2013 AMC 10B Problem 24

An integer with exactly four positive divisors is either $p^3$ or $pq$, where $p$ and $q$ are distinct primes. If $m=p^3$, the divisor sum is $1+p+p^2+p^3$. The values for $p=11$ and $p=13$ fall below and above the interval $2010$ to $2019$, so this case gives none. In the $m=pq$ case, the divisor sum is $1+p+q+pq=(p+1)(q+1)$. If one prime is $2$, the sum is divisible by $3$; only $2010$ and $2016$ qualify, but $\frac{2010}{3}-1=669$ and $\frac{2016}{3}-1=671$ are not prime. If both primes are odd, then the sum is divisible by $4$, leaving $2012$ and $2016$. The factorization $2012=4\cdot503$ would give primes $3$ and $502$, impossible, while $2016=4\cdot504$ $=(3+1)(503+1)$ works. Thus exactly one number is nice, and the correct answer is A .