- Books Name
- Mathematics Book for CBSE Class 10

- Publication
- Carrier Point

- Course
- CBSE Class 10

- Subject
- Mathmatics

**Euclid’s Division Lemma**

Consider the following folk puzzle.

*A trader was moving along a road selling eggs. An idler who didn’t have*

*much work to do, started to get the trader into a wordy duel. This grew into a*

*fight, he pulled the basket with eggs and dashed it on the floor. The eggs broke.*

*The trader requested the Panchayat to ask the idler to pay for the broken eggs.*

*The Panchayat asked the trader how many eggs were broken. He gave the*

*following response:*

*If counted in pairs, one will remain;*

*If counted in threes, two will remain;*

*If counted in fours, three will remain;*

*If counted in fives, four will remain;*

*If counted in sixes, five will remain;*

*If counted in sevens, nothing will remain;*

*My basket cannot accommodate more than 150 eggs.*

So, how many eggs were there? Let us try and solve the puzzle. Let the number

of eggs be *a*. Then working backwards, we see that *a *is less than or equal to 150:

If counted in sevens, nothing will remain, which translates to *a *= 7*p *+ 0, for

some natural number *p*. If counted in sixes, *a = *6*q+*5, for some natural number *q*.

If counted in fives, four will remain. It translates to *a *= 5*w *+ 4, for some natural

number *w*.

If counted in fours, three will remain. It translates to *a *= 4*s *+ 3, for some natural

number *s*.

If counted in threes, two will remain. It translates to *a *= 3*t *+ 2, for some natural

number *t*.

If counted in pairs, one will remain. It translates to *a *= 2*u *+ 1, for some natural

number *u*.

That is, in each case, we have *a *and a positive integer *b *(in our example,

*b *takes values 7, 6, 5, 4, 3 and 2, respectively) which divides *a *and leaves a remainder

*r *(in our case, *r *is 0, 5, 4, 3, 2 and 1, respectively), that is smaller than *b*. The

* This is modified form of a puzzle given in ‘Numeracy Counts!’ by A. Rampal, and others. moment we write down such equations we are using Euclid’s division lemma,

which is given in Theorem 1.1.

Getting back to our puzzle, do you have any idea how you will solve it? Yes! You

must look for the multiples of 7 which satisfy all the conditions. By trial and error

(using the concept of LCM), you will find he had 119 eggs.

In order to get a feel for what Euclid’s division lemma is, consider the following

pairs of integers:

17, 6; 5, 12; 20, 4

Like we did in the example, we can write the following relations for each such

pair:

17 = 6 × 2 + 5 (6 goes into 17 twice and leaves a remainder 5)

5 = 12 × 0 + 5 (This relation holds since 12 is larger than 5)

20 = 4 × 5 + 0 (Here 4 goes into 20 five-times and leaves no remainder)

That is, for each pair of positive integers *a *and *b*, we have found whole numbers

*q *and *r*, satisfying the relation:

*a *= *bq *+ *r*, 0 £ *r *< *b*

**Note **that *q *or *r *can also be zero.

Why don’t you now try finding integers *q *and *r *for the following pairs of positive

integers *a *and *b*?

(i) 10, 3; (ii) 4, 19; (iii) 81, 3

Did you notice that *q *and *r *are unique? These are the only integers satisfying the

conditions *a *= *bq *+ *r*, where 0 £ *r *< *b*. You may have also realised that this is nothing

but a restatement of the long division process you have been doing all these years, and

that the integers *q *and *r *are called the *quotient *and *remainder, *respectively.

A formal statement of this result is as follows:

**Theorem 1.1 (Euclid’s Division Lemma): ***Given positive integers a and b,*

*there exist unique integers q and r satisfying a = bq + r, 0 *£ *r < b.*

This result was perhaps known for a long time, but was first recorded in Book VII

of Euclid’s Elements. Euclid’s division algorithm is based on this lemma.

Euclid’s division algorithm is a technique to compute the Highest Common Factor

(HCF) of two given positive integers. Recall that the HCF of two positive integers *a*

and *b *is the largest positive integer *d *that divides both *a *and *b*.

