problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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
$$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
121. Find the value of the difference
$$
A=\sqrt{|12 \sqrt{5}-29|}-\sqrt{12 \sqrt{5}+29}
$$ | S o l u t i o n. Noting that $|12 \sqrt{5}-29|=29-12 \sqrt{5}$, we get:
$$
A=\sqrt{29-12 \sqrt{5}}-\sqrt{29+12 \sqrt{5}}
$$
Since
$$
29-12 \sqrt{5}=20-12 \sqrt{5}+9=(2 \sqrt{5}-3)^{2}
$$
and $29+12 \sqrt{5}=(2 \sqrt{5}+3)^{2}$, we obtain that
$$
A=2 \sqrt{5}-3-(2 \sqrt{5}+3)=-6
$$
A n s w e r: -6 . | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
123. How many right-angled triangles are there with sides of integer lengths, if one of the legs of these triangles is equal to 15? | Solution. If $x$ is the hypotenuse and $y$ is the unknown leg of a right triangle, then $x^{2}-y^{2}=15^{2} ;(x-y)(x+y)=$ $=3 \cdot 3 \cdot 5 \cdot 5$
Since $(x-y)$ and $(x+y)$ are natural numbers, and $x+y > x-y$, there are only 4 cases:
$$
\left\{\begin{array} { l }
{ x - y = 1 } \\
{ x + y = 225 }
\end{array} \le... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
130. Find the smallest value of the expression $\left|36^{m}-5^{n}\right|$, where $m$ and $n$ are natural numbers. | Solution. The difference $36^{m}-5^{n}$ ends with the digit 1, and the difference $5^{n}-36^{m}$ ends with the digit 9. We will prove that $\left|36^{m}-5^{n}\right|$ cannot be equal to either 1 or 9.
From the equation $36^{m}-5^{n}=1$, it follows that $\left(6^{m}-1\right)\left(6^{m}+1\right)=5^{n}$, and then the pow... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
131. A natural number $n$ is the product of four consecutive natural numbers, each greater than 5. Determine the maximum number of possible last digits of $n$, given that its last digit is not 0. | Solution. Among four consecutive numbers whose product is equal to $n$, there are two even numbers, and therefore the number $n$ is even. Consequently, none of these numbers is divisible by 5 - otherwise, the number $n$ would end in the digit 0. Thus, we have:
$$
\begin{gathered}
n=(5 k+1)(5 k+2)(5 k+3)(5 k+4)=\left(2... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
132. For which $x$ and $y$ is the number $x x y y$ a square of a natural number? | Solution. The number $\overline{x x y y}$ can be represented as $11 \cdot \overline{x 0 y}$ $\overline{x x y y}=1000 x+100 x+10 y+y=1100 x+11 y=11(100 x+y)=11 \times$ $\times x 0 y)$. Therefore, it will be a square only in the case when $x 0 y$ is divisible by 11 and the quotient is also a square of a natural number. B... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
139. Can you measure out 10 liters of water using two containers of 9 and 11 liters? | Solution. Fill the 11-liter vessel (vessel $A$) with water, then use this water to fill the 9-liter vessel (vessel $B$), empty $B$, and transfer the remaining 2 liters from $A$ to $B$. After this, vessel $A$ will be empty, and vessel $B$ will have 2 liters. Repeating these operations three more times, we will end up wi... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
140. Find the greatest common divisor of all nine-digit numbers composed of the digits $1,2,3,4,5,6,7,8,9$ (without repetition). | Solution. Each of the 9! such numbers is divisible by 9, since the sum of the digits of each number, which is 45, is divisible by 9. We will prove that 9 is the greatest common divisor of these numbers.
Let the greatest common divisor of the considered numbers be greater than 9. Denote it by $d$. Then the difference b... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
141. Find the greatest common divisor of all six-digit numbers composed of the digits $1,2,3,4,5,6$ (without repetition). | Re shenie. Each of the 6! numbers is divisible by 3 (since the sum of the digits is 21). We will show that 3 is the greatest common divisor of these numbers.
Using the divisibility rules of numbers, it is easy to show that the greatest common divisor of the considered six-digit numbers cannot be equal to $2, 4, 5, 6, ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
142. Which of the numbers is larger: $2^{3^{2^{3}}}$ or $3^{2^{3^{2}}}$? | Solution. Since, on the one hand, $2^{3^{8}}>2^{3^{7} \cdot 2}=4^{3^{7}}>3^{3^{7}}$, and on the other hand, $3^{7}>3^{6}>729>2^{9}$, then $3^{3^{7}}>3^{2^{9}}$. Therefore, the first of the given numbers in the problem is greater than the second.
143(1171)'. Two cars, moving along a circular road at constant speeds in ... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
144(1172). Two athletes are running on the same closed track. The speed of each is constant, and one takes 5 seconds less than the other to run the entire track. If they start running simultaneously from the same starting line in the same direction, they will be side by side after 30 seconds. How long will it take for ... | Solution. Let $s$ (m) be the length of the closed track, $t$ (s) the time it takes for the first athlete to run the entire track. Then the speed of the first athlete is $\frac{s}{t}(\mathrm{~m} / \mathrm{s})$, and the speed of the second athlete is $\frac{s}{t+5}(\mathrm{m} / \mathrm{s})$.
According to the problem, we... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
145(1173). A pedestrian and a cyclist set off from point $A$ to point $B$ simultaneously. At point $B$, the cyclist turns back and meets the pedestrian 20 minutes after the start of the journey. Without stopping, the cyclist continues to point $A$, turns back, and catches up with the pedestrian 10 minutes after the fir... | Solution. Let the speed of the pedestrian be $v_{1}$ meters per minute, the speed of the cyclist be $v_{2}$ meters per minute, and $t$ minutes be the time the pedestrian spent from the moment of the second meeting with the cyclist until arriving at point $B$.
