problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
85 (954). Prove that the value of the numerical expression $11^{11}+$ $+12^{12}+13^{13}$ is divisible by 10. | Solution. It is sufficient to prove that the value of the numerical expression ends in zero. Indeed, the number $11^{11}$ ends in the digit 1. To determine which digit the number $12^{12}$ ends in, it is advisable to represent $12^{12}$ as $\left(12^{4}\right)^{3}$ and note that $12^{4}$ ends in the digit 6. Therefore,... | 10989 | Number Theory | proof | Yes | Yes | olympiads | false | 35,939 |
89 (958). Prove that:
$$
\text { 1) } \left.\frac{171717}{252525}=\frac{1717}{2525}=\frac{17}{25} ; 2\right) \quad \frac{313131}{757575}=\frac{3131}{7575}=\frac{31}{75} \text {. }
$$ | Solution.
$$
\begin{gathered}
\frac{171717}{252525}=\frac{17 \cdot 10^{4}+17 \cdot 10^{2}+17}{25 \cdot 10^{4}+25 \cdot 10^{2}+25}=\frac{17\left(10^{4}+10^{2}+1\right)}{25 \cdot\left(10^{4}+10^{2}+1\right)}=\frac{17}{25} \\
\frac{1717}{2525}=\frac{17 \cdot 100+17}{25 \cdot 100+25}=\frac{17 \cdot 101}{25 \cdot 101}=\fra... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,940 |
91 (960). Prove the validity of the equality:
1) $(n-2)(n-1) n(n+1)+1=\left(n^{2}-n-1\right)^{2}$;
2) $n(n+1)(n+2)(n+3)+1=\left(n^{2}+3 n+1\right)^{2}$. | Solution.
1) $(n-2)(n-1) n(n+1)+1=(n-2)(n+1)(n-1) n+1=$ $\left(n^{2}-n-2\right)\left(n^{2}-n\right)+1=\left(n^{2}-n\right)^{2}-2\left(n^{2}-n\right)+1=$ $\left(n^{2}-n-1\right)^{2}$
2) $n(n+1)(n+2)(n+3)+1=n(n+3)(n+1)(n+2)+1=$ $=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1=\left(n^{2}+3 n\right)^{2}+2\left(n^{2}+3 ... | 3926341 | Algebra | proof | Yes | Yes | olympiads | false | 35,941 |
## 95(964). Simplify the expression and find its numerical value for the given values of the letters:
1) $x^{2}-\frac{x^{3}-4 x y^{2}}{x(x-2 y)+y^{2}} \cdot \frac{x^{2}-2 x y+y^{2}}{x-2 y}, x=-\frac{1}{2}, y=-5$;
2) $\frac{a^{2}+a b+b^{2}}{a^{3}+2 a^{2} b+a b^{2}}+\frac{b-a}{(a+b)^{2}} \cdot \frac{a}{a-b}-\frac{a}{a^{2... | Solution.
2) $\frac{a^{2}-b c}{(a+b)(a+c)}+\frac{b^{2}-a c}{(b+c)(a+b)}+\frac{c^{2}-a b}{(a+c)(b+c)}$.
$$
\text { 1) } \begin{gathered}
\frac{4}{1+x^{4}}+\frac{2}{1+x^{2}}+\frac{1}{1+x}+\frac{1}{1-x}=\frac{4}{1+x^{4}}+\frac{2}{1+x^{2}}+ \\
+\frac{2}{1-x^{2}}=\frac{4}{1+x^{4}}+\frac{4}{1-x^{4}}=\frac{8}{1-x^{8}}
\end{g... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,943 |
105(974). Determine the value of $b$, if the graph of the function defined by the formula passes through the point with coordinates $(3 ; 10)$:
1) $y=x+b$;
2) $y=3 x+b$
3) $y=-\frac{1}{3} x+b$;
4) $y=-\frac{1}{2} x+b$.
A n s w e r: 1) $b=7$; 2) $b=1$; 3) $b=11$; 4) $b=11.5$.
106(975). Define the function by a formul... | Solution. Let $x$ be the number of turns in the spring. Considering that the number of gaps in the spring is one less than the number of turns, according to the problem, we set up the equation $5x + 8(x-1) = 122$, from which $x = 10$.
Answer: 10 turns. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,944 |
108(977). A person, living in a cottage near a railway station, usually manages to walk from the cottage to the station in time for the train departure in 18 minutes. Once, this person was delayed at home before leaving by several minutes. Although after that, he walked 1.2 times faster than usual, he still missed the ... | Solution. If the usual speed of the gardener is $v$ m/min, then the distance from the garden to the station is $18 v$ meters. When the gardener walked at a speed of $1.2 v$ m/min, it took 15 minutes to travel from home to the garden $(18 v: 1.2 v=15)$. Since the gardener was 2 minutes late, the train left 13 minutes af... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,945 |
112(981). A pedestrian left point $A$ for point $B$. After 1 hour and 24 minutes, a cyclist left point $A$ in the same direction. After one hour, the cyclist was 1 km behind the pedestrian, and another hour later, the distance remaining for the cyclist to $B$ was half the distance remaining for the pedestrian. Find the... | Solution. Let the speed of the pedestrian be $v_{1}$ kilometers per hour, and the speed of the cyclist be $v_{2}$ kilometers per hour. Since in 2 hours and 24 minutes (i.e., $\frac{12}{5}$ hours), the pedestrian walked 1 km more than the cyclist traveled in 1 hour, we have the equation
$$
\frac{12}{5} v_{1}-v_{2}=1
$$... | v_{1}=5,v_{2}=11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,946 |
113(982). A cyclist arrived from point $A$ to point $B$ at the appointed time, moving at a certain speed. If he had increased this speed by 3 km/h, he would have arrived at the destination one hour earlier, and if he had traveled 2 km less per hour than he actually did, he would have been an hour late. Determine the di... | Solution. Let the first pipe fill the pool in $x$ hours, then the second pipe - in $(x+6)$ hours; $\frac{2}{3}$ of the pool the second pipe will fill in $\frac{2}{3}(x+6)$ hours.
According to the problem, we have the equation
$$
x=\frac{1}{2} \cdot \frac{2}{3}(x+6)
$$
from which $x=3, x+6=9$.
Answer: the first pipe... | 1040 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,947 |
118(987). A swimmer is swimming against the current of a river and meets an empty boat floating downstream. Continuing to swim against the current for another $t$ minutes after the moment of meeting, he then turns back and catches up with the boat $s$ meters from the meeting point. Find the speed of the river current. | Solution. Let the speed of the swimmer in still water be $v_{1}$ meters per minute, and the speed of the boat (the speed of the river current) be $v_{2}$ meters per minute. Then the swimmer's speed downstream is $\left(v_{1}+v_{2}\right)$ meters per minute, and his speed upstream is $\left(v_{1}-v_{2}\right)$ meters pe... | \frac{}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,948 |
119(988). In a flask, there is a solution of table salt. From the flask, $\frac{1}{5}$ of the solution is poured into a test tube and evaporated until the percentage of salt in the test tube doubles. After this, the evaporated solution is poured back into the flask. As a result, the salt content in the flask increases ... | Solution. Before solving the problem, it is reasonable to clarify how the expression "percentage content of salt" should be understood (this is the ratio of the mass of salt to the mass of the solution, expressed as a percentage).
