$$What is the greatest prime factor of\ \ \left(2^4\right)^2\ -\ 1?$$
(A) 3
(B) 5
(C) 11
(D) 17
(E) 19
How will i find the solution to this?
OA D
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What is the greatest prime factor
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BTGmoderatorDC
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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
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Brent@GMATPrepNow
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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
This means that (2�)² = 2^8
So, (2�)² - 1 = 2^8 - 1
= (2� + 1)(2� - 1)
= (2� + 1)(2² + 1)(2² - 1)
= (16 + 1)(4 + 1)(4 - 1)
= (17)(5)(3)
As we can see, 17 is the greatest prime factor.
Answer: D
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Scott@TargetTestPrep
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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
2^8 - 1
(2^4 - 1)(2^4 + 1)
(15)(17)
(3)(5)(17)
The largest prime factor is 17.
Answer: D
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