so on. We need to show that no non-integer number is a period of this function. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,630 |
500. Show that the function $y=a x^{3}$ will be increasing in the interval ( $-\infty \ldots+\infty$ ) for $a>0$ and decreasing for $a<0$. | Instruction. $y_{2}-y_{1}=a x_{2}^{3}-a x_{1}^{3}=a\left(x_{2}-x_{1}\right) .\left(x_{2}^{2}+x_{1} x_{2}+x_{1}^{2}\right)$. If $x_{2}>x_{1}$, then $x_{2}-x_{1}>0, x_{2}^{2}+x_{1} x_{2}+x_{1}^{2}$ is always greater than zero, 148
if $x_{1}$ and $x_{2}$ have the same sign. If $x_{1}$ and $x_{2}$ have different signs, the... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,631 |
504. Consider the increase and decrease of the function $y=x^{-n}$, where $n$ - is a natural number. | Instruction. The function $y=x^{-n}$ is defined for any value of $x$ except $x=0$, hence $y=\frac{1}{x^{n}}$. Let $n$ be an even number, but $y_{1}=x^{n}$ increases in the interval ( $0 \ldots+\infty$ ). Therefore, the function $y=\frac{1}{x^{n}}$ will decrease in the interval $(0 \ldots+\infty)$. In the interval $(-\i... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 42,633 |
505. Consider the question of the increase and decrease of the function $y=x^{\frac{m}{n}}$, where $m$ and $n$ are natural numbers. | Proof. Let's take two values of the argument $x_{1}$ and $x_{2}$, where $x_{2}>x_{1}$. Consider the interval of change of the function from 0 to $+\infty$:
$$
\begin{gathered}
x_{2}^{\frac{m}{n}}-x_{1}^{\frac{m}{n}}=\left(x_{2}^{\frac{1}{n}}-x_{1}^{\frac{1}{n}}\right)\left(x_{2}^{\frac{m-1}{n}}+x_{2}^{\frac{m-2}{n}} x... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 42,634 |
506. Show that the function $y=x^{2}$ increases without bound, i.e.
$$
\lim _{x \rightarrow \infty} x^{2}=+\infty
$$ | Instruction. If $x$ increases without bound, then $x$ can become greater than any predetermined positive number, hence $x>M$, but then $x^{2}>M^{2}$, from which it follows that
$$
\lim _{x \rightarrow \infty} x^{2}=+\infty
$$
150 | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,635 |
508. Prove that if $a>1$, then $a^{n}>1$ and if $a<1$, then $a^{n}<1$, where $n-$ is a natural number. | Proof. Since $a>1$, then $a=1+\alpha, \alpha>0$. Thus, $a^{n}=(1+\alpha)^{n}$ or $a^{n}=1+n \alpha+\frac{n(n-1)}{2} \alpha^{2}+\ldots+\alpha^{n}$, but all terms on the right side of this equation are positive, therefore, $a^{n}>1$. If $a<1$, then $a^{n}=\frac{1}{b^{n}}$, but $b^{n}>1$ as proven, therefore $a^{n}<1$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,636 |
509. Prove that if $a>1$, then $\sqrt[n]{a}>1$ and if $a<1$, then $\sqrt[n]{a}<1$ ( $n$ - a natural number). | Proof. Let $a>1$. Set $\sqrt[n]{a}=x$, then $a=x^{n}$. Subtract 1 from both sides of the equation; $a-1=x^{n}-1=$ $=(x-1)\left(x^{n-1}+x^{n-2}+\ldots+1\right)$, but all terms in the second parenthesis are positive, therefore, $(a-1)$ and $(x-1)$ have the same sign, but $a-1>0$, hence $x-1>0$, i.e., $x>1$ or $\sqrt[n]{a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,637 |
510. Prove that $\sqrt[n]{a^{m}}$ is greater than 1 if $a>1$ and less than 1 if $a<1$. | Proof. Let $\sqrt[n]{a}=b$. By the proven fact, $b>1$. We have
$$
a^{\frac{m}{n}}-1=b^{m}-1
$$
or
$$
a^{\frac{m}{n}}-1=(b-1)\left(b^{m-1}+b^{m-2}+\ldots+b+1\right)
$$
But each term in the second parenthesis on the right-hand side is positive, hence $a^{\frac{m}{n}}-1$ and $b-1$ have the same sign. Since $b-1>0$, th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,638 |
511. Prove that
$$
\lim _{n \rightarrow \infty} \sqrt[n]{a}=1, a>0
$$
where $n-$ is a natural number. | Proof. Let $a>1$. Take
$$
a-1=a^{\frac{n}{n}}-1=\left(a^{\frac{1}{n}}-1\right)\left(a^{\frac{n-1}{n}}+a^{\frac{n-2}{n}}+\ldots+1\right)
$$
Any term in the second parenthesis is greater than 1, except for the last term, which is equal to 1. Therefore, $(a-1)>\left(a^{\frac{1}{n}}-1\right) n$, from which $a^{\frac{1}{n... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,639 |
515. Prove that
$$
\lim _{x \rightarrow \infty} a^{x}=\infty,
$$
if $a>1$,
$$
\lim _{x \rightarrow \infty} a^{x}=0
$$
if $a<1$. | Proof. Let $a>1$; $x$ takes positive integer values. We have $a^{n}-1>n(a-1)$, or $a^{n}>1+n(a-1)$, but $1+n(a-1)$ increases without bound as $n$ increases without bound. Therefore, $a^{n}$ can be made greater than any predetermined positive number $M$, i.e., $a^{n}>M$ if $n>\frac{M-1}{a-1}$. Hence,
$$
\lim _{n \right... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,643 |
