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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 be replaced by $987654321-123456789=864197532$. Subtracting 123456789 again, we get 740740743, then 617283954, 493827165, 370370376, 246913587, 123456798, and finally, $123456798-123456789=9$. (In the end, we divided 987654321 by 123456789 with a remainder and got a remainder of 9.) As we have seen, the common divisors of the original pair and the pair 123456789, 9 are the same, and according to the Euclidean algorithm, we need to subtract 9 from 123456789. The number 123456789 is divisible by 9 (it equals $9 \cdot 13717421$), so when subtracting, we get 9 (and then 0). Answer: the greatest common divisor is $9. \triangleleft$
|
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. Rewriting this relation as $y=C-2 x$, from a geometric perspective, we need to find the smallest $C$ such that the line $y=C-2 x$ intersects the given set. Clearly, the desired value of $C$ is the one for which this line passes through the point $\left(\frac{4}{3} ; \frac{4}{3}\right)$. Thus, $C=\frac{8}{3}+\frac{4}{3}=4$.
Answer: 4.
|
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 $|\boldsymbol{q}|=|\boldsymbol{n}+\boldsymbol{m}+\boldsymbol{p}| \leqslant|\boldsymbol{n}|+|\boldsymbol{m}|+|\boldsymbol{p}|$, then
$$
15=\sqrt{12^{2}+9^{2}} \leqslant \sqrt{a^{2}+1}+\sqrt{b^{2}+9}+\sqrt{c^{2}+25}
$$
Equality is achieved if the vectors $\boldsymbol{m}(a ; 1), \boldsymbol{n}(b ; 3)$ and $\boldsymbol{p}(c ; 5)$ are collinear, $\frac{a}{1}=\frac{b}{3}=\frac{c}{5}$, from which it follows that $b=3 a$ and $c=5 a$. Therefore, $a+3 a+5 a=12$, from which $a=\frac{4}{3}, b=4, c=\frac{20}{3}$.
Answer: 15.
|
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
$$
\begin{aligned}
4^{x}+4^{-x}-2^{x+1}-2^{1-x}+5=2^{2 x}+ & 2^{-2 x}-2\left(2^{x}+2^{-x}\right)+5= \\
=2^{2 x}+2 \cdot 2^{x} \cdot 2^{-x}+2^{-2 x} & -2-2\left(2^{x}+2^{-x}\right)+5= \\
& =\left(2^{x}+2^{-x}\right)^{2}-2\left(2^{x}+2^{-x}\right)+3
\end{aligned}
$$
then the substitution $t=2^{x}+2^{-x}$ reduces the problem to finding the minimum value of the quadratic polynomial $g(t)=t^{2}-2 t+3$, which is $g(1)=2$.
Answer: 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 increasing, so its minimum value is achieved at $t=2$.
Answer: 3.
|
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 $k=d-\left[\frac{d}{2}\right]-\left[\frac{d}{3}\right]-\left[\frac{d}{5}\right]$. Thus, for each of the thirty values of $d$, there is a unique value of $k$ such that $x=30 k+d$ is a solution to the given equation.
Answer: 30 solutions.
|
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 triangle $A B K$ is 10.
|
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 vertex $C$ with point $K$ and set $x=S_{B_{1} C K}$ and $y=S_{A_{1} C K}$ (figure).
Then
$$
\frac{10}{8}=\frac{B K}{K B_{1}}=\frac{y+5}{x} \text { and } \frac{10}{5}=\frac{A K}{K A_{1}}=\frac{x+8}{y} .
$$
Thus, $5 x=4 y+20$ and $2 y=x+8$. Solving the obtained system, we get that $x=12$ and $y=10$, so $x+y=22$.
