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.
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 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 20130428 12:41:10 