516. Show that the logarithmic function is increasing when the base is greater than 1, and decreasing when the base is less than 1 but greater than zero. | Proof. Let $a>1$ and $0<x_{1}<x_{2}$. Since $x_{2}-x_{1}>0$ and $a^{\log _{a} x_{1}}>0$, it follows that,
$$
a^{\log _{a} x_{2}-\log _{a} x_{1}}-1>0
$$
or
$$
a^{\log _{a} x_{2}-\log _{a} x_{1}}>1
$$
from which: $\log _{a} x_{2}>\log _{a} x_{1}$. Let $a1$, therefore,
$$
\log _{\frac{1}{a}} x_{2}-\log _{\frac{1}{a}}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,644 |
517. Show that
$$
\lim _{x \rightarrow \infty} \log _{a} x=\infty, \quad \lim _{x \rightarrow 0} \log _{a} x=-\infty \quad \text { for } a>1
$$
and
$$
\lim _{x \rightarrow \infty} \log _{a} x=-\infty, \lim _{x \rightarrow 0} \log _{a} x=\infty \text { for } a<1
$$ | Instruction. If $x \rightarrow \infty$, then we can find such a value of $x$ that can be made and remains greater than any predetermined positive number $M=a^{A}$, i.e., $x>a^{A}$. Then
$$
a^{\log _{a} x}>a^{A}
$$
From this, it follows that
$$
\log _{a} x>A
$$
or
$$
\lim _{x \rightarrow \infty} \log _{a} x=\infty
... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,645 |
519.* Show that the function $y=x^{n}$, where $n$ is a natural number, is a continuous function for each point in its domain. | Proof. Let $a>0$ be any value of $x$ from the domain of admissible values. Consider
$$
\begin{aligned}
& \left|x^{n}-a^{n}\right|=|x-a|\left|x^{n-1}+x^{n-2} a+\ldots+x a^{n-2}+a^{n-1}\right| \leqslant \\
& \leqslant|x-a|\left\{\left|x^{n-1}\right|+\left|x^{n-2} a\right|+\ldots+\left|x a^{n-2}\right|+\left|a^{n-1}\righ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,646 |
520*. Show that the function $y=\sqrt[n]{x^{k}}$ is a continuous function for each point in the domain of admissible values. | Proof. Let $a>0$ be any value of the argument from the domain of admissible values. Let $x>0$. Choose two such positive numbers $B$ and $A$ that
$$
B\left|x^{\frac{1}{n}}-a^{\frac{1}{n}}\right| \cdot n B^{\frac{n-1}{n}}
\end{aligned}
$$
Divide the first equality by the second:
$$
\frac{\left|x^{\frac{k}{n}}-a^{\frac... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,647 |
521*. Show that the exponential function is a continuous function, i.e.
$$
\lim _{x \rightarrow a} a^{x}=a^{\alpha}(a>0)
$$
for each point in the domain of permissible values. | Assumption. Suppose that $x$ approaches $\alpha$, remaining greater than $\alpha$, and $a>1$. Then
$$
|x-\alpha|<\frac{1}{2 \cdot 10^{k}}
$$
or
$$
\alpha-\frac{1}{2 \cdot 10^{k}}<x<\alpha+\frac{1}{2 \cdot 10^{k}}
$$
from which
$$
a^{\alpha-\frac{1}{2 \cdot 10^{k}}}<a^{\alpha}<a^{x}<a^{\alpha+\frac{1}{2 \cdot 10^{k... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,648 |
522*. Show that the logarithmic function is continuous at each point of the interval (0... $)$ ), i.e.
$$
\lim _{x \rightarrow a} \log _{b} x=\log _{b} a
$$ | Let
$$
\lim _{x_{n} \rightarrow a} x_{n}=a
$$
then
$$
\lim _{x_{n} \rightarrow a} \frac{x_{n}}{a}=1
$$
Therefore,
$$
b^{\log _{b} \frac{x_{n}}{a}}=\frac{x_{n}}{a}, \lim _{x_{n} \rightarrow a} b^{\log _{b} \frac{x_{n}}{a}}=\lim _{x_{n} \rightarrow a} \frac{x_{n}}{a}=1
$$
but then
$$
\lim _{x_{n} \rightarrow a} \l... | proof | Calculus | proof | Yes | Yes | olympiads | false | 42,649 |
1. What will be the result if we add:
a) the smallest three-digit and the largest two-digit number;
b) the smallest odd one-digit and the largest even two-digit number. | a) The smallest three-digit number is 100, and the largest two-digit number is 99.
$$
100+99=199
$$
b) The smallest odd one-digit number is 1, and the largest even two-digit number is 98.
$$
1+98=99
$$
Answer: a) 199 ; b) 99 . | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,650 |
2. On a balance that is in equilibrium, on one scale pan lies one apple and two identical pears. On the other scale pan, there are two such apples and one such pear. Which is lighter - the apple or the pear? How did you find out? | 2. Based on the problem statement, let's draw a diagram
Remove one apple from each side of the scales. The scales will remain in balance because the apples were identical.
 and their surnames (Ivanov, Petrov, Sergeev). Let's draw a table.
| | I | $\Pi$ | S |
| :---: | :---: | :---: | :---: |
| $I$ | | | |
| $\Pi$ | | | |
| $S$ | | | |
Capital letters on the left denote the boys' names, and capital letters at th... | IvanPetrov,PetrSergeev,SergeiIvanov | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,652 |
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