Let the initial mass of the solution in the flask be $m$ grams, the mass of salt in it b... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,949 |
120 (989). The road from point $A$ to point $B$ is 11.5 km long and goes first uphill, then on flat ground, and finally downhill. The pedestrian spent 2 hours and 54 minutes on the journey from $A$ to $B$, and 3 hours and 6 minutes on the return journey. The pedestrian's walking speed uphill was 3 km/h, on flat ground ... | Solution. Let the length of the road uphill be $x$ kilometers, on the plain be $y$ kilometers, and downhill be $z$ kilometers. According to the problem, we form a system of three linear equations with three unknowns:
88
$$
\left\{\begin{array}{l}
\frac{x}{3}+\frac{y}{4}+\frac{z}{5}=2.9 \\
\frac{z}{3}+\frac{y}{4}+\fra... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,950 |
125. A steamship from Kyiv to Kherson takes three days, and from Kherson to Kyiv - four days (without stops). How long will it take for rafts to float from Kyiv to Kherson? | Solution. To answer the question of the problem, the distance from Kyiv to Kherson needs to be divided by the speed of the Dnieper current. If $s$ is the distance from Kyiv to Kherson, then the speed of the steamboat downstream is $-\frac{s}{3}$ (km/day), and the speed of the steamboat upstream is $-\frac{s}{4}$ (km/da... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,951 |
127. We collected 100 kg of mushrooms. It turned out that their moisture content was $99 \%$. When the mushrooms were dried, the moisture content decreased to $98 \%$. What was the mass of these mushrooms after drying? | Solution. According to the condition, in 100 kg of mushrooms, there is 1 kg of dry matter $(100-0.99 \cdot 100=1)$. Since the mass of dry matter in the total mass of mushrooms is constant (1 kg) and after drying it became $2 \%(100-98=2)$, the mass of mushrooms after drying became 50 kg (if $2 \%-1$ kg, then $100 \%-50... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,952 |
130. How many four-digit numbers can be formed from the digits $0,1,2,3,4,5$, if the numbers must be odd and digits may be repeated? | S o l u t i o n. For the first position, we can choose any of the five digits (excluding zero). For the second and third positions, we can choose any of the six digits, as digits can be repeated. For the last position, we can choose any of the three odd digits. Therefore, the total number of four-digit numbers is $5 \c... | 540 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,953 |
135. All natural numbers from 1 to 100 are divided into two groups: even and odd numbers. Determine in which group the sum of all digits used to write the numbers is greater and by how much.
Answer: The sum of the digits of the odd numbers is greater by 49.
$\square$ 136. Given two consecutive natural numbers $a$ and... | Proof. $b=a+1, \quad c=a \cdot(a+1)$, then $x=$ $=\left(a^{2}+a+1\right)^{2}$.
$\triangle$ 137. When the father was 27 years old, the son was only 3 years old, and now the son is three times younger than the father. How old is each of them now?
Solution. Let the son be $x$ years old now, then the father is $3 x$ year... | 12,36 | Number Theory | proof | Yes | Yes | olympiads | false | 35,954 |
139. On a meadow, grass grows. If 9 cows were let onto the meadow, they would empty it in 4 days. If 8 cows were let onto the meadow, they would eat all the grass in 6 days. How many cows can graze on the meadow all the time while the grass grows? | Solution. Let $x$ kilograms of grass grow on the meadow per day, and one cow eats $y$ kilograms of grass per day. We have the system of equations:
$$
\left\{\begin{array}{l}
a+4 x-9 \cdot 4 y=0 \\
a+6 x-8 \cdot 6 y=0
\end{array}\right.
$$
where $a$ is the amount of grass that grew on the meadow before the cows were l... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,955 |
142. It is said that in the 19th century, every tenth man in Russia was Ivan, and every twentieth was Petr. If this is true, who was more numerous in Russia: Ivan Petrovich or Petr Ivanovich? | Solution. First, let's calculate the number of Ivan Petroviches. According to the problem, the number of Ivans in Russia constituted $\frac{1}{10}$ of all men. Their fathers also constituted $\frac{1}{10}$ of all men, as each Ivan has one father. Since every twentieth man was named Peter, among the men whose sons were ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,957 |
144. A magic square is a square table of positive integers in which the sums of the numbers in each column, each row, and on the diagonals are equal. This sum itself is called the sum of the magic square. Prove that the sum of a $3 \times 3$ magic square is always divisible by 3. | I n d i c a t i o n. Let the first two numbers in the first row of the square be denoted by $a$ and $b$, and the first number in the second row by $c$, the sum of the magic square by $s$. After this, we can find all the other numbers in the square and prove that $s$ is divisible by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,958 |
157. Prove that for any natural number $n$ the number $6^{2(n+1)}-2^{n+3} \cdot 3^{n+2}+36$ is divisible by 900. | Solution.
$$
6^{2(n+1)}-2^{n+3} \cdot 3^{n+2}+36=36\left(6^{2 n}-2 \cdot 2^{n} \cdot 3^{n}+1\right)=36\left(6^{n}-1\right)^{2}
$$
The difference $6^{n-1}$ is divisible by 5, so the product $36 \cdot\left(6^{n}-1\right)^{2}$ is divisible by $900(36 \cdot 5 \cdot 5=900)$, which is what we needed to prove. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,959 |
158. It is known that the number $a$ is $n$ times greater than the number $b$, and the sum of the numbers $a$ and $b$ is $m$ times greater than their difference. Find the sum of the numbers $m$ and $n$, if $m$ and $n$ are natural numbers.
94 | Given the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
\frac{a}{b}=n \\
\frac{a+b}{a-b}=m
\end{array}\right.
$$
From the second equation, express $\frac{a}{b}$. We have:
$$
\begin{gathered}
a+b=m a-m b, a(1-m)=-b(m+1) \\
\frac{a}{b}=\frac{m+1}{m-1}=\frac{m-1+2}{m-1}=1+\frac{2}{m-1} .
\end{g... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,960 |
160. Which is greater: $5^{15}$ or $3^{23}$? | Solved. $3^{23}=9\left(3^{3}\right)^{7}, 5^{15}=5\left(5^{2}\right)^{7}$, therefore $3^{23}>5^{15}$. | 3^{23}>5^{15} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,961 |
168. Let $A$ be a natural number. Denote by $b$ the number formed by the last two digits of $A$, and by $a$ the number formed by the remaining digits of $A$. Prove that the number $A$ is divisible by 7 if and only if:
a) $2a + b$ is divisible by 7; b) $5a - b$ is divisible by 7. | Instruction. a) It follows from the equality $2 a+b=(100 a+b)-$ $-98 a ;$ b) it follows from the equality $5 a-b=105 a-(100 a+b)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,962 |
170. During spring, Zhenya lost $20 \%$ of his weight, then gained $30 \%$ during summer, lost $20 \%$ again during autumn, and gained $10 \%$ during winter. Did Zhenya gain or lose weight over the year? | Solution. Let's assume Zhenya's mass was 100 (units of mass). Then after spring, his mass became $100-20=80$ (units of mass), after summer $-80+\frac{30 \cdot 80}{100}=104$ (units of mass), after autumn $-104-$ $-\frac{20 \cdot 104}{100}=104-20.8=83.2$ (units of mass). After winter, Zhenya's mass became $83.2+8.32=91.5... | 91.52 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,963 |
171. A jeweler's diamond cracked during polishing, as a result, its value decreased by $32 \%$. What part of the diamond broke off if the value of the diamond is proportional to the square of its mass? | Solution. Let a diamond of mass $p$ during polishing split into two parts: of mass $p_{1}$ and mass $p_{2}$. Since the cost of the diamond is proportional to the square of its mass, we have the system of equations:
$$
\left\{\begin{array}{l}
p_{1}+p_{2}=p \\
k p_{1}^{2}+k p_{2}^{2}=k \cdot 0.68 \cdot p^{2}
\end{array}... | \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,964 |
172. In the store, there are candies priced at 2 p. per kilogram and 3 p. per kilogram, in equal amounts. At what price should the mixture of these candies be sold? | S o l u t i o n. Let the cost of each type of candy be $s$ rubles. Then the amount of candy priced at 2 p. per kilogram is $\frac{s}{2}$ kilograms, and the amount of candy priced at 3 p. per kilogram is $\frac{s}{3}$ kilograms. Therefore, the price of the mixture is $\frac{2 s}{\frac{s}{2}+\frac{s}{3}}=2.4 \mathrm{p}$.... | 2.4\mathrm{p} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,965 |
173. In the Olympiad, 55 schoolchildren participated. All of them submitted their work. During the checking of each problem, one of three grades was given: «+» - the problem was solved, «-» - the problem was attempted, «0» - the problem was not attempted. After checking all the works, it turned out that in no two works... | Instruction. If the number of tasks was $a$, then
$$
\frac{(a+1)(a+2)}{2}=55 \text { when } a=9 .
