CAT 2023 Slot 2 · Quant · Question 6

Two factors other than 1 and the number itself means four factors in total.

If a number has prime factorisation {p}_{1}^{{e}_{1}}{p}_{2}^{{e}_{2}}\cdots, its factor count is \left({e}_{1}+1 \right)\left({e}_{2}+1 \right)\cdots.

For that product to be 4, the only options are 4=4 or 4=2\times 2.

The first gives e+1=4, that is a cube of a prime, {p}^{3}.

The second gives two primes each to the first power, that is p\times q with p\ne q.

Prime cubes below 50: {2}^{3}=8 and {3}^{3}=27; the next is {5}^{3}=125, far too large — so 2 numbers.

Products of two distinct primes below 50, taking 2 as the smaller: 6, 10, 14, 22, 26, 34, 38 and 46 — that is 8 numbers.

Taking 3 as the smaller: 15, 21, 33 and 39 — that is 4 numbers, since 3\times 17=51 is too large.

Taking 5 as the smaller: only 5\times 7=35, since 5\times 11=55 exceeds 50 — that is 1 number.

Nothing survives from 7 onwards, as 7\times 11=77.

So the semiprimes number 8+4+1=13, and the total is 13+2=15.

Shortcut. Four factors means either {p}^{3} or pq — nothing else. Prime cubes are rare, so the count is almost entirely the semiprimes, and those are listed quickly by fixing the smaller prime and sweeping the larger one up to \frac{50}{p}.