CAT 2025 Slot 2 · Quant · Question 1

Put a=m+2n and b=2m+n, so that ab=27.

Inverting the substitution, m=\frac{2b-a}{3} and n=\frac{2a-b}{3}.

Therefore 2m-3n=\frac{2\left( 2b-a \right)-3\left( 2a-b \right)}{3}=\frac{7b-8a}{3}.

The factor pairs of 27, with both signs allowed, are \left( 1,27 \right), \left( 3,9 \right), \left( 9,3 \right), \left( 27,1 \right) and their negatives.

For m and n to be integers, 2b-a and 2a-b must be divisible by 3, which rules out \left( 1,27 \right), \left( 27,1 \right) and their negatives.

Evaluating \frac{7b-8a}{3} on the survivors: \left( 3,9 \right) gives 13, \left( 9,3 \right) gives -17, \left( -3,-9 \right) gives -13, and \left( -9,-3 \right) gives 17.

The maximum is 17, reached at m=1 and n=-5.

Shortcut.

From 2m-3n=\frac{7b-8a}{3}, the value grows as a falls and b rises. Since ab=27>0, both have the same sign, so test the negative pairs first — the most negative a with the least negative b is \left( -9,-3 \right), which is the answer straight away.