A certain 10-base integer with distinct digits can be converted to base b by merely reversing its digits.
Find the lowest value of b.

Let 10x+y be a two digit base 10 number where x<b and y<b.

It is to be equal to its reversal in base b: by+x

10x+y=by+x
9x = (b-1)y
We want b to be as small as possible.
b cannot equal 2 because that would require x=1 and y=9, which violates y<b.  (91 is not proper in base 2, although in a sense it is 19 in base 10.)
For b to equal 3, again we would require x=2 and y=9.
It turns out b=4 is possible with x=1 and y=3 (there are other false solutions.)
So we have the reversal 13 (base 10) equals 31 (base 4.)

This is the smallest possible two digit solution, now to show there aren't smaller bases with more digits.

A base 2 solution is not possible, as the problem requires distinct digits and there are only two binary digits.

A base 3 solution would only be feasible with three digits.   With a 2 or 1 in the 10^2 place, the reversal can only be 1 or 2 times 3^2 plus 0 times 3 plus 2 or 1 which is no where near big enough.  So we are done.

Posted by Jer on 2011-03-30 22:39:18