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XXXIX OM - I - Problem 1
For each positive number $ a $, determine the number of roots of the polynomial $ x^3+(a+2)x^2-x-3a $. | Let's denote the considered polynomial by $ F(x) $. A polynomial of the third degree has at most three real roots. We will show that the polynomial $ F $ has at least three real roots - and thus has exactly three real roots (for any value of the parameter $ a > 0 $).
It is enough to notice that
If a continuous func... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
XX OM - II - Task 2
Find all four-digit numbers in which the thousands digit is equal to the hundreds digit, and the tens digit is equal to the units digit, and which are squares of integers. | Suppose the number $ x $ satisfies the conditions of the problem and denote its consecutive digits by the letters $ a, a, b, b $. Then
The number $ x $ is divisible by $ 11 $, so as a square of an integer, it is divisible by $ 11^2 $, i.e., $ x = 11^2 \cdot k^2 $ ($ k $ - an integer), hence
Therefore,
... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
V OM - I - Task 2
Investigate when the sum of the cubes of three consecutive natural numbers is divisible by $18$. | Let $ a - 1 $, $ a $, $ a + 1 $, be three consecutive natural numbers; the sum of their cubes
can be transformed in the following way:
Since one of the numbers $ a - 1 $, $ a $, $ a + 1 $ is divisible by $ 3 $, then one of the numbers $ a $ and $ (a + 1) (a - 1) + 3 $ is also divisible by $ 3 $. Therefore, the sum $ ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
LIX OM - II - Task 1
Determine the maximum possible length of a sequence of consecutive integers, each of which can be expressed in the form $ x^3 + 2y^2 $ for some integers $ x, y $. | A sequence of five consecutive integers -1, 0, 1, 2, 3 satisfies the conditions of the problem: indeed, we have
On the other hand, among any six consecutive integers, there exists a number, say
$ m $, which gives a remainder of 4 or 6 when divided by 8. The number $ m $ is even; if there were a representation
in... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XV OM - I - Problem 11
In triangle $ ABC $, angle $ A $ is $ 20^\circ $, $ AB = AC $. On sides $ AB $ and $ AC $, points $ D $ and $ E $ are chosen such that $ \measuredangle DCB = 60^\circ $ and $ \measuredangle EBC = 50^\circ $. Calculate the angle $ EDC $. | Let $ \measuredangle EDC = x $ (Fig. 9). Notice that $ \measuredangle ACB = \measuredangle $ABC$ = 80^\circ $, $ \measuredangle CDB = 180^\circ-80^\circ-60^\circ = 40^\circ $, $ \measuredangle CEB = 180^\circ - 80^\circ-50^\circ = \measuredangle EBC $, hence $ EC = CB $. The ratio $ \frac{DC}{CE} $ of the sides of tr... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXVIII - II - Task 3
In a hat, there are 7 slips of paper. On the $ n $-th slip, the number $ 2^n-1 $ is written ($ n = 1, 2, \ldots, 7 $). We draw slips randomly until the sum exceeds 124. What is the most likely value of this sum? | The sum of the numbers $2^0, 2^1, \ldots, 2^6$ is $127$. The sum of any five of these numbers does not exceed $2^2 + 2^3 + 2^4 + 2^5 + 2^6 = 124$. Therefore, we must draw at least six slips from the hat.
Each of the events where we draw six slips from the hat, and the seventh slip with the number $2^{n-1}$ ($n = 1, 2, ... | 127 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
XLII OM - I - Problem 8
Determine the largest natural number $ n $ for which there exist in space $ n+1 $ polyhedra $ W_0, W_1, \ldots, W_n $ with the following properties:
(1) $ W_0 $ is a convex polyhedron with a center of symmetry,
(2) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) is obtained from $ W_0 $ by ... | Suppose that polyhedra $W_0, W_1, \ldots, W_n$ satisfy the given conditions. Polyhedron $W_1$ is the image of $W_0$ under a translation by a certain vector $\overrightarrow{\mathbf{v}}$ (condition (2)). Let $O_0$ be the center of symmetry of polyhedron $W_0$ (condition (1)); the point $O_1$, which is the image of $O_0$... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXIV OM - III - Task 2
Let $ p_n $ be the probability that a series of 100 consecutive heads will appear in $ n $ coin tosses. Prove that the sequence of numbers $ p_n $ is convergent and calculate its limit. | The number of elementary events is equal to the number of $n$-element sequences with two values: heads and tails, i.e., the number $2^n$. A favorable event is a sequence containing 100 consecutive heads. We estimate the number of unfavorable events from above, i.e., the number of sequences not containing 100 consecutiv... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
XXII OM - III - Problem 5
Find the largest integer $ A $ such that for every permutation of the set of natural numbers not greater than 100, the sum of some 10 consecutive terms is at least $ A $. | The sum of all natural numbers not greater than $100$ is equal to $1 + 2 + \ldots + 100 = \frac{1 + 100}{2} \cdot 100 = 5050$. If $a_1, a_2, \ldots, a_{100}$ is some permutation of the set of natural numbers not greater than $100$ and the sum of any $10$ terms of this permutation is less than some number $B$, then in p... | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XV OM - II - Task 3
Prove that if three prime numbers form an arithmetic progression with a difference not divisible by 6, then the smallest of these numbers is $3$. | Suppose that the prime numbers $ p_1 $, $ p_2 $, $ p_3 $ form an arithmetic progression with a difference $ r > 0 $ not divisible by $ 6 $, and the smallest of them is $ p_1 $. Then
Therefore, $ p_1 \geq 3 $, for if $ p_1 = 2 $, the number $ p_3 $ would be an even number greater than $ 2 $, and thus would not be a pri... | 3 | Number Theory | proof | Yes | Yes | olympiads | false |
XLIII OM - I - Problem 2
In square $ABCD$ with side length $1$, point $E$ lies on side $BC$, point $F$ lies on side $CD$, the measures of angles $EAB$ and $EAF$ are $20^{\circ}$ and $45^{\circ}$, respectively. Calculate the height of triangle $AEF$ drawn from vertex $A$. | The measure of angle $ FAD $ is $ 90^\circ - (20^\circ + 45^\circ) = 25^\circ $. From point $ A $, we draw a ray forming angles of $ 20^\circ $ and $ 25^\circ $ with rays $ AE $ and $ AF $, respectively, and we place a segment $ AG $ of length $ 1 $ on it (figure 2).
