Question 106386
I'm a little unsure if this is the complete answer to your problem, but perhaps it will give you a start.

First off, you know that 0 < a < b, since the fraction is < 1, and 0/333 reduces to 0.

Second, we know that 0 < a < 167, and 166 < b < 333.

Third, we know that a = 333 - b.

Our fraction can be represented as {{{(333-b)/b}}}, or put another way, {{{(333/b)-(b/b)}}}

Now the prime factorization of 333 is 3 X 3 X 37.

This tells me that any {{{a/b}}} where {{{a}}} has 3 or 37 as a factor would be reduceable.

There are 55 numbers in the range 0 < a < 167 that have 3 as a factor, and 4 numbers in that range that have 37 as a factor.  However, these two sets have one number in common, namely 111.  Hence the number of numerators that result in a reduceable fraction is 55 plus 4 minus 1, or 58.  Since there are 166 possible fractions excluding the non-reduceable condition, the number of non-reduceable fractions would be 108.

I think.