$$
A n s w e r: 9 tasks. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,966 |
175. I thought of two two-digit numbers starting with the digit 6, and the other digits of these numbers are not 6. If the digits in each of the thought-of numbers are swapped, the value of the product of the two-digit numbers does not change. What numbers did I think of | Solution. Method I. Let the units digits of the two-digit numbers be $x$ and $y$, with $x \neq 6$ and $y \neq 6$. According to the problem, $\overline{6 x} \cdot \overline{6 y}=\overline{x 6} \cdot \overline{y 6}$, i.e., we have the equation
$$
(6 \cdot 10 + x)(6 \cdot 10 + y) = (10 x + 6)(10 y + 6)
$$
Solving this e... | 6964 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,967 |
176. Among 150 schoolchildren, only boys collect stamps. 67 people collect stamps of the USSR, 48 people - of Africa, and 32 people - of America, 11 people only collect stamps of the USSR, 7 people - only of Africa, 4 people - only of America, and only Ivanov collected stamps of the USSR, Africa, and America. Find the ... | Solution. Let's represent the condition of the problem using Euler circles (Fig. 17).
We have a system of three equations:
\[
\begin{cases}
x + y + 1 = 56 & (56 = 67 - 11), \\
x + z + 1 = 41 & (41 = 48 - 7), \\
y + z + 1 = 28 & (28 = 32 - 4)
\end{cases}
\]
, then in total they brought 15 plants ($1+2+3+4+5=15$). Therefore, the remaining 15 plants ($30-15=15$) were brought by the other 15 students ($20-5=15$), i.e., one plant each. Thus, two plants were brought by one student.
## § 8. ADVANCED PROBLEMS IN ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,969 |
2. Does the equation $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1=0$ have negative roots ( $[4]$, № 564 ) ?
口 5(1081). Find the common fraction with a denominator of 21, lying between the fractions $\frac{5}{14}$ and $\frac{5}{12}$. | Solution. Let's bring these fractions to a common denominator:
$$
\begin{aligned}
& \frac{5}{14}=\frac{5}{2 \cdot 7}=\frac{5 \cdot 6}{2 \cdot 7 \cdot 6}=\frac{30}{84} \\
& \frac{5}{12}=\frac{5}{2 \cdot 6}=\frac{5 \cdot 7}{2 \cdot 6 \cdot 7}=\frac{35}{84} .
\end{aligned}
$$
Between the fractions $\frac{30}{84}$ and $\... | \frac{8}{21},2,1,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,970 |
13(1089). The function $y$ of $x$ is given by the formula $y=\frac{a x+b}{c x+d}$, where $a d-b c \neq 0$. Let the values of the argument $x_{1}, x_{2}, x_{3}$, and $x_{4}$ correspond to the function values $y_{1}, y_{2}, y_{3}$, and $y_{4}$. Prove that
$$
\frac{y_{3}-y_{1}}{y_{3}-y_{2}}: \frac{y_{4}-y_{1}}{y_{4}-y_{2... | Dok azatel'stvo.
$$
\begin{gathered}
y_{3}-y_{1}=\frac{a x_{3}+b}{c x_{3}+d}-\frac{a x_{1}+b}{c x_{1}+d}= \\
=\frac{c a x_{1} x_{3}+b c x_{1}+a d x_{3}+b d-a c x_{3} x_{1}-b c x_{3}-a d x_{1}-b d}{\left(c x_{3}+d\right)\left(c x_{1}+d\right)}= \\
=\frac{-b c\left(x_{3}-x_{1}\right)+a d\left(x_{3}-x_{1}\right)}{\left(c... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,972 |
15(1091). Prove that the value of the expression $\sqrt{11+6} \sqrt{ } 2+$ $+\sqrt{11-6 \sqrt{2}}$ is a natural number. | Proof. I method. Let
$$
\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}=A
$$
Then
$$
A^{2}=11+6 \sqrt{2}+11-6 \sqrt{2}+2 \sqrt{121-72}=36
$$
from which
$$
\sqrt{A^{2}}=|A|=6, \text { i.e. }|\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}|=6
$$
Since under the modulus sign is the sum of two radicals, each of which is a positi... | 6 | Algebra | proof | Yes | Yes | olympiads | false | 35,973 |
16 (1092). Prove that for $a \geqslant 0, b \geqslant 0, c \geqslant 0, d \geqslant 0$ the inequality
$$
\sqrt{(a+c)(b+d)} \geqslant \sqrt{a b}+\sqrt{c d}
$$
holds. | Solution. Let's compare the squares of the given non-negative expressions:
$$
\begin{gathered}
(a+c)(b+d) \vee(\sqrt{a b}+\sqrt{c d})^{2} \\
a b+a d+c b+c d \vee a b+c d+2 \sqrt{a b c d}
\end{gathered}
$$
Subtract the sum $a b+c d$ from both sides of the inequality:
$$
\begin{aligned}
& b c+a d \vee 2 \sqrt{a b c d}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,974 |
17(1093). Prove that the sum, difference, product, and quotient of numbers of the form $a+b \sqrt{2}$, where $a$ and $b$ are rational numbers, can also be represented in the same form. | Solution. Let $x=a_{1}+b_{1} \sqrt{2}, \quad y=a_{2}+b_{2} \sqrt{2}$. Then $x \pm y=\left(a_{1} \pm a_{2}\right)+\left(b_{1} \pm b_{2}\right) \cdot \sqrt{2}$, where $a_{1} \pm a_{2} \in \boldsymbol{Q}, b_{1} \pm b_{2} \in \boldsymbol{Q}$.