From the equality $ |AG| =|AB| = 1 $, $ | \measureda... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
L OM - I - Problem 11
In an urn, there are two balls: a white one and a black one. Additionally, we have 50 white balls and 50 black balls at our disposal. We perform the following action 50 times: we draw a ball from the urn, and then return it to the urn along with one more ball of the same color as the drawn ball. ... | Let $ P(k,n) $, where $ 1 \leq k\leq n-1 $, denote the probability of the event that when there are $ n $ balls in the urn, exactly $ k $ of them are white. Then
Using the above relationships, we prove by induction (with respect to $ n $) that $ P(k,n) = 1/(n-1) $ for $ k = 1,2,\ldots,n-1 $. In particular
T... | 51 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XLVI OM - III - Problem 2
The diagonals of a convex pentagon divide this pentagon into a pentagon and ten triangles. What is the maximum possible number of triangles with equal areas? | om46_3r_img_12.jpg
Let's denote the considered pentagon by $ABCDE$, and the pentagon formed by the intersection points of the diagonals by $KLMNP$ so that the following triangles are those mentioned in the problem:
$\Delta_0$: triangle $LEM$; $\quad \Delta_1$: triangle $EMA$;
$\Delta_2$: triangle $MAN$; $\quad \Delta... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
LIX OM - I - Task 9
Determine the smallest real number a with the following property:
For any real numbers $ x, y, z \geqslant a $ satisfying the condition $ x + y + z = 3 $
the inequality holds | Answer: $ a = -5 $.
In the solution, we will use the following identity:
Suppose that the number $ a \leqslant 1 $ has the property given in the problem statement. The numbers
$ x = a, y = z = 2\cdot (\frac{3-a}{2}) $ satisfy the conditions $ x, y, z \geqslant a $ and $ x+y+z = 3 $, thus
by virtue of (1),... | -5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
XLIV OM - I - Problem 11
In six different cells of an $ n \times n $ table, we place a cross; all arrangements of crosses are equally probable. Let $ p_n $ be the probability that in some row or column there will be at least two crosses. Calculate the limit of the sequence $ (np_n) $ as $ n \to \infty $. | Elementary events are determined by six-element subsets of the set of $n^2$ cells of the table; there are $\binom{n^2}{6}$ of them. Let $\mathcal{Z}$ be the complementary event to the event considered in the problem. The configurations favorable to event $\mathcal{Z}$ are obtained as follows: we place the first cross i... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XV OM - I - Problem 7
Given a circle and points $ A $ and $ B $ inside it. Find a point $ P $ on this circle such that the angle $ APB $ is subtended by a chord $ MN $ equal to $ AB $. Does the problem have a solution if the given points, or only one of them, lie outside the circle? | Suppose that point $ P $ of a given circle $ C $ with radius $ r $ is a solution to the problem (Fig. 7).
Since points $ A $ and $ B $ lie inside the circle $ C $, points $ M $ and $ N $ lie on the rays $ PA $ and $ PB $ respectively, and angle $ APB $ coincides with angle $ MPN $. Triangles $ APB $ and $ MPN $ have eq... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
XXVI - I - Task 1
At the ball, there were 42 people. Lady $ A_1 $ danced with 7 gentlemen, Lady $ A_2 $ danced with 8 gentlemen, ..., Lady $ A_n $ danced with all the gentlemen. How many gentlemen were at the ball? | The number of ladies at the ball is $ n $, so the number of gentlemen is $ 42-n $. The lady with number $ k $, where $ 1 \leq k \leq n $, danced with $ k+6 $ gentlemen. Therefore, the lady with number $ n $ danced with $ n+ 6 $ gentlemen. These were all the gentlemen present at the ball. Thus, $ 42-n = n + 6 $. Solving... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
L OM - II - Task 5
Let $ S = \{1, 2,3,4, 5\} $. Determine the number of functions $ f: S \to S $ satisfying the equation $ f^{50} (x) = x $ for all $ x \in S $.
Note: $ f^{50}(x) = \underbrace{f \circ f \circ \ldots \circ f}_{50} (x) $. | Let $ f $ be a function satisfying the conditions of the problem. For numbers $ x \neq y $, we get $ f^{49}(f(x)) = x \neq y = f^{49}(f(y)) $, hence $ f(x) \neq f(y) $. Therefore, $ f $ is a permutation of the set $ S $. Denote by $ r(x) $ ($ x \in S $) the smallest positive integer such that $ f^{r(x)}(x) = x $. Then ... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
XXII OM - III - Task 3
How many locks at least need to be placed on the treasury so that with a certain distribution of keys among the 11-member committee authorized to open the treasury, any 6 members can open it, but no 5 can? Determine the distribution of keys among the committee members with the minimum number of ... | Suppose that for some natural number $ n $ there exists a key distribution to $ n $ locks among an 11-member committee such that the conditions of the problem are satisfied. Let $ A_i $ denote the set of locks that the $ i $-th member of the committee can open, where $ i = 1, 2, \ldots, 11 $, and let $ A $ denote the s... | 462 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Consider the set $M$ of integers $n \in[-100 ; 500]$, for which the expression $A=n^{3}+2 n^{2}-5 n-6$ is divisible by 11. How many integers are contained in $M$? Find the largest and smallest of them? | Answer: 1) 164 numbers; 2) $n_{\text {min }}=-100, n_{\text {max }}=497$. | 164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Solution. According to the problem, the sum of the original numbers is represented by the expression:
$$
\begin{aligned}
& \left(a_{1}+2\right)^{2}+\left(a_{2}+2\right)^{2}+\ldots+\left(a_{50}+2\right)^{2}=a_{1}^{2}+a_{2}^{2}+\ldots+a_{50}^{2} \rightarrow \\
& {\left[\left(a_{1}+2\right)^{2}-a_{1}^{2}\right]+\left[... | Answer: will increase by 150. | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A set of 60 numbers is such that adding 3 to each of them does not change the value of the sum of their squares. By how much will the sum of the squares of these numbers change if 4 is added to each number? | Answer: will increase by 240. | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A set of 70 numbers is such that adding 4 to each of them does not change the magnitude of the sum of their squares. By how much will the sum of the squares of these numbers change if 5 is added to each number? | Answer: will increase by 350. | 350 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A set of 80 numbers is such that adding 5 to each of them does not change the magnitude of the sum of their squares. By how much will the sum of the squares of these numbers change if 6 is added to each number? | Answer: will increase by 480. | 480 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the fraction $\frac{p}{q}$ with the smallest possible natural denominator, for which $\frac{1}{2014}<\frac{p}{q}<\frac{1}{2013}$. Enter the denominator of this fraction in the provided field | 5. Find the fraction $\frac{p}{q}$ with the smallest possible natural denominator, for which
$\frac{1}{2014}<\frac{p}{q}<\frac{1}{2013}$. Enter the denominator of this fraction in the provided field
