2021 AMC 10B Problem 25

For a fixed slope $m,$ the number of points in $S$ on or below $y=mx$ is $$\sum_{x=1}^{30}\lfloor mx\rfloor,$$ for the slopes near the answer. At $m=\frac23,$ grouping $x=3k+1,3k+2,3k+3$ for $k=0,1,\ldots,9$ gives $$\lfloor \frac{2x}{3}\rfloor=2k,\ 2k+1, \ 2k+2,$$ whose sum over each block is $6k+3.$ Thus the total is $$\sum_{k=0}^9(6k+3)=270+30=300.$$ If $m<\frac23,$ the ten points with ratios $\frac{y}{x}=\frac{2}{3}$ are no longer counted, so the count is less than $300.$ Therefore the lower end is $\frac23.$ The next possible ratio $\frac{y}{x}$ greater than $\frac23,$ with $1\le x,y\le30,$ is minimized by checking $x$ modulo $3.$ The best candidates are $$\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10},$$ and the smallest is $\frac{19}{28}.$ Hence the interval length is $$\frac{19}{28}-\frac23=\frac1{84}.$$ Thus $a+b=1+84=85.$ Thus, the answer is E .