Division of two-digit numbers by one digit number
A simple division procedure in arithmetic is performed by doing repeated subtraction of the divisor from the dividend, till we reach 0 or a number which is less than the divisor. It allows you to manually divide multi-digit numbers. The result obtained is called a quotient.
What are partial quotients?
A partial quotient is one of the quotients that we obtain when using the method of solving large division problems. In this process the final quotient can be found by adding up each partial quotient in the repeated subtraction division method. The relation between the dividend, divisor, quotient and remainder can be seen as follows:
Quotient \( \times \) Divisor + Remainder = Dividend
How to write the partial quotients?
Solving simple division problems using the partial quotients method (also known as chunking) employs repeated subtraction. Here are three steps that can help us understand the process:
- Step 1: Deduct an easy multiple of the divisor from the dividend
(for example, \( \times~100, \times~10, \times~5, \times~2, \) etc.).
- Step 2: Subtract the large number until the result equals zero or the remainder is less than the divisor.
- Step 3: To find the answer to the division, add up the divisor’s multipliers (partial quotients) used in repeated subtraction.
Let’s understand this through an example.
Divide 86 with 2
Solution:
We need to find \( 86\div~2 \),
Step 1: Remove 80 from 86 and when we do that, 6 is left. Now let us divide 80. Subtract multiples of 2 until you get to zero. Because \( 2\times~40~=~80 \), a multiple of 2 that equals 80 is 40.
Step 2: Now let us divide 6.
Because \( 2\times~3~=~6 \), you can use 3. You’ve nothing left.
Step 3: You divided 86 into two parts, 80 and 6. Now, add up the number of times we added 2 to get to 80 and 6 respectively.
For \( 80\rightarrow 40\) and for \( 6\rightarrow 3\)
And, 40 + 3 = 43
Therefore, \( 86\div~2~=~43 \)
As a result, \( 86\div~2~=~43 \).
Using regrouping to divide two digit numbers by one digit numbers:
Find \( 75\div~3\) using regrouping.
75 consists of 7 ‘tens’ and 5 ‘ones’.
Step 1: We have to divide the ‘tens’ at first.
Division: 7 ‘tens’ \( \div~3\)
Multiplication: 2 ‘tens’ \( \times~3\)
Subtraction: 7 ‘tens’ – 6 ‘tens’
There is ‘1’ ten left after dividing the ‘tens’.
Step 2: We have to do a regrouping of 1 ten as 10 ‘ones’ and add the remaining ‘ones’ from 75 (5 ‘ones’).
10 ‘ones’ +5 ‘ones’ = 15 ‘ones’
Step 3: We have to divide the ‘ones’.
Division: = 15 ‘ones’ \( \div~3\)
Multiplication: = 5 ‘ones’ \( \times~3\)
Subtraction: = 15 ‘ones’ – 15 ‘ones’
There are ‘0’ ‘ones’ left after dividing the ‘ones’.
Therefore, \( 75\div~3~=\)
= 2 ‘tens’+5 ‘ones’
= 25
- Finding fewer partial quotients (with larger numbers divided in each round) is more efficient but any combination is correct as long as the sum of the partial quotients remains the same.
Hence, the partial quotients are 20, 5. The remainder is ‘0’, which means that 3 divides 75 completely and hence the quotient is 25.
Solved Examples
1. For a party, Mary purchased 34 favors. She wants to split the favors evenly between the two tables. What is the maximum number of favors Mary can place on each table?
Solution: Let’s divide \( 34\div~2\) using regrouping.
We have to think that 34 is 3 ‘tens’ and 4 ‘ones’.
Step 1: we have to divide the ‘tens’ at first.
Division: 3 ‘tens’ \(\div~2\)
Multiplication: 1 ten \(\times~2\)
Subtraction: 3 ‘tens’ – 2 ‘tens’
There is ‘1’ ten left after dividing the ‘tens’.
Step 2: We have to do a regrouping of 1 ten as 10 ‘ones’.
