<s>[INST] Number of divisors of the number $2079000$ which are even and divisible by $15$ 
Find the number of divisors of $2079000$ which are even and divisible by $15$?

My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11$
Therefore, the number of divisors should be 
$2 \cdot 2 \cdot (3+1) \cdot (1+1) \cdot (1+1)$
But however this answer is wrong.
Any help would be appreciated.
 [/INST] To count the number of factors of $2079000 =2^3\cdot3^3\cdot5^3\cdot7\cdot11$, we would compute
$$
(3+1)(3+1)(3+1)(1+1)(1+1)=256
$$
However, we only want to count the number of factors divisible by $30=2\cdot3\cdot5$. Therefore, we want to compute the number of factors of $\frac{2079000}{30}=2^2\cdot3^2\cdot5^2\cdot7\cdot11$, because each of these factors times $30$ is a factor of $2079000$ which is divisible by $30$, and this would be
$$
(2+1)(2+1)(2+1)(1+1)(1+1)=108
$$
I am not sure exactly why your computation of the number of factors is wrong because you did not explain why you computed
$$
2\cdot2\cdot(3+1)\cdot(1+1)\cdot(1+1)
$$
instead.
</s>