<s>[INST] 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?
 [/INST] 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.
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