AoPS - Problem

Processing math: 31%

Problem

Source: Romanian TST 2002

Tags: pigeonhole principle, number theory proposed, number theory

WakeUp

05.02.2011 16:46

Let $a,b$ be positive real numbers. For any positive integer $n$ , denote by $x_n$ the sum of digits of the number $[an+b]$ in it's decimal representation. Show that the sequence $(x_n)_{n\ge 1}$ contains a constant subsequence. Laurentiu Panaitopol

JoeBlow

05.02.2011 17:15

Let $M=\max \lbrace \text{sum of digits of }n:0\leq n\leq [a]\rbrace$ , and for each $k\in \mathbb{N}$ let $n_k$ be the smallest integer such that $[an_k+b]\geq 10^k$ . Then $10^k\leq an_k+b<10^k+a$ , so $x_{n_k}\leq M+1$ for all $k$ , hence by the pigeonhole principle infinitely many elements from the sequence $\lbrace x_{n_k}\rbrace$ are equal.

Rust

05.02.2011 17:23

Consider $n_k=[\frac{10^k}{a}]+1$ . Obviosly $x_{n_k}$ bounded, therefore there is constant subsequence $x_{n_k}$ .