12 can be written as a product of three numbers in 18 ways, like (1, 2, 6),…

Concept: To count the number of ordered triples (a, b, c) of positive integers whose product equals N, factorise N into prime powers. For each prime, its exponent e must be split into an ordered sum of three non-negative integers — one contribution to each of a, b, c — which by stars-and-bars can be done in C(e+2, 2) ways. Because different primes are independent of each other, the total number of ordered triples is the product of these per-prime counts across every prime in N's factorisation.

Application: 3600 factorises as 24 × 32 × 52:

1. Split 24's exponent (4) across the three numbers: C(4+2, 2) = C(6, 2) = 15 ways.

2. Split 32's exponent (2) across the three numbers: C(2+2, 2) = C(4, 2) = 6 ways.

3. Split 52's exponent (2) across the three numbers: C(2+2, 2) = C(4, 2) = 6 ways.

4. Multiply the three independent counts: 15 × 6 × 6 = 540.

Cross-check: Apply the same method to the given example, 12 = 22 × 31, before trusting it on 3600:

1. Split 22's exponent (2): C(2+2, 2) = C(4, 2) = 6 ways.

2. Split 31's exponent (1): C(1+2, 2) = C(3, 2) = 3 ways.

3. Total: 6 × 3 = 18 ways — exactly matching the "18 ways" stated in the question, confirming the method is correctly calibrated.

Result: So 3600 can be written as a product of three numbers in 540 ways.