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stringlengths 2
5.64k
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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 at the equality $n=2$, and then the desired number is 178.
Answer: 178.
|
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 by the square of a natural number $a$. Since $392=7^{2} \cdot 2^{3}$, 392 is divisible only by the squares of the numbers 2, 7, and 14. It is clear then that $a=7$.
Answer: $a=7$.
|
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 $-x_{0}^{2}-x_{0}=0$, from which $x_{0}=0$ or $x_{0}=-1$. If $x_{0}=0$, then from the first equality it follows that $m=0$, which contradicts the condition. If $x_{0}=-1$, then we get $m=-2$. (Verification shows that the value $m=-2$ indeed satisfies the condition of the problem.)
Answer: $m=-2$.
|
-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 time?
|
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 ended in draws during regular time.
A n s w e r: 34 matches.
|
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, $x(x-1) y=448$. Listing the divisors of 448, we notice that only two of them (7 and 8) differ by 1. From the obtained equation, it is easy to find $x=8, y=8$.
Answer: The tournament was held in 8 rounds.
|
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 "Minsk". In how many ways can the cashier choose the order of visiting the hotels to deliver the tickets?
|
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, there are 60 ways $(5 \cdot 4 \cdot 3=60)$. Finally, the fourth hotel can be chosen in 2 ways. Thus, there are 120 ways $(2 \cdot 4 \cdot 3 \cdot 2=120)$ to visit all the hotels.
The problem requires the cashier to go to "Russia" before the fourth hotel; this reduces the number of ways by exactly half, as the number of ways in which "Russia" is visited earlier is the same as the number of ways in which it is visited later. Similarly, the second restriction of the problem also reduces the number of possible ways by half. Therefore, the cashier has 30 ways $(120: 2: 2=30)$ to visit all the hotels.
Answer: 30 ways.
|
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 total number of such tickets is 7500 $(5 \cdot 4 \cdot 3 \cdot 5 \cdot 5 \cdot 5=7500)$.
Among the considered numbers, those that do not satisfy the third condition of the problem are the numbers where the third digit is 7 and the fourth digit is 8. The number of such numbers is calculated in the same way as above: it is equal to $300(4 \cdot 3 \cdot 5 \cdot 5=300)$.
Therefore, the number of different lucky numbers is $7200(7500-300=7200)$.
Answer: 7200.
|
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.
Answer: 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 $2\left(\frac{2}{x}+\frac{x}{2}\right)+5$ takes its minimum value, which is $9(2 \cdot 2+5=9)$.
Answer: 9.
|
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 sum
$$
1-x+\frac{1}{1-x}+1
$$
and using the fact that the sum of two positive reciprocals is not less than two, we get that the given fraction is not less than 3, i.e., its minimum value is 3.
Answer: 3.
$10^{*}$
|
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
$$
x=\frac{1}{2}(7 \pm \sqrt{17})
$$
A n s w e r: 5 .
|
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:
$$
\left\{\begin{array} { l }
{ x - y = 1 } \\
{ x + y = 225 }
\end{array} \left\{\begin{array} { l }
{ x - y = 3 } \\
{ x + y = 75 }
\end{array} \quad \left\{\begin{array}{l}
x-y=5 \\
x+y=45
\end{array} ;\left\{\begin{array}{l}
x-y=9 \\
x+y=25
\end{array}\right.\right.\right.\right.
$$
By solving these 4 systems of equations, we find that there are 4 right triangles that satisfy the given property. Their sides are: 1) $15 ; 112 ; 113 ; 2) 15 ; 36 ; 39 ; 3) 15 ; 20$; $25 ; 4) 15 ; 8 ; 17$.
Answer: 4 triangles.
|
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 power of the number 5 is divisible by the number $6^{m}+1$, which ends with the digit 7, which is impossible. From the equation $5^{n}-36^{m}=9$, it follows that $5^{n}$ is divisible by 9, which is incorrect.
Since $36-5^{2}=11$, then 11 is the smallest value of the given expression.
Answer: 11.
|
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:
$$
\begin{gathered}
n=(5 k+1)(5 k+2)(5 k+3)(5 k+4)=\left(25 k^{2}+25 k+4\right) \times \\
\times\left(25 k^{2}+25 k+6\right)=(25 k(k+1)+4)(25 k(k+1)+6)= \\
=(50 m+4)(50 m+6)
\end{gathered}
$$
where $m=\frac{k(k+1)}{2}$ is an integer. From this, we get:
$$
\begin{gathered}
n=2500 m^{2}+500 m+24=2000 m^{2}+500 m(m+1)+24= \\
=1000\left(2 m^{2}+p\right)+24
\end{gathered}
$$
where $p=\frac{m(m+1)}{2}$ is an integer.
Thus, the last three digits of the number $n$ are determined - they are 024. The fourth digit from the end of the number is not uniquely determined; for example, $6 \cdot 7 \cdot 8 \cdot 9=3024,11 \cdot 12 \cdot 13 \cdot 14=24024$.
|
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. But $\overline{x 0 y}=100 x+y=(x+y)+99 x$, 152
so the sum $x+y$ must be divisible by 11, i.e., $x+y=11$. By simple enumeration, we get the solution: $x=7, y=4$.
Answer: the number 7744 is a square of a natural number.
|
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 with an empty vessel $A$ and 8 liters in vessel $B$. Now, fill vessel $A$ from the tap and use it to top up vessel $B$, leaving 10 liters of water in vessel $A$.
|
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 between any two of the considered numbers should be divisible by $d$. However, the difference between the numbers 123456798 and 123456789, which is 9, is not divisible by $d$. Therefore, 9 is the greatest common divisor of these numbers.
|
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 $2, 4, 5, 6, 7, 8, 9$, since among these numbers there is at least one number that does not divide by these numbers (none of the considered numbers is divisible by 9). We will prove that the considered numbers do not have a greatest common divisor greater than 9. Indeed, if $d$ is the greatest common divisor of these numbers, greater than 9, then the difference of any two of them is divisible by $d$. However, the difference between the numbers 123465 and 123456, which is 9, is not divisible by $d$. Therefore, the greatest common divisor of the considered numbers is 3.
|
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 the same direction, are side by side every 56 minutes. When moving at the same speeds in opposite directions, the cars meet every 8 minutes. How long will it take each car to travel the entire circular road?
Solution. Let $s$ (m) be the length of the circular road, $v_{1}$ (m/min) be the speed of the first car, and $v_{2}$ (m/min) be the speed of the second car. According to the problem, we can set up the following system of equations:
' For problems 143-146, the numbers in parentheses are the exercise numbers in textbook $|8|$.
$$
\left\{\begin{array}{l}
56\left(v_{1}-v_{2}\right)=s \\
8\left(v_{1}+v_{2}\right)=s
\end{array}\right.
$$
from which $v_{1}=\frac{s}{14}$ (m/min), $v_{2}=\frac{6 s}{112}$ (m/min).
Then the first car will travel the entire circular road in 14 minutes $\left(s: \frac{s}{14}=14\right)$, and the second car will take 18 minutes 40 seconds $\left(s: \frac{6 s}{112}=18 \frac{2}{3}\right)$.
Answer: 14 minutes; 18 minutes 40 seconds.
|
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 them to meet if they start running simultaneously from the same starting line in opposite directions?
|
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 set up the equation
$$
30 \cdot \frac{s}{t}-30 \cdot \frac{s}{t+5}=s,
$$
from which $t^{2}+5 t-150=0, t=10$ s. Therefore, the speeds of the athletes are $\frac{s}{10} \mathrm{m} / \mathrm{s}$ and $\frac{s}{15} \mathrm{m} / \mathrm{s}$, respectively, and the desired time is $6 \mathrm{s}\left(s:\left(\frac{s}{10}+\frac{s}{15}\right)=6\right)$.
Answer: 6 s.
|
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 first meeting. How long will it take the pedestrian to walk from $A$ to $B$?
|
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:
$$
\left\{\begin{array}{l}
\frac{20 v_{1}+10 v_{1}+t v_{1}+10 v_{1}+t v_{1}}{v_{2}}=20 \\
\frac{20 v_{1} \cdot 2+10 v_{1}}{v_{2}}=10
\end{array}\right.
$$
156
$$
\left\{\begin{array}{l}
20 v_{2}=40 v_{1}+2 t v_{1} \\
10 v_{2}=50 v_{1}
\end{array}\right.
$$
from which $t=30$.
Therefore, the required time (the time it takes for the pedestrian to travel from $A$ to $B$) is 1 hour $(20+10+30=60$ (min)).
Answer: 1 hour.
|
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 cyclist on the road
|
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:
$$
\left\{\begin{array}{l}
v_{1}=\frac{v_{2} t}{2} \\
v_{2}=\frac{v_{1} t}{4.5}
\end{array}\right.
$$
from which $t^{2}=9, t=3$.
