# Don't understand casting out nines

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

#### Solutions Collecting From Web of "Don't understand casting out nines"

$100a+10b+c=(99a+a)+(9b+b)+c=(99a+9b)+(a+b+c)=9(11a+b)+9k=9(11a+b+k)$.
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