Question

Why the divisibility test for 9 is valid and a way to determine whether a three-digit counting number ABC is divisible by 9?

Answer 1

If the sum of digits add up to 9 then the number is divisible by 9 as for example the digits of 450 add up to 9 and so 450/9 = 50----------------------------------------------------------Consider a single digit number A.If A is 9 (or 0), then the number is divisible by 9, otherwise the remainder is A Now consider a two digit number BA.This is the same as 10B + A = (9 + 1)B + A = 9B + B + A This is only divisible by 9 if B + A is a multiple of 9, ie B + A is divisible by 9.Now consider a three digit number CBAThis is the same as: 100C + 10B + A= (99 + 1)C + (9 + 1)B + A= 99C + C + 9B + B + A= 99C + 9B + C + B + A= 9(11C + B) + C + B + AThis is only divisible by 9 if C + B + A is a multiple of 9, ie C + B + A is divisible by 9Now consider a number with n digits:Let the digits be D{i} where i = 0, 1, ..., n - 1 Then the number is the same as 10ⁿ⁻¹D{n-1} + 10ⁿ⁻²D{n-2} + ... + 10¹D{1} + 10⁰D{0} = Σ10ⁱD{i} for i = 0, 1, ..., n-1The question is what is the remainder when 10ⁱ is divided by 9.10⁰ ÷ 9 = 1 ÷ 9 = 0 r 110ⁿ ÷ 9= (10)10ⁿ⁻¹ ÷ 9= (9 + 1)10ⁿ⁻¹ ÷ 9= (9×10ⁿ⁻¹) ÷ 9 + 10ⁿ⁻¹ ÷ 9= 10ⁿ⁻¹ + (10ⁿ⁻¹ ÷ 9)→ remainder is the same as the remainder of of 10ⁿ⁻¹ ÷ 9As n = 0 → remainder 1, 10 to any power has a remainder of 1 when divided by 9.→ The remainder of Σ10ⁱD{i} for i = 0, 1, ..., n-1 divided by 9= Σ1×D{i} for i = 0, 1, ..., n-1= ΣD{i} for i = 0, 1, ..., n-1So the remainder when a number is divided by 10 is the same as the sum of the digits of that number divided by 10.→ A number is divisible by 9 if the remainder is 0, ie it is a multiple of 9.→ A number is divisible by 9 if the sum of its digits is a multiple of 9.To check if the sum of the digits of a number is a multiple of 9 (divisible by 9), you can repeat the summing on its digits.By repeating the summing of digits of the sums until a single digit remains, the original number is divisible by 9 if this final single digit is 9; otherwise this final single digit gives the remainder when the original number is divided by 9. This single digit is known as the digital root of the number.To determine if a 3 digit number ABC is divisible by 9, add its digits A+B+C; if this results in a two digit sum (DE) add the digits again (D + E). You will have a single digit. If this single digit sum is 9, then the original number is divisible by 9, otherwise it is the remainder when the original number is divided by 9.