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Product Digit Poser (Posted on 2010-06-13) Difficulty: 3 of 5
Determine all possible values of a positive integer N such that the product of the nonzero digits in the base-N representation of 2009 (base ten) is equal to 18 (base ten).

No Solution Yet Submitted by K Sengupta    
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Some Thoughts Analytical observations (partial spoiler) | Comment 4 of 5 |
1) By the way, it is clear than the lowest possible base is 4, because bases 2 and 3 do not involve 3 as a digit, and we need a 3 or a 9 to get a product of 18.

2009 base 4 is 33121, so the number base N must be 5 digits or less.  It is just a coincidence that base 4 actually works, in addition to giving us a lower bound.

2) Also, the largest base that can possibly work is (2009 minus 18), because 2009 (base 1991) is 1 18.  This was guaranteed to work.  

3) In general, if M is a positive integer and P the product of its non-zero digits, then base 10 will work.
Base (M - P) will also work if M >= 10.
(M-P) is different from 10 if M = 10 or M > 19.



Edited on June 13, 2010, 11:51 pm
  Posted by Steve Herman on 2010-06-13 15:22:41

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