problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 16. Solve the equation
$$
\log _{4}(x+3)+\log _{4}(x-1)=2-\log _{4} 8
$$ | Solution. The domain of the equation is determined by the system
$$
\left\{\begin{array}{l}
x+3>0 \\
x-1>0
\end{array}\right.
$$
the solution of which is the interval $11$, i.e., wider than the domain of the equation (12).
The solution of equation (12) can be briefly written as:
$$
\begin{aligned}
& \Leftrightarrow... | \sqrt{6}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,558 |
Example 17. Solve the equation
$$
\frac{3}{2} \log _{1 / 4}(x+2)^{2}-3=\log _{1 / 4}(4-x)^{3}-\log _{4}(x+6)^{3}
$$ | Solution. The domain of the equation is determined by the system
$$
\left\{\begin{array}{l}
x+2 \neq 0 \\
4-x>0 \\
x+6>0
\end{array}\right.
$$
the solution of which consists of two intervals: $-60, a \neq 1, b>0, b \neq 1
$$
can lead to the loss of roots or the appearance of extraneous roots, as the left and right s... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,559 |
Example 18. Solve the equation
$$
\log _{x / 2} x^{2}-14 \log _{16 x} x^{3}+40 \log _{4 x} \sqrt{x}=0
$$ | Solution.
The first method. Let's find the domain of the equation (16). It is defined by the system of inequalities
$$
x>0 ; \quad x \neq 1 / 16 ; \quad x \neq 1 / 4 ; \quad x \neq 2
$$
The equation (16) on this domain is equivalent to the equation
$$
2 \log _{x / 2} x-42 \log _{16 x} x+20 \log _{4 x} x=0
$$
It is... | x_{1}=1,\quadx_{2}=\sqrt{2}/2,\quadx_{3}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,560 |
Example 19. Solve the equation
$$
x^{\log x(x+3)^{2}}=16
$$ | Solution. The domain of definition (DOD) of the equation is determined by the system of inequalities $x>0 ; x \neq 1$. Equation (19) on its DOD is equivalent to the equation
$$
(x+3)^{2}=16
$$
Equation (20) has two roots: $x_{1}=1, x_{2}=-7$, which do not belong to the DOD of equation (19). Therefore, equation (19) h... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,561 |
Example 20. Solve the equation
$$
3 x^{\log _{5} 2}+2^{\log _{5} x}=64
$$ | Solution. The domain of the equation: $x>0$. On this set
$$
x^{\log _{5} 2}=2^{\log _{5} x}
$$
therefore, the given equation is equivalent to the equation
$$
3 \cdot 2^{\log _{5} x}+2^{\log _{5} x}=64
$$
i.e., the equation
$$
2^{\log _{5} x}=16
$$
From this, we get $\log _{5} x=4$, i.e., $x=625$. The number 625 b... | 625 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,562 |
Example 22. Solve the system
$$
\left\{\begin{array}{l}
\log _{y} x-\log _{x} y=8 / 3 \\
x y=16
\end{array}\right.
$$ | Solution. The set of admissible values of $x$ and $y$ in the given system is defined by the system of inequalities: $x>0, x \neq 1, y>0$, $y \neq 1$. Letting $z=\log _{y} x$ and considering that for $x$ and $y$ from the domain of admissible values
$$
\log _{x} y=\frac{1}{\log _{y} x}
$$
we obtain the equation
$$
z-1... | (8,2),(\frac{1}{4},64) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,564 |
Example 23. Solve the system
$$
\left\{\begin{array}{l}
\left|\log _{2}(x+y)\right|+\left|\log _{2}(x-y)\right|=3 \\
x y=3
\end{array}\right.
$$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,565 | |
Example 1. Solve the inequality
$$
\left|x^{2}-2 x\right|<x .
$$ | Solution. From the properties of a quadratic trinomial, it follows that $x^{2}-2 x<0$, which for the specified values of $x$ is satisfied only when $1<x<2$. Therefore, the interval $(1 ; 2)$ is part of the solution set of inequality (1).
For $x \geqslant 2$, we have the inequality $x^{2}-2 x<x$, i.e., the inequality $... | (1;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,566 |
Example 2. Solve the inequality
$$
x^{2}-2|x|<3
$$ | Solution. The given inequality is equivalent to the combination of two systems:
$$
\left\{\begin{array} { l }
{ x ^ { 2 } - 2 x - 3 < 0 , } \\
{ x \geqslant 0 , }
\end{array} \left\{\begin{array}{l}
x^{2}+2 x-3<0 \\
x<0
\end{array}\right.\right.
$$
Since $x^{2}-2 x-3=(x+1)(x-3)$, the set of all solutions to the ineq... | -3<x<3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,567 |
Example 3. Solve the inequality
$$
|x|<-x^{2}+x+6 .
$$ | Solution. The given inequality is equivalent to the combination of two systems:
$$
\left\{\begin{array} { l }
{ x 0$ it is equivalent to the system
$$
\left\{\begin{array}{r}
f(x)<a \\
-f(x)<a
\end{array}\right.
$$ | not\found | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,568 |
Example 5. Solve the inequality
$$
|x-6|<x^{2}-5 x+9
$$ | Solution. The given inequality is equivalent to the system
$$
\left\{\begin{array} { l }
{ x - 6 - ( x ^ { 2 } - 5 x + 9 ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}-6 x+15>0 \\
x^{2}-4 x+3>0
\end{array}\right.\right.
$$
The inequality $x^{2}-6 x+15>0$ holds for any $x$. Since $x^{2}-4 x+3=(x-1)(x-... | (-\infty;1)\cup(3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,570 |
Example 6. Solve the inequality
$$
3|x-1|+x^{2}>7
$$ | Solution. The given inequality can be rewritten as
$$
3|x-1|>7-x^{2}
$$
and, consequently, it is equivalent to the system of inequalities
$$
\left[\begin{array}{l}
3 ( x - 1 ) > 7 - x ^ { 2 } , \\
3 ( x - 1 ) < - ( 7 - x ^ { 2 } )
\end{array}\right.
$$
Since $x^{2}+3 x-10=(x+5)(x-2)$, and $x^{2}-3 x-4=(x+1)(x-4)$, ... | (-\infty;-1)\cup(2;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,571 |
Example 8. Solve the inequality
$$
\left|\frac{x^{2}-5 x+4}{x^{2}-4}\right| \leqslant 1
$$ | Solution. The given inequality is equivalent to the system
$$
\left\{\begin{array}{l}
\frac{x^{2}-5 x+4}{x^{2}-4} \leq 1 \\
\frac{x^{2}-5 x+4}{x^{2}-4} \geqslant-1
\end{array}\right.
$$
The first inequality of the system is equivalent to the inequality
$$
\frac{x-8 / 5}{(x-2)(x+2)} \geqslant 0
$$
Applying the metho... | [0;8/5]\cup[5/2;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,573 |
Example 9. Solve the inequality
$$
\left|x^{3}-x\right| \geqslant 1-x
$$ | Solution. The given inequality is equivalent to the system
$$
\left[\begin{array} { l }
{ x ^ { 3 } - x \geqslant 1 - x , } \\
{ x ^ { 3 } - x \leqslant - ( 1 - x ) }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x^{3}-1 \geqslant 0 \\
x^{3}-2 x+1 \leqslant 0
\end{array}\right.\right.
