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Let a_{1}, \ldots, a_{n} be an infinite sequence of strictly positive integers, so that a_{k} < a_{k+1} for any k. Prove that there exists an infinity of terms a_m, which can be written like a_m = x \cdot a_p + y \cdot a_q with x,y strictly positive integers and p \neq q.

Slični zadaci

For what real values of x is \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A given

a) A=\sqrt{2};

b) A=1;

c) A=2,

where only non-negative real numbers are admitted for square roots?
Given any integer n \geq 2, assume that the integers a_1, a_2, \ldots, a_n are not divisible by n and, moreover, that n does not divide \sum^n_{i=1} a_i. Prove that there exist at least n different sequences (e_1, e_2, \ldots, e_n) consisting of zeros or ones such \sum^n_{i=1} e_i \cdot a_i is divisible by n.
An infinite sequence \,x_{0},x_{1},x_{2},\ldots \, of real numbers is said to be bounded if there is a constant \,C\, such that \, \vert x_{i} \vert \leq C\, for every \,i\geq 0. Given any real number \,a > 1,\, construct a bounded infinite sequence x_{0},x_{1},x_{2},\ldots \, such that
\vert x_{i} - x_{j} \vert \vert i - j \vert^{a}\geq 1
for every pair of distinct nonnegative integers i, j.
For any positive integer x define g(x) as greatest odd divisor of x, and f(x) =\begin{cases}\frac{x}{2}+\frac{x}{g(x)}&\text{if\ \(x\) is even},\\ 2^{\frac{x+1}{2}}&\text{if\ \(x\) is odd}.\end{cases}
Construct the sequence x_1 = 1, x_{n + 1} = f(x_n). Show that the number 1992 appears in this sequence, determine the least n such that x_n = 1992, and determine whether n is unique.
Let R_1,R_2, \ldots be the family of finite sequences of positive integers defined by the following rules: R_1 = (1), and if R_{n - 1} = (x_1, \ldots, x_s), then

R_n = (1, 2, \ldots, x_1, 1, 2, \ldots, x_2, \ldots, 1, 2, \ldots, x_s, n).

For example, R_2 = (1, 2), R_3 = (1, 1, 2, 3), R_4 = (1, 1, 1, 2, 1, 2, 3, 4). Prove that if n > 1, then the kth term from the left in R_n is equal to 1 if and only if the kth term from the right in R_n is different from 1.
Find all pairs (a,b) of positive integers that satisfy the equation: a^{b^2} = b^a.