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

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Question Stats: 70% (01:06) correct 30% (01:12) wrong based on 1301 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58973
Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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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

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

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

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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.

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

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1
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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1
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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GMAT 1: 770 Q50 V46 Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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1
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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1
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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1
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
Hence D
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Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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1
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.

Why is it (2^6-1)? $$4^{17}-2^{28}=2^{34}-2^{28}=2^{28}*2^6-2^{28}=2^{28}(2^6-1)=2^{28}*63=2^{28}*3^2*7$$ --> 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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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.

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

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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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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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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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dbiersdo wrote:
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??

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

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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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

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

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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

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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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

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--
$$\frac{4^{17}}{4^{14}}= 4^{17-14} =4^3$$
$$4^{17} - 4^{14} = 4^{14}(4^3-1) = 4^{14} *(64-1) = 4^{14}*63= 4^{14}*7*9..$$
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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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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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.

Why is it (2^6-1)? Intern  Joined: 20 Aug 2018
Posts: 25
Re: What is the greatest prime factor of 4^17 - 2^28?  [#permalink]

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A video explanation can be found here:

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
_________________ Re: What is the greatest prime factor of 4^17 - 2^28?   [#permalink] 24 Nov 2018, 15:18

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