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4. Two non-intersecting circles are arranged such that one of their common internal tangents is perpendicular to one of their common external tangents. Find the area of the triangle formed by these tangents and the third common tangent of the circles, if their radii are $r_{1}$ and $r_{2}$. | 5.I.4. We will use the following notations and formulas. $S$ - the area of the triangle, $p$ - its semiperimeter, $r$ - the radius of the inscribed circle in the triangle. It is known that $S=p r$. If $r_{a}$ is the radius of the circle touching the side $B C$ of triangle $A B C$ and the extensions of its two other sid... | \frac{r_{1}r_{2}(r_{1}+r_{2})}{|r_{1}-r_{2}|} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,282 |
5. Find all natural values of $n$ such that the number $n^{4}+64^{n}$ is composite. | 5.I.5. Let's use the identity
$$
x^{4}+4 y^{4}=\left(x^{2}+2 y^{2}\right)^{2}-4 x^{2} y^{2}=\left(x^{2}-2 x y+2 y^{2}\right)\left(x^{2}+2 x y+2 y^{2}\right) .
$$
If the numbers $x$ and $y$ are natural, then the number $x^{2}-2 x y+2 y^{2}=(x-y)^{2}+y^{2}$ can be equal to 1 only if $x=y=1$. Therefore, for all other va... | n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,283 |
6. The sphere touches all the edges of the tetrahedron, two opposite edges of which are equal to $a$ and $b$, and all other edges are equal to each other. Find the radius of this sphere. | Solution 1. Let $BC = a, AD = b, AB = BC = BD = CD = x$. Let $K$ and $M$ be the midpoints of edges $BC$ and $AD$ (figure).
Planes $ADK$ and $BMC$ are perpendicular to edges $BC$ and $AD$, ... | \frac{\sqrt{2ab}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,284 |
7. The function $f(x)$ is continuous and positive, and $f(x+1) = f(x)$ for all real $x$.
a) Prove that $\int_{0}^{1} \frac{f(x+0.5)}{f(x)} d x \geqslant 1$.
b) Find all values of $\alpha$ such that $\int_{0}^{1} \frac{f(x+\alpha)}{f(x)} d x \geqslant 1$.
## Methodological Block
A. Below are the solutions to problem... | 5.I.7. a) We have,
$$
\begin{aligned}
\int_{0}^{1} \frac{f\left(x+\frac{1}{2}\right)}{f(x)} d x & =\int_{0}^{\frac{1}{2}} \frac{f\left(x+\frac{1}{2}\right)}{f(x)} d x+\int_{\frac{1}{2}}^{1} \frac{f\left(x+\frac{1}{2}\right)}{f(x)} d x= \\
= & \int_{0}^{\frac{1}{2}} \frac{f\left(x+\frac{1}{2}\right)}{f(x)} d x+\int_{0}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 43,285 |
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 | 43,286 |
9. From two groups of skiers with a total of 100 people, a team of 15 people was formed. The first group contributed $p \%$ of its members, and the second group contributed $-10 \%$ of its members. How many skiers are there in each group?
Answer: 50 and 50, or 20 and 80, or 10 and 90.
Solution: Let $x$ and $y$ be the... | 5.I.9. Comment. The error in the given solution is that, by the meaning of the problem, the numbers $x$ and $p x$ are natural, while the number $p$ can be rational. Therefore, the problem has more solutions than those provided in its solution. | 5050,or2080,or1090 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,287 |
10. On the hypotenuse $A B$ of the right triangle $A B C$, a square $A B D E$ is constructed in the half-plane that does not contain triangle $A B C$. Find the distance from the vertex $C$ of the right angle of the triangle to the center of the square, given that $B C=a$ and $A C=b$. | Solution 1. Introduce a coordinate system, the origin of which coincides with the vertex $C$ of the right angle of the triangle, and the ends of the hypotenuse lie on the coordinate axes at points $A(b ; 0)$ and $B(0 ; a)$. The point $E$ - one of the vertices of the square (left figure) - has coordinates $(a+b ; b)$.
... | \frac{+b}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,288 |
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.
