Division with Multi-Digit-Divisors

This article is an account of a teacher’s work with class 5 post her reading of the article ‘Thoughts on the Division Operation’, from the July 2015 issue of At Right Angles [1].

Brief Recap: Division is different from addition, subtraction and multiplication in many ways. It is also the most complicated of these four operations mainly because the algorithms for the other three do not require any estimation no matter how large the numbers dealt with are. However, the standard algorithm of long division requires estimation and demands a deeper engagement with a process that navigates many iterations of “if this, then do that”.

The current NCERT textbooks do not deal with multi-digit divisors at the preparatory level (till Class 5). Neither do they deal with it later, i.e., at the middle stage (Class 6-8). So, should we teach this at all? Is it needed when calculators are everywhere, even on the ‘un’smart phones?

There are two reasons to still teach these:

1. While we may not need to divide by a big number like 365, it is important to know how we can if needed. The process (involving estimation) to divide by a 2-digit number extends for any larger divisor. So, learners should be exposed to division involving 2-digit divisors.
2. A more practical reason, however, is that learners are often expected to divide by a 2-digit number in school. Learners are not allowed to use calculators. Here are some examples:

	7. A milkman sold two of his buffaloes for ¥ 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)
10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum i. find the population in 2001. ii. what would be its population in 2005?

Chapter 8: Comparing Quantities, Class 8, NCERT textbook Ex. 82,Q7 Ex. 8.3, Q10

Other examples may be found in:-

• Mensuration – Finding the radius of a circle given its circumference, e.g., finding the radius of acircle made by bending a 40cm wire.
• Data Handling – While calculating the mean, e.g., finding the mean of 23 items totalling 4178

Given below is a short narrative from the teacher’s experience in doing the problem on Data Handling in her Class 8 mathematics class.

Teacher	We have 23 items with a total weight of 4178 kgs, can anyone estimate the mean weight? You need to explain how you estimated.
Student A	Less than 200 kgs. Because 23 × 200 is 4600 kg.
Student B	More than 150 kgs. Because 23 × 100 is 2300 kg, and so the average must be closer to 200 kg than 100 kg.
Teacher	Excellent thinking! I notice that you are using products of multiples of 10s and 100s, to make your estimates. Now, let’s see if you can use this to your advantage in dividing by 23. Why don’t you round off 23 to the nearest multiple of 10, i.e. 20? Now make an estimate for the first digit of the quotient, i.e., divide 41 by 20.
Student C	We get 2, I can see that we are already going away from the estimated answer!
Teacher	That’s observant. So let’s check by multiplying 23 by 2, we get 46 and as you said, that’s more than 41.
Student B	So, the first digit is 1 and the remainder is 41 − 23 which is 18. Do we bring down the next digit as we do for single-digit division?
Teacher	Yes, we get 187 and we again try 20 × 9 which is 180.
Student A	Oh, so we try 23 × 9 which is 207 and so we then try 23 × 8 which is 184, that’s so close!
Teacher	So, the first two digits of the quotient are 1 and 8 and we get a new remainder of 3, we bring down the next digit 8 and divide 38 by 23.
Student B	I get a quotient of 181 and a remainder of 15. So, my average is about 181 kg, in fact about 181.5 kg because 15 is more than half of 23!
Teacher	Yes, we can continue with decimal division, but for now, this is a great idea of the mean weight. What do you think these 23 items are?
Student D	Well… maybe some sort of animal? Like dolphins? My favourite animals!
Teacher	That’s a great suggestion! I know that some motorcycles weigh about 200 kg. Try and find out some other ‘items’ that weigh 200 kg and let’s discuss tomorrow why they would need to find the average weight of these 23 items! Create your reasons! Can we debate the ethics of the scenarios you create?

Hearing this narrative, we at Math Space decided to write down the basic recipe for tackling a 2-digit divisor. Here it is:

Since there are so many possible cases, (and later we will see just how many there are), here are some examples. In this article, we will restrict ourselves to 3-digit ÷ 2-digit. The rest can be generalized as discussed later.

There are multiple possibilities for 3-digit ÷ 2-digit involving:

• Rounding up

Example 1: 672 ÷ 19

Think about this: How will this change if the divisor is 17 instead of 19, i.e., 672 ÷ 17?

Example 2: 867 ÷ 16

What if the dividend is 863 instead of 867, i.e., 863 ÷ 16?

• Rounding down

Example 3: 772 ÷ 31

How will this change if the dividend is 779 instead of 772, i.e., 779 ÷ 31?

Example 4: 805 ÷ 21

Find the difference if the dividend is 604 instead of 805, i.e., 604 ÷ 21?

Examples of divisions with single-digit quotients were discussed in the previous article. The following table summarizes all \(2 \times 2 \times (1 + 2) = 12\) possibilities with one numerical case each.

EQ = estimated quotient, FQ = final quotient

Step 1 of the recipe can be modified as follows for bigger divisors:
If the divisor has n-digits, then round it off to the nearest multiple of \(10^n\) .
The rest of the steps remain as they are.
For example, let us consider, \(8397\) ÷ \(365\)

We hope that this will help learners navigate the complexity generated by multi-digit divisors, especially the estimation and the nuances it brings along. Notice that as each digit of the quotient is obtained, the recipe focuses on its place value, something that the teacher chose perhaps, to ignore during her discussion with the students.