problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
18*. Which three-digit number has the greatest number of divisors? | 18. 840. This number has 32 divisors. | 840 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,421 |
2. Prove that there do not exist such values of \(m\) and \(n\) for which the polynomials \(3 m^{2}+4 m n-2 n^{2}\) and \(-m^{2}-4 m n+3 n^{2}\) simultaneously take negative values. | 2. Solution. The sum of the given polynomials is \(3 m^{2}+4 m n-2 n^{2}-m^{2}-4 m n+3 n^{2}=2 m^{2}+n^{2} \geqslant 0\),
i.e., it cannot be a negative number. This is only possible when at least the value of one of the polynomials is non-negative for the given values of \(m\) and \(n\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,422 |
3. In Figure 10, the measure of the first angle is \(50^{\circ}+x-y\), and the measure of the second angle is less than the measure of the first by \(10^{\circ}+2 x-2 y\). Prove that line \(D B\) is perpendicular to line \(B K\). | 3. R e e e i e. The magnitude of the second angle is
\[
\left(50^{\circ}+x-y\right)-\left(10^{\circ}+2 x-2 y\right)=40^{\circ}-x+y
\]
Then the sum of the magnitudes of the first and second angles is \(90^{\circ}\). Therefore, the magnitude of angle \(K B D\) is \(180^{\circ}-90^{\circ}=90^{\circ}\) (see Fig. 10). | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,423 |
4. The sum of several addends is equal to \(2 b-1\). To one addend, \(3 b-8\) was added, and from the second, \(-b-7\) was subtracted. How much must be subtracted from the third addend so that the sum becomes equal to zero? | 4. Solution. If \(x\) is the desired number, then
\(2 b-1+(3 b-8)-(-b-7)-x=0 ; x=6 b-2\). | 6b-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,424 |
6. In a three-digit number, the digit in the hundreds place is 2 more than the digit in the units place. Find the difference between this number and the number formed by the same digits but in reverse order. | 6. Solution. Let \(x\) be the tens digit and \(y\) be the hundreds digit of a three-digit number. Then the units digit will be \(y-2\), and thus the desired number is
\(100 y + 10 x + y - 2 - 100(y - 2) - 10 x - 2 = \ldots = 198\). | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,426 |
7. Find \(k\), if it is known that for any \(x\):
\[
\begin{array}{r}
a x^{2}+b x+c \\
+b x^{2}+a x-7 \\
k x^{2}+c x+3 \\
\hline x^{2}-2 x-5
\end{array}
\] | 7. Solution.
\[
c=-5-(-7+3)=-1
\]
\[
\begin{aligned}
a+b & =-2-(-1) \\
k & =1-(a+b)=1-(-1)=2
\end{aligned}
\] | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,427 |
8. Given three different non-zero digits. All possible three-digit numbers are formed from them. Prove that their sum is divisible by 37. | 8. Solution. Let \(a, b\) and \(c\) be three different digits. Then the sum of all the specified three-digit numbers is:
\[
\begin{gathered}
(100 a+10 b+c)+(100 a+10 c+b)+(100 b+10 a+c)+(100 b+ \\
+10 c+a)+(100 c+10 a+b)+(100 c+10 b+a)=222(a+b+ \\
+c)=6 \cdot 37(a+b+c)
\end{gathered}
\]
and therefore it is divisible ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,428 |
9. The country's football cup was played according to the Olympic system: the losing team was eliminated from the competition, and in case of a draw, a replay was scheduled. A total of \(m\) matches were played, including \(n\) replays. How many teams participated in the cup draw? | 9. Solution. There were \(m-n\) matches in total, each with a winner. Therefore, there were \(m\) - \(\eta\) teams that were eliminated from the competition. Only one team, the winner, did not suffer any defeats. Thus, the number of teams that participated in the competition is \(m-n+1\) ( \(m>n\) ). | -n+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 43,429 |
10(!). The first number is \(a\), the second is \(b\), the third is the difference between the second and the first, the fourth is the difference between the third and the second, and so on. What number is in the 124th place? | 10. Solution. The third number is \(b-a\), the fourth \(b-\) \(-a-b=-a\), the fifth \(-a-(b-a)=-b\), the sixth \(-b-(-a)=\) \(=a-b\), the seventh \(a-b-(-b)=a\), i.e., equal to the first number, the eighth \(a-(a-b)=b\), i.e., equal to the second number, and so on. The number at the \(124-\mathrm{th}\) position is equa... | - | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,430 |
11. Find the sum of the values of the polynomial \(x^{5}-1.7 \cdot x^{3}+2.5\) at \(x=19.1\) and \(x=-19.1\). | 11. Solution. If the sign of \(x\) is changed to the opposite, the given polynomial will take the form: \(-x^{5}+1.7 x^{3}+2.5\). The sum of the given polynomial and the obtained one is 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,431 |
12. Prove that if the numbers \(m\) and \(n\) are natural, then numbers of the form \((5 m+n+1)(3 m-n+4)\) are always divisible by two. | 12. Solution. Since the difference
\[
(5 m+n+1)-(3 m-n+4)=\ldots=2 m+2 n-3
\]
is an odd number, one of the numbers \(5 m+n+1\) or \(3 m-\) \(-n+4\) is even, and the other is odd. Therefore, their product is an even number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,432 |
14. Given:
\[
\begin{aligned}
x+y-z & =a-b \\
x-y+z & =b-c \\
-x+y+z & =c-a
\end{aligned}
\]
Prove that \(x+y+z=0\). | 14. Instruction. It is necessary to add all the given equalities term by term.
Add all the given equalities term by term. | Algebra | proof | Yes | Yes | olympiads | false | 43,434 | |
15*. Show that if one of the terms of the fraction
\[
\frac{k^{2}-5 k+8}{k^{2}+6 k+19}
\]
is divisible by 11, then the fraction can be reduced by \(11 .(k \in N\).) | 15. Solution. Let for some natural \(k\) one of the terms of the fraction is divisible by 11. Since, in addition, the difference
\[
\left(k^{2}+6 k+19\right)-\left(k^{2}-5 k+8\right)=\ldots=11 k+11
\]
is divisible by 11, then the second term of the fraction is also divisible by 11. Therefore, the fraction can be redu... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,435 |
16(!). Several numbers are written in a row. The first one is equal to \(a\), and the last one is equal to \(b\). In the second row, the differences between the second and the first, the third and the second, ..., the last and the penultimate members of the finite sequence are written. Find the sum of all these differe... | 16. \(b-a\). Hint. It is useful to first consider a few particular cases.
