# Remainder

In integer division, the **remainder** is the amount that is left over after dividing one integer by another. In polynomial division, the remainder is the polynomial that is left over after dividing one polynomial by another.

## Integer Division

An integer \(a\) is *divisible* by another integer \(b\) \((\)or is a *multiple* of \(b)\) if \(a\) can be written as \(b\) times another integer:

\[ a = b \times \mbox{(integer)}.\]

For example, \(14 = 7 \times 2 = 2 \times 7,\) so \(14\) is divisible by \(2\) and \(7\). However, dividing 14 by any other integer greater than 1 does not produce an integer result. For example, \( 14 \div 4 \) leaves a remainder because 4 does not go evenly into 14. The multiples of the **divisor** \(4\) are

\[4, 8, 12, 16, 20, \ldots, \]

so \( 12 \div 4 = 3 \) is the largest number of fours that go into 14. However, since \( 12 < 14\), we have \( 14 - 12 = 2 \) left over, which we call a "remainder of 2." We say that \( 14 \div 4 = 3 \text{ R } 2 \), read as "fourteen divided by four equals three remainder 2."

Note that the remainder will always be less than the divisor; in this case, dividing by 4 will always leave a remainder that is either 0, 1, 2, or 3. To find the remainder of a number \(n\) upon division by a divisor \(d\), we first find the largest multiple of \(d\) that goes into \(n\), and the remainder \(r\) is the amount left over:

\[ n = kd + r \mbox{ with } r < d. \]

What is the remainder when 48 is divided by 7?

To find the remainder when 48 is divided by 7, we consider the multiples of 7:

\[ \begin{align} 7 \times 1 &= 7\\ 7 \times 2 &= 14\\ 7 \times 3 &= 21\\ 7 \times 4 &= 28\\ 7 \times 5 &= 35\\ 7 \times 6 &= 42\\ 7 \times 7 &= 49. \end{align} \]

Since \( 7 \times 6 = 42 \) and \( 7 \times 7 = 49 \), only six 7's can go into 48. Then the amount left over is \( 48-42=6 \). Thus the remainder when 48 is divided by 7 is 6. \( _\square \)

## Visual Interpretation

One way to think about remainders is to consider dividing up squares into groups. For instance, in the above example, to find \( 14 \div 4 \), start with 14 squares and line them up 4 at a time. If we stop at 4, 8, or 12 squares, the squares would form a rectangle:

However, since we have 14 squares, there are 2 squares left over:

The leftover squares are the remainder and show why the division of 14 by 4 leaves a remainder of 2. Note that this also shows why the remainder is always either 0, 1, 2, or 3 upon dividing by 4.

An

evennumber is an integer that leaves a remainder of \(0\) upon division by \(2\). Anoddnumber is an integer that leaves a remainder of \(1\) upon division by \(2\). Use the above visual interpretation to explain why the following patterns are true:\[\begin{align} \mbox{(odd number)} + \mbox{(odd number)} &= \mbox{even number}\\ \mbox{(even number)} + \mbox{(odd number)} &= \mbox{odd number}\\ \mbox{(even number)} + \mbox{(even number)} &= \mbox{even number}.\\ \end{align}\]

Since an odd number leaves a remainder of \(1\) upon division by \(2\), if we line up an odd number of squares into rows of 2 squares each, there will be one square left over. Now, if we consider combining two such groups, we have the following pattern:

The result is a rectangle with no squares left over, i.e. an even number.

Similarly, if we combine an odd number of squares (with one leftover square) with an even number of squares (with no leftover squares), we have the following pattern:

This has one square left over, which shows the result is an odd number.

Finally, if we combine an even number of squares with another even number of squares (both with no leftover squares), we have the following pattern, which results in no leftover squares:

## Problem Solving

If \(n\) is an integer that leaves a remainder of 2 upon division by 6, what is the remainder of \(n\) upon division by 3?

Since \(n\) leaves a remainder of 2 upon division by 6, we have

\[ n = 6 k + 2,\]

where \(6k\) is the largest multiple of \(6\) that goes into \(n\). Now, this can also be written as

\[ n = 3 (2k) + 2,\]

and since \(2 < 3\), we have \(3 (2k) \) is the largest multiple of \(3\) that goes into \(n\). This shows that the remainder of \(n\) upon division by \(3\) is also \(2\). \(_\square\)

If \(n\) is an integer that leaves a remainder of 4 upon division by 6, what is the remainder of \(n\) upon division by 3?

Note that this example is similar to the above problem, but with a larger remainder. Now, since \(n\) leaves a remainder of 4 upon division by 6, we have

\[ n = 6 k + 4,\]

where \(6k\) is the largest multiple of \(6\) that goes into \(n\). This can also be written as

\[\begin{align} n &= 3 (2k) + 4\\ &= 3(2k + 1) + 1.\\ \end{align}\]

In comparison to the example above, now \(3(2k+1)\) is the largest multiple of \(3\) that goes into \(n\). This shows that the remainder of \(n\) upon division by \(3\) is \(1\). \(_\square\)

## Polynomial Division

Main Article: Polynomial Division

We have seen that to divide an integer \(n\) by another non-zero integer \(d,\) we find the largest multiple of \(d\) that goes into \(n\) and the remainder \(r\) is the amount left over:

\[ n = kd + r. \]

This idea can also be applied to polynomials. That is, we can divide one polynomial \(f(x)\) by another polynomial \(g(x)\) \(\big(\)for \(g(x)\) non-zero\(\big).\) Then there exists a unique polynomial quotient \(q(x)\) and remainder \(r(x)\) such that

\[f(x)=g(x)\cdot q(x)+r(x).\]