Answer: 22.
|
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
$$
\sqrt[n+1]{\frac{n+1}{n}}=\sqrt[n+1]{\left(1+\frac{1}{n}\right) \cdot 1^{n}}<\frac{n+1+\frac{1}{n}}{n+1}=1+\frac{1}{n(n+1)}=1+\frac{1}{n}-\frac{1}{n+1}
$$
## Therefore
$$
\begin{aligned}
& \sqrt{2}+\sqrt[3]{\frac{3}{2}}+\sqrt[3]{\frac{4}{3}}+\ldots+\sqrt[2009]{\frac{2009}{2008}}< \\
&<2008+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2008}-\frac{1}{2009}= \\
&=2009-\frac{1}{2009}<2009
\end{aligned}
$$
Therefore, the integer part of the given sum is 2008.
Remark. The inequality $\sqrt[n+1]{1+\frac{1}{n}}<1+\frac{1}{n(n+1)}$ also follows from an analogue of Bernoulli's inequality for exponents less than 1.
|
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:
$$
\begin{aligned}
\operatorname{tg} 65^{\circ}-2 \operatorname{tg} 40^{\circ}=\operatorname{ctg} 25^{\circ}-2 \operatorname{ctg} 50^{\circ} & =\frac{1}{\operatorname{tg} 25^{\circ}}-\frac{2}{\operatorname{tg} 50^{\circ}}= \\
& =\frac{1}{\operatorname{tg} 25^{\circ}}-\frac{1-\operatorname{tg}^{2} 25^{\circ}}{\operatorname{tg} 25^{\circ}}=\operatorname{tg} 25^{\circ} .
\end{aligned}
$$
Answer: $25^{\circ}$.
|
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 belongs to the set $S$ if and only if its ternary representation consists only of 0 and 1.
It remains to note that $3 \cdot 729>2009>2 \cdot 729=2 \cdot 3^{6}$. From this, it follows that in the ternary representation, the number 2009 is a 7-digit number, with the first (most significant) digit being 2. The largest number obtained on the 7th step has the representation $1111111_{3}$, so it is less than 2009. Therefore, the total number of such numbers is $2^{7}=128$.
|
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
$$
S_{ABC} = S = \frac{a r_{3}}{2} + \frac{b r_{3}}{2} - \frac{c r_{3}}{2}
$$
from which $r_{3} = \frac{2 S}{a + b - c}$. The radii of the other two excircles are found by analogous formulas: $r_{1} = \frac{2 S}{b + c - a}$ and $r_{2} = \frac{2 S}{a + c - b}$. Since the radius of the circle inscribed in the given triangle is found by the formula $r = \frac{2 S}{a + b + c}$, the following relationship holds:
$$
\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{b + c - a}{2 S} + \frac{a + c - b}{2 S} + \frac{a + b - c}{2 S} = \frac{a + b + c}{2 S} = \frac{1}{r}
$$
According to the problem, $r_{1} = 2$, $r_{2} = 3$, and $r_{3} = 6$, from which $r = 1$. A triangle with the given radii of the excircles exists; it is a right triangle with legs 3 and 4, known as the "Egyptian" triangle.
|
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,
$$
\begin{aligned}
& 3 \operatorname{tg} \alpha - \sin \alpha + 4 \cos \alpha = 12 \\
& \Longleftrightarrow \operatorname{tg} \alpha (3 - \cos \alpha) - 4 (3 - \cos \alpha) = 0 \Longleftrightarrow \\
& (\operatorname{tg} \alpha - 4)(3 - \cos \alpha) = 0 \Longleftrightarrow \operatorname{tg} \alpha = 4
\end{aligned}
$$
Answer: 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,
$$
32 x^{2} y^{2}\left(x^{2}+y^{2}\right) \leqslant(x+y)^{6}
$$
Since $x+y=2$, then $x^{2} y^{2}\left(x^{2}+y^{2}\right) \leqslant 2$. It is clear that when $x=y=1$ we get equality.
Answer: 2.
|
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 $x=2$ is a solution to the equation.
|
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 the number $x=\frac{1}{2}$.
|
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 are the numbers $-1, -2, 3$. Since $3^{x}>0$, then $3^{x}=3$, hence $x=1$.
