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.
 A: Any factors that are even and divisible by $15$ are divisible by $30$.
Effectively you need to find the number of factors of $2079000 /30 = 69300 = 2^2 \cdot 3^2\cdot 5^2\cdot 7\cdot 11$
A: 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 
$$(2 + 1)(2 + 1)(2 + 1)(1 + 1)(1 + 1) = 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 108$$
A: 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.
A: 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.
A: You need to have it divisible by $3$ as well to make it divisible by $15$.  
A: $$2079000 = 2^3*3^3*5^3*7*11$$ 
You want to have at least one of each numbers $2$,$3$,and $5$
Thus  you should have 
$$3*3*3*(1+1)*(1+1)=108$$
A: You should simplify a 2, a 3, and a 5. This is equivalent to find the number of positive divisors of
$$
\frac{2079000}{30} = 69300 = 2^2 \cdot 3^2\cdot 5^2\cdot 7\cdot 11
$$
which is 
$
3^3 \cdot 2^2.
$
A: They have to be divisible by $15$, $2$, and $3$, or divisible by $30$, but from your method, the divisors can be divisible by $10$. What I mean mathematically is:
$$30(2^2\cdot3^3\cdot5^2\cdot7\cdot11)\ne 2079000$$
Any even number must be divisible by $2$, and any number divisible by $15$ and $2$ must be divisible by $30$; as $15$ and $2$ are relatively prime numbers.
Can you find the amount of numbers which are divisible by $30$ and divide into $2079000$?
Hint: You have to perform an operation with $2079000$ and $30$.
