2018 AMC 10B Problem 11

A "never prime" expression must be composite for every prime $p$, so it should carry a divisibility argument, and the natural modulus is $3$.

If $p \ne 3$, then $p$ is not divisible by $3$, so $p^2 \equiv 1 \pmod 3$. Then
$$p^2 + 26 \equiv 1 + 26 = 27 \equiv 0 \pmod 3,$$
and $p^2 + 26 > 3$, so it is composite. If $p = 3$, then $p^2 + 26 = 35 = 5 \cdot 7$, also composite. Hence (C) is never prime.

The other constants ($16, 24, 46, 96$) are not $2 \bmod 3$, so this argument fails for them, and indeed each produces a prime: $5^2 + 16 = 41$, $7^2 + 24 = 73$, $5^2 + 46 = 71$, $19^2 + 96 = 457$.

The answer is $\boxed{\textbf{(C)}\ p^2+26}$.

"Always composite" is a modular statement: find a modulus $m$ such that $p^2 + c \equiv 0 \pmod m$ for every prime $p$ except possibly the primes dividing $m$, then check those finitely many exceptions by hand. Since $p^2 \equiv 1 \pmod 3$ for $p \ne 3$, the constant must be $\equiv 2 \pmod 3$; only $26$ qualifies.

Plug in primes systematically. With $p = 5$: $41$ (prime), $49$, $51$, $71$ (prime), $121$. With $p = 7$: $65$, $73$ (prime), $75$, $95$, $145$. After two primes only (C) and (E) survive; a mod-3 check kills (C), and $p = 19$ gives the prime $457$ for (E).

• Testing only p = 2 and p = 3 and picking the first expression that looks composite, without checking the other choices produce a prime for some p.
• Forgetting to check $p = 3$ separately (where $p^2 \equiv 0$, not $1$, mod $3$).