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I was experimenting with an algorithm for generating random numbers from a discrete distribution and came across an interesting observation. Suppose that you have any finite set of rational numbers in simplest form that sums to one, such as

$\frac{1}{2} + \frac{1}{3} + \frac{1}{6}$

$\frac{1}{4} + \frac{1}{5} + \frac{11}{20}$

$\frac{1}{3} + \frac{1}{3} + \frac{1}{3}$

I noticed that in each case, there is some fraction in each sum whose denominator is the least common multiple of all the denominators collectively. In the first example, $6 = LCM(2, 3, 6)$; in the second, $20 = LCM(4, 5, 20)$; in the third, $3 = LCM(3, 3, 3)$.

In all of the examples I've tried, this pattern occurs, but I'm not sure if this is just a coincidence or not. I've attempted to prove that this is true, but I can't make much progress on it.

Am I completely wrong that one of the fractions must have a denominator that's the LCM of the denominators? Or is there a proof that might help shed light on why this is?


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Take the second one for instance, it can also be written as:$\frac{4}{16}+\frac{5}{25}+\frac{11}{20}$. So it's only true when the fraction can't be simplified any further. – Raskolnikov Dec 25 '11 at 21:38
@Raskolnikov- Ah yes, thanks for pointing that out. I've updated the question accordingly. – templatetypedef Dec 25 '11 at 21:41
1/15+1/10+5/6 perhaps? – Lopsy Dec 25 '11 at 21:43
@Lopsy- D'oh! Thanks for pointing that out! If you promote that to an answer, I'll accept. – templatetypedef Dec 25 '11 at 21:45
Sure, done! Interestingly enough, I might remember an old IMOSL problem that asked something similar: given any set of arithmetic sequences on the integers, such that every integer is in exactly one sequence, prove that the period of at least one of the sequences is equal to the the lcm of all the periods. I'll be thinking about that question too - this one is true, unless of course I remembered it wrong. – Lopsy Dec 25 '11 at 21:51
up vote 8 down vote accepted

It's not always true: for example,


is a counterexample.

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It's also not necessarily true even if all the numerators equal 1. See for a particularly egregious counterexample. – Greg Martin Dec 26 '11 at 6:18
@GregMartin your tax dollars at work! – Zackkenyon Jun 19 '13 at 5:03

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