Long Division

Section 3.3 Long Division

Algorithm 3.2.2 and Algorithm 3.2.8 provide a way to find the quotient and remainder of a division problem by repeatedly subtracting or adding the divisor. For larger numbers these methods become hard to use.

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Long division, which is often introduced in early mathematics, is an efficient alternative to the division algorithm. In Example 3.3.1, we show how to read the quotient and remainder from this process. We return to long division in Section 11.1, where we cover the decimal representation of integers. There, Example 11.1.4 guides you through the steps of long division.

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Example 3.3.1. \(300 \fdiv 16\) and \(300\fmod 16\) with long division.

With long division we find the quotient and remainder from the division of \(300\) by \(16\text{.}\)

\begin{equation*} \begin{array}{r} 18\\ \mylongdiv{16}{300}\!\\ \underline{16\phantom{0}}\\ 140\\ \underline{128}\\ 12 \end{array} \end{equation*}

So the quotient is \(300 \fdiv 16 = 18\) and the remainder is \(300\fmod 16 = 12\text{.}\)

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The section Calculator Long Division gives a second way to use a basic calculator to find the quotient and remainder of a division problem. This way is often more practical.

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Subsection Calculator Long Division

A calculator can shorten the process of finding \(q=a\fdiv b\) and \(r=a\fmod b\) for an integer \(a\) and natural number \(b\text{.}\) With the methods below, the size of the numbers is limited only by the number of digits your calculator can handle. The video in Figure 3.3.2 introduces this method. A detailed strategy and more examples follow.

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Figure 3.3.2. Long division by Matt Farmer and Stephen Steward

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Now we describe in detail how to find quotients and remainders with a basic calculator.

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Strategy 3.3.1. Calculator Long Division.

Suppose we are given an integer \(a\) and a natural number \(b\text{.}\) We give a strategy for finding the quotient \(a\fdiv b\) and the remainder \(a\fmod b\) with a calculator.

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(a)

To find \(a\fdiv b\) and \(a\fmod b\) we set up the long division problem:

\begin{equation*} \mylongdiv{b}{a} \end{equation*}

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(b)

Compute \(a \div b\) with a calculator.

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(c)

The quotient \(q\) is the biggest integer that is less than or equal to the numerical value of \(a \div b\text{.}\) If \(a\div b\) is an integer, then set \(q:=a\div b\text{,}\) otherwise, \(q\) is the integer to the left of \(a \div b\) on the number line. Place the entire quotient \(q\) on top of the long division:

\begin{equation*} \begin{array}{r} q\\ \mylongdiv{b}{a}\! \end{array} \end{equation*}

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(d)

Multiply \(b \cdot q\) and place that value under the long division:

\begin{equation*} \begin{array}{r} q\\ \mylongdiv{b}{a}\!\\ b\!\cdot\!q \end{array} \end{equation*}

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(e)

Subtract to get the remainder \(r=a-(b\cdot q)\) (using Theorem 3.2.13):

\begin{equation*} \begin{array}{r} q\\ \mylongdiv{b}{a}\!\\ \underline{b\!\cdot\!q}\\ r \end{array} \end{equation*}

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We use calculator long division to carry out the division from Example 3.3.1.

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Example 3.3.3. Divide \(300\) by \(16\) with calculator long division.

We use Strategy 3.3.1 to compute the quotient \(q\) and remainder \(r\) of the division of \(300\) by \(16\text{.}\)

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(a)

Set up the long division problem:

\begin{equation*} \mylongdiv{16}{300}\! \end{equation*}

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(b)

A calculator gives us that \(300 \div 16 = 18.75\text{.}\)

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(c)

The integer on the number line to the left of \(18.75\) is \(18\text{,}\) so we set \(q:=18\text{.}\) Place the entire quotient on top of the long division:

\begin{equation*} \begin{array}{r} 18\\ \mylongdiv{16}{300}\!\\ \end{array} \end{equation*}

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(d)

Multiply \(16 \cdot 18 = 288\) and place that value under the long division:

\begin{equation*} \begin{array}{r} 18\\ \mylongdiv{16}{300}\!\\ 288 \end{array} \end{equation*}

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(e)

Subtract to get the remainder \(r = 300-288 = 12\text{:}\)

\begin{equation*} \begin{array}{r} 18\\ \mylongdiv{16}{300}\!\\ \underline{288}\\ 12 \end{array} \end{equation*}

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We have found \(q=300 \fdiv 16 = 18\) and \(r=300\fmod 16=12\text{.}\)

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This strategy also works for division of a negative integer.

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Example 3.3.4. Divide \(-457\) by \(24\) with calculator long division.

We use Strategy 3.3.1 to compute the quotient \(q\) and remainder \(r\) of the division of \(a=-457\) by \(b=24\text{.}\)

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(a)

Set up the long division problem:

\begin{equation*} \mylongdiv{24}{-457}\! \end{equation*}

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(b)

A calculator gives us that \(-457 \div 24 = -19.0416\dots\text{.}\)

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(c)

The integer on the number line to the left of \(-19.0416\dots\) is \(-20\) so we set \(q := -20\text{.}\) Place the entire quotient on top of the long division:

\begin{equation*} \begin{array}{r} -20\\ \mylongdiv{24}{-457}\! \end{array} \end{equation*}

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(d)

Multiply \(24 \cdot (-20) = -480\) and place that value under the long division:

\begin{equation*} \begin{array}{r} -20\\ \mylongdiv{24}{-457}\!\\ -480 \end{array} \end{equation*}

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(e)

Subtract to get the remainder \(r = -457 - (-480) = 23\text{:}\)

\begin{equation*} \begin{array}{r} -20\\ \mylongdiv{24}{-457}\!\\ \underline{-480}\\ 23 \end{array} \end{equation*}

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Then \(q=-20\) and \(r=23\text{.}\) So we have \(-457 \fdiv 24 = -20\) and \(-457\fmod 24= 23\text{.}\)

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When we divide a positive integer \(a\) by an integer \(b\) that is greater than \(a\) we can easily read off \(a\fdiv b\) and \(a\fmod b\text{.}\) Namely we have \(a\fdiv b=0\) and \(a\fmod b=a\text{.}\) Still, we consider this case an example, to see how our strategy does in this case.

