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Problem 10.5. In each cell of a square table of size $200 \times 200$, a real number not exceeding 1 in absolute value was written. It turned out that the sum of all the numbers is zero. For what smallest $S$ can we assert that in some row or some column, the sum of the numbers will definitely not exceed $S$ in absolute value? | Answer: 100.
Solution. First, we show that $S<40000$.
$$
This means that one of the numbers $A$ or $D$ in absolute value exceeds 10000. However, each of the corresponding squares contains only 10000 cells, and the numbers in them do not exceed 1 in absolute value. Contradiction.
## Criteria
Any correct solution to the problem is worth 7 points. In the absence of such a solution, the following criteria are summed:
## 16. There is a correct answer.
1 6. "Example". It is proven that $S<100$ is impossible (i.e., an example is provided or its existence is proven).
5 6. "Estimate". It is proven that $S=100$ works for any arrangement of numbers satisfying the condition. In the absence of a complete proof of the "estimate", the following partial progress is credited: | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. Through the point with coordinates $(2,2)$, lines (including two parallel to the coordinate axes) are drawn, dividing the plane into angles of $18^{\circ}$. Find the sum of the abscissas of the points of intersection of these lines with the line $y=2016-x$. | Answer: 10080. Solution. The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(2,2)$, 10 lines are drawn, and the line $y=2016-x$ intersects all of them. For each point on the line $y=2016-x$, the sum of the coordinates is 2016, so the total sum of the abscissas and ordinates is 20160, and the sum of the abscissas is half of that.
Criterion. 2 points for the correct answer without a correct justification. 5 points if all ideas are found, but the answer is incorrect due to a wrong count of the number of intersection points (for example, the student counts 9 or 18 points). | 10080 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Positive numbers $a, b, c, d$ are greater than 1. Find the smallest possible value of the expression
$$
\log _{a}\left(a b^{2}\right)+\log _{b}\left(b^{2} c^{3}\right)+\log _{c}\left(c^{5} d^{6}\right)+\log _{d}\left(d^{35} a^{36}\right)
$$ | Answer: 67.
Solution. From the properties of logarithms, it follows that $\log _{a} b \cdot \log _{b} c \cdot \log _{c} d \cdot \log _{d} a=1$. Also, all these four factors are positive, since all numbers $a, b, c, d$ are greater than 1.
Transform and estimate the given expression
$$
\begin{gathered}
S=\log _{a}\left(a b^{2}\right)+\log _{b}\left(b^{2} c^{3}\right)+\log _{c}\left(c^{5} d^{6}\right)+\log _{d}\left(d^{35} a^{36}\right)= \\
=\left(\log _{a} a+\log _{a} b^{2}\right)+\left(\log _{b} b^{2}+\log _{b} c^{3}\right)+\left(\log _{c} c^{5}+\log _{c} d^{6}\right)+\left(\log _{d} d^{35}+\log _{d} a^{36}\right)= \\
=\left(1+2 \log _{a} b\right)+\left(2+3 \log _{b} c\right)+\left(5+6 \log _{c} d\right)+\left(35+36 \log _{d} a\right) \geqslant \\
\geqslant(1+2+5+35)+4 \sqrt[4]{2 \log _{a} b \cdot 3 \log _{b} c \cdot 6 \log _{c} d \cdot 36 \log _{d} a}=43+4 \cdot 6=67,
\end{gathered}
$$
here in the last transition, the inequality between the arithmetic and geometric means for four positive numbers $2 \log _{a} b, 3 \log _{b} c, 6 \log _{c} d, 36 \log _{d} a$ was used.
Also note that the value $S=67$ is achieved, for example, when $a=2, b=8, c=d=64$, since all four numbers $2 \log _{a} b, 3 \log _{b} c, 6 \log _{c} d, 36 \log _{d} a$ will be equal to 6.
## Criteria
Points for the estimate and example are summed.
Estimate. The largest suitable criterion is used:
5 p. The estimate $S \geqslant 67$ is proven.
1 p. The problem is reduced to finding the minimum of the sum $2 \log _{a} b+3 \log _{b} c+6 \log _{c} d+36 \log _{d} a$.
0 p. The positivity of the numbers $\log _{a} b, \log _{b} c, \log _{c} d, \log _{d} a$ is mentioned.
Example. The largest suitable criterion is used:
2 p. It is proven that the value $S=67$ is achieved for some $a, b, c, d$, greater than 1.
0 p. Only the correct answer is given. | 67 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Tetrahedron $ABCD$ with acute-angled faces is inscribed in a sphere with center $O$. A line passing through point $O$ perpendicular to the plane $ABC$ intersects the sphere at point $E$ such that $D$ and $E$ lie on opposite sides relative to the plane $ABC$. The line $DE$ intersects the plane $ABC$ at point $F$, which lies inside triangle $ABC$. It turns out that $\angle ADE = \angle BDE$, $AF \neq BF$, and $\angle AFB = 80^\circ$. Find the measure of $\angle ACB$. | Answer: $40^{\circ}$.
Note that point $E$ is equidistant from points $A, B, C$, so its projection onto the plane $A B C$ coincides with the projection of point $O$ onto this plane and is the center of the circumscribed circle of triangle $A B C$.
Consider triangles $A D E$ and $B D E$. They have a pair of equal sides $A E$ and $B E$, a common side $D E$, and equal angles $A D E$ and $B D E$. From the Law of Sines, it follows that these triangles are either equal or the angles $D A E$ and $D B E$ are supplementary to $180^{\circ}$. The first situation is impossible, as in the case of equality of triangles $A D E$ and $B D E$, points $A$ and $B$ are equidistant from any point on side $D E$, but by the condition $A F \neq B F$. Therefore, $\angle D A E + \angle D B E = 180^{\circ}$.
Consider the point $X$ of intersection of the ray $A F$ with the sphere $\Omega$ circumscribed around the tetrahedron $A B C D$. Note that the ray $A F$ lies in the planes $A B C$ and $A E D$, so point $X$ lies on the circumscribed circles of triangles $A B C$ and $A E D$. Point $E$ is equidistant from all points of the circumscribed circle of triangle $A B C$; in particular, $A E = X E$. From the inscribed quadrilateral $A E X D$, it follows that $\angle D A E + \angle D X E = 180^{\circ}$. Since $A E = X E$, $E$ is the midpoint of the arc $A X$ of the circumscribed circle of triangle $A D E$, and thus $\angle A D E = \angle X D E$.
Using the previously derived angle equalities, we conclude that triangles $D B E$ and $D X E$ are equal by the second criterion: $\angle D B E = 180^{\circ} - \angle D A E = \angle D X E, \angle X D E = \angle A D E = \angle B D E$, side $D E$ is common. Since triangles $D B E$ and $D X E$ are equal, vertices $B$ and $X$ are equidistant from any point on side $D E$; in particular, $B F = F X$.
It remains to calculate the angles in the plane $A B C$. Sequentially using the inscribed quadrilateral $A B X C$, the isosceles triangle $B F X$, and the exterior angle theorem for triangle $B F X$, we write
$$
\angle A C B = \angle A X B = \frac{1}{2} \cdot (\angle F X B + \angle F B X) = \frac{1}{2} \cdot \angle A F B = 40^{\circ}
$$
Another solution. Let the ray $A F$ intersect the sphere $\Omega$ circumscribed around the tetrahedron $A B C D$ at point $X$. By construction, the point $E$ satisfies the relation $E X = E A$, which implies that $\angle A D E = \angle E A F$. Similarly, we obtain that $\angle B D E = \angle E B F$, and thus $\angle E A F = \angle E B F$.
Denote the point of intersection of the line $O E$ with the plane $A B C$, which is the center of the circumscribed circle of triangle $A B C$, as $O_{1}$. Then $\angle O_{1} A E = \angle O_{1} B E$.
Consider the trihedral angles $A O_{1} E F$ and $B O_{1} E F$. In them, the planar angles $E A F$ and $E B F$, the planar angles $O_{1} A E$ and $O_{1} B E$, and the dihedral angles at the edges $A O_{1}$ and $B O_{1}$ are right. Therefore, the corresponding trihedral angles are equal. And thus the planar angles $\angle F A O_{1} = \angle F B O_{1}$ are equal. Note that this equality can also be derived from the Law of Cosines for trihedral angles.
The specified equality is possible in two cases: either point $F$ lies on the perpendicular bisector of $A B$ (points $A$ and $B$ are symmetric relative to $F O_{1}$), or point $F$ lies on the circumscribed circle of triangle $A B O_{1}$. The first case is forbidden by the condition $A F \neq B F$, so the second case must hold. Then $\angle A O B = \angle A F B = 80^{\circ}$ and is the central angle for the angle $A C B$ in the circumscribed circle of triangle $A C B$. As a result, we conclude that $\angle A C B = 40^{\circ}$.
## Criteria
The following points are summed. | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Buses from Moscow to Voronezh depart every hour, at 00 minutes. Buses from Voronezh to Moscow depart every hour, at 30 minutes. The trip between the cities takes 8 hours. How many buses from Voronezh will the bus that left from Moscow meet on its way? | Answer: 16.
It is clear that all buses from Moscow will meet the same number of buses from Voronezh, and we can assume that the bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Oryol at 4:30, 5:30, ..., 18:30, 19:30 and only them. There are 16 such buses.
$\pm$ Correct reasoning with an arithmetic error leading to an incorrect answer. 4-5 points
干 The frequency of encounters (every half hour) was calculated, but the answer is incorrect. 3 points
- Correct answer without explanation. 1 point | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex? | Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, continuing this sequence in a cycle will include all numbers of the form $x_{3 k+1}$. Similar sequences can be constructed starting from the numbers $x_{2}$ and $x_{3}$. Thus, the numbers at the vertices that have the same remainder when divided by 3 are equal.
Since each set of nine consecutive numbers contains three sets of vertices with different remainders when divided by 3, the sum of the numbers in each such set is 100. Note that 19 gives a remainder of 1 when divided by 3, 20 gives a remainder of 2, and 2019 gives a remainder of 0, so the number written at vertex 2019 is $100 - 19 - 20 = 61$.
## Criteria
The largest suitable criterion is used:
7 points. Any complete solution to the problem.
4 points. Proven that numbers that are 2 apart are equal.
1 point. Noted that numbers that are 8 apart are equal. | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let's call a natural number an almost palindrome if it can be transformed into a palindrome by changing one of its digits. How many nine-digit almost palindromes exist? (20 points) | Solution: After replacing one digit in the almost palindrome, we get a number of the form $\overline{a b c d e d c b a}$. Let's divide the digits into pairs by place value: 1 and 9, 2 and 8, 3 and 7, 4 and 6. In any almost palindrome, the digits in three of the specified pairs of place values must be the same, and in exactly one pair they must be different.
International School Olympiad URFU "Emerald" 2022, 2nd stage
We can turn this number into a palindrome by replacing one of the two digits in the remaining pair with the other digit in that pair. We can choose three pairs of place values in 4 ways (3 of them include the pair 1 and 9, and exactly one way does not include this pair), in each pair (except for 1 and 9) the digits can be any and the same, in the pair 1 and 9 (if it is chosen) the digits are the same and not equal to zero, in the remaining pair the digits are different (any, if it is not the pair 1 and 9, and any except zero in the first place, if the pair 1 and 9 remains), and the middle digit is arbitrary. The total number of such numbers is
$$
3 \cdot 10^{3} \cdot 9 \cdot 10 \cdot 9+10^{4} \cdot 9 \cdot 9=3240000
$$
Answer: 3240000 | 3240000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. If in an acute scalene triangle three medians, three angle bisectors, and three altitudes are drawn, they will divide it into 34 parts. | Write the answer in digits in ascending order, without spaces (for example, 12345).
## Mathematics $8-9$ grade
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 2. (20 points)
Find the maximum possible value of the ratio of a three-digit number to the sum of its digits.
# | # Solution.
Let $N=\overline{a b c}$, where $a, b, c$ are the digits of the number. Clearly, for "round" numbers $N=$ $100, 200, \ldots, 900$, we have $\frac{N}{a+b+c}=100$. Furthermore, if the number $N$ is not round, then $b+c>0$ and $a+b+c \geq a+1$. Since the leading digit of the number $N$ is $a$, we have $N<(a+1) \cdot 100$ and
$$
\frac{N}{a+b+c}<\frac{(a+1) \cdot 100}{a+1}=100
$$
Thus, the maximum value of the considered ratio is 100. This value is achieved only for "round" numbers. | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the language of the "Tekimar" tribe, there are only 7 letters: A, E, I, K, M, R, T, but the order of these letters in the alphabet is unknown. A word is defined as any sequence of seven different letters from the alphabet, and no other words exist in the language. The chief of the tribe listed all existing words in alphabetical order and noticed that the word "Metrika" has the number 3634 in this list. What is the number of the word "Materik" in this list? (20 points) | Solution: The total number of words in the tribe's language is $7!=5040$. Note that the number of words starting with any particular letter is the same for any first letter, which is $7!\div 7=6!=720$. If a word starts with the first letter in the alphabet, the numbering of any such word starts from number 1 and ends at number 6!. If it starts with the second letter in the alphabet, the numbering starts from number $6!+1$ and ends at number $2 \cdot 6$!, and so on. In general, if a word starts with the $k$-th letter in the alphabet, its numbering varies from number $(k-1) \cdot 6!+1$ to number $k \cdot 6$!. The number 3634 satisfies the inequality $5 \cdot 6$! $<3634 \leq 6 \cdot 6$!, so M is the sixth letter in the alphabet. Now, we will repeat a similar process for the remaining six letters. Specifically, we will forget about the existence of the first letter by removing it from the word. Then the word's number will decrease by $5 \cdot 6!=3600$. If we now form all six-letter words and arrange them in alphabetical order, the word "Etrika" would have the number 34. The number of words starting with any particular letter is the same for any first letter, which is $6!\div 6=$ $5!=120$. If the first letter of the word among the remaining six letters is the first in alphabetical order, the numbering of any such word starts from number 1 and ends at number 5!. If it is the second letter in alphabetical order among the remaining, the numbering starts from number $5!+1$ and ends at number $2 \cdot 5$!, and so on. In general, if the first letter of the word is the $k$-th letter in alphabetical order among the remaining, its numbering varies from number $(k-1) \cdot 5$! to number $k \cdot 5$!. The number 34 satisfies the inequality above, which implies $34 \leq 5$!, so E is the first in alphabetical order among the remaining, and thus the first letter of the alphabet. Continuing this similar process, we get the order of the letters: E - first, A - second, T - third, R - fourth, I - fifth, M - sixth, K - seventh.