According to the problem, we form a system of equations:
$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
146(1174). A motorcyclist left point $A$ for point $B$ and at the same time a cyclist left point $B$ for point $A$. The motorcyclist arrived at point $B$ 2 hours after meeting the cyclist, while the cyclist arrived at point $A$ 4.5 hours after meeting the motorcyclist. How many hours were the motorcyclist and the cycli... | Solution. Let the motorcyclist and the cyclist have been on the road for $t$ hours before meeting. If $v_{1}$ kilometers per hour is the speed of the motorcyclist, and $v_{2}$ kilometers per hour is the speed of the cyclist, then we have the system of equations:
$$
\left\{\begin{array}{l}
v_{1}=\frac{v_{2} t}{2} \\
v_... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4(1132). For what values of $a$ do the quadratic trinomials $x^{2}+$ $+a x+1$ and $x^{2}+x+a$ have a common root? | Solution. Let $x=x_{1}$ be the common root of the given quadratic trinomials. Then $x_{1}^{2}+a x_{1}+1=0$ and $x_{1}^{2}+x_{1}+a=0$, i.e., $x_{1}^{2}+a x_{1}+1=x_{1}^{2}+x_{1}+a, \quad a \cdot\left(x_{1}-1\right)=x_{1}-1, \quad\left(x_{1}-1\right) \cdot(a-1)=0$. From this, $a=1$ or $x_{1}=1$.
When $a=1$, each of the ... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5 (1133). For what value of $a$ does the sum of the squares of the roots of the quadratic trinomial $x^{2}-(a-2) x-a-1$ take the smallest value? | Solution. Based on Vieta's theorem, we have $x_{1}+x_{2}=$ $=a-2, x_{1} \cdot x_{2}=-a-1$. Then
$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(a-2)^{2}-2(-a-1)= \\
=a^{2}-2 a+6=(a-1)^{2}+5
\end{gathered}
$$
Since $(a-1)^{2} \geqslant 0$, $x_{1}^{2}+x_{2}^{2}$ takes its minimum val... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
22(1150). How many hours can each of the three workers complete the work if the productivity of the third worker is equal to half the sum of the productivities of the first and second workers? It is known that if the third worker worked for 48 hours, the first would need 10 hours to finish the work, and the second woul... | Solution. Let the entire work be denoted by $A$, the labor productivity of the first worker by $x_{1}$, and the second worker by $x_{2}$. Then, according to the condition, the productivity of the third worker is $\frac{x_{1}+x_{2}}{2}$ and $10 x_{1}=$ $=15 x_{2}$, i.e., $x_{1}=\frac{3}{2} x_{2}, x_{2}=\frac{2}{3} x_{1}... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
46(1174). Prove that the maximum value of the expression $\sin x + \sqrt{2} \cos x$ is $\sqrt{3}$. | Solution. Let's factor out $\sqrt{3}$ in the given expression. We will have:
$$
\sin x+\sqrt{2} \cos x=\sqrt{3}\left(\frac{1}{\sqrt{3}} \sin x+\sqrt{\frac{2}{3}} \cos x\right)
$$
Since $\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\sqrt{\frac{2}{3}}\right)^{2}=1$, there exists an angle $\alpha$ such that $\cos \alpha=\f... | 28 | Algebra | proof | Yes | Yes | olympiads | false |
60(1188). Find the smallest four-digit number which, after being multiplied by 21, becomes a square of a natural number. | Solution. Since $21=3 \cdot 7$ (3 and 7 are prime numbers), for the product $21 m$ (where $m$ is the desired number) to satisfy the condition of the problem, it is necessary that the equality $21 m=21 \cdot 21 a$ holds, where $a$ is the square of the smallest natural number such that $21 a$ is a four-digit number. By t... | 1029 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
78(1188). Check that the equality is correct:
a) $\sqrt{2 \frac{2}{3}}=2 \sqrt{\frac{2}{3}}$
b) $\sqrt{5 \frac{5}{24}}=5 \sqrt{\frac{5}{24}}$.
Specify the condition under which the observed pattern holds. Provide examples. | Solution. a) Method I. Transform the left side of the equation:
$$
\sqrt{2 \frac{2}{3}}=\sqrt{\frac{8}{3}}=\sqrt{\frac{4 \cdot 2}{3}}=2 \sqrt{\frac{2}{3}}
$$
Method II. Transform the right side of the equation:
$$
2 \sqrt{\frac{2}{3}}=\sqrt{\frac{8}{3}}=\sqrt{2 \frac{2}{3}}
$$
Method III. Square both sides of the e... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
82 (1192). The distance between points $A$ and $B$ is 60 km. A car leaves from $A$ to $B$, and at the same time, a second car leaves from $B$ in the same direction. If the speed of the first car is increased by 10 km/h, and the speed of the second car is increased by 8 km/h, the first car will catch up with the second ... | Solution. Method I. Let $t$ hours be the time it takes for one car to catch up with the other when they are driving at their usual speed. Then their rate of approach is $\frac{60}{t}$ kilometers per hour. If the speed of one car increases by 10 km/h and the other by 8 km/h, their rate of approach will increase by 2 km/... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
91 (1201). Find the smallest integer that needs to be added to the expression $(a+2)(a+5)(a+8)(a+11)$ so that the resulting sum is positive for any value of $a$. | Solution. Let $b$ be the smallest integer such that the expression $(a+2)(a+5)(a+8)(a+11)+b$ is positive for any value of $a$. Transform the obtained expression into a quadratic trinomial:
$$
\begin{gathered}
(a+2)(a+5)(a+8)(a+11)+b= \\
=(a+2)(a+11)(a+5)(a+8)+b= \\
=\left(a^{2}+13 a+22\right)\left(a^{2}+13 a+40\right)... | 82 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
99(1209). How many consecutive natural numbers, starting from 1, need to be added to get a three-digit number written with identical digits? | Solution. Let $1+2+3+\ldots+n=\overline{b b b}$. Then $S_{n}=$ $=\frac{(1+n) \cdot n}{2}$. We have $\frac{n(n+1)}{2}=100 b+10 b+b$, from which
$$
n(n+1)=2 \cdot 111 b, n(n+1)=2 \cdot 3 \cdot 37 b=6 b \cdot 37
$$
where $n<45(111 b<1000,222 b<2000, n<45)$.
