# 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 can 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.

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.

There is also a technique for quickly finding out the digit sum of a number called “casting out 9s”. If you add 9 to a number like 5, for example, you get 14 and 1 + 4 = 5. If you add another 9 to 14, you will get 23 and 2 + 3 = 5. Now if you add 90 to 23, you will get 113 and 1 + 1 + 3 is also 5. Adding 9 or a multiple of 9 to a number would not affect its digit sum.

We can then say that subtracting 9 or a multiple of 9 from any number would also not affect its digit sum.

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 arithmetic and polynomial operations. 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 in solving some non-routine problems.

Digit sums and its use to check Arithmetic and polynomial operations will be discussed by Veronica S. Prudente, training director of MATH-Inic Philippines on the opening day of the Inspirational Maths from India (Year 4) webinars. (Please send message to https://www.facebook.com/MATHInicPhils for more details on IMI 4)

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 (digit) sums is the (digit)sum of the product”

Let us look at our example

(524 + 973) x (875 – 257) = 1497 x 618

We can use a calculator to find that 1497 x 618 = 925, 146 and 925,146 ÷ 9 = 102,794 with no remainder

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

1 + 4 + 9 + 7 = 21; 2 + 1 = 3

6 + 1 + 8 = 15; 1 + 5 = 6

Now 3 x 6 = 18; 1 + 8 = 9 → 0

9 + 2 + 5 + 1 + 4 + 6 = 27; 2 + 7 = 9 → 0

We can see that the product of the digit sums of 1497 and 618, which are 3 and 6 respectively is 18 and the digital root of 18 is 0. This is equal to the digit sum of the product 925, 146. Now if you divide this product by 9, we will get exactly 102, 794. This confirms that the remainder is 0.

We can also take the digit sums of 524 (5 + 2 + 4 = 11; 1 + 1 = 2) and 973 ( 9 + 7 + 3 = 19; 1 + 9 = 10 and 1 + 0 = 1) and add them 2 + 1 to get 3 the same as the digit sum of 1497.

Similarly, we can get the digit sum of 875 ( 8 +7 + 5 = 20; 2 + 0 = 2) and 257 ( 2 + 5 + 7 = 14; 1 + 4 = 5 ). To avoid getting negative digit sums, we will use the values 20 and 14 to get the difference of 6.

We can speed up the determination of the digital root by casting out 9s.

**( 524 + 973) x (875 – 257) **cast out 9 and 4 + 5

**(2 + 73) x (875 – 257) **cast out 2 + 7 across addends

**(3) x (8 75 – 257) **cast out the 5s and the 7s

**(3) x (8 – 2) **only 3 x 6 remains

3 x 6 = 18; 1 + 8 = 9 → 0

The remainder is 0.