Maximum Among Divisors

Find all the divisors of the given number $N$:

$\\$To find the divisors of $N$, we can iterate from $1$ to the square root of $N$. If $d$ is a divisor of $N$, then $N/d$ is also a divisor of $N$. We need to consider both $d$ and $N/d$ as divisors. If $d$ and $N/d$ are the same, we only need to consider one of them. $\\$$\\$Calculate the sum of digits for each divisor:

$\\$For each divisor, we can calculate the sum of its digits by repeatedly extract the last digit by using the remainder of division by $10$, add it to the sum and remove the last digit by dividing the number by $10$. Repeat this process while number greater than $0$.

$\\$Don’t forget to keep track of the maximum sum of digits encountered. The time complexity of this approach is O($\sqrt{n}$) for each test cases.