Since the left side of the equation is the product of two cons... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
101(1211). There are two arithmetic progressions $\left(a_{n}\right): 1 ; 5 ; 9 ; 13 ; 17 ; \ldots$ and $\left(b_{n}\right): 2 ; 5 ; 8 ; 11 ; 14 ; \ldots$. Prove that if you write down all the common terms of both progressions, the resulting sequence will also be an arithmetic progression. What is the common difference... | Solution. Method I. Let's find the $n$-th and $k$-th terms of the given arithmetic progressions. We have:
$$
\begin{aligned}
& a_{n}=a_{1}+d(n-1)=1+4(n-1)=4 n-3 \\
& b_{k}=b_{1}+d(k-1)=2+3(k-1)=3 k-1
\end{aligned}
$$
We need to find all such $n$ for which $a_{n}=b_{k}$, i.e., $4 n-3=3 k-1$, from which $k=\frac{4 n-2}... | 12 | Algebra | proof | Yes | Yes | olympiads | false |
109. In the equation $x^{2}-2 x+a=0$, the square of the difference of the roots is 20. Find $a$. | If $x_{1}$ and $x_{2}$ are the roots of the given quadratic equation, then we have the system of equations:
$$
\left\{\begin{array}{l}
\left(x_{1}-x_{2}\right)^{2}=20 \\
x_{1}+x_{2}=2
\end{array}\right.
$$
Squaring the second equation, we get:
$$
\left\{\begin{array}{c}
x_{1}^{2}-2 x_{1} x_{2}+x_{2}^{2}=20 \\
x_{1}^... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
128. How many solutions in integers does the equation
$$
\sqrt{x}+\sqrt{y}=9000 ?
$$ | Solution. Transferring $\sqrt{x}$ to the right side and squaring both sides of the equation, we get the equality $y=9000^{2}-18000 \sqrt{x}+x$, from which it follows that $\sqrt{x}$ is a rational number. However, if the square root of an integer is rational, then the number itself is a perfect square. Therefore, $x$ is... | 9001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
138. Calculate the sum for any $\alpha$
$$
\sin ^{2} \alpha+\sin ^{2}\left(\alpha+1^{\circ}\right)+\sin ^{2}\left(\alpha+2^{\circ}\right)+\ldots+\sin ^{2}\left(\alpha+179^{\circ}\right)
$$ | S o l u t i o n. By combining the first term with the ninety-first, the second with the ninety-second, and so on, and using the reduction formula, we get:
$$
\begin{gathered}
\left(\sin ^{2} \alpha+\sin ^{2}\left(\alpha+90^{\circ}\right)\right)+\left(\sin ^{2}\left(\alpha+1^{\circ}\right)+\sin ^{2}\left(\alpha+91^{\ci... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
140. Find the maximum value of the function
$$
y(x)=3 \sin x+4 \cos x
$$ | Solution. $y(x)=5\left(\frac{3}{5} \sin x+\frac{4}{5} \cos x\right)$. Since $\left(\frac{3}{5}\right)^{2}+$ $+\left(\frac{4}{5}\right)^{2}=1$, there exists an angle $x_{0}$ such that $\frac{3}{5}=\cos x_{0}, \frac{4}{5}=\sin x_{0}$. Then $y(x)=5 \sin \left(x+x_{0}\right)$. It is obvious that the maximum value of $y(x)$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
143. For which $n \in \boldsymbol{N}$ does the equality
$$
\sqrt[n]{17 \sqrt{5}+38}+\sqrt[n]{17 \sqrt{5}-38}=\sqrt{20} ?
$$ | S o l u t i o n. Denoting the first term by $a$, we will have $a+\frac{1}{a}=\sqrt{20}$, from which $a=\sqrt{5}+2$. Since $(\sqrt{5}+2)^{3}=17 \sqrt{5}+$ $+38=(\sqrt{5}+2)^{n}$, then $n=3$.
A n s w e r: $n=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
167. To transport 60 tons of cargo from one place to another, a certain number of trucks were required. Due to the poor condition of the road, each truck had to carry 0.5 tons less than originally planned, which is why 4 additional trucks were required. How many trucks were initially required? | Solution. Let $x$ be the original number of trucks. Then, each truck was supposed to carry $\frac{60}{x}$ tons of cargo, but they carried $\left(\frac{60}{x}-0.5\right)$ tons each. Since all 60 tons were loaded with the help of four additional trucks, we have the equation
$$
\left(\frac{60}{x}-0.5\right)(x+4)=60, \tex... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
168. Two pedestrians set out towards each other at the same time: the first from point $A$, the second from point $B$. The first pedestrian walked 1 km more before the meeting than the second. The first pedestrian arrived at point $B$ 45 minutes after the meeting. The second pedestrian arrived at point $A$ 1 hour and 2... | S o l u t i o n. Let the speed of the first pedestrian who left point $A$ be $v_{1}$ kilometers per hour, and the speed of the second pedestrian be $v_{2}$ kilometers per hour. Since after meeting, the first pedestrian walked $\frac{3}{4} v_{1}$ kilometers, and the second $-\frac{4}{3} v_{2}$ kilometers, and the first ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A person has 12 pints of wine in a barrel (pint - an old French unit of volume, 1 pint ≈ 0.568 liters) and wants to give away half of the wine, but does not have a 6-pint container. However, there are two empty containers with capacities of 8 pints and 5 pints. How can one use them to measure out exactly 6 pints of ... | $\triangle$ The solution to the problem can be written as follows:
| Vessel with a capacity of 8 pints | 0 | 8 | 3 | 3 | 0 | 8 | 6 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Vessel with a capacity of 5 pints | 0 | 0 | 5 | 0 | 3 | 3 | 5 | 0 |
This should be understood as follows. Initiall... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. There is no less than 10 liters of milk in the bucket. How can you pour exactly 6 liters of milk from it using an empty nine-liter bucket and a five-liter bucket? | $\triangle$ Let's denote the initial amount of milk in the first bucket as $a$ liters. Let's think about how to use the fact that the number $a$ is not less than 10. The difference $a-10$ can be used, but the difference $a-11$ cannot. The solution is written as follows:
| Bucket with volume $a$ l | $a$ | $a-5$ | $a-5$... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
12*. In how many ways can milk be transferred from a 12-liter barrel, filled with milk, to another empty barrel of the same volume using two empty cans of 1 liter and 2 liters? Transferring milk from one can to another is not allowed.