Then the time of the motorcyclist's movement is 5 hours $(3+2=5)$, and the time of the cyclist's movement is 7.5 hours $(3+4.5=7.5)$.
Answer: 5 hours; 7.5 hours.
## § 9. PROBLEMS OF INCREASED DIFFICULTY IN THE ALGEBRA COURSE FOR THE 9TH GRADE
$\triangle 1(1129)^{\prime}$. Find the roots of the polynomial $2 x^{5}+x^{4}-10 x^{3}-$ $-5 x^{2}+8 x+4$. (A root of a polynomial in one variable is a value of the variable for which the value of the polynomial is zero.)
Solution. Factorize the given polynomial:
$$
2 x^{5}+x^{4}-10 x^{3}-5 x^{2}+8 x+4=
$$
$$
=x^{4}(2 x+1)-5 x^{2}(2 x+1)+4(2 x+1)=(2 x+1)\left(x^{4}-5 x^{2}+4\right) .
$$
Set each factor equal to zero. We have $2 x+1=0$, from which $x_{1}=-\frac{1}{2}$, or $x^{4}-5 x^{2}+4=0$, from which $x^{2}=1, x^{2}=4$, i.e., $x_{2}=1, x_{3}=-1, x_{4}=2, x_{5}=-2$.
Answer: $-\frac{1}{2} ;-1 ;-2 ; 1 ; 2$.
|
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 given quadratic trinomials has the form $x^{2}+x+1$, which has no solutions in the set of real numbers. Therefore, the given quadratic trinomials have a common root $x=1$. Find the value of $a$ for which $x=1$. When $x=1$, we have $1+a+1=0$, from which $a=-2$.
Answer: $a=-2$.
Remark. It is advisable to do a check. When $a=-2$, we have the quadratic trinomials $x^{2}-2 x+1$ and $x^{2}+x-2$, which have a common root $x=1$.
|
-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 value when $a=1$.
Remark 1. This problem can also be solved without using Vieta's theorem, by calculating the roots of the given quadratic trinomial using formulas. We have
158
$$
\begin{gathered}
x_{1}=\frac{a-2+\sqrt{ } a^{2}+8}{2}, x_{2}=\frac{a-2-\sqrt{ } a^{2}+8}{2}, \\
x_{1}^{2}+x_{2}^{2}=\left(\frac{a-2+\sqrt{ } a^{2}+8}{2}\right)^{2}+\left(\frac{a-2-\sqrt{ } a^{2}+8}{2}\right)^{2}= \\
=\frac{4 a^{2}-8 a+24}{4}=a^{2}-2 a+6=(a-1)^{2}+5 .
\end{gathered}
$$
Remark 2. The value of $a$ at which the quadratic trinomial $a^{2}-2 a+6$ obtained during the solution process takes its minimum value can also be found differently, by calculating the x-coordinate of the vertex of the corresponding parabola using the known formula. In our case, $a=\frac{2}{2}=1$.
|
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 would need 15 hours.
|
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., $x_{1}=\frac{3}{2} x_{2}, x_{2}=\frac{2}{3} x_{1}$.
The first worker can complete the entire work in $\frac{A}{x_{1}}$ hours, or in
$$
\frac{48 \cdot \frac{x_{1}+x_{2}}{2}+10 x_{1}}{x_{1}}=\frac{24\left(x_{1}+\frac{2}{3} x_{1}\right)+10 x_{1}}{x_{1}}=\frac{50 x_{1}}{x_{1}}=50(\text{h})
$$
The second worker can complete the entire work in $\frac{A}{x_{2}}$ hours, or in
$$
\frac{48 \cdot \frac{x_{1}+x_{2}}{2}+15 x_{2}}{x_{2}}=\frac{24\left(\frac{3}{2} x_{2}+x_{2}\right)+15 x_{2}}{x_{2}}=\frac{75 x_{2}}{x_{2}}=75
$$
The third worker can complete the entire work in
$$
\frac{A}{x_{3}}=\frac{A}{\frac{x_{1}+x_{2}}{2}}=\frac{A}{\frac{1}{2}\left(\frac{A}{50}+\frac{A}{75}\right)}=60(\text{h})
$$
Answer: 50 h, 75 h, 60 h.
|
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 $\cos \alpha=\frac{1}{\sqrt{3}}$ and $\sin \alpha=\sqrt{\frac{2}{3}}$. Therefore,
$$
\sin x+\sqrt{2} \cos x=\sqrt{3}(\sin x \cos \alpha+\cos x \sin \alpha)=\sqrt{3} \sin (x+\alpha)
$$
where $\alpha$ is the angle whose sine is $\sqrt{\frac{2}{3}}$ and cosine is $\frac{1}{\sqrt{3}}$. Since $\sin (x+\alpha) \leqslant 1$, then $\sin x+\sqrt{2} \cos x \leqslant \sqrt{3}$, which is what we needed to prove.
Remark 1. Other notations can be introduced: $\sin \alpha=\frac{1}{\sqrt{3}}, \cos \alpha=\sqrt{\frac{2}{3}}$. Then
$$
\begin{gathered}
\sin x+\sqrt{2} \cdot \cos x=\sqrt{3}(\sin \alpha \sin x+\cos \alpha \cos x)= \\
=\sqrt{3} \cos (\alpha-\beta) \leqslant \sqrt{3}
\end{gathered}
$$
Remark 2. After solving this problem, it is advisable (in extracurricular activities) to consider the question of transforming the sum $a \sin x+b \cos x$:
$a \sin x+b \cos x=\sqrt{a^{2}+b^{2}}\left(\frac{a}{\sqrt{a^{2}+b^{2}}} \sin x+\frac{b}{\sqrt{a^{2}+b^{2}}} \cos x\right)=$
$=\sqrt{a^{2}+b^{2}}(\cos \alpha \sin x+\sin \alpha \cos x)=\sqrt{a^{2}+b^{2}} \sin (x+\alpha)$,
where $\alpha$ is the angle whose tangent is $\frac{b}{a}$.
$\triangle$ 47(1175). Given that $\operatorname{tg} \alpha+\operatorname{ctg} \alpha=5$, find the value of the expression $\operatorname{tg}^{2} \alpha+\frac{1}{\sin \alpha} \cdot \frac{1}{\cos \alpha}+\operatorname{ctg}^{2} \alpha$.
Solution. Square the equation $\operatorname{tg} \alpha+\operatorname{ctg} \alpha=5$:
$$
\operatorname{tg}^{2} \alpha+\operatorname{ctg}^{2} \alpha+2 \operatorname{tg} \alpha \operatorname{ctg} \alpha=25
$$
from which $\operatorname{tg}^{2} \alpha+\operatorname{ctg}^{2} \alpha=23$.
Find the value of the product $\frac{1}{\sin \alpha} \cdot \frac{1}{\cos \alpha}$. We have:
$$
\frac{1}{\sin \alpha} \cdot \frac{1}{\cos \alpha}=\frac{1}{\sin \alpha \cos \alpha}=\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{\sin \alpha \cos \alpha}=\operatorname{tg} \alpha+\operatorname{ctg} \alpha=5
$$
Therefore,
$$
\begin{gathered}
\operatorname{tg}^{2} \alpha+\frac{1}{\sin \alpha} \cdot \frac{1}{\cos \alpha}+\operatorname{ctg}^{2} \alpha= \\
=\left(\operatorname{tg}^{2} \alpha+\operatorname{ctg}^{2} \alpha\right)+(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)=23+5=28
\end{gathered}
$$
Answer: 28.
$\triangle$ 48(1176). Prove that if in a triangle with angles $\alpha, \beta$ and $\gamma$ the angle $\gamma$ is obtuse, then the product of the tangents of the angles $\alpha$ and $\beta$ is less than 1.
Proof. By the condition $\alpha+\beta+\gamma=180^{\circ}$, $\gamma>90^{\circ}$. Therefore, $\alpha+\beta<90^{\circ}$, i.e., $\alpha<90^{\circ}-\beta$, from which $\operatorname{tg} \alpha<\operatorname{tg}\left(90^{\circ}-\beta\right)$. We have:
$$
\operatorname{tg} \alpha \operatorname{tg} \beta<\operatorname{tg}\left(90^{\circ}-\beta\right) \operatorname{tg} \beta=\operatorname{ctg} \beta \operatorname{tg} \beta=1 .
$$
The statement is proved.
|
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 trial, we find $a: a=7^{2} ; m$ is found from the equality $m=21 a$.
Answer: 1029.
Remark. After solving this problem, it is advisable to propose a more general one: “The product $21 m$, where the factor $m$ is a four-digit number, can be represented as the square of a natural number. Find $m$.”