$$
\begin{aligned}
& x y=\left(a_{1} a_{2}+2 b_{1} b_{2}\right)+\left(a_{1} b_{2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,975 |
18(1094). The pair of numbers $x=3, y=2$ is a solution to the equation $(x+y \sqrt{2})(x-y \sqrt{2})=1$. Show that there are infinitely many other pairs of natural numbers that satisfy this equation. | Solution. If the pair $(3 ; 2)$ is a solution to the equation $(x+y \sqrt{2})(x-y \sqrt{2})=1$ or its equivalent equation $x^{2} - 2 y^{2}=1(*)$, then the equality $(3+2 \sqrt{2})(3-2 \sqrt{2})=1$ is true. Squaring this equality term by term (to find other solutions to the given equation), we get:
$$
\begin{gathered}
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,976 |
23(1099). A bus left city $M$ for city $N$ at a speed of 40 km/h. A quarter of an hour later, it met a car traveling from city $N$. This car reached city $M$, 15 minutes later it set off back to city $N$ and overtook the bus 20 km from city $N$. Find the distance between cities $M$ and $N$, if the speed of the car is 5... | Solution. Since the bus traveled 10 km in a quarter of an hour $(40: 4=10)$, the bus met the car 10 km from city $M$. If the distance between the cities is $s$ kilometers, then by the time the car overtook the bus, the bus had traveled $(s-30)$ kilometers from the moment of the meeting, spending $\frac{s-30}{40}$ hours... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,977 |
25(1101). The distance between the ports $A$ and $B$ is covered by the steamboat downstream in 5 hours, and upstream in 6 hours. How many hours will it take for a raft to drift downstream this distance? | Solution. I method. To answer the question of the problem, the distance $AB$ needs to be divided by the speed of the current. Let $AB = s$ kilometers. Then the speed of the steamer downstream is $\frac{s}{5}$ kilometers per hour, the speed upstream is $-\frac{s}{6}$ kilometers per hour, the speed of the current is $-\l... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,978 |
28(1104). Two motorcyclists set off simultaneously from $A$ to $B$ and from $B$ to $A$. The first arrived in $B$ 2.5 hours after the meeting, while the second arrived in $A$ 1.6 hours after the meeting. How many hours was each motorcyclist on the road? | Solution. Let $AB = s$ (in km), the time until the motorcyclists meet $-t$ (in hours). Then the time of movement of the first motorcyclist (from $A$ to $B$) is $(t+2.5)$ hours, its speed is $\frac{s}{t+2.5}$ kilometers per hour, the time of movement of the second motorcyclist is $(t+1.6)$ hours, its speed is $\frac{s}{... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,979 |
29(1105). Two cars set off simultaneously from $A$ to $B$ and from $B$ to $A$, and they met after 3 hours. The first car arrived in $B$ 1.1 hours later than the second car arrived in $A$. How many times greater is the speed of the second car compared to the speed of the first car? | Solution. Let the speed of the first car be $v_{1}$ kilometers per hour, and the speed of the second car be $v_{2}$ kilometers per hour. The first car traveled $3 v_{1}$ kilometers before the meeting, and the second car traveled $3 v_{2}$ kilometers. After the meeting, the first car traveled for $\frac{3 v_{2}}{v_{1}}$... | 1.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,980 |
33(1109). Find the members of the proportion $x_{1}: x_{2}=x_{3}: x_{4}$, where the first term is 6 more than the second, and the third is 5 more than the fourth. The sum of the squares of all terms is 793. | Solution. By the condition $x_{1}=x_{2}+6, x_{3}=x_{4}+5,\left(x_{2}+6\right): x_{2}=$ $=\left(x_{4}+5\right): x_{4}$.
We have a system of two equations with two unknowns:
$$
\left\{\begin{array}{l}
\left(x_{2}+6\right)^{2}+x_{2}^{2}+\left(x_{4}+5\right)^{2}+x_{4}^{2}=793 \\
\left(x_{2}+6\right) \cdot x_{4}=x_{2} \cd... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,981 |
36(1112). Prove that if $a+c=2b$ and $2bd=c(b+d)$, and $b \neq 0$ and $d \neq 0$, then $\frac{a}{b}=\frac{c}{d}$. | Proof. Multiply both sides of the given identity \(a+c=2b\) by \(d (d \neq 0)\). We will have \(2bd = ad + cd\).
Since the left side of the obtained identity is equal to the left side of the given identity \(2bd = c(b+d)\), then their right sides are also equal: \(c(b+d) = ad + cd\), or \(cb = ad\), from which by divi... | \frac{}{b}=\frac{}{} | Algebra | proof | Yes | Yes | olympiads | false | 35,982 |
37(1113). Construct the graph of the function given by the formula: a) $y=\frac{1}{|x|}$; b) $y=-\frac{1}{\sqrt{x}}$; c) $y=x|x|$. | Solution. a) Based on the definition of the modulus of a number, we have:
$$
y=\frac{1}{|x|}=\left\{\begin{array}{l}
\frac{1}{x} \text { when } x>0 \\
-\frac{1}{x} \text { when } x<0
\end{array}\right.
$$
Therefore, for positive values of $x$, we have a branch of the hyperbola $y=\frac{1}{x}$ located in the first qua... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,983 |
38(1114). Construct the graph of the function given by the formula:
a) $y=\sqrt{x^{2}}+x$;
c) $y=|x+2|$;
b) $y=\sqrt{x^{2}}-x$
d) $y=|x-1|$. | Solution.
a) $y=\sqrt{x^{2}}+x=|x|+x=\left\{\begin{array}{l}2 x, \text { when } x \geqslant 0, \\ 0, \text { when } x<0 .\end{array}\right.$
The graph is shown in Figure 22.
b) $y=\sqrt{x^{2}}-x=|x|-x=\left\{\begin{array}{l}0 \text { when } x \geqslant 0, \\ -2 x \text { when } x<0 .\end{array}\right.$
The graph is... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,984 | |
39(1115). Plot the graph of the function given by the formula $y=x+\frac{1}{x}$. | Solution. The domain of the function is the set of all real numbers except zero. It is clear that for $x>0$ the graph of the function is located above the line $y=x$ (since $\left.x+\frac{1}{x}>x\right)$, and for $x>0 x+\frac{1}{x} \geqslant 2$, while for $x<0(x \neq-1)$ it is below the line $y=-2$. When $x=1$ and $x=-... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,985 |
41(1117). Construct the graph of the function:
a) $y=\frac{2 x+3}{x}$;
c) $y=\frac{12}{x-4}$
b) $y=\frac{4-5 x}{x}$;
d) $y=-\frac{6}{x+3}$. | Solution. a) Since $y=\frac{2 x+3}{x}=2+\frac{3}{x}$, the graph of the function will be a hyperbola $y=\frac{3}{x}$, shifted by a parallel translation along the y-axis by 2 units upwards (Fig. 28).
b) $y=\frac{4-5 x}{x}=\frac{4}{x}-5$. The graph of the function will be a hyperbola $y=\frac{4}{x}$, shifted by a paralle... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,986 |
42(1118). Prove that the graph of the equation $x y - 2 x + 3 y - 6 = 0$ is a pair of intersecting lines.
120 | Proof. Let's represent the left side of the equation as a product of two factors. We have:
$$
x y-2 x+3 y-6=y(x+3)-2(x+3)=(x+3)(y-2) \text{. }
$$
The graph of the equation $(x+3)(y-2)=0$ is a pair of intersecting lines $x=-3$ and $y=2$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,987 |
43(1119). Prove that the graph of the equation $(y-2) \times$ $\times(y+3)=0$ is a pair of parallel lines. | The principle of zero product states that the product is equal to zero when at least one of the factors is equal to zero, while the other has a defined value.
If $y-2=0$, then $y=2$.
If $y+3=0$, then $y=-3$.
