Long Division With Remainders
Basics on the topic Long Division With Remainders
Long Division with Remainders
In this text, you will learn how to do long division with remainders, how do remainders work, what to do with remainders in division, and how to write remainders. What are Remainders in Division? The remainder is the amount left over after dividing. Why are there remainders in division? Sometimes, items cannot be divided equally, so a remainder occurs. Why are reminders important? When you have to work with long division and remainders, you must look at what each number in the problem represents in order to solve it. Before we look at an example of dividing with remainders, let’s review the parts of division and set up our problem.
Long Division – Parts of Division
What Are the Parts of Division? To set up a long division problem, we write our division symbol and place the dividend, or the number being divided, on the inside. We place the divisor, or the number we are dividing by on the outside.
Long Division with Remainders – Example
The problem asks if the bears will share the berries evenly, in order to find the answer you have to practice division with remainders and be able to identify if there is a remainder or not. Now that the problem is set up we can see how to divide with remainders by following these simple steps:
- First, start by seeing how many times the divisor, or three, goes into the first digit. Since three does not go into two, so now we look at twenty-six Three goes into twenty-six eight whole times, so we write it above the one place.
Second, multiply three times eight to get the product twenty-four, and subtract twenty-four from twenty-six to get two.
Last, three does not go into two, so two is our remainder. If you’re wondering how to write remainders in division, all you have to do is write an “R” and the remainder at the top next to the answer like in the illustration below.
Long Division with Remainders – Summary
Remember, sometimes, items cannot be divided equally. The remainder is the amount left over after dividing. When finding remainders in math, including two digit division with remainders and three digit division with remainders, you simply follow the steps for solving a long division problem, writing the remainder at the top next to the answer labeling it with the letter “R”.
| Step # | What to do |
|---|---|
| 1 | Start by seeing how many times the divisor goes into the first digit. |
| 2 | Multiply the divisors to get the product and subtract the product from the dividend to find out the remainder. |
| 3 | Write the product and the remainder at the top. Label the remainder with “R”. |
Have you practiced yet? You can practice long division with remainders with our interactive exercises, worksheets and more activities after watching the video.
Transcript Long Division With Remainders
Mr. Squeaks and Imani just arrived at Boulders and Bears National Park. They are using some binoculars to watch the bears sort some food for hibernation at a safe distance. They are observing bears to see if they can divide berries, moths, and fish evenly amongst themselves. In order to do this they will use "Long Division with Remainders". Sometimes, items cannot be divided equally(...) the "remainder" is "the amount left over after dividing". For example, Mr. Squeaks and Imani see there are twenty-six berries that need to be divided between three bears. Remember, to set up a long division problem, we write our division symbol (...) and place the DIVIDEND, or the number being divided, HERE. We place the DIVISOR, or the number we are dividing by, HERE. Now, we can start by seeing how many times the DIVISOR, or three, goes into the first digit. Does three go into two? Three DOES NOT go into two, so now we look at both digits. How many times does three go into twenty-six? Three goes into twenty-six eight whole times, so we write it above the ONES place. Three times eight is twenty-four, (...) so we subtract twenty-four from twenty-six. What is twenty-six minus twenty-four? The answer is two (...) three does not go into two and there are no more numbers to bring down... so this is our remainder. We label the remainder with a capital "R" and write it along with the two at the top beside the eight. Out of the twenty-six berries, all three bears will get eight with two leftover. Next, Mr. Squeaks and Imani see the bears will share some moths. There are eighty-one moths to be divided between six bears. We start by seeing how many times six goes into eight. How many times does six go into eight? (...) Six goes into eight one whole time (...) so write one above the TENS place. Six times one is six (...) so subtract six from eight. What is eight minus six? (...) Eight minus six is two. Next, bring down the ONE and see how many times six goes into