Answer: $m \in \left\{21 \cdot 7^{2} ; 21 \cdot 8^{2} ; 21 \cdot 9^{2} ; \ldots ; 21 \cdot 21^{2}\right\}$.
|
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 equation:
$$
2 \frac{2}{3}=4 \cdot \frac{2}{3} ; 2 \frac{2}{3}=2 \frac{2}{3} .
$$
b) Method I. $\sqrt{5 \frac{5}{24}}=\sqrt{\frac{125}{24}}=\sqrt{\frac{25 \cdot 5}{24}}=5 \sqrt{\frac{5}{24}}$.
Method II. $5 \sqrt{\frac{5}{24}}=\sqrt{\frac{25 \cdot 5}{24}}=\sqrt{\frac{125}{24}}=\sqrt{5 \frac{5}{24}}$.
Method III. $5 \frac{5}{24}=25 \cdot \frac{5}{24} ; 5 \frac{5}{24}=5 \frac{5}{24}$.
Derive the condition under which
$$
\sqrt{k \frac{k}{n}}=k \sqrt{\frac{k}{n}}, \text { i.e., } \sqrt{\frac{k n+k}{n}}=k \sqrt{\frac{k}{n}}
$$
Method I. Square the obtained equation:
$$
\frac{k n+k}{n}=\frac{k^{3}}{n}
$$
from which $\frac{k(n+1)}{n}=\frac{k^{\prime}}{n}$, i.e., $n+1=k^{2}, k=\sqrt{n+1}$.
Thus, if $n+1=k^{2}$, then $\sqrt{k \frac{k}{n}}=k \sqrt{\frac{k}{n}}$. For example, if $n=15$, we get $k=4$ and $\sqrt{4 \frac{4}{15}}=4 \sqrt{\frac{4}{15}}$, if $n=80$, we get $k=9$ and $\sqrt{9 \frac{9}{80}}=9 \sqrt{\frac{9}{80}}$.
Method II. Let the integer part of the number $a$ be $x([a]=x)$, and the fractional part of the number $a$ be $y(\{a\}=y)$. Find the relationship between the integer and fractional parts of the number $a$, under which
$$
\sqrt{[a]+\{a\}}=[a] \cdot \sqrt{|a|} .
$$
## Square the equation (1):
$$
[a]+\{a\}=[a\}^{2} \cdot\{a\}
$$
from which $[a]=\{a\} \cdot\left([a]^{2}-1\right)$, i.e., $\{a\}=\frac{[a]}{[a]^{2}-1}$.
For example, if $[a]=6$, then $\{a\}=\frac{6}{36-1}=\frac{6}{35}$. We have $\sqrt{6 \frac{6}{35}}=$ $=6 \sqrt{\frac{6}{35}}$.
If $[a]=10$, then $\{a\}=\frac{10}{100-1}=\frac{10}{99}$. We have $\sqrt{10 \frac{10}{99}}=10 \sqrt{\frac{10}{99}}$.
$\triangle 79(1189)$. Find the value of the expression
$$
\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}
$$
Solution. Method I.
$$
\begin{aligned}
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{99+\sqrt{100}}}= \\
= & \frac{2-1}{\sqrt{2}+\sqrt{1}}+\frac{3-2}{\sqrt{3}+\sqrt{2}}+\frac{4-3}{\sqrt{4}+\sqrt{3}}+\ldots+\frac{100-99}{\sqrt{100}+\sqrt{99}}=
\end{aligned}
$$
$$
\begin{gathered}
=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\ldots+\sqrt{100}-\sqrt{99}= \\
=\sqrt{100}-\sqrt{1}=9
\end{gathered}
$$
Method II. By rationalizing the denominator of each fraction, we get that the denominator of each fraction will be equal to one, and the problem reduces to adding the numerators:
$$
\begin{gathered}
(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\ldots+\sqrt{100}-\sqrt{99}= \\
=\sqrt{100}-\sqrt{1}=9
\end{gathered}
$$
$80(1190)$. From the first hundred natural numbers, 51 numbers are taken arbitrarily. Prove that among the taken numbers, there are definitely two such numbers, one of which is a multiple of the other.
Solution. Any natural number can be represented in the form $2^{n} \cdot p$, where $n$ is a non-negative integer, and $p$ is an odd number. Represent each of the first hundred natural numbers in the form $2^{n} \cdot p$, where $n$ is a non-negative integer, and $p$ is an odd number (for odd numbers, $n=0$). For example: $15=$ $=2^{0} \cdot 15 ; 8=2^{3} \cdot 1 ; 24=2^{3} \cdot 3 ; 98=2 \cdot 49$.
Obviously, from two numbers $2^{n} \cdot p$ and $2^{n} \cdot p$, having the same factor $p$, one is a multiple of the other. Indeed, since the first hundred contains 50 odd numbers, the factor $p$ in the expression $2^{n} \cdot p$ for the numbers of the first hundred can take only 50 different values. Therefore, if we take 51 numbers from the first hundred, at least two of them will have the same factor $p$, and thus one of them is a multiple of the other.
|
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 at the same place but one hour earlier. What is the speed of each car?
|
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/h $(10-8=2)$ and will be $\frac{60}{t-1}$ kilometers per hour.
We have the equation
$$
\frac{60}{t-1}-\frac{60}{t}=2, \text { from which } t=6
$$
If the speed of the first car is $v_{1}$ kilometers per hour, then in 6 hours it will travel a distance of $6 v_{1}$ kilometers. The same distance with an increased speed of 10 km/h, this car will take 1 hour less, i.e., we have the equation
$$
6 v_{1}=5\left(v_{1}+10\right), \text { from which } v_{1}=50
$$
Similarly, if $v_{2}$ is the speed of the second car, then we have the equation $6 v_{2}=5\left(v_{2}+8\right)$, from which $v_{2}=40$.
Method II. Let the difference in the speeds of the cars be $v$ kilometers per hour. If the speeds of the cars are increased by 10 km/h and 8 km/h, respectively, the difference in speeds will be $(v+2)$ kilometers per hour. Since with the increased speeds, one car will catch up with the other 1 hour earlier, we have the equation
$$
\frac{60}{v}-\frac{60}{v+2}=1, \text { from which } v=10
$$
Since the difference in the speeds of the cars is 10 km/h, one car will catch up with the other in 6 hours $(60: 10=6)$, and with the increased speeds, in 5 hours $(60:(10+2)=5)$. Since the first car will catch up with the second in the same place, we have the equation $6 v_{1}=5\left(v_{1}+10\right)$, where $v_{1}$ is the speed of the first car. From this, $v_{1}=50.50-10=40$ km/h is the speed of the second car.
Answer: 50 km/h; 40 km/h.
$\square 83(1193)$. Construct the graph of the equation:
a) $|x|+|y|=4$; b) $|x|-|y|=4$.
Solution. a) Method I. Based on the definition of the absolute value of a number, we have:
198
Fig. 33
Fig. 34
Fig. 35
1) For $x \geqslant 0, y \geqslant 0$, we have the equation $x+y=4$.
2) For $x \geqslant 0, y<0$, we have the equation $x-y=4$.
3) For $x<0, y \geqslant 0$, we have the equation $-x+y=4$.
4) For $x<0, y<0$, we have the equation $-x-y=4$.
We construct the graphs of the four obtained equations in the corresponding quadrants (Fig. 33).
Method II. Move $|x|$ to the right side of the given equation, we get $|y|=4-|x|$. We will construct the graph of this equation in the following sequence:
1) $y=4-x$ (Fig. 34)
2) $y=4-|x|$ (Fig. 35)
3) $|y|=4-|x|$ (Fig. 33).
Note. It is advisable to explain to students in advance the rules for constructing graphs of functions $y=f(|x|)$ and graphs of dependencies $|y|=f(x)$.
b) Method I. Based on the definition of the absolute value of a number, we have four equations:
1) $x-y=4$ for $x \geqslant 0, y \geqslant 0$;
2) $x+y=4$ for $x \geqslant 0, y<0$
3) $-x-y=4$ for $x<0, y \geqslant 0$;
4) $-x+y=4$ for $x<0, y<0$.
Fig. 36
Fig. 37
Fig. 38
By constructing the graph of each of the four equations for the corresponding quadrants, we obtain the graph of the given equation (Fig. 36).
