How many factors?

The prime factorisation of 2079000 is $2^3 \times 3^3 \times 5^3 \times 7 \times 11$ so any divisors of 2079000 have a prime factorisation composing of any combination of the numbers in the prime factorisation of 2079000. We are also given that the divisors have to be an even multiple of 15 which means they are a multiple of 30 and therefore have a 2 x 3 x 5 in their prime factorisation.

Factoring this out of the prime factorisation of 2079000, we get:

2079000= (2 x 3 x 5) $(2^2 \times 3^2 \times 5^2 \times 7 \times 11)$

So we then need to work out how many ways there are of choosing numbers from $(2^2 \times 3^2 \times 5^2 \times 7 \times 11)$ . 2 can be chosen 0, 1 or 2 times because there is a $2^2$ so there are 3 choices for the number of 2s. In the same way there are 3 choices for number of 5s, 3 choices for the number of 3s, 2 choices for the number of 7s and 2 choices for the number of 11s, giving 3 x 3 x 3 x 2 x 2 = 108 combinations so the answer is $\boxed{108}$ .