The translation is provided while maintaining the original text's line breaks and format. | b- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,436 |
17*. Vovochka wrote two numbers on the board. The third one he wrote as the sum of the first two, the fourth as the sum of the third and the second, and so on. Then Vovochka told Sasha the sum of six consecutive numbers, starting from one of the written ones. Upon learning the sum, Sasha immediately determined one of t... | 17. Solution. Let the first of the six addends be \(a\), and the second be \(b\). Then the third, fourth, fifth, and sixth addends will be \(a+b\), \(a+2b\), \(2a+3b\), and \(3a+5b\) respectively. The sum of all six numbers is
\[
\begin{gathered}
a+b+(a+b)+(a+2b)+(2a+3b)+(3a+5b)= \\
= \ldots = 4(2a+3b)
\end{gathered}
... | 2a+3b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,437 |
1. Using the same drawing, find on the coordinate plane the points whose coordinates \(x\) and \(y\) satisfy the conditions: 1) \(x=3\); 2) \(x<3\); 3) \(x>3\); 4) \(y=2\); 5) \(y>2\). | 1. 1) Points on the line \(a\) (Fig. 13); 2) all points in the plane lying to the left of the line \(a\); 3) all points in the plane lying to the right of the line \(a\); 4) points on the line \(b\); 5) points lying above the line \(b\). | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,438 |
4(!). On the coordinate plane, find the sets of points whose coordinates \(x\) and \(y\) satisfy the following conditions:
1) \((y+x)\left(y-x^{2}\right)=0\)
2) \(x(y-2)=0\)
3) \(y=|x|\)
4) \((x-3)\left(x^{2}+y^{2}\right)=0\) | 4. 1) Points of the graphs of the functions \(y=-x\) or \(y=x^{2}\) (Fig. 19);
2) points of the lines \(x=0\) or \(y=2\) (Fig. 20);
3) \(y=x\) and \(x \geqslant 0\) or \(y=-x\) and \(x<0\) (Fig. 21);
4) points of the line \(x=3\) or the point ( \(0 ; 0\) ) (Fig. 22). | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,441 |
8. Water begins to flow into an empty bucket at a rate of 4 kilograms per minute through a tap. The bucket can hold 10 kilograms of water. Construct a graph showing the dependence of the mass of water \(m\) (kg) in the bucket on time \(t\) in minutes, if the observation lasted 5 minutes. Write the relationship between ... | 8. \(m=4t\) and \(0 \leqslant t<2.5\), or \(m=10\) and \(2.5 \leqslant t \leqslant 5\) (Fig. 24). | =4t0\leqslant<2.5,or=102.5\leqslant\leqslant5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,443 |
2. Prove that:
1) \(a^{3}+a^{2} c-a b c+b^{2} c+b^{3}=0\), if \(a+b+c=0\);
2) \(a x+2 x+a y+2 y+4=a^{2}\), if \(a-2=x+y\);
3)* \(M x-M=x^{50}-1\), if \(M=1+x+x^{2}+\ldots+x^{49}\). | 2. 1) Instruction. Use the substitution: \(c=-a-b\).
2) \(\mathrm{P}\) solution. Since \(y=a-2-x\), the given expression is equal to
\[
a x+2 x+a(a-x-2)+2(a-x-2)+4=\ldots=a^{2}
\]
3) \(\mathrm{P}\) solution.
\[
\begin{aligned}
M x & =\quad x+x^{2}+x^{3}+x^{4}+\ldots+x^{49}+x^{50} \\
-M & =-1-x-x^{2}-x^{3}-x^{4}-\ldot... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,448 |
3. Let \(T_{0}=1, T_{1}=x, T_{k+1}=2 x T_{k}-T_{k-1}\), where \(k \in N\). Write in standard form \(T_{5}\) and \(T_{0}+T_{1}+T_{2}+T_{3}\).
The polynomials mentioned above are known as "Chebyshev polynomials" (1821 - 1894). The discovery of these polynomials and a series of their remarkable properties was prompted by... | 3. Solution. \(T_{2}=2 x T_{1}-T_{0}=2 x \cdot x-1=2 x^{2}-1\),
\[
T_{3}=4 x^{3}-3 x, \quad T_{4}=8 x^{4}-8 x^{2}+1, \quad T_{5}=16 x^{5}-20 x^{3}+5 x
\]
\[
\begin{aligned}
& T_{0}+T_{1}+T_{2}+T_{3}=1+x+2 x^{2}-1+4 x^{3}- \\
& -3 x=4 x^{3}+2 x^{2}-2 x
\end{aligned}
\] | 4x^{3}+2x^{2}-2x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,449 |
4. Each of the \(k\) numbers is equal to \(2 p+3\), and each of the \(p\) numbers is equal to \(5-2 k\). The arithmetic mean of all \(k+p\) numbers is 4. Prove that \(k=p\). | 4. Solution. Since from the conditions of the problem,
\[
\begin{gathered}
k(2 p+3)+p(5-2 k)=4(p+k), \text { then } \\
2 k p+3 k+5 p-2 p k=4 p+4 k, k=p
\end{gathered}
\] | p | Algebra | proof | Yes | Yes | olympiads | false | 43,450 |
5. Find the set of points in the coordinate plane, the coordinates \(x\) and \(y\) of which satisfy the condition:
1) \(y^{2}-y x^{2}=0\)
2) \(x^{5}+x^{4}=0\). | 5. The sets of desired points are determined by the conditions:
1) \(y=0\) or \(y=x^{2}\) (parabola or x-axis);
2) \(x=0\) or \(x=-1\) (a pair of parallel lines). | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,451 | |
6. Given a five-digit number. If you prepend a seven to the number, the resulting six-digit number will be 5 times larger than the six-digit number obtained by appending a seven to the end. Find this five-digit number. | 6. Solution. Let \(x\) be a five-digit number. If we prepend a seven to it, we get the number \(7 \cdot 10^{5} + x\). If we append a seven to it, we get the number \(10 x + 7\). According to the problem,
\[
7 \cdot 10^{5} + x = 5(10 x + 7)
\]
Solving this equation, we get \(x = 14285\). | 14285 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,452 |
9. The difference between a three-digit number and the sum of its digits is always divisible by 9. Prove or disprove.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 9. Solution. Let \(c\) be the units digit, \(b\) be the tens digit, and \(a\) be the hundreds digit. Then the difference between the three-digit number and the sum of its digits is:
\[
100 a+10 b+c-(a+b+c)=\ldots=9(11 a+b)
\]
Therefore, this difference is divisible by 9. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,454 |
10. Prove that the number
1) \(a b b a\) is divisible by 11;
2) \(\overline{a a b b b}\) is divisible by 37;
3) \(\overline{a b a b a b}\) is divisible by 7;
4) \(\overline{a b a b}-\overline{b a b a}\) is divisible by 9 and 101. | 10. 1) \( \mathrm{P} \) solution and \( \mathrm{P} \). \( \overline{a b b a}=1000 a+100 b+10 b+a=\ldots \) \(\ldots=11(91 a+10 b)\).
Therefore, the given number is divisible by 11.
