Let n be a positive integer. If the sum of the digits of n is divisible by 9, then n is divisible by 9.
I got upto here,
100a + 10b + c = n a + b + c = 9k; k exists in the set Z (all integers)
I didn't know what to do after this, so I consulted the solution
The next step is:
100a + 10b + c = n = 9k + 99a + 9b = 9(k + 11a + b):
I don't get how you can add
99a + 9b randomly, can someone please explain this for me