# Subtraction by Parts

This is the term we apply to the 3^{rd} method of avoiding “borrowing” in subtraction as discussed in our book, ** 25 Math Short Cuts,** which is very helpful when the last few digits of the subtrahend is a little larger than the corresponding digits than the corresponding digits in the minuend

In the first technique, **Subtraction by Steps**, we “over-subtract” while here, we “under-subtract”.

The procedure here is to partition the subtrahend so that subtracting the first part from the minuend will create zeroes in the initial result. Let us apply this to a question in the Juniors category of the 1^{st} Math2shine International Vedic Mathematics Competition:

**What is 74365 – 24369?**

Here we notice that the last four digits of the subtrahend is only 4 more than the last four digits of the minuend. We can write the subtrahend as **24365 + 4**. The subtraction now becomes 74365 – (24365 + 4) = 50000 – 4.

Applying ** All from 9 and the Last from 10(Nikhilam)**, we have 50000 – 4 = 49996.

We will try another way of using **subtraction by parts **by solving a question given in the 2^{nd} International Vedic Mathematics Olympiad (IVMO 2022)**:**

**11.1 – 1.11 = ?**

Since we have two decimal places in the subtrahend we will suffix a zero in the minuend to express it with two decimal places also.

**11.10 → 1 1.10 → 10.00**

** – 1. 11 **

**→ –**

__→__~~1.1~~1 __– 0.01__

** 9.99**

Since **1.1** is common in both terms, we can cross them out leaving a simple 10.00 minus 0.01 which can be easily solved using ** Nikhilam**.

Try these examples:

- 2.22 – 0.222 =
- 9.99 – 0.999 =
- 13.33 – 3.333 =
- 5432 – 2435 =
- 7884 – 5889 =
- 6777 – 4888 =
- 87878 – 77879 =
- 78756 – 59768 =
- 345678 – 135579 =
- 578987 – 278999 =

Answers to last week’s exercises:

- 163 – 128 =35
- 134 – 89 = 45
- 216 – 77 =139
- 544 – 268 = 276
- 1452 – 967 = 485
- 3241 – 1789 = 1452
- 54,123 – 27,886 = 26,237
- 83,623 – 64,765 = 18,858
- 444,555 – 255,666 = 188,889
- 724,313 – 535,997 = 188,316