2) \( \mathrm{P} \) solution. \( \overline{a a a b b b}=1000 \overline{a a a}+\overline{b b b}=\) \(=1000 \cdot 111 a+111 b=37 \cdot 3 \cd... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,455 |
11. A two-digit number, when added to the number written with the same digits but in reverse order, results in the square of a natural number. Find all two-digit numbers with this property. | 11. Solution. Let \(a\) be the tens digit, \(b\) be the units digit. Then for \(k \in N\) according to the condition
\[
10 a+b+10 b+a=k^{2} \text { or } 11(a+b)=k^{2}
\]
Obviously, \(a+b\) must be a multiple of 11 and \(a+b \leqslant 18\). Therefore, \(a+b=11\). Consequently, the conditions of the problem are satisfi... | 29,38,47,56,65,74,83,92 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,456 |
13*. Calculate:
1) \(x^{4}-2 x^{3}+3 x^{2}-2 x+2\) given \(x^{2}-x=3\)
2) \(2 x^{4}+3 x^{2} y^{2}+y^{4}+y^{2} \quad\) given \(x^{2}+y^{2}=1\). | 13. 1) \( \mathrm{P} \) is a solution. \( x^{2}\left(x^{2}-x\right)-x\left(x^{2}-x\right)+2\left(x^{2}-x\right)+2= \) \( =3 x^{2}-3 x+2 \cdot 3+2=3\left(x^{2}-x\right)+8=3 \cdot 3+8=17 \);
2) Solution. \( 2 x^{4}+2 x^{2} y^{2}+x^{2} y^{2}+y^{4}+y^{2}=2 x^{2}\left(x^{2}+\right. \) \( \left.+y^{2}\right)+y^{2}\left(x^{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,458 |
14. Find the value of the polynomial \(x(1+y)-y(x y-1)-x^{2} y\) given \(x+y=-p\) and \(x y=q\). | 14. Solution. Expanding the brackets, we get:
\(x+x y-x y^{2}+y-x^{2} y=(x+y)+x y(-x-y+1)=\) \(=q(p+1)-p=p q+q-p\). | pp | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,459 |
17*. Prove that \(6 n^{2}-7 n+2\) is greater than zero for any integer \(n\) | 17. Solution. Let's write the polynomial in the form: \(n(6 n-7)+2\). If \(n \neq 0\), then \(6 n^{2}-7 n+2>0\), if \(n=0\), the value of the trinomial is 2, if \(n=1\), the value of the trinomial is 1, if \(n \geqslant 2\), then \(6 n-7>0\), therefore, \(n(6 n-7)>0\), \(n(6 n-7)+2>0\). Thus, for any integer \(n\), the... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,461 |
18*. Lyuda proved that the number \(7^{17}+17 \cdot 3-1\) is divisible by 9. Can we use this conclusion to prove that the number \(7^{18}+18 \cdot 3-1\) is also divisible by 9? | 18. Solution. Let \(7^{17} + 17 \cdot 3 - 1 = 9k\), where \(k \in \mathbb{N}\). Using substitution: \(7^{17} = 9k - 17 \cdot 3 + 1\). We get:
\[
\begin{gathered}
7^{18} + 18 \cdot 3 - 1 = 7 \cdot 7^{17} + 53 = 7(9k - 50) + 53 = 63k - \\
- 297 = 9(7k - 33).
\end{gathered}
\]
Therefore, the given number is divisible by... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,462 |
1. Prove that for any values of the variables:
1) \((3 b-1)(4 b+1)>(2 b+1)(5 b-3)\);
2) \((y-1)(y+2)=4+(y-2)(y+3)\). | 1. 1) Solution. Since
\[
(3 b-1)(4 b+1)-(2 b+1)(5 b-3)=\ldots=2 b^{2}+2>0
\]
the inequality holds for any \(b\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,463 |
2(!). Prove that if \(a(b+1)+b(a+1)=(a+1)(b+1)\), then \(a b=1\). Formulate and prove the converse statement. | 2. Solution. Since \(a(b+1)+b(a+1)-(a+1) \times\) \(\times(b+1)=0, a b+a+a b+b-a b-a-b-1=0\), then \(a b=1\).
To solve the second part of the problem, it is required to prove that if \(a b=1\), then the first equality holds. To prove this, it is sufficient to show that the last equality implies the previous one, the s... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,464 |
3. Prove that:
1) if \(2(a+1)(b+1)=(a+b)(a+b+2)\), then \(a^{2}+b^{2}=2\);
2) if \(a^{2}+c^{2}=2 b^{2}\), then the equality \((a+b)(a+c)+(c+a)(c+b)=2(b+a)(b+c)\) holds. | 3. 2) Solution. We need to prove that
\[
\begin{gathered}
(a+b)(a+c)+(c+a)(c+b)-2(b+c)(b+a)=0, \\
a^{2}+a c+a b+b c+c^{2}+c b+c a+a b-2 b^{2}-2 a b- \\
-2 a c-2 b c=0, \\
a^{2}+c^{2}-2 b^{2}=0, a^{2}+c^{2}=2 b^{2} .
\end{gathered}
\]
Since the fourth equality implies the third, the third implies the second, and the s... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,465 |
6. Prove the equality:
\[
\begin{aligned}
& 55554 \cdot 55559 \cdot 55552-55556 \cdot 55551 \cdot 55558= \\
= & 66665 \cdot 66670 \cdot 66663-66667 \cdot 66662 \cdot 66669
\end{aligned}
\] | 6. Solution. Let \(a=55555, b=66666\). The solution to the problem reduces to proving the identity
\[
\begin{aligned}
& (a-1)(a+4)(a-3)-(a+1)(a-4)(a+3)= \\
& =(b-1)(b+4)(b-3)-(b+1)(b-4)(b+3)
\end{aligned}
\]
Each part of this equality is independent of the variable and equals 24. | 24 | Algebra | proof | Yes | Yes | olympiads | false | 43,468 |
7. Calculate in the most rational way:
\[
3 \frac{1}{117} \cdot 4 \frac{1}{119}-1 \frac{116}{117} \cdot 5 \frac{118}{119}-\frac{5}{119}
\] | 7. Solution. Let \(a=\frac{1}{117}\) and \(b=\frac{1}{119}\). Then \(3 \frac{1}{117}=3+a\),
\[
4 \frac{1}{119}=4+b, \quad 1 \frac{116}{117}=2-a, \quad 5 \frac{118}{119}=6-b
\]
therefore, the given expression can be written as:
\[
(3+a)(4+b)-(2-a)(6-b)-5 b
\]
After some simple transformations, we get that this algeb... | \frac{10}{117} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,469 |
8. Find natural \(k\) and \(n\), if it is known that
\[
k n^{2}-k n-n^{2}+n=94
\] | 8. Solution. Factoring the left and right sides of the equation, we get:
\[
(n-1) n(k-1)=1 \cdot 2 \cdot 47
\]
This equality in the domain of natural numbers is possible, obviously, only when \(n=2\) and \(k=48\). | k=48,n=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,470 |
9. Solve the equation: \(x^{3}+x^{2}+x+1=0\). | 9. Solution. Factoring the left side of the equation, we get:
\[
(x+1)\left(x^{2}+1\right)=0
\]
Since \(x^{2}+1>0\), then \(x+1=0, x=-1\). | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,471 |
10. If \(a+b+c=2 p\), then the following equality holds:
\[
(2 a p+b c)(2 b p+a c)(2 c p+a b)=(a+b)^{2}(a+c)^{2}(b+c)^{2}
\]
Prove. | 10. Instruction. Show that \(2 a p+b c=(a+b)(a+c), 2 p b+\) \(+a c=(b+a)(b+c), 2 c p+a b=(c+a)(c+b)\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,472 |
14*. Factorize:
1) \(x^{4}-3 x^{2}+1\)
2) \(a^{5}+a^{4}-2 a+1\)
3) \(m^{5}-2 m^{3}-m-1\). | 14. 1) \(\left(x^{2}+x-1\right)\left(x^{2}-x-1\right)\)
2) \(\left(a^{2}+a-1\right)\left(a^{3}+a-1\right)\)
3) \(\left(m^{3}+m^{2}+1\right)\left(m^{2}-m-1\right)\). | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,475 |
15*. Find all values of the natural number \(n\), for which numbers of the form: 1) \(n^{3}-n^{2}+n-1\); 2) \(n^{3}-6 n+4\), 3) \(n^{5}-2 n^{3}-n-\) -1 - would be prime. | 15. 1) \(\mathrm{P}\) is a prime number. Factoring the polynomial, we get:
\[
n^{3}-n^{2}+n-1=(n-1)\left(n^{2}+1\right)
\]
If \(n=1\), the product is zero; if \(n=2\), the product is a prime number 5; if \(n>2\), each factor is greater than 1, so the product will be a composite number. The required number is 2; 2) \(... | 2,3,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,476 |
16*. Prove that the polynomial \(x^{12}-x^{7}-x^{5}+1\) cannot take negative values for any \(x\). | 16. Solution. Transform the polynomial to the form: \(\left(x^{5}-1\right)\left(x^{7}-1\right)\).
