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.

Your factorization of $2079000$ is incorrect.
\begin{align*}
20790000 & = 2079 \cdot 1000\\
         & = 2079 \cdot 10^3\\
         & = 2079 \cdot 2^3 \cdot 5^3\\
         & = 3 \cdot 693 \cdot 2^3 \cdot 5^3\\
         & = 3 \cdot 3 \cdot 231 \cdot 2^3 \cdot 5^3\\
         & = 3 \cdot 3 \cdot 3 \cdot 77 \cdot 2^3 \cdot 5^3\\
         & = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11
\end{align*}
If a divisor of $2079000$ is a multiple of $2$ and $15 = 3 \cdot 5$, it must be a multiple of $2 \cdot 15  = 30$ since $2$ and $15$ are relatively prime.  If a divisor of $2079000$ is a multiple of $30 = 2 \cdot 3 \cdot 5$, then $\frac{1}{30}$ of it must be a factor of
$$\frac{2079000}{30} = \frac{2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11}{2 \cdot 3 \cdot 5} = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11$$
Hence, the number of such divisors is 
$$(2 + 1)(2 + 1)(2 + 1)(1 + 1)(1 + 1) = 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 108$$
