TITLE: Prove that every number ending in a $3$ has a multiple which consists only of ones.
QUESTION [19 upvotes]: Prove that every number ending in a $3$ has a multiple which consists only of ones.
Eg. $3$ has $111$, $13$ has $111111$.

Also, is their any direct way (without repetitive multiplication and checking) of obtaining such  multiple of any given number( ending with $3$) ?

REPLY [2 votes]: Find a more generalized solution for any natural number in base here: 
All odd primes except $5$ divide a number made up of all $1$s