## a very quick Riddle

**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

January 23, 2020 at 2:07 pm

I see (at Riddler page) that solution by way of continued fractions is recommended.

January 23, 2020 at 8:13 am

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