2025 CSAT — Q5

We need three distinct factors of $N$ whose sum is $N$ itself. The largest proper factor of $N$ is $N/2$ (and it exists only if $N$ is even), so any sum of three factors that reaches the full $N$ must lean on the large ones. The clean structural fact:

So every multiple of $6$ automatically has the property: take $p=N/2,\ q=N/3,\ r=N/6$, three distinct factors summing to $N$. Below $50$ the multiples of $6$ are

Is there any non-multiple of $6$ that also works? The only triples of unit fractions $\tfrac1a+\tfrac1b+\tfrac1c = 1$ with $a<b<c$ are $\{2,3,6\}$ — the divisor triple must be $\{N/2,N/3,N/6\}$, which forces $6 \mid N$. (E.g. $\tfrac12+\tfrac14+\tfrac14$ repeats a factor; $\tfrac12+\tfrac13+\tfrac1{12} = \tfrac{11}{12} \ne 1$.) No non-multiple of $6$ qualifies.

Answer: (c) 8.

The whole problem collapses once you see it as a unit-fraction identity: $p+q+r=N$ with $p,q,r \mid N$ means $\tfrac{N}{p}+\tfrac{N}{q}+\tfrac{N}{r}$ are three distinct positive integers whose reciprocals sum to $1$. The only such reciprocal triple is $\tfrac12+\tfrac13+\tfrac16$, so the property is exactly equivalent to ”$N$ is a multiple of $6$.” That reframing turns a hunt-and-check item into a one-line count ($\lfloor 49/6\rfloor = 8$) and protects you from both boundary slips. Under the clock, recognising the $\{2,3,6\}$ identity is worth far more than testing numbers one by one.