# Dividing a 2-digit number by a 1-digit number

## In a nutshell

Division is the opposite of multiplication. Two-digit numbers are whole numbers between $10$ and $99$. They can primarily be divided by a one-digit number using two methods: chunking and short division.

## Chunking

The chunking method uses multiplication to quickly perform repeated subtraction. The number of total subtractions will be the answer.

##### Example 1

*Andy wants to share *$45$* oranges among *$5$* people. How many oranges will each person get? *

$\begin{aligned}5{\overline{\smash{)}4~\!5}}\\&&-[5\times3=15]\\30\\&&-[5\times4=20]\\10\\&&-[5\times2=10] \\0 \end{aligned}$

$\underline{3+4+2 = 9 \space oranges\space each }$

*Note: **Keep subtracting until you reach *$0.$

## Short division

Short division is also known as the "bus stop" method. To perform short division:

#### PROCEDURE

- Write the dividend (first number) under the "bus stop" and the divisor (second number) as follows: $48\div4 \rightarrow 4\overline{\smash{)}48}$
- Starting from the left-hand side of the number being divided, find how many times the divisor goes into the first digit.

- Write the answer above the "bus stop". Then repeat for the next digit.

##### Example 2

*Use the "bus stop" method to solve: *$48\div4$*.*

$4\overset{\,\,1\,\,\,\,2}{\overline{\smash{)}4~\!8}}$

$4\div 4 = 1 \space and \space 8\div4 =2$

$\underline{48\div4=12}$

## Remainders

A one-digit number does not always fit exactly into a two-digit number. The amount which is left over after division is known as a remainder.

##### Example 3

*Use the chunking method to solve: *$52\div8.$

$\begin{aligned}6 {\overline{\smash{)}5~\!2}}\\&&-[6\times5=30]\\22\\&&-[6\times3=18]\\4 \end{aligned}$

$5 +3 = 8$

$\underline{52\div6 = 8\space remainder\space 4}$

##### Example 4

*Use short division to solve: $51\div2$.*

*$2\overset{\,\,2\,\,\,\,5}{\overline{\smash{)}5~^1\!1}}$*

$2\times5 = 10$

$11- 10 = 1$

$\underline{51\div2 = 25\space remainder\space 1}$

**Note:** In short division, carry over remainders until the last digit. Then find the largest whole division you can perform. The left over is the remainder.