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What is the greatest prime factor

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by DavidG@VeritasPrep » Mon Oct 30, 2017 6:58 am
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
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 D
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by Brent@GMATPrepNow » Mon Oct 30, 2017 8:43 am
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
Power of a power rule: (b^x)^y = b^(xy)

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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by Scott@TargetTestPrep » Thu Nov 07, 2019 6:38 pm
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
Simplifying, we have:

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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