Question

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

Answer 1

Let $a=a_n........a_3{a}_2{a}_1$ be an integer
[Note a is not the product of $a_1,a_2,a_3,........a_{n}but{a}_1,a_2,a_3,........a_n$ 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
$=8+6\times 10+3\times (10{)}^2$ ]
$a=a_n....a_3{a}_2{a}_1$
$=a_1+(10{)}^1{a}_2+(10{)}^2{a}_3+(10{)}^3{a}_4+.....+(10{)}^{n-1}{a}_n$
$=a_1+10a_2+100a_3+1000a_4+.........$
$=a_1+(a_2+9a_2)+(a_3+99a_3)+(a_4+999a_4)+..........$
$=(a_1+a_2+a_3+a_4+.........)+(9a_2+99a_3+999a_4+........)$ .....(1)
or $a=S+(a_2+11a_3+111a_4+.....)$
$\therefore 9|(a-S)$ .......(2)
Part 1:
a is divisible by 9
i.e., $9|a$ .........(3)
$\therefore 9|[a-(a-S\left)\right]$ [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), $9|\left[\right(a-S)+S]$
i.e., $9|a$
i.e., the integer a is divisible by 9.