$$
Then $y^{3}-z^{3}=7(z-y)$, from which it follows that $y=z$, since $y^{2}+y z+z^{2}$ cannot be equal to -7. As a result, we get the equation $y^{3}-7 y-6=0$, the roots of which ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,289 |
12. Prove the inequality $P>4 R$, where $P$ is the perimeter, and $R$ is the radius of the circumscribed circle of an acute triangle. | Solution 1. It is clear that if the triangle is a right triangle, then $P=a+b+c>2 c=4 R$. Now consider an acute triangle $ABC$ with perimeter $P$ and the radius $R$ of the circumscribed circle around it. Inscribed in this circle is a right triangle $DBC$ (left figure). Since $\angle ACB = \angle ADB$ and $\angle ADB = ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,290 |
1. Prove that the number $1 \cdot 3 \cdot 5 \cdot \ldots \cdot 2009+2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2010$ is divisible by 2011. | 5.II.1. Note that $2 \equiv -2009 \pmod{2011}$, $4 \equiv -2007 \pmod{2011}$, $\ldots$, $2010 \equiv -1 \pmod{2011}$. Therefore,
$2 \cdot 4 \cdot \ldots \cdot 2010 \equiv (-2009) \cdot (-2007) \cdot \ldots \cdot (-1) \equiv -1 \cdot 3 \cdot \ldots \cdot 2009 \pmod{2011}$, hence
$$
2 \cdot 4 \cdot \ldots \cdot 2010 + ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,291 |
2. Find all right-angled triangles with integer side lengths, in which the hypotenuse is one unit longer than one of the legs. | 5.II.2. Let the legs of a right triangle have lengths $a$ and $b$, and the length of its hypotenuse be $a+1$. Then $a^{2}+b^{2}=(a+1)^{2}$, from which $b^{2}=2 a+1$. Therefore, $b=2 k+1$, so $4 k^{2}+4 k+1=2 a+1$, or $a=2 k(k+1)$.
Answer: all triangles with legs $2 k+1$ and $2 k(k+1)$ and hypotenuse $2 k^{2}+2 k+1$, w... | all\triangles\with\legs\2k+1\\2k(k+1)\\hypotenuse\2k^2+2k+1,\where\k\is\any\natural\ | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,292 |
3. The base of the pyramid $P A B C D$ is a trapezoid $A B C D$, the base $A D$ of which is twice the length of the base $B C$. The segment $M N$ is the midline of the triangle $A B P$, parallel to the side $A B$. Find the ratio of the volumes of the solids into which the plane $D M N$ divides this pyramid. | 5.II.3. Let's complete the pyramid $P A B C D$ to the tetrahedron $P A D F$ (figure).
96
The section of the tetrahedron by the plane $D M N$ intersects its edge at some point $O$. The quad... | \frac{5}{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,293 |
4. Prove that for any integers $x$ and $y$, the number $x^{4}+y^{4}+(x+y)^{4}$ is twice the square of some integer. | 5.II.4. The statement of the problem follows from the transformations
\[
\begin{aligned}
& x^{4}+y^{4}+(x+y)^{4}=2 x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+2 y^{4}= \\
& =2\left(x^{4}+2 x^{3} y+3 x^{2} y^{2}+2 x y^{3}+y^{4}\right)= \\
& =2\left(x^{4}+y^{4}+(x y)^{2}+2 x \cdot x y+2 y \cdot x y+2 x^{2} y^{2}\right)=2\le... | 2(x^{2}+xy+y^{2})^{2} | Algebra | proof | Yes | Yes | olympiads | false | 43,294 |
5. Two people toss a coin: one tossed it 10 times, and the other 11 times. Find the probability that the second person gets more heads than the first. | Solution 1. Let $\mathcal{A}$ be the event that the second participant gets heads more times than the first participant.
Let $\mathcal{B}$ be the event that the second participant gets tails more times than the first participant. It is clear that these events are equally probable and do not intersect. Indeed, if the s... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 43,295 |
7. Solve the following problem using as many methods as possible. Methods are considered different if they use different mathematical ideas, as well as various technical approaches to implementing the same idea. Indicate the place of each method used in the school mathematics curriculum.
Compare the numbers $2+\log _{... | Solution 1. First option. We have,
$$
2+\log _{2} 6=3+\log _{2} 3>3+\log _{2} 2 \sqrt{2}=\frac{9}{2}>2 \sqrt{5}
$$
Alternatively, the estimates can be written as,
$$
2 \sqrt{5}-2<2 \cdot \frac{9}{4}-2=\frac{5}{2}<\log _{2} 4 \sqrt{2}<\log _{2} 6
$$
Topic: "Powers with Real Exponents. Logarithms" | 2+\log_{2}6>2\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,297 |
1. Determine for which values of $a$ there exist real numbers $x$ and $y$ that satisfy the equation $\sqrt{2 x y+a}=x+y+17$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 6.I.1. The given equation is equivalent to the system
$$
\left\{\begin{array}{l}
2 x y + a = (x + y + 17)^{2} \\
x + y + 17 \geqslant 0
\end{array}\right.
$$
The first equation can be transformed into the form \((x + 17)^{2} + (x + 17)^{2} = a + 17^{2}\), which, for \(a > -289\), represents the equation of a circle w... | \geqslant-\frac{289}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,298 |
2. Solve the equation $10 x^{3}+12 x^{2}+6 x+1=0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution 1. The equation can be rewritten as $2 x^{3}+(2 x+1)^{3}=0$, so $x \sqrt[3]{2}=-2 x-1$, from which it follows that $x=-\frac{1}{2+\sqrt[3]{2}}$.
