Prove that for any positive integer n, there exists at least one multiple of 5^n that doesn't have any perfect cube digits (0, 1, or 8) in its decimal representation.

Set K=5^N
Set L=K
Set I=1
while I isn't greater than N
if the I-th digit from the right of K is 0, 1, 8
add L to K
add 1 to I
multiply L by 10
the answer is the N rightmost digits of K