Prove that an integer is divisible by 9 if and only if the sum of its digits is divisible by 9.

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Solution

Let $a = a_{n} . . . . . . . . a_{3} a_{2} a_{1}$ be an integer
[Note a is not the product of $a_{1}, a_{2}, a_{3}, . . . . . . . . a_{n} b u t a_{1}, a_{2}, a_{3}, . . . . . . . . a_{n}$ are digits in the value of a. For example 368 is not the product of 3, 6 and 8 rather 3, 6, 8 are digits in value of 368
$= 8 + 6 \times 10 + 3 \times (10)^{2}$ ]
$a = a_{n} . . . . a_{3} a_{2} a_{1}$
$= a_{1} + (10)^{1} a_{2} + (10)^{2} a_{3} + (10)^{3} a_{4} + . . . . . + (10)^{n - 1} a_{n}$
$= a_{1} + 10 a_{2} + 100 a_{3} + 1000 a_{4} + . . . . . . . . .$
$= a_{1} + (a_{2} + 9 a_{2}) + (a_{3} + 99 a_{3}) + (a_{4} + 999 a_{4}) + . . . . . . . . . .$
$= (a_{1} + a_{2} + a_{3} + a_{4} + . . . . . . . . .) + (9 a_{2} + 99 a_{3} + 999 a_{4} + . . . . . . . .)$ .....(1)
or $a = S + (a_{2} + 11 a_{3} + 111 a_{4} + . . . . .)$
$∴ 9 | (a - S)$ .......(2)
Part 1:
a is divisible by 9
i.e., $9 | a$ .........(3)
$∴ 9 | [a - (a - S)]$ [From (2) and (3)]
i.e., $9 | S$ i.e., sum of digits is divisible by 9
Part 2:
S (sum of digits) is divisible by 9
i.e., $9 | S$ .... (4)
From (2) and (4), $9 | [(a - S) + S]$
i.e., $9 | a$
i.e., the integer a is divisible by 9.

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