In the word "Materik," the letters are in the following order: $6,2,3,1,4,5,7$, so its number is 5 . $6!+1 \cdot 5!+1 \cdot 4!+0 \cdot 3!+0 \cdot 2!+0 \cdot 1!+0 \cdot 0!+1=3745$. (The multiplier before $k!$ indicates the number of digits to the right and less than the digit in the ($k+1$)-th position, counting from the end of the word. A one is added at the end because the numbering of words starts with it. | 3745 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of triples of natural numbers $m, n, k$, which are solutions to the equation $m+$ $\sqrt{n+\sqrt{k}}=2023$. (20 points) | Solution: For the left side to be an integer, the numbers $k$ and $n+\sqrt{k}$ must be perfect squares, and since $n+\sqrt{k} \geq 2$, it follows that $\sqrt{n+\sqrt{k}} \geq 2$ and thus $m \leq 2021$. Since $1 \leq m \leq$ 2021, $\sqrt{n+\sqrt{k}}$ can take any value from 2 to 2022 - this value uniquely determines the number $m$. Let $k=x^{2}$ and $n+\sqrt{k}=y^{2}$, where $1 \leq x \leq y^{2}-1$ and $2 \leq y \leq 2022$, then the number $n$ is uniquely determined, specifically $n=y^{2}-x$. Therefore, we need to count the number of
admissible pairs $(x, y)$. There are a total of $\left(2^{2}-1\right)+\ldots+\left(2022^{2}-1\right)=1^{2}+2^{2}+\ldots+2022^{2}-2022$. The formula for the sum of the squares of the first $n$ natural numbers is known:
$$
1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$
Applying this formula, we get $1^{2}+2^{2}+\ldots+2022^{2}-2022=\frac{2022 \cdot 2023 \cdot 4045}{6}-2022=$ 27575680773.
## Answer: 27575680773 | 27575680773 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. A printing house determines the cost of printing a book as follows: it adds the cost of the cover to the cost of each page, and then rounds the result up to the nearest whole number of rubles (for example, if the result is 202 rubles and 1 kopeck, it is rounded up to 203 rubles). It is known that the cost of a book with 104 pages is 134 rubles, and a book with 192 pages costs 181 rubles. How much does the printing of the cover cost, if it is a whole number of rubles, and the cost of one page is a whole number of kopecks? (20 points) | Solution: Let's convert all costs into kopecks. Let one page cost $x$ kopecks, and the cover cost $100 y$ kopecks. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
13300<100 y+104 x \leq 13400 \\
18000<100 y+192 x \leq 18100
\end{array}\right.
$$
Subtracting the first inequality from the second, we get $4600<88 x<4800$. The integer solutions to the inequality are $x=53$ and $x=54$.
If $x=53$, then $7788<100 y \leq 7888$ and $y=78$. But then $100 y+192 x=17976$, which contradicts the second inequality.
If $x=54$, then $7684<100 y \leq 7784$ and $y=77$. Then $100 y+192 x=18068$.
## Answer: a page costs 54 kopecks, and the cover costs 77 rubles | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Maxim came up with a new way to divide numbers by a two-digit number $N$. To divide any number $A$ by the number $N$, the following steps need to be performed:
1) Divide $A$ by the sum of the digits of the number $N$;
2) Divide $A$ by the product of the digits of the number $N$;
3) Subtract the second result from the first.
For which numbers $N$ will Maxim's method give the correct result? (20 points) | Solution: Let $N=\overline{x y}=10 x+y$, where $x, y$ are digits. We need to find all pairs $(x, y)$ such that $\frac{A}{10 x+y}=\frac{A}{x+y}-\frac{A}{x y}$, which means $\frac{1}{10 x+y}=\frac{1}{x+y}-\frac{1}{x y}$. Note that $y \neq 0$. Transform the equation to the form
$$
\begin{gathered}
x y(x+y)=x y(10 x+y)-(x+y)(10 x+y) \\
(x+y)(10 x+y)=x y \cdot 9 x \\
10 x^{2}+11 x y+y^{2}=9 y x^{2}
\end{gathered}
$$
Solve the equation as a quadratic in terms of $x$. For this, rewrite the equation as
$$
\begin{gathered}
(10-9 y) x^{2}+11 y x+y^{2}=0 \\
D=121 y^{2}-4 y^{2}(10-9 y)=81 y^{2}+36 y^{3}=9 y^{2}(9+4 y)
\end{gathered}
$$
For the root to be an integer, the discriminant must be a perfect square, which means $9+4 y$ must also be a perfect square. Trying values of $y$ from 1 to 9, only $y=4$ works. Then the roots of the equation are $x_{1}=\frac{-11 \cdot 4+3 \cdot 4 \cdot 5}{2 \cdot(10-9 \cdot 4)}=-\frac{4}{13}$ and $x_{2}=\frac{-11 \cdot 4-3 \cdot 4 \cdot 5}{2 \cdot(10-9 \cdot 4)}=2$. Only $x=2$ is valid, so $N=24$. For this number, Maksim's method gives the correct result.
Answer: 24 | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (20 points)
It is known that the only solution to the equation
$$
\pi / 4=\operatorname{arcctg} 2+\operatorname{arcctg} 5+\operatorname{arcctg} 13+\operatorname{arcctg} 34+\operatorname{arcctg} 89+\operatorname{arcctg}(x / 14)
$$
is a natural number. Find it. | # Solution.
Consider the equation:
$$
\operatorname{arcctg} a-\operatorname{arcctg} b=\operatorname{arcctg} y \text {. }
$$
Let $\alpha=\operatorname{arcctg} a, \beta=\operatorname{arcctg} b$. Now $y=\operatorname{ctg}(\alpha-\beta)=\frac{1+\operatorname{ctg} \alpha \operatorname{ctg} \beta}{\operatorname{ctg} \beta-\operatorname{ctg} \alpha}=\frac{1+a b}{b-a}$
$$
\begin{aligned}
\operatorname{arcctg} x / 14 & =\operatorname{arcctg} 1-\operatorname{arcctg} 2-\operatorname{arcctg} 5-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{2+1}{2-1}-\operatorname{arcctg} 5-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} 3-\operatorname{arcctg} 5-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{15+1}{5-3}-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} 8-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{104+1}{13-8}-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} 21-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{814+1}{34-21}-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} 55-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{89 * 55+1}{89-55}=\operatorname{arcctg} 144
\end{aligned}
$$
From this, we have $x=144 * 14=2016$. | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Before you is a segment-digit. For displaying time on electronic clocks, each digit uses seven segments, each of which can be lit or not; the lit segments form the digit as shown in the figure
## 8723456789
That is, to display zero, six segments are used, to display one - two segments, and so on. On electronic clocks, only hours and minutes are displayed. How many minutes in a day does it take to display a time that uses more lit segments than the time one minute later? (The day starts at 00:00 and ends at 23:59) (20 points) | Solution: Let's write down the number of segments required to display each digit:
\section*{8723456789
$$
\begin{array}{llllllllll}
6 & 2 & 5 & 5 & 4 & 5 & 6 & 3 & 7 & 6
\end{array}
$$
Consider several types of displayed time:
1) The last digit of the minutes is not 9. Then the next displayed time differs by only the last digit. To reduce the number of segments, the last digit of the minutes at this moment should be 0, 3, 6, or 8. In total, such time is displayed for $24 \cdot 6 \cdot 4 = 576$ minutes in a day.
2) The last digit of the minutes is 9, but the number of minutes is not 59. Then in the next minute, the last digit will become zero, and the number of segments for its display will not change, meaning this number should decrease in the tens of minutes. That is, the digit in the tens of minutes can only be 0 or 3. In total, such time is displayed for $24 \cdot 2 = 48$ minutes in a day.
3) The number of minutes is 59, and the number of hours does not end in 9. Then in the next minute, the number of minutes will be zero, which means the number of segments for displaying the minutes will increase by 1. Therefore, the number of segments for displaying the last digit of the hours should decrease by at least 2, meaning the last digit of the hours can only be 0 or 6. In total, such time is displayed for 5 minutes in a day.
4) The number of minutes is 59, and the number of hours ends in 9. Then in the next minute, one of the times 10:00, 20:00 will be displayed - only the time 09:59 fits.
In total, suitable time is displayed for $576 + 48 + 5 + 1 = 630$ minutes in a day.
## Answer: 630 minutes | 630 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Masha chose a natural number $n$ and wrote down all natural numbers from 1 to 6 n on the board. Then, Masha halved half of these numbers, reduced a third of the numbers by a factor of three, and increased all the remaining numbers by a factor of six. Could the sum of all the resulting numbers match the sum of the original numbers? (20 points) | Solution: Yes, this could have happened. Let $n=2$, then the sum of all numbers from 1 to 12 is
$$
1+2+3+\ldots+12=78
$$
Suppose Masha halved the numbers 2, 3, 5, 9, 10, 11, reduced the numbers 4, 6, 8, 12 by a factor of three, and increased the numbers 1, 7 by a factor of six. Then the sum of the resulting numbers will be
$$
\frac{2+3+5+9+10+11}{2}+\frac{4+6+8+12}{3}+6(1+7)=78
$$
Answer: could have | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In how many ways can the numbers from 1 to 9 be arranged in a $3 \times 3$ table (each number appearing exactly once) such that in each column from top to bottom and in each row from left to right, the numbers are in increasing order? (20 points) | Solution: Let's number the cells of the table as shown in the figure. It is clear that the number 1 is in the upper left cell, and the number 9 is in the lower right cell.
| 1 | $a_{2}$ | $a_{3}$ |
| :---: | :---: | :---: |
| $a_{4}$ | $a_{5}$ | $a_{6}$ |
| $a_{7}$ | $a_{8}$ | 9 |
By the condition, $a_{5}>a_{2}, a_{5}>a_{4}, a_{5}<a_{6}, a_{5}<a_{8}$, so $4 \leq a_{5} \leq 6$. Let's consider the cases.
1) If $a_{5}=4$, then the numbers $a_{2}$ and $a_{4}$ are 2 and 3. There are 2 ways to arrange them. Now let's calculate the number of ways to choose the numbers $a_{3}$ and $a_{6}$. Any of the remaining pairs of numbers can be placed in their positions, and $a_{3}<a_{6}$, so the arrangement of each pair is unique. There are a total of $C_{4}^{2}=6$ such pairs. The remaining two numbers are arranged uniquely. In total, we have $2 \cdot 6=12$ ways of arranging.
2) If $a_{5}=6$, then the numbers $a_{6}$ and $a_{8}$ are 7 and 8, and the case is similar to the previous one. We get another 12 ways of arranging.
3) If $a_{5}=5$, then let's see which numbers can be in the cells $a_{3}$ and $a_{7}$. The numbers 2 and 8 cannot be placed in these cells, as these numbers must be neighbors of 1 and 9, respectively. If $a_{3}=3$, then $a_{2}=2$ and $a_{4}=4$. Any of the remaining numbers can be placed in the cell $a_{6}$ in 3 ways, and the remaining numbers are placed uniquely. The considered case is similar to the cases where $a_{3}=7, a_{7}=3$, and $a_{7}=7$ - in each we get 3 ways of arranging, but the cases where the numbers $a_{3}$ and $a_{7}$ are 3 and 7 were counted twice. There are a total of two such cases:
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 1 | 4 | 7 |
| :---: | :---: | :---: |
| 2 | 5 | 8 |
| 3 | 6 | 9 |
In the end, we get $3 \cdot 4-2=10$ ways.
If neither of the numbers in the cells $a_{3}$ and $a_{7}$ is 3 or 7, then the cells $a_{3}$ and $a_{7}$ can only contain the numbers 4 and 6 in any order. Then the cells $a_{2}$ and $a_{4}$ contain the numbers 2 and 3 in any order, and the cells $a_{6}$ and $a_{8}$ contain the numbers 7 and 8 in any order. In total, there are 8 ways of arranging.
All cases have been considered, and the desired number of ways is $24+10+8=42$.