Answer: 4027 | 4027 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. If $\quad a=\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}, \quad$ then $\quad P(a)=\overline{a_{6} a_{1} a_{2} a_{3} a_{4} a_{5}}$, $P(P(a))=\overline{a_{5} a_{6} a_{1} a_{2} a_{3} a_{4}} \quad$ with $\quad a_{5} \neq 0, a_{6} \neq 0, a_{1} \neq 0 . \quad$ From the equality $P(P(a))=a$ it follows that $a_{1}=a_{5},... | Answer: 1) 81 is the number; 2) $a=\overline{t u t u t u}, t, u$, where $t, u$ - are any digits, not equal to zero. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Square the numbers $a=10001, b=100010001$. Extract the square root of the number $c=1000200030004000300020001$. | 1) $a^{2}=100020001$; 2) $b^{2}=10002000300020001$; 3) $\sqrt{c}=1000100010001$. | 1000100010001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. By what natural number can the numerator and denominator of the ordinary fraction of the form $\frac{5 n+3}{7 n+8}$ be reduced? For which integers $n$ can this occur? | Answer: it can be reduced by 19 when $n=19k+7, k \in Z$. | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. By what natural number can the numerator and denominator of the ordinary fraction of the form $\frac{4 n+3}{5 n+2}$ be reduced? For which integers $n$ can this occur? | Answer: can be reduced by 7 when $n=7 k+1, k \in Z$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The angle at vertex $B$ of triangle $A B C$ is $130^{\circ}$. Through points $A$ and $C$, lines perpendicular to line $A C$ are drawn and intersect the circumcircle of triangle $A B C$ at points $E$ and $D$. Find the acute angle between the diagonals of the quadrilateral with vertices at points $A, C, D$ and $E$.
P... | Solution. Let $a$ be the number of students in the first category, $c$ be the number of students in the third category, and $b$ be the part of students from the second category who will definitely lie in response to the first question (and say "YES" to all three questions), while the rest of the students from this cate... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a convex quadrilateral $A B C D$, the lengths of sides $B C$ and $A D$ are 2 and $2 \sqrt{2}$ respectively. The distance between the midpoints of diagonals $B D$ and $A C$ is 1. Find the angle between the lines $B C$ and $A D$. | Answer: $\alpha=45^{\circ}$. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Let's introduce the notation: $A B=2 c, A C=2 b, \measuredangle B A C=\alpha$. The feet of the perpendicular bisectors are denoted by points $P$ and $Q$. Then, in the right triangle $\triangle A M Q$, the hypotenuse $A M=\frac{b}{\cos \alpha}$. And in the right triangle $\triangle A N P$, the hypotenuse $A N=\frac{c... | Answer: $60^{\circ}$ or $120^{\circ}$. | 60 | Geometry | proof | Yes | Yes | olympiads | false |
5. In triangle $A B C$, the perpendicular bisectors of sides $A B$ and $A C$ intersect lines $A C$ and $A B$ at points $N$ and $M$ respectively. The length of segment $N M$ is equal to the length of side $B C$ of the triangle. The angle at vertex $C$ of the triangle is $40^{\circ}$. Find the angle at vertex $B$ of the ... | Answer: $80^{\circ}$ or $20^{\circ}$.
## Final round of the "Rosatom" Olympiad, 9th grade, CIS, February 2020
# | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The sum $b_{6}+b_{7}+\ldots+b_{2018}$ of the terms of the geometric progression $\left\{b_{n}\right\}, b_{n}>0$ is 6. The sum of the same terms taken with alternating signs $b_{6}-b_{7}+b_{8}-\ldots-b_{2017}+b_{2018}$ is 3. Find the sum of the squares of the same terms $b_{6}^{2}+b_{7}^{2}+\ldots+b_{2018}^{2}$. | Answer: $b_{6}^{2}+b_{7}^{2}+\ldots+b_{2018}^{2}=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Let $A=\overline{a b c b a}$ be a five-digit symmetric number, $a \neq 0$. If $1 \leq a \leq 8$, then the last digit of the number $A+11$ will be $a+1$, and therefore the first digit in the representation of $A+11$ should also be $a+1$. This is possible only with a carry-over from the digit, i.e., when $b=c=9$. Then... | Answer: eight numbers of the form $\overline{a 999 a}$, where $a=1,2, \ldots, 8$. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 513315 is symmetric, while 513325 is not. How many six-digit symmetric numbers exist such that adding 110 to them leaves them symmetric? | Answer: 81 numbers of the form $\overline{a b 99 b a}$, where $a=1,2, \ldots, 9, b=0,1,2, \ldots, 8$. | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 12 horizontal and 16 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100m. Each block has an address consisting of two integers $(i ; j), i=1,2,... | Answer: 165 blocks; $c_{\min }=4$ coins, $c_{\max }=8$ coins. | 165 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 5134315 is symmetric, while 5134415 is not. How many seven-digit symmetric numbers exist such that adding 1100 to them leaves them symmetric? | Answer: 810 numbers of the form $\overline{a b c 9 c b a}$, where $a=1,2, \ldots, 9$, $b=0,1,2, \ldots, 9, c=0,1,2, \ldots, 8$. | 810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 7 horizontal and 13 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100 m. Each block has an address consisting of two integers $(i ; j), i=1,2,... | # Answer: 72 blocks; $c_{\min }=8$ coins, $c_{\max }=12$ coins. | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Integers, whose decimal notation reads the same from left to right and from right to left, we will call symmetric. For example, the number 513151315 is symmetric, while 513152315 is not. How many nine-digit symmetric numbers exist such that adding 11000 to them leaves them symmetric? | Answer: 8100 numbers of the form $\overline{a b c d 9 d c b a}$, where $a=1,2, \ldots, 9$, $b=0,1,2, \ldots, 9, c=0,1,2, \ldots, 9, d=0,1,2, \ldots, 8$. | 8100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In city "N", there are 10 horizontal and 12 vertical streets, of which a pair of horizontal and a pair of vertical streets form the rectangular boundary of the city, while the rest divide it into blocks that are squares with a side length of 100 meters. Each block has an address consisting of two integers $(i ; j), ... | Answer: 99 blocks; $c_{\min }=10$ coins, $c_{\max }=14$ coins. | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Kuzya the flea can make a jump in any direction on a plane for exactly 19 mm. Her task is to get from point $A$ to point $B$ on the plane, the distance between which is 1812 cm. What is the minimum number of jumps she must make to do this? | Answer: $n_{\min }=954$ jumps. | 954 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. A plot of land in the form of a right-angled triangle with legs of 4 and 3 needs to be divided by a line $L$ into two plots such that 1) the plots have equal area; 2) the length of the common boundary (fence) of the plots is minimized. Indicate the points on the sides of the triangle through which the desired line $... | Problem 6. Answer: 1) The line intersects the larger leg $B C$ (angle $\measuredangle C=90^{\circ}$) at point $M: B M=\sqrt{10}$ and the hypotenuse $B A$ at point $N: B N=\sqrt{10}$.