The graph of this equation is a pair of parallel lines $y=2$ and $y=-3$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,988 |
46(1122). Several points are marked on a plane, no three of which lie on the same line. A line is drawn through every two points. How many points are marked on the plane if it is known that a total of 45 lines have been drawn? | Solution. The problem reduces to solving the equation $\frac{x(x-1)}{2}=45$ in natural numbers, where $x$ is the number of points marked on the plane. We have $x^{2}-x-90=0$, from which $x_{1}=10, x_{2}=-9$. Only the root $x=10$ satisfies the condition of the problem.
Answer: 10 points.
$\triangle 47(969)$ '. Prove t... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,989 |
49(971). A two-digit number, when added to the number written with the same digits but in reverse order, gives the square of a natural number. Find all such two-digit numbers. | Solution. Let's have a two-digit number $\overline{a b}=10 a+b$. Then, according to the condition, $\overline{a b}+\overline{b a}=10 a+b+10 b+a=11(a+b)$. Since 11 is a prime number, the number $11(a+b)$ can only be a square of a natural number if $a+b=11$. Considering that $a$ and $b$ are digits, from the condition $a+... | 29,38,47,56,65,74,83,92 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,990 |
50 (973). A sixty-digit number is written using 30 zeros and 30 ones. Prove that it cannot be a square of a natural number. | Solution. The sum of the digits of the given number is 30, so it is divisible by 3 and not divisible by 9 (according to the divisibility rules for 3 and 9). Since a natural number greater than 1 can be a square of a natural number only if its canonical factorization includes prime factors with only even exponents, the ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,991 |
51 (974). What is the last digit of the difference $43^{43}-17^{17}$? | Solution. Let's determine the last digits of various powers of the numbers 43 and 17. The number 43 ends with the digit $3, 43^2$ ends with the digit $9, 43^3$ ends with the digit $7, 43^4$ ends with the digit 1. Therefore, the last digits in subsequent powers of the number 43 will repeat: $3, 9, 7, 1$, and so on. Sinc... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,992 |
52 (975). Find a two-digit number, the doubled sum of whose digits is equal to their product. | Solution. Let the desired two-digit number be $\overline{a b}=10 a+b$. By the condition $2(a+b)=a b$. To solve this equation, we express one of the unknowns (for example, $a$) in terms of the other ($b$). We will have:
$$
\begin{aligned}
& 2 a-a b=-2 b, a(2-b)=-2 b \\
& a=\frac{2 b}{b-2}=\frac{2(b-2)+4}{b-2}=2+\frac{4... | 63,44,36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,993 |
55 (978). There are 4 balls of different masses. Using a balance scale without weights, how many weighings are needed to arrange these balls in order of decreasing mass? | Solution. Let $m_{A}, m_{B}, m_{C}, m_{D}$ be the masses of four different balls. We will use a graph to show how the weighings are performed (the arrow points from the heavier ball to the lighter one).
Suppose, for example, after two weighings, we have $m_{A}>m_{B}$ and $m_{C}>m_{D}$ (Fig. $29, a$ ). Next, we will co... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,994 |
56 (979). After writing the test, the students Volodya, Sasha, and Petya reported at home:
Volodya: “I got a ‘5’.”
Sasha: “I got a ‘3’.”
Petya: “I did not get a ‘5’.”
After the test was checked, it turned out that one of the boys received a ‘4’, another a ‘3’, and the third a ‘5’. What grade did each receive, given... | Solution. If Vovochka had made a mistake (received not "5"), and the others had correctly named their grades, then Sasha would have received "3", Petya - "4". It turns out that Vovochka received the same grade as Sasha or Petya, which contradicts the condition of the problem: all received different grades. Therefore, V... | Vovochka-5,Sasha-4,Petya-3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,995 |
57 (980). Two people play such a game: the first one calls out any integer from 1 to 10 inclusive, the second one mentally adds another number from the first ten to this number and calls out the sum. To this sum, the first one adds again some number from the first ten and calls out the sum, and so on. The one who calls... | Solution. To win (by naming the number 100), one needs to force the opponent to name a natural number from the interval $[90 ; 99]$. For this, one needs to name the number 89 oneself. To be able to name the number 89 oneself, one needs to force the opponent to name some natural number from the interval $[79 ; 88]$, for... | 1,12,23,34,45,56,67,78,89 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,996 |
58 (981). In the lower left corner of a chessboard stands a piece. In one move, it is allowed to move to one of the three adjacent squares: to the right, up, or diagonally to the right-up. The player who occupies the upper right corner wins (the moves are made alternately). Who wins with correct play: the one who start... | Solution. The problem boils down to moving a figure from a cell with odd coordinates $(1 ; 1)$ to a cell with even coordinates $(8 ; 8)$ in a specified manner. Since from any cell, a move to the right, up, or diagonally to the top-right can increase one or both coordinates by one, the player who starts the game can alw... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,997 |
59(982). In the cabinet, there are books that need to be packed. If they are bundled in groups of 4, 5, or 6, there will always be one extra book left each time, but if they are bundled in groups of 7, there will be no extra books left. What is the number of books in the cabinet, given that there are no more than 400 b... | Solution. The problem reduces to finding a natural number that is a multiple of seven, not exceeding 400, which gives a remainder of 1 when divided by 4, 5, and 6. This number, obviously, has the form $60 n+1$, where 60 is the least common multiple of the numbers $4,5,6, n \in \boldsymbol{N}$, $n<7$. By trial, we find ... | 301 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,998 |
60(983). There is a sheet of paper. It is cut into 4 pieces, then some (or all) of the resulting pieces are again cut into 4 pieces, and so on. Prove that it is impossible to obtain 50 pieces of the sheet in this way. | Solution. It is easy to understand that after each cut, the number of pieces of the sheet increases by 3. Thus, after $m$ cuts, the number of pieces will be $1+3 m$, where $m \in \boldsymbol{N}$. Since $1+3 m \neq 50$, it is impossible to obtain 50 pieces of the sheet in the manner specified in the problem.
$61(984)$.... | 7875 | Combinatorics | proof | Yes | Yes | olympiads | false | 35,999 |
62 (985). In a chess tournament, more than 9 but fewer than 25 grandmasters and masters participated. At the end of the tournament, it turned out that each participant scored half of their points against grandmasters. How many people participated in the tournament? How many of them were masters? | Solution. Let $n$ be the number of chess players in the tournament, of which $m$ are grandmasters, and thus $n-m$ are masters. Therefore, all participants played $\frac{n(n-1)}{2}$ games and scored $\frac{n(n-1)}{2}$ points (both games and points). Among them, the grandmasters played $\frac{m(m-1)}{2}$ games (and score... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,000 |
63 (996). Prove that among all rectangles with a given perimeter, the square has the largest area. | Solution. I method. Let the perimeter of the rectangle be $p$, and the length of one of its sides be $a$. Then the length of the other side, not equal to the first, is $\frac{p}{2}-a$, and the area of the rectangle is $a\left(\frac{p}{2}-a\right)$.