Answer: 42 | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. How many units are in the number $S=9+99+999+\cdots+\overbrace{9 \ldots 90}^{1000}$? | $$
\begin{aligned}
& S=10-1+100-1+1000-1+\cdots+10^{1000}-1=10\left(1+10+100+\cdots+10^{999}\right)- \\
& -100=10 \frac{10^{1000}-1}{9}-1000=\underbrace{111 \ldots 10}_{999 \text { ones }}-1000 \Rightarrow
\end{aligned}
$$
the number has 998 ones. Answer: $\{998\}$. | 998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $n+S(n)=1964$, where $S(n)$ is the sum of the digits of the number $n$. | 1. We have $S(n) \leq 9 * 4=36 \Rightarrow n \geq 1964-36=1928 \Rightarrow n=1900+10 k+l$, where $2 \leq k \leq 9$ $0 \leq l \leq 9 ; 1900+10 k+l+10+k+l=1964,11 k+2 l=54,2 \leq 2 l \leq 18,36 \leq 11 k \leq 54$ $3 \frac{3}{11} \leq k \leq 4 \frac{10}{11} \Rightarrow k=4, l=5, n=1945$. Answer: $\{1945\}$. | 1945 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The denominator of an irreducible fraction is less than the square of the numerator by one. If 2 is added to both the numerator and the denominator, the value of the fraction will be greater than $\frac{1}{3}$. If 3 is subtracted from both the numerator and the denominator, the fraction will be less than $\frac{1}{10}$. Find this fraction, given that the numerator and denominator are single-digit numbers. Write the answer as a two-digit number, where the first digit is the numerator and the second digit is the denominator. | 4. Let $\frac{m}{n}$ be the desired irreducible fraction, where $m, n$ are single-digit numbers. According to the problem, we have $n=m^{2}-1, \frac{m+2}{n+2}>\frac{1}{3} \Rightarrow 3 m+6>m^{2}+1, m^{2}-3 m-5<0$, then $\frac{3-\sqrt{29}}{2}<m<\frac{3+\sqrt{29}}{2} \Rightarrow$ $\Rightarrow 0<m<\frac{3+5.4}{2} \Rightarrow 0<m<4.2$. Therefore, $m \in\{1,2,3,4\}$. Then $m=1$ does not work because $n=0$; similarly, $m=4$ does not work because $n=15$ is a two-digit number. We check $m=2$, then $n=3$ and the fraction will be $\frac{2}{3}$, which is impossible according to the problem. The only option left is $m=3$, then $n=8$, The fraction $\frac{3}{8}$ satisfies all the conditions of the problem. Answer: $\{38\}$. | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. By the property of absolute value, replacing $x$ with $-x$ does not change this relation. This means that the figure defined by the given inequality is symmetric with respect to the OY axis. Therefore, it is sufficient to find the area of half of the figure for $x \geq 0$. In this case, we obtain the inequality $\left|x-2 y^{2}\right|+x+2 y^{2} \leq 8-4 y$. By removing the absolute value sign, we get two regions: 1: $\left\{\begin{array}{c}x \geq 2 y^{2} \\ x-2 y^{2}+x+2 y^{2} \leq 8-4 y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 2 y^{2} \\ y \leq 2-\frac{x}{2}\end{array}\right.\right.$
Region II: $\left\{\begin{array}{c}x \leq 2 y^{2} \\ -x+2 y^{2}+x+2 y^{2} \leq 8-4 y\end{array} \Leftrightarrow\left\{\begin{array}{c}x \leq 2 y^{2} \\ y^{2}+y-2 \leq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}x \leq 2 y^{2} \\ -2 \leq y \leq 1\end{array}\right.\right.\right.$
Next, on the coordinate plane xOy, we plot the graphs of the obtained inequalities for $x \geq 0$, taking into account that $x=2 y^{2}$ is the graph of a parabola with its vertex at the origin, and the branches of this parabola are directed along the positive direction of the Ox axis. The union of these regions gives a figure which is a trapezoid $M N A B$, the area of which is equal to $S=\frac{1}{2} * 3(8+2)=15$. Then the doubled area is 30 (see Fig. 2). | Answer: $\{30\}$.
Fig. 2 | 30 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Since you have to approach each apple and return to the basket, the number of meters walked will be equal to twice the sum of the first hundred numbers, or 101 taken a hundred times, i.e., $10100 \mathrm{M}$. | Answer: $\{10100$ meters $\}$ | 10100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the area of the region defined by the inequality: $|y-| x-2|+| x \mid \leq 4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. Consider three intervals of change for $x$: 1) $x \leq 0$; 2) $0 \leq x \leq 2$; 3) $x \geq 2$.
1) $x \leq 0$; then $|y+x-2|-x \leq 4, |y+x-2| \leq 4+x \Rightarrow -4 \leq x \leq 0$. Squaring the inequality, we get $(y+x-2)^{2} \leq (4+x)^{2}$, from which $(y-6)(y+2x+2) \leq 0$
2) $0 \leq x \leq 2$, then $|y+x-2| \leq 4-x \Leftrightarrow (y+x-2)^{2} \leq (4-x)^{2} \Leftrightarrow (y+2x-6)(y+2) \leq 0$
3) $x \geq 2$, then $|y-x+2| \leq 4-x \Rightarrow 2 \leq x \leq 4$. Therefore, in this region $(y-x+2)^{2} \leq (4-x)^{2} \Leftrightarrow (y-2)(y-2x+6) \leq 0$. On the coordinate plane, we construct the corresponding regions taking into account the change in $x$. We obtain the region $A B C D E F$, which consists of two right triangles $A B F$ and $C D E$, as well as trapezoid $B C E F$. The coordinates of the points are easily found: $A(-4 ; 6), B(0 ; 6), C(2 ; 2), D(4 ; 2), E(2 ;-2)$ and $F(0 ;-2)$. Taking this into account, the area of the region will be 32. Answer: $\{32\}$ | 32 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Find all roots of the equation
$1-\frac{x}{1}+\frac{x(x-1)}{2!}-\frac{x(x-1)(x-2)}{3!}+\frac{x(x-1)(x-2)(x-3)}{4!}-\frac{x(x-1)(x-2)(x-3)(x-4)}{5!}+$
$+\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{6!}=0 . \quad($ (here $n!=1 \cdot 2 \cdot 3 . . . n)$
In the Answer, indicate the sum of the found roots. | 4. Note that by substituting the numbers $1,2,3,4,5,6$ sequentially into the equation, we will get the equality: $0=0$. This means these numbers are roots of the given equation. Since the equation is of the sixth degree, there are no other roots besides the numbers mentioned above. Answer: $\{21\}$ | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Two acute angles $\alpha$ and $\beta$ satisfy the condition $\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin}(\alpha+\beta)$. Find the sum of the angles $\alpha+\beta$ in degrees. | 6. Using the formula for $\operatorname{Sin}(\alpha+\beta)$, we get:
$\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin} \alpha \operatorname{Cos} \beta+\operatorname{Cos} \alpha \operatorname{Sin} \beta \Leftrightarrow \operatorname{Sin} \alpha(\operatorname{Sin} \alpha-\operatorname{Cos} \beta)=\operatorname{Sin} \beta(\operatorname{Cos} \alpha-\operatorname{Sin} \beta)$. Since $\alpha, \beta$ are acute angles, the brackets $\operatorname{Sin} \alpha-\operatorname{Cos} \beta$ and $\operatorname{Cos} \alpha-\operatorname{Sin} \beta$ have the same sign. Let them be positive, then:
$\left\{\begin{array}{l}\operatorname{Sin} \alpha-\operatorname{Cos} \beta>0 \\ \operatorname{Cos} \alpha-\operatorname{Sin} \beta>0\end{array} \Leftrightarrow \Leftrightarrow\left\{\begin{array}{l}\operatorname{Sin} \alpha>\operatorname{Cos} \beta \\ \operatorname{Cos} \alpha>\operatorname{Sin} \beta\end{array} \Leftrightarrow\left\{\begin{array}{l}\operatorname{Sin}^{2} \alpha>\operatorname{Cos}^{2} \beta \\ \operatorname{Cos}^{2} \alpha>\operatorname{Sin}^{2} \beta\end{array}\right.\right.\right.$ $\Rightarrow \operatorname{Sin}^{2} \alpha+\operatorname{Cos}^{2} \alpha>\operatorname{Cos}^{2} \beta+\operatorname{Sin}^{2} \beta=1$, which is impossible. Similarly, it is impossible that $\operatorname{Sin}^{2} \beta+\operatorname{Cos}^{2} \beta<1$. Therefore, $\operatorname{Sin} \alpha=\operatorname{Cos} \beta, \operatorname{Cos} \alpha=\operatorname{Sin} \beta \Rightarrow \operatorname{Sin} \alpha=\operatorname{Sin}(90-\beta) \Rightarrow \alpha+\beta=90^{\circ}$. A similar proof applies if the brackets $\operatorname{Sin} \alpha-\operatorname{Cos} \beta$ and $\operatorname{Cos} \alpha-\operatorname{Sin} \beta$ are both negative. Answer: $\left\{90^{\circ}\right\}$ | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Find $\max x^{2} y^{2} z$ subject to the condition that $x, y, z \geq 0$ and $2 x+3 x y^{2}+2 z=36$. | 7. From the inequality for the arithmetic mean and geometric mean of three numbers, we have:
$\sqrt[3]{2 x \cdot 3 x y^{2} \cdot 2 z} \leq \frac{2 x+3 x y^{2}+2 z}{3}=12 \Rightarrow 3 x y^{2} \cdot 2 z \cdot 2 x \leq 12^{3}, x^{2} y^{2} z \leq 144$
This inequality becomes an equality if the numbers are equal, i.e., $2 x=3 x y^{2}=2 z=12 \Rightarrow z=6, x=6, y=\sqrt{2 / 3}$. Therefore, the maximum value is 144.
Answer: $\{144\}$ | 144 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Masha has an integer number of times more toys than Lena, and Lena has the same number of times more toys than Katya. Masha gave Lena 3 toys, and Katya gave Lena 2 toys. After that, the number of toys the girls had formed an arithmetic progression. How many toys did each girl originally have? Indicate the total number of toys the girls had. | 1. Let initially Katya has $a$ toys, then Lena has $k a$, and Masha has $\kappa^{2} a$, where $\kappa-$ is a natural number $\geq 2$. After the changes: Katya has $a-2$, Lena has $a \kappa+5$, and Masha has $a \kappa^{2}-3$. Then these numbers form an arithmetic progression in some order. Let's consider the possible cases. If $a \kappa^{2}-3$ is the middle number in the arithmetic progression, then $2 a \kappa^{2}-6=a-2+a \kappa+5 \Rightarrow a\left(2 \kappa^{2}-\kappa-1\right)=9$. But $\kappa \geq 2$, so $2 \kappa^{2}-\kappa-1 \geq 5$ and therefore the possible equality: $2 \kappa^{2}-\kappa-1=9$. This equation does not have natural positive integer solutions. If $a \kappa+5$ is the middle number ($a-2$ is obviously not the middle number), then $2 a \kappa+10=a-2+a \kappa^{2}-3$. From this, $a(k-1)^{2}=15$. Therefore, $(\kappa-1)^{2} \in\{1,3,5,15\}$. If $(\kappa-1)^{2}=1$, then $\kappa=2, a=15 \Rightarrow 15,30,60$ are the required numbers. If $(\kappa-1)^{2}=3$, then $\kappa-$ is not an integer. The same statement is true in the other cases. Thus, Katya has 15 toys, Lena has 30, and Masha has 60. Therefore, the answer is: $\{105\}$ | 105 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the minimum value of the sum
$$
\left|x-1^{2}\right|+\left|x-2^{2}\right|+\left|x-3^{2}\right|+\ldots+\left|x-10^{2}\right|
$$ | 4. Grouping the terms, we get $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)+\left(\left|x-2^{2}\right|+\left|x-9^{2}\right|\right)+\left(\left|x-3^{2}\right|+\left|x-8^{2}\right|\right)+\left(\left|x-4^{2}\right|+\left|x-7^{2}\right|\right)+\left(\left|x-5^{2}\right|+\left|x-6^{2}\right|\right)$. It is known that the pair $|x-a|+|x-b|$ takes the minimum value when $a \leq x \leq b$ and this value is equal to $|b-a|$. Therefore, the minimum value of the first bracket $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)$ will be $10^{2}-1^{2}$, the second bracket respectively $9^{2}-2^{2}$, the third $8^{2}-3^{2}$, and so on. However, the intervals $5^{2} \leq x \leq 6^{2}, 4^{2} \leq x \leq 7^{2}, \ldots$ are nested within each other, since the numbers $1^{2}<2^{2}<3^{2} \ldots<10^{2}$. Therefore, the interval $\left[1,10^{2}\right]$ contains all other intervals, on each of which all terms of the form $|x-a|+|x-b|$ simultaneously take the minimum value. Then our expression takes the minimum value, equal to $10^{2}-1^{2}+9^{2}-2^{2}+8^{2}-3^{2}+7^{2}-4^{2}+6^{2}-5^{2}$, (i.e., the sum of the lengths of the nested intervals). We find this sum: $11 \cdot 9+11 \cdot 7+11 \cdot 5+11 \cdot 3+11 \cdot 1=11(9+7+5+3+1)=$ $=11 \cdot 25=275$ Answer: $\{275\}$ | 275 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Solve the system
$\left\{\begin{array}{l}a+c=4 \\ a d+b c=5 \\ a c+b+d=8 \\ b d=1\end{array}\right.$
In the answer, write the sum of all solutions of the given system. | 7. We have
$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=$
$=x^{4}+4 x^{3}+6 x^{2}+5 x+2=(x+1)(x+2)\left(x^{2}+x+1\right)$. Since the polynomial $x^{2}+x+1$ does not have
real roots, the identity $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}+3 x+2\right)\left(x^{2}+x+1\right)$
means that $\left\{\begin{array}{l}x^{2}+a x+b=x^{2}+3 x+2 \\ x^{2}+c x+d=x^{2}+x+1\end{array}\right.$, or $\left\{\begin{array}{l}x^{2}+a x+b=x^{2}+x+1 \\ x^{2}+c x+d=x^{2}+3 x+2\end{array}\right.$.
