The three biggest divisors are N divided by the three smallest divisors. Let's suppose the three smallest divisors are d₁, d₂ and d₃ (d₁ < d₂ < d₃), then the following inequality holds...

N / d₁ + N / d₂ + N / d₃ > N

N * (1 / d₁ + 1 / d₂ + 1 / d₃) > N

1 / d₁ + 1 / d₂ + 1 / d₃ > 1

Let's suppose that d₁ > 2, then the maximum value that the summand can take is 1 / 3 + 1 / 4 + 1 / 5 = 47 / 60, which is less than 1. So now we know that d₁ is 2. Let's see what happens when d₂ > 3. The maximum value the summand can take is 1 / 2 + 1 / 4 + 1 / 5 = 19 / 20. So we know as well that d₂ is 3. Knowing the values of d₁ and d₂, let's find out the possible values of d₃...

1 / 2 + 1 / 3 + 1 / d₃ > 1

1 / d₃ > 1 - 1 / 2 - 1 / 3

1 / d₃ > 1 / 6

d₃ < 6

Finally, d₃ can be either 4 or 5. So N is of the form 2² * 3 * k or 2 * 3 * 5 * k. In either case, N is a multiple of 6. So the correct answer is C.