If n = 10^{2020} + 10^{2019} + 10^{2018} + 10^{2017} + ... + 10^2 + 10^1, what is the sum of digits of the integer n?
+289 Looks hard but is actually easy!
The number, 10^n is just 1000... with n zeroes. Now suppose we have the 1 on the digits place next to it when we add 10^{n-1}. What do you get?
So by this observation, we can see that your whole expression is equal to:
11111... with 2020 1's.
Therefore, the sum of the digits of that epxession is 2020 x 1 = 2020.
