2022 AMC 10B Problem 13

The two primes differ by $2$, so call them $p$ and $p + 2$. Write the cube difference in terms of $p$.

1. Expand. $(p + 2)^3 = p^3 + 6p^2 + 12p + 8$, so
$$(p + 2)^3 - p^3 = 6p^2 + 12p + 8 = 31106.$$
2. Simplify. Subtract $8$: $6p^2 + 12p = 31098$. Divide by $6$: $p^2 + 2p = 5183$.
3. Make a perfect square. Adding $1$ to both sides turns the left side into $(p + 1)^2$:
$$(p + 1)^2 = 5184.$$
4. Take the square root. $70^2 = 4900$ and $72^2 = 5184$, so $p + 1 = 72$ and $p = 71$. The primes are $71$ and $73$, and both really are prime.
5. Read the question again. It asks for the smallest prime bigger than both, so find the next prime after $73$:
- $74, 76, 78$ are even;
- $75$ ends in $5$;
- $77 = 7 \times 11$;
- $79$ has no factor $2, 3, 5$ or $7$ (and $9^2 = 81 > 79$, so no more checks are needed). It is prime.
6. Digit sum. $7 + 9 = 16$.

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

When two numbers differ by a fixed amount, the difference of their cubes is only a quadratic in the smaller one, and here it becomes a perfect square after adding $1$. Knowing squares near $70$ makes the root quick. Last-step traps are common, so always reread what the question actually asks for.

Estimate first. The difference $(p + 2)^3 - p^3$ is about $6p^2$, so $6p^2 \approx 31106$, giving $p^2 \approx 5184$ and $p$ near $72$. The prime pair differing by $2$ near there is $71, 73$. Check: $73^3 - 71^3 = 389017 - 357911 = 31106$.