The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$, respectively, and each graph has two positive integer x-intercepts. Find $h$.
Begin by setting $x$ to 0, then set both equations to $h^2=\frac{2013-j}{3}$ and $h^2=\frac{2014-k}{2}$, respectively. Notice that because the two parabolas have to have positive x-intercepts, $h\ge32$.
We see that $h^2=\frac{2014-k}{2}$, so we now need to find a positive integer $h$ which has positive integer x-intercepts for both equations.
Notice that if $k=2014-2h^2$ is -2 times a square number, then you have found a value of $h$ for which the second equation has positive x-intercepts. We guess and check $h=36$ to obtain $k=-578=-2(17^2)$.
Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is $\boxed{036}$.