From this, we find that $a=3, b=2, c=1, d=1$ or $a=1, b=1, c=3, d=2$ (since the equality of polynomials means the equality of coefficients of corresponding powers). Answer: $\{14\}$ | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1.
Since the pedestrian covers 1 km in 10 minutes, his speed is 6 km/h. There are more oncoming trams than overtaking ones because, relative to the pedestrian, the speed of the former is greater than that of the latter. If we assume the pedestrian is standing still, the speed of the oncoming trams is the sum of the tram's own speed $V$ and the pedestrian's speed. Therefore, the relative speed of the oncoming trams is $V+6$, and the relative speed of the overtaking trams is $V-6$. It is clear that the number of trams passing a given point in a certain time is proportional to their speed, from which we have:
$$
\frac{V+6}{V-6}=\frac{700}{300} \rightarrow 3 V+18=7 V-42 \text {, i.e. } V=15 \text {. }
$$ | Answer: 15 km $/$ hour.
# | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have? | 3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have?
$\square$ We will prove by induction that the polynomial $P_{n}(x)=x^{n}+\ldots$ has exactly $n$ distinct real roots, and that between any two consecutive roots of $P_{n}(x)$ there is exactly one root of the polynomial $P_{n-1}(x)$. Indeed, $P_{2}(x)=x^{2}-1, P_{3}(x)=x\left(x^{2}-2\right)$, and this statement is true for $n=3$. Suppose $x_{1}<x_{2}<\ldots<x_{n}$ are the roots of $P_{n}(x)$, and $y_{1}<y_{2}<\ldots<y_{n-1}$ are the roots of $P_{n-1}(x)$. By the induction hypothesis, $x_{1}<y_{1}<x_{2}<y_{2}<\ldots<x_{n-1}<y_{n-1}<x_{n}$. Consider the interval $\left(x_{i}, x_{i+1}\right)$. On this interval, $P_{n}(x)$ changes sign, and $P_{n-1}(x)$ does not change sign. Therefore, $P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x)$ changes sign on this interval, and hence has a root. Since $P_{n}(x)$ changes sign between $x_{n}$ and $a$ if $a$ is sufficiently large, there is a root of $P_{n+1}(x)$ on the interval $\left(x_{n}, a\right)$. Similarly, the interval $\left(-\infty, x_{1}\right)$ is considered. Thus, the polynomial $P_{n+1}(x)$ has exactly $n+1$ roots. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Misha, over the course of a week, picked an apple each day and weighed it. Each apple weighed a different amount, but the weight of each apple was a whole number of grams and ranged from 221 grams to 230 grams (inclusive). Misha also calculated the average weight of all the apples he picked, and it was always a whole number. The apple picked on the seventh day weighed 225 grams. How much did the apple picked on the sixth day weigh? | Answer: 230 grams.
Solution: Each apple weighed 220 grams plus an integer from 1 to 10. From the numbers 1 to 10, we need to choose 7 numbers such that their sum is divisible by 7. One of these numbers is 5, so we need to choose 6 from the numbers $1,2,3,4,6,7,8,9,10$. The smallest sum of six of these numbers is 23, and the largest is 44. Adding 5, we get numbers from 28 to 49. In this range, there are 4 numbers divisible by 7: 28, 35, 42, 49. Meanwhile, the sum of the weights for the first 6 days must be divisible by 6. From the numbers 23, 30, 37, 44, only the number 30 satisfies this condition. Thus, the total weight for the first 6 days was $220 \cdot 6 + 30$ (we will simply write 30, and add the multiple of 220 at the end). The sum of the weights for the first 5 days must be divisible by 5. The possible range for 5 days is from 16 to 40, and the numbers that fit are $20, 25, 30, 40$. However, the total weight for 5 days is less than that for 6 days, so it must be less than 30, i.e., 20 or 25. However, the difference between 30 and 25 is 5, and the number 5 is already used. Therefore, the weight of the apples for 5 days is $220 \cdot 5 + 20$, meaning the apple picked on the 6th day weighed $220 \cdot 6 + 30 - (220 \cdot 5 + 20) = 230$ grams. This value is valid, as the apples picked over the week could have weighed 221, 223, 222, 226, 228, 210, 205 grams. | 230 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Senya thought of two numbers, then subtracted the smaller from the larger, added both numbers and the difference, and got 68. What was the larger of the numbers Senya thought of? | Answer: 34.
Solution. The subtrahend plus the difference equals the minuend. Therefore, the doubled minuend equals 68.
Comment. Correct solution - 20 points. The result is obtained based on examples and the pattern is noticed but not explained - 15 points. The result is obtained based on one example - 10 points. The solution is started, but the progress is insignificant - 1 point. The correct answer is given without explanation - 1 point. | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The little squirrel has several bags of nuts. In two bags, there are 2 nuts each, in three bags, there are 3 nuts each, in four bags, there are 4 nuts each, and in five bags, there are 5 nuts each. Help the little squirrel arrange the bags on two shelves so that there are an equal number of bags and nuts on each shelf. | Solution. For example, $5+5+5+4+4+2+2=27$ nuts in 7 bags - the first shelf, $5+5+4+4+3+3+3=27$ nuts in 7 bags - the second shelf.
Comment. Correct solution - 20 points. Solution started, some progress made - 5 points. Solution started, but progress insignificant - 1 point. Solution incorrect or absent - 0 points. | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In a row, all natural numbers from 1 to 100 inclusive are written in ascending order. Under each number in this row, the product of its digits is written. The same procedure is applied to the resulting row, and so on. How many odd numbers will be in the fifth row? | Answer: 19.
Solution: Note that if a number contains an even digit in its notation, then the product of the digits will be even. However, then in all subsequent rows, the product will also be even. Let's find all the products of two digits in the multiplication table that are written using only odd digits:
$$
\begin{array}{rr}
1=1 \cdot 1 ; \quad 3=1 \cdot 3 ; & 5=1 \cdot 5 ; \quad 7=1 \cdot 7 \\
9=1 \cdot 9 ; \quad 9=3 \cdot 3 ; \quad 15=3 \cdot 5 ; \quad 35=5 \cdot 7
\end{array}
$$
All other products contain even digits. Therefore, products consisting of odd digits will only be under numbers that consist of the specified digits. The list of numbers: $1,3,5,7,9,11,13,15,17,19,31,33,35,51,53,57,71,75,91$ - a total of 19 numbers.
Comment: Correct solution - 20 points. If the correct approach is followed but one extra number is written or one number is missing - 18 points. The solution is started, and there is some progress - 5 points. The solution is started, but the progress is insignificant - 1 point. The solution is incorrect or missing - 0 points. | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Sixth-graders were discussing how old their principal is. Anya said: "He is older than 38 years." Borya said: "He is younger than 35 years." Vova: "He is younger than 40 years." Galya: "He is older than 40 years." Dima: "Borya and Vova are right." Sasha: "You are all wrong." It turned out that the boys and girls made the same number of mistakes. Can we find out how old the principal is? | Solution. Note that Anya and Vova cannot be wrong at the same time, so Sasha is wrong. Also, at least one of the pair "Anya-Borya" and at least one of the pair "Vova-Galya" is wrong. Thus, there are no fewer than three wrong answers, and due to the evenness, there are exactly four: two boys and two girls. This means that no more than one girl and no more than two boys are correct. Dima essentially confirms Borya's words. However, Borya, Dima, and Vova cannot be correct at the same time, so Borya and Dima are wrong, and among the boys, only Vova is right, and among the girls, only Anya is right. Therefore, the principal is 39 years old. | 39 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. How many irreducible fractions exist where the sum of the numerator and denominator is 100? | Answer: 20.
Solution: Consider the equation $a+b=100$. If $a$ and $b$ have a common divisor, then this divisor also divides 100, meaning that the only possible prime common divisors of $a$ and $b$ satisfying the equation can be 2 and 5. We will pair all numbers as ( $a, 100-a$ ). Since the fraction $\frac{a}{b}$ is irreducible, we exclude even numbers from 1 to 100 (leaving 50 numbers), and numbers divisible by 5 (leaving 40 numbers). Among these 40 numbers, each number $a$ corresponds to the number $100-a$, because if $a$ is odd and not divisible by 5, then $100-a$ also has these properties. We will write the smaller of the numbers $a, 100-a$ in the numerator of the fraction. This results in 20 pairs.
Comment: Correct solution - 20 points. Incomplete justification - 15 points. Correct solution with one calculation error - 18 points. Solution not completed or incorrect, but with good progress - 10 points. Solution started, with some progress - 5 points. Solution started, but with insignificant progress - 1 point. Only the correct answer without a solution - 1 point. | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Two squirrels had the same number of pine cones and the same number of cedar cones. In total, each squirrel had fewer than 25 cones. The first squirrel gathered as many pine cones as it already had and 26 cedar cones. It ended up with more pine cones than cedar cones. The second squirrel gathered as many cedar cones as it already had and ate 4 pine cones. It ended up with more cedar cones than pine cones. How many pine and cedar cones did each squirrel have initially? | Answer: 17 pine cones, 7 cedar cones.
Solution. Let the number of pine cones be $x$ and the number of cedar cones be $y$. We can write the conditions as: $\left\{\begin{array}{l}x+y=26 \\ 2 y>x-4\end{array}\right.$
Add the second and third inequalities: $2(x+y)>x+y+22$, or $x+y>22$. But $x+y=26$, so $x+y>22$ is always true. Now, we need to solve the system: $\left\{\begin{array}{l}x+y=26 \\ 2 y>x-4\end{array}\right.$
From the first equation, we have $y=26-x$. Substitute this into the second inequality: $2(26-x)>x-4$, or $52-2x>x-4$, which simplifies to $56>3x$, or $x<\frac{56}{3}$. Since $x$ and $y$ are integers, we need to find integer solutions for $x$ and $y$.
We also have $2y>x-4$, or $2(26-x)>x-4$, which simplifies to $52-2x>x-4$, or $56>3x$, or $x<\frac{56}{3}$. This gives us the inequality $\left\{\begin{array}{l}x>\frac{49}{3} \\ x<\frac{56}{3}\end{array}\right.$
In the interval $\left(\frac{49}{3} ; \frac{56}{3}\right)$, the only integer is 17. Therefore, $x=17, y=7$.
Comment. Correct solution - 20 points. Correct solution with one calculation error - 18 points. Total number of cones found, but solution not completed - 15 points. Solution started, some progress made - 5 points. Solution started, but progress insignificant - 1 point. Only correct answer without solution - 1 point. | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A $7 \times 7$ table is filled with non-zero integers. First, the border of the table is filled with negative numbers. Then, the cells are filled in any order, and the next number is equal to the product of the previously placed numbers that are closest to it in the same row or column. What is the maximum number of positive numbers that can be in the table? | # Answer. 24.
Solution. Evaluation. We will prove that there must be at least one negative number in the shaded area. Suppose this is not the case. Consider the corner cell of the shaded area where the number was placed last among the four corners (it is shaded black). By assumption, if some other numbers are placed in the cells of its row and column, they are positive. Therefore, a negative number must be placed in the black cell. There are at least \(4 \cdot 6 + 1 = 25\) negative numbers, so there are at most 24 positive numbers.
Example. We will place numbers in the shaded area in the order indicated by the numbers. In cells 1 and 2, we will choose factors by column, in cells 3 and 4, we will choose factors by row, resulting in positive numbers. In cell 5, there will be a negative number. All other cells can be filled with positive numbers, totaling 24.
Comment. Correct solution - 20 points. A filling method that is not optimal is proposed - 5 points. If the sequence of filling the cells is not indicated - 1 point.
## Variant 4 | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A circle passing through the vertices $L$ and $M$ of trapezoid $K L M N$ intersects the lateral sides $K L$ and $M N$ at points $P$ and $Q$ respectively and touches the base $K N$ at point $S$. It turns out that $\angle L S M=50^{\circ}$, and $\angle K L S=\angle S N M$. Find $\angle P S Q$. | Answer: $65^{\circ}$.
Solution. Since $K S$ is a tangent to the circle, then
$$
\angle K S P=\angle S Q P=\angle S L P=\angle K L S=\angle S N M .
$$
But $\angle K S P$ and $\angle S N M$ are corresponding angles for the lines $P S$ and $M N$, so $P S \| M N$ and the inscribed quadrilateral $S Q M P$ is a trapezoid. Therefore, it is an isosceles trapezoid and, hence, $\angle P S Q=\angle S P M=\angle S L M$, the latter by the inscribed quadrilateral $S P L M$. From the parallelism of the lines $K N$ and $L M$ we have
$$
\angle S L M=\angle K S L=\angle K S P+\angle P S L=\angle S M P+\angle P M L=\angle L M S .
$$
Thus
$$
2 \angle P S Q=2 \angle S L M=\angle S L M+L M S=180^{\circ}-\angle L S M=130^{\circ}
$$
Comment. Correct solution - 20 points. Solution not completed, but with good progress - 10 points. Solution started, with some progress - 2 points. Incorrect or no solution - 0 points. | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. On 8 balls, numbers are written: $2,3,4,5,6,7,8,9$. In how many ways can the balls be placed into three boxes so that no box contains a number and its divisor? | Answer: 432.