Answer: $x=-\frac{1}{2+\sqrt[3]{2}}$. | -\frac{1}{2+\sqrt[3]{2}} | Algebra | proof | Yes | Yes | olympiads | false | 43,299 |
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 | 43,300 |
4. Prove that for any $a \geqslant 2, b \geqslant 2$ and $c \geqslant 2$ the inequality
$$
\log _{b+c} a^{2}+\log _{a+c} b^{2}+\log _{a+b} c^{2} \geqslant 3
$$
holds. | 6.I.4. Note that if $x \geqslant y \geqslant 2$, then $x y \geqslant 2 x \geqslant x+y$. Therefore, the inequalities $a b \geqslant a+b$, $b c \geqslant b+c$, and $a c \geqslant a+c$ hold. Consequently,
$$
\log _{b+c} a^{2}=\frac{2 \lg a}{\lg (b+c)} \geqslant \frac{2 \lg a}{\lg (b c)}=\frac{2 \lg a}{\lg b+\lg c}
$$
S... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,301 |
5. Find the conditions on the coefficients of the polynomial $x^{3}+a x^{2}+b x+c$, under which this polynomial has three distinct real roots that are consecutive terms of some geometric progression. | 6.I.5. Let $x_{1}, x_{2}, x_{3}$ be the roots of the polynomial, which are consecutive terms of a geometric progression. Since $x_{2}^{3}=x_{1} x_{2} x_{3} = -c$, then $x_{2}=-\sqrt[3]{c}$. Substituting the found value into the given polynomial, we get $-c + a \sqrt[3]{c^{2}} - b \sqrt[3]{c} + c = 0$, from which $a \sq... | ^{3}b^{3}, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,302 |
6. Prove that the sum of the squares of the distances from the vertices of a regular heptagon to an arbitrary line passing through its center does not depend on the position of this line. | 6.I.6. Let's choose a coordinate system with the origin at the center of the heptagon, and the x-axis coinciding with the considered line. For convenience, we will assume that the radius of the circle circumscribed around this heptagon is 1. The coordinates of the vertices \( A_{k} \) of the heptagon are pairs \(\left(... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,303 |
7. A circle and a parabola have exactly two common points, one of which is the point of tangency of the given parabola and the given circle. Is it true that the second common point is also their point of tangency?
## Methodological Block
A. The solutions to problems $8-9$ are provided below. Evaluate each solution an... | 6.I.7. Of course, we can arrange a circle so that it touches a parabola at two different points. For this, we can take a circle of small radius inside the parabola and lower it so that it touches the parabola. It is clear in this case that the center of this circle will lie on the axis of symmetry of the parabola. Howe... | No,itisnotcorrect | Geometry | proof | Yes | Yes | olympiads | false | 43,304 |
10. Point $P$ lies on the arc $A B$ of the circumcircle of an equilateral triangle $A B C$. Prove that $P C = P A + P B$.
---
Note: The translation maintains the original text's formatting and structure. | Solution 1. Without loss of generality, we can assume that the radius of the circle circumscribed around the triangle is 1. Place triangle $ABC$ on the coordinate plane so that its vertices have coordinates: $A\left(\cos \frac{\pi}{3} ;-\sin \frac{\pi}{3}\right), B\left(\cos \frac{\pi}{3} ; \sin \frac{\pi}{3}\right)$, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,307 |
11. Construct a quadratic trinomial $p(x)$ such that $p(a)=a^{4}$, $p(b)=b^{4}$ and $p(c)=c^{4}$ (here $a, b$ and $c$ are distinct given numbers). | Solution 1. Let $q(x)=x^{4}-x(x-a)(x-b)(x-c)$. It is clear that this is a cubic polynomial such that $q(a)=a^{4}, q(b)=b^{4}$ and $q(c)=c^{4}$. Its leading coefficient is the number $a+b+c$. Therefore, the desired quadratic trinomial is the trinomial
$$
\begin{gathered}
p(x)=x^{4}-x(x-a)(x-b)(x-c)-(a+b+c)(x-a)(x-b)(x-... | p(x)=(^{2}+b^{2}+^{2}+++)x^{2}-(+b)(b+)(+)x+(+b+) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,308 |
12. Let $A, B$ and $C$ be points in space that do not lie on the same line. Denote by $S$ the area of triangle $ABC$. Prove that
$$
S^{2}=S_{x y}^{2}+S_{y z}^{2}+S_{x z}^{2}
$$
where $S_{x y}, S_{y z}$ and $S_{x z}$ are the areas of the projections of triangle $ABC$ onto the coordinate planes $O x y, O y z$ and $O x ... | 6.I.12. Proof 1. Without loss of generality, we can assume that vertex $A$ coincides with the origin. Let the other two vertices have coordinates $B\left(x_{1} ; y_{1} ; z_{1}\right)$ and $C\left(x_{2} ; y_{2} ; z_{2}\right)$. Since
$$
|\overline{O B}||\overline{O C}| \cos \angle B O C=\overline{O B} \cdot \overline{O... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,309 |
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 | 43,310 |
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 | 43,311 |
3. Construct a line passing through a vertex of a convex quadrilateral and dividing it into two equal-area parts.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided translation is already in the requested for... | 6.II.3. Let $E$ be the midpoint of the diagonal $BD$ of quadrilateral $ABCD$. If $E \in AC$, then $AC$ is the desired line. Let $E \notin AC$. Without loss of generality, assume that points $B$ and $E$ lie on the same side of diagonal $AC$ (figure).
