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Cubeless? (Posted on 2004-08-29) Difficulty: 3 of 5
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.

See The Solution Submitted by Federico Kereki    
Rating: 3.4000 (5 votes)

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Solution Similar solution | Comment 2 of 7 |
This can be solved in a similar fashion as "Spirit of '76 with no products". As an algorithm:

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

  Posted by e.g. on 2004-08-30 08:35:32
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