The answer is only $2$. The idea is that if $p\ne 2,5$, the system of congruences $n\equiv 2015\pmod {10000}$ and $n\equiv 0\pmod p$ has a solution by the Chinese Remainder Theorem equal to some residue modulo $10000p$, and thus there is a positive solution. $p=5$ clearly has a multiple ending in $2015$, and $p=2$ clearly doesn't (as all numbers ending in $2015$ are odd).