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4.4. In triangle $A B C$, a point $D$ is marked on side $B C$ such that $B D: D C=1: 3$, and on side $A C$ - points $E$ and $K$, with point $E$ lying between points $A$ and $K$. Segment $A D$ intersects segments $B E$ and $B K$ at points $M$ and $N$ respectively, with $B M: M E=7: 5$, $B N: N K=2: 3$. Find the ratio $M... | Answer: $11: 45$. Solution. We use Menelaus' theorem 1)
\[
\begin{aligned}
& \frac{1}{3} \frac{C A}{A K} \frac{3}{2}=1 \\
\frac{C A}{A K}= & 2 \Rightarrow \frac{C K}{K A}=1
\end{aligned}
\]
\[
\frac{A N}{N D}=\frac{4}{1}
\]
4)
\[
\begin{gathered}
\frac{A M}{M D} \frac{D B}{B C} \frac{C E}{E A}=1 \\
\frac{A M}{M D}... | \frac{11}{45} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,861 |
5.1. For the sequence $\left\{a_{n}\right\}$, it is known that $a_{1}=1.5$ and $a_{n}=\frac{1}{n^{2}-1}$ for $n \in \mathbb{N}, n>1$. Are there such values of $n$ for which the sum of the first $n$ terms of this sequence differs from 2.25 by less than 0.01? If yes, find the smallest one. | Answer: yes, $n=100$.
Solution. The general formula for the terms of the sequence (except the first) can be written as $(n \geqslant 2)$:
$$
a_{n}=\frac{1}{n^{2}-1}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)
$$
As a result, the sum of the first $n$ terms of the sequence, except the first, takes the form:
$... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,862 |
6.1. Find all solutions to the inequality
$$
\sin ^{2018} x+\cos ^{-2019} x \geqslant \cos ^{2018} x+\sin ^{-2019} x
$$
belonging to the interval $\left[-\frac{\pi}{4} ; \frac{7 \pi}{4}\right]$. | Answer: $\left[-\frac{\pi}{4} ; 0\right) \cup\left[\frac{\pi}{4} ; \frac{\pi}{2}\right) \cup\left(\pi ; \frac{5 \pi}{4}\right] \cup\left(\frac{3 \pi}{2} ; \frac{7 \pi}{4}\right]$.
Solution. The function $f(t)=t^{2018}-t^{-2019}$ is increasing on each of the half-intervals $[-1,0)$ and $(0,1]$. Indeed, the derivative $... | [-\frac{\pi}{4};0)\cup[\frac{\pi}{4};\frac{\pi}{2})\cup(\pi;\frac{5\pi}{4}]\cup(\frac{3\pi}{2};\frac{7\pi}{4}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,863 |
6.2. Find all solutions of the inequality
$$
\cos ^{2018} x+\sin ^{-2019} x \geqslant \sin ^{2018} x+\cos ^{-2019} x
$$
belonging to the interval $\left[-\frac{4 \pi}{3} ; \frac{2 \pi}{3}\right]$. | Answer: $\left[-\frac{4 \pi}{3} ;-\pi\right) \cup\left[-\frac{3 \pi}{4} ;-\frac{\pi}{2}\right) \cup\left(0 ; \frac{\pi}{4}\right] \cup\left(\frac{\pi}{2} ; \frac{2 \pi}{3}\right]$. | [-\frac{4\pi}{3};-\pi)\cup[-\frac{3\pi}{4};-\frac{\pi}{2})\cup(0;\frac{\pi}{4}]\cup(\frac{\pi}{2};\frac{2\pi}{3}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,864 |
6.3. Find all solutions of the inequality
$$
\sin ^{2018} x+\cos ^{-2019} x \leqslant \cos ^{2018} x+\sin ^{-2019} x
$$
belonging to the interval $\left[-\frac{5 \pi}{4} ; \frac{3 \pi}{4}\right]$. | Answer: $\left[-\frac{5 \pi}{4} ;-\pi\right) \cup\left[-\frac{3 \pi}{4} ;-\frac{\pi}{2}\right) \cup\left(0 ; \frac{\pi}{4}\right] \cup\left(\frac{\pi}{2} ; \frac{3 \pi}{4}\right]$. | [-\frac{5\pi}{4};-\pi)\cup[-\frac{3\pi}{4};-\frac{\pi}{2})\cup(0;\frac{\pi}{4}]\cup(\frac{\pi}{2};\frac{3\pi}{4}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,865 |
6.4. Find all solutions of the inequality
$$
\cos ^{2018} x+\sin ^{-2019} x \leqslant \sin ^{2018} x+\cos ^{-2019} x
$$
belonging to the interval $\left[-\frac{\pi}{3} ; \frac{5 \pi}{3}\right]$. | Answer: $\left[-\frac{\pi}{3} ; 0\right) \cup\left[\frac{\pi}{4} ; \frac{\pi}{2}\right) \cup\left(\pi ; \frac{5 \pi}{4}\right] \cup\left(\frac{3 \pi}{2} ; \frac{5 \pi}{3}\right]$ | [-\frac{\pi}{3};0)\cup[\frac{\pi}{4};\frac{\pi}{2})\cup(\pi;\frac{5\pi}{4}]\cup(\frac{3\pi}{2};\frac{5\pi}{3}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,866 |
7.1. How many values of the parameter $a$ exist for which the equation
$$
4 a^{2}+3 x \lg x+3 \lg ^{2} x=13 a \lg x+a x
$$
has a unique solution | Answer: 2.
Solution. The given equation is equivalent to the equation $(3 \lg x-a)(x+\lg x-4 a)=0$, from which we have the system
$$
\left[\begin{array}{l}
\lg x=\frac{a}{3} \\
x+\lg x=4 a
\end{array}\right.
$$
Each equation in this system has a unique positive solution for any $a$, since the continuous functions $f... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,867 |
8.1. The lateral edge of a regular pyramid is 2. Can its volume be equal to $3.25?$ | Answer: No.
Solution. The volume of a regular pyramid is less than the volume of a cone circumscribed around it. If we denote the lateral edge by $b$, and the angle between the generatrix of the cone and the base by $\alpha$, then the radius of the base of the cone is $b \cos \alpha$, and the height of the cone is $b ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,868 |