 Prove that any section of a regular tetrahedron by a plane passing through the midpoints of two of its skew edges divides the tetrahedron into equal volumes.
b) Prove that the statement formulated in the previous part is true for any (not necessarily regular) tetrahedron. | 6.II.4a. The line passing through the midpoints of two skew edges of a regular tetrahedron is the perpendicular bisector of each of these edges. Therefore, when rotated by an angle of $180^{\circ}$ around this line, each of these edges, and hence the entire tetrahedron, is mapped onto itself. Any plane containing this ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,313 |
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 | 43,314 |
6. Evaluate the three solutions and answers provided below. Identify all errors and shortcomings, and provide the correct solution to this problem.
From a deck of cards, six cards were selected - three of black suits and three of red suits. The selected cards were shuffled and stacked. Find the probability that the th... | 6.II.6. Comment. In all the given solutions, the same mistake has been made. Namely, the systems of events considered in them do not consist of equally probable events.
Example of a correct solution. The number of all possible arrangements of the cards is $6!=720$. In $36=3!\cdot 3!$ of these arrangements, the red car... | \frac{1}{10} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 43,315 |
7. Solve the following problem using as many methods as possible. Methods are considered different if they use different mathematical ideas, as well as different technical approaches to implementing the same idea. Indicate the place of each method used in the school mathematics curriculum.
A quadrilateral is inscribed... | Solution 1. We will sequentially perform three axial symmetries: relative to the line $B C$, the line $C D$, and the line $A_{1} D_{1}$ - the image of the line $A D$ under the first symmetry (figure).
.
The perimeter of quadrilateral $K L M N$ coincides with the length of the broken line $K_{1} L M N K_{2}$, which is n... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,317 |
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 | 43,318 |
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 | 43,321 |
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 | 43,324 |
8*. Replace the asterisks with digits so that the seven-digit number $30 * 0 * 03$ is divisible by 13. Find all solutions. | 8. $3000803,3020303,3030703,3050203,3060603,3080103$, 3090503. | 3000803,3020303,3030703,3050203,3060603,3080103,3090503 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,325 |
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 | 43,326 |
10*. A student passed 31 exams over 5 years of study. In each subsequent year, he passed more exams than in the previous year. In the fifth year, the number of exams was three times that of the first year. How many exams were there in the fourth year?
Could the student have had only one exam in the first year? Check a... | 10. Solutions. The student could not have even two exams in the first year, as in this case, the total number of exams over five years would be less than 31.4 exams, and he also could not have more than three exams initially. Indeed, in this case, he would have to have no less than $4+$ $+5+6+7+12=34$ exams, i.e., more... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,327 |
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 | 43,334 |
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 | 43,338 |
11. Do there exist such integers $x$ and $y$ for which the following equalities hold simultaneously:
$$
x y=4747 \text { and } x-y=-54 ?