Method II. The graph of the equation $|y|=|x|-4$ will be constructed in the following sequence:
$\begin{array}{ll}\text { 1) } y=x-4 & \text { (Fig. } 37) \text {; } \\ \text { 2) }\end{array}$
2) $y=|x|-4$ (Fig. 38);
3) $|y|=|x|-4$ (Fig. 36).
|
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:
$$
\begin{gathered}
(a+2)(a+5)(a+8)(a+11)+b= \\
=(a+2)(a+11)(a+5)(a+8)+b= \\
=\left(a^{2}+13 a+22\right)\left(a^{2}+13 a+40\right)+b= \\
=c(c+18)+b=c^{2}+18 c+b, \text { where } c=a^{2}+13 a+22 .
\end{gathered}
$$
Find the discriminant of the trinomial $c^{2}+18 c+b: D=324-4 b$. Since the quadratic trinomial $c^{2}+18 c+b$, which has a positive leading coefficient, takes positive values for any value of the variable only if its discriminant is negative, we require that the inequality $324-4 b < 0$ holds, i.e., $b > 81$.
The smallest integer satisfying the condition $b > 81$ is 82. Therefore, $b=82$ is the smallest integer for which the trinomial $c^{2}+18 c+b$ takes positive values for any $c$, and thus the expression $(a+2)(a+5)(a+8)(a+11)+82$ is positive for any value of $a$.
The solution to the problem can be simplified if we choose $c$ to be the expression $a^{2}+13 a+31$. Then $\left(a^{2}+13 a+22\right) \times$ $\times\left(a^{2}+13 a+40\right)+b=(c-9)(c+9)+b=c^{2}-81+b>0, \quad$ from which $b>81$.
|
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 consecutive natural numbers, the right side of the equation must also be the product of two consecutive natural numbers. One of them is prime (37), and the other is a multiple of 6, so it must be 36. We have $n(n+1)=36 \cdot 37, n=36$.
It is advisable to check:
$$
1+2+\ldots+36=\frac{(1+36) \cdot 36}{2}=666
$$
Answer: 36 numbers need to be added.
|
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 of this progression?
|
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 $k=\frac{4 n-2}{3}=n+\frac{n-2}{3}$.
We require that $n-2$ be divisible by three, i.e., $n-2=3 m$, where $m=0,1,2, \ldots$. Hence, $n=3 m+2$.
Thus, the common terms of both progressions are those terms of the progression $\left(a_{n}\right)$ that have numbers $n=3 m+2$, where $m=0,1,2, \ldots$, i.e., $a_{2}, a_{5}, a_{8}, a_{11}, \ldots$, or $5,17,29,41, \ldots$.
Clearly, the common terms of both progressions form an arithmetic progression, the first term of which is 5, and the common difference is 12.
Method II. We have:
$$
a_{n}=1+4(n-1), b_{n}=2+3(n-1)
$$
Then
$$
a_{n+1}=a_{n}+d=a_{n}+4=5+4(n-1)
$$
212
$$
\begin{gathered}
b_{n+1}=b_{n}+d=b_{n}+3=5+3(n-1) \\
c_{n}=a_{n+1}=b_{n+1}=5+4(n-1)=5+3(n-1)
\end{gathered}
$$
Therefore, $4(n-1)=3(n-1)$, from which we get that $n-1$ is divisible by 4 and by 3, i.e., $n-1$ is divisible by 12.
$$
n-1=12(k-1) ; c_{k}=5+12(k-1)
$$
|
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:
$$
\left\{\begin{array}{c}
x_{1}^{2}-2 x_{1} x_{2}+x_{2}^{2}=20 \\
x_{1}^{2}+2 x_{1} x_{2}+x_{2}^{2}=4
\end{array}\right.
$$
from which $4 x_{1} x_{2}=-16 ; a=x_{1} x_{2}=-4$.
Answer: $a=-4$.
|
-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 a perfect square. Similarly, it can be shown that the number $y$ is also a perfect square.
Thus, the given equation has as many solutions in integers as there are ways to represent the number 9000 as the sum of two addends (taking into account the order). Since there are 9001 such ways, the given equation has 9001 solutions.
Answer: 9001.
|
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:
$$
\begin{gathered}
\left(\sin ^{2} \alpha+\sin ^{2}\left(\alpha+90^{\circ}\right)\right)+\left(\sin ^{2}\left(\alpha+1^{\circ}\right)+\sin ^{2}\left(\alpha+91^{\circ}\right)\right)+\ldots+ \\
+\left(\sin ^{2}\left(\alpha+89^{\circ}\right)+\sin ^{2}\left(\alpha+179^{\circ}\right)\right)=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+ \\
+\left(\sin ^{2}\left(\alpha+1^{\circ}\right)+\cos ^{2}\left(\alpha+1^{\circ}\right)\right)+\ldots+ \\
+\left(\sin ^{2}\left(\alpha+89^{\circ}\right)+\cos ^{2}\left(\alpha+89^{\circ}\right)\right)=90
\end{gathered}
$$
A n s w e r: 90 .
|
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 $y(x)$ is 5.
Answer: 5.
$\square$ 141. Prove the identity $\cos 25^{\circ}+\cos 47^{\circ}=\cos 11^{\circ}+\cos 61^{\circ}+\sin 7^{\circ}$.
Solution. Method I. Multiply both sides of the identity by $2 \cos 18^{\circ}$. In the left part, we will have:
$2 \cos 18^{\circ}\left(\cos 25^{\circ}+\cos 47^{\circ}\right)=\cos 43^{\circ}+\cos 7^{\circ}+\cos 65^{\circ}+\cos 29^{\circ}$.
In the right part, we will get:
$$
\begin{aligned}
& 2 \cos 18^{\circ}\left(\cos 11^{\circ}+\cos 61^{\circ}+\sin 7^{\circ}\right)= \\
&= \cos 29^{\circ}+\cos 7^{\circ}+\cos 79^{\circ}+\cos 43^{\circ}+\sin 25^{\circ}-\sin 11^{\circ}
\end{aligned}
$$
Since $\cos 79^{\circ}=\sin 11^{\circ}$, and $\sin 25^{\circ}=\cos 65^{\circ}$, the identity is proven.
Method II. Apply the formula for the sum of cosines of two angles to the left part of the identity and to the sum $\cos 61^{\circ}+\cos 83^{\circ}\left(\sin 7^{\circ}=\right.$ $=\cos 83^{\circ}$ ) in the right part of the identity. We will have:
$$
2 \cos 36^{\circ} \cos 11^{\circ}=\cos 11^{\circ}+2 \cos 72^{\circ} \cos 11^{\circ}
$$
Move the product $2 \cos 72^{\circ} \cos 11^{\circ}$ to the left part and divide both sides of the obtained identity by $\cos 11^{\circ} \neq 0$. We will get the identity
$$
\begin{aligned}
& 2 \cos 36^{\circ}-2 \cos 72^{\circ}=1 \\
& 2 \cdot 2 \sin 54^{\circ} \sin 18^{\circ}=1
\end{aligned}
$$
To prove the obtained identity (1), multiply and divide its left part by $\cos 18^{\circ}$ (to use the formula for the sine of a double argument). We will have:
$$
\begin{gathered}
2 \cdot 2 \sin 54^{\circ} \sin 18^{\circ}=\frac{2 \cdot 2 \sin 54^{\circ} \sin 18^{\circ} \cos 18^{\circ}}{\cos 18^{\circ}}= \\
=\frac{2 \sin 54^{\circ} \sin 36^{\circ}}{\cos 18^{\circ}}=\frac{2 \cos 36^{\circ} \sin 36^{\circ}}{\cos 18^{\circ}}= \\
=\frac{\sin 72^{\circ}}{\cos 18^{\circ}}=1
\end{gathered}
$$
The identity (1), and thus the given identity, is proven.
Remark. It is advisable to solve this problem simultaneously with problems 1217 and 1223 from [3].
|
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
$$
\left(\frac{60}{x}-0.5\right)(x+4)=60, \text{ or } x^{2}-4 x-480=0 \text{. }
$$
Solving the obtained quadratic equation, we find $x_{1}=-24$, $x_{2}=20$.
Since only the positive root satisfies the condition of the problem, we have $x=20$.
Answer: 20 trucks.
|
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 20 minutes after the meeting. Find the distance from $A$ to $B$.
|
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 pedestrian walked 1 km less than the second, we have the equation
$$
\frac{4}{3} v_{2}-\frac{3}{4} v_{1}=1
$$
Since before the meeting, the pedestrians spent the same amount of time, we have
$$
\frac{\frac{4}{3} v_{2}}{v_{1}}=\frac{\frac{3}{4} v_{1}}{v_{2}}
$$
Thus, we have the system of equations:
$$
\left\{\begin{array} { l }
{ \frac { 4 } { 3 } v _ { 2 } - \frac { 3 } { 4 } v _ { 1 } = 1 , } \\
{ \frac { \frac { 4 } { 3 } v _ { 2 } } { v _ { 1 } } = \frac { \frac { 3 } { 4 } v _ { 1 } } { v _ { 2 } } , }
\end{array} \text { or } \quad \left\{\begin{array}{l}
v_{2}=\frac{12+9 v_{1}}{16}, \\
\frac{4}{3} v_{2}^{2}=\frac{3}{4} v_{1}^{2},
\end{array}\right.\right.
$$
from which $7 v_{1}^{2}-24 v_{1}-16=0, v_{1}=4$. Therefore, the distance from $A$ to $B$ is 7 km $\left(2 \cdot 4 \cdot \frac{3}{4}+1=7\right)$.