8.4. The volume of a regular pyramid is $8 \pi$. Can its lateral edge be equal to 3.96? | Answer: No.
LOMONOSOV - 2019. MATHEMATICS. Grading Criteria (10-11 grades)
| Problem № 1 = $\mathbf{1 5}$ points | Pluses-minuses | Points |
| :--- | :---: | :---: |
| Correct equation, but arithmetic errors | $\mp$ | 5 |
| Note. Graphical solution is considered a solution. | | |
| Problem № 2 = 15 points | Pluse... | No | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,869 |
1. In rectangle $A B D F$, on sides $B D=2$ and $D F=3$, points $C$ and $E$ are chosen respectively, such that triangle $A F E$ is equal to triangle $E D C$. Then, from rectangle $A B D F$, triangles $A B C$, $C D E$, and $A F E$ are cut off. Find the angles of the remaining triangle. | Answer: $45^{\circ}, 45^{\circ}, 90^{\circ}$. Solution. Since triangles $A F E$ and $E D C$ are equal, then $\angle F A E=\angle D E C$, $\angle F E A=\angle D C E, A E=E C$. At the same time, since $\angle F E A+\angle F A E=90^{\circ}$, then $\angle F E A+\angle D E C=90^{\circ}$. Therefore, $\angle A E C=90^{\circ}$... | 45,45,90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,870 |
2. Can non-negative integers be placed on the faces of two cubes so that when they are randomly thrown, the sum of the points showing can be equal to any divisor of the number 36? If this is possible, indicate the sum of all 12 numbers on the faces; if it is impossible - indicate 0. | Answer: The sum is 111. Solution. The numbers $1,2,3,4,5$ and 6 can be placed on the faces of one die, and $0,6,12,18,24,30$ on the faces of the other. Then the sum of the points rolled can be equal to any of the numbers from 1 to 36. The sum of the numbers on all faces is 111. This sum does not depend on how the dice ... | 111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,871 |
3. Find all three-digit numbers $\overline{L O M}$, composed of distinct digits $L, O$ and $M$, for which the following equality holds:
$$
\overline{L O M}=(L+O+M)^{2}+L+O+M
$$ | Answer: 156. Instructions. Let $x=L+O+M$. Then $\overline{L O M}=x(x+1)$. In this case, $x \geq 10$ (otherwise $x(x+1)<100)$ and $x \leq 24$ (the sum of digits does not exceed $9+8+7=24$). Therefore, $x \in[10 ; 24]$. From the relation $100 \cdot L+10 \cdot O+M=x^{2}+L+O+M$ it follows that $x^{2}=99 \cdot L+9 \cdot O$,... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,872 |
4. Pete was given a new electric jigsaw for his birthday, with a function to count the length of the cuts made. To try out the gift, Pete took a square piece of plywood with a side of $50 \mathrm{cm}$ and cut it into squares with sides of 10 cm and squares with sides of 20 cm. How many squares in total were obtained, i... | Answer: 16. Solution. Each protruding edge is part of the perimeter of two figures, in addition, the perimeter of the original square must be taken into account, from which we get that the total perimeter of the resulting small squares is $280 \cdot 2+200=760$. Now we can denote the number of squares through $x$ and $y... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,873 |
5. How many four-digit numbers exist that have the following properties: all digits of the number are even; the number is divisible by four, if the last digit is erased, the resulting three-digit number is not divisible by four? | Answer: 120. Solution. The last digit of the number must be divisible by 4 (that is, it can be 0, 4 or 8), and the second to last - not (2 or 6). In addition, the first digit is not zero. Therefore, we get $4 \cdot 5 \cdot 2 \cdot 3=120$ variants. | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,874 |
1.1. $[5-6.1$ (20 points)] In a certain family, dad works on a "2 out of 2" schedule (2 days working, 2 days off), mom works on a "1 out of 2" schedule (1 day working, 2 days off), and the children attend school on a five-day workweek (Monday to Friday). On Saturday, September 2, mom visited grandma in the village, and... | Answer: September 23 (Saturday).
Solution. According to the condition, September 2 and 3 were holidays for mom, so her working days later in September are the 4th, 7th, 10th, etc. On Sunday, September 3, it was a holiday for dad, but it is unknown whether he had a holiday the previous day. Let's consider 2 cases.
1) ... | September\23\(Saturday) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,875 |
2.1. [5-6.2 (20 points), 7-8.2 (15 points), 9.2 (15 points)] In how many ways can the word «РОТОР» be read from the letters in the diagram, if it is not allowed to return to the letters already passed, and readings that differ only in direction are considered the same?
| Р O Т O | дохо |
| :--- | :--- |
| O Т O P | О ... | Answer: 25.
Solution. The number of ways to read starting from the upper left letter "P" is $2^{4}=16$, as 4 steps are required to move through the letters, and each step is directed either to the right or down. If we strike out the considered letter "P", it remains to count the number of remaining ways in Fig. 1.
| ... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,876 |
3.1. [5-6.3 (20 points)] In three flasks, there is concentrated acid: 10 g in the first, 20 g in the second, and 30 g in the third. There is also a fourth flask with water. If some amount of water from the fourth flask is added to the first flask, and the remaining water is poured into the second flask, then in the fir... | Answer: $\frac{21}{200}=0.105$.
Solution. If the acid constitutes $\frac{1}{20}$ of the first flask, then it contains $10: \frac{1}{20}=200$ g of the solution, so there are 190 g of water in it. Similarly, for the second flask, we find that the total solution in it is $20: \frac{7}{30}=\frac{600}{7}$ g, and the water ... | \frac{21}{200}=0.105 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,877 |
5.1. $[5-6.4$ (а) - 10 points, б) - 20 points)] On a grid paper, a figure is drawn (see the picture). It is required to cut it into several parts and assemble a square from them (parts can be rotated, but not flipped). Is it possible to do this under the condition that a) there are no more than four parts; b) there are... | Answer: a) yes; b) yes.
Solution. Possible cutting options for parts a) and b) are shown in Fig. 5, a) and b).
Fig. 5, a)
 - 20 points, б) - 20 points)] Vovochka adds numbers in a column in the following way: he does not remember the tens, and under each pair of digits in the same place value, he writes their sum, even if it is a two-digit number. For example, for the sum $248+208$, he would get the value 4416.
a) In how m... | Answer: a) 244620 b) 1800.
Solution. a) Vovochka will get the correct answer if and only if the sum of the last and second-to-last digits also results in a digit, i.e., there is no carry-over. Let's count how many pairs of digits have a sum that does not exceed 9. If one digit is 0, the other can be any of 10 digits (... | 244620 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,882 |
9.1. $\left[7-8.1\right.$ (15 points)] Find all solutions to the numerical riddle AB $=$ B $^{V}$ (different letters correspond to different digits; on the left side, there is a two-digit number, not the product of digits A and B).
Answer: $32=2^{5}, 36=6^{2}, 64=4^{3}$. | Solution. If $\mathrm{B}=1$, then Б ${ }^{\mathrm{B}}=$ Б - is not a two-digit number. Let's list the two-digit numbers obtained if $\mathrm{B} \geqslant 2$:
1) $\mathrm{B}=2: 4^{2}=16,5^{2}=25,6^{2}=36,7^{2}=49,8^{2}=64,9^{2}=81$. In this case, the solution to the puzzle is only the equality $36=6^{2}$.
2) $\mathrm{B... | 32=2^{5},36=6^{2},64=4^{3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,884 |
9.2. Find all solutions to the numerical puzzle $\mathrm{A}^{\mathrm{B}}=\mathrm{BA}$ (different letters correspond to different digits; in the right part, a two-digit number is given, not the product of the digits B and A). | Answer: $2^{5}=32,6^{2}=36,4^{3}=64$. | 2^{5}=32,6^{2}=36,4^{3}=64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,885 |
11.1. [7-8.6 (20 points)] On the bisector of angle $B A C$ of triangle $A B C$, a point $M$ is marked, and on the extension of side $A B$ beyond point $A$, a point $N$ is marked such that $A C=A M=1$ and $\angle A N M=\angle C N M$. Find the length of segment $A N$. | Answer: 1.
