n is a two digit integer

[AJWILL]

if n is a two digit integer and [(.3)(.009)(.000243)/(.81] *10^n] is an integer. How many values of n are possible ?

[A] 89
90
[C] 91
[D] 92
[E] 93

how to go about it?

OA B

[eagleeye]

if n is a two digit integer and [(.3)(.009)(.000243)/(.81] *10^n] is an integer. How many values of n are possible ?

[A] 89
90
[C] 91
[D] 92
[E] 93

how to go about it?
OA B

I am not sure about the brackets, so I am assuming the expression is:
[(.3)(.009)(.000243)/(.81] * (10^n)

First of all simplify:
[(.3)(.009)(.000243)/(.81] = (3*(10^-1))*(9*10^-3)*(243*10^-5)/(81*10^-2)
= (3*9*243/81)*(10^(-1-3-5+2) = (3*9*3)*10^(-7)= 3^4*10^(-7)

Now the complete expression becomes 3^4*(10^(n-7)). Since 3 and 10 have no factors in common, and this expression is an integer, we need that n-7>=0
=> n >= 7.

Now n is a two digit number. Therefore 10 <= n <=99
Hence range of values of n = 99-10+1 = 89+1 = 90.

B is correct.