## a very quick Riddle

Posted in Books, Kids, pictures, R with tags , , , , , , on January 22, 2020 by xi'an

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<t) sol=c(a,t)}
```

leading to 2/4039 as the target. Note that

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