Note that the question in this problem is different from the previous problems. | Let's denote by $a_{n}$ the number of ways to transfer milk from a full container of $n$ liters (where $n \in \boldsymbol{N}$) using two empty cans of 1 liter and 2 liters.
If we first use the 1-liter can, there will be $n-1$ liters left in the first container, and it can be transferred in $a_{n-1}$ ways.
If we first... | 233 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
66. In a Bashkir village, every resident speaks either Bashkir, or Russian, or both languages. 912 residents of the village speak Bashkir, 653 - Russian, and 435 people speak both languages. How many residents are there in this village? | $\triangle$ Let's apply Euler circles. Let $A$ denote the set of villagers who speak Bashkir, and $B$ the set of villagers who speak Russian (Fig. 26).
Fig. 26
=28, \quad n(B)=13, \quad n(C)=10, \quad n(A \cap B)=8 \\
n(A \cap C)=6, \quad n(B \cap C)=5, \quad n(A \cap B \cap C)=2
\end{gathered}
\]
First, we... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
92. There are 9 coins, 8 of which are genuine and of the same weight, and one is counterfeit and heavier than the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the counterfeit coin? | $\triangle$ Let's number the coins with natural numbers from 1 to 9. Here, it is better to weigh the coins by placing three on each pan of the balance. Place coins numbered $1,2,3$ on the left pan, and coins numbered $4,5,6$ on the right pan. There are two possible outcomes, as in problem 90.
Suppose the balance is in... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
93. There are 10 coins, 9 of which are genuine and of the same weight, and one is counterfeit and lighter than the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the counterfeit coin? | $\triangle$ Let's number the coins. Place coins numbered $1,2,3$ on the left pan and coins numbered $4,5,6$ on the right pan. We can use Figure 30 for further steps.
If the scales balance, the counterfeit coin is among the remaining four. It takes two more weighings to determine the counterfeit coin from these four (s... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
97. There are 5 parts that are indistinguishable in appearance, 4 of which are standard and of the same mass, and one is defective, differing in mass from the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the defective part? | $\triangle$ Let's number the parts. Now try to figure out the weighing scheme presented below (Fig. 32).
As can be seen from this, it took three weighings to find the defective part.
Answer: in three. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
99. There are 6 identical-looking coins, but 4 are genuine, of the same weight, while 2 are counterfeit, lighter, and also weigh the same. What is the minimum number of weighings on a balance scale without weights that are needed to find both counterfeit coins? | $\triangle$ As usual, let's number the coins. During the first weighing, we will place coins numbered $1,2,3$ on the left pan, and all the others on the right (Fig. 33).
The scales may balance. Then each of the triplets contains one counterfeit coin. To find them, we need two more weighings (see the solution to proble... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
111. How many five-digit natural numbers can be formed using the digits 1 and 0 if the digit 1 appears exactly three times in each number? | $\triangle$ We will search for the specified numbers by enumeration, ensuring that we do not miss any number. It is simpler to start by finding the positions for the two zeros, because if the positions for the zeros are determined, the three remaining positions will be filled with ones uniquely.
Let's fix one of the z... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
113. How many two-digit natural numbers are there in which the first digit is greater than the second? | $\triangle$ If the first digit of a two-digit number is 1, then there is only one such number - 10. If the first digit of the number is 2, then there are two such numbers - 20 and 21. If the first digit is 3, then there are already three such numbers - 30, 31, and 32. And so on. Finally, if the first digit is 9, then t... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
124*. A checker can move in one direction along a strip divided into cells, moving either to the adjacent cell or skipping one cell in a single move. In how many ways can it move 10 cells? | $\triangle$ A checker can move one cell in 1 way, two cells in 2 ways, three cells in $-1+2=3$ ways, four cells in $-3+2=5$ ways, five cells in $-3+5=8$ ways, and so on. In fact, the Fibonacci sequence (see the solution to problem 12) is formed, which is defined by the recurrence relation
$$
a_{n}=a_{n-1}+a_{n-2}(n \i... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
128. The trainer brings three lions A, B, C and two tigers D and E onto the circus arena and seats them in a row on pedestals. At the same time, the tigers cannot be placed next to each other, otherwise a fight between them is inevitable. How many ways are there to arrange the animals? | $\triangle$ First, let's calculate the number of ways to arrange the tigers. If tiger G takes the first place, then tiger D can take the third, fourth, or fifth place. If G takes the second place, then D can only take the fourth or fifth place. If $\Gamma$ is placed on the third or fourth place, then in each of these c... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
134. How many four-digit natural numbers without repeating digits can be written using only the digits $1,2,3,4,5$ and 6? | $\triangle$ A four-digit number can be represented as $\overline{x y z t}$, where $x, y, z, t$ are digits. Here, the digit $x$ can take 6 values, after which the digit $y-5$ (repetition of digits is not allowed), then the digit $z-4$, and the digit $t-3$. Therefore, the number of such four-digit numbers is $6 \cdot 5 \... | 360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
146. From a group of 15 people, four participants for the relay race $800+400+200+100$ are selected. In how many ways can the athletes be arranged for the relay stages? | $\triangle$ First, let's choose an athlete for the 800 m stage, then for the 400 m, followed by 200 m, and finally for the 100 m stage. Since the order in which the selected athletes compete is significant, we are dealing with permutations—permutations of 4 out of 15. We get:
