2020 AMC 12A Problem 25

There are no negative solutions, while $x=0$ is always a solution. For $n\ge1$ and $x\in[n,n+1),$ put $y=\frac{x}{n}.$ The equation becomes $ay^2-y+1=0.$ Its roots must be real, so $0\lt a\le\tfrac14.$ If the two roots are $\alpha\le\beta,$ their sum and product are both $\frac{1}{a},$ so $(\alpha-1)(\beta-1)=1.$ Write $$\alpha=1+\dfrac1u,\qquad \beta=1+u$$ with $u\ge1.$ Then $a=\dfrac{u}{(u+1)^2}.$ The root $x=n\alpha$ lies in $[n,n+1)$ exactly when $n\lt u,$ while $n\beta$ never lies there for a positive integer $n.$ The required positive total ensures $u\gt1.$ Let $N$ be the largest positive integer less than $u,$ so $N\lt u\le N+1.$ The sum of all solutions is therefore $$\dfrac{u+1}{u}\cdot\dfrac{N(N+1)}2=420.$$ Because $\dfrac{u+1}{u}$ decreases with $u,$ these inequalities imply $$\dfrac{N(N+2)}2\le420\lt\dfrac{(N+1)^2}{2},$$ which forces $N=28.$ Substitution gives $406\cdot\dfrac{u+1}{u}=420,$ so $u=29.$ Hence $a=\dfrac{29}{30^2}=\dfrac{29}{900}.$ Indeed the positive solutions are $x=\dfrac{30n}{29}$ for $1\le n\le28,$ and their sum is $420.$ Therefore $p+q=29+900=929.$ Thus, C is the correct answer.