9

Divisibility by 9

  • If the sum of all digits of the number is divisible by nine then the number is divisible by nine else not.

Proof:

Let the number be abcd.

abcd can be written as
a×1000 + b×100 + c×10 + d

which can also be written as
a×(9×111+1) + b×(9×11+1) + c×(9+1) + d

which can also be written as
a×(9×111)+ a + b×(9×11)+ b + c×(9)+c + d

Every power of 10 i.e. 10x when divided by 9 has remainder 1.

So a×(9×111)+ b×(9×11)+ c×(9) is divisible by 9 as it has 9 as factor or common factor.

The number is divisible by 9 if a+b+c+d is divisible by 9.
Remainder

  • The remainder of the number, formed by the sum of the digits of the number, when divided by 9 is the remainder of the original number.

Examples

  1. Is 348 divisible by 9.
    The sum of digits is 3+4+8 = 15. Repeating the process again 1+5=6. 6 is the remainder.

  2. Is 3483 divisible by 9.
    Sum of digits is 3+4+8+3 = 18. 1+8 = 9. Hence the number is divisible and remainder is 0.


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