$$
A_{15}^{4}=15 \cdot 14 \cdot 13 \cdot 1... | 32760 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
155. On a plane, there are 10 lines, and among them, there are no parallel lines and through each point of their intersection, exactly two lines pass. How many intersection points do they have? | $\triangle$ Each intersection point is determined by a pair of lines $a$ and $b$ that intersect at this point. Since the order of these lines does not matter, each pair of such lines is a combination - a combination of 10 elements taken 2 at a time. We get:
$$
C_{10}^{2}=\frac{10 \cdot 9}{1 \cdot 2}=45
$$
Answer $: 4... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
161. How many six-digit natural numbers exist, each of which has its digits arranged in ascending order? | $\triangle$ Let's take the nine-digit number 123456789, whose digits are in ascending order. To obtain any six-digit number with the same property, it is sufficient to erase any three digits from this nine-digit number. Since the order of the digits being erased does not matter, we are dealing with combinations. We hav... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
165. On a straight line, there are 6 points, and on a parallel line, there are 8 points. How many triangles exist with vertices at these points? | $\triangle$ First, let's calculate the number of triangles where two vertices lie on the first line, and one vertex lies on the second line:
$$
C_{6}^{2} \cdot C_{8}^{1}=15 \cdot 8=120
$$
Similarly, let's calculate the number of triangles where one vertex lies on the first line, and two vertices lie on the second lin... | 288 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
206. Find all two-digit numbers, each of which is 13 more than the sum of the squares of its digits. | $\triangle$ Let the desired number be denoted as $\overline{x y}$. We obtain:
$$
10 x + y = x^2 + y^2 + 13, \quad x^2 - 10 x + (y^2 - y + 13) = 0
$$
We will consider the last equation as a quadratic equation in terms of $x$. Its discriminant must be non-negative:
$$
\frac{1}{4} D = 25 - y^2 + y - 13 \geq 0, \quad -y... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
215. The sequence $\left(x_{n}\right)$ is defined by the recurrence relations
$$
x_{1}=1, x_{2}=2, x_{n}=\left|x_{n-1}-x_{n-2}\right|(n>2) .
$$
Find $x_{1994}$. | $\triangle$ First, let's compute the first few terms of the sequence: $x_{1}=1, x_{2}=2, x_{3}=|2-1|=1, x_{4}=|1-2|=1, x_{5}=|1-1|=0, x_{6}=$ $=|0-1|=1, x_{7}=1, x_{8}=0, x_{9}=1, x_{10}=1, x_{11}=0$.
Starting from the third term, the terms of this sequence are either 1 or 0. Only the terms with indices
$$
5,8,11, \l... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
218. Find the last two digits of the power $7^{1995}$. | $\triangle$ Let's find the last two digits of the first few terms of the sequence ($7^{n}$): $07,49,43,01,07,49,43,01,07,49, \ldots$.
From this, we can see that these digits repeat when the term number increases by 4, generally by $4k (k \in N)$. Since
$$
1995=4 \cdot 498 + 3
$$
the last two digits of $7^{1995}$ coi... | 43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
234*. Once in a room, there were several inhabitants of an island where only truth-tellers and liars live. Three of them said the following.
- There are no more than three of us here. All of us are liars.
- There are no more than four of us here. Not all of us are liars.
- There are five of us. Three of us are liars.
... | $\triangle$ Let's consider the first of the three speakers. Suppose he is a truth-teller. Then both of his statements, including the second one, "All of us are liars," would be true, which means he is a liar. We have reached a contradiction. Therefore, he can only be a liar.
In this case, both statements of the first ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
299. There are 8 balls: 2 red, 2 blue, 2 white, and 2 black. Players A and B take turns attaching one ball to one of the vertices of a cube. Player A aims to achieve a vertex such that this vertex and its three adjacent vertices each have a ball of a different color, while Player B aims to prevent this. Who will win wi... | $\triangle$ The problem does not specify who starts the game - A or B. Therefore, let's consider two cases.
1) Suppose A starts the game.
If A, for example, attaches a red ball to one of the vertices of the cube, then B's best response is to attach a red ball to one of the adjacent vertices, and so on (Fig. 36). Cons... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
331. There is a $10 \times 10$ square table, in the cells of which natural numbers from 1 to 100 are written in sequence:
in the first row - numbers from 1 to 10, in the second row - from 11 to 20, and so on. Prove that the sum $S$ of any 10 numbers in the table, none of which are in the same row and none of which are ... | $\triangle$ Let's denote the term of the original sum $S$ from the first row as $a_{1}$, from the second row as $10+a_{2}$, from the third row as $20+a_{3}$, and so on, finally from the tenth row as $90+a_{10}$.
Here, each of the natural numbers $a_{1}, a_{2}, \ldots, a_{10}$ is within the range from 1 to 10, and thes... | 505 | Combinatorics | proof | Yes | Yes | olympiads | false |
337. A sheet of paper was torn into 5 pieces, some of these pieces were torn into 5 parts, some of these new parts were torn again into 5 parts, and so on. Can this method result in 1994 pieces of paper? What about 1997? | $\triangle$ Each time a sheet or a piece of paper is torn into 5 parts, the total number of pieces increases by 4. Therefore, the number of paper pieces at each step can only be of the form $4k+1 (k \in N)$. This expression is the invariant.
Since $1994 \neq 4k+1 (k \in N)$, it is impossible to obtain 1994 pieces, whi... | 1997 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
350. Among 18 coins, one is counterfeit. Genuine coins weigh the same, while the counterfeit coin differs in weight from the genuine ones. What is the minimum number of weighings on correct balance scales without weights needed to determine whether the counterfeit coin is lighter or heavier than the genuine ones? (Ther... | $\triangle$ Let's number the coins. Divide the set of coins into three piles, with 6 coins in each.
