Counting Divisors From a Prime Factorisation

"How many divisors does 360 have?" is a common interview warm-up, and listing them by hand is slow and error-prone once the number gets past a couple dozen. The clean approach is to first write the number as a product of prime powers, then use a formula that converts that factorisation directly into a divisor count.

If a number $N$ factors as $p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k}$, where each $p_i$ is a distinct prime, then the number of positive divisors of $N$ is $(a_1+1)(a_2+1)\cdots(a_k+1)$. The reasoning is that any divisor of $N$ is built by picking, independently for each prime, one of $a_i + 1$ possible exponents (0 through $a_i$), and multiplying the choices together — the count of divisors is just the count of these independent choices multiplied out.

For $N = 360$: factorise it as $2^3 \times 3^2 \times 5^1$. The exponents are 3, 2, and 1, so the divisor count is $(3+1)(2+1)(1+1) = 4 \times 3 \times 2 = 24$. Indeed, 360 has exactly 24 positive divisors — a fact that would take real patience to verify by listing them one at a time.

The same trick extends naturally: interviewers sometimes ask for the number of divisors that are themselves perfect squares, or the sum of all divisors (a related but different formula, $\prod \frac{p_i^{a_i+1}-1}{p_i-1}$), so recognising that the whole family of divisor questions reduces to "factorise first" is the transferable skill, not memorising one specific formula.