The no. can be written as 4^17 - 4^14 = 4^14 (4^3 -1) = 4^14(63) = 4^14(3^2 x 7)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
Hence [spoiler]the ans would be D. (7)[/spoiler]
What is the greatest prime factor of 4^17 - 2^28?
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harsh.champ
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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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Osirus@VeritasPrep
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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.
(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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harsh.champ
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Hey abhi,abhi332 wrote:4(17-14) = 4*3=12
factors are 2, 2 , 3
Ans is B
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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ajith
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4^17 - 2^28imane81 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
= 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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thephoenix
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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
-------->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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beat_gmat_09
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Its 4^17 and not 2^17.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)?
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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Thank you!beat_gmat_09 wrote:Its 4^17 and not 2^17.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)?
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)
