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 Dark Divisibility (Posted on 2013-04-27)
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: 5.0000 (2 votes)

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 re: Proof | Comment 15 of 18 |
(In reply to Proof by Math Man)

Math Man,

The statement:
'That means that n/w+n/x+n/y+n/z=n, so 1/w+1/x+1/y+1/z=1.'
implies that the set {w,x,y,z} is closed and comprises the list:

{2,3,7,42}, {2,3,8,24}, {2,3,9,18}, {2,3,10,15}, {2,3,12,12}, {2,4,5,20}, {2,4,6,12}, {2,4,8,8}, {2,5,5,10}, {2,6,6,6}, {3,3,4,12}, {3,3,6,6}, {3,4,4,6}, {4,4,4,4}.

Comparing the solutions I have found:

{1, 1, 4, 6, 12, 24, 12, 3}
{1, 2, 2, 5, 10, 20, 5, 5}
{1, 2, 6, 9, 18, 108, 3, 9}
{1, 4, 5, 10, 20, 200, 5, 4} implies x=4w;  (no solution available in list)

{1, 3, 8, 12, 24, 288, 8, 3} implies x=3w; (no solution available in list)
{1, 6, 14, 21, 42, 1764, 7, 3} implies x=6w; (no solution available in list)

{2, 3, 10, 15, 30, 900, 10, 3}

there does not seem to be any obvious correspondence.

Edited on May 10, 2013, 2:17 am
 Posted by broll on 2013-05-10 02:15:34

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