Solution. The numbers 5 and 7 can be placed in any box, the number of ways is $3 \cdot 3=9$. The numbers $2, 4, 8$ must be in different boxes, the number of ways is $3 \cdot 2 \cdot 1=6$. Thus, the numbers $2, 4, 5, 7, 8$ can be arranged in 54 ways. Suppose the numbers 2 and 3 are placed in the same box (this box is already chosen for the number 2). Then there are two choices left for the number 9, and 2 choices for the number 6, in total $54 \cdot 2 \cdot 2=216$ ways. Suppose the numbers 2 and 3 are placed in different boxes, the number of choices for the box for 3 is 2. Then the box for the number 6 is chosen in a unique way, and the box for the number 9 is any of the two where there is no number 3, in total $54 \cdot 2 \cdot 2=216$ ways, thus $216+216=432$. | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. What is the maximum number of cells in an $8 \times 8$ square that can be colored so that the centers of any four colored cells do not form the vertices of a rectangle with sides parallel to the edges of the square? | Answer: 24 cells.
Solution: Suppose that no less than 25 cells are marked. We will say that a pair of marked cells in the same row covers a pair of columns in which these cells are located. A prohibited quartet arises when a pair of columns is covered twice. If in one row there are $m$ marked cells, and in another $n>m+1$ marked cells, then by moving one marked cell from the second row to the first, the number of covered pairs in the first row will increase by $m$, and in the second row it will decrease by $n-1>m$, i.e., the total number of pairs will decrease. Thus, the minimum number of pairs for a given number of marked cells is achieved when the maximum number of cells in a row differs from the minimum by no more than 1. Let's consider two cases.
1) There is a column in which no more than two cells are marked. Let's strike it out, in which case seven columns will remain, i.e., 21 pairs. One of the possible distributions of 23 cells is seven rows with three cells and one row with two cells, which gives $7 \cdot 3+1=22$ marked pairs. By the pigeonhole principle, there will be a pair of columns covered twice. As shown above, with a different distribution of marked pairs, there will be more.
2) In each column, no fewer than three cells are marked. Since $8 \cdot 3<25$, there will be a row in which no fewer than four cells are marked. Let these be the first four cells in the bottom row. Above them, in each of the first four columns, there are at least two more cells - a total of at least eight. They are located in seven rows, therefore, two of them are in the same row. Together with the corresponding cells in the bottom row, they form a prohibited quartet. An example for 24 cells is shown in the figure.
 | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In a rectangle of size $7 \times 9$ cells, some cells contain one baby squirrel each, such that in every rectangle of size $2 \times 3$ (or $3 \times 2$) there are exactly 2 baby squirrels. Draw how they can be seated. | Answer. For example, like this (21 squirrels sit in the shaded cells).
Comment. A correct example - 20 points. | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Diana wrote a two-digit number, and appended to it a two-digit number that was a permutation of the digits of the first number. It turned out that the difference between the first and second numbers is equal to the sum of the digits of the first number. What four-digit number was written? | # Answer: 5445.
Solution. The difference between such numbers is always divisible by 9, since $10a + b - (10b + a) = 9(a - b)$. This difference equals the sum of two digits, so it is no more than 18. However, it cannot be 18, because then both the first and second numbers would be 99. Therefore, the difference between the two numbers is 9, and the tens digits of the two numbers differ by no more than 1. But the sum of the tens digits of the two numbers is the sum of the digits of the first number, and it equals 9. Clearly, only the digits 4 and 5 fit.
Comment. Correct solution - 20 points. Incomplete proof - 10 points. Solution by trial and error, without proving that there are no other options - 5 points. The answer includes the number 4554 - 1 point is deducted. Some progress in the solution - 5 points. The solution is started, but there is no noticeable progress - 1 point. The correct answer is given without explanation - 0 points. | 5445 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. For a children's party, pastries were prepared: 10 eclairs, 20 mini tarts, 30 chocolate brownies, 40 cream puffs. What is the maximum number of children who can each take three different pastries? | Answer: 30.
Solution: From eclairs, baskets, and brownies, at least 2 pastries must be taken, and there are 60 of them in total, meaning no more than 30 children can take three different pastries. They can do this as follows: 10 children will take an eclair, a brownie, and a roll, and 20 children will take a basket, a brownie, and a roll.
Comment: Correct solution - 20 points. Proof that 30 is the maximum value - 10 points, example - 10 points. If the proof is insufficiently justified, 5 points are deducted. Solution started, with some progress - 5 points. Solution started, but no significant progress - 1 point. Correct answer given without explanation - 0 points. | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On 900 cards, all natural numbers from 1 to 900 are written. Cards with squares of integers are removed, and the remaining cards are renumbered, starting from 1.
Then the operation of removing squares is repeated. How many times will this operation have to be repeated to remove all the cards | Answer: 59.
Solution: During the first operation, 30 cards will be removed, leaving $900-30=30 \cdot 29$ cards. Since $30 \cdot 29 > 29^2$, all squares except $30^2$ remain. During the second operation, 29 cards will be removed. There will be $30 \cdot 29 - 29 = 29^2$ cards left. Thus, in two operations, we transition from the number $30^2$ to the number $29^2$. Let's write this in a general form: during the first operation, $n$ cards will be removed, leaving $n^2 - n = n(n-1) > (n-1)^2$ cards. During the second operation, $n-1$ cards will be removed. There will be $n^2 - n - (n-1) = (n-1)^2$ cards left, meaning that in two operations, we always transition from the number $n^2$ to the number $(n-1)^2$. In total, there are 30 squares among the numbers from 1 to 900, and 29 transitions between them. To reach a single card, it will take $2 \cdot 29$ operations, and one more to remove the last card with the number 1.
Comment: Correct solution - 20 points. There is a gap in the justification - 15 points. Errors in the calculation of the number of options with a correct idea - 5 points are deducted. The solution is started, with some progress - 5 points. The solution is started, but there is no noticeable progress - 1 point. The correct answer is given without explanation - 0 points.
## Variant 4 | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Squirrel Pusistik and Lohmatik ate a basket of berries and a bag of seeds, which contained more than 50 but less than 65 seeds, starting and finishing at the same time. At first, Pusistik ate berries, and Lohmatik ate seeds, then (at some point) they switched. Lohmatik ate berries six times faster than Pusistik, and seeds only three times faster. How many seeds did Lohmatik eat if Lohmatik ate twice as many berries as Pusistik? | # Answer: 54.
Solution. Divide the berries into 3 equal parts. Each part, Lomhatic ate 6 times faster than Pushistik, but there are two parts, so he spent only 3 times less time on the berries than Pushistik. Therefore, Pushistik ate the seeds in one-third the time of Lomhatic. Since Pushistik eats three times slower, he ate 9 times fewer seeds than Lomhatic. Dividing the seeds in the ratio of 9:1, we see that Pushistik got one-tenth of the seeds, from which it is not difficult to understand that there were 60 seeds in the bag. Therefore, Pushistik ate 6 seeds, and Lomhatic - 54.
Comment. Correct solution - 20 points. Solution started, with some progress - 5 points. Solution started, but no significant progress - 1 point. Incorrect or no solution - 0 points. | 54 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Three squirrels usually eat porridge for breakfast: semolina (M), buckwheat (B), oatmeal (O), and millet (R). No porridge is liked by all three squirrels, but for each pair of squirrels, there is at least one porridge that they both like. How many different tables can be made where each cell contains a plus (if it is liked) or a minus (if it is not liked)?
| | M | B | O | R |
| :--- | :--- | :--- | :--- | :--- |
| Squirrel 1 | | | | |
| Squirrel 2 | | | | |
| Squirrel 3 | | | | | | Answer: 132.
Solution: If two different pairs like the same porridge, then the porridge is liked by three squirrels, violating the first condition. From three squirrels, three different pairs can be formed, and these three pairs like different porridges. There are 4 ways to choose these 3 porridges, and $3!=6$ ways to distribute 3 porridges among the three pairs. In total, there are 24 ways to distribute the porridges that the pairs like. The fourth porridge can have the following scenarios: 1) it is not liked by anyone, 2) it is liked by only one squirrel, 3) it is liked by only one pair.
In case 1), the total number of options remains 24. In case 2), the number 24 is multiplied by the number of ways to choose the squirrel who likes the fourth porridge, which is 3, resulting in 72. In case 3), the situation is as follows: there is a pair that likes two porridges, and two pairs that like two other porridges. The first two porridges can be chosen in $\frac{4 \cdot 3}{2}=6$ ways. The pair that likes the first two porridges can be chosen in three ways (since we choose one out of three). The two remaining porridges can be distributed among the two remaining pairs in two ways. In total, there are $6 \cdot 3 \cdot 2=36$ options in this case. Therefore, the total number of different tables can be $24+72+36=132$.
Comment: Correct solution - 20 points. Correct solution with computational errors - 15 points. Partial case considered - 5 points. Solution started but progress is insignificant - 1 point. Only the correct answer without a solution - 0 points.
## Variant 3 | 132 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On the border of a circular glade, points $A, B, C, D$ are marked clockwise. At point $A$ is a squirrel named An, at point $B$ is a squirrel named Bim, at point $C$ stands a pine tree, and at point $D$ stands an oak tree. The squirrels started running simultaneously, An towards the pine tree, and Bim towards the oak tree. They collided at point $M$, which is closer to the pine tree than to the oak tree. Is it true that if An had run from point $A$ to the oak tree, and Bim from point $B$ to the pine tree, An would have arrived first? Each squirrel runs in a straight line and at a constant speed. | Answer: No.
Solution: Let the speed of Ana be $-v$, and the speed of Bim be $-u$. They both reached point $M$ at the same time, so $\frac{A M}{v}=\frac{B M}{u}$. Triangles $D A M$ and $C B M$ are similar (by three angles).
Therefore, $\frac{A D}{B C}=\frac{A M}{B M}$. But $\frac{A M}{B M}=\frac{v}{u}$, so $\frac{A D}{B C}=\frac{v}{u}$, which means $\frac{A D}{v}=\frac{B C}{u}$. Thus, the squirrels will arrive at the same time.
Comment: Correct solution - 20 points. Correct solution not fully completed - 15 points. Solution with inaccuracies - 15 points. Solution based on an example - 10 points. Some progress - 5 points. Incorrect or no solution - 0 points.
3. In how many ways can 30 apples be distributed into 3 baskets such that the first basket has fewer apples than the second, the second has fewer apples than the third, and no basket is empty?
## Answer: 61.
Solution: Consider the equation $a+b+c=30$. Arrange 30 units in a row, choose two gaps between the units, and place a dash in each. The number of units to the left of the first dash is $a$ (the number of apples in the first basket), and the number to the right of the second dash is $c$ (the number of apples in the third basket). The number of ways to choose two gaps is $\frac{29 \cdot 28}{2}=406$. We need to subtract the number of cases where among the numbers $a, b, c$ there are equal numbers. Let $a=b$. Then $2 a+c=30$, this equation has 14 non-zero solutions ( $c$ is even and can vary from 2 to 28). Similarly, there will be 14 cases when $a=c$ or $b=c$. Thus, $406-3 \cdot 14=364$. The case $a=b=c$ is counted once in the total number of ways and subtracted three times, but it should be excluded only once, so we need to add 2. The number $364+2=366$ is the number of ordered triples of distinct $a, b, c$. But we need the order $a<b<c$, so we divide by the number of permutations of three elements: $\frac{366}{6}=61$.
Comment: Correct solution - 20 points. Correct solution not fully completed - 15 points. Significant progress in the solution - 10 points. Only the correct answer without a solution - 0 points. | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (7-8 grade) Maria Ivanovna is a strict algebra teacher. She only puts twos, threes, and fours in the grade book, and she never gives the same student two twos in a row. It is known that she gave Vovochka 6 grades for the quarter. In how many different ways could she have done this? Answer: 448 ways. | Solution. Let $a_{n}$ be the number of ways to assign $n$ grades. It is easy to notice that $a_{1}=3$, $a_{2}=8$. Note that 3 or 4 can be placed after any grade, while 2 can only be placed if a 3 or 4 preceded it. Thus, a sequence of length $n$ can be obtained by appending 3 or 4 to a sequence of length $n-1$ or by appending 32 or 42 to a sequence of length $n-2$, from which we get the recurrence relation $a_{n}=2 a_{n-1}+2 a_{n-2}$. From this, it is easy to find that $a_{3}=22, a_{4}=60, a_{5}=164, a_{6}=448$. | 448 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. 7-8 grade Excellent student Kolya found the sum of the digits of all numbers from 0 to 2012 and added them all together. What number did he get? Answer: 28077. | Solution. Note that the numbers 0 and 999, 1 and 998, ..., 499 and 500 complement each other to 999, i.e., the sum of their digit sums is 27. Adding their digit sums, we get $27 \times 500 = 13500$. The sum of the digits of the numbers from 1000 to 1999, by similar reasoning, is 14500 (here we account for the fact that each number has an additional one, so the sum increases by a thousand). The numbers 2000, 2001, ... 2012 remain, and their digit sum is 77. In total, we get $13500 + 14500 + 77 = 28077$. | 28077 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. A line parallel to the selected side of a triangle with an area of 27 cuts off a smaller triangle with an area of 12. Find the area of the quadrilateral, three vertices of which coincide with the vertices of the smaller triangle, and the fourth lies on the selected side. Choose the answer option with the number closest to the one you found. | Answer: 18. (E)
 | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In how many ways can a coach form a hockey team consisting of one goalkeeper, two defenders, and three forwards if he has 2 goalkeepers, 5 defenders, and 8 forwards at his disposal? Among the proposed answer options, choose the one closest to the correct one. | Answer: 1120. (B)
$$
\begin{array}{|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 915 & \mathbf{B} & 1120 & \mathbf{C} & 1400 & \mathbf{D} \\
1960 & \mathbf{E} & 2475 & \mathbf{F} \\
\hline
\end{array}
$$ | 1120 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. In how many ways can a team be selected from a group consisting of 7 boys and 8 girls, so that the team has 4 boys and 3 girls? Among the proposed answer options, choose the one closest to the correct one. | Answer: 1960. (D)
$$
\begin{array}{|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} 915 & \mathbf{B} & 1120 & \mathbf{C} & 1400 & \mathbf{D} & 1960 & \mathbf{E} \\
\hline
\end{array}
$$ | 1960 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. In how many ways can a coach form a basketball team consisting of two guards and three forwards if he has 6 guards and 11 forwards at his disposal? Among the options provided, choose the one closest to the correct answer.
| Answer: 2475. (E)
 | 2475 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. In how many ways can a team be assembled consisting of 3 painters and 4 plasterers, if there are 6 painters and 8 plasterers? Among the options provided, choose the one closest to the correct answer. | Answer: $1400 .(\mathrm{C})$
 | 1400 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. Ivan Semenovich leaves for work at the same time every day, drives at the same speed, and arrives exactly at 9:00. One day he overslept and left 40 minutes later than usual. To avoid being late, Ivan Semenovich drove at a speed 60% faster than usual and arrived at 8:35. By what percentage should he have increased his usual speed to arrive exactly at 9:00? | Answer: By $30 \%$.
