2024 CSAT — Q55

For a fixed sum $k$, count the pairs $(p,q)$ of positive integers with $p < q$ (strict, so equal halves are excluded). We want the smallest $k$ giving more than one such pair.

• $k = 3$: only $(1,2)$. Unique.
• $k = 4$: only $(1,3)$ — note $(2,2)$ is barred by $p < q$. Unique.
• $k = 5$: $(1,4)$ and $(2,3)$ — two pairs. Not unique.

So $k = 5$ is the first sum that fails to pin $(p,q)$.

Answer: (c) 5.

The number of strict pairs summing to $k$ is $\left\lfloor \frac{k-1}{2} \right\rfloor$, which first equals $2$ at $k = 5$. The two traps are (i) miscounting $(2,2)$ as a valid pair under $p<q$ (which would wrongly flag $k=4$), and (ii) walking past $k=5$ to a “more obviously multiple” sum. “Smallest value that does not determine uniquely” means find the first $k$ with $\ge 2$ partitions — test small $k$ in order and stop at the first failure. Strictness of $p<q$ is the whole subtlety.