A n s w e r: 7 km.
## CONTENTS
Preface . . . . . . . . . . . . . . . . . 3
§1. Why do we solve problems in school (on the functions of problems in mathematics education) . . . . . . . . . . 5
§ 2. On the methodology of teaching students to solve non-standard algebraic problems . . . . . . . . . . 11
§3. On the role of observations and induction in finding ways to solve non-standard algebraic problems . . . . . . . . . 19
§4. On finding different ways to solve problems . . . . 24
§ 5. On constructing graphs of functions and dependencies containing the modulus sign . . . . . . . . . . 30
§6. Methodological recommendations for using challenging problems in the process of teaching algebra . . . . . . . . . 35
§ 7. Challenging problems in the algebra course for 7th grade . . . 40
§8. Challenging problems in the algebra course for 8th grade . . . 99
§9. Challenging problems in the algebra course for 9th grade . . . 157
## Educational Publication Kostrikina Nina Petrovna CHALLENGING PROBLEMS IN THE ALGEBRA COURSE $7-9$ grades
Editorial Board Chair $T$. A. Burmistrova Editor L. M. Kotoeva
Assistant Editor L. I. Zasedateleva Artists L. B. Nikolaev, V. V. Kostin Art Editor Yu. V. Pakhomov Technical Editor N. T. Rudnikova
Proofreader M. Yu. Sergeeva
IB № 12646
Submitted for typesetting 25. 12. 90. Signed for printing from transparencies 12. 08. 91. Format $60 \times 90^{1 / 16}$. Offset paper. Font. Offset printing. Usual printing sheets $15+0.25$ f. Usual printed sheets 15.75. Usual publication sheets $13.34+0.34$ f. Price 1 r. 90 k.
Order of the Red Banner of Labor publishing house "Prosveshchenie" of the Ministry of Printing and Mass Information of the RSFSR. 129846, Moscow, 3rd passage of Marinskaya Roshcha, 41.
Transparencies made by the Order of the Red Banner of Labor Saratov Polygraphic Combine of the Ministry of Printing and Mass Information of the RSFSR. 410004, Saratov, ul. Chernyshevskogo, 59.
Printed through V/O "Vneshpechat"
Printed by Graphischer Großbetrieb Pößneck GmbH - Ein Mohndruck-Betrieb
## SOME "REMARKABLE"
$$
\begin{aligned}
& (1+3+5+\ldots+(2 n-1))^{2}=n^{2} \\
& 1^{3}+2^{3}+3^{3}+\ldots+n^{3}=(1+2+3+\ldots+n)^{2} \\
& \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(n-1) n}=\frac{n-1}{n} \\
& \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}
\end{aligned}
$$
## EXAMPLES OF RULES
1) $\overline{a b} \cdot 11=100 a+10(a+b)+b$
$36 \cdot 11=100 \cdot 3+10(3+6)+b=391$
$\overline{a b c} \cdot 11=1000 a+100(a+b)+10(b+c)+c$
$243 \cdot 11=1000 \cdot 2+100(2+4)+10(4+3)+3=$
2) $\overline{a 5}^{2}=a(a+1) \cdot 100+25$
$65^{2}=6 \cdot 7 \cdot 100+25=4225$
3) If $b+c=10$, then $\overline{a b} \cdot \overline{a c}=100 a(a+1)+b c$ $38 \cdot 32=3 \cdot 4 \cdot 100+8 \cdot 2=1216$
## INEQUALITIES
$$
\begin{aligned}
& a+\frac{1}{a} \geqslant 2 \text { for } a>0 \\
& \frac{a+b}{2} \geqslant \sqrt{a b}, \frac{a+b+c}{3} \geqslant \sqrt{a b c} \text { for } a \geqslant 0, b \geqslant 0, c \geqslant 0 \\
& (a+b)(b+c)(c+a) \geqslant 8 a b c \quad \text { for } a>0, b>0, c>0 \\
& \sqrt{(a+c)(b+d)} \geqslant \sqrt{a b}+\sqrt{c d} \text { for } a \geqslant 0, b \geqslant 0, c \geqslant 0, d \geqslant 0
\end{aligned}
$$
## MENTAL CALCULATION
$$
\begin{aligned}
& \text { 4) }(a+b)(a-b)=a^{2}-b^{2} \\
& 71 \cdot 69=(70+1)(70-1)= \\
& =4900-1=4899 \\
& 111 \cdot 89=(100+11)(100-11)= \\
& =10000-121=9879 \\
& \text { 5) } a^{2}=a^{2}-b^{2}+b^{2}=(a+b)(a-b)+b^{2} \\
& 27^{2}=(27+3)(27-3)+3^{2}=729
\end{aligned}
$$
[^0]: ' Pólya G. Mathematical Discovery - M. Science, $1970-$ P. 16
[^1]: ' Problems $1-47$ in parentheses indicate the exercise numbers in the textbook $|4|$ 40
[^2]: ' Problems $48-79$ in parentheses indicate the exercise numbers in the textbook |I| 60
[^3]: ' Problems $80-120$ in parentheses indicate the exercise numbers in the textbook [7].
[^4]: ' Problems $1-64$ in parentheses indicate the exercise numbers in the textbook [6].
[^5]: ${ }^{1}$ Problems $65-107$ in parentheses indicate the exercise numbers in the textbook [3].
|
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 wine?
This is the most famous problem of its kind; it is known as Poisson's problem. The famous French mathematician, mechanician, and physicist Siméon Denis Poisson (1781-1840) solved it in his youth and later said that it was precisely this problem that inspired him to become a mathematician.
|
$\triangle$ 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. Initially, both vessels are empty (first column). We fill the vessel with a capacity of 8 pints (second column), then we fill the vessel with a capacity of 5 pints from it (third column), then we pour these 5 pints from the smaller vessel into a barrel that already contains 12 pints (fourth column), then we transfer 3 pints of wine from the vessel with a capacity of 8 pints to the vessel with a capacity of 5 pints (fifth column), and so on until the larger of the two vessels contains 6 pints of wine.
Try to solve the problem differently, by first filling the vessel with a capacity of 5 pints. Will the solution be shorter?
Another method for solving such problems - using graphs - is discussed in the reference [16].
|
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 | $a$ | $a-5$ | $a-5$ | $a-10$ | $a-10$ | $a-1$ | $a-1$ | $a-6$ | $a-6$ |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Bucket with volume 9 l | 0 | 0 | 5 | 5 | 9 | 0 | 1 | 1 | 6 |
| Bucket with volume 5 l | 0 | 5 | 0 | 5 | 1 | 1 | 0 | 5 | 0 |
|
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 use the 2-liter can, there will be $n-2$ liters left, and it can be transferred in $a_{n-2}$ ways. Therefore, we get:
$$
a_{n}=a_{n-1}+a_{n-2}
$$
This formula can be used only when $n>2$. (In mathematics, a formula that expresses a general term of a sequence in terms of one or several of its preceding terms is called a recurrence formula.)
If initially there were 1 or 2 liters of milk in the container, it can be transferred to another container in one or two ways, respectively:
$$
a_{1}=1, \quad a_{2}=2
$$
Let's introduce the zeroth term of the sequence, assuming $a_{0}=1$. Then the obtained formula can be used even when $n=2$:
$$
a_{2}=a_{1}+a_{0}=1+1=2
$$
Now let's calculate all terms from $a_{3}$ to $a_{12}$ in sequence:
$$
\begin{gathered}
a_{3}=a_{2}+a_{1}=2+1=3, \quad a_{4}=a_{3}+a_{2}=3+2=5, \quad a_{5}=a_{4}+a_{3}=5+3=8, \\
a_{6}=a_{5}+a_{4}=8+5=13, \quad a_{7}=21, a_{8}=34, \\
a_{9}=55, \quad a_{10}=89, \quad a_{11}=144, \quad a_{12}=233 .
\end{gathered}
$$
Answer: 233.
