Source: *BrainDen.com*, 9/7/2011

**Hint:** Try some valuies of n...

For n = 2, [(2^{2}-1)(2^{2}-2^{1}] ÷ 2! = ?

Now try n = 3, 4, 5...Is there a pattern?

If not, what is another way to attack the problem?

**Solution Commentary:** This problem is troublesome, as a pattern of "yes" is initially established for n = 1, 2, 3, ....10. Then, trouble arises, as n = 11 leads to a non-integer.

Why?

When the problem was posed, user "thoughtfulfellow" responded with a good analysis: "Reversed order of numerator and factored out 2's to get:

2^{(n-1)}*(2-1)*2^{(n-2)}*(2^{2}-1)....2^{2}*(2^{(n-2)}-1)*2*(2^{(n-1)}-1)*(2^{n}-1).

=2^{(n2-n)}*1*3*7*15*31*63*127*255*511*1023*2047...(2^{n}-1)

Factoring out the 2's from the denominator, we will have a maximum of 2^{(n-1)} so we only need worry about the odd factors in n!.

n! with 2's factored out =1*3*5*3*7*9*5*11 ....

If n=11, then 11 must be a factor of the numerator for the results to be an integer but this is not the case so this does not always result in an integer."

Nice thoughts, "Thoughtfulfellow"!