Like in the case of multiplication, division of numbers is generally done with either short division, or long division.
Here we look at how to effectively approach dealing with short division examples.
Short Division Method
If we wanted to divide 462 by 2. 462 ÷ 2462 is the DIVIDEND , 2 is the DIVISOR
The eventual answer to the sum is the QUOTIENT.
Generally a division sum is laid out in the following way:
QUOTIENT
DIVISOR DIVIDEND
So for 462 ÷ 2, we set up as follows.
\begin{array}{r} \\[-2pt] 2 {\bf{|}} {\overline{\space 462 \space\space}} \end{array}
Then carry out the division in stages from left to right, to find the answer, the QUOTIENT.
1)
2 goes into 4 twice with no remainder, so a 2 goes above the 4 in the QUOTIENT section.
\begin{array}{r} 2 \space\space\space\space\space\space\\[-2pt] 2 {\bf{|}} {\overline{\space 462 \space\space}} \end{array}
2)
2 goes into 6 three times with no remainder, so next we have a 3 in the QUOTIENT above the 6
\begin{array}{r} 23 \space\space\space\space\\[-2pt] 2 {\bf{|}} {\overline{\space 462 \space\space}} \end{array}
3)
Lastly 2 goes into 2 once, so a 1 is the final entry in the QUOTIENT, above the 2.
\begin{array}{r} 231 \space\space\\[-2pt] 2 {\bf{|}} {\overline{\space 462 \space\space}} \end{array}
Now have the answer to the division sum. 462 ÷ 2 = 231
Short Division Examples
1.1
660 ÷ 4 ?
Solution
\begin{array}{r} \\[-2pt] \bf 4 {\bf{|}} {\overline{\space 660 \space\space}} \end{array}
1)
4 goes into 6 once, with a remainder of 2.
When this happens, the 1 goes above in the answer, while the remainder 2 gets moved alongside the next number to the right in the dividend, which here is 6
The effect of the 2 is that it turns 6 into 26 .
\begin{array}{r} 1 \space\space\space\space\space\space\space\\[-2pt] 4 {\bf{|}} {\overline{\space 6{\tiny{2}}60 \space\space}} \end{array}
2)
Now 4 goes into 26 six times, with a remainder of 2.
\begin{array}{r} 1\space6 \space\space\space\space\space\\[-2pt] 4 {\bf{|}} {\overline{\space 6{\tiny{2}}6{\tiny{2}}0 \space\space}} \end{array}
3)
Lastly 4 goes into 20 exactly five times, with no remainder.
\begin{array}{r} 1\space6 \space5 \space\space\\[-2pt] 4 {\bf{|}} {\overline{\space 6{\tiny{2}}6{\tiny{2}}0 \space\space}} \end{array}
660 ÷ 4 = 165
1.2
232 ÷ 8 ?
Solution
\begin{array}{r} \\[-2pt] \bf 8 {\bf{|}} {\overline{\space 232 \space\space}} \end{array}
1)
8 doesn’t go into 2, so 0 goes above in the answer.
\begin{array}{r} 0 \space\space\space\space\space\space\\[-2pt] 8 {\bf{|}} {\overline{\space 232 \space\space}} \end{array}
2)
We now consider how many times 8 goes into 23, which is two times, with a remainder of 7.
\begin{array}{r} 02 \space\space\space\space\space\\[-2pt] 8 {\bf{|}} {\overline{\space 23{\tiny{7}}2 \space\space}} \end{array}
3)
Lastly 8 goes into 72 exactly eight times, with no remainder.
\begin{array}{r} 02\space8 \space\space\\[-2pt] 8 {\bf{|}} {\overline{\space 23{\tiny{7}}2 \space\space}} \end{array}
660 ÷ 4 = 165
Short Division with Remainder
Sometimes with short division examples, final answers can have a remainder rather than working out as exact whole numbers.When this happens the remainder is written as part of the final answer.
Example
2.1
397 ÷ 3 ?
Solution
\begin{array}{r} \\[-2pt] 3 {\bf{|}} {\overline{\space 397 \space\space}} \end{array}
1)
3 goes into 3 once, with no remainder.
\begin{array}{r} 1 \space\space\space\space\space\space\\[-2pt] 3 {\bf{|}} {\overline{\space 397 \space\space}} \end{array}
2)
3 goes into 9 three times, with no remainder.
\begin{array}{r} 13 \space\space\space\space\\[-2pt] 3 {\bf{|}} {\overline{\space 397 \space\space}} \end{array}
3)
3 goes into 7 two times, but there is a remainder of 1.
The 3 goes above in the answer, while the remainder of 1 is part of the final answer.
\begin{array}{r} 132 \space{\tiny{r1}}\space\\[-2pt] 3 {\bf{|}} {\overline{\space 397 \space\space}} \space\space \end{array}
397 ÷ 3 = 132 r1