CAT 2022 Slot 1 · Quant · Question 21

For A, compute the first few values and take their common divisor.

At k=1: 3+4+5=12.

At k=2: 9+16+25=50.

The greatest common divisor of 12 and 50 is 2, so A can be at most 2.

And 2 always divides: {3}^{k}+{5}^{k} is a sum of two odd numbers, hence even, and {4}^{k} is even too.

So A=2.

For B, simplify the expression before testing anything.

Since {4}^{k+2}=16\times{4}^{k}, the three terms are {4}^{k}+3\times{4}^{k}+16\times{4}^{k}.

Factoring out {4}^{k}, the expression is {4}^{k}\left( 1+3+16 \right)=20\times{4}^{k}.

Every such number is a multiple of 20\times 4=80, since k is at least 1.

And the smallest value, at k=1, is exactly 80, so nothing larger can divide them all.

So B=80.

Therefore A+B=2+80=82.

Shortcut. The second expression collapses to a single term once {4}^{k+2} is rewritten, so B is just its value at k=1. For A, two values are enough: \gcd \left( 12,50 \right)=2, and parity confirms 2 always works.