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Example 3.3.5. Divide \(10\) by \(55\) with calculator long division.

We find \(10 \fdiv 55\) and \(10\fmod 55\text{.}\) Let \(a = 10\) and \(b = 55\text{.}\) We use Strategy 3.3.1 to compute the quotient \(q=10\fdiv 55\) and remainder \(r=10\fmod 55\) of the division of \(10\) by \(55\text{.}\)

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(a)

Set up the long division problem:

\begin{equation*} \mylongdiv{55}{10}\! \end{equation*}

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(b)

A calculator gives us that \(10 \div 55 = 0.18\dots\text{.}\)

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(c)

The integer on the number line to the left of \(0.18\dots\) is \(q = 0\text{.}\) Place the entire quotient on top of the long division:

\begin{equation*} \begin{array}{r} 0\\ \mylongdiv{55}{10}\! \end{array} \end{equation*}

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(d)

Multiply \(55 \cdot 0 = 0\) and place that value under the long division:

\begin{equation*} \begin{array}{r} 0\\ \mylongdiv{55}{10}\!\\ 0 \end{array} \end{equation*}

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(e)

Subtract to get the remainder \(r = 10 - 0 = 10\text{:}\)

\begin{equation*} \begin{array}{r} 0\\ \mylongdiv{55}{10}\!\\ \underline{\phantom{0}0}\\ 10 \end{array} \end{equation*}

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Then, \(q = 0\) and \(r = 10\text{,}\) and we have \(10 \fdiv 55 = 0\) and \(10\fmod 55=10\text{.}\)

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When \(a\div b\) is an integer then the quotient \(a\fdiv b\) is equal to \(a\div b\) and the remainder is zero. We show that we also obtain this result using our calculator long division strategy.

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Example 3.3.6. Divide \(480\) by \(160\) with calculator long division.

We find \(480\fdiv 160\) and \(480\fmod 160\text{.}\) We use Strategy 3.3.1.

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(a)

Set up the long division problem:

\begin{equation*} \mylongdiv{160}{480}\! \end{equation*}

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(b)

A calculator gives us that \(480 \div 160 = 3\text{.}\)

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(c)

As \(480 \div 160\) is an integer we set \(q:=480 \div 160 = 3\text{.}\) Place the entire quotient on top of the long division:

\begin{equation*} \begin{array}{r} 3\\ \mylongdiv{160}{480}\! \end{array} \end{equation*}

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(d)

Multiply \(160 \cdot 3 = 480\) and place that value under the long division:

\begin{equation*} \begin{array}{r} 3\\ \mylongdiv{160}{480}\!\\ 480 \end{array} \end{equation*}

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(e)

Subtract to get the remainder \(r:= 480-480=0\text{:}\)

\begin{equation*} \begin{array}{r} 3\\ \mylongdiv{160}{480}\!\\ \underline{480}\\ 0 \end{array} \end{equation*}

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So \(q = 3\) and \(r = 0\text{,}\) and we have \(480 \fdiv 160 = 3\) and \(480\fmod 160=0\text{.}\)

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In practice we often do not explicitly write down all steps of the strategy.

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Example 3.3.7. \(9087 \fdiv 87\) and \(9087\fmod 87\) with calculator long division.

We compute \(9087 \fdiv 87\) and \(9087 \fmod 87\text{.}\) A calculator gives us:

\begin{equation*} 9087 \div 87 = 104.4482...\text{.} \end{equation*}

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The closest integer to the left of \(104.4482..\text{.}\) on the number line is \(104\text{.}\) This is the quotient. Now we use it to compute the remainder:

\begin{equation*} 9087 - 104 \cdot 87 = 39 \end{equation*}

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We have found \(9087 \fdiv 87 = 104\) and \(9087 \fmod 87 = 39\text{.}\)

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We can shorten the process for a negative number in the same way.

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Example 3.3.8. \(-107 \fdiv 72\) and \(-107\fmod 72\) with calculator long division.

We compute \(-107 \fmod 72\) and \(-107 \fdiv 72\) with a calculator. We have

\begin{equation*} -107 \div 72 = -1.48611...\text{.} \end{equation*}

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The integer on the number line to the left of \(-1.48611..\text{.}\) is \(-2\) (starting at -1.48611 we go left on the number line until we find an integer). So \(-2\) is the quotient.

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Now we compute the remainder by subtracting the quotient \(-2\) times \(72\) from \(-107\text{.}\)

\begin{equation*} -107 - (-2)\cdot 72 = -107 - (-144) = -107+144 = 37 \end{equation*}

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We have computed \(-107 \fdiv 72 = -2\) and \(-107 \fmod 72 = 37\text{.}\)

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