Find the minimal positive integer $m$, so that there exist positive integers $n>k>1$, which satisfy $11...1=11...1.m$, where the first number has $n$ digits $1$, and the second has $k$ digits $1$.
Problem
Source: First Saudi Arabia JBMO TST 2019, P1
Tags: number theory
kaede
07.04.2020 13:59
Suppose that $( m,n,k) \in \mathbb{N}^{3}$ satisfies the condition.
Then $11\equiv 11m\pmod{100}$, which implies $m\equiv 1\pmod{100}$.
So we must have $m\geq 101$.
Note that $1111=11\cdot 101$.
Hence the answer is $m=101$.
Manteca
16.06.2020 05:15
As $11...1 = 11...1.m$ $\frac{10^n-1}{9} = \frac{10^k-1}{9} . m$ $m = \frac{10^n-1}{10^k-1}$ This implies $10^k-1|10^n-1$, and : $10^k-1|10^n-1-10^{n-k}(10^k-1) = 10^{n-k} - 1$ In fact, if we continue this way, we have that: If $n \equiv A\; (mod\; k)$ such that $0\leq A\leq k-1$ Then $10^k-1|10^A -1$ This implies $A=0$, otherwise the numerator will not be $0$ and as it is less than the denominator, is impossible. Then $n=Bk$ and $m = \frac{(10^k)^B-1}{10^k-1} = \sum_{i=0}^{B-1}(10^k)^i$ In order to reduce the terms and the amount of them, the least $B$ we can take is $2$ (if $B=1$ then $m=1$) and the least $10^k$ is $10^2$, then the least value of $m$ is $m=101$ Where $1111 = 11.101$ and we are done.