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No. Properties

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Senior Manager
Senior Manager
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Joined: 05 Oct 2008
Posts: 270

Kudos [?]: 515 [0], given: 22

No. Properties [#permalink]

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New post 27 Jan 2009, 03:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Attachment:
PS_NoProp_E.JPG
PS_NoProp_E.JPG [ 50.01 KiB | Viewed 796 times ]

Kudos [?]: 515 [0], given: 22

Manager
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Joined: 28 Apr 2008
Posts: 110

Kudos [?]: 12 [0], given: 0

Re: No. Properties [#permalink]

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New post 27 Jan 2009, 05:26
E

h(100)+1= 2^50*(1*2*3...50)+1, so all integers between 1 and 50 are factors of [2^50*(1*2*3...*50)], in which case they can not be factors of [2^50*(1*2*3...*50)+1].

thus, the smallest prime factor must be above 50--> E

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Intern
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Joined: 19 Jan 2009
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Re: No. Properties [#permalink]

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New post 27 Jan 2009, 08:08
Given h(100) = 2 x 4 x 6 x 8 ..........100
= 2(1) x 2(2) x 2 (3) x 2 (4) ........2(50)
= 2^50 x 1 x2 x 3 x4 ........50
Which implies h(100) is divisible by all numbers till 50

h(100) + 1 cannot be divisible by numbers till 50 ..as they would leave a remainder 1.
so the smallest prime factor of h(100) + 1 has to be greater than 50 .

Answer E.

www.graduatetutor.com

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Kudos [?]: 832 [0], given: 19

Re: No. Properties [#permalink]

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New post 27 Jan 2009, 17:58
study wrote:
Attachment:
PS_NoProp_E.JPG


Detail work-out: 7-p556064?t=74417&hilit=smallest+prime#p556064
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Kudos [?]: 832 [0], given: 19

Re: No. Properties   [#permalink] 27 Jan 2009, 17:58
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