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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?

Thanks!

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    @N.S. oh sorry, I thought you meant n denominators. and yes that would be true. It follows from the answer for http://math.stackexchange.com/questions/95720/bound-on-lcm-of-denominators-of-rational-numbers-that-sum-to-1/989032012-01-20

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It's not always true: for example,

$\frac{1}{15}+\frac{1}{10}+\frac{5}{6}=1$

is a counterexample.

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    @GregMartin your tax dollars at work!2013-06-19