Solution: By increasing the speed by $60 \%$, i.e., by 1.6 times, Ivan Semenovich reduced the time by 1.6 times and gained 40+25=65 minutes. Denoting the usual travel time as $T$, we get $\frac{T}{1.6}=T-65$, from which $T=\frac{520}{3}$. To arrive in $T-40=\frac{400}{3}$, the speed needed to be increased by $\frac{520}{3} \div \frac{400}{3}=1.3$ times, i.e., by $30 \%$. | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The English club is attended by 20 gentlemen. Some of them are acquainted (acquaintances are mutual, i.e., if A knows B, then B knows A). It is known that there are no three gentlemen in the club who are pairwise acquainted.
One day, the gentlemen came to the club, and each pair of acquaintances shook hands with each other (once). What is the maximum number of handshakes that could have been made? | Answer: 100 handshakes.
Solution: Choose a gentleman with the maximum number of acquaintances (if there are several, choose any one). Suppose he has $n$ acquaintances. These acquaintances cannot be pairwise acquainted with each other. Consider the remaining $(20-n-1)$ gentlemen, each of whom has no more than $n$ acquaintances, so the number of handshakes they make does not exceed $(20-n-1) \cdot n$. Therefore, the total number does not exceed $(20-n) \cdot n$. This number is maximal when $n=10$. It can be shown that 100 handshakes are possible - by dividing the gentlemen into 2 groups of 10 people each and ensuring that each person in the first group is acquainted with each person in the second (while there are no acquaintances within the groups). | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. From point $M$, lying inside triangle $A B C$, perpendiculars are drawn to the sides $B C, A C, A B$, with lengths $k, l$, and $m$ respectively. Find the area of triangle $A B C$, if $\angle C A B=\alpha$ and $\angle A B C=\beta$. If the answer is not an integer, round it to the nearest integer.
$$
\alpha=\frac{\pi}{6}, \beta=\frac{\pi}{4}, k=3, l=2, m=4
$$ | Answer: 67.
Solution. Denoting the sides of the triangle by $a, b, c$, using the sine theorem we get $S=\frac{k a+l b+m c}{2}=R(k \sin \alpha+l \sin \beta+m \sin \gamma)$.
Since, in addition, $S=2 R^{2} \sin \alpha \sin \beta \sin \gamma$, we can express $R=\frac{k \sin \alpha+l \sin \beta+m \sin \gamma}{2 \sin \alpha \sin \beta \sin \gamma}$.
Therefore, $S=\frac{(k \sin \alpha+l \sin \beta+m \sin \gamma)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}$.
After substituting the numerical values of $k, l, m, \alpha, \beta, \gamma$ we get
$$
S=\frac{(3 \sin \alpha+2 \sin \beta+4 \sin \gamma)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}=\frac{(3+4 \sqrt{2}+2 \sqrt{6})^{2}}{\sqrt{3}+1} \approx 67
$$ | 67 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.1. The master's day shift lasts $10 \%$ longer than the apprentice's shift. If the apprentice worked as long as the master, and the master worked as long as the apprentice, they would produce the same number of parts. By what percentage does the master produce more parts per day than the apprentice? | Solution. Let $p=10 \%$. Suppose the productivity of the apprentice is $a$ parts per hour, the master's productivity is $b$ parts per hour, the apprentice's workday is $n$ hours, and the master's workday is $m$ hours. Then from the condition it follows:
$$
m=n\left(1+\frac{p}{100}\right), \quad \text { and } \quad m a=n b \Rightarrow n\left(1+\frac{p}{100}\right) a=n b \Rightarrow b=\left(1+\frac{p}{100}\right) a .
$$
In a day, the master makes $b m-a n=\frac{p}{100}\left(2+\frac{p}{100}\right)$ an more parts than the apprentice, which as a percentage of the apprentice's daily output is
$$
\frac{p}{100}\left(2+\frac{p}{100}\right) a n \cdot \frac{100}{a n}=p\left(2+\frac{p}{100}\right)
$$
If $p=10 \%$, then $p\left(2+\frac{p}{100}\right)=21$.
Answer: 21. (C)
 | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 37 more than the product of the same digits. | Solution. 1st method. For a two-digit number $\overline{a b}$, the condition means that
$$
a^{2}+b^{2}-a b=37
$$
Since the equation is symmetric, i.e., with each solution $(a, b)$, the pair $(b, a)$ is also a solution, we can assume without loss of generality that $a \geqslant b$.
- Suppose $a \leqslant 6$. Then the equation (1) has no solutions, because $b \leqslant a$ and $37=a^{2}+b^{2}-a b=a^{2}+b(b-a) \leqslant a^{2} \leqslant 36$.
- For $a=7$, the equation (1) becomes $b^{2}-7 b+12=0 \Longleftrightarrow b=3, b=4$. This gives us the solutions 73, 74, and considering the symmetry, 37, 47.
- For $a=8$ and $a=9$, the equation (1) becomes $b^{2}-8 b-27=0$ and $b^{2}-9 b-44=0$ respectively. Both equations have no integer solutions.
It remains to sum all the obtained numbers: $73+74+37+47=231$.
2nd method. Another solution to this problem. For a two-digit number $\overline{a b}$, the condition means that
$$
a^{2}+b^{2}-a b=37 \Longleftrightarrow(2 a-b)^{2}+3 b^{2}=148
$$
Then $b \leqslant 7$ and a complete enumeration of the cases
| $b$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $3 b^{2}$ | 0 | 3 | 12 | 27 | 48 | 75 | 108 | 147 |
| $(2 a-b)^{2}$ | 37 | 145 | 126 | 121 | 100 | 73 | 40 | 1 |
leaves only three options:
$$
\left[\begin{array}{l}
b=3 \\
2 a-b= \pm 11
\end{array} \Rightarrow a=7 ; \quad\left[\begin{array}{l}
b=4 \\
2 a-b= \pm 10
\end{array} \Rightarrow a=7 ; \quad\left[\begin{array}{l}
b=7 \\
2 a-b= \pm 1
\end{array} \Rightarrow a=3, a=4\right.\right.\right.
$$
Thus, these are the numbers $37, 47, 73, 74$. Their sum is 231.
Answer: 231. (C)
 | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.2. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 57 more than the product of the same digits. | Solution. Such two-digit numbers are $18, 78, 81, 87$. Their sum is 264.
Answer: 264. (D)
\section*{| A | 165 | $\mathbf{B}$ | 198 | $\mathbf{C}$ | 231 | $\mathbf{D}$ | 264 | $\mathbf{E}$ | 297 | $\mathbf{F}$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | 264 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.3. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 73 more than the product of the same digits. | Solution. Such two-digit numbers are $19, 89, 91, 98$. Their sum is 297.
Answer: 297. ( ( )
 | 297 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.4. Find the sum of all two-digit numbers for each of which the sum of the squares of the digits is 31 more than the product of the same digits. | Solution. Such two-digit numbers are $16, 56, 61, 65$. Their sum is 198.
Answer: 198. (B)
 | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. The segment connecting the lateral sides of the trapezoid and parallel to its bases, which are 3 and 21, divides the trapezoid into two parts of equal area. Find the length of this segment. | Solution. 1st method. Let $a=3$ and $b=21$ be the lengths of the bases of the trapezoid. If $c$ is the length of the segment parallel to the bases, and $h_{1}$ and $h_{2}$ are the parts of the height of the trapezoid adjacent to the bases $a$ and $b$ respectively, then, by equating the areas, we get:
$$
\frac{a+c}{2} h_{1}=\frac{c+b}{2} h_{2}=\frac{1}{2} \cdot \frac{a+b}{2}\left(h_{1}+h_{2}\right) \Longleftrightarrow\left\{\begin{array}{l}
\frac{b+c}{a+c}=\frac{h_{1}}{h_{2}} \\
\frac{2(b+c)}{a+b}=1+\frac{h_{1}}{h_{2}}
\end{array}\right.
$$
From which
$$
\frac{2(b+c)}{a+b}=\frac{a+b+2 c}{a+c}
$$
Finally, we get $2 c^{2}=a^{2}+b^{2}$. Therefore
$$
c=\sqrt{\frac{a^{2}+b^{2}}{2}}=\sqrt{\frac{9+441}{2}}=15
$$
2nd method. Let $A B C D$ be a trapezoid with bases $A D=a=3, B C=b=21$ (see Fig. 1). Let $M N$ be the segment parallel to the bases. Let $F$ be the intersection point of the lines $A B$ and $C D$, the area of triangle $B C F$ is $a^{2} S$ and $M N=x$. Then
- from the similarity $M N F \sim B C F$ with the coefficient $k_{1}=x / a$ it follows: $S_{M N F}=k_{1}^{2} S_{B C F}=x^{2} S$;
- from the similarity $A D F \sim B C F$ with the coefficient $k_{2}=b / a$ it follows: $S_{A D F}=k_{2}^{2} S_{B C F}=b^{2} S$.
It remains to note that
$$
S_{A M N D}=S_{M B C N} \Longrightarrow\left(x^{2}-a^{2}\right) S=\left(b^{2}-x^{2}\right) S \Longrightarrow 2 x^{2}=a^{2}+b^{2}
$$
From which $x=\sqrt{\left(a^{2}+b^{2}\right) / 2}=15$.
Answer: 15. (C)
Fig. 1:
Fig. 2:
 | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.2. The segment connecting the lateral sides of the trapezoid and parallel to its bases, which are 7 and 17, divides the trapezoid into two parts of equal area. Find the length of this segment. | Solution. $c=\sqrt{\left(a^{2}+b^{2}\right) / 2}=\sqrt{(49+289) / 2}=13$.
Answer: 13. (E)
$$
\begin{array}{|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 11 & \mathbf{B} & 11.5 & \mathbf{C} & 12 & \mathbf{D} \\
12.5 & \mathbf{E} & 13 & \mathbf{F} \\
\hline
\end{array}
$$ | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.3. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{2} 36-\log _{3} 16\right)-\log _{2} 9-4 \log _{3} 8
$$
do not exceed 11. | Solution. Let $a=\log _{2} 3$. Then the condition of the problem will turn into the inequality
$$
x^{2}+2\left(a-\frac{2}{a}+1\right) x-\left(2 a+\frac{12}{a}+11\right) \leqslant 0
$$
Considering that $a \in\left(\frac{3}{2}, 2\right)$, we get $x \in\left[-2 a-3, \frac{4}{a}+1\right]$. Since $-7<-2 a-3<-6,3<\frac{4}{a}+1<4$, the integer solutions will be the numbers $-6,-5,-4,-3,-2,-1,0,1,2,3$.
Answer: $-15 .(\mathrm{B})$
$$
\begin{array}{|l|l|l|l|l|l|l|l|}
\hline \mathbf{A}-20 & \mathbf{B}-15 & \mathbf{C} & -10 & \mathbf{D} & -5 & \mathbf{E} 0 & \mathbf{F} \\
\hline
\end{array}
$$ | -15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. Three pirates, Joe, Bill, and Tom, found a treasure containing 70 identical gold coins, and they want to divide them so that each of them gets at least 10 coins. How many ways are there to do this? | Solution. Let the treasure consist of $n=70$ coins and each pirate should receive no less than $k=10$ coins.
Give each pirate $k-1$ coins, and lay out the remaining $n-3 k+3$ coins in a row. To divide the remaining coins among the pirates, it is sufficient to place two dividers in the $n-3 k+2$ spaces between the coins. Thus, Joe will receive the coins to the left of the first divider, Bill the coins between the two dividers, and Tom the coins to the right of the second divider. The number of ways to place these two dividers is $C_{n-3 k+2}^{2}=\frac{(n-3 k+2)(n-3 k+1)}{2}$.
If $n=70, k=10$, then we get $C_{42}^{2}=861$ ways.