Let us see how the algorithm works, through an example first. Suppose we need

to find the HCF of the integers 455 and 42. We start with the larger integer, that is,

455. Then we use Euclid’s lemma to get

455 = 42 × 10 + 35

Now consider the divisor 42 and the remainder 35, and apply the division lemma

to get

42 = 35 × 1 + 7

Now consider the divisor 35 and the remainder 7, and apply the division lemma

to get

35 = 7 × 5 + 0

Notice that the remainder has become zero, and we cannot proceed any further.

**We claim **that the HCF of 455 and 42 is the divisor at this stage, i.e., 7. You can easily

verify this by listing all the factors of 455 and 42. Why does this method work? It

works because of the following result.

So, let us state **Euclid’s division algorithm **clearly.

*To obtain the HCF of two positive integers, say c and d, with c > d, follow*

*the steps below:*

**Step 1: **Apply Euclid’s division lemma, to *c *and *d*. So, we find whole numbers, *q *and

*r *such that *c *= *dq *+ *r*, 0 £ *r *< *d*.

**Step 2: **If *r *= 0, *d *is the HCF of *c *and *d*. If *r *¹ 0, apply the division lemma to *d *and *r*.

**Step 3: **Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

This algorithm works because HCF (*c*, *d*) = HCF (*d*, *r*) where the symbol

HCF (*c, d*) denotes the HCF of *c *and *d*, etc.

**Example 1: **Use Euclid’s algorithm to find the HCF of 4052 and 12576.

**Solution:**

**Step 1: **Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get

12576 = 4052 × 3 + 420

**Step 2: **Since the remainder 420 ¹ 0, we apply the division lemma to 4052 and 420, to

get

4052 = 420 × 9 + 272

**Step 3: **We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get

420 = 272 × 1 + 148

We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get

272 = 148 × 1 + 124

We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get

148 = 124 × 1 + 24

We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get

124 = 24 × 5 + 4

We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get

24 = 4 × 6 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

Notice that 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) =

HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052).

Euclid’s division algorithm is not only useful for calculating the HCF of very

large numbers, but also because it is one of the earliest examples of an algorithm that

a computer had been programmed to carry out.

**Remarks:**

1. Euclid’s division lemma and algorithm are so closely interlinked that people often

call former as the division algorithm also.

2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be

extended for all integers except zero, i.e., *b *¹ 0. However, we shall not discuss this

aspect here.

Euclid’s division lemma/algorithm has several applications related to finding

properties of numbers. We give some examples of these applications below:

**Example 2: **Show that every positive even integer is of the form 2*q*, and that every

positive odd integer is of the form 2*q *+ 1, where *q *is some integer.

**Solution: **Let *a *be any positive integer and *b *= 2. Then, by Euclid’s algorithm,

*a *= 2*q *+ *r*, for some integer *q *³ 0, and *r *= 0 or *r *= 1, because 0 £ *r *< 2. So,

*a *= 2*q *or 2*q *+ 1.

If *a *is of the form 2*q*, then *a *is an even integer. Also, a positive integer can be

either even or odd. Therefore, any positive odd integer is of the form 2*q *+ 1.

**Example 3: **Show that any positive odd integer is of the form 4*q *+ 1 or 4*q *+ 3, where

*q *is some integer.

**Solution: **Let us start with taking *a*, where *a *is a positive odd integer. We apply the

division algorithm with *a *and *b *= 4.

Since 0 £ *r *< 4, the possible remainders are 0, 1, 2 and 3.

That is, *a *can be 4*q*, or 4*q *+ 1, or 4*q *+ 2, or 4*q *+ 3, where *q *is the quotient.

However, since *a *is odd, *a *cannot be 4*q *or 4*q *+ 2 (since they are both divisible by 2).

Therefore, any odd integer is of the form 4*q *+ 1 or 4*q *+ 3.

**Example 4: **A sweet seller has 420 *kaju barfis *and 130 *badam barfis*. She wants to

stack them in such a way that each stack has the same number, and they take up the

least area of the tray. What is the number of that can be placed in each stack for this purpose?

**Solution: **This can be done by trial and error. But to do it systematically, we find

HCF (420, 130). Then this number will give the maximum number of *barfis *in each

stack and the number of stacks will then be the least. The area of the tray that is used

up will be the least.

Now, let us use Euclid’s algorithm to find their HCF. We have:

420 = 130 × 3 + 30

130 = 30 × 4 + 10

30 = 10 × 3 + 0

So, the HCF of 420 and 130 is 10.

Therefore, the sweet seller can make stacks of 10 for both kinds of *barfi*.