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# prime factor problem

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Intern
Joined: 28 Jun 2008
Posts: 45

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09 Jun 2009, 20:46
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Hey guys was wondering if anyone could offer a good way to do this problem:

What is the greatest prime factor of 4*17-2*28? (by * I mean raised to the 17th power and raised to the 28th power.)

A. 2
B. 3
C. 5
D. 7
E. 11

Current Student
Joined: 03 Aug 2006
Posts: 112

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09 Jun 2009, 23:19
1
KUDOS
$$4^{17}-2^{28}$$

$$=2^{2^{17}}-2^{28}$$

$$= 2^{34}-2^{28}$$

$$= 2^{28}\times2^6-2^{28}$$

$$=2^{28}(2^6-1)$$

$$=2^{28}(64-1)$$

$$=2^{28}(63)$$

$$=2^{28}\times 7 \times 3 \times 3$$

Hence the largest prime factor is 7 i.e. answer is D
Intern
Joined: 28 Jun 2008
Posts: 45

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10 Jun 2009, 10:14
Hi and thanks for your help. I understand how you got to 2*28x7x3x3 but I dont quite understand why seven is definitively the largest prime factor. Couldnt 2*28 have a prime factor larger than 7?
Current Student
Joined: 03 Aug 2006
Posts: 112

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10 Jun 2009, 10:24
1
KUDOS
A Prime factor is a number that can be divided evenly only by 1, or itself.

$$2^{28} = 2 \times 2 \times 2 \times 2 \times 2 .....28 times.$$

Hence here the largest prime factor here is just 2 as any other factor such as 8

$$2\times 2\times 2 = 8$$

is divisible by 1,2,4 and 8. Hence it is not prime. The only prime factor in $$2^{28}$$ is 2.

So overall the largest prime factor is 7.
Re: prime factor problem   [#permalink] 10 Jun 2009, 10:24
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