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 Cubeless? (Posted on 2004-08-29)
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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 Puzzle Solution Comment 7 of 7 |

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.

 Posted by K Sengupta on 2007-06-12 11:13:15

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