$$
The first inequality of... | (-\infty;\frac{-1-\sqrt{5}}{2}]\cup[\frac{\sqrt{5}-1}{2};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,574 |
Example 11. Solve the inequality
$$
\left|1-\frac{|x|}{1+|x|}\right| \geqslant \frac{1}{2}
$$ | Solution. The domain of admissible values of this inequality consists of all real numbers. The inequality is equivalent to the combination of two systems:
Let's solve the first system:
$$... | 0\leqslantx\leqslant1orx<0orx>2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,576 |
Example 12. Solve the inequality
$$
|x-1|+|2-x|>3+x
$$ | Solution. The points $x=1$ and $x=2$ divide the number line (the domain of inequality (10)) into three intervals: $x<1$, $1 \leqslant x \leqslant 2$, and $x>2$. We will solve the given inequality on each of these intervals.
If $x<1$, then $1-x>0$ and $2-x>0$. Inequality (10) becomes $1-x+2-x>3+x$, i.e., $x < 0$.
$$
\b... | (-\infty;0)\cup(6;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,577 |
Example 14. Solve the inequality
$$
\frac{|x-3|}{x^{2}-5 x+6} \geq 2
$$ | Solution. Inequality (12) is equivalent to the combination of two systems:
For the first system of this combination, we get:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x - 3 \geqs... | 3/2\leqslantx<2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,579 |
Example 15. Solve the system of inequalities
$$
\left\{\begin{array}{l}
|2 x-3| \leqslant 3 \\
\frac{1}{x}<1
\end{array}\right.
$$ | Solution. The domain of the system consists of all real numbers except zero. The first inequality of the system is equivalent to the double inequality
$$
-3 \leqslant 2 x-3 \leqslant 3 \Leftrightarrow 0 \leqslant 2 x \leqslant 6 \Leftrightarrow 0 \leqslant x \leqslant 3
$$
If $x>0$, then the inequality $1 / x < 1$.
... | (1,3] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,580 |
Example 16. Solve the inequality
$$
-|y|+x-\sqrt{x^{2}+y^{2}-1} \geqslant 1
$$ | Solution. The domain of the inequality (13) is determined by the condition $x^{2}+y^{2}-1 \geqslant 0$. Rewrite inequality (13) as
$$
x-|y| \geqslant 1+\sqrt{x^{2}+y^{2}-1}
$$
From this, it follows that $x-|y| \geqslant 0$. Under this condition, both sides of the obtained inequality are non-negative on its domain. Th... | (1;0) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,581 |
Example 17. For all $a$ solve the inequality
$$
\left|x^{2}-5 x+4\right|<a
$$ | Solution. Since $\left|x^{2}-5 x+4\right| \geqslant 0$ for any $x$, the inequality (14) has no solutions when $a \leqslant 0$.
Let $a>0$. Since $x^{2}-5 x+4=(x-1)(x-4)$, the number line (domain of the inequality (14)) is divided into three intervals: $x<1$, $1 \leqslant x \leqslant 4$, and $x>4$.
We will solve the in... | x\in(\frac{1}{2}(5-\sqrt{9+4}),\frac{1}{2}(5+\sqrt{9+4}))for\frac{9}{4};\,x\in(1,\frac{1}{2}(5+\sqrt{9-4}))\cup(4,\frac{1}{2}(5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,582 |
Example 18. Find all values of $a \neq 0$, for which the inequality
$$
a^{2}\left|a+x / a^{2}\right|+|1+x| \leqslant 1-a^{3}
$$
has at least four different solutions that are integers. | Solution. The left side of inequality (17) is non-negative for any values of $a$ and $x$; therefore, inequality
Fig. 3.1 can have a solution only when its right side is non-negative, i.e.... | \in(-\infty;-\sqrt{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,583 |
Example 2. Solve the inequality
$$
\frac{\sqrt{6+x-x^{2}}}{2 x+5} \geqslant \frac{\sqrt{6+x-x^{2}}}{x+4}
$$ | Solution. The domain of definition of the original inequality is determined by the system
$$
\left\{\begin{array}{l}
6+x-x^{2} \geqslant 0 \\
2 x+5 \neq 0 \\
x+4 \neq 0
\end{array}\right.
$$
from which we find: $-2 \leqslant x \leqslant 3$.
For the values $x=-2$ and $x=3$, the inequality (2) is satisfied; therefore,... | -2\leqslantx\leqslant-13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,584 |
Example 4. Solve the inequality
$$
\sqrt{x+2}>\sqrt{8-x^{2}}
$$ | Solution. Inequality (4) is equivalent to the system of inequalities
$$
\left\{\begin{aligned}
8-x^{2} & \geqslant 0 \\
x+2 & >8-x^{2}
\end{aligned}\right.
$$
The solutions to the first inequality of this system are all $x$ for which $|x| \leqslant 2 \sqrt{2}$, i.e., all numbers in the interval $-2 \sqrt{2} \leqslant... | 2<x\leq2\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,586 |
Example 8. Solve the inequality
$$
\sqrt{-x^{2}+6 x-5}>8-2 x
$$ | Solution. The quadratic trinomial $-x^{2}+6 x-5$ has roots $x_{1}=1$ and $x_{2}=5$; therefore, inequality (10) is equivalent to the combination of two systems
From the second system of thi... | 3<x\leqslant5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,590 |
Example 9. Solve the inequality
$$
\frac{\sqrt{2-x}+4 x-3}{x} \geqslant 2
$$
## 148 CH. 3. INEQUALITIES WITH ONE UNKNOWN | Solution. From the conditions $2-x \geqslant 0$ and $x \neq 0$, we find the domain of the inequality (11): $x<2$, and $x \neq 0$. If $x>1$, then by the properties of exponents we have
$$
\sqrt{2-x} 0 , } \\
{ \sqrt { 2 - x } \geqslant 3 - 2 x , } \\
{ \{ \begin{array} { l }
{ x 0, \\
3-2 x0, \\
3-2 x \geqslant 0, \\
... | (-\infty;0)\cup[1;2] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,591 |
Example 12. Solve the inequality
$$
\sqrt{25-x^{2}}+\sqrt{x^{2}+7 x}>3
$$ | Solution. The domain of definition of the given inequality is determined by the system
$$
\left\{\begin{array}{l}
25-x^{2} \geqslant 0 \\
x^{2}+7 x \geqslant 0
\end{array}\right.
$$
from which $0 \leqslant x \leqslant 5$. Both sides of the inequality (17) are non-negative; therefore, it is equivalent to the system
$... | 0\leqslantx\leqslant5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,593 |
Example 13. Solve the inequality
$$
\sqrt{x^{2}-8 x+15}+\sqrt{x^{2}+2 x-15}>\sqrt{4 x^{2}-18 x+18}
$$ | Solution. Since
$$
\begin{gathered}
x^{2}-8 x+15=(x-3)(x-5) \\
x^{2}+2 x-15=(x+5)(x-3) \\
4 x^{2}-18 x+18=4(x-3)(x-3 / 2)
\end{gathered}
$$
the domain of the inequality consists of $x=3, x \leqslant-5$ and $x \geqslant 5$.