Answer: 1.
Topic: "Methods for solving irrational equations"
|
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 $n=p^{12}$. From the following table, it is clear that the required numbers are $3^{12}$ and $5^{12}$.
| $p$ | 2 | 3 | 5 | 7 |
| :---: | :---: | :---: | :---: | :---: |
| $p^{12}$ | 4096 | 531441 | 244140625 | 13841287201 |
Answer: two numbers.
|
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 $3^{x^{2}-4} \leqslant 1$, so $f(x) \leqslant 1$, hence the equation has no solutions on the interval $(-2; 0]$. Now notice that the function $f(x)$ is increasing on the interval $[0; +\infty)$, so the equation has at most one solution. It remains to note that $f(2)=80$, so $x=2$ is a solution to the equation. Answer: $\{2\}$.
|
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 similar reasoning, this number is not divisible by 5. Further, since $n^{2}+n+11=$ $(n-3)^{2}+2+7n$, this number is not divisible by 7 because no perfect square can have a remainder of 5 when divided by 7. Finally, note that when $n=0$, the number $n^{2}+n+11$ is divisible by 11. Therefore, 11 is the smallest prime divisor among all divisors of numbers of the form $n^{2}+n+11$.
|
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 lengths of 1 cm, 2 cm, ... 6 cm be formed? Why? Can a square with a side length of 7 cm be formed? How can it be done? In how many different ways can it be formed? Consider all possible cases. Can a square with a side length of 12 cm or more be formed? Why?
|
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 (Why?). Since the total length of all the sticks is 45 cm, the perimeter of the square cannot be more than 44 cm, and the side length of the square cannot be more than $11 \mathrm{~cm}$. (MvSh)
|
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 books bought was 6 more than the number of albums. How many books were bought if the price of a book, which is more than one ruble, exceeds the price of an album?
The adults delved into calculations. But the little brother, a fifth-grader, suddenly blurted out:
- There's nothing to solve here. 8 books!
How did he solve the problem?
If solving this problem caused you difficulty, then think it over again and again, delve into the conditions so that they are clearly imprinted in your memory. Look for and seek new approaches to solving the problem. Try to "guess" the answer, find it by making a rough estimate. How many solutions does the problem have? Write down the solution and check your answer.
|
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 numbers corresponding to the "old" points $A, B, C$, and $K$ will not change.
|
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 product can be divisible by seven to a power greater than three. It is not difficult to show that the given product contains the number seven to the power of $142+20+2=164$.
|
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 sum ends in six.
|
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 and second, and so on, of the second sequence. Write down a few terms of the corresponding fourth sequence. What are the terms of the fifth, sixth, ... sequences?
|
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|c|c|c|c|c|c|c|c|c|}
\hline\(m\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \(\ldots\) \\
\hline\(r_{1}\) & 1 & 1 & 6 & 1 & 6 & 6 & \(\ldots\). & \(\ldots\) & \(\ldots\) & \(\ldots\) \\
\hline\(r_{2}\) & 3 & 2 & 6 & 4 & 5 & 1 & \(\ldots\) & \(\ldots\) & \(\ldots\). & \(\ldots\) \\
\hline\(r_{1}+r_{2}\) & 4 & 3 & 12 & 5 & 4 & 7 & \(\ldots\) & \(\ldots\) & \(\ldots\) & \(\ldots\) \\
\hline
\end{tabular}
From the table, it is clear that \(r_{1}+r_{2}=7\) when \(m=6\).
|
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 that this approach to solving problems was often used by many mathematicians, and in particular one of the greatest mathematicians in the world - Leonhard Euler (1707-1783). He writes: “... properties of numbers known today were for the most part discovered by observation and were discovered long before their truth was confirmed by rigorous proofs. There are even many properties of numbers with which we are well acquainted, but which we are still unable to prove; only observation has led us to their knowledge.”
|
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 "Minsk". In how many ways can the cashier choose the order of visiting the hotels to deliver the tickets?
