Casting Out Nines

Inspired by a recent Numberphile video I thought I would share with you a cool trick to check your arithmetic is correct called ‘Casting Out Nines‘.

To check the result of a calculation using this technique, take the digital root of each number in the calculation. Then perform the calculation on these digital roots, and take the digital root of this answer. If no mistake has been made, the digital root of the result of this calculation should be the same as the digital root of the result of the original calculation.

What is a digital root?

The digital root is the value (a single digit) you get by completing an iterative process of summing the digits of the number, where you use the result from the previous iteration to compute a digit sum on each iteration.

Example: Multiplication

12345 x 67890 = 838102050

The digital roots of 12345 and 67890 are 6 and 3 respectively. The product of these digital roots is 18. The digital root of 18 is 9.

The sum of the digits of 838102050 is 27, whose digital root is also 9.

Hence, this confirms that the calculation is correct. Pretty cool eh?

How does this work?

For any number

{\displaystyle 10^{n}d_{n}+10^{n-1}d_{n-1}+\cdots +d_{0}}

the digit sum is {\displaystyle d_{n}+d_{n-1}+\cdots +d_{0}}. The difference between the original number and its digit sum is

{\displaystyle {\begin{aligned}&10^{n}d_{n}+10^{n-1}d_{n-1}+\cdots +d_{0}-\left(d_{n}+d_{n-1}+\cdots +d_{0}\right)\\={}&\left(10^{n}-1\right)d_{n}+\left(10^{n-1}-1\right)d_{n-1}+\cdots +9d_{1}.\end{aligned}}}

Noting that numbers of the form {\displaystyle 10^{i}-1} are always divisible by 9, replacing the original number by its digit sum has the effect of casting out

{\displaystyle {\frac {10^{n}-1}{9}}d_{n}+{\frac {10^{n-1}-1}{9}}d_{n-1}+\cdots +d_{1}}

lots of 9.

M x

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