# a very quick Riddle

January 21, 2020
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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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