I am currently working on an Olympiad math problem, and I am struggling to find a solution. I would greatly appreciate your help in solving this problem. I was unable to solve the problem because I don't know the exact rules for solving the problem.
Problem Source: Bangladesh Math Olympiad
A small help will be enough for me to proceed.
Here is a number (444…888…9) that has the 2018
digits of 4, followed by 2017 digits of 8
and one digit of 9. Find the sum of the
digits of the square root of this number.
Best Answer
Start by noting the pattern $$49=7^2 \\4489=67^2 \\444889=667^2$$
Now, try to prove using induction (or otherwise) that $$\underbrace{4\dots 4}_{n\text{ times}}\;\underbrace{8\dots 8}_{n-1\text{ times}}\;9=\left(\underbrace{6\dots 6}_{n-1\text{ times}}\;7\right)^2$$ Once you do that, the rest is easy.