SAUDI ARABIAN IMO Booklet 2023

The given number can be expressed as $$\begin{array}{rl}\sum _{i=1}^{2023}\underset{i\text{ 1’s}}{\underbrace{11\dots 11}}& =\sum _{i=1}^{2023}\frac{10^i-1}{9}=\frac{{\sum }_{i=1}^{2023}10^i-2023}{9}\\ & =\frac{\underset{2023\text{ 1’s}}{\underbrace{111\dots 1110}}-2023}{9}=\frac{\underset{2019\text{ 1’s}}{\underbrace{111\dots 111}}09087}{9}.\end{array}$$ We further notice that $$\underset{2019\text{ 1’s}}{\underbrace{111\dots 111}}09087=\sum _{i=0}^{223}\underset{9\text{ 1’s}}{\underbrace{11\dots 11}}\cdot 10^{9i+8}+11109087.$$ On the other hand, with respect to $\frac{111111111}{9}=12345679$ and $\frac{11109087}{9}=1234343$, upon dividing the expression above by 9, we get the number equals to $$\sum _{i=0}^{223}12345679\cdot 10^{9i+8}+1234343.$$ Therefore, we need to calculate the sum of digits of this number. In each $12345679$ separated block, the sum of digits equals $37$, thus the answer is $$224\cdot 37+(1+2+3+4+3+4+3)=8308.$$