In solving this problem, we obtained the sequence
$$
1,1,2,3,5,8,13,21,34, \ldots
$$
where the first two terms are 1, and each subsequent term, starting from the third, is the sum of the two preceding terms. This sequence plays a significant role in number theory and is called the Fibonacci sequence, named after the Italian scholar Leonardo Fibonacci (1180-1240).
|
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
C
10
Let's denote the number of elements in any finite set $A$ by $n(A)$. Then, according to the problem,
$$
n(A)=912, \quad n(B)=653, \quad n(A \cap B)=435
$$
We need to find the number of elements in the union of sets $A$ and $B$.
First, let's add the numbers $n(A)$ and $n(B)$. However, the elements that belong to the intersection of sets $A$ and $B$ are counted twice. Therefore, we need to subtract $n(A \cap B)$ from this sum. We get:
$$
n(A \cup B)=n(A)+n(B)-n(A \cap B).
$$
Substitute the values of $n(A)$, $n(B)$, and $n(A \cap B)$ into formula (1):
$$
n(A \cup B)=912+653-435=1130
$$
Answer: 1130.
Clearly, formula (1) is valid not only under the conditions of problem 66, but also for any finite sets $A$ and $B$.
|
1130
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
68. A large group of tourists set off on an international trip. Out of them, 28 people speak English, 13 speak French, 10 speak German, 8 speak both English and French, 6 speak both English and German, 5 speak both French and German, 2 speak all three languages, and 41 people do not speak any of the three languages. How many tourists are in the group
|
Let's denote the set of tourists in the group who speak English, French, or German, respectively, as $A$, $B$, and $C$. According to the problem,
\[
\begin{gathered}
n(A)=28, \quad n(B)=13, \quad n(C)=10, \quad n(A \cap B)=8 \\
n(A \cap C)=6, \quad n(B \cap C)=5, \quad n(A \cap B \cap C)=2
\end{gathered}
\]
First, we need to find the number of tourists who speak at least one of the three foreign languages, i.e., $n(A \cup B \cup C)$. For this, we will use Venn diagrams (Fig. 27).
Let's calculate the sum $n(A) + n(B) + n(C)$. Since each of the numbers $n(A \cap B)$, $n(A \cap C)$, and $n(B \cap C)$ is included twice in this sum, we need to subtract the sum $n(A \cap B) + n(A \cap C) + n(B \cap C)$ from it.
Now, let's determine how many times the number $n(A \cap B \cap C)$ is included in the expression
\[
n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C)
\]
It is included three times with a positive sign (in each of the terms $n(A)$, $n(B)$, and $n(C)$) and three times with a negative sign (in each of the terms $n(A \cap B)$, $n(A \cap C)$, and $n(B \cap C)$). Therefore, to avoid losing the tourists who belong to the set $A \cap B \cap C$, we need to add the number $n(A \cap B \cap C)$. We get:
\[
\begin{gathered}
n(A \cup B \cup C) = n(A) + n(B) + n(C) - \\
- n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C).
\end{gathered}
\]
Then we have:
\[
n(A \cup B \cup C) = 28 + 13 + 10 - 8 - 6 - 5 + 2 = 34.
\]
Thus, the total number of tourists in the group is $34 + 41 = 75$.
Answer: $75$.
Formula (2) is valid for any three finite sets $A$, $B$, and $C$.
Now, using formulas (1) and (2), let's try to guess how, in general, for any finite sets $A_{1}, A_{2}, A_{3}, \ldots, A_{k}$, the value of $n\left(A_{1} \cup A_{2} \cup \ldots \cup A_{k}\right)$ is calculated.
The following formula is valid:
\[
\begin{aligned}
& n\left(A_{1} \cup A_{2} \cup \ldots \cup A_{k}\right) = n\left(A_{1}\right) + n\left(A_{2}\right) + \ldots + n\left(A_{k}\right) - n\left(A_{1} \cap A_{2}\right) - \\
& \quad - n\left(A_{1} \cap A_{3}\right) - \ldots - n\left(A_{k-1} \cap A_{k}\right) + n\left(A_{1} \cap A_{2} \cap A_{3}\right) + \\
& \quad + n\left(A_{1} \cap A_{2} \cap A_{4}\right) + \ldots + (-1)^{k-1} \cdot n\left(A_{1} \cap A_{2} \cap \ldots \cap A_{k}\right).
\end{aligned}
\]
This formula is called the principle of inclusion and exclusion.
In particular, when $k$ is even, the last term in the right-hand side of formula (3) has a negative sign (as in formula (1)), and when $k$ is odd, it has a positive sign (as in formula (2)).
|
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 equilibrium (Fig. 30). Then the counterfeit coin is in the triplet with numbers $7,8,9$, and it can be found using one more weighing (see the solution to problem 90).
Suppose the balance is not in equilibrium. In this case, the counterfeit coin is in one of the first two triplets, which is heavier, and it can also be found using one more weighing.
Thus, two weighings were needed. Could the goal have been achieved in one? To answer this question, we need to change the weighing system, placing not three, but four, or two, or even one coin on each pan of the balance. But each of the mentioned options is less advantageous than the one we discussed.
Answer: two.
|
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 (see problem 91).
If the scales are unbalanced, the counterfeit coin is in the lighter of these two sets of three. One more weighing is needed to identify the counterfeit coin.
In general, three weighings are required. But could we have managed with two? Let's try placing four coins on each pan (other options are clearly worse). However, the counterfeit coin could be in one of these sets of four, which would still require two more weighings, totaling three.
Answer: three.
|
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 problem 90).
The scales may be unbalanced. In this case, both counterfeit coins are contained in one triplet - the lighter one (in this case - on the left pan). One more weighing will be required.
Answer: in three.
|
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 zeros in the second position; then the other zero can be placed in the third, fourth, or fifth positions. If we now fix one zero in the third position, the second zero can be placed in the fourth or fifth positions (the case where zeros are in the third and second positions has already been considered). Finally, if we fix one zero in the fourth position, the other zero can only be placed in the fifth position. This gives us the following 6 numbers:
$$
10011,10101,10110,11001,11010,11100 .
$$
## Answer: 6 .
|
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 there are nine such two-digit numbers - from 90 to 98. Therefore, the total number of numbers is $1+2+3+\ldots+9=45$.
Answer: 45.
48
|
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
$$
a_{n}=a_{n-1}+a_{n-2}(n \in N, n>2)
$$
and initial conditions $a_{1}=1$ and $a_{2}=2$. We find its 10th term.
Answer: $89$.
|
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 cases, D can be placed in one of two places. If G takes the fifth place, then there are three places left for D. Therefore, the number of ways to arrange the pair of tigers is
$$
3+2+2+2+3=12
$$
Now, let's deal with the arrangement of the lions. Lion A can be placed in any of the three remaining places after the tigers are arranged; then lion B can be placed in any of the two remaining places; lion C will get the last free place. Thus, the number of ways to arrange the lions, according to the rule of product, is $3 \cdot 2 \cdot 1=6$.
Therefore, the total number of ways to arrange the animals, based on the same rule, is $12 \cdot 6=72$.
Answer: 72.
|
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 $6 \cdot 5 \cdot 4 \cdot 3=360$.
Answer: 360.
|
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:
$$
A_{15}^{4}=15 \cdot 14 \cdot 13 \cdot 12=32760
$$
Answer: 32760.
|
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 $: 45$.
|
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 have:
$$
C_{9}^{3}=\frac{9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3}=84
$$
Answer: 84.
|
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 line:
$$
C_{6}^{1} \cdot C_{8}^{2}=6 \cdot 28=168
$$
This gives us $120+168=288$ triangles.
Answer: 288.
|
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:
$$
\frac{1}{4} D = 25 - y^2 + y - 13 \geq 0, \quad -y^2 + y + 12 \geq 0, \quad y^2 - y - 12 \leq 0, \quad -3 \leq y \leq 4
$$
Since $y$ is a digit, we find more precise bounds for $y$:
$$
0 \leq y \leq 4
$$
Now we will check the cases $y=0,1,2,3,4$.
Answer: 54.
|
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
$$
5,8,11, \ldots, 5+3 k=2+(3+3 k)=2+3(k+1)
$$
are equal to 0, i.e., the terms whose indices, when divided by 3, give a remainder of 2. (The latter can be strictly proven by mathematical induction).
Let's check if the remainder of the division of the number 1994 by 3 is 2:
$$
1994=3 \cdot 664+2
$$
Therefore, $x_{1994}=0$.
Answer: 0.
|
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 $7^{1995}$ coincide with the last two digits of $7^{3}$, and therefore form the number 43. Answer: 43.
|
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.
How many people are in the room and how many of them 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 person are false: in reality, there are more than three people in the room, and not all of them are liars.
Let's examine the statements of the second person. His statement "Not all of us are liars" is true, so he is a truth-teller. This means there are no more than four people in the room. Considering the previous conclusion, we get that there are exactly four people.
