how to calculate digit sum of \({({10}^{1998}+5})^{2}\)
(10^1998+5)^2 = (infinity)*5+(infinity)*(infinity)+25+5*(infinity)
how to calculate digit sum of (10^1998 + 5)^2=1 x 10^3996. The addition of 5 has insignificant effect on the size of this number. It just means the last 2 digits will be 25.