For the first weighing, place all the coins from the first pile on one scale pan, and all the coins from the second pile on the other. There are two possible cases.
1) Suppose the scales balance during this weighing. T... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
364. Once I decided to take a ride on a chairlift. At some point, I noticed that the chair coming towards me had the number 95, and the next one had the number 0, followed by 1, 2, and so on. I looked at the number on my chair; it turned out to be 66. Have I passed the halfway point? At which chair will I pass the half... | $\triangle$ Total number of chairs is 96, and half the journey consists of 48 chairs. I will be in the middle of the cable car when the number of chairs ahead and behind me is the same. For this to happen, the number of the oncoming chair must be equal to $66-48=18$. Since the encounter with chair number 18 is ahead of... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
390. There are 200 candies. What is the minimum number of schoolchildren to whom these candies can be distributed so that among them, there will always be at least two who receive the same number of candies (which could be zero)? | $\triangle$ Let's find the minimum $n$ such that $200400$. There is no need to solve this quadratic inequality in the usual way. Since the product $n(n-1)$ increases with $n$, the minimum $n$ satisfying the inequality can be found by trial and error: $n=21$.
This means that if there are 21 schoolchildren, there will a... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
444. Eight hockey teams play against each other in a round-robin tournament to determine the final four. What is the minimum number of points that guarantees a team's advancement to the final four? | $\triangle$ All eight teams together will score $\frac{8 \cdot 7}{2} \cdot 2=56$ points. Therefore, 7, 8, and even 9 points do not guarantee a team's advancement to the final four.
What about 10 points? Imagine that there are five teams that played all their matches against each other to a draw, while the other three ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
454. Several identical boxes together weigh 10 tons, with each of them weighing no more than 1 ton. What is the minimum number of three-ton trucks needed to haul away all this cargo in one trip? | $\triangle$ Four trucks may not be enough. For example, if there are 13 identical boxes weighing $\frac{10}{13}$ tons each, then in one of the trucks, you cannot place more than three boxes, because with four boxes, the weight is $4 \cdot \frac{10}{13}$ tons, which is more than 3 tons. In this case, with four trucks, y... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 7.
The following very simple problem is one of many entertaining problems that have gained widespread popularity. In a dark room stands a wardrobe, in the drawer of which there are 24 red and 24 blue socks. How many socks should be taken from the drawer to ensure that at least one pair of socks of the same color ca... | 7. Usually, the question of the problem is given an incorrect answer: 25 socks. If the problem asked how many socks should be taken from the drawer to ensure that at least 2 socks of different colors are among them, then the correct answer would indeed be such: 25 socks. But in our problem, the question is about ensuri... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Alice returned the rattle to its rightful owner, but a few days later, another brother broke the rattle again. This time, the raven did not come to scare the brothers, and they began to beat each other with all their might. Alice grabbed the broken rattle and ran out of the forest.
After some time, Alice met the White... | 60. The chances are zero. Suppose the statement of the brother Alice met is true. Then the owner of the rattle should have been lying on the day of the meeting and, consequently, could not have been the brother Alice met. On the other hand, suppose the statement of the brother Alice met is false. Then the owner of the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 107.
On another island of knights, liars, and normal people, the king held opposite views and gave his daughter different paternal advice: “My dear, I don’t want you to marry any knight or liar. I would like your husband to be a solid, reputable normal person. You should not marry a knight because all knights are h... | 107. In both cases, a single statement would suffice. The King could be convinced by the true statement “I am not a knight” (such a statement could not belong to either a knight or a liar) and the false statement “I am a liar.” | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
132.
In one museum, I had the chance to see a pair of caskets adorned with the following inscriptions:
On the gold one
Both caskets in this set were made by members of the Cellini family
On the silver one
Neither of these caskets was made by either a son of Bellini or a son of Cellini
Whose work is each of the tw... | 132. The statement engraved on the lid of the golden casket cannot be true, for otherwise we would have arrived at a contradiction. Therefore, the golden casket was made by someone from the Cellini family. Since the inscription on the golden casket is false, both caskets could not have been made by members of the Celli... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 28. How Many Favorites?
- "Here's a different kind of puzzle," said the Gryphon. "Once, the Queen of Hearts held a reception for thirty guests. She needed to distribute one hundred gingerbread cookies among the guests. Instead of cutting the cookies into pieces, the Queen preferred to give four cookies to each of h... | 28. How many favorites? This problem, usually solved using algebra, is very simple if approached in the following way. First, let's distribute 3 pretzels to each of the 30 guests of the Queen. We will have 10 pretzels left. At this point, all non-favorites will have received all the pretzels they are entitled to, while... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 29. Pretzels and Pretzelkins
- Here's another problem,—began the Gryphon. One day, the Mock Turtle went to the store to buy pretzels for the next tea party.
- How much are your pretzels?—he asked the store owner.
- The price depends on the size: I can offer you small pretzelkins and large pretzels. One pretzel cost... | 29. Pretzels and pretzelkins. Since each pretzel costs as much as one pretzelkin, 7 pretzels cost as much as 21 pretzelkins, and 7 pretzels and 4 pretzelkins cost as much as 25 pretzelkins. On the other hand, 4 pretzels and 7 pretzelkins cost as much as 19 pretzelkins (since 4 pretzels cost as much as 12 pretzelkins). ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
30. At the Duchess's, the Cook's, and the Cheshire Cat's
- I want to offer you a most intriguing task,- said the Gryphon. - One day, the Mock Turtle, the March Hare, and the Dormouse decided to visit the Duchess, the Cook, and the Cheshire Cat. Upon arriving, they found that no one was home. On the kitchen table, they... | 30. At the Duchess's, the Cook's, and the Cheshire Cat's. The Cheshire Cat must find 2 pretzels on the tray: after eating half of the pretzels and one more pretzel, nothing will be left on the tray. Sonya must find 6 pretzels on the tray: after eating half of the pretzels and one more pretzel, 2 pretzels will be left f... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
31. How many days did the gardener work?
- I want to give you a task, - said the Griffin. - Usually it is solved using algebra, but if you use my method, you will do wonderfully without it!