Solution. Let's place point $K$ on the extension of segment $NC$ beyond point $C$. Point $M$ lies on the bisectors of angles $BAC$ and $ANC$, which means it is equidistant from lines $AC$ and $NC$, and therefore lies on the bisector of angle $ACK$. Then, considering the equality of angles $M$ and $C$ in the... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,887 |
13.1. [7-8.7 (20 points), 9.8 (15 points), 10.8 (20 points)] There is a rotating round table with 16 sectors, on which numbers $0,1,2, \ldots, 7,8,7,6, \ldots, 2,1$ are written in a circle. 16 players are sitting around the table, numbered in order. After each rotation of the table, each player receives as many points ... | Answer: 20.
Solution. Players No. 5 and No. 9 together scored $72+84=156=12 \cdot 13$ points. In one spin, they can together score no more than 12 points. Therefore, in each of the 13 spins, they together scored 12 points. Note that the 12 points they score can be one of the sums $8+4, 7+5$, $6+6, 5+7$ or $4+8$, when ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,889 |
14.1. $[9.1$ (15 points), 10.1 (15 points)] Calculate
$$
\sqrt{1+2 \sqrt{1+3 \sqrt{1+\ldots+2017 \sqrt{1+2018 \cdot 2020}}}}
$$
Answer: 3. | Solution. Since $2018 \cdot 2020=2019^{2}-1, 2017 \cdot 2019=2018^{2}-1, \ldots$, we get
$$
\begin{gathered}
\sqrt{1+2 \sqrt{1+3 \sqrt{1+\ldots+2017 \sqrt{1+2018 \cdot 2020}}}}= \\
=\sqrt{1+2 \sqrt{1+3 \sqrt{1+\ldots+2016 \sqrt{1+2017 \cdot 2019}}}}=\ldots=\sqrt{1+2 \cdot 4}=3
\end{gathered}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,890 |
15.1. $[9.3$ (15 points)] In trapezoid $ABCD$, diagonal $AC$ is equal to 1 and is also its height. Perpendiculars $AE$ and $CF$ are drawn from points $A$ and $C$ to sides $CD$ and $AB$ respectively. Find $AD$, if $AD=CF$ and $BC=CE$. | Answer: $\sqrt{\sqrt{2}-1}$.
Solution. Let $A D=C F=x, B C=C E=y$. Then, from the right triangles $A B C$, $A C E$ and $A C D$, by the Pythagorean theorem, we find $A B=\sqrt{1+y^{2}}, A E=\sqrt{1-y^{2}}, C D=\sqrt{1+x^{2}}$. The area of triangle $A B C$ is $S_{A B C}=\frac{1}{2} A C \cdot B C=\frac{1}{2} C F \cdot A ... | \sqrt{\sqrt{2}-1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,891 |
16.1. [9.4 (15 points), 10.4 (15 points)] On the graph of the function $y=x+\frac{1}{x}$, where $x>0$, find the point closest to the origin. | Answer: $\left(\frac{1}{\sqrt[4]{2}} ; \frac{1+\sqrt{2}}{\sqrt[4]{2}}\right)$
Solution. The square of the distance from the point $\left(x, x+\frac{1}{x}\right)$ to the origin is
$$
r^{2}(x)=x^{2}+\left(x+\frac{1}{x}\right)^{2}=2 x^{2}+2+\frac{1}{x^{2}}
$$
By the inequality between the arithmetic mean and the geomet... | (\frac{1}{\sqrt[4]{2}};\frac{1+\sqrt{2}}{\sqrt[4]{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,893 |
17.1. [9.6 (15 points), 10.3 (15 points)] Find the prime factorization of the smallest natural number that has exactly 2020 distinct natural divisors. | Answer: $2^{100} \cdot 3^{4} \cdot 5 \cdot 7$.
Solution. The number of $\tau(n)$ different natural divisors of a number $n$, which admits a factorization into prime factors
$$
n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}
$$
(here $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k} \geqslant ... | 2^{100}\cdot3^{4}\cdot5\cdot7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,894 |
18.1. $[10.2$ (15 points)] A wooden parallelepiped, all sides of which are expressed in whole centimeters, was painted red, and then sawn parallel to the faces into cubes with a side of 1 cm. It turned out that one third of the resulting cubes have at least one red face, while the remaining two thirds have all faces un... | Answer: 18 cm.
Solution. Let the length of the parallelepiped be $n$ cm, then the width is $n-2$, and the height is $n-4$. This gives $n(n-2)(n-4)$ cubes. The number of uncolored cubes will be $(n-2)(n-4)(n-6)$ (a layer one cube wide is "removed" from each side). This results in the equation $(n-2)(n-4)(n-7)=\frac{2}{... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,895 |
20.1. [10.7 (15 points)] The function $f$, defined on the set of integers, satisfies the following conditions:
1) $f(1)+1>0$
2) $f(x+y)-x f(y)-y f(x)=f(x) f(y)-x-y+x y$ for any $x, y \in \mathbb{Z}$;
3) $2 f(x)=f(x+1)-x+1$ for any $x \in \mathbb{Z}$.
Find $f(10)$. | Answer: 1014.
Solution. If $h(x)=f(x)+x$, then from condition 2) we get $h(x+y)=h(x) h(y)$. Then, for $x=y=0$, this equality takes the form $h(0)^{2}=h(0)$, i.e., $h(0)=0$ or $h(0)=1$. In the first case, $h(x) \equiv 0$, which is impossible due to condition 1). If $a=h(1)$, then $h(x)=a^{x}$ for any $x \in \mathbb{Z}$... | 1014 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,896 |
20.2. The function $g$, defined on the set of integers, satisfies the conditions:
1) $g(1)>1$
2) $g(x+y)+x g(y)+y g(x)=g(x) g(y)+x+y+x y$ for any $x, y \in \mathbb{Z}$;
3) $3 g(x)=g(x+1)+2 x-1$ for any $x \in \mathbb{Z}$.
Find $g(5)$. | Answer: 248.
Solution. Let $g(1)=a$, then from condition 3) we sequentially find $g(2)=3 a-1, g(3)=$ $9 a-6, g(4)=27 a-23, g(5)=81 a-76$. From condition 2), substituting $x=4$ and $y=1$, we obtain after simplifications the equation $a^{2}-5 a+4=0$, from which $a=1$ or $a=4$. The case $a=1$ contradicts condition 1). Th... | 248 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,897 |
# Problem 1.
B-1
Find $x$ :
$$
\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}}=\frac{16}{37}
$$ | Answer: $-0.25$
Solution. $\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}}=\frac{16}{37} \Leftrightarrow 2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}=\frac{37}{16} \Leftrightarrow \frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}=\frac{5}{16} \Leftrightarrow 3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}=\frac{1... | -0.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,899 |
# Task 2.
In a school test, there are 5 sections, each containing the same number of questions. Anton answered 20 questions correctly. The percentage of his correct answers was more than 60 but less than 70. How many questions were there in the test in total? | Answer: 30
Solution. According to the condition $\frac{60}{100}<\frac{20}{x}<\frac{70}{100}$, that is, $\frac{200}{7}<x<\frac{100}{3}$, so $29 \leqslant x \leqslant 33$. Since the number of questions must be divisible by 5, then $x=30$.