2) $L_{\text {min }}=2$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. When purchasing goods for an amount of no less than 1000 rubles, the store provides a discount on subsequent purchases of $50 \%$. Having 1200 rubles in her pocket, Dasha wanted to buy 4 kg of strawberries and 6 kg of sugar. In the store, strawberries were sold at a price of 300 rubles per kg, and sugar - at a price... | First purchase: 3 kg of strawberries, 4 kg of sugar. Its cost is $300 \times 3 + 4 \times 30 = 1020$ rubles. Second purchase: 1 kg of strawberries, 2 kg of sugar. With a $50\%$ discount, its price is $(300 + 60) \cdot 0.5 = 180$ rubles. The total amount of both purchases is $1200$ rubles. | 1200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. According to the properties of logarithms, after transformations we get
$$
\begin{aligned}
& \log _{2}\left(b_{2} b_{3} \ldots b_{n}\right)=\log _{2} b_{1}^{2} \\
& b_{2} b_{3} \ldots b_{n}=b_{1}^{2}
\end{aligned}
$$
Using the formula for the general term of a geometric progression, we get
$$
b_{2}=b_{1} q, b_{3}... | Answer: -12 when $n=4$ and $n=9$. | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $p_{1}=x-a$ can have a root $x=a$ coinciding with one of the roots of the product $p(x)=p_{1}(x) \cdot p_{2}(x)$.
Case $1 \quad a=1$
Then the polynomial $p_{2}(x)=(x-1)^{r}(x-2)^{s}(x+3)^{t}$, where $r \geq 1, s \geq 1, t \geq 1-$ are integers, $r+s+t=4$, satisfies the condition of the problem. The ... | Answer: $p_{1}(x)=x+3, p_{2}(x)=(x-1)(x-2)(x+3)^{2} ; a_{0}=21$ | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number $a$ is divisible by 21 and has 105 different divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\min }=2^{6} \cdot 3^{4} \cdot 7^{2}=254016$ | 254016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number $a$ is divisible by 35 and has 75 different divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\text {min }}=2^{4} \cdot 5^{4} \cdot 7^{2}=490000$. | 490000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number $a$ is divisible by 55 and has 117 distinct divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\min }=2^{12} \cdot 5^{2} \cdot 11^{2}=12390400$. | 12390400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The border of a square with a side of 9, cut out of white cardboard, is painted red. It is necessary to cut the square into 6 equal-area parts, the boundaries of which contain segments painted red with the same total length.
## Solutions
Option 1
Problem 1 | Answer: 2 km.
Solution
$S-$ the length of the path, $S_{1}$ - the length of the path on descents, $S_{2}$ - the length of the path on ascents, $S / 2=S_{1}+S_{2}$
$t=15-8-6=1-$ the time of the journey there and back. Then
$1=\frac{S_{1}}{6}+\frac{S_{2}}{3}+\frac{S}{2 \cdot 4}+\frac{S_{1}}{3}+\frac{S_{2}}{6}+\frac{S... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.17. Final round of the "Rosatom" Olympiad, 7th grade
# Answers and solutions
Problem 1 Answer: 9
There exists a set of 8 buttons in which there are no three buttons of the same color: each color has two buttons. In any set of 9 buttons, there will be at least one triplet of buttons of the same color.
If we assum... | Answer: 9 buttons.
Problem 2 Answer: 1261
\[
\left\{\begin{array}{l}
a=35 n+1 \\
a=45 m+1
\end{array} \rightarrow 35 n=45 m \rightarrow 7 n=9 m \rightarrow\left\{\begin{array}{c}
n=9 t \\
m=7 t, t \in Z
\end{array} \rightarrow\right.\right.
\]
\[
a=315 t+1 \geq 1000 \rightarrow t \geq 4 \rightarrow a_{\text {min }}=... | 1261 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In the decimal representation of a six-digit number $a$, there are no zeros and the sum of its digits is 18. Find the sum of all different numbers obtained from the number $a$ by cyclic permutations of its digits. In a cyclic permutation, all digits of the number, except the last one, are shifted one place to the ri... | 3. Solution. Case 1. The number $a=333333$. This number does not change under cyclic permutations, so it is the only one and the sum of the numbers is the number itself, that is, 333333.