If \(x > 1\), then \(x^{5}\) and \(x^{7}\) are also greater than 1, each factor is positive, and therefore the product is positive.
The statement is proven for all possible values of the variable. 17. Solution. Write the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,477 |
18*. Factorize:
1) \(x^{8}+x^{6}+x^{4}+x^{2}+1\)
2) \(a^{5}-a^{4}-a^{3}-a^{2}-a-2\). | 18. 1) \(\left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{4}-x^{3}+x^{2}-x+1\right)\).
2) \(\mathrm{P}\) is an expression. \(a^{5}-a^{4}-a^{3}-a^{2}-a-2=\)
\[
\begin{aligned}
& =a^{5}-2 a^{4}+ \\
& +a^{4}+2 a^{3}+ \\
& \quad+a^{3}-2 a^{2}+
\end{aligned}
\]
\[
+a^{2}-2 a+7=
\]
\(=a^{4}(a-2)+a^{3}(a-2)+a^{2}(a-2)+a(a-2)+(a-... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,478 | |
20*. Prove that there does not exist a natural number \(k\) for which the following equality would hold:
\[
k^{6}+k^{4}+k^{2}=10^{k+1}+9
\] | 20. Solution. Let such a natural \(k\) exist. Then, obviously, \(k\) must be an odd number. Moreover, from the given equality, it follows that \(\left(k^{4}+1\right)\left(k^{2}+1\right)=10^{k+1}+10\). The left side of this equality is divisible by four, while the right side is not divisible by 4. Contradiction.
60 | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,480 |
2. Prove or disprove that the following relations are true for any values of the variables:
1) \((a-b)^{2}+(a b+1)^{2}=\left(a^{2}+1\right)\left(b^{2}+1\right)\)
2) \((x-2)(x-3)(x+3)(x+2)+x^{2}=\left(x^{2}-6\right)^{2}\);
3) \((m-1)^{2}(m+2)>m\left(m^{2}-4\right)\);
4) \(2(3 a-1)^{2}+3(2 a+1)^{2}>(5 a+1)(5 a-1)\). | 2. Instruction. Consider the differences between the left and right sides of the relations. For the third example, this difference is \(m+2\) and can be negative, for example, at \(m=-5\). Therefore, the inequality is not true for all values of the variable \(m\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,481 |
3. If \(a+c=2 b\), then \(a^{2}+8 b c=(2 b+c)^{2}\) always. Prove or disprove.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | 3. \(\mathrm{P}\) is an identity. Let's use the substitution \(a=2 b-c\). We get: \((2 b-c)^{2}+8 b c-(2 b+c)^{2}=\ldots=0\).
Therefore, the statement is true. | proof | Algebra | MCQ | Yes | Yes | olympiads | false | 43,482 |
5. Find: 1) \(x^{6}+3 x^{2} y^{2}+y^{6}\), if \(x^{2}+y^{2}=1\); 2) \(x^{4}+\) \(+x^{2} y^{2}+y^{4}\), if \(x^{2}+y^{2}=a\) and \(x y=b\). | 5. 1) \(\mathrm{P}\) is the solution. First method. The given polynomial is equal to
\[
\left(x^{2}+y^{2}\right)\left(x^{4}-x^{2} y^{2}+y^{4}\right)+3 x^{2} y^{2}=\ldots=\left(x^{2}+y^{2}\right)^{2}=1
\]
Second method. Since \(y^{2}=1-x^{2}\), then, performing the substitution, we get that the given polynomial is ide... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,484 |
6*. If \(a+b=ab\), then the identity holds:
\[
\left(a^{3}+b^{3}-a^{3} b^{3}\right)^{3}+27 a^{6} b^{6}=0 . \text { Prove. }
\] | 6. Solution. Since \(a^{3} b^{3}=(a+b)^{3}\), then
\[
\begin{gathered}
a^{3}+b^{3}-a^{3} b^{3}=a^{3}+b^{3}-(a+b)^{3}=\ldots=-3 a b(a+b)= \\
=-3 a^{2} b^{2}
\end{gathered}
\]
Therefore, the left side of the equation will be equal to \(\left(-3 a^{2} b^{2}\right)^{3}+27 a^{6} b^{6}=0\). 62 | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,485 |
7. If \(a+b+c=2 x\), then \(x^{2}+(x-a)^{2}+(x-b)^{2}+(x-\) \(-c)^{2}=a^{2}+b^{2}+c^{2}\). Prove. | 7. Solution. \(x^{2}+(x-a)^{2}+(x-b)^{2}+(x-c)^{2}=4 x^{2}-\) \(-2 x(a+b+c)+a^{2}+b^{2}+c^{2}=4 x^{2}-4 x^{2}+a^{2}+b^{2}+c^{2}=\)
\[
=a^{2}+b^{2}+c^{2}
\]
since \(a+b+c=2 x\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,486 |
8. Reduce the following polynomials to standard form:
1) \((a-b)(a+b)\left(a^{2}+a b+b^{2}\right)\left(a^{2}-a b+b^{2}\right)\)
2) \((x-1)^{3}(x+1)^{2}\left(x^{2}+1\right)\left(x^{2}+x+1\right)\)
3) \(\left(x^{4}-x^{2}+1\right)\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)\). | 8. 1) \(a^{6}-b^{6}\); 2) \(\left(x^{2}-1\right)\left(x^{3}-1\right)\left(x^{4}-1\right)=\ldots=x^{9}-x^{7}-\) \(-x^{8}-x^{5}+x^{4}+x^{3}+x^{2}-1\); 3) \(x^{8}+x^{4}+1\). | x^{9}-x^{7}-x^{8}-x^{5}+x^{4}+x^{3}+x^{2}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,487 |
10. Prove that for any natural \(n\) the number \((2 n-1)^{3}-\) \(-(2 n)^{2}+2 n+1\) is divisible by 16 | 10. The given polynomial is equal to \(8 n(n-1)^{2}\). Since the number \(n(n-1)\) is even, the polynomial \(8 n(n-1)^{2}\) is divisible not only by 8, but also by 16. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,489 |
12(!). Show that the number \(a-b\) for any values of \(a\) and \(b\) is a root of the equation: \(x^{3}+3 a b x+b^{3}-a^{3}=0\).
