Can there exist two functions \(f\) and \(g\) that take only integer values such that for any integer \(x\) the following relations hold:
a) \(f (f (x)) = x\), \(g (g (x)) = x\), \(f (g (x)) > x\), \(g (f (x)) > x\)?
b) \(f (f (x)) < x\), \(g (g (x)) < x\), \(f (g (x)) > x\), \(g (f (x)) > x\)?
Find all functions \(f (x)\) such that \(f (2x + 1) = 4x^2 + 14x + 7\).
For each pair of real numbers \(a\) and \(b\), consider the sequence of numbers \(p_n = \lfloor 2 \{an + b\}\rfloor\). Any \(k\) consecutive terms of this sequence will be called a word. Is it true that any ordered set of zeros and ones of length \(k\) is a word of the sequence given by some \(a\) and \(b\) for \(k = 4\); when \(k = 5\)?
Note: \(\lfloor c\rfloor\) is the integer part, \(\{c\}\) is the fractional part of the number \(c\).
Prove that for any natural number \(a_1> 1\) there exists an increasing sequence of natural numbers \(a_1, a_2, a_3, \dots\), for which \(a_1^2+ a_2^2 +\dots+ a_k^2\) is divisible by \(a_1+ a_2+\dots+ a_k\) for all \(k \geq 1\).
How many rational terms are contained in the expansion of
a) \((\sqrt 2 + \sqrt[4]{3})^{100}\);
b) \((\sqrt 2 + \sqrt[3]{3})^{300}\)?
Suppose that in each issue of our journal in the “Quantum” problem book there are five mathematics problems. We denote by \(f (x, y)\) the number of the first of the problems of the \(x\)-th issue for the \(y\)-th year. Write a general formula for \(f (x, y)\), where \(1 \geq x \geq 12\) and \(1970 \geq y \geq 1989\). Solve the equation \(f (x, y) = y\). For example, \(f (6, 1970) = 26\). Since \(1989\), the number of tasks has become less predictable. For example, in recent years, half the issues have 5 tasks, and in other issues there are 10. Even the number of magazine issues has changed, no longer being 12 but now 6.
Author: V.A. Popov
On the interval \([0; 1]\) a function \(f\) is given. This function is non-negative at all points, \(f (1) = 1\) and, finally, for any two non-negative numbers \(x_1\) and \(x_2\) whose sum does not exceed 1, the quantity \(f (x_1 + x_2)\) does not exceed the sum of \(f (x_1)\) and \(f (x_2)\).
a) Prove that for any number \(x\) on the interval \([0; 1]\), the inequality \(f (x_2) \leq 2x\) holds.
b) Prove that for any number \(x\) on the interval \([0; 1]\), the \(f (x_2) \leq 1.9x\) must be true?
The function \(f (x)\) for each real value of \(x\in (-\infty, + \infty)\) satisfies the equality \(f (x) + (x + 1/2) \times f (1 - x) = 1\).
a) Find \(f (0)\) and \(f (1)\). b) Find all such functions \(f (x)\).
Take any two non-equal numbers \(a\) and \(b\), then we can write \[a^2 - 2ab + b^2 = b^2 - 2ab + a^2\] Using the formula \((x-y)^2 = x^2 - 2xy + y^2\), we complete the squares and rewrite the equality above as \[(a-b)^2 = (b-a)^2.\] As we take a square root from the both sides of the equality, we get \[a-b = b-a.\] Finally, adding to both sides \(a+b\) we get \[\begin{aligned} a-b + (a+b) &= b-a + (a+ b)\\ 2a&= 2b\\ a&=b. \end{aligned}\] Therefore, All NON-EQUAL NUMBERS ARE EQUAL! (This is gibberish, isn’t it?)
Consider equation \[x-a=0\] Dividing both sides of this equation by \(x-a\), we get \[\frac{x-a}{x-a} = \frac{0}{x-a}.\] But \(\frac{x-a}{x-a}=1\) and \(\frac{0}{x-a}=0\). Therefore, we get \[1=0.\]