# Long Division Using Partial Quotients

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Basics on the topic
**Long Division Using Partial Quotients **

## Long Division Using Partial Quotients – Introduction

Long division can seem like a complex process, but it becomes more manageable when we break it down into smaller steps. The method of **long division using partial quotients** simplifies the task by allowing us to take it one piece at a time, making it a user-friendly approach for students learning division.

## Long Division with Partial Quotients – Explanation

**Long division with partial quotients** is a systematic way of dividing larger numbers (the dividend) by smaller numbers (the divisor) to find the quotient and remainder. This method is similar to standard long division but focuses on estimating and subtracting multiples of the divisor from the dividend.

The **partial quotients method** involves dividing the dividend into parts that the divisor can easily multiply into, then subtracting these parts from the dividend step by step until what's left is too small to be divided by the divisor. The final quotient is the sum of all these partial quotients, and any remaining number becomes the remainder.

This method is particularly helpful for students who are intimidated by the traditional long division algorithm or have difficulty with multiplication facts.

## Long Division Using Partial Quotients – Example

Let's solve a division problem using partial quotients with the following example:

**Example: Divide 452 by 3 using the partial quotients method.**

Step | Action | Result |
---|---|---|

1. Set up the problem | Place $452$ inside the division box and $3$ outside. | |

2. Estimate a first partial quotient | How many times does $3$ go into $452$? Estimate $100$. | $3 \times 100 = 300$ |

3. Subtract this first product from $452$ | $452 - 300 = 152$ | Remaining $152$ |

4. Estimate a second partial quotient | $3$ goes into $152$ about $50$ times. | $3 \times 50 = 150$ |

5. Subtract this second product | $152 - 150 = 2$ | Remaining $2$ |

6. Finish with a remainder | $2$ is too small for $3$ to go into, so it remains. | Remainder $2$ |

7. Add the partial quotients | Sum of $100$ and $50$ gives the final quotient. | Final quotient $150$ |

The answer is 150 with a remainder of 2, which we write as 150 R2.

## Long Division Using Partial Quotients – Guided Practice

Let's go through another example together:

**Example: Divide 627 by 4 using the partial quotients method.**

Step | Action | Result |
---|---|---|

1. Set up the problem | Place $627$ inside the division box and $4$ outside. | |

2. Estimate and subtract | Consider $4$ multiplying into $627$. Start with $100$. | $4 \times 100 = 400$ |

3. Subtract the first product from $627$ | $627 - 400 = 227$ | Remaining $227$ |

4. Estimate the next partial quotient | $4$ goes into $227$ about $50$ times. | $4 \times 50 = 200$ |

5. Subtract the second product | $227 - 200 = 27$ | Remaining $27$ |

6. Continue with smaller partial quotients | $4$ goes into $27$ about $6$ times. | $4 \times 6 = 24$ |

7. Subtract and find the remainder | $27 - 24 = 3$, which is our remainder. | Remainder $3$ |

8. Add the partial quotients | Sum of $100$, $50$, and $6$ gives the final quotient. | Final quotient $156$ |

Thus, 627 divided by 4 is 156 with a remainder of 3, or 156 R3.

*Try some division on your own using this method!*

## Long Division Using Partial Quotients – Summary

**Key Learnings from this Text:**

- Partial quotients break down the division process into smaller, more manageable steps.
- This method requires estimation and subtraction to find the quotient.
- The final answer is the sum of the partial quotients, with any leftover amount as the remainder.
- It's important to go through the steps carefully to ensure accuracy.

Dive into other division strategies and boost your math skills with our interactive practice problems, engaging videos, and helpful worksheets on our educational platform.

## Long Division Using Partial Quotients – Frequently Asked Questions

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Transcript
**Long Division Using Partial Quotients **

The victory in the Great Beetle War would not have been possible without the help of hometown heroes like Airman Squeaks, who used the strategy of partial quotients to help deliver groups of supplies behind enemy lines. Another method for solving long-division problems is to use partial quotients. This method divides a problem into smaller, more manageable parts using multiples of ten or simple numbers. We'll use Airman Squeaks' supply deliveries to demonstrate how to divide three and four-digit numbers. He has four hundred thirty-two corn kernels to take to six battalions. We need to calculate how many kernels would be given to each. We set up the problem exactly as we would with the traditional long-division method. First, estimate how many times six goes into four hundred thirty-two. Ask yourself, what multiple of ten can we multiply by six to get close to this number? Six times seventy equals four hundred and twenty. Seventy is a partial quotient and goes here. Four hundred twenty is known as the partial dividend and is written here. Now we subtract the partial dividend from the original dividend. Subtract four hundred thirty-two from four hundred twenty to get twelve. Now ask ourselves, what do we multiply six by to get twelve? The partial quotient two goes here, and the partial dividend twelve goes here. We subtract again and get zero. We add the partial quotients to get the final answer. Seventy-two. Airman Squeaks will deliver seventy-two corn kernels to each battalion. He must also ensure that the troops are adequately supplied with insect repellent. He has seven hundred sixty-five cans that need to be delivered to five different locations. We set up the problem like this. What multiple of ten can we multiply five by to get close to seven hundred, sixty-five? Although there are many multiples we could choose, here we can say five times one hundred equals five hundred. What do we get when we subtract five hundred from seven hundred sixty-five? Two hundred, sixty-five. What multiple of ten can we multiply five by to get close to two hundred sixty-five? Five times fifty equals two hundred fifty. Subtract again. What is left? Fifteen. How many groups of five are in fifteen? Three. Three times five is fifteen. What is the sum of the partial quotients? One hundred fifty-three. Finally, Airman Squeaks had to deliver two thousand, three hundred sixty-seven sticks to build three barracks. How many sticks would each one receive? Pause the video and solve this one on your own and press play when you are ready to review it. Although there are many ways to begin multiplying this problem by using different multiples, we chose the largest number that would get us the closest. Three times seven hundred is two thousand one hundred. After we subtract, we have two hundred sixty-seven. Three times eighty gives us two hundred forty. We are left with twenty-seven. And three times nine equals twenty-seven. The sum of the partial quotients is seven hundred eighty-nine. While our hero returns from his daring mission, let's review. Remember, another method for solving long-division problems is to use partial quotients. This method divides a problem into smaller, more manageable parts using multiples of ten or simple numbers. Using multiples of ten or other simple numbers, we estimate how many times the divisor goes into the dividend. Airman Harry Squeaks has returned from a daring mission deep in enemy territory after weeks of anticipation. Once again, our valiant airmen have risked it all for our freedom.