|
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 \(z^{2} \leqslant 18\), and therefore \(z \leqslant 4\). By checking, we find that there is only one number that satisfies the conditions of the problem: 27.
|
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. \( 2 x^{4}+2 x^{2} y^{2}+x^{2} y^{2}+y^{4}+y^{2}=2 x^{2}\left(x^{2}+\right. \) \( \left.+y^{2}\right)+y^{2}\left(x^{2}+y^{2}\right)+y^{2}=2 x^{2} \cdot 1+y^{2} \cdot 1+y^{2}=2\left(x^{2}+y^{2}\right)=2 \).
If students are already familiar with the formulas for abbreviated multiplication, the exercise can be solved by substitution \( x^{2}=1-y^{2} \).
|
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 identically equal to 1.
|
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 \((2 y+1)^{3}=2 p+1\), where \(y \in N\). Then the equation \(y\left(4 y^{2}+6 y+3\right)=p\) must hold.
For a prime \(p\), this is true only when \(y=1\), and consequently, \(p=13\).
|
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-negative, therefore, \(900-2 y \geqslant 0, y \leqslant 450\). The maximum area is 450 square meters; the width of the site is 15 m, and the length is \(30 m\).
Second method. We can use the method of completing the square:
\[
\begin{aligned}
y=2\left(30 x-x^{2}\right)= & 2\left[225-\left(225-30 x+x^{2}\right)\right]=2[225- \\
& \left.-(15-x)^{2}\right]
\end{aligned}
\]
When \(x=15\), we get the maximum value of \(y\), which is 450.
Other methods of solution are possible. In particular, the problem can be solved based on geometric considerations.
|
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:
\[
a\left(\frac{m}{2}+x\right)^{2}+a\left(\frac{m}{2}-x\right)^{2}=a\left(\frac{m^{2}}{4}+x^{2}\right)
\]
Obviously, this sum will be the smallest if \(x=0\), i.e., if both pieces have the same mass.
|
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: $\quad \vec{r}_{1} \cdot \vec{r}_{2}=(\vec{i}+\vec{j}) \cdot (\vec{j}+\vec{k})=1=$ $=r_{1} r_{2} \cos \varphi$. But $\left|r_{1}\right|=\left|r_{2}\right|=\sqrt{2}$.
$$
\text { Hence } \cos \varphi=\frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2} ; \varphi=60^{\circ} \text {. }
$$
|
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:
$$
\begin{gathered}
\cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha=1 ; \cos ^{2} \alpha=\frac{1}{3} \\
\cos \alpha= \pm \frac{1}{\sqrt{3}} ; \alpha \approx 54^{\circ} 44^{\prime} 8^{\prime \prime} \text { or } \alpha=125^{\circ} 15^{\prime} 52^{\prime \prime}
\end{gathered}
$$
The coordinate planes divide the space into 8 octants. Each of the desired lines passes through two octants, and therefore there are 4 such lines.
|
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, returns along the same straight line. What is the magnitude of the dihedral angle between the mirrors?
(Remember, according to the laws of reflection, the incident and reflected rays lie in a plane perpendicular to the mirror plane and form equal angles with the mirror plane.)
|
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 isosceles triangle $A_{1} A_{2} O$ is $2 \alpha$ ( $\alpha$ is the given angle). Therefore, in the right triangle $A_{2} A_{3} O$, one angle is $\alpha$, the other is $2 \alpha$. Thus, $3 \alpha=90^{\circ} ; \alpha=30^{\circ}$.
|
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 $P C^{\prime}$. The planes of symmetry of the rays $P A$ and $P B, P B$ and $P C$ intersect along the line $l$, through which the plane of symmetry of the rays $P C$ and $P A$ also passes (why?). The line $l$ forms congruent angles with the rays $P A, P B$, and $P C$. This same line $l$ forms congruent angles with the opposite rays $P A^{\prime}, P B^{\prime}$, and $P C^{\prime}$. The plane passing through $P$ perpendicular to $l$ forms congruent angles with the lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$, and thus also with the lines $a, b$, and $c$.
