Hi there, still fairly new posting reply so please excuse the format:
Our goal is to find all m's such that sum of the digits of \(10^{60} - m = 533\)
For a simpler case: Let's assume \(m = 99 \) and \(10^4\) instead of \(10^{60}\); \(10^4- m\) we would have \(10000-99 = 9901\),
From that we can see for every n in \(10^n\), we will have at least \(n-2\) 9's (in our case \(n = 4 \) and number of 9's is \(n-2\))
so for \(10^{60} \) we would have 58 9's and the sum of digits so far would be \( (58 * 9) = 522,\) we are still missing \(533-522 = 11.\)
Keep in mind we are
not looking for all m's such that the digits of m add up to 11, we are looking for all m's such that the digits of \(100-m\) is 11. (Observe that I'm using 100 here since we are only concerned with the last 2 digits of the sum that add up to 11, the rest are all 9's as stated above)
so we need the sum of digits of \(100-m\) equal to 11, m could be
\(m = 17 -> 100 - 17 = 83, 8+3=11\)
\(m = 26 -> 100 - 26 = 74, 7+4 =11\), and so on. Also observe the reverse of the digits also works (e,g 71, 62)
in the end we have 7 cases (17, 26, 35, 44, 53, 62, 71). Answer is C