Minimum number of states in DFA accepting strings where the numbers of a and b are divisible by X and Y respectively?
I think the minimum number of states is $X\cdot Y$.  In particular, for $X=6$ and $Y=8$, I think 48 states are required.

One can prove this using Myhill-Nerode.  The equivalence class (in Myhill-Nerode) of string $s$ is uniquely determined by the pair $\langle n_a(s) \bmod X, n_b(s) \bmod Y \rangle$, where $n_a(s)$ counts the number of $a$'s in $s$, and $n_b(s)$ the number of $b$'s.  Two different equivalence classes $\langle i,j \rangle$, $\langle i',j' \rangle$ can be separated by the suffix $a^{X-i} b^{Y-j}$, since that'll take the former to an accepting state and the latter to a non-accepting state (if $i \ne i'$ or $j \ne j'$).  This gives us $X \cdot Y$ different equivalence classes, so any DFA accepting this language must have at least $X \cdot Y$ states.

Are you sure you can do it in 15 states, for $X=6$ and $Y=8$?  Is it possible you might have made a mistake somewhere in your DFA?