12a (!). Are there such values of \(a\) for which the number \(a-1\) is a root of the equation: \(x^{3}-a x^{2}+1=0\) ? | 12. Solution. Let \(x=a-b\). Then the left side of the given equation will be equal to
\[
(a-b)^{3}+3 a b(a-b)+b^{3}-a^{3}=\ldots=0
\]
Therefore, when \(x=a-b\), the given equation becomes an identity. This means that \(a-b\) is a root of the equation for any \(a\) and \(b\).
12a. Solution. If the number \(a-1\) is ... | =0or=2 | Algebra | proof | Yes | Yes | olympiads | false | 43,491 |
13(!). It is known that \(a=x+y\) and \(b=x y\) and \(c=x^{2}+y^{2}\). Find the dependence between \(a, b\) and \(c\). | 13. Solution. Since \(a^{2}=(x+y)^{2}=\left(x^{2}+y^{2}\right)+2(x y)\), then \(a^{2}=c+2 b\). | ^{2}=+2b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,492 |
14*. Prove that if \(p\) is a prime number and \(p>2\), then the number \(p^{2}-5\) is not divisible by 8. | 14. Solution. Suppose that \(p^{2}-5=8 k\), where \(k \in \mathbb{N}\). Since \(p\) cannot be equal to 2, \(p\) is an odd number. Let \(p=2 m+1, m \in \mathbb{N}\). Then the equation \((2 m+1)^{2}-5=8 k\) must hold, and consequently, \(m(m+1)=2 k+1\).
In the left part of the obtained equation, there is an even number,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,493 |
15*. If \(n\) is a natural number greater than 1, then the number \(2^{n}-1\) cannot be a square of a natural number. Prove it. | 15. Solution. Suppose that the number \(2^{n}-1\) is the square of some natural number. Then this number must obviously be odd. Therefore, for \(m \in N\), the following equalities must hold:
\[
\begin{aligned}
2^{n}-1 & =(2 m+1)^{2} \\
2^{n} & =4 m^{2}+4 m+2 \\
2^{n-1} & =2 m^{2}+2 m+1
\end{aligned}
\]
The left side... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,494 |
16*. Prove that there do not exist integers \(x\) and \(y\) satisfying the equation: \(x^{2}+1974=y^{2}\). | 16. Solution. Obviously, the numbers \(x\) and \(y\) are either both even or both odd. Since \(1974 = (x+y)(x-y)\), the left side of this equation is not divisible by 4, while the right side is divisible by 4 in both cases. We have reached a contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,495 |
2. Solve the equations:
1) \(4 x^{2}-12 x+9=0\)
2) \(x^{2}-\pi^{2}=0\)
3) \(a^{3}-a=0\);
4) \(x^{2}+1+\pi=2 x\)
5) \(x^{2}=\pi^{2}-2 \pi+1\)
6) \(x^{2}+y^{2}-4 x+4=0\). | 2. 1) 1.5 ; 2) \(\pi\) or \(-\pi\); 3) 0 or 1 or -1.
4) Solution. Since \(\left(x^{2}+1-2 x\right)+\pi=(x-1)^{2}+\) \(\pi>0\), the equation has no roots.
5) Since \(x^{2}=(\pi-1)^{2}\), the roots of the equation are \(x_{1}=\pi-1\) or \(x_{2}=1-\pi\);
6) \(x=2\) and \(y=0\). | 0or1or-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,496 |
4. Find the geometric locus of points on the coordinate plane, the coordinates \(x\) and \(y\) of which satisfy one of the following conditions:
1) \(y^{2}-x^{2}=0\)
3) \(x^{2}+y^{2}=4(y-1)\)
2) \(x^{2}-2 x+1=0\)
4) \(x^{2}-2 x y+y^{2}=-1\). | 4. 1) \(\mathrm{P}\) is a solution. Since \((y-x)(y+x)=0\), then \(y=x\) or \(y=-x\). Therefore, the desired set of points will be the union of the sets of points of the graphs of these functions.
2) Points of the line \(x=1.3\) ) Point \(A(0 ; 2)\).
4) Empty set. | y=xory=-x,x=1,A(0,2),\text | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,498 |
5. In a right-angled triangle, \(a\) and \(b\) are the legs; \(S\) is the area of the triangle. Find the angles of the triangle if it is known that \((a+b)^{2}=8 S\) | 5. Solution. Noting that \(8 S=8 \cdot 0.5 a b=4 a b\), and using the condition of the problem, we can write: \((a+b)^{2}-4 a b=0\), \((a-b)^{2}=0\), and therefore \(a=b\). This means that the right triangle is isosceles, and its angles are respectively \(45^{\circ}, 45^{\circ}\) and \(90^{\circ}\). | 45,45,90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,499 |
6. If the decimal representations of two natural numbers end with three respectively identical digits, then the cubes of these numbers also end with respectively identical digits. Prove it. | 6. Solution. Let \(a\) and \(b\) be natural numbers, and \(a-b=1000 m\) where \(m \in N\). Then the difference of the cubes of these numbers is:
\(a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)=1000 m\left(a^{2}+a b+b^{2}\right)\).
