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prove if we want that the sum of some fractions be $1$ and the denominators of one of them is $d$ then another denominators should divisible by $d$ or $d$ should be divisible to another denominators.

It seems to be easy I tried to prove it.I first tried some cases.

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

Here we can see that $6$ is divisible by $3$. Also here $6$ is divisible by $2$ .But I want to prove one of the denominators but here two of them is possible.After trying a lot I cannot found any proofs.Any hints?

update1: the numerator should be prime

Taha Akbari
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  • Multiply across by denominators. Then all terms except one have $d$ as a factor. So $d$ must divide the remaining term. But note that you need to be careful what you state as true. Neither 2 nor 3 are divisible by the denominator by 6. – almagest Jun 20 '16 at 08:02
  • Could you please post your answer with more explain? – Taha Akbari Jun 20 '16 at 08:03

1 Answers1

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As stated, the claim offered for proof is not true. For example: $$\frac {7}{12} +\frac{4}{15} +\frac{3}{20} = 1$$

An alternative, which avoids composite numerators:$$\frac {1}{12} +\frac{13}{15} +\frac{1}{20} = 1$$

Joffan
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