## B-2
In a school test, there are 4 sections, each containing the same number o... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,900 |
# Problem 3.
B-1
At the time when a lion cub, located 6 minutes away, set off for a drink, the second, having already quenched its thirst, headed back along the same path at 1.5 times the speed of the first. At the same time, a tortoise, located 32 minutes away, set off along the same path for a drink. After some tim... | Answer: 2.4
Solution. Let the entire path of the turtle be 1, and let $x$ be the speed of the first cub. Then the speed of the second cub is $1.5x$, and the speed of the turtle is $1/32$. The entire path to the watering hole for the first cub is $6x$. Therefore, the initial distance between him and the turtle was $6x-... | 2.4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,901 |
# Problem 4.
Given a square with a side length of 5 cm. On its four sides, the vertices of a second square are located; on the four sides of the second square, the vertices of a third square, and so on. For what smallest natural $n$ will the sum of the areas of the first $n$ squares be guaranteed to be greater than 49... | Answer: 6
Solution. A square inscribed in the given square has the smallest area if its vertices are the midpoints of the sides of the original square (this follows, for example, from the fact that the area of a square is half the square of its diagonal). In this case, the area of the inscribed square is half the area... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,902 |
# Task 5.
Around a round table, 1001 people are sitting, each of whom is either a knight (always tells the truth) or a liar (always lies). It turned out that next to each knight sits exactly one liar, and next to each liar there is a knight. What is the minimum number of knights that can sit at the table? | Answer: 502
Solution. From the condition, it follows that a knight cannot sit between two knights or two liars, and a liar cannot sit between two liars. Thus, when moving around the table, knights will be encountered in pairs, while liars will be encountered singly or in pairs. From this, it follows that each liar can... | 502 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,903 |
# Problem 6.
Find the smallest natural number $n$, for which the number $n+2018$ is divisible by 2020, and the number $n+2020$ is divisible by 2018. | Answer: 2034142
Solution. By the condition $n+2018=2020 m, n+2020=2018 k$, hence $1009 k-1010 m=1$. The solution to this Diophantine equation is: $k=-1+1010 p, m=-1+1009 p$. Therefore, $n+2018=$ $2020(-1+1009 p)=-2020+2020 \cdot 1009 p$, from which $n=-2018-2020+2020 \cdot 1009 p$. The smallest natural $n$ is $n=2020 ... | 2034142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,904 |
# Problem 7.
Inside a convex 13-gon, 200 points are placed such that no three of these 213 points (including the vertices of the polygon) are collinear. The polygon is divided into triangles, with the vertices of each triangle being any three of the 213 points. What is the maximum number of triangles that could result... | Answer: 411
Solution. If each of the points is a vertex of some triangle, then the total sum of the angles will be equal to $180^{\circ} \cdot(13-2)+360^{\circ} \cdot 200=180^{\circ} \cdot(11+400)=180^{\circ} \cdot 411$. This means 411 triangles are formed. If some points do not participate in the cutting, then the nu... | 411 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,905 |
# Problem 8.
Let $A(n)$ denote the greatest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9, A(64)=1$. Find the sum $A(111)+A(112)+\ldots+A(218)+A(219)$. | Answer: 12045
Solution. The largest odd divisors of any two of the given numbers cannot coincide, as numbers with the same largest odd divisors are either equal or differ by at least a factor of 2. Therefore, the largest odd divisors of the numbers $111, 112, \ldots, 218, 219$ are distinct. This means that the largest... | 12045 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,906 |
9.2. Two spheres touch the plane of triangle $A B C$ at points $B$ and $C$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 7, and the distance between their centers is 17. The center of a third sphere with radius 8 is at point $A$, and it touches each of the first two spheres e... | Answer. $2 \sqrt{15} \approx 7.75$. | 2\sqrt{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,907 |
1.1. (4 points) When going from the first to the third floor, Petya walks 36 steps. When going from the first floor to his own floor in the same building, Vasya walks 72 steps. On which floor does Vasya live? | Answer: 5.
Solution. From the first to the third floor, there are 36 steps - the same number as from the third to the fifth floor. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,908 |
2.1. (5 points) The ratio of the sides of a rectangle is $3: 4$, and its diagonal is 9. Find the smaller side of the rectangle. | Answer: 5.4.
Solution. If $5 x=9-$ diagonal of the rectangle, then the smaller side is $3 x=\frac{27}{5}=5.4$. | 5.4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,909 |
3.1. (13 points) Ani has blue, green, and red paints. She wants to paint a wooden cube so that after painting, the cube has two faces of each color. In how many different ways can she do this? Ways of painting that can be obtained by rotating the cube are considered the same. | Answer: 6.
Solution. Red opposite red, blue opposite blue, green opposite green is one way. Red opposite red, blue opposite green is one way and there are two similar ways. Opposite each color, there is a face of one of the two remaining colors - two ways. In total, there are 6 ways to paint. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,910 |
4.1. (13 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offer... | Answer: 199.
Solution. If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If th... | 199 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,911 |
5.1. (13 points) On the picture, the entrance and exit of the maze are marked with arrows. You can move through it such that on this picture you can only move right, down, or up (you cannot turn around). How many different ways are there to navigate this maze?
---
The translation maintains the original text's formatt... | Answer: 16.
Solution. Let's call a place in the maze a fork if from it you can move in two possible directions.
Moving up or to the right after entering, we arrive at one of the forks marked... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,912 |
6.1. (13 points) How many minutes after 17:00 will the angle between the hour and minute hands be exactly the same again?
Answer: $54 \frac{6}{11} \approx 54.55$. | Solution. At 17:00, the angle between the hour and minute hands is $\frac{5}{12} \cdot 360^{\circ}=150^{\circ}$. The next moment when the angle will be the same will occur within the next hour, after the minute hand has overtaken the hour hand. Let this happen after $x$ minutes. We set up the equation: $\left(\frac{x}{... | 54\frac{6}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,913 |
6.2. How many minutes before $16:00$ was the angle between the hour and minute hands exactly the same in the previous instance? | Answer: $21 \frac{9}{11} \approx 21.82$. | 21\frac{9}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,914 |
7.1. (13 points) Find $\frac{S_{1}}{S_{2}}$, where
$$
S_{1}=\frac{1}{2^{18}}+\frac{1}{2^{17}}-\frac{1}{2^{16}}+\ldots+\frac{1}{2^{3}}+\frac{1}{2^{2}}-\frac{1}{2}, \quad S_{2}=\frac{1}{2}+\frac{1}{2^{2}}-\frac{1}{2^{3}}+\ldots+\frac{1}{2^{16}}+\frac{1}{2^{17}}-\frac{1}{2^{18}}
$$
(in both sums, the signs of the terms ... | Answer: -0.2.