$$ | 11. Solution. The product $x y$ of integers $x$ and $y$ can only equal 4747 in four cases: $1 \cdot 4747, 47 \cdot 101$, $(-1) \cdot(-4747)$, and $(-47) \cdot(-101)$. From the second condition, it follows that $x$ is less than $y$. By checking, we find that both conditions are only satisfied in the cases where $x=47$ a... | 47,101or-101,-47 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,339 |
12. Find the set of integer values of the letter $a$, for which the product $(5+a) \cdot(3-a)$ is positive. | 12. $\{-4 ; -3 ; \ldots ; 1 ; 2\}$. | {-4;-3;\ldots;1;2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,340 |
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 | 43,341 |
15*. Vovochka wrote on the board: $1 * 2 * 3 * \ldots * 9=21$, placing either a plus or a minus instead of each asterisk. Sasha changed some of the signs to their opposites, and as a result, instead of the number 21, he wrote the number 20. Prove that one of the boys made a mistake. | 15. Solution. The maximum value of the sum is 45. When the sign of any addend $a$ is changed, the sum changes by $2a$ and, consequently, remains odd. Obviously, the sum of all numbers with any combination of plus and minus signs can only take odd values, positive or negative. Therefore, by writing 20 in the answer, Sas... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 43,343 |
16*. Among the numbers $a, b$, and $c$, one is positive, one is negative, and one is zero. Additionally, it is known that $|a|=b^{2}(b-c)$. Which of the numbers is positive, which is negative, and which is zero? | 16. Solution. If $b=0$, then $a=0$, which contradicts the condition. If $a=0$, then either $b=a=0$ or $b=c$, which in both cases contradicts the condition. The only remaining case is $c=0$. Then the given equation takes the form: $|a|=b^{3}$. Since $|a|>0$, $b$ must be a positive number. Therefore, $a \neq 0, b>0, c=0$... | \neq0,b>0,=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,344 |
18*. Each of the variables $a, b, c, d, e, f$ is defined on the set of numbers $\{1; -1\}$. On which set is the variable $x$ defined, if $x=a-b+c-d+e-f$? | 18. Solution. The maximum value of the sum is 6, and the minimum is -6. When the sign of any number is changed, the sum changes by 2. Therefore, the sum can only take one of the following values: $\{-6; -4; -2; 0; 2; 4; 6\}$. It is not difficult to provide examples confirming that the sum can take each of these values. | {-6;-4;-2;0;2;4;6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,346 |
1. Represent as a product of powers of prime numbers $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \ldots \cdot 14 \cdot 15$ | 1. $2^{11} \cdot 3^{5} \cdot 5^{3} \cdot 7^{2} \cdot 11 \cdot 13$ | 2^{11}\cdot3^{5}\cdot5^{3}\cdot7^{2}\cdot11\cdot13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,347 |
2. Is there a natural number $n$ such that the number $2^{n}+15$ is composite? | 2. For $n=7$ we get: $2^{7}+15=143=11 \cdot 13$ - a composite number. | 143=11\cdot13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,348 |
3. A sum of $a$ rubles was deposited in a savings bank at the beginning of the year. After one year, this amount increased by $2 \%$. The newly obtained amount increased again by $2 \%$ after another year, and so on. How many rubles will be received after $n$ years? $(n \in N)$ | 3. S o l u t i o n. After one year, $102\% = 1.02$ of the invested amount will be received, i.e., $1.02a$ rubles. This amount will increase by two percent after another year, and the deposit will amount to $1.02 \cdot 1.02a = 1.02^2a$ rubles. After $n$ years, $1.02^n a$ rubles will be received. | 1.02^n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,349 |
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 | 43,350 |
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 | 43,354 |
1. Write the set of all distinct divisors of the number 660. Write the set of all divisors of the number 72. Form the set of pairs of numbers that satisfy the following conditions: one number is taken from the first set and one number from the second set, and such that the difference between the first and the second nu... | 1. Divisors of the number 660: $1,2,3,4,5,6,10,11,12,15,20,22$, $30,33,44,55,60,66,110, \ldots$.
Divisors of the number 72: $1,2,3,4,6,8,9,12,18,24,36,72$.
The desired pairs of numbers: 6 and 2, 10 and 6, 12 and 8, 22 and 18. | (6,2),(10,6),(12,8),(22,18) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,355 |
3. Sixth-grader Sasha wrote on the board: $(-2)-(+2)+5=$ $=1$. What other numerical equalities could he have written, using all four numbers: $-2, 2, 1$, and 5 each time? (We will consider, for example, $5+(-2)-(+2)=1$ and $1=(-2)+5-(+2)$ to be the same as the first record.) | 3. More than 15 examples can be provided. Here are some of them:
$$
\begin{array}{ll}
2-(-2)+1=5, & 5-(-2)^{2}=1 \\
2+1-5=-2, & (1-5):(-2)=2 \\
5+2 \cdot(-2)=1, & 5^{2+(-2)}=1 \text { and so on. }
\end{array}
$$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,357 |
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 | 43,358 |
11. At a railway station, there are $k$ traffic lights, each of which can transmit three signals: red, yellow, and green. How many different signals can be transmitted using all the traffic lights? | 11. Solution. Using two semaphores, one can transmit $3 \cdot 3=3^{2}$ signals, using three $-3^{2} \cdot 3=$ $=3^{3}$ different signals, using four $-3^{3} \cdot 3=3^{4}$ signals, and so on. Using $k$ semaphores, one can transmit $3^{k}$ different signals. | 3^k | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 43,360 |
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 | 43,361 |
1. Let $A$ be the set of numbers of the form $6 n-1$, and $P$ be the set of prime numbers. Is the statement $A \subset P$ true or false, if $n \in N$? | 1. Solutions. By constructing a table
| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $\ldots$. |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $6 n-1$ | 5 | 11 | 17 | 23 | 29 | $35=5 \cdot 7$ | $\ldots$ | $\ldots$ | $\ldots$. |
we see that already at $n=6$ we obtain a composite number. ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,362 |
3. Let $n$ be a natural number. Are numbers of the form $n^{2}+$ $+3 n+1$ always prime? | 3. Let's construct a table:
| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $n^{2}+3 n+1$ | 5 | 11 | 19 | 29 | 41 | $55=11 \cdot 5$ | $\ldots$ | $\ldots$ | $\ldots$. |
Not all numbers of the given form are prime. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,364 |
5. If a number is a square of a natural number, then the sum of its distinct divisors is odd. Formulate the converse theorem. Is it true? | 5. Solution. The converse theorem: "If the sum of the divisors of a natural number is odd, then this number is a perfect square." This statement is refuted by the following example. The sum of the divisors of the number 2 is $1+2=3$, i.e., an odd number. However, the number 2 is not a perfect square. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,366 |
6. Is the equality $\left(a^{4}-1\right)^{6}=\left(a^{6}-1\right)^{4}$ an identity if: 1) $a$ is any number; 2) $a=\{-1 ; 0 ; 1\}$?