For $x=3$, both sides of the inequality (18) are equal to zero; therefore, the number $x=3$ is... | x>\frac{17}{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,594 |
Example 16. For all $a$ solve the inequality
$$
\sqrt{\frac{3 x+a}{x-a}}<a-1
$$ | Solution. The left side of inequality (22) is non-negative on the domain of definition, so $a-1>0$, i.e., $a>1$. Let's find the domain of definition of the given inequality. We have
$$
\frac{3 x+a}{x-a} \geqslant 0
$$
from which we obtain two intervals: $-\infty < x \leq -\frac{a}{3}$ and $x > a$. The inequality $a>1... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,596 | |
Example 1. Solve the inequality
$$
25^{x}>125^{3 x-2}
$$ | Solution. Since
$$
25^{x}=\left(5^{2}\right)^{x}=5^{2 x}, \quad 125^{3 x-2}=\left(5^{3}\right)^{3 x-2}=5^{9 x-6}
$$
the given inequality is equivalent to the inequality
$$
5^{2 x}>5^{9 x-6} \Leftrightarrow 2 x>9 x-6 \Leftrightarrow x<6 / 7
$$
Therefore, the interval ( $-\infty ; 6 / 7$ ) is the set of all solutions... | x<\frac{6}{7} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,597 |
Example 2. Solve the inequality
$$
(0.1)^{4 x^{2}-2 x-2} \leqslant(0.1)^{2 x-3}
$$ | Solution. The given inequality is equivalent to the inequality
$$
4 x^{2}-2 x-2 \geqslant 2 x-3 \Leftrightarrow 4 x^{2}-4 x+1 \geqslant 0 \Leftrightarrow(2 x-1)^{2} \geqslant 0
$$
Thus, the original inequality is satisfied by all real numbers.
When solving some exponential inequalities, a transformation is used that... | (2x-1)^{2}\geqslant0 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,598 |
Example 4. Solve the inequality
$$
2^{x+2}-2^{x+3}-2^{x+4}>5^{x+1}-5^{x+2}
$$ | Solution. Since
$$
\begin{gathered}
2^{x+2}=4 \cdot 2^{x}, \quad 2^{x+3}=8 \cdot 2^{x}, \quad 2^{x+4}=16 \cdot 2^{x} \\
5^{x+1}=5 \cdot 5^{x}, \quad 5^{x+2}=25 \cdot 5^{x}
\end{gathered}
$$
then the given inequality is equivalent to the inequality
$$
2^{x}(4-8-16)>5^{x}(5-25) \Leftrightarrow 2^{x}(-20)>5^{x}(-20) \L... | (0;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,600 |
Example 5. Solve the inequality
$$
3^{72}\left(\frac{1}{3}\right)^{x}\left(\frac{1}{3}\right)^{V \bar{x}}>1
$$ | Solution. The domain of the inequality is $x \geqslant 0$. By reducing the left part of the inequality to a power with base 3, we get
$$
3^{72}\left(\frac{1}{3}\right)^{x}\left(\frac{1}{3}\right)^{\sqrt{x}}=3^{72-x-\sqrt{x}}
$$
Thus, the given inequality is equivalent to the inequality
$$
72-x-\sqrt{x}>0
$$
Let $t=... | 0\leqslantx<64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,601 |
Example 6. Solve the inequality
$$
\sqrt{6-x}\left(5^{x^{2}-7.2 x+3.9}-25 \sqrt{5}\right) \geqslant 0
$$ | Solution. The domain of definition of inequality (1) is determined by the condition $6-x \geqslant 0$, i.e., $x \leqslant 6$. Inequality (1) is equivalent to a combination consisting of an equation and a system of two inequalities:
$$
\sqrt{6-x}=0, \quad\left\{\begin{array}{l}
5^{x^{2}-7.2 x+3.9}-25 \sqrt{5} \geqslant... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,602 |
Example 7. Solve the inequality
$$
\frac{1}{5} \cdot 5^{2 x} 7^{3 x+2} \leq \frac{25}{7} \cdot 7^{2 x} 5^{3 x}
$$ | Solution. Dividing both sides of the inequality (2) by the expression $\frac{25}{7} \cdot 7^{2 x} 5^{3 x}$, which is positive for any real $x$, we obtain the inequality
$$
\frac{5^{2 x-1} 7^{3 x+2}}{5^{2} 7^{2 x-1} 5^{3 x}} \leqslant 1
$$
equivalent to (2), i.e., the inequality
$$
7^{x+3} 5-x-3<1
$$
The last inequa... | x\leqslant-3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,603 |
Example 10. Solve the inequality
$$
3^{2 x-1}<11^{3-x}
$$ | Solution. Both sides of the inequality are positive for any value of $x$. Taking the logarithm of both sides of the inequality to the base 3, we obtain the inequality
$$
2 x-10$, we find all solutions of the original inequality - the interval
$$
-\infty<x<\frac{1+3 \log _{3} 11}{2+\log _{3} 11}
$$ | -\infty<x<\frac{1+3\log_{3}11}{2+\log_{3}11} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,606 |
Example 11. Solve the inequality
$$
\sqrt{\frac{9}{10}} \cdot\left(\frac{9}{10}\right)^{x-1}>\frac{10^{(3 / 4) x-1}}{\sqrt{10}}
$$ | Solution. Both sides of the given inequality are positive for any $x$. Taking the logarithm of it to the base 10, we obtain the inequality
$$
(x-1) \lg \frac{9}{10}+\frac{1}{2} \lg \frac{9}{10}>\left(\frac{3}{4} x-1\right)-\frac{3}{2}
$$
i.e., the inequality
$$
x\left(\lg \frac{9}{10}-\frac{3}{4}\right)>\frac{1}{2} ... | x<4\frac{\lg3-2}{8\lg3-7} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,607 |
Example 12. Solve the inequality
$$
3 \cdot 7^{2 x}+37 \cdot 140^{x}<26 \cdot 20^{2 x}
$$ | Solution. Let's write the inequality in the form
$$
3 \cdot 49^{x}+37 \cdot 140^{x}-26 \cdot 400^{x} \leqslant 0
$$
In this inequality, the numbers 49, 140, and 400 form three consecutive terms of a geometric progression with a common ratio of 20/7. Dividing both sides by $400^{x}$, we obtain the inequality
$$
3(7 /... | x\geqslant\log_{7/20}\frac{2}{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,608 |
Example 14. Solve the inequality
$$
10^{7 x-1}+6 \cdot 10^{1-7 x}-5 \leqslant 0
$$ | Solution. Using the properties of exponents, rewrite the given inequality as
$$
10^{7 x-1}+\frac{6}{10^{7 x-1}}-5 \leq 0
$$
Let $t=10^{7 x-1}$, we get
$$
\left\{\begin{array} { l }
{ t > 0 , } \\
{ t + 6 / t - 5 \leqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t>0, \\
t^{2}-5 t+6 \leqslant 0
\end{a... | \frac{1}{7}(1+\lg2)\leqslantx\leqslant\frac{1}{7}(1+\lg3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,609 |
Example 15. Solve the inequality
$$
2^{2 x^{2}-6 x+3}+6^{x^{2}-3 x+1} \geqslant 3^{2 x^{2}-6 x+3}
$$ | Solution. Inequality (11) is equivalent to the inequality
$$
8 \cdot 2^{2\left(x^{2}-3 x\right)}+6 \cdot 2^{x^{2}-3 x} 3^{x^{2}-3 x}-27 \cdot 3^{2}\left(x^{2}-3 x\right) \geqslant 0,
$$
which is a homogeneous inequality of the form
$$
8 \cdot f^{2}(x)+6 f(x) \cdot g(x)-27 g^{2}(x) \geqslant 0
$$
where $f(x)=2^{x^{2... | [\frac{3-\sqrt{5}}{2};\frac{3+\sqrt{5}}{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,610 |
Example 16. Solve the inequality
$$
6 \sqrt[x]{9}-13 \sqrt[x]{3} \sqrt[x]{2}+6 \sqrt[x]{4} \leqslant 0
$$ | Solution. The domain of admissible values of this inequality consists of all natural numbers greater than 1.