If the rectangle with perimeter $p$ is a square, then the length of it... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,001 |
65 (998). An alloy consists of zinc and copper, entering into it in a ratio of $1: 2$, and another alloy contains the same metals in a ratio of $2: 3$. From how many parts of both alloys can a third alloy be obtained, containing the same metals in a ratio of 17:27? | S o l u t i o n. Let the third alloy contain $x$ kg of the first alloy and $y$ kg of the second alloy. We need to find the ratio $\frac{x}{y}$. Since zinc and copper are in the ratio $1:2$ in the first alloy and in the ratio $2:3$ in the second alloy, $x$ kg of the first alloy contains $\frac{1}{3} x$ (kg) of zinc and ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,003 |
67 (1001). A tourist left from $A$ to $B$ at a speed of 4 km/h. An hour later, a second tourist, who walked 5 km per hour, set out from $A$ in pursuit. Another hour later, a cyclist set out from $A$, who, overtaking one tourist, overtook the other after 10 minutes. Find the speed of the cyclist. | S o l u t i o n. Let the speed of the cyclist be $v$ kilometers per hour. Then, the first tourist, who walked 8 km in 2 hours, will be caught by the cyclist in $\frac{8}{v-4}$ hours. The second tourist will be caught by the cyclist in $\frac{5}{v-5}$ hours. If the cyclist first overtakes the second tourist, we have the... | 6.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,004 |
68(1002). Two speed skaters started simultaneously on a 10000 m distance. The first, moving at a speed exceeding the second's by $\frac{1}{3} \mathrm{m} / \mathrm{s}$, completed each lap 2 seconds faster than the second and after 10 minutes overtook the second by half a lap. Find the speeds of these skaters and calcula... | Solution. Let the speed of the second skater be $v$ meters per second, then the speed of the first is $\left(v+\frac{1}{3}\right)$ meters per second. Since the first skater overtakes the second by half a lap in 10 minutes (600 seconds), half the circumference of the track is 200 meters $\left(\frac{1}{3} \cdot 600\righ... | 8\frac{1}{3}\mathrm{}/\mathrm{},8\mathrm{~}/\mathrm{};20\text{,}20\text{}50\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,005 |
69(1003). A cyclist and a motorcyclist are moving towards an intersection along two mutually perpendicular roads. At a certain moment, the cyclist is 8 km away, and the motorcyclist is 15 km away from the intersection. After how many minutes will the distance between them be 5 km, if the cyclist's speed is $\frac{1}{3}... | Solution. Let $OA=8$ km, $OB=15$ km (Fig. 30). If the distance of 5 km between the motorcyclist and the cyclist will be in $t$ minutes $\left(A_{1} B_{1}=5\right.$ km), then $A A_{1}=\frac{1}{3} t$ (km), $B B_{1}=t$ (km). Then $O A_{1}=O A-\frac{1}{3} t=$
. From two settlements $A$ and $B$, two tourists set out towards each other at the same time. Upon meeting, it turns out that the tourist who left from $A$ has walked 2 km more than the second tourist. Continuing their movement at the same speed, the first tourist arrives in $B$ after 1 hour 36 minutes, and the... | Solution. Method I. Let the speed of the first pedestrian (who left from $A$) be $v_{1}$ kilometers per hour, and the speed of the second pedestrian be $v_{2}$ kilometers per hour. Then, before the meeting, the first pedestrian walked $2 v_{1}$ kilometers, and the second pedestrian walked $2 v_{2}$ kilometers, and afte... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,007 |
73(1007). A master and his apprentice were supposed to complete a job by a certain deadline. However, when half of the work was done, the apprentice fell ill, and the master, left alone, finished the job 2 days late. How many days would it take for each of them to complete the entire job working alone, if the master wo... | Solution. Let the master be able to complete the entire job in $x$ days, then the apprentice - in $(x+5)$ days. In one day, the master completed $\frac{1}{x}$ of the work, the apprentice completed $\frac{1}{x+5}$ of the work; when working together, the master and the apprentice completed $\frac{2 x+5}{x(x+5)}$ of the w... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,008 |
74 (1008). A tourist who set out from $A$ to $B$ walked the first half of the distance at a speed of $v_{1}$, and the second half at a speed of $v_{2}\left(v_{1} \neq v_{2}\right)$. On the return trip, he walked half of the total time at a speed of $v_{1}$, and the other half at a speed of $v_{2}$. In which case did th... | Solution. I method. Let the time spent by the tourist on the journey from $B$ to $A$ be $2 t$. Then half of the time ($t$) he walked at a speed of $v_{1}$, and the second half of the time at a speed of $v_{2}$. The entire distance, therefore, is $v_{1} t + v_{2} t$, and half the distance is $\frac{v_{1} t + v_{2} t}{2}... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,009 |
76(1010). A collective farm team must harvest a potato crop by a certain deadline. After 60% of the potatoes were harvested, a combine was sent to assist the team, which reduced the harvesting time by 5 days. How many days would it have taken to harvest the potatoes without the combine's help, if it is known that the c... | Let's solve it. Let the brigade need $t$ days to harvest potatoes without the help of a combine, then the combine needs $(t-8)$ days to complete the entire work.
In one day, the brigade can complete $\frac{1}{t}$ of the entire work, and the combine can complete $\frac{1}{t-8}$ of the entire work in one day. With the h... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,010 |
79. Calculate the sum
$$
1^{2}-2^{2}+3^{2}-4^{2}+\ldots+99^{2}-100^{2}+101^{2}
$$ | $$
\begin{gathered}
1^{2}-2^{2}+3^{2}-4^{2}+\ldots+101^{2}= \\
=1+(101^{2}-100^{2})+(99^{2}-98^{2})+\ldots+(3^{2}-2^{2})= \\
=201+197+\ldots+9+5+1= \\
=\frac{1}{2}(201+1) \cdot 51=101 \cdot 51=5151
\end{gathered}
$$
Answer: 5151. | 5151 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,011 |
80. Prove that the sum of natural numbers from 1 to 1000 is divisible by 143. | Solution. The sum of the first 1000 natural numbers can be written as $(1+1000)+(2+999)+\ldots+(500+501)$. Each term of this sum is equal to $1001=143 \cdot 7$, i.e., divisible by 143, and therefore the entire sum is divisible by 143. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,012 |
81. Prove that the sum $2+2^{2}+2^{3}+\ldots+2^{99}+2^{100}$ is divisible by 3. | Solution. Grouping the terms of the given sum in pairs, we will have:
$$
\begin{gathered}
\left(2+2^{2}\right)+\left(2^{3}+2^{4}\right)+\ldots+\left(2^{99}+2^{109}\right)= \\
=2(1+2)+2^{3}(1+2)+\ldots+2^{99}(1+2)=3\left(2+2^{3}+\ldots+2^{99}\right)
\end{gathered}
$$
Therefore, the given sum is divisible by 3, which i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,013 |
82. Prove the inequality
$$
\sqrt{6+\sqrt{6+\sqrt{6+\ldots+\sqrt{6+\sqrt{6}}}}}<3
$$ | Solution. If the last digit 6 in the left part of the given inequality is replaced by 9, then after extracting all roots we will get the number 3. Therefore, the left part is less than 3, which was to be proved. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,014 |
83. Prove that if $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$, then $a^{2} b^{2} c^{2}=1$ or $a=b=c$.
138 | Solution. From the condition, it follows that
$$
a-b=\frac{b-c}{b c}, b-c=\frac{c-a}{a c}, c-a=\frac{a-b}{a b} .
$$
Multiplying the obtained equalities, we get:
$$
(a-b)(b-c)(c-a)=\frac{(b-c)(c-a)(a-b)}{a^{2} b^{2} c^{2}}
$$
If $a \neq b \neq c$, then $a^{2} b^{2} c^{2}=1$.