There are cases where it is quite difficult to find an example that refutes a general statement. More often, it is easier to prove that such an example exists than to specify it. For inst... | 6. The equality does not hold, for example, when $a=2$, and therefore is not an identity. In the second case, the equality is an identity on the given set.
When solving problems 7-10, the so-called "Dirichlet's principle," named after the German scientist Peter Dirichlet (1805-1859), is used. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,367 |
7. Take any two-digit number and square it. From the result, only the tens and units digits are kept. The new two-digit number obtained is squared again, and so on. Prove that, starting from some number, the results of the calculations will begin to repeat. | 7. Solution. The square of any number can end with no more than 100 "combinations" of the last two digits: $00,01,02, \ldots$ ..., 99. Therefore, no later than on the hundredth step, the obtained result should coincide with one of the previous ones. Naturally, all subsequent calculations will coincide with some of the ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,368 |
8. Prove that when dividing 41 by 61, the resulting infinite decimal fraction will be periodic. | 8. Solution. When dividing 41 by 61, the remainders can only be: $1, 2, \ldots, 60$. Therefore, at least the 61st remainder will be equal to one of the previous ones. Naturally, subsequent remainders will repeat. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,369 |
1. The members of the sequences are defined by the formulas:
1) \(x=|n-3|\)
4) \(t=2^{|n-2|+1}\)
2) \(y=(-1)^{n}\)
5) \(v=(-1)^{n} \cdot n\)
3) \(z=(-1)^{n}-|n-5|\)
6) \(w=\left|2^{n}-n^{2}\right|\).
Does each of these sequences have a greatest member? A least member?
32 | 1) The smallest term is the third. It equals zero.
2) The smallest terms have odd indices, they equal -1. The largest terms have even indices, each equals 1.
3) The largest terms are the fourth and sixth, each equals zero. There is no smallest term.
4) The smallest term is the second, there is no largest term.
5) T... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,371 |
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 | 43,372 |
5. Can we claim that the intersection of the set of numbers of the form \(2^{n}-n\) and the set of squares of natural numbers is an empty set? ( \(n>1, n \in N\) ) | 5. Solution. For \(n=7\) we get: \(2^{7}-7=121=11^{2}\). Therefore, there is at least one value of \(n\) for which the number of the given form is a square of a natural number. | 2^{7}-7=121=11^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,375 |
6. The German mathematician M. Stifel (1487-1567) claimed that numbers of the form \(2^{2 n+1}-1, n \in N\) are prime. Is he correct? | 6. No. For \(n=4\) we get: \(2^{9}-1=511=7 \cdot 73\). | 511=7\cdot73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,376 |
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 | 43,379 |
11*(!). Given the sum: \(2^{8}+2^{7}+2^{6}+2^{5}+2^{4}+2^{3}+2^{2}+2+1\). Is it possible to obtain any natural number from 1 to 511 inclusive by crossing out some of the addends? Find a method (algorithm) to do this, if it exists, of course. | 11. Possible. Solution. In this case, we will limit ourselves to a few examples to show how to do it.
\[
\begin{aligned}
& 405=2^{8}+149 \\
& 149=2^{7}+21 \\
& 21=2^{4}+5 \\
& 5=2^{2}+1 \\
& 405=2^{8}+2^{7}+2^{4}+2^{2}+1
\end{aligned}
\]
Similarly: \(57=2^{5}+2^{4}+2^{3}+1\). All calculations are simplified if you us... | 405=2^{8}+2^{7}+2^{4}+2^{2}+1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,380 |
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 | 43,381 |
14. It is known that the first term of the sequence is zero, and the sequence of differences (see problem 7 of this section) is: \(2,4,-1,0,-5,-3,3\). Construct a diagram. How can you find the last term of the corresponding numerical sequence without calculating the "non-edge" terms? | 14. 0. Instruction. Show that the last term of the sequence is equal to: \(0+(2+4-1+0-5-3+3)\). | 0+(2+4-1+0-5-3+3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,382 |
16*. A numerical sequence consists only of ones and zeros. If all terms standing at odd positions are erased, the remaining numbers form exactly the same numerical sequence.
