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What is the greatest prime factor of 4^17 - 2^28?

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What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 13 Nov 2010, 10:55
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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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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 13 Nov 2010, 11:03
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 05 Jun 2012, 15:22
3
Val1986 wrote:
What is the greatest prime factor of 4^17 - 2^28?

A. 2
B. 3
C. 5
D. 7
E. 11

Is there a shortcut to solving this question? :?


I'm happy to help with this. :)

We know 4 = 2^2, so 4^17 = (2^2)^17 = 2^(2*17) = 2^34

That takes advantage of a law of exponents that says (a^n)^m = a^(n*m)

So, 4^17 - 2^28 = 2^34 - 2^28 = 2^(28 + 6) - 2^28 = (2^28)*(2*6) - 2^28 = (2^6 - 1) *(2^28)
= (64 - 1)*(2^28) = 63*(2^28)

The prime factors of 63 are 3*3*7, so the largest prime factor is 7, answer choice D.

Here's a blog you may find helpful.
http://magoosh.com/gmat/2012/gmat-math-factors/

Does all that make sense? Please let me know if you have any further questions.

Mike :)
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 13 Nov 2010, 12:40
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I have a hard time ignoring that 2^28 and trusting that breaking down 63 into its prime factors will give me the right answer.

Is there a hard and fast rule for why you can forget about 2^28 and trust that the 7 derived from 63 is correct?
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 13 Nov 2010, 13:10
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rockzom wrote:
I see. So essentially any time that you have an expression where all of the bases are prime, you can assume that the highest base would be the greatest prime factor?

For example, the expression \(7^7 * 13^3 * 17\) would have 17 as the greatest prime factor. Correct?


How else? Exponentiation does not "produce" primes: if p is a prime number then p^12 or p^10000 will still have only one prime - p.
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 05 Jun 2012, 15:24
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4^17 - 2^28
= 2^34 - 2^28
= 2^28 * (2^6 - 1)
= (2*2*2*...) * (63)
= (2*2*2*...) * (3 * 3 * 7)

Greatest prime factor is 7
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 30 Aug 2012, 11:46
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Let's simplify (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{3x3x7}

clearly 7
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 12 Nov 2012, 08:30
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carcass wrote:
What is the greatest prime factor of \(4^17 - 2^28\)?

(A) 2
(B) 3
(C) 5
(D) 7
(E) 11


4^17 can be written as 2^34.
Hence we have to find out the greatest prime factor of 2^34-2^28.
Take 2^28 as common.
2^28(2^6-1)
It will become, 2^28 * (64-1)
2^28 * 63
Greatest prime factor of 2^28=2
Greatest prime factor of 63 is 7.
2^28 * 7*9
Therefore the answer is 7.
Hence D
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 16 Mar 2017, 07:07
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Aline2289 wrote:
Bunuel wrote:
student26 wrote:
What is the greatest prime factor of 4^17 - 2^28?

A. 2
B. 3
C. 5
D. 7
E. 11


\(4^{17}-2^{28}=2^{34}-2^{28}=2^{28}(2^6-1)=2^{28}*63=2^{28}*3^2*7\) --> greatest prime factor is 7.

Answer: D.



Why is it (2^6-1)? :oops:


Added extra step:

\(4^{17}-2^{28}=2^{34}-2^{28}=2^{28}*2^6-2^{28}=2^{28}(2^6-1)=2^{28}*63=2^{28}*3^2*7\) --> greatest prime factor is 7.

So, we are factoring out \(2^{28}\) from \(2^{34}-2^{28}\).

Hope it's clear.
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 21 Mar 2017, 06:10
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student26 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

Note that the common factor in each term is 2^28; thus, the expression can be simplified as follows:

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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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 13 Nov 2010, 12:47
rockzom wrote:
I have a hard time ignoring that 2^28 and trusting that breaking down 63 into its prime factors will give me the right answer.

Is there a hard and fast rule for why you can forget about 2^28 and trust that the 7 derived from 63 is correct?


\(4^{17}-2^{28}\) equals to \(2^{28}*3^2*7\), which means that the prime factors of this number are 2, 3, and 7, so the greatest prime factor is 7 (2^28=2*2*...*2, so this expression has only one prime: 2).
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 13 Nov 2010, 13:05
I see. So essentially any time that you have an expression where all of the bases are prime, you can assume that the highest base would be the greatest prime factor?

For example, the expression \(7^7 * 13^3 * 17\) would have 17 as the greatest prime factor. Correct?
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 12 Nov 2012, 12:04
thevenus wrote:

=(2)^34-(2)^28
=(2)^28{(2)^6-1}


I don't get the transition here... what's the process involved that these two equal one another??
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 19 Apr 2016, 18:17
Flexxice wrote:
What is the greatest prime factor of 4^17 - 2^28?

