Let a and b be the number of digits of 4⁵⁰ and 25⁵⁰ respectively\(\left(a,b\in Z^+\right)\),then :

\(\Rightarrow10^{a-1}< 4^{50}< 10^a;10^{b-1}< 25^{50}< 10^b\)

\(\Rightarrow10^{a-1}.10^{b-1}< 4^{50}.25^{50}< 10^a.10^b\)

\(\Rightarrow10^{a+b-2}< 100^{50}< 10^{a+b}\Rightarrow10^{a+b-2}< 10^{100}< 10^{a+b}\)

\(\Rightarrow a+b-2< 100< a+b\)

\(\Rightarrow\left\{{}\begin{matrix}a+b< 102\\a+b>100\end{matrix}\right.\)\(\Rightarrow a+b=101\)

So,the answer is 101