At the outset, we proceed to determine a multiple of 5^n which does not possess any cube digit in the last n digits. Suppose that, we have a multiple of p*(5^n) whose last cube digit is in the rth place from the right.

Then, p*(10^r + 1) has the same digits in places 0 to r-1 and a

non-cube digit in place r, and hence no zeros in places 0 to r.

Repeating the above procedure, we find a multiple of p*(5^n) with no cube digits in the last n digits.

Now, let d be the remainder when p is divided by 2^n, and therefore:

p = s*.2^n + d, and hence:

d*(.5^n) = p*(5^n) - s*(10^n)

Accordingly, d*(5^n) has the same last n digits as p*(5^n). But it has less than (n+1) digits and consequently, it possesses exactly n digits with no cube digits.