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MSC # 14: Digital Root

MSC # 14: Digital Root

What is the square root of x^4  − 6x^3 +17 x^2 − 24x + 16 ?

A) x^2 − 4x +16;  B) x^2 + 4x + 4;  C)  x^2 − 3x – 4;  D) x^2 − 3x + 4;  E) x^2 + 3x − 4

This is a question given in the Senior level of the 2nd International Vedic Mathematics Olympiad (IVMO 2022). This can be answered in less than 10 seconds if you know the applications of the subject of the 14th of our 30 MATH-Inic Specials for Christmas.

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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. We only have to get the digital root or repeated digital sum of that number. 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”. This was also taught to us by our grade-school teachers at the San Roque (now Manuel Roxas) Elementary School in Cavite city in the early 1960s.

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. This shows that adding 9 or a multiple of 9 to a number would not affect its digit sum. Similarly, subtracting 9 or a multiple of 9 from any number would also not affect its digit sum.

Therefore, we can disregard or cast out 1) any 9s or 2) any digits adding up to 9 or a multiple of 9, in finding the digital root of the number.

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.

In the IVMO 2022 question, we note that the square root of x^4 is x^2 and that of 16 is ± 4. By applying the Vedic Math sutra, the first by the first and the last by the last, we can say that the square root is in the form of ( x^2 + Bx  ± 4). We only need to find the value of B and the correct sign of the constant.

By using the sutra the product of the sums is the sum of the product, we can get the digit sum of the coefficients of x^4  − 6x^3 +17 x^2 − 24x + 16. (1 – 6 + 17 – 24 + 16) = 4 and the square root of 4 is ± 2. We therefore should look for an answer with the digit sum of ± 2.

Now looking at the choices,

A) x^2 − 4x +16;  B) x^2 + 4x + 4;  C)  x^2 − 3x – 4;  D) x^2 − 3x + 4;  E) x^2 + 3x − 4

We can immediately eliminate A, because the constant term is 16, and also choices B, C, and E because the digit sums of their coefficients are 9, – 6, and 0, respectively. The coefficients of choice D, 1 – 3 + 4 = 2.
Our featured 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.

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:

What is the remainder when 4523 x (734 + 888) is divided by 9?

We can use a calculator to find that 4523 x (734 + 888) is divided by 9 is 815,145. 111..

The decimal part of this quotient is equivalent to 1/9 which means that the remainder is 1.

Using casting out 9s and repeated summing of digits, we can also get the same result.

We can cast out the 3 8s in 888 and the 3 in 734 since (3×8) + 3 = 27. We can also disregard 4, 2 and 3 in 4523 since they add up to 9.

4,523 x (734 + 888)

We are left with 5 x (7 + 4).

Now 7 + 4 = 11 and 1 + 1 = 2.

5 x 2 = 10 and 1 + 0 = 1.

We can also get the same result in 5 x 11 = 55; 5 + 5 = 10 and 10 – 9 = 1.

More applications of digits sums in checking polynomial operations, factoring polynomials and solving quadratic equations can be found in our book Algebra Made Easy as Arithmetic.

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