Provide examples of such sequences. Provide examples of sequences possessing the same property, but if the terms standing at even positions are r... | 16. 1) \(1,1,0,1,1,0,1,1,0, \ldots\),
\(1,1,1,1,0,1,1,1,1,0, \ldots\)
2) \(1,0,0,1,0,0,1,0,0, \ldots\),
\(1,0,0,1,0,1,1,1,0,0,1,1, \ldots\). | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 43,384 |
1. From 22 matches, form a rectangle with the largest possible area.
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。 | 1. Dimensions of the rectangle: 5 6. Instruction. Consider all possible cases. (KvN, 1971, No. 4, p. 83.) | 5\times6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,385 |
2. In a certain month, three Saturdays fell on even dates. What day of the week was the 25th of the month? | 2. Monday. Instruction. Show that Saturday can only be the 2nd, 9th, 16th, 23rd, and 30th of the month. (MSHII, 1972, No. 6, p. 82.) | Monday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 43,386 |
3. There are 4 sticks 1 cm long; 4 sticks 2 cm long, 7 sticks 3 cm long, and 5 sticks 4 cm long. Can a rectangle be formed using all the sticks in the set? | 3. Solution. Suppose such a rectangle exists. Then its semi-perimeter, equal to the sum of the lengths of two adjacent sides, must be a non-integer. Contradiction. ( (KvANT, 1971, No. 5, p. 82.\()\) | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,387 |
4. Form the largest square using the sticks mentioned in the previous problem.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. The side of the square is \(13 \mathrm{~cm}\). One side of the square can be formed from sticks whose lengths in cm are 4, 4, 2, and 3. The second side can be formed from sticks whose lengths in cm are \(4,4,3,1,1\). The set for the third side: \(4,3,3,2,1\), and for the fourth: \(3,3,3,2,2\). (KvN, 1971, No. 5, p. ... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 43,388 | |
5. Given 173 numbers, each of which is either 1 or -1. Can they be divided into two groups such that the sums of the numbers in each group are equal? | 5. Solution. Suppose that such two groups exist. Then one group will have an odd number of addends, and the other - an even number. The sum of the numbers in one group will be odd, and in the other - even. Contradiction. (KvN, 1971, No. 2, p. 75.) | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 43,389 |
6. Three fractions with numerators of 1 and different denominators (natural numbers) sum up to 1. Find these fractions. | 6. \(\frac{1}{2}, \frac{1}{3}, \frac{1}{6}\). Hint. Show that the greatest of the sought fractions is \(\frac{1}{2}\). (KvN, 1972, No. 2, p. 72.) | \frac{1}{2},\frac{1}{3},\frac{1}{6} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,390 |
7*. There is a rectangular plate weighing 10 g. In what ways can it be cut into three parts, each weighing an integer number of grams, so that with their help, using scales, one could determine the weight of any object weighing 1, 2, ..., 10 g? | 7. 1 g, 2 g, 7 g or 1 g, 3 g, 6 g. Hint. First consider all cases of determining the mass of an object weighing 9 g. (KvN, 1972, No. 1, p. 74.) | 1,2,7or1,3,6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 43,391 |
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 | 43,392 |
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 | 43,393 |
13*. What digits are denoted by the letters \(x, y\) and \(z\), if the product of the number \(\overline{x y}\) and the number \(x\) equals \(\overline{z z z}\)? | 13. \(x=3, y=7, z=1\). Hint. Show that the number \(\overline{x y}\) is divisible by 37. | 3,7,1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,397 |
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 | 43,398 |
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 | 43,400 |
17*. For what values of the natural number \(n\) is the sum \(n^{2}+\) \(+(n+1)^{2}+(n+2)^{2}+(n+3)^{2}\) divisible by ten? | 17. \(n=5 k+1, k=\{0 ; 1 ; 2 ; \ldots\}\). Hint. Show that the last digit of the decimal representation of the number \(n\) can only be 1 or 6. | n=5k+1,k={0;1;2;\ldots} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,401 |
18*. Find all possible rectangles with integer sides, the perimeter of which is numerically equal to their area. | 18. 3 and 6 or 4 and 4. Hint. Show that if a rectangle is divided into squares with a side length of 1, then there should be exactly 4 squares that do not share any points with the sides of the rectangle. | 36or44 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,402 |
19. Find the smallest number, written only using ones and zeros, that is divisible by 225. | 19. 11111111 100. Note. The desired number should be divisible by 9 and 25. | 11111111100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,403 |
20. Can a billion (1000000 000) be written as the product of two numbers, neither of which contains a zero in its representation? | 20. It is possible: \(5^{9} \cdot 2^{9}\). | 5^{9}\cdot2^{9} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,404 |
2. Are the equalities correct: 1) \(96 \cdot 98 \cdot 189=81 \cdot 343 \cdot 2^{6}\); 2) \(12^{18}=\) \(\left.=27^{6} \cdot 16^{9} ; 3\right) 25^{28} \cdot 0.008^{19}=0.25\). | 2. 1) Solution. Since \(96=3 \cdot 2^{5} ; 98=2 \cdot 7^{2} ; 189=\) \(=3^{3} \cdot 7\), then \(96 \cdot 98 \cdot 189=3 \cdot 2^{5} \cdot 2 \cdot 7^{2} \cdot 3^{3} \cdot 7=2^{6} \cdot 7^{3} \times\) \(\times 3^{4}=81 \cdot 343 \cdot 2^{6}\).