Another solution is obtained by considering the rays $P A, P B$, and $P C^{\prime}$ and their opposite rays $P A^{\prime}, P B^{\prime}$, and $P C$. Two more solutions correspond to the triples of rays $P B, P C$, and $P A^{\prime}$ (with their opposites $P B^{\prime}, P C^{\prime}, P A$) and $P A, P C, P B^{\prime}$ (with their opposites $P A^{\prime}, P C^{\prime}$, $P B$). Thus, there are 4 planes that satisfy the condition of the problem.
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4
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Geometry
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math-word-problem
|
Yes
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Yes
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olympiads
| false
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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 can be constructed on a separate drawing. We will then know the length of the slant $|A B|$, and consequently, the length of its projection $\left|A^{\prime} B\right|$ onto the second face. The desired point $B$ must lie on the circle with center at point $A^{\prime}$ and radius $\left|A^{\prime} B\right|$, and at the same time on the circle with center at point $M$ and radius $|M A|$. By constructing point $B$, we will determine the desired ray. Determine how many solutions the problem can have.
142
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2
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Geometry
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math-word-problem
|
Yes
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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 belong to the plane of their centers and coincide with the centers of homothety of the circles of section of the spheres by this plane. Therefore, the plane tangent to all three spheres passes through the axes of homothety of these circles (see problem 15 from §2).
Since the spheres are symmetric relative to the plane of their centers, through each axis of homothety there pass pairs of mutually symmetric planes tangent to the three spheres.
Thus, pairs of symmetric planes tangent to three spheres intersect along lines lying in the plane of the centers and form a configuration of 4 lines and 6 points, with 3 points on each line.
|
4
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Geometry
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math-word-problem
|
Yes
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Yes
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olympiads
| false
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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 possible number of their types, using the following considerations.
Divide the entire space into $f$ congruent regular $n$-sided angles with a common vertex $S$, with $m$ angles at each edge, and express the magnitudes of their dihedral angles in terms of $n$ and $f$. If a regular $n$-sided pyramid is constructed from the found magnitude of the dihedral angle, with the base being a regular $n$-sided polygon - a face of the desired polyhedron, then from such congruent pyramids, a regular polyhedron can be assembled (for example, a cube is assembled from six regular quadrilateral pyramids with a common vertex at the center of the cube). It can be proven that any regular polyhedron can be obtained by such a construction. (The last remark can be used without proof in solving this and the following problems.)
## 44
|
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. Then the magnitude of each of these $f$ polyhedral angles is $\frac{4 \pi}{f}$. Since this magnitude is the difference between the sum of the magnitudes of its dihedral angles and the number $\pi(n-2)$, the sum of all dihedral angles of the $n$-gonal angle is $\frac{4 \pi}{f} + \pi(n-2) = \frac{4 + nf - 2f}{f} \pi$. One dihedral angle has a magnitude that is $n$ times smaller, i.e., $\frac{4 + nf - 2f}{nf} \pi$.
On the other hand, since at each of the $p$ edges of our $n$-gonal angles there are $m$ congruent dihedral angles, the magnitude of each of them is $\frac{2 \pi}{m}$. Thus,
$$
\begin{gathered}
\frac{2 \pi}{m} = \frac{4 + nf - 2f}{nf} \pi \\
2nf + 2mf - mnf = 4m \\
f = \frac{4m}{2m + 2n - mn}
\end{gathered}
$$
## 168
Since $f > 0, m > 0$, from (1) we have $2m + 2n - mn > 0$,
$$
\frac{2}{m} + \frac{2}{n} > 1
$$
Thus, the division of space into congruent polyhedral angles is possible only if the numbers $m \geqslant 3$ and $n \geqslant 3$ satisfy the inequality (2). Clearly, only five cases are possible: 1) $m=3$ and $n=3: \frac{2}{3} + \frac{2}{3} > 1$. From (1) we get $f=4$, the polyhedron is bounded by 4 triangles - this is a tetrahedron. The corresponding values of the numbers $a$ and $p$ are obtained from the formulas in problems 27 and 28: for the tetrahedron $a=6, p=4$.
