Skip to main content
Logo image

Section 3.1 Quotients and Remainders

To find how often a natural number \(b\) goes into a natural number \(a\text{,}\) we subtract \(b\) from \(a\text{.}\) Then we subtract \(b\) from the result, subtract \(b\) from that result, and continue. We repeat this process until \(b\) no longer goes into the previous result. This happens when the previous result is less than \(b\text{.}\) The number of times we have subtracted \(b\) is called the quotient (of the division of \(a\) by \(b\)). The number left from \(a\) after subtracting \(b\) as often as possible is called the remainder (of the division of \(a\) by \(b\)).
We show the process described above with an example.

Example 3.1.1. How often does \(7\) go into \(26\) ?

To find out how often \(7\) goes into \(26\) we repeatedly subtract \(7\text{.}\) We stop this process when \(7\) no longer goes into the result of the previous subtraction. In other words, we stop when we reach a number that is less than seven. We get:
\begin{align*} 26 - 7 \amp = 19\\ 19 - 7 \amp = 12\\ 12 - 7 \amp = 5 \end{align*}
We stop here because \(7\) does not go into \(5\text{.}\) We have subtracted \(7\) three times. This means that \(7\) goes into \(26\) three times. So the quotient is \(3\text{.}\) We have \(5\) left over. This is the remainder. We have found that
\begin{equation*} 26 - 3\cdot 7 = 5 \mbox{ or } 26 = 3\cdot 7+5\text{.} \end{equation*}
When we divide a positive integer \(a\) by an integer \(b\text{,}\) \(q\) is the number of times \(b\) goes into \(a\text{.}\) So \(q\) is the quotient when \(a\) is divided by \(b\text{.}\) Because \(b\) goes into \(a\) exactly \(q\) times and \(b\) cannot go into \(a\) more than \(q\) times, we have
\begin{equation*} a-q\cdot b\lt b \end{equation*}
So the remainder is \(r:=a-q\cdot b\) with \(0\le r \lt b\text{.}\) We have
\begin{equation*} a=q\cdot b+r. \end{equation*}
If \(a\) is written in the form \(a=q\cdot b+r\) with \(0\le r \lt b\text{,}\) then it is easy to find the quotient and remainder when \(a\) is divided by \(b\text{.}\)

Example 3.1.2.

We have
\begin{equation*} 283416405986=490687\cdot 577590+501656. \end{equation*}
We find the quotient and remainder of the division of \(283416405986\) by \(577590\text{.}\)
To match the notation from above we set \(a:=283416405986\) and \(b:=577590\text{.}\) With this notation the equation above becomes
\begin{equation*} a=490687\cdot b+501656. \end{equation*}
As \(501656 \lt 577590\) this is of the form \(a=q\cdot b+r\) with \(0\le r \lt b\text{.}\) Now we read off the quotient \(q=490687\) and the remainder \(r=501656\text{.}\)
When we divide a positive integer \(a\) by an integer \(b\) that is greater than \(a\) then \(b\) does not go into \(a\text{.}\) So \(b\) goes into \(a\) zero times and what is left over, the remainder, is \(a\text{.}\)

Example 3.1.3. How often does \(7\) go into \(3\) ?

Clearly \(7\) does not go into \(3\text{,}\) that is, \(7\) goes into \(3\) zero times. We have \(3\) left over. So the remainder is \(3\text{.}\)
Watch the video in Figure 3.1.4 for a review of what we have done so far and to see more examples.
Figure 3.1.4. Quotients and Remainders by Matt Farmer and Stephen Steward
We write the process above as an algorithm called the Division Algorithm, and then extend it to negative integers. We introduce the operations \(\fdiv\) and \(\fmod\) as notation for quotients and remainders. We also describe how to do integer division with a calculator. We study properties of the operation \(\fmod\) and give an example of using it to validate ISBN numbers.