2) The equality is true.
3) Solution. \(25^{28} \cdot 0.008^{19}=\left(5^{2}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,405 |
4. Prove or disprove that if \(a=x^{3} y, b=x^{2} y^{2}\), \(c=x y^{3}\), then for any \(x\) and \(y\) the following relations hold:
1) \(a c+b^{2}-2 x^{4} y^{4}=0\)
3) \(a b c+b^{3}>0\).
2) \(a y^{2}+c x^{2}=2 x y b\) | 4. 1) Solution. Since \(a c+b^{2}-2 x^{4} y^{4}=x^{3} y \cdot x y^{3}+\) \(+x^{4} y^{4}-2 x^{4} y^{4}=0\), the first equality is an identity.
2) Solution. \(a y^{2}+c x^{2}-2 x y b=x^{3} y y^{2}+x y^{3} x^{2}-2 x y x^{2} y^{2}=\) \(=0\).
The equality is an identity.
3) Solution. \(a b c+b^{3}=x^{3} y \cdot x^{2} y^{2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,407 |
5. Are the following equalities identities:
1) \(\left[-a^{3}(-a)^{3}\right]^{2}+\left[-a^{2}(-a)^{2}\right]^{3}=0\)
2)* \((-1)^{n} a^{n+k}=(-a)^{n} \cdot a^{k}\) ? | 5. Both equalities are identities. T h e y a r e.
1) \(\left(a^{3} \cdot a^{3}\right)^{2}+\left(-a^{2} \cdot a^{2}\right)^{3}=a^{12}-a^{12}=0\);
2) \((-1)^{n} a^{n+k}=(-1)^{n} \cdot a^{n} \cdot a^{k}=(-1 \cdot a)^{n} \cdot a^{k}=(-a)^{n} \cdot a^{k}\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,408 |
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 | 43,409 |
9. Each of the numbers \(a, b\) and \(с\) is not equal to zero. It is known that the numbers \((-2)^{8} a^{3} b^{3} c^{n-1}\) and \((-3)^{3} a^{2} b^{5} c^{n+1}\) have the same sign. Determine the sign of the number \(a\). | 9. Solution. The numbers \((-2)^{8}\) and \((-3)^{9}\) have different signs, the numbers \(b^{3}\) and \(b^{5}\) have the same sign, and the numbers \(c^{n-1}\) and \(c^{n+1}\) also have the same sign, since the difference between the exponents is two, and therefore both exponents are either even or odd simultaneously.... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,412 |
10. Find such natural values of \(x\) for which the following equalities hold:
1) \(2^{x-5}=2\)
2) \(2^{x}=512\)
3) \(x^{5}=243\)
4) \(x^{4}=625\). | 10. 1) \(x=6\)
3) \(x=3\)
2) \(x=9\)
4) \(x=5\). | 1)6,2)9,3)3,4)5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,413 |
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 | 43,415 |
13*. Given four numbers \(a, b, c, d\), each of which is not zero. Each of the first three numbers is multiplied by the one following it, and the fourth is multiplied by the first. The same operation is performed with the new set. Prove that eventually a set of positive numbers will always be obtained. | 13. Instruction. Show that the fifth set consists of squares of some numbers. | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,416 |
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 | 43,417 |
15. Each of the numbers, starting from the second, is equal to the previous one multiplied by the square of the first number. Prove that the product of the first, third, fifth, and seventh terms of the resulting sequence is equal to the fourth power of the fourth number. | 15. Instruction. The solution of the problem boils down to proving the identity: \(a \cdot a^{5} \cdot a^{9} \cdot a^{13}=a^{28}=\left(a^{7}\right)^{4}\).
42 | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,418 |
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 | 43,419 |
17(!). How many different divisors do the following numbers have:
1) \(2^{7}\);
2) \(5^{4}\);
3) \(2^{7} \cdot 5^{4} ;\)
4) \(2^{m} \cdot 5^{n} \cdot 3^{k}\);
5) 3600
6) \(42^{5}\) ? | 17. 5) \((4+1)(2+1)(2+1)=45\) divisors. 6) 216 divisors. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,420 |
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