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Notice that (2^4)^2 - 1, can be written as 16^2 - 1^2. This is the difference of squares. So if x^2 - y^2 = (x+y(x-y), then 16^2 - 1^2 = (16 + 1)(16 -1) = 17*15 = 17*5*3. So the largest prime factor is 17. The answer is Dlheiannie07 wrote:$$What is the greatest prime factor of\ \ \left(2^4\right)^2\ -\ 1?$$
(A) 3
(B) 5
(C) 11
(D) 17
(E) 19
How will i find the solution to this?
OA D
Power of a power rule: (b^x)^y = b^(xy)lheiannie07 wrote:$$What is the greatest prime factor of\ \ \left(2^4\right)^2\ -\ 1?$$
(A) 3
(B) 5
(C) 11
(D) 17
(E) 19
How will i find the solution to this?
OA D
Simplifying, we have:BTGmoderatorDC wrote:$$What is the greatest prime factor of\ \ \left(2^4\right)^2\ -\ 1?$$
(A) 3
(B) 5
(C) 11
(D) 17
(E) 19
How will i find the solution to this?
OA D