Answer: 861. | 861 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.2. Three pirates, Joe, Bill, and Tom, found a treasure containing 80 identical gold coins, and they want to divide them so that each of them gets at least 15 coins. How many ways are there to do this? | Solution. Since $n=80, k=15$, it results in $C_{37}^{2}=666$ ways.
Answer: 666. | 666 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.3. Three pirates, Joe, Bill, and Tom, found a treasure containing 100 identical gold coins, and they want to divide them so that each of them gets at least 25 coins. How many ways are there to do this? | Solution. Since $n=100, k=25$, it results in $C_{27}^{2}=351$ ways.
Answer: 351. | 351 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.4. Three pirates, Joe, Bill, and Tom, found a treasure containing 110 identical gold coins, and they want to divide them so that each of them gets at least 30 coins. How many ways are there to do this? | Solution. Since $n=110, k=30$, we get $C_{22}^{2}=231$ ways.
Answer: 231.
Solution. Let $A=\underbrace{11 \ldots 1}_{1007}$. Then
$$
\sqrt{\underbrace{111 \ldots 11}_{2014}-\underbrace{22 \ldots 2}_{1007}}=\sqrt{A \cdot 10^{1007}+A-2 A}=\sqrt{A(9 A+1)-A}=3 A=\underbrace{33 \ldots 3}_{1007} .
$$
Answer: 3021
Solution. The number is $\underbrace{66 \ldots 6}_{1007}$.
Answer: 6042. | 231 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. In the triangular pyramid $S A B C$, the edges $S B, A B$ are perpendicular and $\angle A B C=120^{\circ}$. Point $D$ on edge $A C$ is such that segment $S D$ is perpendicular to at least two medians of triangle $A B C$ and $C D=A B=44 \sqrt[3]{4}$. Find $A D$ (if the answer is not an integer, round it to the nearest hundredth). | Solution. Since segment $SD$ is perpendicular to two medians of triangle $ABC$, it is perpendicular to the plane $(ABC)$ (see Fig. 3). By the theorem of three perpendiculars, it follows that $DB \perp AB$.
Fig. 3:
Fig. 4:
Let $\alpha = \angle BAC, x = AD$. Then, applying the Law of Sines in triangle $ABC$ (see Fig. 4), we have:
$$
\begin{aligned}
\frac{x \cos \alpha}{\sin (\pi / 3 - \alpha)} & = \frac{x + x \cos \alpha}{\sin (2 \pi / 3)} \Longleftrightarrow \sin \alpha = \frac{\sqrt{3} \cos^2 \alpha}{1 + \cos \alpha} \Longleftrightarrow (1 - \cos^2 \alpha)(1 + \cos \alpha)^2 = 3 \cos^4 \alpha \Longleftrightarrow \\
\Longleftrightarrow 4 \cos^4 \alpha + 2 \cos^3 \alpha - 2 \cos \alpha - 1 = 0 & \Longleftrightarrow (\cos 2\alpha + 1)(2 \cos^3 \alpha - 1) = 0
\end{aligned}
$$
Since $\alpha < \pi / 2$, from the last equation: $\cos \alpha = 1 / \sqrt[3]{2}$. Therefore, $AD = AB / \cos \alpha = 88$.
## Answer: 88. | 88 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. For the function $f(x)=2013-8 x^{3}+12 x^{2}-14 x-a-\sin 2 \pi x$ find the number of integer values of $a$, for each of which the equation
$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x-1
$$
on the interval $[50 ; 51]$ has a unique solution. | Solution. Since
$$
2013-8 x^{3}+12 x^{2}-14 x-a-\sin 2 \pi x=2008-(2 x-1)^{3}-4(2 x-1)-a+\sin \pi(2 x-1)
$$
then after the substitution of the variable $t=2 x-1$, we get a new problem: "For the function $F(t)=2008-t^{3}-4 t- a+\sin \pi t$, find the number of integer values of $a$, for each of which the equation
$$
\underbrace{F(F(\ldots F}_{2013 \text { times }}(t) \ldots))=t
$$
on the interval $[99 ; 101]$ has a unique solution."
The function $F(t)=2008-t^{3}-4 t-a+\sin \pi t$ is monotonically decreasing over the entire number line (which can be verified either by calculating $F^{\prime}(t)$ or by directly using the definition of a monotonically decreasing function). Therefore,
$$
\underbrace{F(F(\ldots F}_{2013 \text { times }}(t) \ldots))=t \Longleftrightarrow F(t)=t .
$$
The function $g(t)=t-F(t)$ is monotonically increasing. Consequently, the equation $g(t)=0$ has a unique solution on $[99 ; 101]$ if and only if
$$
\left\{\begin{array} { l }
{ g ( 1 0 1 ) \geqslant 0 , } \\
{ g ( 9 9 ) \leqslant 0 , }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
101-2008+101^{3}+404+a \geqslant 0 \\
99-2008+99^{3}+396+a \leqslant 0
\end{array}\right.\right.
$$
From the obtained two-sided estimates for $a$, it follows that the number of integer values of $a$ is
$$
g(101)-g(99)+1=2+101^{3}-99^{3}+404-396+1=60013
$$
## Answer 60013 . | 60013 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.2 For the function $f(x)=2013-a+12 x^{2}-\cos 2 \pi x-8 x^{3}-16 x$ find the number of integer values of $a$, for each of which the equation
$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x-1
$$
on the interval $[50 ; 51]$ has a unique solution. | Solution. After substituting $t=2 x-1$, we obtain a new problem for the function $F(t)=2007-t^{3}-5 t-a+\cos \pi t$. The number of integer values of $a$ is
$$
g(101)-g(99)+1=2+101^{3}-99^{3}+505-495+1=60015, \quad \text { where } g(t)=t-F(t) .
$$
## Answer 60015 . | 60015 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3 For the function $f(x)=2013+\sin 2 \pi x-8 x^{3}-12 x^{2}-18 x-a$, find the number of integer values of $a$ for each of which the equation
$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x+1
$$
has a unique solution on the interval $[49,50]$. | Solution. After substituting $t=2 x+1$, we obtain a new problem for the function $F(t)=2020-t^{3}-6 t-a-\sin \pi t$. The number of integer values of $a$ is
$$
g(101)-g(99)+1=2+101^{3}-99^{3}+606-594+1=60017, \quad \text { where } g(t)=t-F(t)
$$
Answer 60017. | 60017 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4 For the function $f(x)=2013-a+\cos 2 \pi x-12 x^{2}-8 x^{3}-20 x$ find the number of integer values of $a$, for each of which the equation
$$
\underbrace{f(f(\ldots f}_{2013 \text { times }}(x) \ldots))=2 x+1
$$
has a unique solution on the interval $[49,50]$. | Solution. After substituting $t=2 x+1$, we obtain a new problem for the function $F(t)=2021-t^{3}-7 t-a-\cos \pi t$. The number of integer values of $a$ is
$$
g(101)-g(99)+1=2+101^{3}-99^{3}+707-693+1=60019, \quad \text { where } g(t)=t-F(t)
$$
Answer 60019. | 60019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. An apple, a pear, an orange, and a banana were placed in four boxes (one fruit per box). Inscriptions were made on the boxes:
On the 1st: Here lies an orange.
On the 2nd: Here lies a pear.
On the 3rd: If in the first box lies a banana, then here lies an apple or a pear.
On the 4th: Here lies an apple.
It is known that none of the inscriptions correspond to reality.
Determine which fruit lies in which box. In your answer, write down sequentially, without spaces, the numbers of the boxes in which the apple, pear, orange, and banana lie, respectively (you should get a 4-digit number). | Answer: 2431
Solution: The inscription on the 3rd box is incorrect, so in the first box lies a banana, and in the third - not an apple and not a pear, therefore, an orange. From the inscription on the 4th box, it follows that there is no apple there, so since the banana is in the 1st, and the orange is in the 2nd, then in the 4th lies a pear. The apple remains, which lies in the second box. | 2431 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Beginner millionaire Bill buys a bouquet of 7 roses for $20 for the entire bouquet. Then he can sell a bouquet of 5 roses for $20 per bouquet. How many bouquets does he need to buy to earn a difference of $1000? | Answer: 125
Solution. Let's call "operation" the purchase of 5 bouquets (= 35 roses) and the subsequent sale of 7 bouquets (= 35 roses). The purchase cost is $5 \cdot 20=\$ 100$, and the selling price is $7 \cdot 20=\$ 140$. The profit from one operation is $\$ 40$.
Since $\frac{1000}{40}=25$, 25 such operations are needed. Therefore, Bill needs to buy $5 \cdot 25=125$ bouquets. | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. Find a natural number $N(N>1)$, if the numbers 1743, 2019, and 3008 give the same remainder when divided by $N$. | Answer: 23.
Solution. From the condition, it follows that the numbers $2019-1743=276$ and $3008-2019=989$ are divisible by $N$. Since $276=2^{2} \cdot 3 \cdot 23$, and $989=23 \cdot 43$, then $N=23$. | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 4. Find the smallest natural number $n$ such that $n^{2}$ and $(n+1)^{2}$ contain the digit 7. | Answer: 26.
Solution. There are no squares ending in the digit 7. There are no two-digit squares starting with 7. Therefore, $n \geq 10$. The first square containing 7 is $576=24^{2}$. Since $25^{2}=625,26^{2}=676,27^{2}=729$, the answer is $n=26$.
Problem 4a. Find the smallest natural number $n$ such that $n^{2}$ and $(n+1)^{2}$ contain the digit 7, but $(n+2)^{2}$ does not.
Answer: 27.
Solution. There are no squares ending in the digit 7. There are no two-digit squares starting with 7. Therefore, $n \geq 10$. The first square containing 7 is $576=24^{2}$. Since $25^{2}=625,26^{2}=676,27^{2}=729,28^{2}=784,29^{2}=841$, the answer is $n=27$. | 26 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. A square with an integer side length was cut into 2020 squares. It is known that the areas of 2019 squares are 1, and the area of the 2020th square is not equal to 1. Find all possible values that the area of the 2020th square can take. In your answer, provide the smallest of the obtained area values. | Answer: 112225.
 | 112225 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Master Li Si Qing makes fans. Each fan consists of 6 sectors, painted on both sides in red and blue (see fig.). Moreover, if one side of a sector is painted red, the opposite side is painted blue and vice versa. Any two fans made by the master differ in coloring (if one coloring can be transformed into another by flipping the fan, they are considered the same). What is the maximum number of fans the master can make? | Answer: 36.
## Solution:
The coloring of one side can be chosen in $2^{6}=64$ ways. It uniquely determines the coloring of the opposite side. However, some fans - those that transform into each other when flipped, we have counted twice. To find their number, let's see how many fans transform into themselves when flipped. There are 8 such fans in total. So, we need to divide the remaining 64 $8=56$ by two - which gives us 28. We need to add those 8 that transform into themselves. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. In a convex quadrilateral $A B C D$, side $A B$ is equal to diagonal $B D, \angle A=65^{\circ}$, $\angle B=80^{\circ}, \angle C=75^{\circ}$. What is $\angle C A D$ (in degrees $) ?$ | Answer: 15.
Solution. Since triangle $ABD$ is isosceles, then $\angle BDA = \angle BAD = 65^{\circ}$. Therefore, $\angle DBA = 180^{\circ} - 130^{\circ} = 50^{\circ}$. Hence, $\angle CBD = 80^{\circ} - 50^{\circ} = 30^{\circ}$, $\angle CDB = 180^{\circ} - 75^{\circ} - 30^{\circ} = 75^{\circ}$. This means that triangle $DBC$ is isosceles, so $BC = BD = AB$. Therefore, $\angle BCA = \angle BAC = \frac{1}{2}(180^{\circ} - 80^{\circ}) = 50^{\circ}$. Hence, $\angle CAD = 65^{\circ} - 50^{\circ} = 15^{\circ}$. | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.
In the test, there are 4 sections, each containing the same number of questions. Andrey answered 20 questions correctly. The percentage of his correct answers was more than 60 but less than 70. How many questions were in the test? | Answer: 32.
Solution. According to the condition $\frac{60}{100}<\frac{20}{x}<\frac{70}{100}$, hence $28 \frac{4}{7}=\frac{200}{7}<x<\frac{100}{3}=33 \frac{1}{3}$, that is $29 \leq x \leq 33$. From the first condition of the problem, it follows that the number of questions must be divisible by 4. | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The test consists of 5 sections, each containing the same number of questions. Pavel answered 32 questions correctly. The percentage of his correct answers turned out to be more than 70 but less than 77. How many questions were in the test?
ANSWER: 45. | Solution: from the condition $0.7<32 / x<0.77$ it follows that $41<x<46$, but $x$ is a multiple of 5, so $x=45$. | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest natural N such that N+2 is divisible (without remainder) by 2, N+3 by 3, ..., N+10 by 10.
ANSWER: 2520. | Solution: Note that $N$ must be divisible by $2,3,4, \ldots, 10$, therefore, $N=$ LCM $(2,3,4, . ., 10)=2^{3} \times 3^{2} \times 5 \times 7=2520$. | 2520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petrov lists the odd numbers: $1,3,5, \ldots, 2013$, while Vasechkin lists the even numbers: $2,4, \ldots, 2012$. Each of them calculated the sum of all digits of all their numbers and told the excellent student Masha. Masha subtracted Vasechkin's result from Petrov's result. What did she get? | Answer: 1007.
