If the units digit of a positive integer n is 2 and there are 4 factor

[quote="Bunuel"]If the units digit of a positive integer n is 2 and there are 4 factors of n that are greater than 1 and less than the square root of n, what is the total number of factors of n?

A. 8
B. 9
C. 10
D. 11
E. Cannot be determined


1. It contains a factor 2^k (but that is irrelevant), even number

There are two cases:

1. If it's a perfect square it will contain ( P/2 - 1 ) factors that are less than the square root of N and ( p/2 + 1 ) factors more than square root
imagine N = X^12 then 1, X, X^2 .... X^5 factors less than X^6
in that case if we know number of factors above 1 and below square root = 4, therefore there are a total of 1 + 4 + (4 + 1) factors

2. If it's anything other than a perfect square , imagine 2^6*3^3
It will contain (6 + 1)(3 + 1) = 28 total factors , such as
1, 2, 3, 6, 12, 16, 24, ... 2^6 * 3^4
These will be equally divided between the square root of N , that is 14 factors below \sqrt{N}, and 14 above it.
We know there are 4 + 1 ( including 1) factors below \sqrt{N}, and then 5 will be above \sqrt{N}

So in both cases 10 total factors will be present irrespective of whatever the number is