Solution. For the sum $S_{1}$ we have
$$
\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2} S_{1}-\frac{1}{2^{19}}\right)-\frac{1}{2^{19}}\right)+\frac{1}{2^{19}}=S_{1}-\frac{1}{2^{3}}-\frac{1}{2^{2}}+\frac{1}{2}
$$
Solving this equation for $S_{1}$, we get
$$
S_{1}=\frac{1-2^{18}}{7 \cdot 2^{18}}
$$
Simi... | -0.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,915 |
8.1. (13 points) Find the smallest natural number such that after multiplying it by 9, the result is a number written with the same digits but in some different order. | Answer: 1089.
Solution. Note that the number must start with one, otherwise multiplying by 9 would increase the number of digits. After multiplying 1 by 9, we get 9, so the original number must contain the digit 9. The number 19 does not work, so two-digit numbers do not work. Let's consider three-digit numbers. The s... | 1089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,916 |
9.1. (13 points) The surface of a round table is divided into 9 identical sectors, in which the numbers from 1 to 9 are written sequentially clockwise. Around the table sit 9 players with numbers \(1, 2, \ldots, 9\), going clockwise. The table can rotate around its axis in both directions, while the players remain in p... | Answer: 57.
Solution. If, as a result of spinning the table, one coin went to someone with numbers from 5 to 8, then player No. 8 received 5 fewer coins than player No. 9, and if one coin went to someone else, then No. 8 received 4 more coins than player No. 4. Let the number of spins where one coin went to someone wi... | 57 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,917 |
1. Several boys and girls gathered around a round table. It is known that exactly for 7 girls, the person sitting to their left is a girl, and for 12 - a boy. It is also known that for $75\%$ of the boys, the person sitting to their right is a girl. How many people are sitting at the table? | Answer: 35 people
Solution: From the condition, it is clear that there are exactly 19 girls.
Note that the number of girls who have boys sitting to their left is equal to the number of boys who have girls sitting to their right. Thus, $75\%$ of the boys equals 12, i.e., there are 16 boys in total sitting at the table... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,918 |
3. Fashionista Elizabeth has 100 different bracelets, and every day she wears three of them to school. Could it be that after some time, every pair of bracelets has been worn on Liz's wrist exactly once? | Answer: No.
Solution: Consider the first bracelet. It must be paired with each of the other 99 exactly once. Suppose Lisa wears it $n$ days. Then it will be paired with exactly $2 n$ bracelets, which cannot equal 99. | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,920 |
4. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Answer: 18108.
Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+... | 18108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,921 |
5. A mad painter runs through the cells of a $2012 \times 2013$ board, initially painted in black and white. At the very beginning, he runs into a corner cell. After the painter leaves a cell, the color of that cell changes. Can the painter always
run across the board and jump off from one of the border cells so that a... | Answer: Yes, always.
Solution: If the painter runs from a corner cell to an arbitrary white cell, then returns to the corner cell along the same route, the specified cell will change color, while all other cells will remain the same color. This way, the color of all cells except the corner cell can be changed. If the ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 5,922 |
6. Sasha and Maxim (out of boredom) wrote non-zero digits in the cells of a $100 \times 100$ table. After this, Sasha said that among the 100 numbers formed by the digits in each row, all are divisible by 9. To this, Maxim replied that among the 100 numbers formed by the digits in each column, exactly one is not divisi... | Solution: If we believe Sasha, then the sum of the digits in each row is divisible by 9, so the total sum of the digits in the table will also be divisible by 9. But if we believe Maksim, then the sum of the digits in all columns except one is divisible by 9, which means the sum of the digits in the table is not divisi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 5,923 |
7. Point $P$ lies inside triangle $A B C$. It is connected to all vertices of the triangle, and perpendiculars are dropped from it to the sides, forming 6 triangles. It turns out that 4 of them are equal. Does this always mean that the triangle is isosceles? | Answer: Not necessarily.
Solution: An example of such a triangle that is not isosceles. Here, $\triangle A B C$ is a right triangle, and point $P$ is obtained by the intersection of the angle bisector of $\angle B C A$ and the perpendicular bisector of side $A C$.
The chord $... | Point\M\should\be\the\foot\of\the\perpendicular\dropped\from\point\B | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,926 |
10. Solve the equation:
$$
\frac{8}{\{x\}}=\frac{9}{x}+\frac{10}{[x]}
$$
where $[x]$ is the greatest integer less than or equal to $x$, and $\{x\}=x-[x]$. | Answer: $\frac{3}{2}$.
Solution: Note that $x$ cannot be negative, because in such a case the right-hand side would be negative. Moreover, $x$ cannot be an integer, because the fractional part would be 0, and it cannot belong to the interval $[0,1)$, because the integer part would be 0. Let's consider the possible cas... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,927 |
5. The function $f(t)$ with domain $D(f)=[1,+\infty)$ satisfies the equation $f\left(\frac{t^{y}-4^{-3}}{2}\right)=y$ for any $y \geq 0$. For each value of $a \neq 0$, find all solutions to the inequality $f\left(\frac{a}{x-2 a}\right) \leq 1$. Solution: The function $g(y)=\frac{4^{y}+4^{-y}}{2}$ is monotonically incre... | Answer: $\quad$ For $a>0: x \in\left[-\frac{26 a}{17} ;-a\right]$. For $a<0: x \in\left[-a ;-\frac{26 a}{17}\right]$. | x\in[-\frac{26}{17};-]for>0,x\in[--\frac{26}{17}]for<0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,928 |
8. In the pyramid $FABC$, $AB=BC$, $FB=FK$, where $K$ is the midpoint of the segment $AC$, and the tangent of the angle between the planes $FAB$ and $ABC$ is in the ratio of 1:3 to the tangent of the angle between the planes $FBC$ and $ABC$. The plane $\pi$ is parallel to $AB$, divides the edge $FC$ in the ratio $1:4$,... | Answer: $5: 11$ or $1: 19$. | 5:11or1:19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,929 |
1.1. (4 points) Mitya is 11 years older than Shura. When Mitya was as old as Shura is now, he was twice as old as she was. How old is Mitya? | Answer: 33.
Solution. 11 years ago, Shura was half as old as she is now. So, she is 22 years old, and Mitya is 33. | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,930 |
2.1. (5 points) One side of a rectangle is $20 \%$ smaller than the other, and its area is $4 \frac{1}{20}$. Find the larger side of the rectangle. | Answer: 2.25.
Solution. If $a-$ is the side we are looking for, then from the condition of the problem $0.8 a^{2}=\frac{81}{20}$, from which $a^{2}=\frac{81}{16}, a=\frac{9}{4}$. | 2.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,931 |
3.1. (13 points) Find the largest natural number $n$ satisfying the inequality $n^{300}<3^{500}$. Answer: 6. | Solution. The inequality is equivalent to the following: $n^{3}<3^{5}=243$. Since $6^{3}=216<243<$ $7^{3}=289$, the desired value is $n=6$. | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,932 |
4.1. (13 points) The numerator of an irreducible fraction was cubed, and the denominator was increased by 3. As a result, the fraction doubled. Find the value of the original fraction, if its numerator and denominator are natural numbers. | Answer: $\frac{2}{3} \approx 0.67$.
