2021 AMC 12B Problem 25

For slope $m,$ column $x$ (with $1\le x\le 30$) contributes $\min(30,\lfloor mx\rfloor)$ points on or below $y=mx,$ and we need the total to equal $300.$ At $m=\tfrac23,$ the cap at $30$ is inactive and $\sum_{x=1}^{30}\lfloor \frac{2x}{3}\rfloor=300.$ The count remains fixed until the next larger slope $\frac{y}{x}$ with $x\le30.$ If $\frac{y}{x}>\frac{2}{3},$ then $3y-2x$ is a positive integer. The closest possibility has $3y-2x=1;$ maximizing $x\le30$ in this congruence gives $(y,x)=(19,28).$ Any numerator at least $2$ gives a larger gap. Hence the interval is $[\tfrac23,\tfrac{19}{28}),$ whose length is $\tfrac{19}{28}-\tfrac23=\tfrac1{84}.$ Since $\gcd(1,84)=1,$ $a+b=1+84=85.$ Thus, the correct answer is E.