2023 CSAT — Q26

Two moves: (1) choose the four largest distinct single-digit numbers, (2) split them so the two pair-sums are as equal as possible (a product of two numbers with fixed total is largest when they are equal).

Largest four distinct single digits: $9, 8, 7, 6$, total $= 30$.

To maximise $(p+q)(r+s)$ with $(p+q)+(r+s) = 30$ fixed, make each pair sum $15$:

Any other split (e.g. $9+8=17,\ 7+6=13 \Rightarrow 17\times 13 = 221$) gives less.

Answer: (b) 225.

Two layered optimisations: maximise the total (pick $9,8,7,6$), then balance the two sums (fixed-sum products peak at equality, the AM–GM idea). The deadliest decoy is $221$ from the “obvious” pairing $9+8$ with $7+6$; the scoring move is to pair high-with-low ($9+6,\ 8+7$) to hit $15\times 15$. Both ideas are needed — a candidate who balances but uses the wrong four digits, or picks the right digits but pairs them lopsidedly, loses the mark.