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Composite Contemplation (Posted on 2013-02-16) Difficulty: 3 of 5
S=a+b+c+d+e+f, where:
each of a,b,c,d,e and f is a positive integer.

If S divides each of:
a*b*c + d*e*f and:
a*b + b*c + c*a - d*e - e*f - d*f
then, is S always composite?

If so, prove it. If not, provide a counter example.

No Solution Yet Submitted by K Sengupta    
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Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): spoiler Comment 3 of 3 |
(In reply to re: spoiler by Jer)

Great proof, xdog.  I disagree with Jer.  Your proof is complete and correct even if there does not exist an S that meets all conditions.

Similarly, I have discovered a truly marvelous proof that if x = 1 and x = 2, then 1 = 2.  (Unfortunately, the proof is too short to fit in this space.)  The proof is complete and valid, even though no x exists that meets the initial conditions.  That turns out to be fortunate, as I would not want to live in a world where 1 = 2.

But back to S.  Here is another set that works:
(a,b,c,d,e,f) = (6,6,6,6,6,6) and S = 36.  As xdog proved, 36 must be composite. 

Edited on February 18, 2013, 10:23 am
  Posted by Steve Herman on 2013-02-18 10:11:27

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