Case 2. The number $a$ consists of three identical cycles of two digits each, for example, $a=242424$. Such numbers have two differe... | 1999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On a sheet of paper, 14 parallel lines $L$ and 15 lines $P$ perpendicular to them are drawn. The distances between adjacent lines from $L$ from the first to the last are given: 2;4;6;2;4;6;2;4;6;2;4;6;2. The distances between adjacent lines from $P$ are also known: 3;1;2;6;3;1;2;6;3;1;2;6;3;1. Find the greatest leng... | 5. Solution. We will prove that the maximum length of the side of the square is 40. Calculate the distance from the first to the last line in $P: 3+1+2+6+3+1+2+6+3+1+2+6+3+1=40$. Therefore, the side length of the square cannot be more than 40. On the other hand, in $L$ the distance from the second line to the third lin... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(29)=P(37)=2022$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\min }=949$. | 949 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(11)=P(13)=2021$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\text {min }}=19$. | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(19)=P(21)=2020$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\min }=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The number $A$ in decimal form is represented as $A=\overline{7 a 631 b}$, where $a, b$ are non-zero digits. The number $B$ is obtained by summing all distinct six-digit numbers, including $A$, that are formed by cyclic permutations of the digits of $A$ (the first digit moves to the second position, the second to th... | 2. Solution. The sum of the digits of number $A$ and the numbers obtained from $A$ by cyclic permutations of its digits is $a+b+17$. After summing these numbers (there are 6 of them), in each digit place of number $B$ we get
$a+b+17$, so $B=(a+b+17)\left(10^{5}+10^{4}+10^{3}+10^{2}+10+1\right)=(a+b+17) \cdot 111111$. ... | 796317 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The real numbers $x_{1}, x_{2}$, and $x_{3}$ are the three roots of the equation $x^{3}-3 x^{2}+2(1-p) x+4=0$, considering their possible multiplicities. Find the smallest value of the expression $\left(x_{1}-1\right)^{2}+\left(x_{2}-1\right)^{2}+\left(x_{3}-1\right)^{2}$ under these conditions. For which $p$ is it ... | The smallest value 6 is achieved when $p=1$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. How many different pairs of integers $a$ and $b$ exist such that the equation $a x^{2}+b x+1944=0$ has positive integer roots? | Answer: $\quad 108+24=132$ pairs. | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. How many different pairs of integers $a$ and $b$ exist such that the equation $a x^{2}+b x+432=0$ has positive integer roots | Answer: $\quad 78+20=98$ pairs. | 98 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. By the condition
$$
T(t)=\frac{270-s(t)}{s(t) / t}=\frac{t(270-s(t))}{s(t)}=C>1, t \in[0.5 ; 1]
$$
Then $s(t)=\frac{270 t}{t+C}$ on this interval. The speed of movement
$$
\begin{aligned}
& v(t)=s^{\prime}(t)=\frac{270 C}{(t+C)^{2}}=60 \text { when } t=1 \text {, i.e. } \\
& \qquad 2 c^{2}-5 c+2=0 \rightarrow C_{... | Answer: 1) 90 km; 2) 86.4 km/hour | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. If $m+2019 n$ and $n+2019 m$ are divisible by $d$, then the number
$$
2019(m+2019 n)-(n+2019 m)=(2019^2-1) n
$$
is also divisible by $d$. If $n$ is divisible by $d$, and $m+2019 n$ is divisible by $d$, then $m$ is divisible by $d$ and the numbers $m$ and $n$ are not coprime. Therefore, $d$ divides the number
$$
2... | Answer: $d_{\text {min }}=101$. | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Kostya is making a car trip from point A to point B, which are 320 km apart. The route of the trip is displayed on the computer screen. At any moment in time $t$ (hours), Kostya can receive information about the distance traveled $s(t)$ (km), the speed of movement $v(t)$ (km/hour), and the estimated time $T=T(t)$ (h... | Answer: 1) 128 km; 2) 38.4 km/h. | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that for some positive coprime numbers $m$ and $n$, the numbers $m+2024 n$ and $n+2024 m$ have a common prime divisor $d>7$. Find the smallest possible value of the number $d$ under these conditions. | Answer: $d_{\min }=17$.
For example,
$$
\begin{aligned}
& m=16, n=1 \rightarrow 2024 m+n=2024 \cdot 16+1=32385=17 \cdot 1905 \\
& m+2024 n=16+2024=17 \cdot 120
\end{aligned}
$$ | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that for some positive coprime numbers $m$ and $n$, the numbers $m+1941 n$ and $n+1941 m$ have a common prime divisor $d>8$. Find the smallest possible value of the number $d$ under these conditions. | Answer: $d_{\min }=97$.
For example,
$$
\begin{aligned}
& m=96, n=1 \rightarrow 1941 m+n=1941 \cdot 96+1=97 \cdot 1921 \\
& m+1941 n=96+1941=97 \cdot 21
\end{aligned}
$$ | 97 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Danya is making a car trip from point A to point B, which are 300 km apart. The route of the trip is displayed on the computer screen. At any moment in time $t$ (hours), Danya can receive information about the distance traveled $s(t)$ (km), the speed of movement $v(t)$ (km/hour), and the estimated time $T=T(t)$ (hou... | Answer: 1) 180 km; 2) 48 km/h. | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that for some positive coprime numbers $m$ and $n$, the numbers $m+1947 n$ and $n+1947 m$ have a common prime divisor $d>9$. Find the smallest possible value of the number $d$ under these conditions. | Answer: $d_{\min }=139$.