Since the difference of the cubes of the numbers \(a\) and \(b\) is divisible by 1000, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,500 |
7. Prove that if the sum or difference of two natural numbers is divisible by 50, then the squares of these numbers have the same tens and units digits. | 7. Instruction. It is necessary to show that the difference of the squares of numbers \(a\) and \(b\) is divisible by 100. For this, it is sufficient to factorize this difference of squares and show that one of the factors is divisible by 50, and the second is divisible by 2. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,501 |
8. If \(q > p^{2}\), then the polynomial \(x^{2} + 2px + q\) cannot take negative values. Prove it. | 8. Solution. Using the inequality \(q > p^{2}\), we get:
\[
x^{2} + 2 p x + q > x^{2} + 2 p x + p^{2} = (x + p)^{2} \geqslant 0
\]
Therefore, for any \(x\), the values of the quadratic polynomial are non-negative. | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,502 |
10. Factorize:
1) \(\left(x^{2}-1+x\right)\left(x^{2}-1+3 x\right)+x^{2}\)
2)* \((x-a)^{4}+4 a^{4}\)
3) \((a+1)^{4}+2(a+1)^{3}+a(a+2)\);
4) \((p+2)^{4}+2\left(p^{2}-4\right)^{2}+(p-2)^{4}\). | 10. 1) Hint. Use the substitution \(y=x^{2}+x-1\). 2) \(\mathrm{P}\) solution. Let \(y=x-a\). Then
\[
\begin{gathered}
(x-a)^{4}+4 a^{4}=y^{4}+4 a^{4}=\left(y^{2}+2 a^{2}\right)^{2}-4 a^{2} y^{2}=\ldots \\
\ldots=\left(x^{2}+a^{2}\right)\left(x^{2}-4 a x+5 a^{2}\right)
\end{gathered}
\]
3) Hint. Use the substitution ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,504 |
11. Prove the inequalities:
1) \((x+1)(x-2 y+1)+y^{2} \geqslant 0\)
2) \((x-2)(x-3)(x-4)(x-5)+1 \geqslant 0\);
3) \(\left(a^{2}-a-1\right)\left(3 a^{2}+a+1\right)-4 a^{4} \leqslant 0\). | 11. 1) Solution. We use the following substitution: \(z=x+1\). We obtain: \(z(z-2 y)+y^{2}=(z-y)^{2} \geqslant 0\).
2) Solution. Multiplying respectively the first and fourth, second and third binomials, we get:
\[
\begin{gathered}
\left(x^{2}-7 x+10\right)\left(x^{2}-7 x+12\right)+1=z(z+2)+1= \\
=(z+1)^{2} \geqslant... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,505 |
13(!). Let \(x+y=-p\) and \(x y=q\). Prove that:
1) \(x^{2}+y^{2}=p^{2}-2 q\)
2) \(x^{3}+y^{3}=-p^{3}+3 p q\)
3) \(x^{4}+y^{4}=p^{4}-4 p^{2} q+2 q^{2}\). | 13. 1) Solution. \(x^{2}+y^{2}=(x+y)^{2}-2xy=p^{2}-2q\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,507 |
14*. Using the result of the previous problem, prove the following identity:
\[
x^{4}+y^{4}+(x+y)^{4}=2\left(x^{2}+x y+y^{2}\right)^{2}
\] | 14. Solution. Performing the substitutions, we get:
\[
\begin{gathered}
p^{4}-4 p^{2} q+2 q^{2}+p^{4}=2\left(p^{2}-q\right)^{2}=\ldots \\
\ldots=2\left(x^{2}-x y+y^{2}\right)^{2}
\end{gathered}
\]
The solution to exercises 13 and 14 can serve as a pretext for a brief story about symmetric polynomials.
If in the poly... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,508 |
1. Factorize the polynomials:
1) \(a^{4}+2 a^{3}+1\)
2) \(m^{3}+2 m-3\);
3) \(2 a^{4}-a^{2}-1\). | 1. 1) Solution. First method. \(a^{4}+2 a^{3}+1=\left(a^{4}+a^{3}\right)+\) \(+\left(a^{3}+1\right)=a^{3}(a+1)+(a+1)\left(a^{2}-a+1\right)=(a+1) \times\) \(\times\left(a^{3}+a^{2}-a+1\right)\).
Second method. \(a^{4}+2 a^{3}+1=a^{4}+a^{3}+\)
\[
\begin{gathered}
+a^{3}+a^{2}- \\
-a^{2}-a+ \\
+a+1= \\
=a^{3}(a+1)+a^{2}... | (-1)(^{2}++3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,509 |
2*. Разложить многочлены на множители:
1) \(a^{7}+a^{5}+1\)
2) \(a^{5}+a b^{4}+b^{5}\)
3) \(a^{7}-1\)
4) \(2 a^{3}-a x^{2}-x^{3}\). | 2. 1) \(\left(a^{2}+a+1\right)\left(a^{5}-a^{4}+\dot{a}^{3}-a+1\right)\)
2) \(\left(a^{2}+a b+b^{2}\right)\left(a^{3}-a^{2} b+b^{3}\right)\)
3) \((a-1)\left(a^{6}+a^{5}+\ldots+a+1\right)\)
4) \((a-x)\left(2 a^{2}+2 a x+x^{2}\right)\). | (-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,510 |
6. If \(m+n=2\), then \(m n \leqslant 1\). Prove.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly. | 6. Solution. First method. Since \(m=2-n\), then 1 \(-m n=1-n(2-n)=\ldots=(1-n)^{2} \geqslant 0\), and therefore \(1 \geqslant m n\). Second method. Let \(m=1+x\), then \(n=2-(1+x)=\) \(=1-x\) and, consequently,
\(1-m n=1-(1+x)(1-x)=\ldots=x^{2} \geqslant 0 ; 1 \geqslant m n\).
Third method. Since \(4 m n=(m+n)^{2}-(... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 43,512 |
7. Prove that among all natural numbers of the form \(2 p+1\), where \(p\) is a prime number, only one is a perfect cube. Find this number. | 7. Solution. First method. Let \(2 p+1=x^{3}, x \in N\). Then \(2 p=(x-1)\left(x^{2}+x+1\right)\). Since \(x^{2}+x+1\) is an odd number, it is not divisible by 2. Therefore, the given equation is possible only when \(x-1=2\). In this case, \(x=3, p=13\), and the desired number is 27.
Second method. Let \((2 y+1)^{3}=2... | 27 | Number Theory | proof | Yes | Yes | olympiads | false | 43,513 |
1(!). The perimeter of a triangle is \(12 \mathrm{~cm}\), and the length of one of its sides is 5 cm. Find the relationship between the lengths of the other two sides and plot the graph of this relationship. | 1. Solution. Let the length of the second side be \(x\) cm, and the third side be \(y\) cm. The length of each side is a positive number that is less than half the perimeter of the triangle. Therefore,
 \(\left\{\begin{array}{l}x+y=10^{20} \\ x-y=10^{19}\end{array}\right.\)
2) \(\left\{\begin{array}{l}2 a^{3}+3 b^{5}=19 \\ 3 a^{3}-4 b^{5}=20\end{array}\right.\)
3) \(\left\{\begin{array}{l}x+y=(a+1)^{2}, \\ x-y=(a-1)^{2} ;\end{array}\right.\)
4) \(\left\{\begin{array}{l}a^{2}-2 a b... | 4. 1) \(x=55 \cdot 10^{18}\) and \(y=45 \cdot 10^{18}\); 2) \(a=2\) and \(b=1\); 3) \(x=\) \(=a^{2}+1\) and \(y=2 a\); 4) \(a=b=3\). | =2,\,b=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,517 |
5. Determine the type of quadrilateral whose vertices are the points of intersection of the lines:
\[
y=x+3, y=x-3, y=-x+3, y=-x-3 .
