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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: spoiler | Comment 2 of 3 |
(In reply to spoiler by xdog)

I like what you supplied for a proof.  It is slightly incomplete since all it does is show S is not prime, not that S can exist that divides the two expressions.

You need to prove there is at least one value of S.

Fortunately (a,b,c,d,e,f) = (6,5,4,3,2,1) works
(so nice when you find one on the first try)
S = 6+5+4+3+2+1 = 21
6*5*4+3*2*1 = 126 = 21*6
6*5 + 5*4 + 6+4 - 3*2 - 3*1 - 2*1 = 63 = 21*3
and of course, 21 is composite.

  Posted by Jer on 2013-02-17 22:13:01

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