For example,
$$
\begin{aligned}
& m=138, n=1 \rightarrow 1947 m+n=1947 \cdot 138+1=139 \cdot 1933 \\
& m+1947 n=138+1947=139 \cdot 15
\end{aligned}
$$ | 139 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of divisors of the number $a=2^{3} \cdot 3^{2} \cdot 5^{2}$, which are divisible by 3. Find the sum of such divisors. | 1. Solution. Among the divisors of the number $a=2^{3} \cdot 3^{2} \cdot 5^{2}$, which are divisible by 3, the number 2 can be chosen with exponents $0,1,2,3$ (4 options), the number 3 - with exponents 1,2 (2 options), and the number 5 - with exponents $0,1,2$ (3 options). Therefore, the total number of divisors is $4 ... | 5580 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. How many integers $b$ exist such that the equation $x^{2}+b x-9600=0$ has an integer solution that is a multiple of both 10 and 12? Specify the largest possible $b$. | 2. Solution. Since the desired integer solution $x$ is divisible by 10 and 12, it is divisible by 60, hence it can be written in the form $x=60 t, t \in Z$. Substitute $x=60 t$ into the original equation:
$3600 t^{2}+60 b t-9600=0$. Express $b: b=\frac{160}{t}-60 t$. For $b$ to be an integer, $t$ must be a divisor of ... | 9599 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How many three-digit positive numbers $x$ exist that are divisible by 3 and satisfy the equation $GCD(15, GCD(x, 20))=5$? Find the largest one. | 3. Solution. Since 20 is not divisible by 3, $NOD(x, 20)$ is also not divisible by 3. Therefore, the equation $NOD(15, NOD(x, 20))=5$ is equivalent to the condition that $NOD(x, 20)$ is divisible by 5, which is possible if and only if $x$ is divisible by 5. Thus, the condition of the problem is equivalent to $x$ being ... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The quadratic trinomial $p(x)=a x^{2}+b x+c, a>0$ when divided by ( $x-1$ ) gives a remainder of 4, and when divided by ( $x-2$ ) - a remainder of 15. Find the maximum possible value of the ordinate of the vertex of the parabola $y=p(x)$ under these conditions. For what value of $x$ is it achieved? | 4. Solution. By the Remainder Theorem, the remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$.
Therefore, the conditions of the problem are equivalent to the system:
$\left\{\begin{array}{l}4 a+2 b+c=15 \\ a+b+c=4\end{array} \Rightarrow\left\{\begin{array}{l}b=11-3 a \\ c=2 a-7\end{array}\right.\righ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the sides of triangle $A B C$ with side lengths 2, 3, and 4, external squares $A B B_{1} A_{1}, B C C_{2} B_{2}, C A A_{3} C_{3}$ are constructed. Find the sum of the squares of the side lengths of the hexagon $A_{1} B_{1} B_{2} C_{2} C_{3} A_{3}$.
## Answers and Solutions
Problem 1 Answer: 165 m. | Solution
$m$ - the number of steps Petya takes, $n$ - the number of steps Vova takes
$0.75 m=0.55 n=L-$ path length
$k$ - the step number of Petya leading to the coincidence of the footprint,
$i$ - the step number of Vova leading to the coincidence of the footprint,
$0.75 k=0.55 i \rightarrow 15 k=11 i \rightarrow... | 116 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Dima drew a parallelogram $A B C D$ and points $M$ and $N$ on sides $B C$ and $C D$ respectively such that $B M: M C=D N: N C=2: 3$. After that, he erased everything except points $A, M$ and $N$ using a cloth. Vova restored the drawing using a ruler and a compass. How did he do it? | 4. Points $B$ and $D$ are the intersection points of lines $C M$ and $C N$ with the line passing through point $O$ and parallel to line $M N$.
## Problem 5
Numbers $A$, when divided by $n$, have a remainder of $r$, when divided by $n+1$ - a remainder of $r+1$, and when divided by $n+2$ - a remainder of $r+2$, have th... | 1710 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Pasha, Masha, Tolya, and Olya ate 88 candies, and each of them ate at least one candy. Masha and Tolya ate 57 candies, but Pasha ate the most candies. How many candies did Olya eat? | Solution. Either Masha or Tolya ate no less than 29 candies, then Pasha ate no less than 30 candies. Then the number of candies eaten by Pasha, Masha, and Tolya is no less than $57+30=87$. Since there are a total of 88 candies, and Olya did not refuse any candies, she gets one candy.
Answer: 1 candy. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Pete is trying to lay out a square on a table using identical cardboard rectangles measuring $14 \times 10$. Will he be able to do this? Propose your version of constructing such a square. What is the minimum number of rectangles he will need? | Solution. Let $n$ be the number of rectangles that form the square. Then the area of the square is $S=14 \cdot 10 \cdot n$. If the side of the square is formed by $m$ sides of length 14 and $k$ sides of length 10, then the length of the side of the square is $14 m + 10 k$, and its area is $S=(14 m + 10 k)^{2}$. Equatin... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a rectangular table, the letters of the word "олимпиада" (olympiada) are arranged in a certain order.
| $\mathrm{O}$ | Л | И | M | П | И | A | Д | A |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Л | И | M | П | И | A | Д | A | O |
| И | M | П | И | A | Д | A | O | Л |
| M | П | И | A | Д |... | Solution. In each cell of the table shown below
| $\mathrm{O}$ | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| 1 | 3 | 6 | 10 | 15 | 21 | 28 | | |
| 1 | 4 | 10 | 20 | 35 | 56 | | | |
the number of different paths leading to... | 93 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Let point $M$ divide edge $A B$ in the ratio $A M: M B=\lambda$, point $N$ divide edge $D C$ in the ratio $D N: N C=\mu$, and point $P$ divide edge $D B$ in the ratio $D P: P B=\theta$. We need to find the ratio $A Q: Q C$.
=325 \rightarrow 11 n+66-k=325 \rightarrow k=11 n-259
$$
Considering the condition $k \in... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A football is sewn from 256 pieces of leather: white ones in the shape of hexagons and black ones in the shape of pentagons. Black pentagons only border white hexagons, and any white hexagon borders three black pentagons and three white hexagons. Find the number of white hexagons on the football. | Answer: 160.
Criteria for checking works, 7th grade
Preliminary round of the sectoral physics and mathematics olympiad for schoolchildren "Rosatom", mathematics
# | 160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. For what least integer $n$ are all solutions of the equation
$$
x^{3}-(5 n-9) x^{2}+\left(6 n^{2}-31 n-106\right) x-6(n-8)(n+2)=0 \text { greater than }-1 \text { ? }
$$ | Solution. Rewrite the equation as
$$
x^{3}-(5 n-9) x^{2}+\left(6 n^{2}-31 n-106\right) x-6 n^{2}+36 n+96=0 .
$$
Notice that $x_{1}=1>-1$ is a solution for all $n$ (the sum of the coefficients of the polynomial on the left side of the equation is zero for all $n$). Divide the equation by $x-1$:
$$
x^{2}-5(n-2) x+6\le... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The midline of a trapezoid is 4. A line parallel to the bases of the trapezoid and dividing its area in half intersects the lateral sides at points $M$ and $N$. Find the smallest possible length of the segment $M N$. | # Solution.
