What is the greatest prime factor of 3^6 - 1?

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What is the greatest prime factor of 3^6 - 1?

by swerve » Mon Dec 11, 2017 6:43 am
What is the greatest prime factor of 3^6 - 1 ?

A. 2
B. 3
C. 7
D. 13
E. 17

The OA is D.

Please, can any expert explain this PS question for me? I don't understand why that is the correct answer. Thanks.

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by Jay@ManhattanReview » Mon Dec 11, 2017 7:38 am
swerve wrote:What is the greatest prime factor of 3^6 - 1 ?

A. 2
B. 3
C. 7
D. 13
E. 17

The OA is D.

Please, can any expert explain this PS question for me? I don't understand why that is the correct answer. Thanks.
3^6 - 1 = 3^6 - 1^6 = (3^3)^2 - (1^3)^2 = (3^3 - 1^3) * (3^3 + 1^3); applying a^2 - b^2 = (a - b)(a + b)

(3^3 - 1^3) * (3^3 + 1^3) = [(3 - 1)(3^2 + 3*1 + 1^2)]*[(3 + 1)(3^2 - 3*1 + 1^2)] = 2*(9 + 3 =1)*4*(9 - 3 +1) = 2*13*4*7

We see that the greatest prime factor is 13.

The correct answer: D

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Mon Dec 11, 2017 7:38 am
swerve wrote:What is the greatest prime factor of 3^6 - 1 ?

A. 2
B. 3
C. 7
D. 13
E. 17
3^6 - 1 is a difference of squares, since 3^6 = (3³)² and 1 = 1²

So, we get: 3^6 - 1 = (3³ + 1)(3³ - 1)
= (27 + 1)(27 - 1)
= (28)(26)
= (2)(2)(7)(2)(13)

The prime factors are 2, 7 and 13
The greatest value is 13

Answer: D

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by Jay@ManhattanReview » Mon Dec 11, 2017 7:42 am
swerve wrote:What is the greatest prime factor of 3^6 - 1 ?

A. 2
B. 3
C. 7
D. 13
E. 17

The OA is D.

Please, can any expert explain this PS question for me? I don't understand why that is the correct answer. Thanks.
You may use brute force to get this one sorted.

We have 3^6 - 1 = 729 - 1 = 728

Since we want the greatest prime factor and the five options are arranged in an ascending order, start dividing 728 one by one starting from option E onwards in the reverse order.

We see that 728 is not divisible by 17 but is divisible by 13, so 13 is the correct answer.

The correct answer: D

Hope this helps!

-Jay
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by ErikaPrepScholar » Mon Dec 11, 2017 8:01 am
3^6 is a little too large of an exponent to deal with in our heads, so we'll try to factor this out. We'll start by expressing $$3^6-1$$ as $$ \left(3^3-1\right)\left(3^3+1\right)$$ 3^3 is 27, so this gives $$\left(27-1\right)\left(27+1\right)=26\cdot28$$
Then we can do the prime factorization of each number:
$$26\cdot28=\left(2\cdot13\right)\left(2\cdot2\cdot7\right)$$
So the prime factors are 2, 7, and 13, giving a greatest prime factor of 13.
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by Scott@TargetTestPrep » Wed Jan 30, 2019 5:58 pm
swerve wrote:What is the greatest prime factor of 3^6 - 1 ?

A. 2
B. 3
C. 7
D. 13
E. 17
We can factor 3^6 - 1 as a difference of squares. Factoring gives us:

3^6 - 1 = (3^3 - 1)(3^3 + 1) = 26 x 28 = 13 x 2 x 7 x 4, so the greatest prime factor is 13.

Answer: D

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by [email protected] » Wed Jan 30, 2019 7:09 pm
Hi All,

A number of the posts have provided elegant solutions to this question. The basic math behind this prompt is Arithmetic and Prime Factorization though, so if you don't immediately "see" the elegant approach, you can still get to the answer....

We're asked to find the LARGEST prime factor of 3^6 - 1

3^6 = 9^3 = (9)(9)(9) = 729

729 - 1 = 728

Now, we can prime factor 728.

You probably immediately see that 728 is divisible by 2, but if you know your 'rules of division', you can see that it's also divisible by 4...

728 =
(4)(182)
(4)(2)(91)
(4)(2)(7)(13)

13 is the largest prime.

Sometimes this type of approach isn't practical (especially if the numbers involved are HUGE), but when you're given 'manageable' numbers, there's nothing wrong with admitting that you don't see the 'hidden pattern.' If you can get to the correct answer in a reasonable amount of time by just doing arithmetic, then it's better to do THAT than waste time staring at the screen.

Final Answer: D

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