If 4^(n + 4) - 4^(n + 2) = 960, What is the value of n?

Question

Find the largest possible value of positive integer N, such that N! can be expressed as the product of (N-4) consecutive positive integers?

Answer 1

For this equation to be true 4^(n+4) should be just greater than 960 or in other words should be as close as possible to 960 and exceeding it. It's known that 4^(5) = 1024 therefore corresponding to this
n = 1 which means the given equation becomes
4^(1+4) - 4^(1+2) = 1024 - 64 = 960.

Answer 2

4 ^(n + 4) = 4^(n+2+2) = 4^2 {4^ (n+2)}
so
4^(n+4) - 4^(n+2) = 4^2 {4^ (n+2)} - 4^ (n+2) = 960........................i
facting out 4^ (n+2) eqn i becomes

4^ (n+2) [(4^2)-1] = 4^ (n+2)[16-1] = 4^ (n+2) [15] = 960................ii

dividing by 15 both sides of eqn ii

4^ (n+2) = 64 but 64 = 4^3, so
4^ (n+2) = 4^3, which makes
n+2 = 3, giving
n = 1