A) 2

B) 3

C) 5

D) 7

E) 11



First of all 4^17 = (2^2)^17 = 2^34

So, 4^17 - 2^28 = 2^34 - 2^28
= 2^28(2^6 - 1) [factored out 2^28]
= 2^28(2^3 + 1)(2^3 - 1) [factored again]
= 2^28(8 + 1)(8 - 1) [evaluated]
= 2^28(9)(7) [evaluated]
= 2^28(3)(3)(7)

So, as you can see, the greatest prime factor is 7

Answer = D

Cheers,
Brent
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 19 Apr 2016, 21:21
GMATPrepNow wrote:
Flexxice wrote:
What is the greatest prime factor of 4^17 - 2^28?

A) 2

B) 3

C) 5

D) 7

E) 11



First of all 4^17 = (2^2)^17 = 2^34

So, 4^17 - 2^28 = 2^34 - 2^28
= 2^28(2^6 - 1) [factored out 2^28]
= 2^28(2^3 + 1)(2^3 - 1) [factored again]
= 2^28(8 + 1)(8 - 1) [evaluated]
= 2^28(9)(7) [evaluated]
= 2^28(3)(3)(7)

So, as you can see, the greatest prime factor is 7

Answer = D

Cheers,
Brent



Thank you! I tried it in another way and I do not get why I got another answer than you did.

4^17 - 2^28 = 4^17 - 4^14 = 4^3 = 2^6

Because of that, I chose answer A.
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 19 Apr 2016, 22:04
Flexxice wrote:
GMATPrepNow wrote:
Flexxice wrote:
What is the greatest prime factor of 4^17 - 2^28?

A) 2

B) 3

C) 5

D) 7

E) 11



First of all 4^17 = (2^2)^17 = 2^34

So, 4^17 - 2^28 = 2^34 - 2^28
= 2^28(2^6 - 1) [factored out 2^28]
= 2^28(2^3 + 1)(2^3 - 1) [factored again]
= 2^28(8 + 1)(8 - 1) [evaluated]
= 2^28(9)(7) [evaluated]
= 2^28(3)(3)(7)

So, as you can see, the greatest prime factor is 7

Answer = D

Cheers,
Brent



Thank you! I tried it in another way and I do not get why I got another answer than you did.

4^17 - 2^28 = 4^17 - 4^14 = 4^3 = 2^6

Because of that, I chose answer A.


Hi,
you are wrong in the highlighted portion--
\(\frac{4^{17}}{4^{14}}= 4^{17-14} =4^3\)
\(4^{17} - 4^{14} = 4^{14}(4^3-1) = 4^{14} *(64-1) = 4^{14}*63= 4^{14}*7*9..\)
so 7 is the answer..
You can not subtract the way you have done, it can be done ONLY if two terms are being divided as shown above..

a simpler examplewill be
\(2^3 -2^1\).. as per you it will be\(2^2=4.\).
BUT \(2^3-2^1=8-2=6..\)
hope it helps
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 21 Apr 2016, 01:57
Flexxice wrote:


Thank you! I tried it in another way and I do not get why I got another answer than you did.

4^17 - 2^28 = 4^17 - 4^14 = 4^3 = 2^6

Because of that, I chose answer A.


We cannot simply subtract the terms.

4^17 - 4^14 = 4^14 (4^3 - 1) = 4^14*(64 - 1) = 4^14 * 63
Here we have taken 4^14 common from both the terms and written the remainder inside the brackets.
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 15 Mar 2017, 17:40
Bunuel wrote:
student26 wrote:
What is the greatest prime factor of 4^17 - 2^28?

A. 2
B. 3
C. 5
D. 7
E. 11


\(4^{17}-2^{28}=2^{34}-2^{28}=2^{28}(2^6-1)=2^{28}*63=2^{28}*3^2*7\) --> greatest prime factor is 7.

Answer: D.



Why is it (2^6-1)? :oops:
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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New post 24 Nov 2018, 15:18
A video explanation can be found here:
https://www.youtube.com/watch?v=gCL6DmaIqK4

The first step is noting that we want to work from a common base, i.e. instead of 4 and 2, we want 2^2 and 2. Now we have

(2^2)^17 - 2^28

What’s tested are your knowledge of exponent rules, factoring, and prime numbers. We have

2^34 – 2^28, which is

2^28(2^6 – 1) [note that 2^6 can be viewed more simply as 2^3*2^3, or 8*8], which is

2^28(64 – 1), which is

2^28 (63), which is

2^28 (3^2)(7)

Prime factors are 2, 3 and 7. Greatest prime factor is 7
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Re: What is the greatest prime factor of 4^17 - 2^28?   [#permalink] 24 Nov 2018, 15:18

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