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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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13 Nov 2010, 09:55
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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
[Reveal] Spoiler: OA

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13 Nov 2010, 10:03
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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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13 Nov 2010, 11:40
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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13 Nov 2010, 11:47
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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13 Nov 2010, 12:05
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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13 Nov 2010, 12:10
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Expert's post
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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17 Jan 2011, 09:01
what A silly mistake.
I have solved it ,but in last I assume that I have to find the greatest n.m. of prime factors.not the n.m. itself

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Re: what is the greatest prime factor? [#permalink]

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05 Jun 2012, 14:22
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Expert's post
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? [#permalink]

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05 Jun 2012, 14:24
1
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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? [#permalink]

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05 Jun 2012, 17:58
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?

Answered this in 40 seconds (is that already considered a short cut? hehe)

= 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)

Now here comes the intuitive part
2^28 = greatest prime factor is 2
63 = 9 * 7 = 3 * 3 * 7 (prime factorization) = greatest prime factor is 7

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Re: what is the greatest prime factor? [#permalink]

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05 Jun 2012, 23:41
1
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mikemcgarry wrote:
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

Wow. I am floored by how great of an explanation you provided. Posts like that make me really think that doing thousands of practice problems with good explanations beats out reading books on math every day of the week.

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What is the greatest prime factor of [#permalink]

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30 Aug 2012, 10:46
1
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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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01 Oct 2012, 05:03
4^17 – 2^28
4^17 can be written as (2^2)^17 = 234
Therefore we get 2^34 - 2^28
Taking 22^8 as common we get 22^8 (26 – 1)

2^28 x 63

2^28 has only one prime factor ie. 2
63 has 3 x 3 x 7 as its prime factors …

Therefore all the factors of this product would be 2^28 x 3 x 3 x 7 ..

Thus 7 is the greatest factor (D)
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Re: What is the greatest prime factor of 4^17 - 2^28 [#permalink]

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12 Nov 2012, 07:30
1
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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
Hence D
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Re: What is the greatest prime factor of [#permalink]

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12 Nov 2012, 11:04
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 [#permalink]

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12 Nov 2012, 11:10
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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12 Nov 2012, 16:37
Great explanation Mod

Those questions seems simple when you master the concepts but indeed are really tough

Quote:
Wow. I am floored by how great of an explanation you provided. Posts like that make me really think that doing thousands of practice problems with good explanations beats out reading books on math every day of the week.

when you master the concept and you know them cold..........In my opinion the only way is to practice questions from all level to see different things from odds angles

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

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26 Feb 2014, 00:39
4^17 - 2^28 =
4^17 - 4^14 =
4^14 . (4^3 - 1) =

4^14 . 63 = 4^14. 9. 7

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

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01 Jul 2014, 05:40
4 is not prime, so break the 4's down into 2's:
4^17 = (2^2)^17 = 2^34
so we have
2^34 - 2^28
at this point, make a common exponent, so that we can factor out the largest possible common factor.
(2^28)(2^6) - (2^28)
(2^28)(2^6 - 1)
(2^28)(63)
finish breaking into primes:
(2^28)(3)(3)(7)
so the 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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14 Mar 2016, 07:21
The Key to all of these questions is => factorise and simplify
here taking 2^28 common and 2 is a prime factor
so we need to only factorise 63
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Re: What is the greatest prime factor of 4^17 - 2^28?   [#permalink] 14 Mar 2016, 07:21

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