Since the statement of the third speaker, "There are five of us," is false, he is a liar. Therefore, there are not three liars in the room. Moreover, there are at least two liars in the room—the first and the third—and the number of liars is less than four, since there is a truth-teller among those present. In this case, there are exactly two liars.
Answer: 4 people, 2 liars.
|
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 with the correct strategy?
|
$\triangle$ 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). Consequently, in this case, B wins.
2) Suppose B makes the first move.
If B first attaches, say, a blue ball to one of the vertices, then A should attach a blue ball, but not to an adjacent vertex, rather to the opposite vertex, and so on (Fig. 37). Clearly, in this case, A wins.
Answer: The one who moves second wins.
|
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 in the same column, is constant. Find this sum.
|
$\triangle$ 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 these numbers are pairwise distinct, because if, for example, $a_{1}=a_{2}$, then the numbers $a_{1}$ and $10+a_{2}$ would be in the same column of the table. We obtain:
$$
\begin{gathered}
S=a_{1}+\left(10+a_{2}\right)+\left(20+a_{3}\right)+\ldots+\left(90+a_{10}\right)= \\
=(10+20+30+\ldots+90)+\left(a_{1}+a_{2}+\ldots+a_{10}\right)=450+\left(a_{1}+a_{2}+\ldots+a_{10}\right) .
\end{gathered}
$$
Since the numbers $a_{1}, a_{2}, \ldots, a_{10}$ are pairwise distinct and take all integer values from 1 to 10, each of the natural numbers from 1 to 10 appears in the sum $a_{1}+a_{2}+\ldots+a_{10}$ exactly once. Therefore,
$$
\begin{gathered}
a_{1}+a_{2}+\ldots+a_{10}=1+2+3+\ldots+10=55 \\
S=450+55=505
\end{gathered}
$$
Where is the invariant here? The sum $S$ is the invariant: if some terms in it are replaced by others, but in such a way that all terms of the new sum are in different rows and different columns of the table, the sum will take the same value $S=505$.
Answer: 505.
|
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, while $1997 = 4k+1$ when $k=499$, so it is possible to obtain 1997 pieces.
Answer: 1994 pieces cannot be obtained, 1997 can.
|
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? (There is no need to find the counterfeit coin.)
|
$\triangle$ 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. Then the counterfeit coin is in the third pile.
Now place the first pile of coins on one scale pan, and the third pile on the other. If, for example, the third pile tips the scales, then the counterfeit coin is heavier than the genuine ones.
2) Suppose the scales were unbalanced during the first weighing. Then the counterfeit coin is either in the first or the second pile. Therefore, all the coins in the third pile are genuine.
Place the first pile of coins on one scale pan, and the third pile on the other. If the scales are unbalanced, then the counterfeit coin is in the first pile, and the last weighing
will show whether it is lighter or heavier than the genuine ones. If the scales balance, then the counterfeit coin is in the second pile, and the first weighing will also determine whether it is lighter or heavier than the genuine ones.
Answer: in two.
|
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 halfway point?
|
$\triangle$ 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 me, I have not yet traveled half the journey.
Answer: not traveled; 18
|
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 always be two of them who receive the same number of candies.
Let's also verify that with $n=20$, it is possible to distribute the candies such that any two schoolchildren receive a different number of candies. For this, we will give 19 out of the 20 schoolchildren $0,1,2, \ldots, 18$ candies respectively. In total, this is $0+1+2+\ldots+18=\frac{18 \cdot 19}{2}=171$ candies, and thus $200-171=29$ candies remain undistributed. We will give these to the last, twentieth schoolchild.
Answer: the minimum number of schoolchildren is 21.
|
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 teams won all their matches. Then each of these teams will score $4+6=10$ points. However, 10 points still do not guarantee a team's advancement to the final, since there are five such teams, not four.
We will prove that if a team scores 11 points, it will advance to the final. Assume the opposite: there are five teams, each of which scores at least 11 points. In this case, they will score at least 55 points together, which means the remaining three teams will score no more than 1 point together. But this is impossible, since these three teams will score at least $\frac{3 \cdot 2}{2} \cdot 2=6$ points in matches against each other.
Answer: 11.
|
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, you can transport no more than 12 boxes out of 13.
Five trucks are sufficient in all cases. Indeed, we can load at least 2 tons of cargo into each truck: if less than 2 tons are loaded, then you can add another box (or boxes) since each box weighs no more than one ton by the condition. Therefore, with five trucks, you can load at least 10 tons of cargo.
Answer: 5.
|
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 can be formed? (In this and the next problem, the question is about the smallest number of socks.)
|
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 ensuring that at least 2 socks of the same color are among the taken socks, so the correct answer to the problem is different: 3 socks. If I take 3 socks from the drawer, then they will either all be of the same color (in which case I will certainly be able to choose at least 2 socks of the same color from them), or 2 socks will be of the same color, and the third sock will be of a different color, which will also allow me to form a pair of socks of the same color.
|
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 King and told him in detail everything that had happened.
- All of this is very, very interesting, - the King assured her. - The most remarkable thing about the whole story is that, although you know to whom you returned the rattle, we still do not know who the owner is, Tralala or Trulala.
- You are quite right, your majesty, - agreed Alice. - But what are we to do with the broken rattle?
- Nonsense, - said the King. - It's nothing for me to fix it again.
And true to his word, the White King repaired the rattle so skillfully that it became as good as new, and a few days later he returned it to Alice. With trepidation, Alice set off for the forest, fearing that the brothers were still fighting. But by this time, Tralala and Trulala had declared a truce, and Alice found one of them resting under a tree. Alice approached him and asked:
- Which of you does the rattle belong to?
He answered enigmatically:
- The true owner of the rattle lies today.
What are the chances that he was the true owner of the rattle?
|
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 rattle should have been lying on the day of the meeting. Therefore, in this case, he also cannot be the owner of the rattle.
|
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 hypocrites. You should not marry a liar either because all liars are treacherous. No, no matter what, a decent normal person would be just right for you!”
Suppose you are a resident of this island and a normal person. Your task is to convince the king that you are a normal person.
a) How many true statements will you need for this?
b) How many false statements will you need for the same purpose?
(In both cases, we are talking about the minimum number of statements.)
|
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 two caskets?
|
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 Cellini family. Therefore, the silver casket was made by someone from the Bellini family. Thus, the statement engraved on the lid of the silver casket is true, so neither casket was made by either the son of Bellini or the son of Cellini. Therefore, the golden casket was made by Cellini, and the silver one by Bellini.
|
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 her favorites, and three cookies to the rest of the guests.
How many favorites did the Queen of Hearts have?"
|
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 each of the favorites still needs to receive 1 more pretzel.
Therefore, all the remaining pretzels are intended for the favorites—one pretzel each. This means there must be 10 favorites.
Verification. Each of the 10 favorites should receive 4 pretzels, which totals 40 pretzels for all the favorites. Each of the remaining 20 guests will receive 3 pretzels, which adds up to another 60 pretzels. \(40 + 60 = 100\). Therefore, our solution is correct.
|
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 costs as much as three pretzelkins.
- How much do seven pretzels and four pretzelkins cost?—asked the Mock Turtle.
- Twelve cents more than four pretzels and seven pretzelkins,—came the mysterious answer.
How much does 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). Thus, the difference in cost between 25 and 19 pretzelkins is 12 cents. Therefore, 6 pretzelkins (25-19=6) cost 12 cents, 1 pretzelkin costs 2 cents, and 1 pretzel costs 6 cents.
Verification. 4 pretzels and 7 pretzelkins cost $24+14=38$ cents, and 7 pretzels and 4 pretzelkins cost $42+8=50$ cents, which is indeed 12 cents more than in the first case.
|
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 saw a plate with pretzels. The Mock Turtle first ate half of the pretzels and then, thinking about it, ate one more pretzel. The March Hare ate half of the remaining pretzels and one more pretzel. Then the Dormouse ate half of the remaining pretzels and one more pretzel. At this point, the Cheshire Cat returned and ate half of the remaining pretzels and one more pretzel, after which there were no pretzels left on the plate.
How many pretzels were there at first?
|
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 for the Cheshire Cat. The March Hare saw 14 pretzels on the tray: after eating 7 pretzels and one more pretzel, 6 pretzels were left. The Mad Hatter saw 30 pretzels: after eating 15 pretzels and one more pretzel, 14 pretzels were left.
Thus, there were 30 pretzels on the tray at first.
|
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 condition that the gardener would receive three gingerbread cookies for each day he worked diligently, and if he slacked off, he would not only receive nothing but would also owe one gingerbread cookie.
- And what if the gardener turns out to be a lazybones and slacks off so many days that by the end of twenty-six days, he would owe several gingerbread cookies? - asked Alice. - What then?
- Then his head will certainly be cut off.
