Let $r$ and $k$ be positive integers such that all prime divisors of $r$ are greater than $50$. A positive integer, whose decimal representation (without leading zeroes) has at least $k$ digits, will be called nice if every sequence of $k$ consecutive digits of this decimal representation forms a number (possibly with leading zeroes) which is a multiple of $r$. Prove that if there exist infinitely many nice numbers, then the number $10^k-1$ is nice as well.
Problem
Source: Baltic Way 2007
Tags: number theory proposed, number theory
Rust
01.12.2010 08:08
Let $x=a_{k+1}*10^{k+1}+a_k*10^k+...+a_0$ is nice number and $A=a_k*10^k+...+a_2, B=a_{k-1}*10^{k-1}+...+a_1, C=a_{k-2}*10^{k-1}+...+a_0$. Then $r|a_{k+1}*10^{k-1}+A,r|a_k*10^{k-1}+B=10A+a_1,r|a_{k-1}*10^{k-1}+C=10B+a_0$. Therefore $r|a_{k+1}*10^k-a_1, r|a_k*10^k-a_0$. If $a_{k+1}=a_1$ or $a_k=a_0$, then $r|10^k-1\to r|\frac{10^k-1}{9}.$ Else $r|a_0a_{k+1}-a_1a_k$. If $x$ had more digits, then $r|a_ia_{k+i+1}-a_{i+1}a_{k+i} \forall i\ge 0.$ If never $a_i=a_{k+i}$ it gives contradition.
Tintarn
12.10.2018 14:06
Rust wrote: Let $x=a_{k+1}*10^{k+1}+a_k*10^k+...+a_0$ is nice number and $A=a_k*10^k+...+a_2, B=a_{k-1}*10^{k-1}+...+a_1, C=a_{k-2}*10^{k-1}+...+a_0$. Then $r|a_{k+1}*10^{k-1}+A,r|a_k*10^{k-1}+B=10A+a_1,r|a_{k-1}*10^{k-1}+C=10B+a_0$. Therefore $r|a_{k+1}*10^k-a_1, r|a_k*10^k-a_0$. If $a_{k+1}=a_1$ or $a_k=a_0$, then $r|10^k-1\to r|\frac{10^k-1}{9}.$ Else $r|a_0a_{k+1}-a_1a_k$. If $x$ had more digits, then $r|a_ia_{k+i+1}-a_{i+1}a_{k+i} \forall i\ge 0.$ If never $a_i=a_{k+i}$ it gives contradition. I cannot see how you are going to use the condition about the prime factors of $r$. And that assumption is crucial to the problem. For instance, if $r=2$ and $k=1$, all numbers with only even digits are nice, but $r \nmid 10^k-1$.
The important fact is that $50>1+2+3+4+5+6+7+8+9$.