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.

Using your method, which probably isn't the best:
$2079000 =2^3*3^3*5^3*7*11$ so all factors are of form $2^a3^b*5^c*7^d*11^e$ where $a,b,c$ are between $0$ and $3$ and $d,e$ are between $0$ and $1$ exclusively.
So there $4*4*4*2*2$ factors. 
But to be even the $a$ must be at least $1$ and divisible by $15$ they must be divisible by $3$ (why did you overlook $3$????) and $5$ so $b$ and $c$ must be at least $1$.  So There are only $3$ options for $a,b,c$; $1$ to $3$ inclusively.
So there are $3*3*3*2*2$ such factors.
But perhaps a better way to do it is  $30*k$ is a factor of $207900$ if and only if $k$ is a factor of $\frac {207900}{30}=69300 = 2^2*3^2*5^2*7*11$ so there are $3*3*3*2*2$ such factors.