- I seem to recall someone telling me about executions that "they never had such a thing" *.
- Of course, his head will remain intact, the Griffin grinned. - All these executions are pure fiction, but everyone is happy.
But let's get back to our problem. When the gardener finished his work, it turned out that the King owed him sixty-two gingerbread cookies.
How many days did the gardener work diligently and how many did he slack off?
- You are clearly fond of gingerbread! - said Alice, looking intently at the Griffin.
- If you are hinting at the trial, miss, - objected the Griffin, - then I repeat, I never stole any gingerbread. They all made it up, that I stole gingerbread!
- I don't understand how you managed to avoid imprisonment, - remarked Alice.
- After the trial, I managed to have a private conversation with the King, - explained the Griffin.
Such an explanation did not fully satisfy Alice.
- I think it's time to change the subject, suggested the Griffin. - We've talked enough about gingerbread. By the way, I know a good problem about the royal clocks.
|
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 could have earned for diligent work and the 1 pretzel that is deducted for idleness). Therefore, the gardener slacked off for 4 days and worked diligently for 22 days.
Verification. For 22 days of diligent work, the gardener earned 66 pretzels. For 4 days of slacking off, the gardener returned 4 pretzels.
Thus, he received a total of 62 pretzels.
|
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 front paws again. After remaining silent for some time, he spoke again:
- Let's say... Nine people got lost in the mountains. They only had enough food for five days. Can you imagine, just five days!
Having reached this point in his story, the Tortoise Quasi was so moved by the tragedy that he could not continue from the overflow of emotions.
- Well, that's enough, there, there! - the Griffin tried to comfort him, patting his back.
- Can you imagine what will happen to them if they are not found! - the Tortoise Quasi sobbed. - But it happened (and this is the most beautiful part of the story!)... Yes, the most beautiful part of the story is that the next day the poor souls met another group of lost people in the mountains...
- What's so beautiful about that? - the Griffin asked.
- The most beautiful part is that the first group generously shared their supplies with the second - they divided the provisions equally among everyone, after which the food lasted for another three days.
How many people were in the second group?
|
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 everyone for 3 days. Therefore, there should have been 12 people in total.
This means that the second group had 3 people.
|
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 for Quasi-Tortoise.
- Alright, - agreed the Griffin, yawning. - We'll do it this way. You tell your problem to Alice, since she hasn't heard it, and I'll take a little nap. Only one condition: wake me up when you're done. Agreed?
The Griffin curled up comfortably and dozed off, while Quasi-Tortoise told Alice the old problem about the frog.
- One morning, a frog fell into a well that was thirty feet deep. During the day, climbing up the walls, it climbed three feet, and at night, while sleeping, it slipped back two feet.
How many days will it take for the frog to get out of the well?
|
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, and so on. Finally, in the morning of the
$28-$th day, the frog is 27 feet above the bottom of the well. By evening of the same day, it will reach the top and climb out of the well, after which it will no longer have to slide back down.
|
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 story is that I kept hearing your voice and couldn't sleep for a minute! Tell us another problem!
- Alright, - the Turtle Quasi agreed. - A man set out on a bicycle for twelve miles to the railway station to catch a train. He reasoned as follows:
- I have one and a half hours to catch the train. Four miles of the road go uphill, and I will have to walk them. I think I can manage it in an hour. Then four miles of the road go downhill. On this section, I can reach a speed of twelve miles per hour. The last four miles of the road are on flat ground. On this section, I can reach a speed of eight miles per hour. On average, this is eight miles per hour, so I will get to the station on time.
Was the cyclist's reasoning correct?
|
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 hill (at 4 miles per hour) and less time descending the hill (at 12 miles per hour).
It is not difficult to calculate how long he was on the road. Climbing the hill took him 1 hour, he spent half an hour (or 30 minutes) traveling on the flat section of the road, and a third of an hour (or 20 minutes) descending the hill. In total, he was on the road for 1 hour and 50 minutes, missing the train by 20 minutes.
|
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 something?
- Everything is correct, - agreed the Tortoise Quasi, - but we don't know what happened afterward! After all, judging by everything, the train could have derailed!
- No, just think about it! - Alice got angry. - First, they come up with all sorts of nonsense, and then they make a sad story out of everything!
The Tortoise Quasi did not answer, only grabbing his head with his front paws.
- Alright, - he said finally. - I will tell you a sad story. One morning, a boy had to go to school...
- That's sad! - agreed the Griffin.
- No, no! That's not sad, - objected the Tortoise Quasi. - The saddest part is still to come.
Alice and the Griffin listened to the whole story attentively but found nothing sad in it.
- So, - continued the Tortoise Quasi, - the father said to the boy: "Hurry up, or you'll be late for school!" - "I won't be late, Dad! - the boy replied. - I have everything calculated precisely. If I walk at a speed of four miles per hour, I will be five minutes late for the start of classes, but if I walk at a speed of five miles per hour, I will arrive ten minutes before the start of classes."
How far is it to school?
|
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 miles per hour). Five miles per hour is one mile in 12 minutes, and 4 miles per hour is one mile in 15 minutes. Therefore, by walking faster, the boy saves 3 minutes per mile, and 15 minutes over a distance of 5 miles.
Thus, the school is 5 miles from home.
Verification. Walking at a speed of 5 miles per hour, the boy spends one hour on the journey, and walking at a speed of 4 miles per hour, he spends one hour and fifteen minutes (in one hour he covers the first 4 miles, and in a quarter of an hour, the last mile), that is, 1 hour and 15 minutes. The time difference is indeed 15 minutes.
|
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 the necessary calculations.
- I got a remainder of two, - she said.
- Wrong! - the Black Queen triumphantly cried. - You don't know division!
- Not a bit of it! - the White Queen confirmed.
Why don't you take a pencil and paper and check if Alice's answer is correct? Just in case, take a look at the solution provided at the end of the book.
|
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
We see that eleven thousand eleven hundred and eleven is 12111, which is a number divisible by 3 without a remainder.
|
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 farmer have?
And Alice was able to answer this question correctly. Can you? (No need to use algebra at all!)
|
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, 60 is half of the total area in acres.
Thus, the farmer had 120 acres of land.
Verification. One third of 120 is 40 (40 acres were used for pumpkins). One quarter of 120 is 30 (30 acres were allocated for peas). One fifth of 120 is 24 (24 acres were planted with beans). Since $40 + 30 + 24 = 94$, corn occupies the remaining $120 - 94 = 26$ acres.
|
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 decimal point are zeros.
|
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 $2^{n}=A^{2 q}-1=$ $=\left(A^{q}-1\right)\left(A^{q}+1\right)$, i.e., both $A^{q}-1$ and $A^{q}+1$ are powers of two, differing by two. Therefore, these are two and four, i.e., $2^{n}=8$.
|
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 their sum?
## 7th grade
|
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 find the number of solutions to the equations $a+b+c=16,17$ in non-negative integers not exceeding 11. There are 216 in total.
|
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 Leningrad style and in the Moscow style, including the ticket 000000?
|
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:
$$
\left\{\begin{array}{l}
x_{0}=x_{0}^{2}+x_{1}^{2}+\ldots+x_{100}^{2} ; \\
x_{1}=2\left(x_{0} x_{1}+x_{1} x_{2}+\ldots+x_{99} x_{100}\right) ; \\
x_{2}=2\left(x_{0} x_{2}+x_{1} x_{3}+\ldots+x_{98} x_{100}\right) \\
x_{3}=2\left(x_{0} x_{3}+x_{1} x_{4}+\ldots+x_{97} x_{100}\right) ; \\
\cdots \cdots \cdots \\
x_{100}=2 x_{1} x_{100}
\end{array}\right.
$$
|
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 second or third) coordinates of any two marked vertices must be different, their sum is no less than \(0+1+2+\ldots+(n-1)=n(n-1) / 2\), and the sum of all coordinates is no less than \(3 n(n-1) / 2\), where \(n\) is the number of all marked vertices of the partition. But, on the other hand, this sum is equal to \(30 n\). Therefore, \(30 n \geqslant 3 n(n-1) / 2\), or \(n-1 \leqslant 20\). An example with 21 vertices is left to the reader as an exercise.
|
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 than 1300 intersections from which it is impossible to exit without violating the movement rules. (The corners of the city are also considered intersections.)
|
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 1500 m path that can be traveled according to the rules.
Now consider the figure shown in Fig. 46, containing eight intersections. By enumeration, it is not difficult to convince oneself that one of these intersections must be a sink. It is left as a not difficult exercise for the reader to cover the city plan with 1300 such figures without overlap.
|
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 expression is 1.
|
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 in these seven vertices can result?
## 9th grade
|
75.18. Answer. There are seven possible arrangements.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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