Why must 2 digits of any 2-digit number that is multiple of 9 (eg 27) add up to 9?
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Because 9 is exactly 1 less than the number at wich we roll over digits. That means whenever you add 9 to any number the digital sum doesn’t change because the last digit reduces by one and the second last digit increases by one.
Since a multiple of 9 is 9+9+…+9 each addition keeps the digital sum at 9
The exception is when your number ends in 0, wich also applies in your example: 99 is a multiple of 9 and it’s digital sum is 18. But the next exception corrects that again, if your second last digits rolls over it reduces the sum again. 108 is a multiple of 9 and it works again.
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