CAT 2025 Slot 2 · Quant · Question 9

Rearranging, 8b=3ac-8a, so b=\frac{a\left( 3c-8 \right)}{8}.

For b to be a natural number we need 3c-8>0, so c\ge 3, and a\left( 3c-8 \right) divisible by 8.

c=3 gives b=\frac{a}{8}, so a=8, b=1 and 3a+2b+c=24+2+3=29.

c=4 gives b=\frac{a}{2}, so a=2, b=1 and 3a+2b+c=6+2+4=12.

c=5 gives b=\frac{7a}{8}, so a=8, b=7 and the total is 43.

c=6 gives b=\frac{5a}{4}, so a=4, b=5 and the total is 28.

c=8 gives b=2a, so a=1, b=2 and the total is 15.

At c=4 the triple is \left( a,b,c \right)=\left( 2,1,4 \right) — three distinct naturals — and 3\left( 2 \right)\left( 4 \right)=24=8\left( 2+1 \right) checks out.

Larger c forces a up as well and the total only grows, so the minimum is 12.

Shortcut. 3a+2b+c is smallest when a is small, so look for the c that makes the fraction \frac{3c-8}{8} reduce most. c=4 gives \frac{4}{8}=\frac{1}{2}, the tamest denominator available, so a=2 — test that case first and it wins.