Notations: $A D=a, B C=b, C K=H, N E=h$ - heights of triangles $N C P$ and $D N Q$, $M N=x$
$$
S_{\text {AMND }}=S_{\text {MBCN }} \rightarrow \frac{x+b}{2} H=\frac{a+x}{2} h \r... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find all $x$ that satisfy the inequality $n^{2} x^{2}-\left(2 n^{2}+n\right) x+n^{2}+n-6 \leq 0$ for any natural $n$. | Solution. The roots of the quadratic trinomial $n^{2} x^{2}-\left(2 n^{2}+n\right) x+n^{2}+n-6$ are
$$
x_{1}=1-\frac{2}{n}, x_{2}=1+\frac{3}{n}
$$
Factorize the left side of the inequality
$$
n^{2}\left(x-\left(1-\frac{2}{n}\right)\right)\left(x-\left(1+\frac{3}{n}\right)\right) \leq 0
$$
Solving the inequality usi... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. On the edges of a trihedral angle with vertex at point $S$, there are points $M, N$, and $K$ such that $S M^{2}+S N^{2}+S K^{2} \leq 12$. Find the area of triangle $S M N$, given that the angle $M S N$ is $30^{\circ}$, and the volume of the pyramid $S M N K$ is maximally possible. | Solution. Let's introduce the notations: $S M=m, S N=n, S K=k$.
The volume of the pyramid $S M N K$ is
$$
V=\frac{1}{3} S_{S M N} \cdot h=\frac{1}{6} m n \sin \alpha \cdot h=\frac{1}{6} m n... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Quadrilateral $ABCD$ is inscribed in a circle, and its diagonals intersect at point $P$. Points $K, L$, and $M$ are the midpoints of sides $AB, BC$, and $CD$ respectively. The radius of the circle circumscribed around triangle $KLP$ is 1. Find the radius of the circle circumscribed around triangle $LMP$.
Problem 1 ... | Solution. Let $y_{k}$ be the number of passengers in the car with number $k, k=1,2,3, \ldots, 10$. According to the problem, $\sum_{k=1}^{10} y_{k}=270$. Additionally, it is stated that $y_{2} \geq y_{1}+2, y_{3} \geq y_{1}+4, \ldots, y_{9} \geq y_{1}+16, \quad y_{10} \geq y_{1}+18$. Adding these inequalities,
$$
y_{2... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
0.35 \cdot 160+0.1 x=0.2 \cdot 160+0.2 x, 0.15 \cdot 160=0.1 x, x=240 \text {. }
$$
In total, it results in $160+240=400$ g of solution. | Answer: 400.
## Problem 9
Points $B_{1}$ and $C_{1}$ are the feet of the altitudes of triangle $ABC$, drawn from vertices $B$ and $C$ respectively. It is known that $AB=7, \quad AC=6, \sin \angle BAC=\frac{2 \sqrt{110}}{21}$. Find the length of the segment $B_{1}C_{1}$.
## Solution:
Let the angle $\angle BAC$ be de... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points). Tourist Nikolai Petrovich was late by $\Delta t=5$ minutes for the departure of his riverboat, which had set off downstream. Fortunately, the owner of a fast motorboat agreed to help Nikolai Petrovich. Catching up with the riverboat and disembarking the unlucky tourist, the motorboat immediately set off... | 1. $s=\left(3 v_{\mathrm{T}}+v_{\mathrm{T}}\right) \cdot\left(t_{1}+\Delta t\right)=\left(5 v_{\mathrm{T}}+v_{\mathrm{T}}\right) \cdot t_{1} \rightarrow 4 \Delta t=2 t_{1} \rightarrow t_{1}=2 \Delta t=10$ min $t_{2}=\frac{s}{5 v_{\mathrm{T}}-v_{\mathrm{T}}}=\frac{6 v_{\mathrm{T}} \cdot t_{1}}{4 v_{\mathrm{T}}}=\frac{3}... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) A satellite is launched vertically from the pole of the Earth at the first cosmic speed. To what maximum distance from the Earth's surface will the satellite travel? (The acceleration due to gravity at the Earth's surface $g=10 \mathrm{m} / \mathrm{c}^{2}$, radius of the Earth $R=6400$ km). | 2.
According to the law of conservation of energy: $\frac{m v_{I}^{2}}{2}-\frac{\gamma m M}{R}=-\frac{\gamma m M}{R+H}$
The first cosmic speed $v_{I}=\sqrt{g R}$
The acceleration due to gravity at the surface $g=\frac{\gamma M}{R^{2}}$
Then $\frac{m g R}{2}-m g R=-\frac{m g R^{2}}{R+H}$
Finally, $H=R=6400$ km.