twenty-one. How many times does six go into twenty-one? (...) Six goes into twenty one three whole times, so write a three above the ONES place. What is our next step? (...) Six times three is eighteen, so we subtract eighteen from twenty-one to get three. Three is our remainder, so we write a capital "R" along with the three next to our answer. Out of eighty-one moths, all six bears will get thirteen with three leftover. Last, Mr. Squeaks and Imani see the bears will share some fish. There are three hundred twenty-eight fish to be divided between nine bears. What is our first step? Calculate how many times nine goes into three. Nine does not go into three, so we calculate how many times nine goes into thirty-two instead. What is our next step? Since nine goes into thirty-two three whole times (...) we write the three above the TENS place... and since nine times three is twenty-seven (...) we subtract twenty-seven from thirty-two. What do we get when we subtract? Thirty-two minus twenty-seven is five. What is our next step? (...) Our next step is to bring down the eight (...) how many times does nine goes into fifty-eight? Nine goes into fifty-eight six whole times (...) so we write a six above the ONES place. What is our next step? (...) Since nine times six is fifty-four, we subtract fifty-four from fifty-eight (...) to get four. Four is our remainder. Last, write the remainder next to our answer using a capital "R" to label it. Out of three hundred twenty-eight fish, all nine bears will get thirty-six with four leftover. Remember (...) sometimes, items cannot be divided equally (...) the "remainder" is "the amount left over after dividing"... and we label our remainder with a capital "R" beside the answer. [shocked/confused] Wait...where did the leftovers go!? I wonder what the bears will do with all their leftovers...
Long Division With Remainders exercise
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What is a remainder?
HintsIf Mr. Squeaks and Imani are going to take the remainder after the bears have shared the food, what could the remainder be?
We find the remainder once we have solved the problem.
SolutionThe remainder in division is what is left over after dividing.
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Can you figure out the answer?
HintsHow many times did 3 go into 4? How many times did 3 go into 14?
If we subtract 12 from 14, what is the remainder we are left with?
SolutionHere is the complete long division problem.
3 goes into 4 1 whole time, so we write 1 at the top.
1 x 3 = 3 so we then subtract 3 from 4 and we are left with 1. We bring the 4 down from the ones place.
3 goes into 14 4 whole times. We write this next to the 1 at the top. 3 x 4 = 12 so we subtract 12 from 14, leaving us with 2.
3 does not go into 2 so 2 is our remainder. We write this after the R at the top.
We can then see that our answer is 14 R 2.
Each bear will get 14 fish and there will be 2 left over.
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Does Mr. Squeaks have the correct remainder?
HintsSolve 93 $\div$ 4 using a pencil and paper. Do you see any differences between yours and this one?
If 4 goes into 9 2 whole times, what should we subtract from 9?
How many times does 4 go into 13?
There are 4 errors to highlight.
SolutionOn the left is the problem solved correctly.
On the right is where the highlighting should be.
_______________________________________________________
First, we need to figure out how many times 4 goes into 9.
4 goes into 9 2 whole times, so we write 2 at the top.
2 x 4 is 8, so we subtract 8 from 9 to get 1.
4 does not go into 1, so we bring the 3 down to make 13.
4 goes into 13 3 whole times.
3 x 4 is 12, so we subtract 12 from 13 to get 1.
The remainder is 1.
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Can you solve the problems?
HintsHow many times does the divisor go into the first number of the dividend? Remember, the number we are dividing is the dividend and the number we are dividing by is the divisor.
Remember to subtract from the dividend.
How many are you left with at the end? This is the remainder.
This is how we would start 87 $\div$ 6. Can you complete the rest of the problem?
SolutionHere is how to solve 87 $\div$ 6.
- 6 goes into 8 1 whole time, so we write 1 at the top and subtract 6 from 8.
- 8 - 6 = 2. 6 does not go into 2, so we bring down the 7.
- 6 goes into 27 4 whole times, so we write 4 at the top.
- 4 x 6 is 24, so we subtract 24 from 27.
- 27 - 24 = 3.
- 6 does not go into 3, so this is our remainder.
- 87 $\div$ 6 = 14 R 3
102 $\div$ 7
- 7 does not go into 1, so we see how many times 7 goes into 10.
- 7 goes into 10 1 whole time, so we write 1 at the top.
- 10 - 7 = 3.
- 7 does not go into 3, so we bring the 2 down to get 32.
- 7 goes into 32 4 whole times, so we write 4 at the top.
