CAT 2023 Slot 3 · Quant · Question 6

If a number factorises as {p}^{a}{q}^{b}\cdots, its factor count is \left( a+1 \right)\left( b+1 \right)\cdots.

For that product to be 15, the possibilities are 15 itself or 3\times 5.

The first gives a+1=15, that is a single prime to the 14th power, {p}^{14}.

The smallest such number is {2}^{14}=16384 — far too large to compete.

The second gives exponents 2 and 4, that is a number of the form {p}^{2}{q}^{4}.

To keep it small, give the larger exponent to the smaller prime: {2}^{4}\times{3}^{2}=16\times 9=144.

Swapping the exponents gives {3}^{4}\times{2}^{2}=81\times 4=324.

The next candidates are {2}^{4}\times{5}^{2}=400 and beyond, all larger than 324.

So the two smallest numbers with exactly 15 factors are 144 and 324.

Their sum is 144+324=468.

Shortcut. Since 15 is odd, every exponent-plus-one must be odd, so every exponent is even — which means the number is a perfect square. That alone narrows the hunt to squares, and 144={12}^{2} and 324={18}^{2} are the first two that work.