Once the King hired one of the spade suit gardeners for twenty-six days to do some work in the garden. The King set the conditio... | 31. How many days did the gardener work? Working diligently, the gardener can earn a maximum of $3 \cdot 26=78$ pretzels. He earned only 62 pretzels. Therefore, he did not receive 16 pretzels because he was slacking off. Each day the gardener slacked off, he lost 4 pretzels (the difference between the 3 pretzels he cou... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
33. How many people got lost in the mountains?
Alice and the Griffin had to wait several minutes before the Tortoise Quasi gathered his strength and could continue.
- You see,一began the Tortoise Quasi.
- I don't see anything!- the Griffin cut in.
The Tortoise Quasi did not respond, only grabbing his head with his fr... | 33. How many people got lost in the mountains? Let's call one portion the amount of supplies one person consumes in a day. Initially, 9 people had 45 portions (a 5-day food supply). On the second day, they had only 36 portions left. On the same day, they met a second group, and the 36 remaining portions were enough for... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
36. How long will it take to get out of the well?
- As you wish, - agreed Quasi-Tortoise. I'll tell you a problem about a frog that fell into a well.
- But that's a very old problem! - objected the Griffin. - It has a very long beard! Don't you know any new problems?
- I haven't heard this problem, - Alice spoke up fo... | 36. How long does it take to get out of the well? Those who think the frog will get out of the well in 30 days are wrong: the frog could have gotten out of the well by evening on the 28th day. Indeed, in the morning of the 2nd day, the frog is 1 foot above the bottom of the well, in the morning of the 3rd day - 2 feet,... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
37. Will the cyclist make it to the train?
- Wasn't the previous problem sad? the Turtle Quasi asked. - Just think! The poor frog spent so many days in a dark well! And to get out of there, she had to undertake a climb like a real mountaineer!
- Nonsense! - the Griffin interrupted him. - The saddest part of the whole ... | 37. Will the cyclist make it to the train? The cyclist reasoned incorrectly: he averaged distances, not time. If he had traveled at 4 miles per hour, 8 miles per hour, and 12 miles per hour for the same amount of time, his average speed would indeed have been 8 miles per hour. However, he spent more time climbing the h... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
39. How far is it to school?
All the time while Alice and the Griffin were solving the previous problem, the Tortoise Quasi was weeping inconsolably.
- Can you tell me what is sad about this problem? - the Griffin barked at him angrily. - After all, the passenger caught up with the train. Or did I misunderstand somet... | 39. How far is it to school? The difference in time between being 5 minutes late and arriving 10 minutes before the start of the lesson is 15 minutes. Therefore, if the boy walks to school at a speed of 5 miles per hour, he will save 15 minutes (compared to how long it would take him to walk if he went at a speed of 4 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 52. The First Question
- Do you know division? - asked the Black Queen.
- Of course! - Alice replied confidently.
- Excellent! Suppose you divide eleven thousand, eleven hundred, and eleven by three. What is the remainder of the division? If you like, you can use a pencil and paper.
Alice set to work and performed... | 52. The first question. Alice made a mistake by recording eleven thousand eleven hundred and eleven as 11111, which is incorrect! The number 11111 is eleven thousand one hundred and eleven! To understand how to correctly write the dividend, let's add eleven thousand, eleven hundred, and eleven in a column:
11000
1100... | 12111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
61. Another task about a piece of land
- Here's another task,- said the Black Queen. Another farmer had a piece of land. On one third of his land, he grew pumpkins, on one fourth he planted peas, on one fifth he sowed beans, and the remaining twenty-six acres he allocated for corn.
How many acres of land did the farm... | 61. Another problem about a plot of land. Let's bring all fractions to a common denominator (equal to 60): $1 / 3 + 1 / 4 + 1 / 5 = {}^{20} / 60 + {}^{15} / 60 + {}^{12} / 60 = {}^{47} / 60$. Corn occupies ${}^{13} / 60$ of the entire area. Therefore, ${}^{13} / 60$ of the plot is 26 acres, and since 13 is half of 26, ... | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
32. Find the first 1963 digits after the decimal point in the decimal representation of the number $(\sqrt{26}+5)^{1963}$. | 63.32. Since the sum $(\sqrt{26}+5)^{1963}+(5-\sqrt{26})^{1963}$ is an integer (this is verified using the binomial theorem), and the inequality $-0.1 < 5 - \sqrt{26} < 0$ holds, then $-10^{-1963} < (5 - \sqrt{26})^{1963} < 0 \quad$ and, consequently, the first 1963 digits of the number $(\sqrt{26}+5)^{1963}$ after the... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
37. Solve the equation $x^{4}-2 y^{4}-4 z^{4}-8 t^{4}=$ $=0$ in integers. | 63.37. Answer. $x=y=z=t=0$. Hint. Consider an integer solution with the smallest absolute value of $x$ and try to find another solution with $x_{1}=x / 2$. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
