Puzzles

If 1/10890 has a finite number of digits in base \(b\), then \(b^n/10890\) is an integer for some integer \(n\). In order for this to be possible \(b\) must be a mutiple of each prime factor of 10890: the smallest such \(b\) will be when the exponent of each prime factor is 1.

The prime factorisation of 10890 is 2×3³×5×11². Therefore the smallest base in which 1/10890 has a finite number of digits is 2×3×5×11=330.