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# (2P)!=|P+1| simply and elegantly, isn't it

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SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 303 [0], given: 0

(2P)!=|P+1| simply and elegantly, isn't it [#permalink]

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20 Jun 2003, 08:04
00:00

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(2P)!=|P+1|

simply and elegantly, isn't it?

Kudos [?]: 303 [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 303 [0], given: 0

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20 Jun 2003, 12:19
seems OK, but post your solution, not guessing!

Kudos [?]: 303 [0], given: 0

Manager
Joined: 25 May 2003
Posts: 54

Kudos [?]: 3 [0], given: 0

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21 Jun 2003, 14:54
no method, just T&E

Kudos [?]: 3 [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 303 [0], given: 0

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23 Jun 2003, 00:07
Nevertheless, a method does exist:

When you see a factorial, think about positive integers no less than one.

(2P)!=|P+1|

draw two graphs and see intersections. there are two intersections P={0 and 1}

Kudos [?]: 303 [0], given: 0

23 Jun 2003, 00:07
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