a very quick Riddle

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A very quick Riddler’s riddle last week with the question

Find the (integer) fraction with the smallest (integer) denominator strictly located between 1/2020 and 1/2019.

and the brute force resolution

for (t in (2020*2019):2021){ a=ceiling(t/2020) if (a*2019

leading to 2/4039 as the target. Note that

\dfrac{2}{4039}=\dfrac{1}{\dfrac{2020+2019}{2}}

\dfrac{2}{4039}=\dfrac{1}{\dfrac{2020+2019}{2}}



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