\] | 5. A square with vertices at points \(A(0 ; 3), B(0 ;-3) ; C(3 ; 0)\), \(D(-3 ; 0)\). | AwithverticesatpointsA(0;3),B(0;-3);C(3;0),D(-3;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,518 |
7. Write the equation of the line passing through the points:
1) \(A(3 ; 0)\) and \(B(3 ; 5)\);
3) \(E(0 ; 4)\) and \(K(2 ; 0)\)
2) \(C(2 ; 2)\) and \(D(-1 ;-1)\);
4) \(M(3 ; 2)\) and \(P(6 ; 3)\). | 7. 1) Equation of the line: \(x=3\). 2) \(y=x\). 3) S o l u t i o n. Equation of the line: \(y=k x+b\). We need to find \(k\) and \(b\). When \(x=0\) and \(y=4\), we get: \(4=0 \cdot x+b\), hence, \(b=4\). When \(x=2\) and \(y=0\), we have: \(0=k \cdot 2+4\), hence, \(k=-2\). Equation of the line: \(y=-2 x+4\).
4) S o... | 3,x,-2x+4,\frac{1}{3}x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,520 |
11. In Figure 28, a system of highways is depicted, with arrows indicating the directions of permitted movement. A convoy of 36 cars passed along \(A B\). It is known that out of these cars, 10 more passed along \(B C\) than along \(D E\), and 2 cars passed along \(C D\). How many cars from the convoy passed along \(B ... | 11. \(24,12,14,22\). Hint. Let \(x\) be the number of cars that passed through segment \(B C\), and \(y\) be the number of cars that passed through segment \(B D\). Show that the solution to the problem can be reduced to solving the following system of equations:
\[
\left\{\begin{array}{l}
x+y=36 \\
x-y-2=10
\end{arra... | 24,12,14,22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 43,524 |
12. Solve the following system of equations:
\[
\left\{\begin{array}{l}
(x-y+1)^{3}+y=10 \\
(x-y+1)^{3}+x=11
\end{array}\right.
\] | 12. Solution. By subtracting the first equation from the second term by term, we find that \(x-y=1\). Consequently, after the appropriate substitution, the system of equations will take the form:
\[
\left\{\begin{array}{l}
(1+1)^{3}+y=10 \\
(1+1)^{3}+x=11 . \text { We get: } x=3 \text { and } y=2
\end{array}\right.
\] | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,525 |
14. Find the distance between the points of intersection of the three lines: \(y = 3x\), \(y = 3x - 6\), and \(y = 1975\). | 14. 2. Instruction. Show that the desired distance is equal to the distance between points \(A(0 ; 0)\) and \(B(2 ; 0)\). | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,527 |
1. Calculate in the most rational way:
\[
\frac{7^{16}-1}{2402000\left(49^{4}+1\right)}
\]
76 | 1. 2,4. Hint. Factor the numerator and reduce the fraction
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note I added for context, and it should not be part of the translation. Here is the requested translation:
1. 2,4. Hint. Factor the numerator and reduce the fraction | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,528 |
2. Plot the graph of the function:
\[
y=\frac{x^{3}-x^{2}-x+1}{x^{2}-1}
\] | 2. S o l u t i o n. After reducing the fraction, we get \(y=x-1\) under the condition that \(x \neq 1\) and \(x \neq -1\). Therefore, the graph of the function will look as shown in Figure 31. | x-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,529 |
3. If \(b\) is an integer not equal to 1, will the following be an integer:
\[
\frac{t^{5}-5 b+4}{b^{2}-2 b+1} ?
\] | 3. Yes. Instruction. Factor the numerator of the fraction and cancel the fraction by \(b^{2}-2 b+1\). | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,530 | |
4. Simplify the expression:
\[
\left(\frac{8^{x}-2^{x}}{4^{x}+2^{x}}\right)^{2}+2^{1+x}
\] | 4. Instruction. Use the substitution \(y=2^{x}\).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,531 |
5*. Show that if \(x^{3}+x-1=0\), then the following equality holds:
\[
\frac{x^{4}-2 x^{3}+x^{2}-3 x+5}{x^{5}-x^{2}-x+2}=3
\] | 5. 3. Instruction. The numerator of the fraction is equal to:
\[
\begin{gathered}
\left(x^{4}+x^{2}-x\right)-\left(2 x^{3}+2 x-2\right)+ \\
+3=3
\end{gathered}
\]
The denominator of the fraction is equal to:
\[
\begin{gathered}
\left(x^{5}+x^{3}-x^{2}\right)-\left(x^{3}+x-1\right)+ \\
+1=1
\end{gathered}
\]
. Simplify the fraction:
\[
\frac{a-b-c+k}{a+b+c+k}
\] | 6. Solution. By multiplying the numerator and the denominator of the fraction by \(\boldsymbol{c} \neq 0\), performing the substitution, and using the equality \(a b=c k\), we get:
\[
\frac{a c-b c-c^{2}+a b}{a c+b c+c^{2}+a b}=\ldots=\frac{a-c}{a+c}
\]
Alternatively, from the given equality, we can express, for exam... | \frac{-}{+} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,533 |
7*. If the equality \(a b(a+b)=1\) holds, where \(a>0\), \(b>0\), then
\[
\frac{a}{a^{3}+a+1}=\frac{b}{b^{3}+b+1}
\]
Prove.
8(!) Simplify the fraction
\[
\frac{x^{2}+x y+y^{2}}{1+x y+x^{2} y^{2}}
\]
given that 1) \(x+y-x y=1\), 2) \(x+y+x y=-1\). | 7. Solution. First method. Since \(1=a^{2} b+a b^{2}\), where \(a>0, b>0\), then
\[
\begin{aligned}
\frac{a}{a^{3}+a+1} & =\frac{a}{a^{3}+a+a^{2} b+a b^{2}}=\frac{1}{a^{2}+1+a b+b^{2}}= \\
= & \frac{b}{a^{2} b+b+a b^{2}+b^{3}}=\frac{1}{b^{3}+b+1} .
