A Factorial Triplet Puzzle (Posted on 2007-01-31)

	Submitted by K Sengupta
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Let S_n = 1! + 2! + 3!+ ….+ n! When k=2, we claim that the equation S_n = m^2 has precisely two solutions n=m=1 and n=m=3. Noting that d! = 0(Mod 10) for all d greater than or equal to 5. and: S_4 = 1+2+6+24 = 33 = 3 (mod 10). However we know that the last digit of a perfect square can never be equal to 3, and accordingly, no solution to the given equation is possible for n>= 4. By Checking for the cases n =1,2,3 we observe that there are precisely two solutions given above, which are, n=m=1 and n=m=3. We will show that n=m=1 is the only solution whenever k is greater than or equal to 3. If n is greater than or equal to 2, then clearly, S_n = 0 (Mod 3). But, m^k = 0 (Mod 3) implies that m = 0(Mod 3, and accordingly, m^k = 0 (mod 27) whenever k is greater than or equal to 3. Now, d! = 0 (Mod 27); but: S_8 = 1+2+6+24+120+720+5040+40320 = 46233, which is not congruent to 0(Mod 27). Since d! is divisible by 27 for all d greater than or equal to 8, it follows that: No solution to the given equation is possible whenever n is greater than or equal to 8. This is succinctly borne out by checking for the values n = 2,3,4,5,6,7 yielding : S_n = 3, 9, 33, 153, 873 and 5913 none of which correspond to a perfect kth power of a positive integer for k greater than or equal to 3. Consequently, n=m=3; k=2 correspond to the only solution to the equation under reference

	Subject	Author	Date
	re(2): some thoughts -- n from 1 to n	Gamer	2007-02-02 01:09:20
	re: some thoughts	Dej Mar	2007-02-02 00:24:59
	Restrictions on m	Gamer	2007-02-01 23:46:35
	Some ideas	Gamer	2007-02-01 23:25:38
	some thoughts	Vishal Gupta	2007-02-01 15:39:07