Inequality (12) is homogeneous. Dividing both parts by $\sqrt[x]{9}$ and setting $t=\sqrt[x]{2 / 3}$, we get
$$
6-13 t+6 t^{2} \leqslant 0
$$
The solution to this inequality is the interval $2 / 3 \leqslant t... | {x:x\geqslant2,x\in{N}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,611 |
Example 17. Solve the inequality
$$
(x-2)^{x^{2}-6 x+8}>1
$$
## 170 CH. 3. INEQUALITIES WITH ONE UNKNOWN | Solution. The domain of admissible values for this inequality is defined by the condition $x>2$. For such $x$ we have
$(13) \Leftrightarrow 10^{\left(x^{2}-6 x+8\right) \lg (x-2)}>1 \Leftrightarrow\left(x^{2}-6 x+8\right) \lg (x-2)>0 \Leftrightarrow$
$$
\begin{aligned}
& \Leftrightarrow\left[\begin{array}{l}
\left\{\... | (2;3)\bigcup(4;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,612 |
Example 18. Solve the inequality
$$
5^{\frac{1}{4} \log _{5}^{2} x} \geqslant 5 x^{\frac{1}{5} \log _{5} x}
$$ | Solution. The domain of admissible values of inequality (14) is defined by the condition $x>0$. For such $x$, both sides of inequality (14) are positive. Taking the logarithm of both sides to the base 5, we obtain the inequality
$$
\frac{1}{4} \log _{5}^{2} x \geqslant 1+\frac{1}{5} \log _{5}^{2} x
$$
equivalent to (... | (0;5^{-2\sqrt{5}}]\cup[5^{2}\sqrt{5};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,613 |
Example 19. Solve the system
$$
\left\{\begin{array}{l}
4^{\left|x^{2}-8 x+12\right|-\log _{4} 7}=7^{2 y-1} \\
|y-3|-3|y|-2(y+1)^{2} \geqslant 1
\end{array}\right.
$$ | Solution. The equation of the given system is equivalent to the equation
$$
4^{\left|x^{2}-8 x+12\right|}=7^{2 y}
$$
Both sides of it are positive; therefore, it is equivalent to the equation
$$
\left|x^{2}-8 x+12\right|=(2 y) \log _{4} 7
$$
Since $\log _{4} 7>0$ and $\left|x^{2}-8 x+12\right| \geqslant 0$ for any ... | (2,0),(6,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,614 |
Example 20. Find all values of $a$ for which the inequality
$$
4^{x^{2}}+2(2 a+1) \cdot 2^{x^{2}}+4 a^{2}-3>0
$$
holds for any $x$. | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,615 | |
Example 1. Solve the inequality
$$
\log _{7} \frac{x-2}{x-3}<0
$$ | Solution. Since the base of the logarithm is greater than one, the given inequality is equivalent to the double inequality
$$
00 \\
\frac{x-2}{x-3}0
$$
is satisfied for all $x$ from the intervals $-\infty<x<2$ and $3<x<$ $<+\infty$.
The inequality
$$
\frac{x-2}{x-3}<1
$$
is equivalent to the inequality
$$
\frac{x... | -\infty<x<2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,616 |
Example 2. Solve the inequality
$$
\lg \left(x^{2}-2 x-3\right) \geqslant 0
$$ | Solution. The base of the logarithm is greater than one; therefore, the given inequality is equivalent to the inequality
$$
\begin{aligned}
x^{2}-2 x-3 \geqslant 1 \Leftrightarrow x^{2}-2 x-4 & \geqslant 0 \Leftrightarrow \\
& \Leftrightarrow(x-(1-\sqrt{5}))(x-(1+\sqrt{5})) \geqslant 0
\end{aligned}
$$
from which it ... | x\in(-\infty,1-\sqrt{5}]\cup[1+\sqrt{5},+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,617 |
Example 3. Solve the inequality
$$
x \log _{1 / 10}\left(x^{2}+x+1\right)>0
$$ | Solution. Since $\log _{1 / 10} A=-\lg A$, the given inequality is equivalent to the inequality
$$
x \lg \left(x^{2}+x+1\right) 0 , } \\
{ \operatorname { l g } ( x ^ { 2 } + x + 1 ) 0
\end{array}\right.\right.
$$
Let's solve the first system. Since $x^{2}+x+1>1$ for $x>0$, then $\lg \left(x^{2}+x+1\right)>0$ for $x>... | x<-1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,618 |
Example 4. Solve the inequality
$$
\log _{2} \frac{5-12 x}{12 x-8}+\log _{1 / 2} x \leqslant 0
$$ | Solution. Since $\log _{1 / 2} x=-\log _{2} x$, the given inequality is equivalent to the inequality
$$
\log _{2} \frac{5-12 x}{12 x-8} \leqslant \log _{2} x
$$
## 182 CH. 3. INEQUALITIES WITH ONE UNKNOWN
which, taking into account that the base of the logarithm is greater than one, is equivalent to the system of in... | \frac{5}{12}<x\leqslant\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,619 |
Example 5. Solve the inequality
$$
\log _{3}\left(x^{2}-2\right)<\log _{3}\left(\frac{3}{2}|x|-1\right)
$$ | Solution. Since the base of the logarithms is greater than one, the given inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}-20
\end{array}\right.
$$
which, taking into account that $x^{2}=|x|^{2}$, can be rewritten as
$$
\left\{\begin{array}{l}
|x|^{2}-20
\end{array}\right.
