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? 

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

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

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

Short division

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

PROCEDURE

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

Example 2

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

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

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

​$$\underline{48\div4=12}$$​​

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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: $$51\div2$$.

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

​$$2\times5 = 10$$​​

​$$11- 10 = 1$$​​

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

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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.