2025 CSAT — Q27

A prime $> 2$ is odd, so $p^2 + k$ (which must be a prime $< 30$, hence odd unless it equals $2$) forces a parity condition. Split on $p$.

Case $p$ odd (so $p^2$ is odd): for $p^2 + k$ to be odd-prime, $k$ must be even, i.e. $k = 2$. Check $p^2 + 2 < 30$ prime:

• $p = 3$: $9 + 2 = 11$ — prime ✓ (gives $k = 2$).
• $p = 5$: $25 + 2 = 27 = 3^3$ — not prime.
• $p \ge 7$: $p^2 + 2 \ge 51 > 30$ — out.

So the odd-$p$ branch yields $k = 2$.

Case $p = 2$ (so $p^2 = 4$): need $4 + k < 30$ prime, $k$ prime:

• $k = 3 \to 7$ ✓, $k = 7 \to 11$ ✓, $k = 13 \to 17$ ✓, $k = 19 \to 23$ ✓.
• $k = 2 \to 6$, $k = 5 \to 9$, $k = 11 \to 15$, $k = 23 \to 27$ — all composite.

So the $p=2$ branch yields $k \in \{3, 7, 13, 19\}$.

Distinct values of $k$: $\{2,\ 3,\ 7,\ 13,\ 19\}$ — five values.

Answer: (b) 5.

The parity gate is the whole solve: $p^2 + k$ must be an odd prime $< 30$, so exactly one of $p^2, k$ must be even. Since $2$ is the only even prime, this splits cleanly into ”$p$ odd $\Rightarrow k = 2$” and ”$p = 2 \Rightarrow k$ odd.” Forgetting the small odd-$p$ branch (it contributes the lone $k=2$) is the most common under-count to (a) $4$; admitting composites is the over-count to (c)/(d). The discipline is: fix the parity, bound by $< 30$, then verify each candidate is genuinely prime — never assume. Five distinct $k$ survive.