$$
... | -2<x<-\sqrt{2}\sqrt{2}<x<2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,620 |
Example 8. Solve the inequality
$$
\log _{0.5}\left(\log _{6} \frac{x^{2}+x}{x+4}\right)<0
$$ | Solution. Since the base of the logarithm is less than one, the given inequality is equivalent to the inequality
$$
\log _{6} \frac{x^{2}+x}{x+4}>1
$$
which, considering that the base of the logarithm is greater than one, is equivalent to the inequality
$$
\frac{x^{2}+x}{x+4}>6
$$
i.e., the inequality
$$
\frac{x^{... | -4<x<-38<x<+\infty | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,622 |
Example 9. Solve the inequality
$$
\frac{\lg ^{2} x-3 \lg x+3}{\lg x-1}<1
$$ | Solution. Let $y=\lg x$, then the given inequality takes the form
$$
\frac{y^{2}-3 y+3}{y-1}<1
$$
This inequality is equivalent to the inequality
$$
\frac{y^{2}-3 y+3}{y-1}-1<0 \Leftrightarrow \frac{y^{2}-4 y+4}{y-1}<0
$$
i.e., the inequality
$$
\frac{(y-2)^{2}}{y-1}<0
$$
The solution to the last inequality is th... | 0<x<10 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,623 |
Example 10. Solve the inequality
$$
\log _{4}\left(3^{x}-1\right) \log _{1 / 4} \frac{3^{x}-1}{16} \leq \frac{3}{4}
$$ | Solution. Let $y=3^{x}-1$; then the given inequality takes the form
$$
\log _{4} y \log _{1 / 4} \frac{y}{16} \leqslant \frac{3}{4}
$$
Since
$$
\log _{1 / 4} \frac{y}{16}=-\log _{4} \frac{y}{16}=-\left(\log _{4} y-\log _{4} 16\right)=2-\log _{4} y
$$
the inequality (4) can be rewritten as
$$
2 \log _{4} y-\log _{4... | 0<x\leqslant12\leqslantx<+\infty | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,624 |
Example 11. Solve the inequality
$$
\log _{x^{2}}(2+x)<1
$$ | Solution.
The first method. The given inequality is equivalent to the inequality
$$
\log _{x^{2}}(2+x)0 \\
x^{2} \neq 1 \\
2+x>0
\end{array}\right.
$$
from which we find the domain of the inequality:
$$
-21$ ), the solution of which on this set are the intervals $-2x^{2}
$$
(since $x^{2}2$.
When solving logarithm... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,625 |
Example 13. Solve the inequality
$$
\log _{1 / 3}\left(x^{2}-6 x+18\right)-2 \log _{1 / 3}(x-4)<0
$$ | Solution. The domain of admissible values of the inequality is determined by the system
$$
\left\{\begin{array}{l}
x^{2}-6 x+18>0 \\
x-4>0
\end{array}\right.
$$
From it, we find the domain of admissible values: the interval $44 \\
x^{2}-6 x+18>(x-4)^{2}
\end{array}\right.
$$
Since
$$
\begin{aligned}
\left\{\begin{a... | 4<x<+\infty | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,627 |
Example 18. Solve the inequality
$$
\log _{x} 2 x \leqslant \sqrt{\log _{x}\left(2 x^{3}\right)}
$$ | Solution. The given inequality is equivalent to the inequality
$$
\log _{x} 2+10$, $x \neq 1$.
If $0 < x < 1$, then we have
$$
\left\{\begin{array} { l }
{ 1 / \operatorname { l o g } _ { 2 } x \geqslant 1 } \\
{ 1 / \operatorname { l o g } _ { 2 } x \leqslant - 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{... | 0<x\leqslant1/\sqrt[3]{2}2\leqslantx<+\infty | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,632 |
Example 20. Solve the inequality
$$
\frac{\log _{5}\left(x^{2}-4 x-11\right)^{2}-\log _{11}\left(x^{2}-4 x-11\right)^{3}}{2-5 x-3 x^{2}} \geqslant 0
$$ | Solution. The domain of definition (ODZ) of the given inequality consists of all $x$ satisfying the system
$$
\left\{\begin{array}{l}
x^{2}-4 x-11>0 \\
-3 x^{2}-5 x+2 \neq 0
\end{array}\right.
$$
i.e., it is the union of three intervals: $-\infty<x<-2$, $-2<x<2-\sqrt{15}, 2+\sqrt{15}<x<+\infty$. Since for such values... | -\infty<x<-2,6\leqslantx<+\infty,-2<x<2-\sqrt{15} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,633 |
Example 21. Solve the inequality
$$
\log _{2}\left(\sqrt{x^{2}-4 x}+3\right)>\log _{1 / 2} \frac{2}{\sqrt{x^{2}-4 x}+\sqrt{x+1}+1}+1
$$ | Solution. The domain of admissible values of the inequality consists of all $x$ satisfying the system
$$
\left\{\begin{array}{l}
x^{2}-4 x \geqslant 0 \\
x+1 \geqslant 0
\end{array}\right.
$$
i.e., consists of the intervals $-1 \leqslant x \leqslant 0$ and $4 \leqslant x\sqrt{x^{2}-4 x}+\sqrt{x+1}+1
$$
i.e., the ine... | -1\leqslantx\leqslant0 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,634 |
Example 22. Solve the inequality
$$
\left(\sqrt{x^{2}-4 x+3}+1\right) \log _{2} \frac{x}{5}+\frac{1}{x}\left(\sqrt{8 x-2 x^{2}-6}+1\right) \leqslant 0
$$ | Solution. The domain of admissible values of the original inequality consists of all $x$ satisfying the system
$$
\left\{\begin{array}{l}
x>0 \\
x^{2}-4 x+3 \geqslant 0 \\
8 x-2 x^{2}-6 \geqslant 0
\end{array}\right.
$$
i.e., the system
$$
\left\{\begin{array}{l}
x>0 \\
x^{2}-4 x+3 \geqslant 0 \\
x^{2}-4 x+3 \leqsla... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,635 |
Example 23. Solve the system
$$
\left\{\begin{array}{l}
4 \log _{2}^{2} x+1=2 \log _{2} y \\
\log _{2} x^{2} \geqslant \log _{2} y
\end{array}\right.
$$ | Solution. The domain of admissible values of the system is defined by the system of inequalities $x>0, y>0$. The second inequality of the system on the domain of admissible values is equivalent to the inequality $2 \log _{2} x \geqslant \log _{2} y$, replacing $2 \log _{2} y$ in which with $4 \log _{2}^{2} x+1$, we obt... | (\sqrt{2};2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,636 |
3. If a number has the form $2 m^{2}+n^{2}$, then its square can also be represented in the same form. | 3. $\left(2 m^{2}+n^{2}\right)^{2}=4 m^{4}+n^{4}-4 m^{2} n^{2}+8 m^{2} n^{2}=\left(2 m^{2}-\right.$ $\left.-n^{2}\right)^{2}+2(2 m n)^{2}=2 m_{1}^{2}+n_{1}^{2}$, where $m_{1}=2 m n, n_{1}=2 m^{2}-n^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,639 |
4. The differences of squares of consecutive numbers form an arithmetic progression; the differences of the differences (second-order differences) of cubes of consecutive numbers also form an arithmetic progression. Prove. | 4. $(n+1)^{2}-n^{2}=2 n+1$, numbers of the form $2 n+1$ form an arithmetic progression with a difference of 2.
