CAT 2023 Slot 3 · Quant · Question 7

First find which natural numbers n satisfy the given condition.

At n=1: {5}^{0}=1 and {3}^{2}=9, so 1<9 holds.

At n=2: {5}^{1}=5 and {3}^{3}=27, so it holds.

At n=3: {5}^{2}=25 and {3}^{4}=81, so it holds.

At n=4: {5}^{3}=125 and {3}^{5}=243, so it holds.

At n=5: {5}^{4}=625 and {3}^{6}=729, so it just holds.

At n=6: {5}^{5}=3125 but {3}^{7}=2187, so it fails — and since 5 grows faster than 3, it fails for every larger n too.

So the allowed values are n=1,2,3,4,5.

The second condition is {3}^{n+1}<{2}^{n+m}, which rearranges to {2}^{m}>\frac{{3}^{n+1}}{{2}^{n}}.

The right side is 3\times{\left( \frac{3}{2}\right)}^{n}, which grows with n, so the hardest case is n=5.

There, {\left( \frac{3}{2}\right)}^{5}=\frac{243}{32}\approx 7.59, so {2}^{m}>3\times 7.59\approx 22.78.

Since {2}^{4}=16 is too small and {2}^{5}=32 clears it, the least integer is m=5.

Verifying at n=5 with m=5: {3}^{6}=729 and {2}^{10}=1024, so the inequality holds.

Shortcut. Only the largest allowed n matters, because dividing by {2}^{n} leaves a quantity that increases with n. So the whole question is {3}^{6}<{2}^{5+m} at n=5, and the familiar {2}^{10}=1024 beating {3}^{6}=729 settles it at once.