2023 CSAT — Q10

To minimise $\dfrac{N}{S(N)}$ (number over digit-sum) among $3$-digit numbers, push the number small and the digit-sum large at the same time — i.e. keep the hundreds digit at its floor ($1$) and load the lower places with $9$s.

Write $N = 100h + 10t + u$ with $S = h + t + u$. The ratio is

This shrinks when $h$ is small and $u$ (which is in the denominator but not the heavy numerator term) is large. Set $h = 1$ and $u = 9$, then maximise $t$: $t = 9$ gives $N = 199$, $S = 19$,

Check neighbours: $189 \to 189/18 = 10.5$; $198 \to 198/18 = 11$; $109 \to 10.9$. None beats $199$. So $D = 199$.

Hundreds digit $= 1$, units digit $= 9$, difference $= 9 - 1 = 8$.

Answer: (c) 8.

The ratio $1 + \dfrac{99h + 9t}{h+t+u}$ exposes the structure: the numerator is dominated by $h$ (weight $99$), so hundreds digit $= 1$ is forced; then the units digit only helps the denominator, so push it to $9$; finally maximise the tens digit. That reasoning lands on $199$ in one pass — far faster than scanning hundreds of candidates. The decoy is $999$ (“biggest digit sum”), but a huge digit-sum on a huge number is no help; it is the ratio that matters.