# MSC #14 – Digit Sums: Casting Out 9s

All grade-school learners are taught that “a number is divisible by 9 if the sum of its digits is divisible by 9”. However, only a few are taught that there is no need to divide. If the digit sum has more than one digit we add again until a single digit remains. For example, the digit sum of the number 12,345 is 1 + 2 + 3 + 4 + 5 or 15. We add the digits of 15, 1 + 5 and get 6, its repeated digital sum or digital root. (We will use the terms digit sum and digital root interchangeably in this discussion.)

Also, very few students are made to realize that the digital root is the remainder when a number is divided by 9. This is unfortunate because the digital root has wide-ranging applications from Arithmetic to higher Mathematics.

A Vedic Math sub-sutra(corollary) “the product of the sums is the sum of the product” enables us to use digit sums to check the results of polynomial multiplication. In combination with another sub-Sutra “The first by the first and the Last by the last”, digit sums provide a convenient method in factoring polynomials and solve some non routine problems.

Our present example not only shows that the digit sum is the remainder when a number is divided by 9 and that we do not really need to divide to get the digit sum, but also

1) how to quickly determine the digit sum using the “casting out 9s” method which I learned also from my grade School teachers at San Roque (now Manuel Roxas) Elementary School in Cavite City in the early 1960s. Sadly, many young teachers are unaware of this method because it is not included in the current Math curriculum.

2) how to “short cut a short cut”. Our example shows that we can “cast out 9s” from different addends.

3) “the product of the sums is the sum of the product”

Let us look at our example

4523 x (734 + 888) = 4523 x 1622 = 7,336, 306

Let us first take the product of the digit sums and compare it with the digit sum of the product

4 + 5 + 2 + 3 = 14; 1 + 4 = 5

1 + 6 + 2 + 2 = 11; 1 + 1 = 2

7 + 3 + 3 + 6 + 3 + 0 + 6 = 28; 2 + 8 = 10; 1 + 0 = 1

We can see that the product of the digit sums of 4523 and 1622, which are 5 and 2 respectively is 10 and the digital root of 10 is 1. This is equal to the digit sum of 7,336,306. Now if you divide this product by 9, we would get 815, 145.111… The decimal part 0.1111… is equivalent to 1/9 and this confirms that our remainder is 1.

We can also take the digit sums of 734 (7 + 3 + 4 = 14; 1 + 4 = 5) and 888 ( 3 x 8 = 24; 2 + 4 = 6) and add them (5 + 6 = 11) to get 2 the same as the digit sum of 1622.

Now we can see that we can cast out 9 or group of numbers totaling to 9 or a multiple of 9 to speed up the determination of digital root. In 4523, we can delete or cast out 4 and 5 so that only 2 and 3 remains.

A further short cut is shown when we cast out one addend (888 and one other digit, 3, from the other addend)

For more discussions on digit sums and its applications, please see Algebra Made Easy as Arithmetic by Virgilio Y. Prudente