# 30MSC 2023 #1 – All from 9 and the Last from 10

As we have mentioned in our announcement yesterday, we have decided to reorganize our book **30 Master Strategies in Computing**, so that Vol I will deal mainly with the elementary principles of Vedic Math. So for our **1 ^{st} MATH-Inic Special for Christmas 2023, **we will discuss this question which was given in the primary age groups of the

**2**on November 26, 2022.

^{nd}International Vedic Mathematics Olympiad (IVMO 2022)**600,000 – 375,072 =**

Questions similar to this one is given in every VM competition because they require the longest solutions using conventional methods but by using the Vedic Math Sutra or word formula, **All from 9 and the last from 10 (Nikhilam Navatascaraman Dasatah**),they can be easily solved mentally. Mastery of this Sutra’s applications naturally lead to quicker computations in other arithmetic operations. **Nikhilam multiplication** and **Nikhilam division** are just two specific techniques which use this Sutra.

Many problems require us to quickly find the 10’s complement of a number, which tells us how far the number is from the next higher **perfect base** or **power of 10**. The 10’s complement is also called the number’s **deficiency** from its base.

**All from 9 and the Last from 10** enables us to quickly do that by simply looking at the subtrahend and deducting, starting from the left, digit by digit from 9 except the rightmost which we will deduct from 10.

In the given problem, we are required to find the difference between a multiple of a power of 10 which is 600,000 and another 6-digit number.

Many young learners would say that this subtraction problem is among the most difficult if the traditional right to left procedure taught in schools is used because it requires a series of “borrowings”.

But this is easily solved by first using another Vedic Math Sutra, **By One Less than the One Before. **Subtract 1 from 6 to get 5 then we can proceed from left to right, reciting the answer digit by digit as we compute.

Since we can recognize that the leading 6 stands for six hundred thousand, we can immediately announce the answer as (5 – 3) **two hundred** and then apply **All from 9 and the Last from 10** on the succeeding digits: (9 – 7) **twenty**

(9 – 5) **four thousand,**

** **(9 – 0) **nine hundred**

** **(9 – 7) **twenty**

(10 – 2) **eight**

More examples and applications can be found in Chapter 1 of our rewritten e-book, **30 Master Strategies in Computing, Volume I,** which will be available before the end of the year.

Follow us on our Facebook page, MATH-Inic Philippines at https://www.facebook.com/MATHInicPhils to see tomorrow’s **30MSC 2023 #2: Addition without Carrying!**