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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)

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re(2): Proof | Comment 17 of 21 |
(In reply to re: Proof by broll)

Bravo, Math Man!  Your proof looks great to me.  Nice work!


All of Broll's results are in fact on the closed set that Broll articulated.

1,1,4,6  corresponds to 2,3,12,12
1,2,2,5  corresponds to 2,5,5,10
1,2,6,9  "                     2,3,9,18
1,3,4,4  "                     3,3,4,12
1,3,8,12 "                    2,3,8,24
1,4,5,10 "                    2,4,5,20
1,6,14,21 "                  2,3,7,42
2,3,10,15 "                  2,3,10,15

The last of these is particularly interesting, because a,b,c,d = z,y,x,w.  This is also the only one where the LCM is divisible by both 3 and 5.  Coincidence?  I think not.

Edited on May 11, 2013, 11:41 am
  Posted by Steve Herman on 2013-05-10 15:02:08

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