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
Xi'an's Og 
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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