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Let $w=\lfloor x\rfloor$ and $f=\{x\}$ denote the whole part and the fractional part of $x,$ respectively, for which $0\le f<1$ and $x=w+f.$ We rewrite the given equation as $$w\cdot f=a\cdot (w+f{)}^2.(1)$$ Since $a\cdot (w+f{)}^2\ge 0,$ it follows that $w\cdot f\ge 0,$ from which $w\ge 0.$ We expand and rearrange $(1)$ as $$a{f}^2+(2a-1)wf+a{w}^2=0,(2)$$ which is a quadratic with either $f$ or $w.$ For simplicity purposes, we will treat $w$ as some fixed nonnegative integer so that $(2)$ is a quadratic with $f.$ By the Quadratic Formula, we have $$f=w\left(\frac{1-2a\pm \sqrt{1-4a}}{2a}\right).(3)$$ If $w=0,$ then $f=0.$ We get $x=w+f=0,$ which does not affect the sum of the solutions. Therefore, we consider the case for $w\ge 1:$ Recall that $0\le f<1,$ so $\frac{1-2a\pm \sqrt{1-4a}}{2a}\ge 0.$ From the discriminant, we require that $0\le 1-4a<1,$ or $$0<a\le \frac{1}{4}.(4)$$ We consider each part of $0\le f<1$ separately: 1. $f\ge 0$ From $(2),$ note that $a>0,(2a-1)w<0,$ and $a{w}^2>0.$ By Descartes' Rule of Signs, we deduce that $(2)$ must have two positive roots, so $f\ge 0$ is always valid. Alternatively, from $(3)$ and $(4),$ note that all values of $a$ for which $0<a\le \frac{1}{4}$ satisfy $1-2a>\sqrt{1-4a}.$ We deduce that both roots in $(3)$ must be positive, so $f\ge 0$ is always valid. 4. $f<1$ We rewrite $(3)$ as $$f=w(\frac{1}{2a}-1\pm \frac{\sqrt{1-4a}}{2a}).$$ From $(4),$ it follows that $\frac{1}{2a}\ge \frac{1}{1/2}=2.$ The larger root is $$f\ge w\left(2-1+2\sqrt{1-4a}\right)\ge 1(2-1+2\sqrt{1-4\cdot \frac{1}{4}})=1,$$ which contradicts $f<1.$ So, we take the smaller root, from which $$f=w(\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a})$$ for some constant $k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}>0.$ We rewrite $f$ as $$f=wk,$$ in which $f<1$ is valid as long as $k<\frac{1}{w}.$ Note that the solutions of $x$ are generated at $$w=1,2,3,\dots ,W,$$ up to some value $w=W$ such that $\frac{1}{W+1}\le k<\frac{1}{W}.$ Now, we express $x$ in terms of $w$ and $k:$ $$x=w+f=w+wk=w(1+k).$$ The sum of all solutions to the original equation is $$\sum _{w=1}^{W}w(1+k)=(1+k)\cdot \sum _{w=1}^{W}w=(1+k)\cdot \frac{W(W+1)}{2}=420.(★)$$ As $1+k<1+\frac{1}{W},$ we conclude that $1+k$ is slightly above $1$ so that $\frac{W(W+1)}{2}$ is slightly below $420,$ or $W(W+1)$ is slightly below $840.$ By observations, we get $W=28.$ Substituting this into $(★)$ produces $k=\frac{1}{29},$ which satisfies $\frac{1}{W+1}\le k<\frac{1}{W},$ as required. Finally, we solve for $a$ in $k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}:$ $$\begin{array}{rl}\frac{1}{29}& =\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}\\ \frac{2}{29}a& =1-2a-\sqrt{1-4a}\\ \frac{60}{29}a-1& =-\sqrt{1-4a}\\ \frac{60^2}{29^2}{a}^2-\frac{120}{29}a+1& =1-4a\\ \frac{60^2}{29^2}{a}^2-\frac{4}{29}a& =0\\ a\left(\frac{60^2}{29^2}a-\frac{4}{29}\right)& =0.\end{array}$$ Since $a>0,$ we obtain $\frac{60^2}{29^2}a-\frac{4}{29}=0,$ from which $$a=\frac{4}{29}\cdot \frac{29^2}{60^2}=\frac{29}{900}.$$ The answer is $29+900=\boxed{929}.$