SAMC

Let $k\in \{1,2,\dots ,2010\}$. If $3\mid k$, then $k\nmid 5^{2010!}-3^{2010!}$. Also, if $5\mid k$, then $k\nmid 5^{2010!}-3^{2010!}$. It follows that any multiple of $3$ or $5$ in the set $\{1,2,\dots ,2010\}$ is not a divisor of $5^{2010!}-3^{2010!}$. Any number $k$ in $\{1,2,\dots ,2010\}$ which is not divisible by $3$ or $5$ is also a divisor of $5^{2010!}-3^{2010!}$. We have $\phi (k)<k<2010$, and from Euler's Theorem we get $$5^{2010!}-3^{2010!}=a^{\phi (k)}-b^{\phi (k)}\equiv 1-1(\mathrm{mod}k).$$ The desired number is $$2010-\left[\frac{2010}{3}\right]-\left[\frac{2010}{5}\right]+\left[\frac{2010}{15}\right]=1072.$$