Solution. According to the condition $2 \cdot \frac{a}{b}=\frac{a^{3}}{b+3}$, from which $a^{2}=\frac{2(b+3)}{b}=2+\frac{6}{b}$. Therefore, 6 is divisible by $b$. If $b=1$, 2 or 6, then $a^{2}=8,5$ or 3 respectively. Such integers $a$ do not exist. If $b=3$, then $a=2$. The required... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,933 |
5.1. (13 points) Find the smallest natural number that is divisible by 11 and whose representation contains 5 zeros and 7 ones. (You can use the divisibility rule for 11: a number is divisible by 11 if the difference between the sum of the digits in the even positions and the sum of the digits in the odd positions is d... | Solution. The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Let's consider the case when the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. To find the smallest number, we should place this digit as far ... | 1000001111131 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,935 |
6.1. (13 points) Misha drew a triangle with a perimeter of 11 and cut it into parts with three straight cuts parallel to the sides, as shown in the figure. The perimeters of the three shaded figures (trapezoids) turned out to be 5, 7, and 9. Find the perimeter of the small triangle that resulted from the cutting. | Answer: 10.
Solution. Since opposite sides of parallelograms are equal, the lateral sides of each resulting trapezoid are equal to the corresponding segments on the sides of the original tri... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,936 |
7.1. (13 points) Yura has unusual clocks with several minute hands moving in different directions. Yura calculated that the minute hands coincided in pairs exactly 54 times in one hour. What is the maximum number of minute hands that can be on Yura's clocks? | Answer: 28.
Solution. Let some two arrows coincide, then after 30 seconds they will coincide again. Therefore, the arrows in each such pair will coincide exactly 2 times per minute. Thus, if $n$ arrows move in one direction, and $m$ arrows move in the other, then $2 m n=54, m n=27$. Therefore, $n$ can be $1,3,9$ or 27... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,937 |
8.1. (13 points) Two people spend time playing a game: they take turns naming prime numbers not exceeding 100 such that the last digit of the number named by one player is equal to the first digit of the number named by the next player (except for the first prime number named in the game). Repeating numbers that have a... | # Answer: 3.
Solution. We will describe a winning strategy for the first player. First, the first player names a prime number ending in 9 and different from 79 (for example, 19). Since among the numbers $90, \ldots, 99$ the only prime is 97, the second player must name this number on their next move. Then, on the thir... | 3 | Number Theory | proof | Yes | Yes | olympiads | false | 5,938 |
9.1. (13 points) In how many ways can eight of the nine digits $1,2,3,4,5,6$, 7,8 and 9 be placed in a $4 \times 2$ table (4 rows, 2 columns) so that the sum of the digits in each row, starting from the second, is 1 more than in the previous one? | Answer: 64.
Solution. The sum of all nine numbers is 45. Let $x$ be the sum of the two numbers in the first row, and let $a$ be the one number out of the nine that we do not place in the figure. Then $x+x+1+x+2+x+3=45-a$, from which $4 x+a=39$. Since $a$ is an integer from 1 to 9, we get 2 possible cases: either $x=9,... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,939 |
1.1. Find all pairs of two-digit natural numbers for which the arithmetic mean is $25 / 24$ times greater than the geometric mean. In your answer, specify the largest of the arithmetic means for all such pairs. | Answer: 75.
Solution: Let $a, b$ be the required numbers (without loss of generality, we can assume that $a > b$) and let $\frac{a+b}{2}=25x$ and $\sqrt{ab}=24x$. Then $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}=98x$ and $(\sqrt{a}-\sqrt{b})^2=$ $a+b-2\sqrt{ab}=2x$. From this, it follows that $\sqrt{a}+\sqrt{b}=7(\sqrt{a}-\... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,941 |
2.1. The vertices of a cube are labeled with numbers $\pm 1$, and on its faces are numbers equal to the product of the numbers at the vertices of that face. Find all possible values that the sum of these 14 numbers can take. In your answer, specify their product. | Answer: -20160.
Solution. It is obvious that the maximum value of the sum is 14. Note that if we change the sign of one of the vertices, the sum of the numbers in the vertices will increase or decrease by 2. On the other hand, the signs of three faces will change. If their sum was $1, -1, 3, -3$, it will become $-1, 1... | -20160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,942 |
3.1. Functions $f(x)$ and $g(x)$ are defined for all $x>0$; $f(x)$ is the greater of the numbers $x$ and $1 / x$, while $g(x)$ is the lesser of the numbers $x$ and $1 / x$. Solve the equation $f(5 x) \cdot g(8 x) \cdot g(25 x)=1$. If there is only one solution, provide it; if there are multiple solutions, provide their... | Answer: 0.09 (exact value: 0.089).
Solution. The following cases are possible:
a) $0<5 x<8 x<25 x \leqslant 1$. Then $f(5 x) \cdot g(8 x) \cdot g(25 x)=\frac{8 x \cdot 25 x}{5 x}=40 x=1$, from which $x=\frac{1}{40}=0.025$.
b) $0<5 x<8 x \leqslant 1<25 x$. Then $f(5 x) \cdot g(8 x) \cdot g(25 x)=\frac{8 x}{5 x \cdot 2... | 0.09 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,943 |
4.1. How many numbers from 1 to 1000 (inclusive) cannot be represented as the difference of two squares of integers? | Answer: 250.
Solution. Note that any odd number $2 n+1$ can be represented as $(n+1)^{2}-n^{2}$. Moreover, an even number that is a multiple of 4 can be represented as $4 n=(n+1)^{2}-(n-1)^{2}$. The numbers of the form $4 n+2$ remain. Note that a square can give remainders of 0 or 1 when divided by 4, so numbers of th... | 250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,944 |
5.1. The sequence is defined by the relations $a_{1}=1$,
$$
a_{2 n}=\left\{\begin{array}{ll}
a_{n}, & \text { if } n \text { is even, } \\
2 a_{n}, & \text { if } n \text { is odd; }
\end{array} \quad a_{2 n+1}= \begin{cases}2 a_{n}+1, & \text { if } n \text { is even, } \\
a_{n}, & \text { if } n \text { is odd. }\en... | Answer: 5.
Solution: The given rules can be easily interpreted in terms of the binary system: if $n$ ends in 0 and 1 is appended to the right, then 1 is appended to the right of $a_{n}$. If $n$ ends in 1 and 0 is appended, then 0 is appended to the right of $a_{n}$. In all other cases, $a_{n}$ does not change (when 0 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,945 |
6.1. On the coordinate plane, an isosceles right triangle with vertices at points with integer coordinates is depicted. It is known that there are exactly 2019 points with integer coordinates on the sides of the triangle (including the vertices). What is the smallest possible length of the hypotenuse of the triangle un... | Answer: 952.
Solution. The smallest length of the hypotenuse corresponds to the case where the distance between the points is the smallest, which is only possible when the legs follow the grid lines (for a rotated triangle, the distance between the points will be no less than $\sqrt{2}$). Then each leg has a length of... | 952 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,946 |
7.1. Natural numbers from 1 to some $n$ are written in a row. When one of the numbers was removed, it turned out that the arithmetic mean of the remaining numbers is $40 \frac{3}{4}$. Find the number that was removed. | Answer: 61.
