2013 AMC 10B Problem 24

Step 1: which numbers have exactly four divisors?

1. Look at examples: $6$ has divisors $1, 2, 3, 6$; $10$ has $1, 2, 5, 10$; $8$ has $1, 2, 4, 8$. There are two kinds:
- a product of two different primes, $m = pq$, with divisors $1, p, q, pq$;
- a prime cubed, $m = p^3$, with divisors $1, p, p^2, p^3$.
2. Their divisor sums factor nicely:
$$1 + p + q + pq = (1 + p)(1 + q), \qquad 1 + p + p^2 + p^3 = (1 + p)(1 + p^2).$$
Check with $6$: $1 + 2 + 3 + 6 = 12 = 3 \times 4$.

Step 2: prime cubes.

1. For $p = 11$ the sum is $1 + 11 + 121 + 1331 = 1464$. For $p = 13$ it is $1 + 13 + 169 + 2197 = 2380$. These jump right over $2010$ to $2019$, so no cube works.

Step 3: two different primes, one of them $2$.

4. The sum is $3(q + 1)$. Here $q$ is an odd prime, so $q + 1$ is even and the sum is a multiple of $6$. In our range the multiples of $6$ are $2010$ and $2016$.
- $2010$: $q + 1 = 670$, $q = 669 = 3 \times 223$. Not prime.
- $2016$: $q + 1 = 672$, $q = 671 = 11 \times 61$. Not prime.

Step 4: two different odd primes.

5. Now $p + 1$ and $q + 1$ are both even, so the sum is a multiple of $4$. In our range that means $2012$ or $2016$.
- $2012 = 4 \times 503$, and $503$ is prime (it is not divisible by $2, 3, 5, 7, 11, 13, 17$ or $19$). The only way to write $2012$ as even times even is $2 \times 1006$, which needs $p + 1 = 2$, so $p = 1$. Not prime, so $2012$ fails.
- $2016$: try $p = 3$. Then $p + 1 = 4$ and $q + 1 = 504$, so $q = 503$, which is prime. So $m = 3 \times 503 = 1509$ has divisors $1, 3, 503, 1509$, adding up to $2016$. Nice!

Exactly one number, $2016$, is nice.

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

"Exactly four divisors" allows only two shapes of number, and both have a divisor sum that factors. Checking even and odd then throws out most of the ten candidates at once, and only a couple need real factoring. When a problem gives a condition on divisors, first work out what the number itself must look like.

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 $\boxed{\textbf{(A)}\ 1}$.