I'm going to answer this question using logic. Let x be Jenny's favorite number. We are going to square root this number,
and then multiply it by
.
This product needs to be an integer. How is that obtainable? To be an integer, we need to get rid of the nasty
part. The only way I can think of to get rid of it, is to multiply it by
, because
. Thus we have some conditions we need to fulfill when choosing Jenny's favorite number.
When we take the square root of Jenny's favorite number, x, it must contain a perfect square and a 2 in its prime factorization. For example, 8 works because 8 = 2 x 2 x 2, or 2² x 2.
You notice 8 is made up of a perfect square multiplied by 2. So when we take the square root of 8, we get:

So 8 is the same thing as 
So when we multiply this by
, we will get an integer! So as long as Jenny's favorite number consists of a perfect square and two in its prime factorization, we will have an integer!
So possible choices are: 2,8,18.
Why does 18 work? Because 
When we multiply this by
, we get 6, which is an integer.
b) Suppose instead of multiplying by
, we divided by
. Is the resulting quotient still an integer?
YES, because we can get rid of the
part by dividing by
as well. This leaves only the "perfect square" part left in our square root, and obviously a perfect square is an integer when we square root it.
I hope that made sense! (⌐■_■)