Solution. The sum of numbers from 1 to $n$ is $S_{n}=\frac{n(n+1)}{2}$. Let the number removed be $m$, where $1 \leqslant m \leqslant n$. Then the condition of the problem can be written as
$$
\frac{S_{n}-m}{n-1}=40 \frac{3}{4}
$$
Transforming: $\frac{n+2}{2}-\frac{m-1}{n-1}=40 \frac{3}{4}$. Note that $\... | 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,947 |
1.1. A ballpoint pen costs 10 rubles, a gel pen costs 50 rubles, and a fountain pen costs 80 rubles. What is the maximum number of gel pens that can be bought given that exactly 20 pens must be purchased in total, and among them there must be pens of all three types, and exactly 1000 rubles must be spent on them? | Answer: 13. Solution. If $x$ ballpoint pens, $y$ gel pens, and $z$ fountain pens are bought, then we have two equations: $x+y+z=20 ; 10 x+50 y+80 z=1000$ (or $x+5 y+8 z=100$ ). Subtracting the first equation from the second, we get $4 y+7 z=80$. It follows that $z$ must be divisible by 4, i.e., $z=4 n$. Therefore, $4 y... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,949 |
1.2. A ballpoint pen costs 10 rubles, a gel pen costs 40 rubles, and a fountain pen costs 60 rubles. What is the maximum number of ballpoint pens that can be bought if you need to buy exactly 15 pens in total, and among them there must be pens of all three types, and you need to spend exactly 500 rubles? | Answer: 6. Comments. The solutions are two triples of integers $(6,5,4)$ and $(4,10,1)$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,950 |
1.3. A ballpoint pen costs 10 rubles, a fountain pen costs 60 rubles, and a gel pen costs 70 rubles. What is the maximum number of fountain pens that can be bought if you need to buy exactly 25 pens in total and among them there must be pens of all three types, and you need to spend exactly 1000 rubles on them? | Answer: 9. Comments. The solutions are two triples of integers $(11,9,5)$ and $(12,3,10)$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,951 |
1.4. A ballpoint pen costs 10 rubles, a gel pen costs 30 rubles, and a fountain pen costs 60 rubles. What is the maximum number of ballpoint pens that can be bought if you need to buy exactly 20 pens in total and among them there must be pens of all three types, and you need to spend exactly 500 rubles on them? | Answer: 11. Comments. The solutions are two triples of integers $(11,5,4)$ and $(8,10,2)$. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,952 |
2.1. Find the smallest value of $a$, for which the sum of the squares of the roots of the equation $x^{2}-3 a x+a^{2}=0$ is $0.28$. | Answer: $-0.2$ Solution. By Vieta's theorem $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(3 a)^{2}-2 a^{2}=7 a^{2}$. According to the condition $7 a^{2}=\frac{28}{100}$, hence $a^{2}=\frac{1}{25}$ and $a= \pm 0.2$. It is important that for $a= \pm 0.2$ the discriminant $D=9 a^{2}-4 a^{2}=5 a^{2}>0$. | -0.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,953 |
3.1. A plane parallel to the base of a quadrilateral pyramid with a volume of 81 cuts off a smaller pyramid with a volume of 3. Find the volume of the pyramid whose four vertices coincide with the vertices of the section, and the fifth vertex lies in the base plane of the larger pyramid. | Answer: 6. Solution. If we denote the first given volume as $V$, and the second as $v$ (in this case $V=81, v=3$), then the larger pyramid and the smaller pyramid are similar with a similarity coefficient of $\sqrt[3]{\frac{V}{v}}$. Therefore, the height $H$ of the larger pyramid relates to the height $h$ of the smalle... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,954 |
4.1. Given the function $f(x)=|x+1|-2$. How many roots does the equation $f(f(\ldots f(f(x)) \ldots))=\frac{1}{2}$ have, where the function $f$ is applied 2013 times? | Answer: 4030. Solution. Considering the graph of the function $f$ (see Fig. 1), we conclude that $|f(x)+1|=f(x)+1$ and $|f(x)+1|-2=f(x)-1$. Therefore, the graph of the function $f(f(x))$
ha... | 4030 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,955 |
5.1. Specify the integer closest to the number $\sqrt{2012-\sqrt{2013 \cdot 2011}}+\sqrt{2010-\sqrt{2011 \cdot 2009}}+\ldots+\sqrt{2-\sqrt{3 \cdot 1}}$. | Answer: 31. Solution. The number given in the condition is equal to
$$
\frac{\sqrt{4024-2 \sqrt{2013 \cdot 2011}}+\sqrt{4010-2 \sqrt{2011 \cdot 2009}}+\ldots+\sqrt{4-2 \sqrt{3 \cdot 1}}}{\sqrt{2}}
$$
$=\frac{\sqrt{(\sqrt{2013}-\sqrt{2011})^{2}}+\sqrt{(\sqrt{2011}-\sqrt{2009})^{2}}+\ldots+\sqrt{(\sqrt{3}-\sqrt{1})^{2}... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,956 |
6. Triangle $L O M$ with angle $\angle L O M=\alpha^{\circ}$ is rotated by some acute angle around point $O$. As a result, point $L$ moves to point $N$, which lies on side $L M$, and point $M$ moves to point $K$ such that $O M \perp N K$. Find the angle of rotation (in degrees).
Note: In different versions, different ... | Answer: $\frac{2}{3} \alpha \cdot$ Solution. Let the angle of rotation be denoted by $\varphi$. Then $\angle L O N=\angle M O K=\varphi$. Since $O N=O L$ (property of rotation), we have
$\angle O L N=\angle O N L=90^{\circ}-\frac{\varphi}{2}$.
