 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Dark Divisibility (Posted on 2013-04-27) 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) Comments: ( Back to comment list | You must be logged in to post comments.) re: another | Comment 7 of 18 | The possible difficulty I see here is as follows:
1=(a+b+c+d)/abcd
1 = 1/(abc)+1/(abd)+1/(acd)+1/(bcd)
(this was my first, incomplete, assessment of the problem)

But also  abcd=n(a+b+c+d)
Now we have:
1/2=1/(abc)+1/(abd)+1/(acd)+1/(bcd)
1/3=1/(abc)+1/(abd)+1/(acd)+1/(bcd)
1/4=1/(abc)+1/(abd)+1/(acd)+1/(bcd)
1/5=1/(abc)+1/(abd)+1/(acd)+1/(bcd) etc.

There are literally thousands of potential solutions to these equations as the sum of 4  fractions with numerator 1 (not to mention the infinite number of potential candidates on LHS) though I accept the mere existence of a solution doesn't guarantee that each of a,b,c,d will always be integral. But I'd be surprised if the answer was as simple as 'make an exhaustive list'.

Edited on April 28, 2013, 10:38 am
 Posted by broll on 2013-04-28 10:37:21 Please log in:

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