A repunit is an integer consisting entirely of ones, such as 1, 11, 111, etc. Let N be any integer not divisible by 2 or 5. Show that there is an integer M such that the product MN is a repunit.
Another way of stating this problem is to show that for any positive integer N coprime to 10 that there is a repunit R which is a multiple of N. Then M is just R/N.
Divide each of the first N+1 repunits by N to find the remainders. By the pigeonhole principle there must be at least one pair of matching remainders.
The difference must be a multiple of N. Also, the difference can be expressed as the product of a repunit R and a power of 10. But N is coprime to 10 which then implies that N is a factor of R.
Solution