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

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SVP
Joined: 03 Feb 2003
Posts: 1608
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Kudos [?]: 76 [0], given: 0

(2P)!=|P+1| simply and elegantly, isn't it [#permalink]  20 Jun 2003, 07:04
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(2P)!=|P+1|

simply and elegantly, isn't it?
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

[#permalink]  20 Jun 2003, 11:19
seems OK, but post your solution, not guessing!
Manager
Joined: 25 May 2003
Posts: 54
Followers: 1

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

[#permalink]  21 Jun 2003, 13:54
no method, just T&E
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

[#permalink]  22 Jun 2003, 23:07
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}
[#permalink] 22 Jun 2003, 23:07
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# (2P)!=|P+1| simply and elegantly, isn't it

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