$$
(n+1)^{3}-n^{3}=3 n^{2}+3 n+1,3(n+1)^{2}+3(n+1)+1-
$$
$-\left(3 n^{2}+3 n+1\right)=6 n+6$; numbers $6 n+6$ form an arithmetic progression with a difference of 6. | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,640 |
5*. Consider the table (see p. 4), containing all odd numbers of the form $4 n+1$, starting from one, arranged in a spiral counterclockwise. Show that the main diagonal $1-9-49 \ldots$ contains the squares of all odd numbers. Also show that the diagonal $5-45 \ldots$, extended infinitely in the direction $5-45 \ldots$,... | 5. The numbers on the diagonal $1-9-49-\ldots$ have the form $4n+1$, where $n$ takes the values $0,2,6,12,20, \ldots$, generally $k(k+1)$; therefore, $4k(k+1)+1=(2k+1)^2$. The numbers on the diagonal $5-45-\ldots$ have the form $4n+1$, where $n$ takes the values $1,11,29,55, \ldots$, generally $2k(2k-1)-1$; therefore, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,641 |
7. The product of four consecutive terms of an arithmetic progression, added to $d^{4}$ (where $d$ is the difference of the progression), is a perfect square.[^0]
| 145 | 141 | 137 | 133 | 129 | 125 | 121 | 221 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 149 | 65 | 61 | 57 | 53 | 49 | 117 | 2... | 7. $a(a+d)(a+2 d)(a+3 d)+d^{4}=a(a+3 d)(a+$ +d) $(a+2 d)+d^{4}=\left(a^{2}+3 a d+d^{2}\right)^{2}$. | (^{2}+3+^{2})^{2} | Algebra | proof | Yes | Yes | olympiads | false | 40,642 |
8. The number $\left(10^{n}+10^{n-1}+\ldots+1\right) \cdot\left(10^{n+1}+5\right)+1$ is a perfect square. | 8. $\left(10^{n}+10^{n-1}+\ldots+1\right)\left(10^{n+1}+5\right)+1=$ $=\frac{10^{n+1}-1}{10-1}\left(10^{n+1}+5\right)+1=\left(\frac{10^{n}+1+2}{3}\right)^{2}$. The number $\frac{10^{n}+1+2}{3}$ is an integer, since the number $10^{n+1}+2=1000 \ldots 02$ is divisible by 3, because the sum of its digits is divisible by 3... | (\frac{10^{n}+3}{3})^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,643 |
10. If the sum (difference) of two numbers is the square of some number, then twice the sum (difference) of their cubes is the sum of the squares of three other numbers. | 10. If we accept $a+b=t^{2}$, then $2\left(a^{3}+b^{3}\right)=(a t)^{2}+$ $+(a t-b t)^{2}+(b t)^{2}$.
## 28 | 2(^{3}+b^{3})=()^{2}+(-)^{2}+()^{2} | Algebra | proof | Yes | Yes | olympiads | false | 40,644 |
11. There do not exist two integer squares that are consecutive numbers. | 11. If $a^{2}-b^{2}=1$, then we have the system of equations:
$$
\left\{\begin{array}{l}
a+b=1 \\
a-b=1
\end{array}\right.
$$
from which $a=1, b=0$ (trivial case). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,645 |
12. Find an integer whose cube is equal to the sum of the cubes of the three preceding consecutive numbers. | 12. Answer. 6. We obtain an interesting identity:
$$
3^{3}+4^{3}+5^{3}=6^{3}
$$ | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,646 |
14. Prove a more general statement: a number of the form $x^{4 m}+2^{4 n+2}$ is composite for $m>0, n \geqslant 0$ and $x$-integer. | 14. $x^{4 m}+2^{4 n+2}=\left(x^{2 m}+2^{2 n+1}+2^{n+1} x^{m}\right)\left(x^{2 m}+2^{2 n+1}-\right.$ $\left.-2^{n+1} x^{m}\right)=\left(x^{2 m}+2^{2 n+1}+2^{n+1} x^{m}\right)\left[\left(x^{m}-2^{n}\right)^{2}+2^{2 n}\right]$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,647 |
15. The number $n^{k}$ for $k \geqslant 2$ is the sum of $n$ consecutive odd numbers. | 15. $(2 a+1)+(2 a+3)+\ldots+(2 a+2 n-1)=n^{k}$, $(2 a+n) n=n^{k}, a=\frac{n^{k-1}-n}{2} ;$ for example, for $3^{5}$ we have: $a=$ $=\frac{3^{4}-3}{2}=39 ; 3^{5}=79+81+83$. | 39 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,648 |
16. Any composite (and only composite) odd number can be represented as the sum of consecutive odd numbers. Which even composite numbers have the same property? | 16. $(2 a+1)+(2 a+3)+\ldots+(2 a+2 n-1)=2 k+$ $+1=n_{1} n_{2},(2 a+n) n=n_{1} n_{2}$, we can take $n=n_{2}$, if $n_{2}<n_{1}$.
Among the even composite numbers, those that have this property are multiples of 4. | multiples\of\4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,649 |
19. The sum of two consecutive triangular numbers is a perfect square. | 19. If the number $k$ is triangular, then $k=\frac{n(n+1)}{2}$, hence $n=\frac{-1+\sqrt{1+8 k}}{2}$. Therefore, $8 k+1$ must be a perfect square; and since a perfect square cannot end in any of the digits $2,3,7,8$, the number $k$ cannot end in any of the digits $2,4,7,9$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,652 |
23. Numbers of the form $\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3}$ are called tetrahedral. Show that the sum of the first $n$ triangular numbers is equal to the $n$-th tetrahedral number. | 23. $\sum_{1}^{n} \frac{n(n+1)}{2}=\frac{1}{2}\left(\sum_{1}^{n} n^{2}+\sum_{1}^{n} n\right)=\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\right.$ $\left.+\frac{n(n+1)}{2}\right]=\frac{n(n+1)(n+2)}{6}$. | \frac{n(n+1)(n+2)}{6} | Algebra | proof | Yes | Yes | olympiads | false | 40,654 |
25. Three integers $x, y, z$ form an increasing arithmetic progression. Can the cubes of these numbers also form an arithmetic progression? | 25. We have the system of equations $\left\{\begin{array}{l}x^{3}+z^{3}=2 y^{3} \\ x+z=2 y\end{array}\right.$, hence $2 y^{3}=2 y\left(x^{2}-x z+z^{2}\right) ; \quad$ therefore,
1) $y=0$, then $x=-z$.
2) $y^{2}=x^{2}-x z+z^{2}$; substituting $y=\frac{x+z}{2}$ here, we get $(x-z)^{2}=0, x=z$; hence, $y=z$. Thus, we obta... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,656 |
26. Investigate the possibility of equality:
$$
k y^{3}=x^{3}+z^{3}
$$
where $k$ is an integer, if $x, y, z$ are integers forming an increasing arithmetic progression. | 26. From the system $\left\{\begin{array}{l}x^{3}+z^{3}=k y^{3} \\ x+z=2 y\end{array}\right.$ we find $k y^{3}=2 y\left(x^{2}-\right.$ $\left.-x z+z^{2}\right) ;$ therefore, 1) $\left.y=0, x=-z ; 2\right)(k-8) x^{2}+$ $+2(k+4) x z+(k-8) z^{2}=0, \frac{x}{z}=\frac{-(k+4) \pm \sqrt{24(k-2)}}{k-8}$, if $k \neq 8, k-2=6 t^... | (1-),,(1+) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,657 |
28. The sum of the squares of five consecutive integers cannot be a perfect square. | 28. $N=(x-2)^{2}+(x-1)^{2}+x^{2}+(x+1)^{2}+(x+2)^{2}=$ $=5\left(x^{2}+2\right)$ : since $x^{2}$ cannot end in the digits 3 and 8, $x^{2}+2$ is not divisible by 5, therefore $N$ cannot be a perfect square. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,658 |
30. If the difference of two numbers $a$ and $b$ is divisible without remainder by a number $c$, then the remainders of the division of the numbers $a$ and $b$ by $c$ are equal to each other. The converse theorem is also true. | 30. Let $a=c t_{1}+m, \quad b=c t_{2}+n, a-b=c \cdot\left(t_{1}-\right.$ $\left.-t_{2}\right)+m-n$ (1); by the condition, $a-b$ is divisible by $c$, so $n$ is divisible by $c$; but since $m<c$ and $n<c$, then $m-n=0$, i.e., $m=n$. Conversely: if $m=n$, then from (1) we find $a-b=$ $=c\left(t_{1}-t_{2}\right) ;$ therefo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,659 |
31. The number 1270, when divided by a certain number, gives a quotient of 74. Find the divisor and the remainder. | 31. From the system of inequalities $\left\{\begin{array}{l}1270-74 x>0 \\ 1270-74 x<x\end{array}\right.$ we find $x=17, r=12$.
30 | 17,r=12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,660 |
33. The sum of the cubes of three consecutive numbers is divisible by 9. | 33. $A=(a-1)^{3}+a^{3}+(a+1)^{3}=3 a\left(a^{2}+2\right)$. If $a$ is divisible by 3, then $A$ is divisible by 9. If $a$ is not divisible by 3, then $a=$ $=3 t \pm 1 ; A=3(3 t \pm 1)\left(9 t^{2} \pm 6 t+3\right)=9(3 t \pm 1)\left(3 t^{2} \pm 2 t+\right.$ $+1)$ is divisible by 9.
Alternatively: $A=a^{3}+(a+1)^{3}+(a+2)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,661 |
35. The number $a(a+1)(2a+1)$ is divisible by 6 for any integer $a$. | 35. Instruction. $a(a+1)(2a+1)=a(a+1)(a+2)+$ $+(a-1)a(a+1)$. | Number Theory | proof | Yes | Yes | olympiads | false | 40,662 | |
43. If the sum of the squares of three numbers is a perfect square, then at least two of these numbers will be even. | 43. Consider the equation $x^{2}+y^{2}+z^{2}=u^{2}$. If we assume that only one of the numbers $x, y, z$ is even, for example $z$, then we get $(2 m+1)^{2}+(2 n+1)^{2}+(2 k)^{2}=4\left(m^{2}+\right.$ $\left.+n^{2}+k^{2}+m+n\right)+2$, which cannot be a perfect square.
If we assume that all numbers $x, y, z$ are odd, t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,666 |
46*. Show that $\underbrace{444 \ldots}_{n \text { times }} \underbrace{488 \ldots 8}_{n-1 \text { times }} 9=(\underbrace{666 \ldots 67)^{2}}_{n-1 \text { times }}$. | 46. $\underbrace{444 \ldots}_{n \text { digits }} \underbrace{488 \ldots 89}_{n-1 \text { digits }}=4 \cdot 10^{2 n-1}+4 \cdot 10^{2 n-2}+\ldots+$
$+4 \cdot 10^{n}+8 \cdot 10^{n-1}+\ldots+8 \cdot 10+9=4 \cdot \frac{10^{2 n}-10^{n}}{9}+$ $+8 \cdot \frac{10^{n}-10}{9}+9=\left(\frac{2 \cdot 10^{n}+1}{3}\right)^{2}$
$(\u... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,667 |
47. The number $A$ consists of $2 n$ digits, all equal to 4; the number $B$ consists of $n+1$ digits, all equal to 2; the number $C$ consists of $n$ digits, all equal to 8. Show that the number $A+B+C+7$ is a perfect square. | 47. Instruction. Show that $A+B+C+7=$ $=\left(\frac{2 \cdot 10^{n}+7}{3}\right)^{2}$ | (\frac{2\cdot10^{n}+7}{3})^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 40,668 |
48. If a number written in the decimal system is read in the ternary system and if the resulting number is divisible by 7, then the number written in the decimal system is also divisible by 7. | 48. Let $a \cdot 3^{n-1}+b \cdot 3^{n-2}+\ldots+l$ be divisible by 7, then $a(10-7)^{n-1}+b(10-7)^{n-2}+\ldots+l=7 k;$ by expanding using the binomial theorem, we find that $a \cdot 10^{n-1}+b \cdot 10^{n-2}+\ldots+l$ is divisible by 7. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,669 |
52. Find the remainder when the number $50^{13}$ is divided by 7. | 52. $50^{13}=(49+1)^{13}=49 n+1$ (by the binomial theorem); therefore, the remainder of the division of $50^{13}$ by 7 (and also by 49) is 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,671 |
53. If $5 m+3 n$ is divisible by 11, then $9 m+n$ is also divisible by 11 ( $m$ and $n$ - integers). | 53. $3(9 m+n)=(5 m+3 n)+22 m$; from this it is clear that $9 m+n$ is divisible by 11. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,672 |
54. If $a-b$ is divisible by $c$, then $a^{n}-b^{n}$ is also divisible by $c$. | 54. $a=c t_{1}+m, b=c t_{2}+m$ (see № 30); from here $a^{n}-$ $-b^{n}=\left(c t_{1}+m\right)^{n}-\left(c t_{2}+m\right)^{n}=c k \quad$ (by the binomial theorem); therefore, $a^{n}-b^{n}$ is divisible by $c$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,673 |
55*. A number consisting of $3^{n}$ identical digits is divisible by $3^{n}$.
The product of $n$ consecutive numbers is divisible without remainder by $1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$. This can be explained, incidentally, as follows:
$$
\frac{a(a-1)(a-2) \ldots[a-(n-1)]}{1 \cdot 2 \cdot 3 \cdot \ldots \cdot n... | 55. For $n=1$ the theorem is true, since the sum of the digits of the number $\overline{a a a}$ is divisible by 3. Let the theorem be true for some value of $n$; we will show that it is true for $n+1$:
$$
\begin{aligned}
& \underbrace{\overline{a a a \ldots a}}_{3^{n+1} \text { digits }}=\overline{a a \ldots a} \under... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,674 |
56. If $p$ is a prime number, then $(a+b)^{p}-\left(a^{p}+b^{p}\right)$ is divisible by $p$. | 56. $(a+b)^{p}-\left(a^{p}+b^{p}\right)=p a^{p-1} b+\frac{p(p-1)}{1 \cdot 2} a^{p-2} b^{2}+$ $+\ldots+\frac{p(p-1)(p-2) \ldots(p-n+1)}{1 \cdot 2 \cdot 3 \ldots n} a^{p-n} b^{n}+\ldots+p a b^{p-1} ;$
all binomial coefficients of this expansion are integers, multiples of the prime number $p$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,675 |
58. If $p$ is an odd prime number, then $2^{p-1}-1$ is divisible by $p$. | 58. $2\left(2^{p-1}-1\right)=2^{p}-2=(1+1)^{p}-2=p+\frac{p(p-1)}{1 \cdot 2}+$ $+\frac{p(p-1)(p-2)}{1 \cdot 2 \cdot 3}+\ldots+p=p N$; therefore, $2\left(2^{p-1}-1\right)$ is divisible by $p$, and since $p$ is odd, $2^{p-1}-1$ is divisible by $p$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,676 |
59*. If $f(x)$ is an integer polynomial with integer coefficients in $x$ and, moreover, $f(0)$ and $f(1)$ are odd numbers, then the equation $f(x)=0$ has no integer roots.
When divided by 5, the remainders can be $0,1,2,3$, 4; therefore, every integer can be represented in one and only one of the following forms: $5 t... | 59. Let $f(x)=a x^{n}+b x^{n-1}+c x^{n-2}+\ldots+k x+l$; therefore, $f(0)=l, f(1)=a+b+c+\ldots+k+l$.
Since by condition $f(0)$ and $f(1)$ are odd numbers, then $l$ is odd, and $a+b+c+\ldots+k$ is even. The equation $f(x)=0$ cannot have an even root, because then $l$ would be even. We will now show that the equation $f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,677 |
60. For what values of $n$ is the number $3\left(n^{2}+n\right)+7$ divisible by 5? | 60. First method. Since $3\left(n^{2}+n\right)+7=3\left(n^{2}+n\right)+2+5$, it is sufficient to find those values of $n$ for which $3\left(n^{2}+n\right)+2$ is divisible by $5:$
$$
3\left(n^{2}+n\right)+2=5 t, \quad n=\frac{-3 \pm \sqrt{15(4 t-1)}}{6}
$$
and if $4 t-1=15 t_{1}^{2}$, then $n=\frac{-1 \pm 5 t_{1}}{2}$... | 5t+2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,678 |
75. A number of the form $4 n+3$ cannot be a perfect square for any integer $n$. | 75. An odd number can only be the square of an odd number, but $(2 t+1)^{2}=4 t(t+1)+1=4 t_{1}+1$; therefore, the number $4 n+3$ cannot be a perfect square. This can also be explained as follows: from the equation $4 n+3=4 t_{1}+$ +1 follows the equation $2\left(t_{1}-n\right)=1$, which is impossible for integers $t_{1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,681 |
76. The number $a^{2}+a+2$ is not divisible by 15 for any integer $a$. | 76. $a^{2}+a+2=15 m, a=\frac{-1 \pm \sqrt{60 m-7}}{2}$, but the number $60 m-7$ cannot be a perfect square, as it ends in the digit 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,682 |
77. Show that if $p \geqslant 5$ and $2 p+1$ are prime numbers, then $4 p+1$ is a composite number.
Problem No. 58 is a special case of Fermat's Little Theorem: if $\boldsymbol{a}$ is not divisible by a prime number $p$, then $a^{p-1}-1$ is divisible by $p$. In Problem No. 58, we proved this theorem for $a=2$. In the ... | 77. Since $p$ is a prime number greater than 3, it can be represented as $3 m \pm 1$. If $p=3 m+1$, then $2 p+1=3(2 m+1)$ - a composite number, which contradicts the condition, hence $p=3 m-1$; then $4 p+1=3(4 m-1)$, i.e., a composite number. | 4p+1=3(4-1) | Number Theory | proof | Yes | Yes | olympiads | false | 40,683 |
86. If the sum of the squares of two numbers is divisible by 7, then each of these numbers is divisible by 7. | 86. The numbers $a$ and $b$ are both multiples of or both not multiples of 7. If we assume that they are both not multiples of 7, then by Fermat's theorem, $a^{6}-1$ and $b^{6}-1$, and therefore $a^{6}+b^{6}-2$ are multiples of 7, but $a^{6}+b^{6}-2=\left(a^{2}+b^{2}\right)\left(a^{4}-a^{2} b^{2}+b^{4}\right)-2$, which... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,687 |
94. For what values of $x$ is the number $2^{x}-1$ divisible by 5? | 94. Instruction. Consider the remainders of the division of $x$ by 6.
Answer. $x=6 t+1, x=6 t+4$, where $t$ is a non-negative integer. | 6t+1,6t+4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,693 |
97. Show that $a^{p}-b$ is divisible by the prime $p$, if $a-b$ is a multiple of $p$. | 97. $a^{p}-b=a^{p}-a+(a-b)$.
The above text translated into English, keeping the original text's line breaks and format, is as follows:
97. $a^{p}-b=a^{p}-a+(a-b)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,694 |
98. For what integer values of $x$ is the number $a\left(x^{3}+a^{2} x^{2}+\right.$ $\left.+a^{2}-1\right)$ divisible by 6 for any integer $a$?
As is known, the difference $a^{n}-b^{n}$ is divisible by $a-b$ for any natural number $n$; $a^{n}-b^{n}$ is divisible by $a+b$ for even $n$; $a^{n}+b^{n}$ is divisible by $a+... | 98. $x=3 t$ and $x=3 t-a^{2}$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,695 |
99. Numbers of the form $2 \cdot 2^{2^{2 n}}+1$ and $3 \cdot 2^{2 n}+1$ are composite for $n>0$. | 99. Since $2 \cdot 2^{2^{n}}=2^{2^{n}+1}$ and, moreover, $2^{n}+1$ is odd, then $2 \cdot 2^{2^{n}}+1$ is divisible by $2+1$ and, therefore, is composite.
Since $N=3 \cdot 2^{2^{2 n}}+1=3 \cdot 2^{4^{n}}-6+7=6\left(2^{4^{n}-1}-\right.$ $-1)+7$ and $4^{n}-1$ is divisible by $4-1$, i.e., 3, then $2^{4^{n}-1}-1=$ $=2^{3 k... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,696 |
101. The number $N=2^{4 n+1}-4^{n}-1$ is divisible by 9 for an integer $n \geqslant 0$. The number $8 N+9$ is a perfect square.
10 | 101. $N=2^{4 n+1}-4^{n}-1=2 \cdot 4^{2 n}-4^{n}-1=\left(2 \cdot 4^{n}+\right.$ $+1)\left(4^{n}-1\right)=\left(2^{2 n+1}+1\right)\left(4^{n}-1\right)$, both factors are multiples of 3, therefore, $N$ is a multiple of 9. In addition, $8 N+9=$ $=\left(4^{n+1}-1\right)^{2}$. | (4^{n+1}-1)^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,697 |
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