(... | 6400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (15 points) Looking down from the edge of the stream bank, Vovochka decided that the height of his rubber boots would be enough to cross the stream. However, after crossing, Vovochka got his legs wet up to his knees ($H=52$ cm). Estimate the height $h$ of Vovochka's boots. Assume the depth of the stream is constant,... | 6. From the figure
$\frac{d}{h}=\operatorname{tg} \alpha ; \quad \frac{d}{H}=\operatorname{tg} \beta ; \quad \frac{H}{h}=\frac{\operatorname{tg} \alpha}{\operatorname{tg} \beta}$
Since all the information about the bottom of the stream falls into the space limited by the eye's pupil, all angles are small: $\operatorn... | 39 | Other | math-word-problem | Yes | Yes | olympiads | false |
1. Dmitry is three times as old as Grigory was when Dmitry was as old as Grigory is now. When Grigory becomes as old as Dmitry is now, the sum of their ages will be 49 years. How old is Grigory? | Solution: Let Gregory be $y$ years old in the past, and Dmitry be $x$ years old. Then currently, Gregory is $x$ years old, and Dmitry is $3 y$ years old. In the future, Gregory will be $3 y$ years old, and Dmitry will be $z$ years old, and according to the condition, $z+3 y=49$. Since $z-3 y=3 y-x ; 3 y-x=x-y$, then $9... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $\frac{4}{\sqrt{\log _{3}(81 x)}+\sqrt{\log _{3} x}}+\sqrt{\log _{3} x}=3$. | Solution: Using the properties of logarithms, our equation can be rewritten as $\frac{4}{\sqrt{4+\log _{3} x}+\sqrt{\log _{3} x}}+\sqrt{\log _{3} x}=3$. Let $t=\log _{3} x$. Then $\frac{4}{\sqrt{4+t}+\sqrt{t}}+\sqrt{t}=3$. Multiplying the numerator and denominator of the first fraction by $\sqrt{t+4}-\sqrt{t}$, we arri... | 243 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. On the sides $B C, C A, A B$ of an equilateral triangle $A B C$ with side length 7, points $A_{1}, B_{1}, C_{1}$ are taken respectively. It is known that $A C_{1}=B A_{1}=C B_{1}=3$. Find the ratio of the area of triangle $A B C$ to the area of the triangle formed by the lines $A A_{1}, B B_{1}, C C_{1}$. | Solution:
Triangles $A B A_{1}, B C B_{1}, C A C_{1}$ are equal by sides and the angle between them, triangles $A D C_{1}, B E A_{1}, C F B_{1}$ are equal by side and angles. Triangle $A D C_{1}$ is similar to $A B A_{1}$ by two angles, triangle $A B C$ is similar to $D E F$. Let $S=S_{A B C}, S_{1}=S_{A B A_{1}}, S_{... | 37 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Ivan is twice as old as Peter was when Ivan was as old as Peter is now. When Peter becomes as old as Ivan is now, the sum of their ages will be 54 years. How old is Peter? | Solution: Let Peter's age in the past be $y$ years, and Ivan's age be $x$ years. Then currently, Peter is $x$ years old, and Ivan is $2 y$ years old. In the future, Peter will be $2 y$ years old, and Ivan will be $z$ years old, and according to the condition, $z+2 y=54$. Since $z-2 y=2 y-x ; 2 y-x=x-y$, then $6 y-x=54$... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $\frac{1}{\sqrt{\log _{5}(5 x)}+\sqrt{\log _{5} x}}+\sqrt{\log _{5} x}=2$. | Solution: Using the properties of logarithms, our equation can be rewritten as $\frac{1}{\sqrt{1+\log _{5} x}+\sqrt{\log _{5} x}}+\sqrt{\log _{5} x}=2$. Let $t=\log _{5} x$. Then $\frac{1}{\sqrt{1+t}+\sqrt{t}}+\sqrt{t}=2$. By multiplying the numerator and denominator of the first fraction by $\sqrt{t+1}-\sqrt{t}$, we a... | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. On the sides $B C, C A, A B$ of an equilateral triangle $A B C$ with side length 11, points $A_{1}, B_{1}, C_{1}$ are taken respectively. It is known that $A C_{1}=B A_{1}=C B_{1}=5$. Find the ratio of the area of triangle $A B C$ to the area of the triangle formed by the lines $A A_{1}, B B_{1}, C C_{1}$. | Solution:
Triangles $A B A_{1}, B C B_{1}, C A C_{1}$ are equal by sides and the angle between them, triangles $A D C_{1}, B E A_{1}, C F B_{1}$ are equal by side and angles. Triangle $A D C_{1}$ is similar to $A B A_{1}$ by two angles, triangle $A B C$ is similar to $D E F$. Let $S=S_{A B C}, S_{1}=S_{A B A_{1}}, S_{... | 91 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. For what values of the parameter $a$ does the equation $x^{4}-40 x^{2}+144=a\left(x^{2}+4 x-12\right)$ have exactly three distinct solutions? | Solution: Factorize the right and left sides of the equation:
$(x-2)(x+2)(x-6)(x+6)=a(x-2)(x+6)$. The equation can be written as
$(x-2)(x+6)\left(x^{2}-4 x-12-a\right)=0$. It is obvious that 2 and -6 are roots of this equation. We are satisfied with the situation where exactly one of the roots of the equation $x^{2}-4... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
From channel A to the Wiki site, $850 * 0.06=51$ people will transition
From channel B to the Wiki site, $1500 * 0.042=63$ people will transition
From channel C to the Wiki site, $4536 / 72=63$ people will transition | Answer: The most people will transition from channels B and V - 63 people each
## 2
Cost of transition from advertising on channel A: $-3417 / 51 = 67$ rubles
Cost of transition from advertising on channel B: $4914 / 63 = 78$ rubles
Answer: The lowest cost of transition to the site from advertising on channel A - 6... | 2964 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.
Expenses for medical services provided to a child (under 18 years old) of the taxpayer by medical organizations
Correct answers: Pension contributions under a non-state pension agreement concluded by the taxpayer with a non-state pension fund in their own favor, Expenses for medical services provided to a child (... | Answer and write it in rubles as an integer without spaces and units of measurement.
Answer: $\qquad$
Correct answer: 16250
Question 12
Score: 5.00
Insert the missing terms from the drop-down list.
Under the insurance contract, one party
insured; insurance premium; beneficiary; insurer; insurance amount
undertak... | 16250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.
the higher the risk of a financial instrument, the higher its return
Correct answers: the higher the reliability of a financial instrument, the higher its return, a financial instrument can be reliable, profitable, and liquid at the same time, risk is not related to the return of a financial instrument
Question ... | Answer write in rubles as an integer without spaces and units of measurement.
Answer: $\qquad$
Correct answer: 1378
Question 16
Score: 5.00
Establish the correspondence between specific taxes and their types.
| personal income tax | federal tax; local tax; regional tax; |
| :---: | :---: |
| land tax | federal ta... | 1378 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 15. Problem 15
What amount of mortgage credit in rubles will a bank client receive if their initial payment of 1800000 rubles amounted to $30 \%$ of the cost of the purchased property? | Answer in the form of a number should bewrittenwithoutspaces,withoutunitsofmeasurementandanycharacters.
## Answer: 4200000
# | 4200000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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