Divisors of $75600$ of the type of $4n+2$ 
Find the total no. of divisors of $75600$ which of the type of $4n+2$
  where $n\in \mathbb{N}$ and 
  $75600=2^4 \cdot 3^3 \cdot 5^2 \cdot 7^1$

Now I think divisors of type $(4n+2)$ should be of type 
$2^1\cdot 3^a\cdot 5^b\cdot 7^c$. Am I correct?
 A: I think you're correct and that's the complete solution. (4n+2)=2(2n+1).
So, the prime factorization of every divisor can have only $2^1$ and not more. Apart from that, the combination of any power or 3,5,7 would be an odd number, and any odd number can be expressed as (2n+1). 
Possible powers of 3 can be from 0 to 3 = 4.
Possible powers of 5 can be from 0 to 2 = 3.
Possible powers of 7 can be from 0 to 1 = 2.
So, the total possibilities = (4).(3).(2) = 24.
Thus, there are 24 such divisors.
Hope it helps :)
A: First of all, you are right.
Therefore, you need to count the number of divisors of $3^3\cdot5^2\cdot7^1$:


*

*In each divisor, the factor $3$ can appear between $0$ and $3$ times, i.e., $4$ different combinations

*In each divisor, the factor $5$ can appear between $0$ and $2$ times, i.e., $3$ different combinations

*In each divisor, the factor $7$ can appear between $0$ and $1$ times, i.e., $2$ different combinations


Hence there are $4\cdot3\cdot2=24$ divisors:


*

*$3^0\cdot5^0\cdot7^0$

*$3^1\cdot5^1\cdot7^1$

*$3^2\cdot5^2\cdot7^0$

*$3^3\cdot5^0\cdot7^1$

*$3^0\cdot5^1\cdot7^0$

*$3^1\cdot5^2\cdot7^1$

*$3^2\cdot5^0\cdot7^0$

*$3^3\cdot5^1\cdot7^1$

*$3^0\cdot5^2\cdot7^0$

*$3^1\cdot5^0\cdot7^1$

*$3^2\cdot5^1\cdot7^0$

*$3^3\cdot5^2\cdot7^1$

*$3^0\cdot5^0\cdot7^0$

*$3^1\cdot5^1\cdot7^1$

*$3^2\cdot5^2\cdot7^0$

*$3^3\cdot5^0\cdot7^1$

*$3^0\cdot5^1\cdot7^0$

*$3^1\cdot5^2\cdot7^1$

*$3^2\cdot5^0\cdot7^0$

*$3^3\cdot5^1\cdot7^1$

*$3^0\cdot5^2\cdot7^0$

*$3^1\cdot5^0\cdot7^1$

*$3^2\cdot5^1\cdot7^0$

*$3^3\cdot5^2\cdot7^1$


Each of those divisors should be multiplied by $2^1$ of course.
