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From: poppers@aecom.UUCP
Newsgroups: net.math
Subject: Re: Factorialize...or Fail! (spoiler - rot13)
Message-ID: <466@aecom.UUCP>
Date: Mon, 2-Apr-84 10:52:35 EST
Article-I.D.: aecom.466
Posted: Mon Apr  2 10:52:35 1984
Date-Received: Thu, 5-Apr-84 01:24:17 EST
References: <195@harvard.UUCP>
Organization: Albert Einstein Coll. of Med., NY
Lines: 15


	Mr. Horton just posted the correct solution to the problem of
1! + 2! +...+ n! = k**2 in integers n and k. I thought that I might put
the solution's logic another way:

	The sum of the first 4 factorials is 33, and since n! for n>4 is
divisible by 10 (don't ask me why! Look up Nick's article!), higher sums
will all have terminal digit 3. But all squares have terminal digit 
0, 1, 4, 5, 6, or 9.

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	| PERITUS CLAVIS |			Michael Poppers
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