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Call N=5^n. Start analyzing its digits from right to left. Each time you find a zero, one or eight (say, at the Mth position) add N times 10^M; this will make the cube digit disappear (it will change to 5, 6 or 3) without affecting the digits to the right of the Mth place. Keep on doing this until the rightmost n digits are non-zero. At that time, you'll have a multiple of N that can be expressed as X.10^n+Y, and Y has no 0's, 1's or 8's in it. As the sum is a multiple of N, and X*10^n is also a multiple of N, it follows that Y must be a multiple of N, so Y is a number as we wanted. |