2) For $m=3, n=4$ we get: $\frac{2}{3} + \frac{2}{4} > 1, f=6$. The polyhedron is bounded by six squares - this is a hexahedron or cube. Here $a=12, p=8$.
3) For $m=4, n=3$ we also get $\frac{2}{4} + \frac{2}{3} > 1 ; f=8$. The polyhedron is bounded by eight triangles - this is an octahedron. In it, $a=12, p=6$.
4) For $m=3, n=5$ we have: $\frac{2}{3} + \frac{2}{5} > 1 ; f=12$. The polyhedron is bounded by twelve pentagons - this is a dodecahedron. In it, $a=30, p=20$.
5) For $m=5, n=3 ; \frac{2}{5} + \frac{2}{3} > 1, f=20$. The polyhedron is bounded by 20 triangles - this is an icosahedron. In it, $a=30, p=12$.
For $m=4$ and $n=4$ we have: $\frac{2}{4} + \frac{2}{4} = 1$, and the inequality (2) is not satisfied. For further increases in the values of $m$ and $n$, the expression on the left side of the inequality (2) becomes even smaller, and it cannot be valid. It can be shown that any regular polyhedron can be obtained by the described construction, and our considerations show that there are only five types of regular polyhedra in space.
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5
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Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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 $(P Q)\|(A D)\|(B C)$, and thus, $A B C D$ is a square. Similarly,
Fig. 101
we will prove that $A B B^{\prime} A^{\prime}, B C C^{\prime} B^{\prime}$, $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, and $A D D^{\prime} A^{\prime}$ are also squares. Thus, the polyhedron $A B C D C^{\prime} D^{\prime} A^{\prime} B^{\prime}$ is a cube, whose edges are the diagonals of the dodecahedron.
If in the pentagon $A M N B P$ we take another diagonal $[A N]$, we get the square $A N E Q$, which is a face of the cube $A N E Q C^{\prime} N^{\prime} E^{\prime} Q^{\prime}$. From the other three diagonals $[B M \mid, [M P]$, and $\mid N P]$, we obtain three more cubes: $B M F^{\prime} Q D^{\prime} M^{\prime} F Q^{\prime}$, $M P D E^{\prime} M^{\prime} F^{\prime} B^{\prime} E$, and $N P C F N^{\prime} P^{\prime} A^{\prime} F^{\prime}$.
It is not difficult to verify that each diagonal of each face of the dodecahedron is a side of one and only one of these cubes. Therefore, there are exactly five such cubes (60 face diagonals are distributed 12 to each of the 5 cubes).
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5
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Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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 radii.
A sphere $S$, touching the given four spheres, can occupy the following positions relative to them: 1) all 4 spheres touch sphere $S$ externally; 2) all 4 spheres touch sphere $S$ internally; 3) 3 spheres touch externally, one internally (4 positions); 4) 3 spheres touch internally, one externally (4 positions); 5) 2 spheres touch internally, two externally (6 positions). In total, for the touching sphere, we get 16 different possibilities.
b) Let us indicate the construction in each case.
1) Let a sphere with center $O$ and radius $R$ touch externally four spheres with centers $O_{1}, O_{2}, O_{3}, O_{4}$ and one radius $r$. Suppose the radii $r$ of the four spheres are reduced by the same length and the radius $R$ is increased by the same length. Since the distances between the centers have not changed, the contact of the spheres is preserved. Reduce the radii of the spheres to zero, then the spheres will "degenerate" into points $O_{1}, O_{2}, O_{3}, O_{4}$, and the radius of the tangent sphere will become equal to $R+r$. The problem is reduced to constructing a sphere through 4 given points. By constructing this sphere and reducing its radius by the length $r$, we obtain the desired sphere, touching the given four spheres.