- 4 x 7 is 28, so we subtract 28 from 32 to get 4.
- 7 does not go into 4, so that is our remainder.
- 102 $\div$ 7 = 14 R 4
249 $\div$ 2
- 2 goes into 2 1 whole time, so we write 1 at the top.
- 2 goes into 4 2 whole times, so we write 2 at the top.
- 2 goes into 9 4 whole times, so we write 4 at the top.
- 4 x 2 is 8, so we subtract 8 from 9 to get 1.
- 2 does not go into 1, so that is our remainder.
- 249 $\div$ 2 = 124 R 1
475 $\div$ 3
- 3 goes into 4 1 whole time, so we write 1 at the top.
- 4 - 3 = 1.
- 3 does not go into 1, so we bring the 7 down to get 17.
- 3 goes into 17 5 whole times, so we write 5 at the top.
- 5 x 3 is 15, so we subtract 15 from 17 to get 2.
- 3 does not go into 2, so we bring the 5 down to get 25.
- 3 goes into 25 8 whole times, so we write 8 at the top.
- 8 x 3 is 24, so we subtract 24 from 25 to get 1.
- 3 does not go into 1, so that is our remainder.
- 475 $\div$ 3 = 158 R 1
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How many berries will Mr. Squeaks get?
HintsEach of the 5 bears is going to get 14 berries. How many are left?
If we subtract 20 from 21, what are we left with? This is the remainder in this problem.
SolutionMr. Squeaks will get 1 berry!
We can see that 5 goes into 7 1 whole time. We then subtract 5 from 7 to get 2.
5 doesn't go into 2, so we bring the 1 down and see how many times 5 goes into 21.
5 goes into 21 4 whole times.
4 x 5 is 20, so we then subtract 20 from 21 to find the remainder.
21 - 20 = 1.
We are left with a remainder of 1.
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What will Mr. Squeaks and Imani be left with?
HintsTry solving the division problem using a pencil and paper if you need to.
Look at the divisor. What could the remainder be?
SolutionThese are the remainders Mr. Squeaks and Imani will be left with.
- 3 fish
- 1 moth
- 2 berries
- 4 insects
To find how many fish there would be for Mr. Squeaks and Imani:
- 5 goes into 6 1 whole time, so we write 1 at the top.
- 6 - 5 = 1.
- 5 does not go into 1, so we bring the 8 down to make 18.
- 5 goes into 18 3 whole times, so we write 3 at the top.
- 3 x 5 is 15, so we subtract 15 from 18 to get 3.
- 5 does not go into 3, so that is our remainder.
- 68 $\div$ 5 = 13 R 3
To find how many moths there would be for Mr. Squeaks and Imani:
- 4 goes into 9 2 whole times, so we write 2 at the top.
- 2 x 4 is 8, so we subtract 8 from 9.
- 9 - 8 = 1.
- 4 does not go into 1, so we bring the 7 down to make 17.
- 4 goes into 17 4 whole times, so we write 4 at the top.
- 4 x 4 is 16, so we subtract 16 from 17 to get 1.
- 4 does not go into 1, so that is our remainder.
- 97 $\div$ 4 = 24 R 1
To find how many berries there would be for Mr. Squeaks and Imani:
- 3 does not go into 1, so we see how many times 3 goes into 12.
- 3 goes into 12 4 whole times, so we write 4 at the top.
- 3 goes into 5 1 whole time, so we write 1 at the top.
- 5 - 3 = 2.
- 3 does not go into 2, so that is our remainder.
- 125 $\div$ 3 = 41 R 2
To find how many insects there would be for Mr. Squeaks and Imani:
- 6 does not go into 1, so we see how many times 6 goes into 19.
- 6 goes into 19 3 whole times, so we write 3 at the top.
- 3 x 6 is 18, so we subtract 18 from 19 to get 1.
- 6 does not go into 1, so we bring the 0 down to make 10.
- 6 goes into 10 1 whole time, so we write 1 at the top.
- 10 - 6 = 4.
- 6 does not go into 4, so that is our remainder.
- 190 $\div$ 6 = 31 R 4
