An algebra problem by Hritesh Mourya

Notice that:

$\log_{10} 10 = 1,\quad\log_{10} 99 \approx 1.996\\ \log_{10} 100 = 2,\quad\log_{10} 999 \approx 2.9996\\ \log_{10} 1000 = 3,\quad \log_{10} 9999 \approx3.99996$

From here, we can see that the number of digits, $n$ in a number $x$ is represented as

$n(x) = \lfloor \log_{10}x \rfloor + 1$

Therefore, the number of digits in $2000^{2000}$ :

$n(2000^{2000})\\ = \lfloor \log_{10} 2000^{2000} \rfloor + 1\\ = \lfloor 2000 \log_{10} 2000 \rfloor + 1\\ =\lfloor 6602.06 \rfloor + 1\\ =6602+1\\=\boxed{6603}$

log2000^2000 = 2000log2000 = 2000×3.301029 =6602.058 here characteristic is 6602 so the no of digits is (characteristic+1) =(6602+1) =6603 .