10 ‘ones’ + 4 ‘ones’ = 14 ‘ones’
Step 3: We have to divide the ‘ones’.
Division: 14 ‘ones’ \(\div~2\)
Multiplication: 7 ‘ones’ \(\times~2\)
Subtraction: 14 ‘ones’ – 14 ‘ones’
There are ‘0’ ‘ones’ left after dividing the ‘ones’.
Therefore,
\(34\div~2\)
= 1 ‘tens’ +7 ‘ones’
= 17
Hence, the partial quotient is 10 + 7. The remainder is 0 and the quotient is 17.
Therefore, the maximum number of favors Mary can place on each table is 17.
2. In preparation for a race, Joseph biked 72 miles in 3 days. Every day, Joseph rode the same number of miles. Joseph rode his bike for how many miles each day?
Solution:
Let’s divide \(72\div~3\) using regrouping.
We have to think that 72 is 7 ‘tens’ and 2 ‘ones’.
Step 1:
We have to divide the ‘tens’ first.
Division: 7 ‘tens’ \(\div~3\)
Multiplication: 2 ‘tens’ \(\times~3\)
Subtraction: 7 ‘tens’ – 6 ‘tens’
There is ‘1’ ten left after dividing the ‘tens’.
Step 2: We have to do a regrouping of 1 ten as 10 ‘ones’.
10 ‘ones’ + 2 ‘ones’ = 12 ‘ones’
Step 3:
Division: 12 ‘ones’ \(\div~2\)
Multiplication : 4 ‘ones’ \(\times~3\)
Subtraction : 12 ‘ones’ – 12 ‘ones’
There are ‘0’ ‘ones’ left after dividing the ‘ones’.
Therefore,
\(72\div~3\)
= 2 ‘tens’ + 4 ‘ones’
= 24
Hence, the partial quotient is 20 + 4. The remainder ‘0’and the quotient is 24.
Therefore, Joseph rode his bike for 24 kilometers each day.
3. John’s farm produced 92 pounds of strawberries. To get his strawberries to the stores, he divides them among four trucks. How many pounds of strawberries can he put inside each truck?
Solution: Let’s divide \(92\div~4\) using regrouping.
We have to think that 92 is 9 ‘tens’ and 2 ‘ones’.
Step 1:
We have to divide the ‘tens’ first.
Division : 9 ‘ones’ \(\div~4\)
Multiplication : 2 ‘ones’ \(\times~4\)
Subtraction : 9 ‘ones’ – 8 ‘ones’
There are ‘1’ ten left after dividing the ‘ones’.
Step 2: We have to do a regrouping of 1 ten as 10 ‘ones’.
10 ‘ones’ + 2 ‘ones’ = 12 ‘ones’
Step 3:
We have to divide the ‘ones’.
Division : 12 ‘ones’ \(\div~2\)
Multiplication : 3 ‘ones’ \(\times~4\)
Subtraction : 12 ‘ones’– 12 ‘ones’
There are ‘0’ ‘ones’ left after dividing the ‘ones’.
Therefore,
\(92\div~4\)
= 2 ‘tens’ + 3 ‘ones’
= 23
Hence, the partial quotient is 20+3. The remainder ‘0’and the quotient is 23.
Therefore, in each truck, 23 pounds of strawberries can be stored.
Long division of numbers is simple to carry out when partial quotients are used. The quotient and remainder will be displayed as a fraction of the second number after the partial quotient method division of the numbers has been carried out. When dividing large numbers, partial quotients come in handy. You can break the problem down into smaller chunks and simplify the division by using partial quotients. The total is then calculated by adding all of the bits back together.
When performing operations like addition and subtraction with two-digit numbers or larger, regrouping is defined as the process of forming groups of ‘tens’. Regrouping is the process of rearranging numbers into groups in order to complete an operation. When the digits in the minuend are smaller than the digits in the same place in the subtrahend, regrouping is used.