All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
A Factorial Triplet Puzzle (Posted on 2007-01-31) Difficulty: 3 of 5
Determine all possible triplets of integers (n,m,k) satisfying 1!+2!+3!+...+n!=mk, where n, m and k are greater than 1.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some ideas | Comment 2 of 5 |

In order to be a perfect power, the expression must be 0 mod m. Factorials from m on will be 0 mod m, so 1!...(m-1)! must be 0 mod m.

Also, the factorial expression can be written as 1*(2*(3*...*(n+1)...+1)+1) which equals 1!+2!+3!+...n!

The inside equals a perfect square, thus = 1!+2!...+(n-3)!+nē(n-2)!

Edited on February 1, 2007, 11:27 pm
  Posted by Gamer on 2007-02-01 23:25:38

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information