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

0

Answer 2

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 CBA

This 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 + A

This is only divisible by 9 if C + B + A is a multiple of 9, ie C + B + A is divisible by 9

• Now 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-1

The question is what is the remainder when 10ⁱ is divided by 9.

10⁰ ÷ 9 = 1 ÷ 9 = 0 r 1

10ⁿ ÷ 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 ÷ 9

As 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-1

So 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.

Answer 3

The divisibility test works because x*10y leaves leaves a remainder of x for any non-negative integer y.So, to test if ABC is divisible by 9, you need to check if A+B+C is divisible by 9, that is if A+B+C = 9, 18 or 27 (it cannot be larger than that).