Solution: Let's break down the numbers of Petrov and Vasechkin into pairs as follows: $(2,3),(4,5), \ldots,(98,99),(100,101), \ldots$ (2012,2013), with 1 left unpaired for Petrov. Notice that in each pair, the sum of the digits of the second number is 1 greater than that of the first (since they differ only in the last digit). There will be a total of $\frac{2012}{2}=1006$ such pairs. Therefore, the difference in the sums of the digits will be 1006, and considering the 1 left unpaired for Petrov, it will be -1007. | 1007 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. There are no fewer than 150 boys studying at the school, and there are $15 \%$ more girls than boys. When the boys went on a trip, 6 buses were needed, and each bus had the same number of students. How many people in total study at the school, given that the total number of students is no more than 400? | Answer: 387.
Solution: The number of boys is a multiple of 6, let's denote it as $6n$, obviously, $n \geqslant 25$. Then the number of girls is $6n \times 1.15 = 6.9n$. The total number of students is $12.9n \leqslant 400$, so $n \leqslant 31$. Considering that $6.9n$ must be an integer, and therefore $n$ must be a multiple of 10, we get that $n=30$, i.e., a total of 387 students. | 387 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Malvina and Buratino play according to the following rules: Malvina writes six different numbers in a row on the board, and Buratino comes up with his own four numbers $x_{1}, x_{2}, x_{3}, x_{4}$ and writes under each of Malvina's numbers one of the sums $x_{1}+x_{2}, x_{1}+x_{3}, x_{1}+x_{4}$, $x_{2}+x_{3}, x_{2}+x_{4}, x_{3}+x_{4}$ (each one exactly once). For each sum that equals the number above it, Buratino gets 3 apples, and for each sum that is greater, he gets 1 apple. What is the maximum number of apples that Buratino can guarantee to get? | # Answer: 14.
## Solution.
Solution. Let Malvina write the numbers $a_{1}>a_{2}>a_{3}>a_{4}>a_{5}>a_{6}$. If Buratino comes up with the numbers $x_{1}=\left(a_{1}+a_{2}-a_{3}\right) / 2, x_{2}=\left(a_{1}+a_{3}-a_{2}\right) / 2, x_{3}=\left(a_{2}+a_{3}-a_{1}\right) / 2, x_{4}=a_{4}-x_{3}$, then by writing $x_{1}+x_{2}=a_{1}$ under $a_{1}, x_{1}+x_{3}=a_{2}$ under $a_{2}, x_{2}+x_{3}=a_{3}$ under $a_{1}, x_{3}+x_{4}=a_{4}$ under $a_{4}$, $x_{1}+x_{4} \geqslant a_{4} \geqslant a_{5}$ under $a_{5}$ and $x_{2}+x_{4} \geqslant a_{4} \geqslant a_{6}$ under $a_{6}$, he will get 14 apples.
To get more than 14 apples, Buratino must ensure at least 5 equalities, which is not always possible: among any 5 sums composed by Buratino, there are such 4 that the sum of two of them equals the sum of the other two. Therefore, if Malvina writes such 6 numbers that the sum of no two of them equals the sum of any other two (for example, $1,10,100,1000,10000,100000$), then Buratino will not be able to get 15 or more apples. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. For what values of $a$ does the equation
$$
[x]^{2}+2012 x+a=0
$$
(where $[x]$ is the integer part of $x$, i.e., the greatest integer not exceeding $x$) have the maximum number of solutions? What is this number? | Answer: 89 solutions at $1006^{2}-20120
\end{array}\right.
$
has the maximum number of integer solutions.
The solution to the first inequality is the interval $\left[-1006-\sqrt{1006^{2}-a} ;-1006+\sqrt{1006^{2}-a}\right]$ under the condition $a \leqslant 1006^{2}$.
The solution to the second inequality for $a>1006^{2}-2012$ is the entire real line, and for $a \leqslant 1006^{2}-2012-$ the set is
$\left(-\infty ;-1006-\sqrt{1006^{2}-2012-a}\right) \cup\left(-1006-\sqrt{1006^{2}-2012-a} ;+\infty\right)$.
Thus, the solutions to the system form:
for $1006^{2}-20121006^{2}$ - 2012. The parameter can be represented as $a=1006^{2}-2012+b, b>0$. The solutions form the interval $[-1006-\sqrt{2012-b} ;-1006+\sqrt{2012-b}]$ and the smaller $b$ is, the larger this interval is.
Since $44^{2}=19360)$. Therefore, for $1006^{2}-2012<a \leqslant$ $1006^{2}-1936$ the number of integer solutions is equal to the number of integer points on the interval $[-1006-44 ;-1006+44]$, i.e., 89.
For $a=1006^{2}-2012$, the point $k=-1006$ disappears from the solution. As $a$ further decreases, new integer points $-1006 \pm 45$ will first appear in the set
$\left[-1006-\sqrt{1006^{2}-a} ;-1006-\sqrt{1006^{2}-2012-a}\right) \cup\left(-1006-\sqrt{1006^{2}-2012-a} ;-1006+\sqrt{1006^{2}-a}\right]$ at $a=1006^{2}-45^{2}=1006^{2}-2025$. However, at this point, integer points lying on the interval $\left[-1006-\sqrt{1006^{2}-2012-a} ;-1006+\sqrt{1006^{2}-2012-a}\right]=$ $[-1006-\sqrt{13} ;-1006+\sqrt{13}]$, i.e., 7 points: $-1006 ;-1006 \pm 1 ;-1006 \pm 2 ;-1006 \pm 3$, must be excluded from the solution. Since further decreasing $a$ reduces the total length of the solution set, the number of integer points will also decrease. | 89 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Determine how many zeros the number $N!$ ends with! | Solution. Let $N=2014$. Among the first 2014 natural numbers, 402 numbers are divisible by 5, of which 80 numbers are divisible by 25. Among these 80 numbers, 16 numbers are divisible by 125, of which 3 numbers are divisible by 625. There are more even numbers among the first 2014 natural numbers than those divisible by 5. Therefore, the number of zeros at the end of the number 2014! is determined by the number of "fives" as divisors in this number. In total: $402+80+16+3=501$ zeros.
## Answer: | 501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. By how much is the sum of the squares of the first hundred even numbers greater than the sum of the squares of the first hundred odd numbers | Answer: 20100.
Solution: Group the terms as $\left(2^{2}-1^{2}\right)+\left(4^{2}-3^{2}\right)+\cdots+\left(200^{2}-199^{2}\right)=$ $(2-1) \cdot(2+1)+(4-3) \cdot(4+3)+\ldots+(200-199) \cdot(200+199)=1+2+\cdots+$ $199+200$. Divide the terms into pairs that give the same sum: $1+200=2+199=\ldots=100+101=201$. There will be exactly 100 such pairs, so the sum is 20100. | 20100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petrov and Vasechkin were repairing a fence. Each had to nail a certain number of boards (the same amount). Petrov nailed two nails into some boards, and three nails into the rest. Vasechkin nailed three nails into some boards, and five nails into the rest. Find out how many boards each of them nailed, if it is known that Petrov nailed 87 nails, and Vasechkin nailed 94 nails. | Answer: 30.
Solution: If Petrov had nailed 2 nails into each board, he would have nailed 43 boards and had one extra nail. If he had nailed 3 nails into each board, he would have nailed 29 boards. Therefore, the desired number lies between 29 and 43 (inclusive). Similarly, if Vasechkin had nailed 3 nails into each board, he would have nailed 31 boards and had 1 extra nail, and if he had nailed 5 nails into each board, he would have nailed 18 boards and had 4 extra nails. This means there were 29, 30, or 31 boards. Note that Vasechkin nailed an odd number of nails into each board, so the number of boards must be even - 30. | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers? | Answer: 18.
Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces.
Thus, the sum cannot be less than 18. We will show that 18 can be achieved - place the number 1 on the top and bottom faces of the cube, 3 on the right and left faces, and 5 on the front and back faces. | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square. | Answer: 183
Solution: Let the required number be a, then $1000a + a + 1 = n^2$. We can rewrite this as: $1001a = (n - 1)(n + 1)$. Factorize $1001 = 7 \times 11 \times 13$, meaning the product (n-1)(n+1) must be divisible by 7, 11, and 13. Additionally, for the square to be a six-digit number, $n$ must be in the interval $[317; 999]$.
Consider the following cases:
a) $n-1$ is divisible by $143, n+1$ is divisible by 7, then we find $n=573$;
b) $n-1$ is divisible by $7, n+1$ is divisible by 143, then $n=428$;
c) $n-1$ is divisible by $77, n+1$ is divisible by 13, then $n=155$ - does not fit;
d) $n-1$ is divisible by $13, n+1$ is divisible by 77, then $n=846$;
e) $n-1$ is divisible by 91, $n+1$ is divisible by 11, then $n=274$ - does not fit;
f) $n-1$ is divisible by $11, n+1$ is divisible by 91, then $n=727$.
The smallest $n=428, n^2=428^2=183184$. | 183 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Petya formed all possible natural numbers that can be formed from the digits $2,0,1$, 8 (each digit can be used no more than once). Find their sum. Answer: 78331 | Solution: First, consider the units place. Each of the digits 1, 2, 8 appears once in this place for single-digit numbers, twice for two-digit numbers, four times for three-digit numbers, and four times for four-digit numbers - a total of 11 times.
In the tens place, each of them appears 3 times for two-digit numbers, 4 times for three-digit numbers, and 4 times for four-digit numbers - also 11 times.
In the hundreds place, each appears 6 times in three-digit numbers and 4 times in single-digit numbers. In the thousands place, each appears 6 times.
In total, we get $11 \times 11 + 11 \times 11 \times 10 + 11 \times 10 \times 100 + 11 \times 1000 \times 6 = 78331$. | 78331 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (5-7,8,9) There are 2014 boxes on the table, some of which contain candies, while the others are empty.
On the first box, it is written: “All boxes are empty.”
On the second - “At least 2013 boxes are empty.”
On the third - “At least 2012 boxes are empty.”
...
On the 2014th - “At least one box is empty.”
It is known that the inscriptions on the empty boxes are false, while those on the boxes with candies are true. Determine how many boxes contain candies. | Answer: 1007.
Solution: Suppose that $N$ boxes are empty, then $2014-N$ boxes contain candies. Note that on the box with number $k$, it is written that there are at least $2015-k$ empty boxes. Therefore, the inscriptions on the boxes with numbers $1,2, \ldots, 2014-N$ are false. Consequently, $N=2014-N$, from which $N=1007$. | 1007 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. $(5-7,8)$ The phone PIN code consists of 4 digits (and can start with zero, for example, 0951). Petya calls "lucky" those PIN codes where the sum of the outer digits equals the sum of the middle digits, for example $1357: 1+7=3+5$. In his phone, he uses only "lucky" PIN codes. Petya says that even if he forgets one digit (but remembers its position), he can easily restore it. And if he forgets two digits (but remembers their positions), he will have to try only a small number of PIN codes.
a) How many PIN codes will Petya have to try in the worst case?
b) How many "lucky" PIN codes exist in total? | Answer: a) $10 ;$ b) 670 .
Solution: a) Obviously, in the worst case, Petya tries all possible values for one digit and reconstructs the PIN code from them. Thus, no more than 10 combinations need to be tried. Using the example of the PIN code 0099, we can see that if you forget the two middle digits, you need to try exactly 10 combinations. b) The sum of a pair of digits can take values from 0 to 18. It is easy to notice that the sum equal to 0 can be obtained in only one way, equal to 1 in two ways, ..., equal to 10 in 11 ways, ..., equal to 18 in one way. In total, the number of lucky PIN codes is $N=1^{2}+2^{2}+\ldots+10^{2}+11^{2}+10^{2}+\ldots+1^{2}=670$. | 670 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (5-7,8) There are 10 segments, the length of each of which is expressed as an integer not exceeding some $N$. a) Let $N=100$. Provide an example of a set of 10 segments such that no three of them can form a triangle. b) Find the maximum $N$ for which it can be guaranteed that there will be three segments that can form a triangle. | Answer: a) $1,1,2,3,5,8,13,21,34,55 ;$ b) $N=54$.
Solution: a) If each new segment is chosen to be equal to the sum of the two largest of the remaining ones, then it is impossible to form a triangle with its participation.
b) From the previous point, it is clear that for $N=55$ such a sequence can be constructed.
We will show that for $N=54$ it is already impossible to construct such a sequence. Let's order the lengths of the segments in ascending order: $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{10} \leqslant 54$. If a triangle cannot be formed from segments of lengths $a_{i}, a_{i+1}, a_{i+2}$, this means that the triangle inequality is violated, i.e., $a_{i+2} \geqslant a_{i}+a_{i+1}$. Thus, since it is obvious that $a_{1}, a_{2} \geqslant 1$, by sequentially applying this inequality for $i=1,2, \ldots, 8$, we get: $a_{3} \geqslant a_{1}+a_{2} \geqslant 2, a_{4} \geqslant a_{2}+a_{3} \geqslant 3, \ldots a_{10} \geqslant a_{8}+a_{9} \geqslant 55$, which leads to a contradiction. | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. $(8,9)$ What is the maximum possible area of quadrilateral $A B C D$, the sides of which are $A B=1, B C=8, C D=7$ and $D A=4$? | Answer: 18.
Solution: Note that $1^{2}+8^{2}=7^{2}+4^{2}=65$. With fixed lengths of $A B$ and $B C$, the area of $\triangle A B C$ will be maximized if $\angle A B C=$ $90^{\circ}$. In this case, $A C=\sqrt{65}$, and thus $\angle B C D=$ $90^{\circ}$ as well, and the area of $\triangle B C D$ is also maximized. | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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