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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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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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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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Each moment of time ought to be put to proper use, either in business, in improving the mind, in the innocent and necessary relaxations and entertainments of life, or in the care of the moral and religious part of our nature.
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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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
Answer: (D)
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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.
Re: What is the greatest prime factor of 4^17 - 2^28 [#permalink]
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12 Nov 2012, 08:30
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 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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12 Nov 2012, 17:37
Great explanation Mod
Those questions seems simple when you master the concepts but indeed are really tough
@ vandygrad11
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
Re: What is the greatest prime factor of 4^17 - 2^28? [#permalink]
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01 Jul 2014, 06:40
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, 08:21
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 hence 7 is the answer
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66% (00:35) correct 
