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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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Some Thoughts Infinite examples, no proof yet. | Comment 8 of 18 |
I'm attempting to prove an equivalent result:  the LCM of a,b,c,d is divisible by 3 or by 5.

Let a+b+c+d = LCM(a,b,c,d) = x.

  • x cannot be prime
  • x cannot be a prime power.  If x=p^n, then p^n would have to be one of the numbers to make the LCM be p^n.
  • x cannot be of the form 2*p if p>5.  This can easily be seen if p=7.  We'd seek a sum 14.  But the only choices for a,b,c,d are in {1,2,7} and you need at least one 2 and one 7.  You can't have another 7 and two more 2's isn't big enough.  Replace 7 with p.
  • If x works then p*x is also works.  If a+b+c+d=lcm(a,b,c,d)=x for any then ap+bp+cp+dp=p(a+b+c+d)=p*lcm(a,b,c,d)=lcm(ap,bp,cp,dp)=px.
The converse of the last statement is almost always true.  The only exception I've seem is that although x can be 12=6+4+1+1=2^3*3 and x can be 18=9+6+2+1=2*3^2, x cannot be 6.

What's left to prove is: 2 is always a factor of x (although this is pretty obvious for the same reason as my third bullet.)

For the record, the only irreducible solutions I have found are
10 = 5+2+2+1 = 2*5
12 = 6+4+1+1 = 2^2*3
18 = 9+6+2+1 = 2*3^2
The sequence of numbers that work starts 10,12,18,20,24,30,36,40,42,48,50...


  Posted by Jer on 2013-04-28 12:41:10
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