Euclid's Division Lemma

Go back to  'Arithmetic-Integers'

What is Euclid's Division Lemma?

Euclid’s Division Lemma (lemma is like a theorem) says that given two positive integers a and b, there exist unique integers q and r such that:

\[a = bq + r,\;0 \le r < b\]

The integer q is the quotient, and the integer r is the remainder. The quotient and remainder are unique. In simple words, this lemma says that whenever you divide an integer by another (non-zero) integer, you will get a unique quotient integer and a unique remainder integer.

Here are some examples of the application of this lemma:

\[\begin{align}&\underline {17}  = \underline 3 \left( 5 \right) + 2\\&\underline {24}  = \underline 7 \left( 3 \right) + 3\\&\underline {61}  = \underline {11} \left( 5 \right) + 6\\&\;\;\underline 7  = \underline {12} \left( 0 \right) + 7\end{align}\]

In each line, the underlined numbers are the dividend and divisor respectively, the number in brackets is the quotient, while the last number is the remainder. Note that the remainder is always less than the divisor.

A visual model for Euclid's Division Lemma

Consider an arbitrary integer, say 37. Suppose that you divide this by 7. What will be the quotient and the remainder? We have:

\[37 = 7\left( 5 \right) + 2\]

Clearly, the quotient is 5 and the remainder is 2. Let us interpret this physically. Let us divide the number line into blocks of 7 units as shown below:

Numbers and number blocks

We have named each block as shown in the figure. In which block will 37 lie? It will lie in B5, which means that between this block and the number 0, there are 5 intervening blocks (B0 to B4). This corresponds to the quotient.

What is the interpretation of the remainder? For that, zoom onto block B5, which will of course be 7 units long:

Example 1: Numbers and number blocks

Cleary, 37 lies 2 units from the start of its block (which is B5). This corresponds to the remainder.

Let us repeat this process again. Suppose that we have to divide 80 by 9. This time, we divide the number line into blocks of 9 units each. The number 80 will lie in block B8, 8 units from the start of its block:

Example 2: Numbers and number blocks

Thus, the quotient in this case is 8, and the remainder is 8.

Finally, let us generalize this. Let a and b be any positive integers. Suppose that we divided a by b. We will obtain a quotient q and a remainder r. We can write this fact as follows:

\[a = bq + r,\;0 \le r < b\]

Physical interpretation: we divide the number line into blocks of b units each. Then, the number a will lie in block number q, at a distance of r units from the start of the block:

Example 3: Numbers and number blocks

This is, as we have seen earlier, Euclid’s Division Lemma: Given positive integers a and b, there exist unique integers q and r such that:

\[a = bq + r,\;0 \le r < b\]

That the quotient and remainder are unique should be obvious. Also, the remainder must be less than the divisor. Can you see why? Refer to the last figure above. Note how each block is of length b units, and how the remainder can only take values from a minimum of 0 (start of the block) to a maximum of b – 1 (one unit less than the end of the block).

A question for you: does Euclid’s Division Lemma hold if a is a negative integer?

Examples on Euclid's Division Lemma

Example 1. An odd positive integer n is written as follows:

\[n = 4k + m\]

What values can m take?

Solution. If you divide n by 4, we will have (by Euclid’s Division Lemma):

\[n = 4k + m,\;m = 0,\;1,\;2,\;3\]

Thus, m can take four values: 0, 1, 2, or 3. However, it should be clear that if m is 0 or 2, then n will be even, whereas we want n to be odd. This means that m can take only two values: 1 and 3. In other words, any odd positive integer n can be written as

\[n = 4k + 1\;\;\;\;\;{\rm{or}}\,\,\,\,\,\,\,\,\,\,\,n = 4k + 3\]

Example 2.  Let n be an odd integer. Show that \({n^2} - 1\) is a multiple of 8.

Solution. By Euclid’s Division Lemma, n can be written as

\[n = 2k + 1,\;k \in \mathbb{Z}\]


\[\begin{align}&{n^2} - 1 \;= {\left( {2k + 1} \right)^2} - 1\\& \qquad\quad= \left( {4{k^2} + 4k + 1} \right) - 1\\& \qquad\quad= 4{k^2} + 4k\\&\qquad\quad = 4k\left( {k + 1} \right) & \end{align}\]

Note that the product \(k\left( {k + 1} \right)\) will always be even. This is because either k is even, or if not, then k + 1 is even. Thus, n must be a multiple of 8.

Example 3. The cube of a positive integer is divided by 9. What can be the possible remainders?

Solution. This is an interesting problem. Somehow, we have to use Euclid's Division Lemma, and we have to figure out how. If we can express the cube of any positive integer in the form 9q + r, then our problem simplifies: we then have to find what values r can take.

Let n be any positive integer. Note that on division of n by 3, we can get three possible remainders: 0, 1 or 2. Thus, n can be written as:

\[\begin{align}&n = 3k\\&\quad{\rm{or}}\\&n = 3k + 1\\&\quad{\rm{or}}\\&n = 3k + 2\end{align}\]

We consider each case separately.

Case 1: In this case, we have:

\[{n^3} = {\left( {3k} \right)^3} = 27{k^3} = 9\left( {3{k^3}} \right)\]

Thus, the cube of n will be perfectly divisible by 9 in this case (the remainder will be 0).

Case 2: In this case, we have:

\[\begin{align}&{n^3} = {\left( {3k + 1} \right)^3}\\&\quad = {\left( {3k} \right)^3} + {\left( 1 \right)^3} + 3\left( {3k} \right)\left( 1 \right)\left( {3k + 1} \right)\\&\quad= 27{k^3} + 1 + 9k\left( {3k + 1} \right)\\&\quad = 9k\left( {3{k^2} + 3k + 1} \right) + 1\\&\quad = 9\left\{ {k\left( {3{k^2} + 3k + 1} \right)} \right\} + 1 & \end{align}\]

Clearly, on dividing the cube of n by 9 in this case, we will be left with a remainder of 1.

Case 3: In this case, we have:

\[\begin{align}&{n^3} = {\left( {3k + 2} \right)^3}\\&\quad= {\left( {3k} \right)^3} + {\left( 2 \right)^3} + 3\left( {3k} \right)\left( 2 \right)\left( {3k + 2} \right)\\&\quad = 27{k^3} + 8 + 18k\left( {k + 2} \right)\\&\quad= 9k\left( {3{k^2} + 2k + 4} \right) + 8\\&\quad = 9\left\{ {k\left( {3{k^2} + 2k + 4} \right)} \right\} + 8 & \end{align}\]

On dividing the cube of n by 9 in this case, we will evidently be left with a remainder of 8.

Thus, we conclude that dividing the cube of a positive integer by 9 can yield three possible remainders: 0, 1 and 8. In other word, the cube of any integer can be expressed as follows:

\[\begin{align}&{n^3} = 9k\\&\quad\;{\rm{or}}\\&{n^3} = 9k + 1\\&\quad\;{\rm{or}}\\&{n^3} = 9k + 8\end{align}\]

You may verify this result by taking some cubes and checking the remainders you obtain upon their division by 9.

Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum