problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|
694. Find the slope of a line that forms an angle of $45^{\circ}$ with the horizontal. | $\triangleright$ We already know that such a line is given by the equation $y=x$, so the slope is $1 . \triangleleft$ | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
735. Find the greatest divisor of the pair of numbers 123456789 and 987654321 (in other words, the greatest common measure of segments of such lengths). | $\triangleright$ In principle, this problem can be solved by brute force (and even in a reasonable time if you have a computer at hand), trying all numbers from 1 to 123456789 (larger numbers clearly do not fit).
However, it can also be done without a computer, using the Euclidean algorithm. The number 987654321 can b... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the minimum value of the expression $2 x+y$, defined on the set of all pairs $(x, y)$ satisfying the condition
$$
3|x-y|+|2 x-5|=x+1
$$ | 1.I.1. The following figure shows a set defined by the equation $3|x-y|+|2 x-5|=x+1$.
Let $C=2 x+y$. The problem requires finding the minimum value of $C$ for all points in the depicted set... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the minimum value of the expression
$$
\sqrt{a^{2}+1}+\sqrt{b^{2}+9}+\sqrt{c^{2}+25}
$$
if it is known that $a+b+c=12$. | Solution 1. Introduce vectors $\boldsymbol{m}(a ; 1), \boldsymbol{n}(b ; 3)$ and $\boldsymbol{p}(c ; 5)$ and set $\boldsymbol{q}=\boldsymbol{n}+\boldsymbol{m}+\boldsymbol{p}$. Since $a+b+c=12$, then $\boldsymbol{q}(12 ; 9)$. Since $|\boldsymbol{q}|=|\boldsymbol{n}+\boldsymbol{m}+\boldsymbol{p}| \leqslant|\boldsymbol{n}... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Find the minimum value of the function
$$
f(x)=4^{x}+4^{-x}-2^{x+1}-2^{1-x}+5
$$
Solution: Since
$$
\begin{aligned}
4^{x}+4^{-x}-2^{x+1}-2^{1-x}+5=2^{2 x}+ & 2^{-2 x}-2\left(2^{x}+2^{-x}\right)+5= \\
=2^{2 x}+2 \cdot 2^{x} \cdot 2^{-x}+2^{-2 x} & -2-2\left(2^{x}+2^{-x}\right)+5= \\
& =\left(2^{x}+2^{-x}\right)^{2... | 1.I.8. Comment. The following error was made in the given solution. Since $2^{x}>0$, the range of the function $y=2^{x}+2^{-x}$ is the interval $[2 ;+\infty)$. Therefore, it is necessary to find the minimum value of the function $g(t)=t^{2}-2 t+3$ on the interval $[2 ;+\infty)$. On this interval, the function $g(t)$ is... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of solutions to the equation $x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{5}\right]$ (here, as usual, $[x]$ is the integer part of the number $x$, i.e., the greatest integer not exceeding the number $x$). | 1.II.3. Let $x=30 k+d$, where $k \in \mathbb{Z}$, and $d \in\{0,1, \ldots, 29\}$. Substituting this expression into the given equation, we get
$$
30 k+d=15 k+\left[\frac{d}{2}\right]+10 k+\left[\frac{d}{3}\right]+6 k+\left[\frac{d}{5}\right]
$$
or $k=d-\left[\frac{d}{2}\right]-\left[\frac{d}{3}\right]-\left[\frac{d}{... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Point $A_{1}$ lies on side $B C$ of triangle $A B C$, point $B_{1}$ lies on side $A C$. Let $K$ be the intersection point of segments $A A_{1}$ and $B B_{1}$. Find the area of quadrilateral $A_{1} C B_{1} K$, given that the area of triangle $A_{1} B K$ is 5, the area of triangle $A B_{1} K$ is 8, and the area of tri... | 1.II.5. We will use the following consideration. If a line passes through the vertex of a triangle, then the ratio of the areas of the parts into which it divides the triangle is equal to the ratio of the lengths of the segments into which it divides the opposite side, $\frac{S_{1}}{S_{2}}=\frac{m}{n}$. Let's connect v... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate $\left[\sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+\ldots+\sqrt[2009]{\frac{2009}{2008}}\right]$ (here $[x]$ is the integer part of the number $x$, i.e., the greatest integer not exceeding the number $x$). | 2.I.3. Since each of the numbers is greater than 1, their sum is greater than 2008. Let's prove that this sum is less than 2009. By the inequality between the geometric mean and the arithmetic mean, we have
$$
\sqrt[n+1]{\frac{n+1}{n}}=\sqrt[n+1]{\left(1+\frac{1}{n}\right) \cdot 1^{n}}<\frac{n+1+\frac{1}{n}}{n+1}=1+\f... | 2008 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Compute $\operatorname{arctg}\left(\operatorname{tg} 65^{\circ}-2 \operatorname{tg} 40^{\circ}\right)$. | 2.II.1. The answer follows from a chain of trigonometric transformations:
$$
\begin{aligned}
\operatorname{tg} 65^{\circ}-2 \operatorname{tg} 40^{\circ}=\operatorname{ctg} 25^{\circ}-2 \operatorname{ctg} 50^{\circ} & =\frac{1}{\operatorname{tg} 25^{\circ}}-\frac{2}{\operatorname{tg} 50^{\circ}}= \\
& =\frac{1}{\operat... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Let $\mathcal{S}$ be the smallest subset of the set of integers satisfying the following properties:
1) $0 \in \mathcal{S}, 2)$ if $x \in \mathcal{S}$, then $3 x \in \mathcal{S}$ and $3 x+1 \in \mathcal{S}$.
Find the number of non-negative integers in the set $\mathcal{S}$ that do not exceed 2009. | Solution 1. We will write the numbers from the set $\mathcal{S}$ in the ternary numeral system. Then, the transition from number $x$ to number $3x$ means appending 0 to the right of the "ternary representation" of the number $x$, and the transition from $x$ to $3x+1$ means appending 1 to the right. Therefore, a number ... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The radii of the excircles of a certain triangle are 2, 3, and 6 cm. Find the radius of the circle inscribed in this triangle. | 3.II.4. First, let's derive the formula for the radius of the excircle. Let $P$ be the center of the circle with radius $r_{3}$, which touches the side $AB$ and the extensions of the sides $CA$ and $CB$ of triangle $ABC$. From the equality $S_{PAB} + S_{ABC} = S_{PAC} + S_{PBC}$, it follows that
$$
S_{ABC} = S = \frac... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Compute $\operatorname{tg} \alpha$, if $3 \operatorname{tg} \alpha-\sin \alpha+4 \cos \alpha=12$. | 4.II.1. We have,
$$
\begin{aligned}
& 3 \operatorname{tg} \alpha - \sin \alpha + 4 \cos \alpha = 12 \\
& \Longleftrightarrow \operatorname{tg} \alpha (3 - \cos \alpha) - 4 (3 - \cos \alpha) = 0 \Longleftrightarrow \\
& (\operatorname{tg} \alpha - 4)(3 - \cos \alpha) = 0 \Longleftrightarrow \operatorname{tg} \alpha = 4... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let $x$ and $y$ be positive numbers whose sum is 2. Find the maximum value of the expression $x^{2} y^{2}\left(x^{2}+y^{2}\right)$. | Solution 1. The idea is suggested by the formula for $(x+y)^{4}$. Since
$(x+y)^{4}-8 x y\left(x^{2}+y^{2}\right)=x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}=(x-y)^{4} \geqslant 0$, then $8 x y\left(x^{2}+y^{2}\right) \leqslant(x+y)^{4}$. It is clear that $4 x y \leqslant(x+y)^{2}$. Therefore,
$$
32 x^{2} y^{2}\left... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Solve the equation $\frac{3}{\log _{2} x}=4 x-5$.
Answer: 2.
Solution: The function $y=\log _{2} x$ is increasing, therefore $y=\frac{3}{\log _{2} x}$ is a decreasing function. On the other hand, the function $y=4 x-5$ is increasing, therefore, the given equation has no more than one root. By trial, we find that $... | 5.I.8. Comment. The error in the given solution is that the function $y=\frac{3}{\log _{2} x}$ is decreasing not on its entire domain, but only on each of the intervals $(0 ; 1)$ and $(1 ;+\infty)$. Therefore, this equation has no more than one root in each of them. In addition to $x=2$, the solution to the equation is... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Solve the equation $27^{x}-7 \sqrt[3]{7 \cdot 3^{x}+6}=6$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution 1. Let $z=\sqrt[3]{7 y+6}$ and transition to the system
$$
\left\{\begin{array}{l}
y^{3}=7 z+6 \\
z^{3}=7 y+6
\end{array}\right.
$$
Then $y^{3}-z^{3}=7(z-y)$, from which it follows that $y=z$, since $y^{2}+y z+z^{2}$ cannot be equal to -7. As a result, we get the equation $y^{3}-7 y-6=0$, the roots of which ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a row, all natural numbers less than a billion that have exactly 13 natural divisors (including one and the number itself) were written down. How many of them have an even sum of digits? | 6.I.3. If $n=p_{1}^{s_{1}} p_{2}^{s_{2}} \ldots p_{k}^{s_{k}}$, then the number of divisors of the number $n$ is $\left(s_{1}+1\right)\left(s_{2}+1\right) \ldots\left(s_{k}+1\right)$. Since the number of divisors of the given number is a prime number, the number itself is a power of a prime number; in this case $n=p^{1... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $3^{x^{2}+x-2}-3^{x^{2}-4}=80$. | 6.II.1. Let's rewrite the equation as $3^{x^{2}-4}\left(3^{x+2}-1\right)=80$ and set $f(x)=3^{x^{2}-4}\left(3^{x+2}-1\right)$. If $x \leqslant -2$, then $f(x) \leqslant 0$, hence the equation has no solutions on the interval $(-\infty; -2]$. If $-2 < x \leqslant 0$, then $0 < 3^{x+2} - 1 \leqslant 8$ and $3^{x^{2}-4} \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Let the function $f(x)$ be defined on $\mathbb{R}$, and for any $x$, the condition $f(x+2) + f(x) = x$ holds. It is also known that $f(x) = x^3$ on the interval $(-2; 0]$. Find $f(2012)$. | 6.II.2. Since $f(x+2)+f(x)=x$, then $f(x+4)+f(x+2)=x+2$, therefore
$$
f(x+4)=x+2-f(x+2)=x+2-(x-f(x))=f(x)+2 .
$$
Hence, $f(2012)=f(0)+2 \cdot 503=1006$. | 1006 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the smallest prime number $p$ such that $n^{2}+n+11$ is divisible by $p$ for some integer $n$.
untranslated text is preserved in terms of line breaks and format, as requested. | 6.II.5. Since $n(n+1)$ is an even number, the number $n^{2}+n+11$ is odd, and therefore this number is not divisible by 2. Since $n^{2}+n+11=$ $(n-1)^{2}+1+3n+9$, the number $n^{2}+n+11$ is not divisible by 3, because no perfect square can have a remainder of 2 when divided by 3. Since $n^{2}+n+11=(n-2)^{2}+2+5n+5$, by... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. There are 9 sticks of different lengths from 1 cm to 9 cm. What are the side lengths of the squares that can be formed from these sticks, and in how many ways can they be formed? Ways of forming a square are considered different if different sticks are used and not necessarily all of them.
Can squares with side len... | 1. Solution. One way to form squares is to have side lengths of $7 \mathrm{~cm}, 8 \mathrm{~cm}, 10 \mathrm{~cm}, 11 \mathrm{~cm}$. In five ways, squares can be formed, each with a side length of $9 \mathrm{~cm}$. Thus, there are 9 ways to form the squares.
Hint. The side length of the square must be greater than 6 cm... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The incident took place in 1968. A high school graduate returned from a written university entrance exam and told his family that he couldn't solve the following problem:
Several identical books and identical albums were bought. The books cost 10 rubles 56 kopecks, and the albums cost 56 kopecks. The number of book... | 4. S o l u t i o n. Since the book is more expensive than a ruble, no more than 10 books were bought. Moreover, it is clear that no fewer than 7 books were bought. By checking, we find that the number 1056 is divisible by 8 and not divisible by $7,9,10$. A n s w e r: 8 books. | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Twelve people are carrying 12 loaves of bread. Each man carries 2 loaves, each woman carries half a loaf, and each child carries a quarter of a loaf, and all 12 people are involved in the carrying. How many men, how many women, and how many children were there? | 7. 5 men, 1 woman, and 6 children. Show that the number of men cannot be, first, less than five, and second, more than five. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. Find the sum of the areas of all different rectangles that can be formed using 9 squares (not necessarily all), if the side of each square is $1 \mathrm{~cm}$. | 9. Solution. Let $a$ and $b$ be the lengths of the sides of the rectangle. If $a=1$, then $b=\{1 ; 2 ; \ldots ; 9\}$. If $a=2$, then $b=\{2 ; 3 ; 4\}$. If $a=3$, then $b=3$. The sum of the areas of all rectangles is $(1+2+\ldots+9)+2 \cdot(2+3+4)+3 \cdot 3=72$ (sq. cm). (MvSh) | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Find the sum: $-100-99-98-97-\ldots+100+101+102$.
. | 5. Solution. Since $-100+100=-99+99=\ldots$ $\ldots=1+1=0$, the considered sum is equal to $102+101=203$. | 203 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. On the number line, four points are marked. Point $A$ corresponds to the number -3, point $B$ corresponds to the number -5, and point $C$ corresponds to the number 6. Find the fourth number corresponding to point $K$ under the following condition: if the direction of the number line is reversed, the sum of the new ... | 10. Solution. When the direction of the number line is changed, the sign of each number (except, of course, zero) changes to the opposite. Since the sum did not change in this case, it can only be equal to zero. Therefore, the fourth number sought is $0-(-5-3+6)=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. Several points are marked on a number line. The sum of the numbers corresponding to these points is $-1.5$. Each of the marked points was moved two units to the left on the number line, and therefore the sum of the numbers became $-15.5$. How many points were there? | 13. Solution. When a point is moved two units to the left on the number line, the number corresponding to this point decreases by 2 units. The sum of all numbers decreased by $-1.5-(-15.5)=$ $=14$, so there were a total of $14: 2=7$ points. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. What is the greatest power of 7 that divides the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 999 \cdot 1000$? | 4. S o l u t i o n. The number of integers divisible by 7 in the first thousand natural numbers is 142, those divisible by $49=7^{2}$ is 20, and those divisible by $343=7^{3}$ is 2. If the power of seven is greater than three, the corresponding power will be greater than 1000. Therefore, none of the factors of the prod... | 164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. What digits do the decimal representations of the following numbers end with:
1) $135^{x}+31^{y}+56^{x+y}$, if $x \in N, y \in N$
2) $142+142^{2}+142^{3}+\ldots+142^{20}$
3) $34^{x}+34^{x+1}+34^{2 x}$, if $x \in N$. | 12. 3) $\mathrm{Solution.} \mathrm{If} \mathrm{a} \mathrm{number} \mathrm{ends} \mathrm{in} \mathrm{four,} \mathrm{then}$ even powers of it end in 6, and odd powers end in 4. Therefore, one of the first two terms ends in four, and the other ends in six. The third term ends in six, so the decimal representation of the s... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. A checker can move in one direction along a strip divided into cells, moving either to the adjacent cell or skipping one cell in a single move. In how many ways can it move 10 cells? 11 cells? | 6. Solution. A checker can move to one cell in one way, to two cells in two ways, to three cells $1+2=3$ ways, to four cells $2+3=5$ ways, and so on. It can move to the tenth cell in 89 different ways, and to the 11th cell in 144 different ways. (VZMSh) | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12*. There are 10 light bulbs, each of which can be either on or off. How many different signals can be transmitted using these light bulbs? | 12. $2^{10}=1024$ different signals. 13. $4^{5} ; 3 \cdot 4^{4}$. | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2(!). Let the formula for the \(n\)-th term of the sequence be \(x=n^{2}\). Write down the first 10 terms of this sequence. Write down the sequence of differences between the second and first terms, the third and second terms, and so on. Write down another sequence of differences between the second and first, the third... | 2. The results of the calculations can be recorded in the form of the following table:
 | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. For what least natural \(m\) are the numbers of the form:
1) \(m^{3}+3^{m}\)
2) \(m^{2}+3^{m}\)
divisible by 7? | 10. Solution. Let \(r_{1}\) be the remainder of the division of \(m^{3}\) by 7, and \(r_{2}\) be the remainder of the division of \(3^{m}\) by 7. Clearly, the number \(m^{3}+3^{m}\) will be divisible by 7 when the sum of the remainders \(r_{1}+r_{2}\) equals 7. Let's construct the following table:
\begin{tabular}{|c|c... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. Simplify the expression:
\[
-(-(-(\ldots-(-1) \ldots)))
\]
which contains 200 pairs of parentheses.
If solving the problem is difficult, consider the numbers:
\[
\begin{aligned}
& -(-1)=\ldots, \\
& -(-(-1))=\ldots
\end{aligned}
\]
notice the pattern and draw a conclusion.
It is appropriate to inform students... | 12. -1. 13. 1) \(x=4n-1\); 2) \(y=(-1)^{n+1}\); 3) \(z=n^2+1\); 4) \(t=2^n-1\). | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8*. Among the numbers from 1 to 1000, how many are divisible by 4 but do not have the digit 4 in their representation? | 8. 162 numbers. Instruction. Count how many numbers are divisible by 4 and have the digit 4 in their representation. Subtract the result from 250 - the number of numbers from 0 to 999 (from 1 to 1000) that are divisible by 4. (KvN, 1972, No. 4, p. 82.) | 162 | Other | math-word-problem | Yes | Yes | olympiads | false |
9. How many six-digit numbers exist where each digit in an even position is one greater than the digit to its left (digits are numbered from left to right)? | 9. \(8 \cdot 9 \cdot 9=648\). Hint. The six-digit numbers in question are determined by the choice of the first, third, and fifth digits. (Kvant, 1972, No. 5, p. 81; MSU, 1971) | 648 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14*. An Aeroflot cashier needs to deliver tickets to five groups of tourists. Three of these groups are staying at the hotels "Druzhba", "Rossiya", and "Minsk". The address of the fourth group will be provided by the tourists from "Rossiya", and the address of the fifth group will be provided by the tourists from "Mins... | 14. 30. Note. If there are no restrictions on the order of the tour, then there are \(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120\) different ways to tour the hotels. Show that each restriction reduces the number of ways by half. (MTG, 1970, No. 6, p. 72.) | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
16*. Among the first ten thousand numbers, how many of them end in 1 and can be represented in the following form: \(8^{m}+5^{n} ?(m \in N, n \in N)\) | 16. Five numbers. Hint. Show that \(m=4\). (KvN, 1972, No. 5, p. 81 .) | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Calculate at \(x=7\):
\((x-4)^{(x-5)^{(x-6)}}{ }^{(x+6)^{(x+5)}}\)
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6. Calculate at \(x=7\):
\((x-4)^{(x-5)^{(x-6)}}{ }^{(x+6)^{(x+5)}}\) | 6. When \(x=7\), the given expression equals \(3^{2^{1}}=9\). | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12(!). Find the numerical value of the monomial \(0.007 a^{7} b^{9}\), if \(a=-5, b=2\). | 12. Solution. Since the second factor is negative and the third is positive, the numerical value of the monomial is negative. We have:
\(-0,007 \cdot 5^{7} \cdot 2^{9}=-0,007 \cdot 4 \cdot(2 \cdot 5)^{7}=-280000\). | -280000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. Given the monomial \((-1)^{n} a^{n-2} b^{9-n}\). Write in a row the set of all possible forms of this monomial for different permissible natural values of the exponents. | 14. Instruction. It is clear that \(n>2\) and \(n<9\), i.e., \(n=\{3 ; 4\); \(5 ; 6 ; 7 ; 8\}\). For these values of \(n\), six different monomials will be obtained. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16*. Find a two-digit number that has the property that the cube of the sum of its digits is equal to the square of the number itself. | 16. Solution. Let \(x\) be the sum of the digits of a two-digit number \(y\). According to the condition, \(x^{3}=y^{2}\). This equality in natural numbers is possible only when \(x=z^{2}\) and \(y=z^{3}\), where \(z \in N\). Since the sum of the digits of a two-digit number is no more than 18, then \(z^{2} \leqslant 1... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18*. Which three-digit number has the greatest number of divisors? | 18. 840. This number has 32 divisors. | 840 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. In a three-digit number, the digit in the hundreds place is 2 more than the digit in the units place. Find the difference between this number and the number formed by the same digits but in reverse order. | 6. Solution. Let \(x\) be the tens digit and \(y\) be the hundreds digit of a three-digit number. Then the units digit will be \(y-2\), and thus the desired number is
\(100 y + 10 x + y - 2 - 100(y - 2) - 10 x - 2 = \ldots = 198\). | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Find \(k\), if it is known that for any \(x\):
\[
\begin{array}{r}
a x^{2}+b x+c \\
+b x^{2}+a x-7 \\
k x^{2}+c x+3 \\
\hline x^{2}-2 x-5
\end{array}
\] | 7. Solution.
\[
c=-5-(-7+3)=-1
\]
\[
\begin{aligned}
a+b & =-2-(-1) \\
k & =1-(a+b)=1-(-1)=2
\end{aligned}
\] | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Find the sum of the values of the polynomial \(x^{5}-1.7 \cdot x^{3}+2.5\) at \(x=19.1\) and \(x=-19.1\). | 11. Solution. If the sign of \(x\) is changed to the opposite, the given polynomial will take the form: \(-x^{5}+1.7 x^{3}+2.5\). The sum of the given polynomial and the obtained one is 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given a five-digit number. If you prepend a seven to the number, the resulting six-digit number will be 5 times larger than the six-digit number obtained by appending a seven to the end. Find this five-digit number. | 6. Solution. Let \(x\) be a five-digit number. If we prepend a seven to it, we get the number \(7 \cdot 10^{5} + x\). If we append a seven to it, we get the number \(10 x + 7\). According to the problem,
\[
7 \cdot 10^{5} + x = 5(10 x + 7)
\]
Solving this equation, we get \(x = 14285\). | 14285 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13*. Calculate:
1) \(x^{4}-2 x^{3}+3 x^{2}-2 x+2\) given \(x^{2}-x=3\)
2) \(2 x^{4}+3 x^{2} y^{2}+y^{4}+y^{2} \quad\) given \(x^{2}+y^{2}=1\). | 13. 1) \( \mathrm{P} \) is a solution. \( x^{2}\left(x^{2}-x\right)-x\left(x^{2}-x\right)+2\left(x^{2}-x\right)+2= \) \( =3 x^{2}-3 x+2 \cdot 3+2=3\left(x^{2}-x\right)+8=3 \cdot 3+8=17 \);
2) Solution. \( 2 x^{4}+2 x^{2} y^{2}+x^{2} y^{2}+y^{4}+y^{2}=2 x^{2}\left(x^{2}+\right. \) \( \left.+y^{2}\right)+y^{2}\left(x^{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Prove the equality:
\[
\begin{aligned}
& 55554 \cdot 55559 \cdot 55552-55556 \cdot 55551 \cdot 55558= \\
= & 66665 \cdot 66670 \cdot 66663-66667 \cdot 66662 \cdot 66669
\end{aligned}
\] | 6. Solution. Let \(a=55555, b=66666\). The solution to the problem reduces to proving the identity
\[
\begin{aligned}
& (a-1)(a+4)(a-3)-(a+1)(a-4)(a+3)= \\
& =(b-1)(b+4)(b-3)-(b+1)(b-4)(b+3)
\end{aligned}
\]
Each part of this equality is independent of the variable and equals 24. | 24 | Algebra | proof | Yes | Yes | olympiads | false |
9. Solve the equation: \(x^{3}+x^{2}+x+1=0\). | 9. Solution. Factoring the left side of the equation, we get:
\[
(x+1)\left(x^{2}+1\right)=0
\]
Since \(x^{2}+1>0\), then \(x+1=0, x=-1\). | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find: 1) \(x^{6}+3 x^{2} y^{2}+y^{6}\), if \(x^{2}+y^{2}=1\); 2) \(x^{4}+\) \(+x^{2} y^{2}+y^{4}\), if \(x^{2}+y^{2}=a\) and \(x y=b\). | 5. 1) \(\mathrm{P}\) is the solution. First method. The given polynomial is equal to
\[
\left(x^{2}+y^{2}\right)\left(x^{4}-x^{2} y^{2}+y^{4}\right)+3 x^{2} y^{2}=\ldots=\left(x^{2}+y^{2}\right)^{2}=1
\]
Second method. Since \(y^{2}=1-x^{2}\), then, performing the substitution, we get that the given polynomial is ide... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Prove that among all natural numbers of the form \(2 p+1\), where \(p\) is a prime number, only one is a perfect cube. Find this number. | 7. Solution. First method. Let \(2 p+1=x^{3}, x \in N\). Then \(2 p=(x-1)\left(x^{2}+x+1\right)\). Since \(x^{2}+x+1\) is an odd number, it is not divisible by 2. Therefore, the given equation is possible only when \(x-1=2\). In this case, \(x=3, p=13\), and the desired number is 27.
Second method. Let \((2 y+1)^{3}=2... | 27 | Number Theory | proof | Yes | Yes | olympiads | false |
14. Find the distance between the points of intersection of the three lines: \(y = 3x\), \(y = 3x - 6\), and \(y = 1975\). | 14. 2. Instruction. Show that the desired distance is equal to the distance between points \(A(0 ; 0)\) and \(B(2 ; 0)\). | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate in the most rational way:
\[
\frac{7^{16}-1}{2402000\left(49^{4}+1\right)}
\]
76 | 1. 2,4. Hint. Factor the numerator and reduce the fraction
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note I added for context, and it should not be part of the translation. Here is the requested translation:
1. 2,4. Hint. Factor the numerator and reduce the fraction | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
26. It is known that in a right-angled triangle \(c^{2}=a^{2}+b^{2}\), where \(a\) and \(b\) are the lengths of the legs; \(c\) is the length of the hypotenuse. Find the acute angles of the triangle if \(\left(\frac{c}{a+b}\right)^{2}=0.5\). | 26. Solution. From the given equality, it follows that \(2 c^{2}=\) \(=(a+b)^{2}\), therefore,
\(2\left(a^{2}+b^{2}\right)=a^{2}+2 a b+b^{2} ;(a-b)^{2}=0, a=b\),
hence the given right triangle is isosceles, and each of its acute angles is \(45^{\circ}\). | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The factory's production over four years has increased by 4 times. By what percentage did the production on average increase each year compared to the previous year? | 3. About \(41 \%\). Hint. If \(x\) is the required percentage, then (see problem 3 of 3): \(\left(1+\frac{x}{100}\right)^{4}=4\) and so on. | 41 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. It is required to fence a rectangular area adjacent to a wall. The fence should have a length of \(60 \mathrm{~m}\). What should be the length and width of this area so that its area is the largest? | 10. Solution. First method. Let the width of the site be \(x \mu\), then its length will be ( \(60-2 x\) ) \(m\), and the area will be \(y=x(60-2 x)\) square meters. Considering the obtained equation as a quadratic equation in terms of \(x\), we get: \(2 x^{2}-60 x+y=0\). The discriminant of this equation is non-negati... | 450 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. The price of a diamond is proportional to the square of its mass. If a diamond is broken into two parts, in which case will the total price of the two parts be the lowest? | 12. Solution. Let the price of a diamond be calculated by the formula \(y=a m^{2}\), where \(m\) is its mass. Let the mass of the first piece be \(\frac{m}{2}+x\). Then the mass of the second piece will be \(m-\frac{m}{2}-x=\frac{m}{2}-x\). The total cost of the two pieces will be:
\[
a\left(\frac{m}{2}+x\right)^{2}+a... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
141. A line passing through the center of the circumscribed circle and the orthocenter of triangle $ABC$ cuts off equal segments $CA_1, CB_1$ from its sides $CA$ and $CB$. Prove that angle $C$ is $60^{\circ}$.
## § 9. Geometric Loci
| 141. Drop a perpendicular $O M_{2}$ from the center $O$ of the circumscribed circle onto $A C$. Triangles $C M_{2} O$ and $C H H_{1}$ are equal, and $C H_{1}=\frac{1}{2} A C$, which means that $\angle C=60^{\circ}$. | 60 | Geometry | proof | Yes | Yes | olympiads | false |
297. The continuations of the medians of a triangle intersect the circumscribed circle at points $A_{1}, B_{1}, C_{1}$. Prove that
$$
\frac{A G}{G A_{1}}+\frac{B G}{G B_{1}}+\frac{C G}{G C_{1}}=3
$$
where $G$ is the centroid of triangle $A B C$. | 297. Using the concept of the power of a point, establish that
$$
\frac{A G}{G A_{1}}+\frac{B G}{G B_{1}}+\frac{C G}{G C_{1}}=\frac{A G^{2}+B G^{2}+C G^{2}}{R^{2}-d^{2}}
$$
But based on Leibniz's theorem, $A G^{2}+B G^{2}+C G^{2}=3 R^{2}-3 d^{2}$. 116 | 3 | Geometry | proof | Yes | Yes | olympiads | false |
38. Calculate the angle between the bisectors of the coordinate angles $x O y$ and $y O z$. | 38. The bisector of angle $x O y$ is defined by the vector $\vec{r}_{1}=\vec{i}+\vec{j}$; the bisector of angle $y O z$ is defined by the vector $\overrightarrow{r_{2}}=\vec{j}+\vec{k}$. Consider the scalar product: $\quad \vec{r}_{1} \cdot \vec{r}_{2}=(\vec{i}+\vec{j}) \cdot (\vec{j}+\vec{k})=1=$ $=r_{1} r_{2} \cos \v... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
45. Write the system of equations of a line passing through the origin and forming equal angles with the three coordinate axes. Determine the magnitude of these angles. How many solutions does the problem have? | 45. According to the problem, $\cos \alpha=\cos \beta=\cos \gamma$ and the equations take the form: $x=y=z$. Further, we have:
$$
\begin{gathered}
\cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha=1 ; \cos ^{2} \alpha=\frac{1}{3} \\
\cos \alpha= \pm \frac{1}{\sqrt{3}} ; \alpha \approx 54^{\circ} 44^{\prime} 8^{\prime... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two plane mirrors serve as the faces of a dihedral angle. A light ray, perpendicular to the edge of the angle and parallel to the first mirror, is reflected from the second mirror, then from the first, then again from the second, again from the first, and finally, reflecting for the fifth time from the second mirror... | 2. Since the light ray falls perpendicular to the edge of the dihedral angle, all reflections occur in one plane perpendicular to the edge (Fig. 81). Since after the third reflection the ray returns along the same path, at this reflection it must be directed perpendicular to the mirror plane. The exterior angle of the ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given three pairwise intersecting lines, not parallel to one plane, and a point $P$ not belonging to any of them, construct a plane through this point so that it forms equal angles with the given lines.
保留源文本的换行和格式，这里直接输出翻译结果。 | 5. Draw through point $P$ the lines $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$, respectively parallel to the given lines $a, b$, and $c$. Point $P$ divides each of the lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$ into two opposite rays $P A$ and $P A^{\prime}, P B$ and $P B^{\prime}, P C$ and $P C^{\prime}$. ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11*. From a point $M$ on the edge of a dihedral angle in one of its faces, a ray is drawn. Draw from the same point $M$ in the other face a ray that forms an angle of a given magnitude with the first ray. | 11. Let's take an arbitrary point $A$ on this ray; let $A^{\prime}$ be the projection of this point onto the other face. Suppose the desired ray is drawn, and we construct a segment $M B$ on it such that $|M B|=|M A|$. Then, in the isosceles triangle $A M B$, the two sides and the angle between them are known, and it c... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
58. Three spheres with different radii and centers not lying on the same straight line lie outside each other. What figure is formed by the lines of intersection of pairs of planes that are symmetric with respect to the plane of the centers of the spheres and simultaneously tangent to all three given spheres? | 58. The common tangent plane of two spheres passes through the center of their homothety, since the radii drawn to the points of tangency are parallel, and the line passing through the points of tangency also passes through the center of homothety (see the previous problem). The centers of homothety of three spheres be... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
36*. A polyhedron is called regular if all its faces are regular and congruent polygons to each other and all its polyhedral angles are also regular and congruent to each other ${ }^{2}$. Investigate the possibility of constructing simple regular polyhedra with $n$-sided faces and $m$-sided angles and determine the pos... | 36. Let the faces of a regular polyhedron be regular $n$-gons, and the polyhedral angles at the vertices be regular $m$-gonal angles; obviously, $m \geqslant 3, n \geqslant 3$. Divide the space into $f$ congruent regular $n$-gonal angles with a common vertex $S$ such that around each edge $[S A)$ there are $m$ angles. ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
46*. Inscribed a cube in a dodecahedron so that all eight vertices of the cube are vertices of the dodecahedron. How many solutions does the problem have? | 46. Consider a dodecahedron with center $S$ (Fig. 101). Due to the congruence of the faces, their diagonals are also congruent: $[A B] \cong|B C| \cong \cong[C D] \cong[D A]$. In the regular triangular pyramid $P A B Q$, the edge $[P Q]$ is perpendicular to the base edge $[A B]$ (see problem 8 from § 5). But $(P Q)\|(A... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
25. Four pairwise non-intersecting spheres of equal radii are given, with centers not lying in the same plane. a) How many spheres exist that touch all four spheres simultaneously? b) How to construct these spheres? | 25. a) The contact of spheres can be of two kinds: 1) external contact, when the spheres lie one outside the other and the distance between their centers is equal to the sum of the radii; 2) internal contact, when one sphere is inside the other and the distance between the centers is equal to the difference of the radi... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In a box, there are 7 red and 4 blue pencils. Without looking into the box, a boy randomly draws pencils. What is the minimum number of pencils he needs to draw to ensure that there are both red and blue pencils among them? | 1. Answer: 8 pencils. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
29. Calculate: $1111 \cdot 1111$. | 29. Mentally perform the multiplication of the given numbers "in a column". A n s w e r. 1234321. | 1234321 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
31. Name the number consisting of 11 hundreds, 11 tens, and 11 units. | 31. $1100+110+11=1221$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
31. $1100+110+11=1221$. | 1221 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
33. The difference between a three-digit number and an even two-digit number is 3. Find these numbers. | 33. The desired even two-digit number can only be 98. Consequently, the three-digit number is 101. | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
38. Find the smallest natural number that, when increased by 1, would be divisible by $4, 6, 10, 12$. | 38. This number will be the least common multiple of the given numbers, decreased by 1. Answer. 59. | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
39. All odd numbers from 27 to 89 inclusive are multiplied together. What is the last digit of the resulting product? | 39. The product ends with the digit 5. Indeed, if in the multiplication of several odd numbers at least one factor ends with the digit 5, then the entire product ends with the digit 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
40. What is one and a half thirds of 100? | 40. One and a half thirds is a third plus half of a third, i.e. $\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$. Answer. 50. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
43. An adult needs 16 kg of salt for consumption over 2 years and 8 months. How much salt does an adult need for one year? | 43. 2 years 8 months make up 32 months. If 32 months require 16 kg of salt, then 12 months require 6 kg of salt. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
46. Calculate: $79 \cdot 81+1$. | 46. $79 \cdot 81+1=(80-1)(80+1)+1=80^{2}-1+1=6400$.
46. $79 \cdot 81+1=(80-1)(80+1)+1=80^{2}-1+1=6400$.
(Note: The original text and the translation are identical as the content is a mathematical expression which is universal and does not change in translation.) | 6400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
59. A number was increased by $25 \%$. By what percentage does the resulting value need to be decreased to return to the original number?
10 | 59. Answer. By $20 \%$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The note is for your understanding and should not be included in the output. Here is the correct format:
59. Answer. By $20 \%$. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
60. Which numbers are more numerous among the first 100 natural numbers: those divisible by 3, or those divisible by at least one of the numbers 5 or 7?
## Problems for the eighth grade | 60. Among the first 100 natural numbers, there are 33 numbers divisible by 3, 20 numbers divisible by 5, and 14 numbers divisible by 7. Since the numbers 35 and 70 were counted both as divisible by 5 and by 7, the total number of numbers that are divisible by at least one of the numbers 5 or 7 will be $20+14-2=32$. The... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
61. Insert arithmetic operation signs between the digits 12345 so that the result equals 1. | 61. There are several solutions. For example: $1+2-3-$ $-4+5=1 ; 1-2+3+4-5=1$. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
72. Diagonals of two faces are drawn from the same vertex of a cube. Find the magnitude of the angle between these diagonals. | 72. By connecting the ends of the diagonals with a segment, we obtain an equilateral triangle (Fig. 7). Therefore, the angle between the specified diagonals contains \(60^{\circ}\).
## Fig. 7
: x=100$ is true for any digit $x$. 102. Since $\operatorname{tg} 45^{\circ}=1$, then $\lg \operatorname{tg} 45^{\circ}=0$. Therefore, the given product is also zero. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
108. The purchase of a book cost 1 ruble and one third of the book's cost. What is the cost of the book? | 108. Answer. 1 ruble 50 kopecks. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
109. What is the last digit of the difference
$$
1 \cdot 2 \cdot 3 \cdot 4 \ldots 13-1 \cdot 3 \cdot 5 \cdot 7 \ldots 13 ?
$$ | 109. The minuend ends in 0, and the subtrahend ends in 5. Therefore, the difference ends in 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
116. How many lines can divide a plane into 5 parts? | 116. Two lines can divide the plane only into 3 or 4 parts (see Fig. 15). Three lines can divide the plane only into 4, 6, and 7 parts (see Fig. 16). Four lines can divide the plane into 5 parts only if these lines are parallel (see Fig. 17). It is obvious that with any arrangement of five or more lines, the number of ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
118. A circle is inscribed in a square, and then a new square is inscribed in the circle. Find the ratio of the areas of these squares. | 118. It is clear that the second square can be placed so that its vertices fall on the points of tangency of the circle with the sides of the first square (see Fig. 18). After this, it is not difficult to verify that the desired ratio of the areas of the squares is 2.
 Prove that the square of an integer cannot end in any of the digits $2, 3, 7, 8$.
b) Prove that for no natural number $n$ can the numbers $5n+2$ and $5n+3$ be perfect squares.
c) Prove that for no integer $n$ is the number $n^2 + 3$ divisible by 5.
d) Find all natural values of $n$ for which the number $123 \l... | 7. a) Any integer can be represented in the form: \( n = 10k + r \), where \( k \) is an integer and \( r = 0, 1, 2, \ldots, 9 \). Based on the equality \( n^2 = 10(10k^2 + 2kr) + r^2 \), we conclude that the numbers \( n^2 \) and \( r^2 \) end with the same digit. It remains to verify directly that \( r^2 \) does not ... | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
17. Find all three-digit numbers starting with the digit 3 and having exactly 3 divisors. | 17. It is easy to notice that only the square of a prime number can have exactly three divisors. The only three-digit number starting with the digit 3 and having exactly three divisors is 361. | 361 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
19. Find all prime numbers $p$ such that the numbers $p+4$ and $p+8$ are also prime. | 19. It is not hard to notice that the number 3 satisfies the requirements of the problem. Let's show that the problem has no other solutions. Indeed, every prime number not equal to 3 has one of two forms: $3k+1$ or $3k+2$, where $k$ is a non-negative integer. But in the case $p=3k+1$, the number $p+8$ is divisible by ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
22. Find all natural $n$ for which the number $n^{2}+3 n$ is a perfect square. | 22. For $n=1$ the number $n^{2}+3 n$ equals 4, i.e., is a perfect square. Let now $n>1$. Obviously, in this case
$$
(n+1)^{2}<n^{2}+3 n<(n+2)^{2}
$$
Since a perfect square cannot exist between the squares of two consecutive natural numbers, the problem has a unique solution $n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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