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

Home > Just Math
Dark Divisibility (Posted on 2013-04-27) Difficulty: 3 of 5
The natural numbers a,b,c,d are such that their least common multiple equals a+b+c+d. Prove that abcd is divisible by 3 or by 5.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Infinite examples, no proof yet. | Comment 12 of 21 |
(In reply to re: Infinite examples, no proof yet. by Charlie)

You are correct, 42 is not irreducible either.
42 = 2*3*7,

I had been thinking that although x=2*3 doesn't work that every multiple of 2*3 would work (after all 2^2*3, 2*3^2, 2*3*5 all work as well.)  It turns out 2*3*11=66 does not work so there is not a pattern based on multiples of 6.

So (if 2 is a factor) the irreducible values are 10,12,18,42.

I can't seem to figure out how to show definitively that 2 must be a factor.


  Posted by Jer on 2013-04-29 12:49:14

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 (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information