Moreover, $\angle O N K=\angle O L N=90^{\circ}-\frac{\varphi}{2}$ (a cons... | \frac{2}{3}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,957 |
7.1. Find the number of all integer solutions of the inequality $\sqrt{1-\sin \frac{\pi x}{4}-3 \cos \frac{\pi x}{2}}-\sqrt{6} \cdot \sin \frac{\pi x}{4} \geq 0$, belonging to the interval [1991; 2013]. | Answer: 8. Solution. Denoting $\sin \frac{\pi x}{4}=t$, we get $\sqrt{1-t-3\left(1-2 t^{2}\right)} \geq \sqrt{6} \cdot t$, or $\sqrt{6 t^{2}-t-2} \geq \sqrt{6} \cdot t$. For non-negative $t$, we obtain $6 t^{2}-t-2 \geq 6 t^{2}$, or $t \leq-2$, which means there are no solutions. For $t<0$, the condition $6 t^{2}-t-2 \... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,958 |
7.2. Find the number of all integer solutions of the inequality $\sqrt{3 \cos \frac{\pi x}{2}-\cos \frac{\pi x}{4}+1}-\sqrt{6} \cdot \cos \frac{\pi x}{4} \geq 0$, belonging to the interval [1991; 2013]. | Answer: 9. Comments. Solution of the inequality: $\frac{8}{3}+8 k \leq x \leq \frac{16}{3}+8 k$. From the given interval, the following numbers will be included in the answer: 1995, 1996, 1997, 2003, 2004, 2005, 2011, 2012, 2013 - 9 numbers. | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,959 |
7.3. Find the number of all integer solutions of the inequality $\sqrt{1+\sin \frac{\pi x}{4}-3 \cos \frac{\pi x}{2}}+\sqrt{6} \cdot \sin \frac{\pi x}{4} \geq 0$, belonging to the interval [1991; 2013]. | Answer: 9. Comments. Solution of the inequality: $\frac{2}{3}+8 k \leq x \leq \frac{10}{3}+8 k$. From the given interval, the following numbers will be included in the answer: 1993, 1994, 1995, 2001, 2002, 2003, 2009, 2010, 2011 - 9 numbers. | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,960 |
7.4. Find the number of all integer solutions of the inequality $\sqrt{3 \cos \frac{\pi x}{2}+\cos \frac{\pi x}{4}+1}+\sqrt{6} \cdot \cos \frac{\pi x}{4} \geq 0$, belonging to the interval $[1991 ; 2013]$. | Answer: 9. Comments. Solution of the inequality: $-\frac{4}{3}+8 k \leq x \leq \frac{4}{3}+8 k$. From the given interval, the following will be included in the answer: 1991, 1992, 1993, 1999, 2000, 2001, 2007, 2008, 2009 - 9 numbers. | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,961 |
8.1. Find the area of the figure defined on the coordinate plane by the inequality $2(2-x) \geq\left|y-x^{2}\right|+\left|y+x^{2}\right|$. | Answer: 15. Solution. In the region above the graphs of the functions $y=x^{2}$ and $y=-x^{2}$, the original inequality takes the form $2-x \geq y$. In the region below the graphs of the functions $y=x^{2}$ and $y=-x^{2}$, the original inequality takes the form $2-x \geq -y$, i.e., $y \geq x-2$. In the region lying abo... | 15 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,962 |
9.1. First-grader Masha, entering the school, each time climbs the school porch stairs, which have 10 steps. Being at the bottom of the stairs or on one of its steps, she can either go up to the next step or jump over one step up (Masha cannot yet jump over two or more steps). What is the minimum number of times Masha ... | Answer: 89. Solution. Note that Masha can climb a porch with one step in one way, and a porch with two steps in two ways: either stepping on each step, or, by stepping over the first step, landing directly on the second.
Let $a_{n}$ be the number of ways Masha can climb a porch with $n$ steps. Since Masha can reach th... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,963 |
10.1. Having found some polynomial of the sixth degree $x^{6}+a_{1} x^{5}+\ldots+a_{5} x+a_{6}$ with integer coefficients, one of the roots of which is the number $\sqrt{2}+\sqrt[3]{5}$, write in the answer the sum of its coefficients $a_{1}+a_{2}+\ldots+a_{6}$. | Answer: -47. Solution. $x=\sqrt{2}+\sqrt[3]{5} \Rightarrow (x-\sqrt{2})^{3}=5 \Rightarrow x^{3}-3 x^{2} \sqrt{2}+3 x \cdot 2-2 \sqrt{2}=5$ $\Rightarrow x^{3}+6 x-5=(3 x^{2}+2) \cdot \sqrt{2} \Rightarrow (x^{3}+6 x-5)^{2}=(3 x^{2}+2)^{2} \cdot 2$ $\Rightarrow P_{6}(x)=(x^{3}+6 x-5)^{2}-2(3 x^{2}+2)^{2}$. In this case, t... | -47 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,964 |
Problem 2. Misha has a set of nine cards with the letters of the word "LOMOMONOSOV". On the reverse side of each card, Misha wrote a digit such that the digits on cards with the same letters are the same, and the digits on cards with different letters are different. It turned out that the equation
$$
L+\frac{O}{M}+O+H... | Answer: $8+\frac{2}{3}+2+9+\frac{2}{6}=20$ (the 8 and 9, as well as 3 and 6, can be swapped).
Solution. The left side of the resulting correct equality is less than the value of $10+9+10=29$, which means the digit corresponding to the letter "O" can be either 1 or 2. In this case, the equality $\frac{\mathrm{O}}{\math... | 8+\frac{2}{3}+2+9+\frac{2}{6}=20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,966 |
Problem 3. On a river, two identical sightseeing boats departed from one pier in opposite directions at 13:00. At the same time, a raft also set off from the pier. After an hour, one of the boats turned around and headed in the opposite direction. At 15:00, the second boat did the same. What is the speed of the current... | Answer: 2.5 km/h.
Solution. Consider the reference frame associated with the river. In this frame, the boats move with equal speeds: initially, they move away from each other in opposite directions for 1 hour, then move in the same direction for another 1 hour (during which the distance between them does not change), ... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,967 |
Problem 4. Each cell of a $3 \times 3$ table is painted in one of three colors such that cells sharing a side have different colors, and not all three colors need to be used. How many such colorings exist? | Answer: 246.
Solution. The central cell can be painted in any of the three colors, let's call this color $a$. Each of the four cells that share a side with the central one can be painted in any of the two remaining colors. Let the cell located above the central one be painted in color $b$. The third color we will call... | 246 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,968 |
Problem 5. From 24 identical wooden cubes, a "pipe" was glued - a $3 \times 3 \times 3$ cube with the "core" of three cubes removed (see figure). Is it possible to draw a diagonal in each square on the surface of the "pipe" so that a closed path is formed, which does not pass through any vertex more than once? | Answer: No.
Solution. The number of diagonals on the surface of the "tube" is the same as the number of faces of the cubes on this surface: $4 \cdot 9 + 2 \cdot 8 + 12 = 64$. A non-self-int... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,969 |
Problem 6. Petya missed the football match "Dynamo"-"Shinnik" and, coming to school in the morning, heard his friends discussing the game results. Petya's neighbor at school, Vasya, refused to name the score of the match but agreed to honestly answer any two questions Petya might have about the game, with answers being... | Answer: Yes.
Solution. From the words of Roman and Oleg, it follows that the total number of goals in the match does not exceed two, and that one of the teams won. The score options are: $0: 1, 1: 0, 0: 2$ and $2: 0$. From Seryozha's words, it is now clear that "Shinnik" won, as the losing team did not score a single ... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,970 |
2.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five... | Answer: 7111765.
Solution. Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive... | 7111765 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,971 |
3.1. From one point on a circular track, a pedestrian and a cyclist started simultaneously in the same direction. The cyclist's speed is $55\%$ greater than the pedestrian's speed, and therefore the cyclist overtakes the pedestrian from time to time. At how many different points on the track will the overtakes occur? | Answer: 11.
Solution: Let's assume the length of the path is 55 (in some units), and the speeds of the pedestrian and the cyclist are $100 x$ and $155 x$. Then, the overtakes will occur every $1 / x$ units of time. During this time, the pedestrian covers 100 units, which means he will be 10 units away in the opposite ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,972 |
4.1. We will call a natural number interesting if all its digits, except the first and last, are less than the arithmetic mean of the two adjacent digits. Find the largest interesting number. | Answer: 96433469.
Solution. Let $a_{n}$ be the n-th digit of the desired number. By the condition, $a_{n}9$. Similarly, there cannot be four negative differences. Therefore, the maximum number of such differences can be 7, and the desired number is an eight-digit number. To make it the largest, the first digit should ... | 96433469 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,973 |
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