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If x, y, and z are consecutive odd integers such that 31

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If x, y, and z are consecutive odd integers such that 31 [#permalink]

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New post 24 Aug 2008, 13:50
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If x, y, and z are consecutive odd integers such that 31 < x < y < z < 101, what is the greatest prime factor of the sum 2^x + 2^y + 2^z ?

A) 2
B) 7
C) 11
D) 37
E) 41
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Re: PS - consecutive exponents [#permalink]

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New post 24 Aug 2008, 14:19
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Hi,

Its not really complex as it looks like. It sounds different and difficult.

2^(anything) will always have one prime factor, i.e. 2. There can be nothing else, eg: 2,4,8,16,32...

now coming to question.
2^x + 2^y + 2^z
= 2^x + 2^(x+2) + 2^(x+4) [They are consecutive odd nos.]
= 2^x * [ 1 + 2^2 + 2^4 ]
= 2^x * [ 1 + 4 + 16 ]
= 2^x * 21

This should have three factors 2 and 3, 7.

So Ans is 7. [Option B]
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Re: PS - consecutive exponents [#permalink]

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New post 24 Aug 2008, 15:34
sarzan wrote:
If x, y, and z are consecutive odd integers such that 31 < x < y < z < 101, what is the greatest prime factor of the sum 2^x + 2^y + 2^z ?

A) 2
B) 7
C) 11
D) 37
E) 41


B

2^x + 2^y + 2^z = 2^x + 2^(x+2) + 2^(x+4) = (2^x)*(1+2^2+2^4)

= (2^x)*(1+4+16) = (2^x)*21 = (2^x)*3*7

The last expression is the prime factorization of 2^x + 2^y + 2^z.

So the greatest prime factor is 7.
Re: PS - consecutive exponents   [#permalink] 24 Aug 2008, 15:34
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If x, y, and z are consecutive odd integers such that 31

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