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a)if 32400 = 1* 32400 = 2*16200 = 3*10800 = ....................(n times)

what is the max value of n
6.

Since 1, 2, 3, 4, 5, 6 are all factors of 32400. But 7 is not. Which breaks the streak.
what if 7 is skipped and from 8 its considered?

what if it was .....what it is the total number of ways it can be split as a multiplication of two numbers
oh Sorry. I misinterpreted the question. In that case, I think it should be 38.

Because prime factorization of 32400 = $2^43^45^2$. From which we can form a total of 5 x 5 x 3 = 75 factors of 32400. Out of these half will be repeated like this: 2*16200, ... 16200*2.

So we get 38.  :)

Am I right??
Here is my approach to the questions you just asked in your comment.

First, find the prime factorization of the number $32400 = 2^4*3^4*5^2$. Now count the number of factors of $32400$ using this factorization. That will be, $5*5*3=75$. This number includes redundant multiplication of type $1*32400 = 32400*1$, $2*16200=16200*2$ etc.. That can be dealt by halving the number of factor - $\left \lceil 75/2 \right \rceil = 38$
so the answer to tha main question  would be 6 or 38??
38. According to your comment "what it is the total number of ways it can be split as a multiplication of two numbers".
okay :)