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What is the largest prime factor of the expression 3^8 − 2

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by [email protected] » Thu Oct 25, 2018 6:43 pm
Hi All,

We're asked to find the LARGEST prime factor of the expression 3^8 - 2^12. Certain Quant questions ultimately come down to 'rewriting' information that you've been given. By rewriting 3^8 and 2^12 as "squared" terms, we'll have a 'difference of squares', which is a Classic Quadratic pattern in Algebra.

3^8 = 9^4 = 81^2
2^12 = 4^6 = 64^2

Thus, we have...
81^2 - 64^2 =
(81 - 64)(81 + 64) =
(17)(145) =
(17)(5)(29)

The largest prime factor is 29.

Final Answer: E

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by Brent@GMATPrepNow » Fri Oct 26, 2018 5:34 am
BTGmoderatorDC wrote:What is the largest prime factor of the expression 3^8 − 2^12?

A. 2
B. 3
C. 5
D. 17
E. 29

OA E

Source: Veritas Prep
There's a difference of squares "hiding" in the expression 3^8 − 2^12
That is 3^8 − 2^12 = (3^4)² − (2^6)²

We know that: x² - y² = (x + y)(x - y)

So, we get: (3^4)² − (2^6)² = (3^4 + 2^6)(3^4 - 2^6)
= (81 + 64)(81 - 64)
= (145)(17)
= (5)(29)(17)

The greatest prime factor is 29

Answer: E

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by GMATGuruNY » Fri Oct 26, 2018 6:00 am
A general rule:
Given x^(even power) - y^(even power), where x and y are distinct positive integers, we can always simplify by applying the difference of two squares.
Each time we apply the difference of two squares to the expression in blue, the exponents are reduced by 1/2.
For example:
x� - y� = (x� + y²)(x� - y²) = (x� + y²)(x² + y¹)(x² - y¹)
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by Scott@TargetTestPrep » Sat Nov 03, 2018 4:59 pm
BTGmoderatorDC wrote:What is the largest prime factor of the expression 3^8 − 2^12?

A. 2
B. 3
C. 5
D. 17
E. 29
We see that the expression 3^8 - 2^12 is a difference of squares. Thus, we have:

3^8 - 2^12 = (3^4 + 2^6)(3^4 - 2^6) = (3^4 + 2^6)(3^2 + 2^3)(3^2 - 2^3)

3^4 + 2^6 = 81 + 64 = 145 = 5 x 29

3^2 + 2^3 = 9 + 8 = 17

3^2 - 2^3 = 9 - 8 = 1

So 3^8 - 2^12 = 5 x 29 x 17, and the largest prime factor is 29.

Answer: E

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