The number of divisors of 55n³ are 55 (Including 1 and the number itself). How many divisors does 7n⁷ have?

Answer: 352

Solution:

Every number has a unique prime factorization.
The number of divisors of a number n = p₁^e1 p₂^e2 … is given by

(e₁+1)(e₂+1)…

Since 55n³ has 55 divisors, it follows that 55n³ = p₁⁵⁴ or else 55n³ = p₁⁴ p₂¹⁰
No other possibility exists, because 55 can't be expressed as the product of three (or more) integers greater than 1.

The first possibility (55n³ = p₁⁵⁴) doesn't work because we know there are at least two prime factors of 55n³ (5 and 11),
so 55n³ = p₁⁴ p₂¹⁰

Furthermore, p₁ and p₂ are 5 and 11 (or 11 and 5), so we can conclude that n³ = p₁³ p₂⁹, so n = p₁ p₂³

So 7n⁷ = p₁⁷ p₂²¹ 7¹, giving us e₁=7, e₂=21, and e₃=1
Thus the number of factors of 7n⁷ is (e₁+1)(e₂+1)(e₃+1) = 8*22*2 = 352