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.

(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