If $a^{2} b^{2} c^{2} \neq 1$, then $(a-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,015 |
84. Nine identical books cost 11 rubles and some kopecks, while 13 such books cost 15 rubles and some kopecks. Determine the exact cost of one book | S o l u t i o n. It is clear that each book costs 1 r. $m$ k., where $m$ is an integer, and $m<100$. From the first condition of the problem, it follows that $200<9 m<300$, i.e., $23 \leqslant m \leqslant 33$. From the second condition, $200<13 m<300$, i.e., $16 \leqslant m \leqslant 23$. Therefore, $m=23$.
A n s w e ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,016 |
85. The percentage of VII grade students attending gymnastics classes is between 2.9 and $3.1 \%$. Determine the smallest possible number of students in this class. | Solution. If there are $x$ students in the class and $y$ of them do gymnastics, then by the condition $2.9 < 100 \cdot \frac{y}{x} < 3.1$, or $29x < 1000y < 31x$. If $x \leqslant 32$, then $31x \leqslant 992 \leqslant 1000y$, so $x \geqslant 33$. Since $29 \cdot 33 < 1000 \cdot y < 31 \cdot 33$, the number of students ... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,017 |
87. Prove that the sum of all natural numbers, in the notation of which each of the digits from 1 to 9 appears exactly once, is divisible by 999999999. | Solution. Since all digits from 1 to 9 are equally important in this problem, each digit appears the same number of times in each place; let's denote this number by $k$. Then the sum of the digits in each place is $k(1+2+3+\ldots+9)=$ $=45k$, and the sum of all numbers is therefore $45k\left(1+10+10^{2}+\right.$ $\left... | 5k\cdot999999999 | Number Theory | proof | Yes | Yes | olympiads | false | 36,018 |
89. Calculate the value of the expression $\frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}$, given that $10 a^{2}-3 b^{2}+5 a b=0$ and $9 a^{2}-b^{2} \neq 0$. | Solution. Since $5 a b=3 b^{2}-10 a^{2}$, then
$$
\begin{aligned}
& \quad \frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}=\frac{3 a^{2}+15 a b-6 b^{2}}{9 a^{2}-b^{2}}=\frac{3 a^{2}+3\left(3 b^{2}-10 a^{2}\right)-6 b^{2}}{9 a^{2}-b^{2}}= \\
& \text { Answer: }-3 . \\
& =\frac{-3\left(9 a^{2}-b^{2}\right)}{9 a^{2}-b^{2}}=-3 .
\... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,019 |
90. If the first digit of a four-digit number, which is a perfect square, is decreased by 3, and the last digit is increased by 3, the result is also a perfect square. Find this number. | S o l u t i o n. Let the given number be $a^{2}$, and the modified number $b^{2}$, then $a^{2}-3000+3=b^{2}$, or $a^{2}-b^{2}=2997$, from which $(a-b) \times$ $\times(a+b)=2997=3 \cdot 3 \cdot 3 \cdot 3 \cdot 37$. Clearly, $a+b$ is greater than 37, and from the problem statement, it follows that $a+b$ is less than 200 ... | 4761 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,020 |
91. Prove the inequality
$$
\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \ldots \cdot \frac{120}{121}>\frac{1}{11}
$$ | Solution. Let $x=\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \ldots \cdot \frac{120}{121}$, and $y=\frac{1}{2} \cdot \frac{3}{4} \times$ $\times \frac{5}{6} \ldots \cdot \frac{119}{120}$. Since $\frac{2}{3}>\frac{1}{2} ; \frac{4}{5}>\frac{3}{4} ; \frac{6}{7}>\frac{5}{6} ; \ldots ; \frac{120}{121}>\frac{119}{1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,021 |
92. If the first digit of a three-digit number is increased by $n$, and the second and third digits are decreased by $n$, then the resulting number will be $n$ times the original number. Find the number $n$ and the original number. | Solution. Let $100 x+10 y+z$ be the original number. According to the condition, we have the equality
$$
100(x+n)+10(y-n)+(z-n)=n(100 x+10 y+z)
$$
from which
$$
100 x+10 y+z=\frac{89 n}{n-1}
$$
Since 89 is a prime number, either $n-1$ must be equal to 1, or $n$ must be divisible by $n-1$. In both cases, we arrive a... | 178 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,022 |
93. Prove that for any integers $m$ and $n$
$$
(5 m+3 n+1)^{5}(3 m+n+4)^{4}
$$
is divisible by 16. | Solution. If numbers $m$ and $n$ have the same parity, then $(3 m+n+4)$ is an even number, and, consequently, the number $(3 m+n+4)^{4}$ is divisible by 16. If $m$ and $n$ have different parity, then $5 m+3 n+1$ is an even number, and, consequently, the number $(5 m+3 n+1)^{4}$ is divisible by 16. Therefore, the given ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,023 |
94. Prove that there do not exist integers $x$ and $y$ for which the equation
$$
\sqrt{x^{2}+x+1}+\sqrt{y^{2}-y+1}=11
$$
holds. | Solution. Since for any integer $x$ the numbers $x^{2}$ and $x$ are either both even or both odd, the sum $x^{2}+x$ is always even. Therefore, the expression under the first square root is an odd number, and consequently, the square root of it is an odd number.
Similarly, it can be proven that the second term on the l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,024 |
95. Can a number of the form $n^{2}+9 n+12$, where $n \in \boldsymbol{N}$, be divisible by 121? | Solution. Let's write the equality $n^{2}+9 n+12=(n-1) \times$ $\times(n+10)+22$. If $n-1$ is not divisible by 11, then $n+10(n+10=$ $=n-1+11$ ) is also not divisible by 11. In this case, the right-hand side of the obtained equality is not divisible by 11, and even more so not by 121. If, however, $n-1$ is divisible by... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,025 |
96. The number 392 was divided by a natural number $a$, and from the quotient, $a$ was subtracted. The same operation was performed with the resulting difference, and the same operation was performed again with the new result. The final answer was the number $-a$. What is the value of $a$? | Solution. According to the problem, we form the equation
$$
\left(\left(\frac{392}{a}-a\right): a-a\right): a-a=-a
$$
from which
$$
\begin{aligned}
\left(\frac{392}{a}-a\right): a-a & =0, \\
a^{3}+a^{2}-392=0, a^{2}(a+1) & =392, \text { where } a \in \boldsymbol{N} .