2) If the radius of the sphere passing through the points $O_{1}, O_{2}, O_{3}, O_{4}$ is increased by $r$, then we obtain a sphere internally tangent to the four spheres.
3) Similarly, "contract" the three spheres of external contact to their centers; then the fourth tangent sphere will have a radius of $2r$. The problem is reduced to constructing a sphere passing through the given points and tangent to the given sphere. For this, we draw a circle through 3 points and draw a perpendicular to its plane through the center. The center of the tangent sphere must lie on this perpendicular. Draw a plane through this line and the center of the fourth sphere. The problem is reduced to a planimetric one: through two given points (these are the points of intersection of the obtained plane with the circle passing through the centers of the three spheres) draw a circle tangent to the given circle (see problem 49 from $\S 2$).
4) In this case, the three spheres are again contracted to points, and the further construction remains the same as in the previous case.
5) In this position, two spheres are contracted to two points, and the radii of the other two spheres are doubled. The center of the tangent sphere lies in the plane of symmetry of the two points and in the plane of symmetry of the two spheres. Since the 4 centers of the given spheres do not lie in the same plane, the indicated planes of symmetry intersect along some line $l$. Any sphere with its center on the line $l$ and passing through one of the points $O_{1}$ or $O_{2}$, due to symmetry, passes through the other as well, and if it touches one of the spheres, it touches the other as well. We again come to the same problem: through two given points, draw a sphere tangent to the given sphere. For its solution, the same planimetric construction can be used.
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16
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Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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?
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1. Answer: 8 pencils.
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8
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Combinatorics
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math-word-problem
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Yes
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Yes
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olympiads
| false
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29. Calculate: $1111 \cdot 1111$.
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29. Mentally perform the multiplication of the given numbers "in a column". A n s w e r. 1234321.
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1234321
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Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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
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Number Theory
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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.
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101
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Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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38. Find the smallest natural number that, when increased by 1, would be divisible by $4, 6, 10, 12$.
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38. This number will be the least common multiple of the given numbers, decreased by 1. Answer. 59.
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59
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Number Theory
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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.
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5
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Number Theory
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math-word-problem
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Yes
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Yes
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olympiads
| false
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40. What is one and a half thirds of 100?
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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.
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50
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Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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?
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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.
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6
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Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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46. Calculate: $79 \cdot 81+1$.
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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
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Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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
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Algebra
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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 $20+14-2=32$. Therefore, the numbers divisible by three will be one more.
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33
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Number Theory
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math-word-problem
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Yes
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Yes
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olympiads
| false
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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
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math-word-problem
|
Yes
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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
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60
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Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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74. From a bag, $60 \%$ of the flour it contained was taken, and then $25 \%$ of the remainder. What percentage of the original amount of flour is left in the bag?
|
74. After taking $60 \%$ of the flour from the bag, $40 \%$ remained. When $25 \%$ of the remainder, i.e., a quarter of it, was taken, $30 \%$ of the original amount of flour remained in the bag.
|
30
|
Algebra
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math-word-problem
|
Yes
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Yes
|
olympiads
| false
|
75. Can the sum of four consecutive integers be a prime number?
## Problems for the ninth grade
|
75. The sum of four consecutive integers is an even number. There is only one even prime number. This number is 2. Since $-1+0+1+2=2$, the answer to the question of the problem is affirmative.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
80. A perfectly smooth sphere, having the same size as the Earth, is encircled by a wire at the equator. This wire was extended by one meter and arranged so that an equal gap was formed between the wire and the surface of the sphere. Would a mouse be able to squeeze through the resulting gap?
|
80. If the circumference of a circle increases by 1 m, then the radius of the circle will increase by $1: 2 \pi \approx 16 \text{ cm}$, i.e., the gap formed will be large enough for not only a mouse but also a cat to pass through.