\end{aligned}
\]
Second method. If \(a=b\), the identity is obvious. ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,534 |
11. In which two-digit number is the doubled sum of the digits equal to their product? | 11. Solution. Let \(a\) be the tens digit and \(b\) be the units digit of a two-digit number. Clearly, \(a=\{1 ; 2 ; \ldots ; 9\}, b=\{0 ; 1 ; 2 ; \ldots ; 9\}\). According to the problem, \(2(a+b)=a b\), therefore,
\[
b=\frac{2}{a-2} \text { and } a \neq 2
\]
By checking the values of \(a\), we find that \(b\) can b... | 63,44,36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,536 |
12*. A student "illegally" simplified the exponents but got the correct answer:
\[
\frac{43^{3}+17^{3}}{43^{3}+26^{3}}=\frac{43+17}{43+26}=\frac{20}{23}
\]
Explain the reason for this "coincidence." | 12. Instruction. If \(b=c\) and \(a+c \neq 0\), then the equality
\[
\frac{a^{3}+b^{3}}{a^{3}+c^{3}}=\frac{a+b}{a+0}
\]
holds; if \(b \neq c\) and \(a+c \neq 0\), then from the above equality it follows that \(a=b+c\). | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,537 | |
13. Prove that the equation
\[
\frac{1}{x^{2}}+\frac{1}{x y}+\frac{1}{y^{2}}=1
\]
has no solutions if \(x \in N\) and \(y \in N\). | 13. Solution. When \(x=1\) or \(y=1\), the value of the expression on the left side of the equation is greater than 1, and if \(x>1\) and \(y>1\), the value of the expression on the left side of the equation is less than one. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 43,538 |
14. (This problem was proposed by academician A. N. Kolmogorov at one of the summer camp olympiads of the Little Academy of Sciences of Crimea "Iscatel'".) Find all integers \(a\) for which the fraction
\[
\frac{a^{3}+1}{a-1}
\]
takes integer values. | 14. \(-1,0,2,3\). Hint. Separate the integer part from the fraction.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task is to tran... | -1,0,2,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 43,539 |
15. Prove that the points on the number line corresponding to numbers of the form
\[
\frac{n^{4}+n^{2}+2}{n^{4}+n^{2}+1}
\]
( \(\boldsymbol{n} \in N\) ) are located on a segment whose length does not exceed \(\frac{1}{3}\). | 15. Instruction. Extract the integer part from the fraction and show that numbers of the given form are greater than 1 and less than \(1 \frac{1}{3}\). | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,540 |
16(!). Solve the equation \(x+y=x y\) in integers. | 16. Solution. First method. Noting that \(y \neq 1\), we get:
\[
x=\frac{y}{y-1}=\frac{(y-1)+1}{y-1}=1+\frac{1}{y-1}
\]
But the fraction \(\frac{1}{y-1}\) can take integer values only when \(y-1=1\) or \(y-1=-1\). In the first case, \(y=2, x=2\). In the second case, \(x=y=0\).
Second method. The simplicity of the an... | 2or0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,541 |
17. Find a two-digit number that is equal to three times the product of its digits. | 17. 24 or 15. Hint. Solve the equation \(10 x+y=3 x y\) in integers, if \(1 \leqslant x \leqslant 9\) and \(0 \leqslant y \leqslant 9\). | 24or15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,542 |
18. Find a two-digit number that is equal to three times the square of its units digit. | 18. 12 or 75. Hint. Solve the equation \(10 x+y=3 y^{2}\) in integers, if \(1 \leqslant x \leqslant 9\) and \(0 \leqslant y \leqslant 9\). | 12or75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,543 |
19. It is known that \(2 b=1+a b\) and \(a \neq 1\) and \(b \neq 1\). Prove that under these conditions, the following equality holds:
\[
\frac{a+1}{a-1}-\frac{b+1}{b-1}=2
\] | 19. Solution. First method.
\[
\begin{aligned}
& (a+1)(b-1)-(b+1)(a-1)=2(a-1)(b-1) \\
& a b+b-a-1-a b-a+b+1=2 a b-2 a-2 b+2 \\
& 2 b=1+a b
\end{aligned}
\]
Since the last equality implies the validity of all preceding ones, the given equality is true under the given conditions.
Second method. After verifying that in... | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,544 |
\(21(!)\). Let \(a, b\) and \(c\) be the sides of a triangle. Prove that
\[
\frac{a+b}{a+b+c}>\frac{1}{2}
\] | 21. Instruction. Show that the difference between the left and right sides of the given inequality is greater than zero. Consider that the sum of the lengths of two sides of a triangle is greater than the length of the third side. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,545 |
22. Let \(a>-1\). Prove that as \(a\) increases, the fraction \(\frac{a}{a+1}\) also increases. | 22. Solution. First method.
\[
\frac{a}{a+1}=\frac{(a+1)-1}{a+1}=1-\frac{1}{a+1}
\]
As the variable increases, the subtrahend decreases, and the difference increases.
Second method. Let \(h>0\). Show that the difference
\[
\frac{a+h}{a+h+1}-\frac{a}{a+1}
\]
is positive. | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,546 |
24. Calculate the following sum for \(x y z=1\):
\[
\frac{1}{1+x+x y}+\frac{1}{1+y+y z}+\frac{1}{1+z+z x}
\] | 24. 1. Instruction. Use the substitution \(z=\frac{1}{x y}\). | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,548 | |
25. If \(b^{2}-n^{2}=a^{2}-k^{2}=c^{2}-m^{2}\), then the following equality holds:
\[
\frac{b m-c n}{a-k}+\frac{c k-a m}{b-n}+\frac{a n-b k}{c-m}=0
\]
Prove.
78 | 25. Instruction. Multiply the numerator and denominator of the first fraction by \(a+k\), the second - by \(b+n\), the third - by \(c+m\).
82 | proof | Algebra | proof | Yes | Yes | olympiads | false | 43,549 |
26. It is known that in a right-angled triangle \(c^{2}=a^{2}+b^{2}\), where \(a\) and \(b\) are the lengths of the legs; \(c\) is the length of the hypotenuse. Find the acute angles of the triangle if \(\left(\frac{c}{a+b}\right)^{2}=0.5\). | 26. Solution. From the given equality, it follows that \(2 c^{2}=\) \(=(a+b)^{2}\), therefore,
\(2\left(a^{2}+b^{2}\right)=a^{2}+2 a b+b^{2} ;(a-b)^{2}=0, a=b\),
hence the given right triangle is isosceles, and each of its acute angles is \(45^{\circ}\). | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,550 |
27*. If \(\frac{a-c}{b+c}+\frac{b-a}{a+c}+\frac{c-b}{b+a}=1\), then the following equality holds:
\[
\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}=4
\] | 27. Instruction. Add one to each fraction on the left side of the first equation, and add 3 to the right side. | Algebra | proof | Yes | Yes | olympiads | false | 43,551 | |
28*. Variables \(x\) and \(y\) are positive, \(x+y=6\). Find the minimum value of the sum \(\frac{1}{x}+\frac{1}{y}\). | 28. Solution. First method. Since \(y=6-x\), the given sum is:
\[
\frac{x+y}{x y}=\frac{6}{x(6-x)}=\frac{6}{9-9+6 x-x^{2}}=\frac{6}{9-(3-x)^{2}}
\]
The sum will be the smallest when \((3-x)^{2}=0\), i.e., when \(x=3\), so the smallest value of the sum is \(\frac{2}{3}\).
Second method. Let \(x=3+a\), then \(y=6-3-a=... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,552 |
29*. What is greater: \(\frac{23^{1873}+1}{23^{1974}+1}\) or \(\frac{23^{1974}+1}{23^{1975}+1}\) ?
30 *. Let \(|y| \neq 1\) and \(y \neq 0\). It is also known that
\[
x_{1}=\frac{y-1}{y+1}, \quad x_{2}=\frac{x_{1}-1}{x_{1}+1}, \quad x_{3}=\frac{x_{2}-1}{x_{2}+1}, \ldots
\]
What is \(y\), if \(x_{1974}=3\) ?
31 *. I... | 29. The first fraction is larger. Hint. You can use the substitution \(m=23^{1973}\) and find the difference between the fractions. | \frac{23^{1873}+1}{23^{1974}+1}>\frac{23^{1974}+1}{23^{1975}+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43,553 |
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