2013 AMC 10B Problem 24

A number has exactly four divisors when it is $p^3$ (divisors $1, p, p^2, p^3$) or $pq$ with distinct primes $p < q$ (divisors $1, p, q, pq$). The divisor sums are
$$1 + p + p^2 + p^3 = (1+p)(1+p^2) \qquad\text{or}\qquad (1+p)(1+q).$$

Cubes. $(1+p)(1+p^2)$ for $p = 11$ is $12 \cdot 122 = 1464$ and for $p = 13$ is $14 \cdot 170 = 2380$; nothing lands in $2010$ to $2019$.

Products $pq$. If $p = 2$, then $n = 3(q + 1)$ with $q$ odd, so $n$ is a multiple of $6$: $2010$ needs $q = 669 = 3 \cdot 223$, and $2016$ needs $q = 671 = 11 \cdot 61$; neither is prime. If both primes are odd, $p + 1$ and $q + 1$ are even, so $n$ is a multiple of $4$: only $2012$ and $2016$ qualify.

• $2012 = 4 \cdot 503$ with $503$ prime: the only way to write it as a product of two even numbers is $2 \cdot 1006$, giving $p = 1$. Not nice.
• $2016 = 4 \cdot 504$: take $p + 1 = 4$, $q + 1 = 504$, i.e. $p = 3$, $q = 503$, both prime. So $m = 1509$ has divisor sum $1 + 3 + 503 + 1509 = 2016$. Nice.

Exactly one number, $2016$, is nice.

The answer is $\boxed{\textbf{(A)}\ 1}$.

"Exactly four divisors" is a strong structural condition with only two shapes, and each shape has a factored divisor sum. Parity then eliminates most candidates at once: a sum $(p+1)(q+1)$ with odd primes is divisible by $4$, and with $p = 2$ it is divisible by $6$. Only the survivors need actual factoring.

• Forgetting that m = p^3 also has exactly four divisors, or accepting a factorization (p+1)(q+1) without checking that both p and q are prime.
• Allowing $p = q$ (then $m = p^2$ has only three divisors) or $p = 1$, which is not prime.