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$r$ is a number such that $r=p^a$. If the sum of some fractions equal to $1$ and one of the denominators is divisible by $r$ then there is another denominators that is exactly divisible $r$.

It seems to be really easy but I cannot prove it for example:

$\frac{7}{12}+\frac{4}{15}+\frac{3}{20}=1$

You can see here we have two denominators that are divisible by $4$ or two denominators that are divisible by $3$.Any hints?

Taha Akbari
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    Related (follow-up?) to http://math.stackexchange.com/questions/1832975/when-the-sum-of-some-fractions-be-1/1832986#1832986 – Joffan Jun 20 '16 at 08:47

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Hint: suppose there is one, and only one, term with denominator divisible by $p^a$. Let $L$ be the least common multiple of all denominators. Multiply both sides by $L/p$. Then the RHS is an integer and the LHS is not.

David
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  • Could you please explain more – Taha Akbari Jun 20 '16 at 08:36
  • Well... you did ask for a hint, not an answer. See what happens if you assume that $\frac7{12}+\frac4{15}+\frac3{20}+\frac18=1$ and follow my suggestion. – David Jun 20 '16 at 08:38
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    @TahaAkbari Do you see why $L/p$ will be an integer? Do you see why $L/p\cdot 1$ is an integer? Now, looking at the LHS, each fraction involved... do you see why $L/p$ times the fractions will result in integers in every case except for the one with denominator divisible by $p^a$? If you add up a bunch of integers and one non-integer... – JMoravitz Jun 20 '16 at 08:41