trick of 9's is not a trick

Suppose for positive integers a, b, c, d, … , we have a + b + c + d + … = 9k for some positive integer k.

For a multi-digit number, say four digits, we can write the number as

1000a + 100b + 10c + d, which can also be written

999a + a + 99b + b + 9c + c + d.

Clearly 999a + 99b + 9c is divisible by 9,

and

a + b + c + d was given to be divisible by 9, so

since all of the individual summands is divisible by 9, then the sum must be divisible by 9.

This same logic can be used on any multiple digit number whose individual digits add to a multiple of 9.

EXAMPLE:

1278

1 + 2 + 7 + 8 = 18 = 9(2).

1278 = 1(1000) + 2(100) + 7(10) + 8

= 1(999) + 1 + 2(99) + 2 + 7(9) + 7 + 8

= 9(111) + 9(22) + 9(7) + 9(2)

= 9(142).