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|A Factorial Triplet Puzzle (Posted on 2007-01-31)
Determine all possible triplets of integers (n,m,k) satisfying 1!+2!+3!+...+n!=mk, where n, m and k are greater than 1.
Submitted by K Sengupta
Rating: 4.0000 (1 votes)
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
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