Saudi Arabian Mathematical Competitions

Let $k \in \{1,2, \ldots, 2010\}$. 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 $\{1,2, \ldots, 2010\}$ is not a divisor of $5^{2010!}-3^{2010!}$. Any number $k$ in $\{1,2, \ldots, 2010\}$ which is not divisible by $3$ or $5$ is also a divisor of $5^{2010!}-3^{2010!}$. We have $\varphi(k)<k<2010$, and from Euler's Theorem we get
$$5^{2010!}-3^{2010!}=a^{\varphi(k)}-b^{\varphi(k)} \equiv 1-1 \quad(\bmod k) .$$
The desired number is
$$2010-\left[\frac{2010}{3}\right]-\left[\frac{2010}{5}\right]+\left[\frac{2010}{15}\right]=1072 .$$