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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: 5.0000 (2 votes)

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re(2): Infinite examples, no proof yet. | Comment 12 of 18 |
(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

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