Question 999014
Question 1:
A number made of a digit x repeated 6 times is x(111111),
so it is divisible by 111111 and its factors, such as
3,  7, 11, 13, 37, and products of those factors.
Since {{{111111=3*7*11*13*37}}} , there are
{{{2^5=32}}} factors of 111111, including 1, and 111111.


Question 2:
A number divisible by 99 is divisible by 9 and by 11.
Adding up all its digits, and adding up the digits of the sum,
and repeating the adding process as needed, the result is 9.
The difference between the sum of the digits in the first, third, fifth, etc place,
and the sum of the other digits is 0, or 11, or 22, etc.