It's a number

$\begin{aligned} \text{The curve}: \quad f(x) & = x^x \\ \text{Gradient of the curve}: \quad f'(x) & = \frac{d}{dx} x^x = \frac{d}{dx} e^{x \ln x} \\ & = e^{x \ln x} \frac{d}{dx} x \ln x \\ & = x^x (\ln x + 1) \\ \text{Gradient at }(20,20^{20}): \quad f'(20) & = 20^{20} (\ln 20 + 1) \\ \text{Gradient of perpendicular}: \quad - \frac{1}{f'(20)} & = - \frac{1}{20^{20} (\ln 20 + 1)} \\ \text{Equation of line}: \quad \frac{y-20^{20}}{x-20} & = - \frac{1}{20^{20} (\ln 20 + 1)} \\ \frac{x-20}{20^{20}-y} & = 20^{20} (\ln 20 + 1) \\ \text{At the x-axis, } y=0: \quad \frac{x_0-20}{20^{20}-0} & = 20^{20} (\ln 20 + 1) \\ \Rightarrow x_0 & = 20^{40} (\ln 20 + 1) + 20 \\ \text{Number of digits of }x_0: \quad n & = \left \lfloor \log_{10} 20^{40} \ln 20 \right \rfloor + 1 \\ & = \left \lfloor 40 \log 20 + \log (\ln 20) \right \rfloor + 1 \\ & = \left \lfloor 52.04119983 + 0.476502998 \right \rfloor + 1 \\ & = 52 + 1 = \boxed{53} \end{aligned}$