The number of factors of 1800 that are multiple of 6 is:

The number of factors of 1800 that are multiples of 6 is:

Prime factorisation of 1800 = $$2^3\times3^2\times\ 5^2$$

We know that the total number of factors of N = $$p^a\ \times q^b\times\ r^c$$ will be $$(a+1)(b+1)(c+1)$$

And, the number of factors that are multiple of 6 will be

N =$$6\times\ \left(2^2\times3^1\times\ 5^2\right)$$

Total factors will be (2+1)*(1+1)*(2+1) = 18