|
16
|
Geometry
|
math-word-problem
|
Yes
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Yes
|
olympiads
| false
|
82. There is a water tap and two containers: a three-liter and a five-liter. How can you get 4 liters of water in the larger one?
|
82. Let's fill a five-liter container and pour 3 liters from it into a three-liter container. Then we will empty the three-liter container and transfer the 2 liters of water remaining in the five-liter container into it. Finally, we will fill the five-liter container from the tap and pour 1 liter from it into the three-liter container. Now we will have 4 liters of water in the five-liter container.
|
4
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Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
86. Three numbers end with different digits. Can their pairwise products end with the same digit?
|
86. They can. For example, the three numbers 2, 5, and 10 end in different digits, but their pairwise products all end in the same digit 0.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
87. The grandfather is three times older than the older grandson and four times older than the younger one. The two grandsons together are 49 years old. How old is the grandfather?
|
87. Let the grandfather be $x$ years old. We need to solve the equation: $x: 3+$ $+x: 4=49$. From this, $x=84$.
|
84
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
88. A chocolate bar consists of 40 pieces. What is the minimum number of breaks required to get all 40 pieces, if breaking several pieces at once is not allowed?
|
88. After the first break, we get two pieces, after the second - three pieces, after the third - four pieces, and so on. Therefore, after 39 breaks, we will have all 40 segments of the chocolate bar separately. The order of the breaks is completely irrelevant.
|
39
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
91. Find the smallest natural number that, when divided by $4, 5, 6$, and 12, gives a remainder that is two units less than the divisor each time.
|
91. From the condition of the problem, it follows that if 2 is added to the required number, then it will be divisible by each of the given numbers. This means that the required number is equal to the least common multiple of the given numbers, decreased by 2. Answer. 58.
|
58
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
97. For the first half of the journey, the motorcyclist traveled at a speed of 30 km/h, and for the second half - at a speed of 60 km/h. Find the average speed of the motorcyclist.
|
97. Let the entire distance be denoted as $a$ km. The motorcyclist spent $0.5a : 30 + 0.5a : 60 = a : 40$ hours on the entire journey. Therefore, the motorcyclist traveled at an average speed of 40 km/h.
## Puc. 9
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
99. How many diagonals does a convex decagon have?
|
99. From each vertex of a decagon, 7 diagonals emanate. Then the product $7 \cdot 10=70$ gives twice the number of diagonals, since each diagonal is counted twice in this calculation. Thus, the total number of diagonals will be $70: 2=35$.
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
101. Write the number 100 using six identical non-zero digits in such a way that the method of writing does not depend on the digits used.
102 Calculate:
$\lg \operatorname{tg} 37^{\circ} \cdot \lg \operatorname{tg} 38^{\circ} \cdot \lg \operatorname{tg} 39^{\circ} \ldots \lg \operatorname{tg} 47^{\circ}$
|
101. Let $x$ be any non-zero digit. Then the expression $(\overline{x x x}-\overline{x x}): x=100$ 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 parts of the plane will be more than five.
Fig. $\quad 16$
|
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.
Fig. $\quad 18$
24
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. a) 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 $123 \ldots n + 97$ is a perfect square.
|
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 end in any of the digits 2, 3, 7, 8.
b) Numbers of this form, when \( n \) is even, end in the digits 2 and 3, and when \( n \) is odd, they end in the digits 7 and 8. As follows from problem 7a, numbers of this form cannot be perfect squares.
c) Since \( n^2 \) cannot end in 2 or 7, \( n^2 + 3 \) cannot end in 5 or 0.
d) Let \( P(n) = 1 \cdot 2 \cdot 3 \ldots n + 97 \). Then \( P(1) = 98 \), \( P(2) = 99 \), \( P(3) = 103 \), \( P(4) = 121 \). Thus, \( P(4) \) is a perfect square. If \( n \geq 5 \), then \( P(n) \) ends in 7 and cannot be a perfect square. Answer: \( n = 4 \).
|
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, and in the case $p=3k+2$, the number $p+4$ is divisible by 3.
|
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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