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

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(2P)!=|P+1| simply and elegantly, isn't it [#permalink] New post 20 Jun 2003, 08:04
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(2P)!=|P+1|

simply and elegantly, isn't it?
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 [#permalink] New post 20 Jun 2003, 12:19
seems OK, but post your solution, not guessing!
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 [#permalink] New post 21 Jun 2003, 14:54
no method, just T&E 8-)
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 [#permalink] New post 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}
  [#permalink] 23 Jun 2003, 00:07
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(2P)!=|P+1| simply and elegantly, isn't it

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