Exponentials and prime factors -OG

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Exponentials and prime factors -OG

by Kuros » Wed Apr 19, 2017 4:02 pm
What is the greatest prime factor of 4^17-2^28?

a)2
b)3
c)5
d)7
e)11

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by Brent@GMATPrepNow » Wed Apr 19, 2017 4:12 pm
Kuros wrote:What is the greatest prime factor of 4^17 - 2^28?

a)2
b)3
c)5
d)7
e)11
To answer this we must find the prime factorization of 4^17 - 2^28
To do this, we'll apply to algebraic factoring techniques (and some exponent rules).

Since 4 is not prime, let's first take 4^17 and replace 4 with 2^2

When we do this, we get (2^2)^17 - 2^28
We can now apply the Power of a Power Rule to rewrite this as 2^34 - 2^28
From here, let's factor out 2^28 to get 2^28(2^6 - 1)
2^6 evaluates to be 64, so we get: 2^28(64 - 1)
This equals 2^28(63)
We can find the prime factorization of 63 to write this as (2^28)(3)(3)(7)

So, 4^17 - 2^28 = (2^28)(3)(3)(7), which means the greatest prime factor is 7

Answer: D
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by [email protected] » Wed Apr 19, 2017 8:29 pm
Hi Kuros,

When dealing with 'exponent rule' questions, you often have to make sure that you're comparing "like" base values.

Here, we have 4^17 and 2^28.... so we're starting with DIFFERENT bases. We can 'rewrite' 4 as (2^2) though, which gives us....

(2^2)^17 - 2^28

Now that we're taking "a power to a power", we multiply the 2 and the 17....

2^34 - 2^28

From here, we can factor out a common element: 2^28

(2^28)(2^6 - 1)

2^6 is relatively easy to calculate.... 64

(2^28)(64 - 1)
(2^28)(63)
(2^28)(7)(3)(3)

Thus, the greatest prime factor here is 7

Final Answer: D

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by Jay@ManhattanReview » Wed Apr 19, 2017 9:10 pm
Kuros wrote:What is the greatest prime factor of 4^17-2^28?

a)2
b)3
c)5
d)7
e)11
Hi Kuros,

4^17 - 2^28 = (2^2)^17 - 2^28

=> 2^34 - 2^28

=> 2^28(2^6 - 1)

=> 2^28(64 -1 )

=> 2^28(63)

=> (2^28)*(3^2)*(7)

We see that there are three prime factors in (2^28)*(3^2)*(7); these are 2, 3, and 7. The greatest among them is 7.

The correct answer: D

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

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by Jeff@TargetTestPrep » Fri Apr 28, 2017 2:04 pm
Kuros wrote:What is the greatest prime factor of 4^17-2^28?

a)2
b)3
c)5
d)7
e)11
We need to determine the greatest prime factor of 4^17 - 2^28. We can start by breaking 4^17 into prime factors.

4^17 = (2^2)^17 = 2^34

Now our equation is as follows:

2^34 - 2^28

We see that 2^28 is a common factor in both terms, so we factor it out:

2^28(2^6 - 1)

2^28(64 - 1)

2^28(63)

2^28 x 9 x 7

2^28 x 3^2 x 7

We see that the greatest prime factor must be 7.

Answer: D

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