40. For what values of $n$ is the expression $2^{n}+1$ a non-trivial power of a natural number? | 63.40. Only for $n=3$. Indeed, if $2^{n}+1=A^{p}$, then $2^{n}=A^{p}-1=(A-1)\left(A^{p-1}+A^{p-2}+\ldots+A+1\right)$. Then $A^{p-1}+A^{p-2}+\ldots+A+1$ is a power of two, not equal to 1, and since $A$ is odd, and the given sum is even, there is an even number of terms in it. Let $p=2 q$. Then $2^{n}=A^{2 q}-1=$ $=\left... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In the cells of a chessboard, natural numbers are placed such that each number is equal to the arithmetic mean of its neighbors. The sum of the numbers in the corners of the board is 16. Find the number standing on the field $e 2$. | 64.3. Consider the largest number standing in one of the cells. Obviously, all adjacent numbers to it are equal to it. Those adjacent to them are also equal to them, and so on. Therefore, all numbers on the board are equal. Hence, the number written in the field $e 2$ is 4. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Front tires of a car wear out after 25000 km of travel, while the rear tires wear out after 15000 km of travel. When should the tires be swapped to ensure they wear out simultaneously? | 65.3. Answer. After 9375 km of travel. | 9375 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A rectangle of 19 cm $\times$ 65 cm is divided by lines parallel to its sides into squares with a side of 1 cm. Into how many parts will this rectangle be divided if we also draw its diagonal? | 65.4. The diagonal intersects $19+65-1$ cells, and, consequently, 83 parts are added to the initial partition. The total number of parts is $19 \cdot 65+83=1318$. | 1318 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Odd numbers from 1 to 49 are written in the form of a table
| 1 | 3 | 5 | 7 | 9 |
| ---: | ---: | ---: | ---: | ---: |
| 11 | 13 | 15 | 17 | 19 |
| 21 | 23 | 25 | 27 | 29 |
| 31 | 33 | 35 | 37 | 39 |
| 41 | 43 | 45 | 47 | 49 |
Five numbers are chosen such that no two of them are in the same row or column. What is ... | 65.6. Answer. Regardless of the choice, the sum will be 125. | 125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
29. Given a cube $12 \times 12 \times 12$, which is cut by planes parallel to its faces into unit cubes. How many parts will the cube be divided into if a section in the form of a regular hexagon is made? | 65.29. Indication. Introduce a Cartesian coordinate system with the origin at one of the cube's vertices. Then the plane $x+y+z=18$ intersects the cube with a corner at point $(a, b, c)$ if and only if $15<a+b+c<18$, i.e., $a+b+c=16$ or 17 (the corner of the cube that is closest to the origin is meant). It remains to f... | 216 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
20. A tram ticket is called lucky in the Leningrad style if the sum of its first three digits equals the sum of the last three digits. A tram ticket is called lucky in the Moscow style if the sum of its digits in even positions equals the sum of its digits in odd positions. How many tickets are lucky both in the Lening... | 69.20. There are 6700 such tickets. Note. The second digit of the number should match the fifth.
保留源文本的换行和格式，翻译结果如下：
69.20. There are 6700 such tickets.
Note. The second digit of the number should match the fifth. | 6700 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. Given the numbers $5^{1971}$ and $2^{1971}$. They are written consecutively. What is the number of digits in the resulting number? | 71.14. Answer. 1972 digits, since the product of these numbers is equal to $10^{1971}$. | 1972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
15. Around a circle, 100 integers are written, the sum of which is 1. A chain is defined as several consecutive numbers. Find the number of chains whose sum of numbers is positive. | 71.15. Answer. 4951. Use the fact that from two chains, one of which complements the other, the sum of the numbers is positive in exactly one of them. | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
19. Solve the system of equations:
$$
\left\{\begin{array}{l}
x_{0}=x_{0}^{2}+x_{1}^{2}+\ldots+x_{100}^{2} ; \\
x_{1}=2\left(x_{0} x_{1}+x_{1} x_{2}+\ldots+x_{99} x_{100}\right) ; \\
x_{2}=2\left(x_{0} x_{2}+x_{1} x_{3}+\ldots+x_{98} x_{100}\right) \\
x_{3}=2\left(x_{0} x_{3}+x_{1} x_{4}+\ldots+x_{97} x_{100}\right) ;... | 71.19. Answer. There are a total of 202 solutions. The main series: $x_{0}=0, x_{k}= \pm 1 / 2, x_{j}=0$ for $j \neq k$ (200 solutions). Another solution: all $x_{i}$ are equal to zero, and the last solution is: $x_{0}=$ $=1, x_{i}=0$ for $i>0$. | 202 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
36. Each side of an equilateral triangle is divided into 30 equal parts. Lines drawn through the division points parallel to the sides of the triangle divide it into 900 small triangles. What is the maximum number of vertices of the partition, no two of which lie on the same drawn line or side? | 72.36. Answer. 21. Indeed, for each vertex, we can define three coordinates: the numbers of lines parallel to one of the sides of the triangle, counted from the vertex not belonging to the corresponding side (see Fig. 45). It is clear that the sum of the coordinates of any vertex is 30. Since the first (as well as the ... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
38 *. The city has the appearance of a square grid $100 \times 100$ with the side of each cell being 500 m. Along each side of each cell, one can move only in one direction. It is known that one can travel no more than 1 km through the city without violating the movement rules.
66
Prove that there will be no fewer tha... | 72.38. We will call an intersection a "source" if you can only leave from it, a "sink" if you cannot leave from it, and a "through" if you can enter and leave it. Then, no two intersections of the same type can be adjacent - for sinks and sources this is obvious, and if two throughs are adjacent, it means there is a 15... | 1300 | Combinatorics | proof | Yes | Yes | olympiads | false |
27. Find the maximum of the expression
$$
\begin{aligned}
& x_{1}+x_{2}+x_{3}+x_{4}-x_{1} x_{2}-x_{1} x_{3}-x_{1} x_{4}-x_{2} x_{3}-x_{2} x_{4}-x_{3} x_{4}+ \\
& +x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}-x_{1} x_{2} x_{3} x_{4}
\end{aligned}
$$ | 74.27. The given expression can be rewritten as follows:
$$
1-\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)\left(1-x_{4}\right)
$$
Since $\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)\left(1-x_{4}\right)$ is a non-negative number that can be equal to zero, the maximum value of the expres... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. In seven consecutive vertices of a regular 100-gon, chips of seven colors are placed. In one move, it is allowed to move any chip 10 fields clockwise to the $11-\mathrm{th}$, if it is free. It is required to collect the chips in the seven vertices following the initial ones. How many different arrangements of chips... | 75.18. Answer. There are seven possible arrangements. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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