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

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Source: — Problem Solving |

by harsh.champ » Thu Feb 18, 2010 12:31 pm
imane81 wrote:Please help on this one

What is the greatest prime factor of 4^17 - 2^28?
A. 2
B. 3
C. 5
D. 7
E. 11
The no. can be written as 4^17 - 4^14 = 4^14 (4^3 -1) = 4^14(63) = 4^14(3^2 x 7)

Hence [spoiler]the ans would be D. (7)[/spoiler]
Last edited by harsh.champ on Thu Feb 18, 2010 12:33 pm, edited 1 time in total.
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by abhi332 » Thu Feb 18, 2010 12:32 pm
4(17-14) = 4*3=12

factors are 2, 2 , 3

Ans is B
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by Osirus@VeritasPrep » Thu Feb 18, 2010 12:33 pm
You have to get like bases

(2^2)^17 - 2^28

Factor this out to get

2^28(2^6 - 1)

reduce this to
2^28 (63)

get 63 into prime factors

2^28 (7*3*3)

Greatest prime is 7 like the cheating troll said.
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by harsh.champ » Thu Feb 18, 2010 12:48 pm
abhi332 wrote:4(17-14) = 4*3=12

factors are 2, 2 , 3

Ans is B
Hey abhi,
I think you misunderstood the ques.
I guess you took the exponential sign to be multiplication sign.
Just check it.
Also in the ques. it is written the greatest prime factor not the no. of factors.
Though if we replace the exponential sign by multiplication sign we will get 12 and 3 as the highest prime factor.
The ans. as given by you would also be three but that would be a case of luck as the soln. approach is varying.
Last edited by harsh.champ on Thu Feb 18, 2010 12:51 pm, edited 1 time in total.
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by ajith » Thu Feb 18, 2010 12:50 pm
imane81 wrote:Please help on this one

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(7*3^2)

What is the greatest prime factor of 4^17 - 2^28 is 7
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by abhi332 » Thu Feb 18, 2010 8:58 pm
hey Harsh,

Thanks, I think I was in half sleep while answering this question ;-)
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by thephoenix » Thu Feb 18, 2010 9:10 pm
4^17=2^34
-------->2^34-2^28=2^28(2^6-1)=2^28(63)=2^28*(9*7)

hence prime factors are 2,3,7
and 7 is thehighest
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by fasterpie » Fri Nov 12, 2010 3:35 pm
I have a question about part of this solution. How do you get from

(2^17) - 2^28
to
2^28 (2^6-1)?
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by beat_gmat_09 » Fri Nov 12, 2010 7:01 pm
fasterpie wrote:I have a question about part of this solution. How do you get from

(2^17) - 2^28
to
2^28 (2^6-1)?
Its 4^17 and not 2^17.

4^17 = (2^2)^17 ; by the indices rule multiple 17 by 2 (exponent 2)
17*2 = 34.
4^17 = 2^34


2^34 - 2^28 ; take 2^28 as common factor, 2^28 (2^6 - 1)
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by fasterpie » Fri Nov 12, 2010 11:31 pm
beat_gmat_09 wrote:
fasterpie wrote:I have a question about part of this solution. How do you get from

(2^17) - 2^28
to
2^28 (2^6-1)?
Its 4^17 and not 2^17.

4^17 = (2^2)^17 ; by the indices rule multiple 17 by 2 (exponent 2)
17*2 = 34.
4^17 = 2^34


2^34 - 2^28 ; take 2^28 as common factor, 2^28 (2^6 - 1)
Thank you!
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