[net.math] Factorialize...or Fail!

poppers@aecom.UUCP (Michael Poppers) (03/21/84)

	Factorials are taking over the Earth! Swarms of...whoops - sorry,
wrong commercial.
	Anyway, wouldst thou like to conquer some factorials today?

	1) Find all solutions, in integers, to the equation
			------------
			'        2 '
			'  n! = k  '
			'----------'

	2) Do likewise for the equation

		     m
		   -----
		   \
		    \           2
		    /     n! = k
		   /
		   -----
		    n=1


	******************
	# PERITUS CLAVIS #                      Michael Poppers
	# MACHINAE VIVET #                   ** poppers @ AECOM **
	******************

poppers@aecom.UUCP (Michael Poppers) (03/23/84)

	Here I am, catching up on net.math from the last few weeks, and what
do I see - the same problem posted by a fellow AECOMer?! I'll just have to
come up with a solution myself (once I've decrypted the "rota 13" answer and see
a different path to the solution!) - but don't hold your breath.

	****************
	PERITUS CLAVIS					Michael Poppers
	MACHINAE VIVIT					** The Yek-Yak **
	****************
>

nike@harvard.UUCP (Nike Horton) (04/01/84)

Gur Ceboyrz:  Fubj gung 1!+2!+...+a! != x^2 sbe nyy a naq x.

Gur Nafjre:

Fvapr a! zbq 5 = 0 sbe a>=5, 1!+2!+...a! zbq 5 = 1!+2!+3!+4! zbq 5 = 3 sbe a>=4
Shegurezber, fvapr  0^2 zbq 5= 0, 1^2 zbq 5 = 1, 2^2 zbq 5 = 4,
3^2 zbq 5 = 4, naq 4^2 zbq 5 = 1, 1!+2!+...a! pnaabg or n cresrpg fdhner
vs a>=4.
Sbe a<4, vg vf eryngviryl pyrne gung bhg bs 1,1+2,1+2+6, naq 1+2+6+24=32, bayl
1 naq 1+2+6 ner cresrpg fdhnerf.

Guhf gur *bayl* fbyhgvbaf ner 1!=1^2 naq 1!+2!+3!=3^2.
_______
                        Tertbel Xhcreoret
                           Ohearq Bhg
"Rherxn!" -Nepuvzrqrf






POSTED BY **************************************
Nicholas Horton		genrad!wjh12!harvard!nike
Harvard Arts and Sciences Computer Services

poppers@aecom.UUCP (04/02/84)

	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.

	==================
	| PERITUS CLAVIS |			Michael Poppers
	| MACHINAE VIVIT |		** poppers @ aecom **
	================== 
>