\end{aligned}
$$
We have that 392 is divisible b... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,026 |
97. Find the proper fraction that will increase by 3 times if its numerator is cubed, and 3 is added to the denominator. | Solution. Let $\frac{a}{b}$ be a proper fraction (where $a$ and $b$ are natural numbers, $a<b$), such that $\frac{a^{3}}{b+3}=3 \cdot \frac{a}{b}$, from which $\frac{a^{2}}{b+3}=\frac{3}{b}$, $a^{2} b=3 b+9, b\left(a^{2}-3\right)=9$. From this, it is clear that the number $b$ must be a divisor of the number 9, i.e., $b... | \frac{2}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,027 |
99. When dividing a five-digit number, consisting of identical digits, by a four-digit number, also consisting of identical digits, the quotient was 16 and some remainder. After discarding one digit from both the dividend and the divisor, the quotient remained unchanged, but the remainder decreased by 2000. Find these ... | Solution. Let the required numbers be $\overline{a a a a a}$ and $\overline{b b b b}$. According to the problem,
$$
\begin{aligned}
& \overline{a a a a a}=\overline{b b b b} \cdot 16+r \\
& \overline{a a a a}=\overline{b b b} \cdot 16+(r-2000)
\end{aligned}
$$
where $r$ is the remainder of the division of $\bar{a} \o... | 555553333 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,028 |
101. Given two quadratic equations $x^{2}-x+m=0, x^{2}-x+$ $+3 m=0, m \neq 0$. Find the value of $m$ for which one of the roots of the second equation is twice the root of the first equation. | Solution. Let $x_{0}$ be the root of the first equation, and $2 x_{0}$ be the root of the second. Then the following numerical equalities hold:
$$
\begin{gathered}
x_{0}^{2}-x_{0}+m=0 \\
4 x_{0}^{2}-2 x_{0}+3 m=0
\end{gathered}
$$
Subtracting the second equality from three times the first, we get $-x_{0}^{2}-x_{0}=0$... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,029 |
105. In a football championship, 16 teams participated. A team receives 2 points for a win; in the case of a draw in regular time, both teams take a series of penalty kicks, and the team scoring more goals gets one point. After 16 rounds, all teams accumulated 222 points. How many matches ended in a draw in regular tim... | S o l u t i o n. In each round, 8 matches are played, so after 16 rounds, 128 matches were played $(8 \cdot 16=128)$. If all matches had ended in wins during regular time, the teams would have accumulated 256 points $(128 \cdot 2=256)$. Consequently, due to draws, 34 points were lost ( $256-222$ ), meaning 34 matches e... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,030 |
106. Several chess players participated in a match tournament, where each participant played several games against every other participant. How many rounds did this competition consist of, if a total of 224 games were played? | Solution. Let $x$ be the number of participants in the competition, and $y$ be the number of rounds. In one round, each chess player plays $(x-1)$ games, and all participants together play $\frac{x(x-1)}{2}$ games. Therefore, the total number of games played in the competition is $\frac{x(x-1) y}{2}$. Consequently, $x(... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,031 |
107. An Aeroflot cashier needs to deliver tickets to five groups of tourists. Three of these groups are staying at the hotels "Druzhba", "Rossiya", and "Minsk". The address of the fourth group will be provided by the tourists from "Rossiya", and the address of the fifth group will be provided by the tourists from "Mins... | Solution. First, let's calculate the number of ways to visit the hotels if there are no restrictions on the order of the visits. Then, the first hotel can be chosen in 5 ways, the second in 4 ways, so the first two hotels can be chosen in 20 ways; the third hotel can be chosen in 3 ways, so for the first three hotels, ... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,032 |
108. Bus tickets have numbers from 000001 to 999999. A number is considered lucky if the first three digits are odd and different, the second three digits are even, and 7 and 8 do not stand next to each other. How many different lucky numbers exist? | Solution. The number of numbers satisfying the first two conditions of the problem (but perhaps not satisfying the third condition) can be calculated as follows: the first digit of such a number can be chosen in five ways, the second in four ways, the third in three ways, and each subsequent digit in five ways, so the ... | 7200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,033 |
109. From the digits $1,2,3,4$, all possible positive decimal fractions with one, two, or three decimal places are formed, each containing each of these digits exactly once. Find the sum of these fractions. | Solution. From the digits $1,2,3,4$, 24 different integers can be formed $(4 \cdot 3 \cdot 2=24)$; in these numbers, each of the given digits appears 6 times in each place value $(24: 4=6)$. Therefore, the sum of the digits in each place value is $60((1+2+3+4) \cdot 6=60)$. Consequently, the sum of all these 24 numbers... | 7399.26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,034 |
110. Each of the 9 forwards and midfielders of the football team "Dynamo" scored at least one goal in the championship, and together they scored 47 goals. Prove that at least two of them scored the same number of goals, given that the most prolific forward scored 12 goals. | Solution. Suppose that all midfielders and forwards of the team scored a different number of goals. In this case, 8 of them scored 35 goals ( $47-12=35$ ), but even if they scored the smallest possible number of goals, the total number of goals scored by them is $36(1+2+\ldots+8=36)$, i.e., more than 35. We have arrive... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,035 |
111. Can a square of size $6 \times 6$ be cut into rectangles of size $1 \times 4$ with sides parallel to the sides of the square | Solution. Let the given square be cut into the required rectangles, and let the shorter side of " $a$ " of them, and the longer side of " $b$ " of them be parallel to a fixed side of the square. Then $a+4 b=6, 4 a+b=6$. After adding these equations, we get $5 a+5 b=12, 5(a+b)=12$. We have obtained that $a+b$ is a fract... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,036 |
112. Prove that
$$
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(n-1) n}=\frac{n-1}{n}
$$ | Solution. Replace each of the addends in the left part of the equation, which is the product of two fractions, with the difference of these fractions. We will have:
$$
\begin{gathered}
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(n-1) n}= \\
=1 \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{... | \frac{n-1}{n} | Algebra | proof | Yes | Yes | olympiads | false | 36,037 |
113. For what least natural $n$ is each of the fractions
$$
\frac{7}{n+9}, \frac{8}{n+10}, \ldots, \frac{31}{n+33}
$$
irreducible
146 | Solution. All the fractions have the form $\frac{k}{k+(n+2)}$, and such a fraction will be irreducible if the numbers $k$ and $n+2$ have no common divisors. Thus, $n+2$ must be coprime with the numbers $7,8, \ldots, 31$, and the smallest natural number with this property is 37. Therefore, the desired number $n$ is 35.
... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,038 |
115. Find the minimum value of the expression $\frac{(4+x)(1+x)}{x}$, where $x$ is a positive number. | Solution. Since
$$
\frac{(4+x)(1+x)}{x}=\frac{4}{x}+x+5=2\left(\frac{2}{x}+\frac{x}{2}\right)+5
$$
and $\frac{2}{x}+\frac{x}{2} \geqslant 2$ (for $c>0$, the inequality $c+\frac{1}{c} \geqslant 2$ holds, i.e., 2 is the minimum value of the sum of two reciprocals (see 627 from [2])), the expression $2\left(\frac{2}{x}+... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,039 |
116. Find the minimum value of the fraction $\frac{x^{2}-3 x+3}{1-x}$, if $x<1$. | Solution. I method. The given fraction can be represented as $\frac{x^{2}}{1-x}+3$. For $x<1$, the first term is non-negative and equals 0 only when $x=0$. Since 0 is included in the set of values for $x$, the given fraction achieves its minimum value of 3 when $x=0$.
II method. Representing the given fraction as the ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,040 |
117. Find the minimum value of the expression
$$
(x-5)(x-1)(x-6)(x-2)+9
$$ | S o l u t i o n. We have:
$$
\begin{gathered}
y=(x-5)(x-1)(x-6)(x-2)+9=\left(x^{2}-7 x+10\right) \times \\
\times\left(x^{2}-7 x+6\right)+9=\left(x^{2}-7 x\right)^{2}+16\left(x^{2}-7 x\right)+69= \\
=\left(x^{2}-7 x+8\right)^{2}+5
\end{gathered}
$$
From here, $y_{\text {min }}=5$, when $x^{2}-7 x+8=0$, i.e., when
$$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,041 |
118. What is the largest and the smallest value that the ratio of a two-digit number to the sum of its digits can take? | S o l u t i o n. Let $x, y$ be the first and second digits of a two-digit number, respectively. Since $x \neq 0$, denoting the considered ratio by $t$, we have:
$$
t=\frac{10 x+y}{x+y}=1+\frac{9 x}{x+y}=1+\frac{9}{1+\frac{y}{x}}
$$
From this, it is clear that $t$ is greater the smaller the ratio $\frac{y}{x}$. Theref... | 10;1.9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,042 |
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