Can the base M number 2009 be a perfect cube, where M is a positive integer ≥ 2?

If the answer is yes, give an example. Otherwise prove that the base M number 2009 can never be a perfect cube.

Would it be correct to say that for 2009 to be a base <b>M </b> number, <b>M</b> would have to be > 9 because any base <= 9 could not have a digit "9" in it?

Assuming that each digit in "2009" represents a multiplier of a different power of <b>M</b>, to get the decimal equivalent one must evaluate:
2*M^3 + 0*M^2 + 0*M^1 + 9*M^0
= 2*M^3 + 9

Posted by Sing4TheDay on 2009-03-09 11:48:37