2022 AMC 10B Problem 13

Call the primes $p$ and $p + 2$. Expand the difference of cubes:
$$(p+2)^3 - p^3 = 6p^2 + 12p + 8 = 31106.$$
So $6p^2 + 12p = 31098$, and dividing by $6$ gives $p^2 + 2p = 5183$. Complete the square:
$$(p+1)^2 = 5184 = 72^2,$$
hence $p = 71$ and the primes are $71$ and $73$ (both prime, as required).

The next prime after $73$: $74, 75, 76$ are even or end in $5$; $77 = 7 \cdot 11$; $78$ is even; $79$ is prime.

The digit sum of $79$ is $7 + 9 = 16$.

The answer is $\boxed{\textbf{(E)}\ 16}$.

A difference of two cubes whose bases differ by a constant is a quadratic in the base, and here it completes the square exactly. Recognizing $5184 = 72^2$ (from $70^2 = 4900$ and $72^2 = 4900 + 280 + 4$) makes the equation instant. Then read the question again: it asks about the next prime, not the pair.

Estimate: $(p+2)^3 - p^3 \approx 6p^2$, so $p^2 \approx 5184$ and $p \approx 72$. The only twin-prime pair near $72$ is $(71, 73)$; a quick check $73^3 - 71^3 = 389017 - 357911 = 31106$ confirms it.

• Stopping at the primes 71 and 73 and reporting a digit sum of one of them (8 or 10), instead of the next prime 79.
• An arithmetic slip in the expansion (e.g. writing $6p^2 + 12p + 4$) that makes the square fail to complete.