CAT 2023 Slot 1 · Quant · Question 4

Both numbers need their prime factorisations before anything else.

For 168:

168=8\times 21={2}^{3}\times 3\times 7.

For 1134:

1134=2\times 567, and 567=7\times 81=7\times{3}^{4}, so 1134=2\times{3}^{4}\times 7.

Raising to the power n, {1134}^{n}={2}^{n}\times{3}^{4n}\times{7}^{n}.

For 168 to divide this, each prime's exponent must be at least as large as in 168: n\ge 3, 4n\ge 1 and n\ge 1.

The binding condition is n\ge 3, so the least value is n=3.

Then {1134}^{3}={2}^{3}\times{3}^{12}\times{7}^{3}.

Also {168}^{m}={2}^{3m}\times{3}^{m}\times{7}^{m}.

For {1134}^{3} to divide {168}^{m}, the requirements are 3m\ge 3, m\ge 12 and m\ge 3.

The binding condition is m\ge 12, so the least value is m=12.

Therefore m+n=12+3=15, the third option.

Shortcut. Each question is settled by a single prime — the one whose exponent ratio is worst. Going in, 1134 is short of 2s, so the power of 2 fixes n=3; coming back, 168 is short of 3s, so the power of 3 fixes m=12. The other primes never bind.