CAT 2023 Slot 2 · Quant · Question 1

Start by factorising the right-hand side into primes.

Since 144=16\times 9={2}^{4}\times{3}^{2}, raising to the 145th power gives {144}^{145}={2}^{580}\times{3}^{290}.

To make n-m as large as possible, push n up and pull m down.

Since m is a natural number, the smallest it can be is m=1.

For n, note that {b}^{n} must divide {2}^{580}\times{3}^{290} with b>1.

The largest exponent available on any single prime is 580, achieved by taking b=2.

So the ceiling on n is 580, and any larger n would need a prime power beyond what is present.

Both extremes are reachable together: take b=2 with n=580, and then a={3}^{290} with m=1.

Checking, {a}^{m}{b}^{n}={3}^{290}\times{2}^{580}={144}^{145}, exactly as required, and a>1.

So the largest value of n-m is 580-1=579 — the third option.

Shortcut. The two bases need not be prime; only b has to be, to squeeze the exponent to its maximum. Lumping the entire leftover {3}^{290} into a single a with m=1 costs almost nothing, so the answer is just 580-1.