UKMT Senior Challenge-XIV

This is one interesting pattern to look at

$9^2 = 81 \\ 99^2 = 9801 \\ 999^2 = 998001 \\ 9999^2 = 99980001 \\ \vdots \\ {\underbrace{999\cdots 999}_{M \text{ nines}}}^2 = \underbrace{999\cdots 999}_{(M-1) \text{ nines}}8\underbrace{000\cdots 000}_{(M-1) \text{ zeroes}}1$

Our number $m$ has $M=99$ nines, thus

$m^2 = \underbrace{999\cdots 999}_{98 \text{ nines}}8\underbrace{000\cdots 000}_{98 \text{ zeroes}}1$

The sum of digits of $m^2$ is: $(98 \times 9)+8+(98 \times 0)+1 = \boxed{891}$ .

if a number that has all digits nines ;
999999......9 ( the number of the nines =n)
then
99999......9² = 9999ْ.....980000.....001 (the number of the nines is n-1)
so the sum of the digits will be
9* (n-1)+8+1
so if we have ninety-nine digits, all of them nines , the sum will be
9*(99-1)+8+1=891