Division in math is used to split larger numbers into sets of smaller ones. It is the opposite of multiplication where sets are combined to find a total. For example, picture five people with 10 apples. When the 10 apples are divided between the five people they each have two. Ordinary division (or short division) in math is used to split numbers into less than 10 sets. Splitting a number into 10 sets or more would require the use of long division...
Instructions
Carry the 1 over again and put it in front of the next 0. Again you have 10 to divide by 3 which is 3, remainder 1. Write the 3 down next in your answer. Now you have 136.33, remainder 1. As you can see, if you carry the 1 over again in front of the next 0, you have a 10 again. So you will continue to end up with the same numbers. You will end up with a string of 3s and there will be no end to the problem. When this happens your answer is 136.33 "recurring." This means there are an infinite number of 3s.
Instructions
- 1
Write down the number of sets you are looking for in front of the total number you want to divide into that many sets. For example, write 3\136 if you want to divide 136 into 3 sets of smaller numbers.
- 2
Figure out how many sets of 3 are in the first number of your longer number. In the case of 136, there are 0 sets of 3 in 1. So now the 1 carries over and you now need to figure out how many sets of 3 are in 13.
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- 3
Figure out how many sets of 3 there are in 13. Because 3 doesn't divide evenly into 13, you will have 4 sets of 3 and a remainder of 1. The remaining 1 carries over to the last number in the original 136. You now need to figure out many sets of 3 are in 16.
- 4
Divide the 16 into sets of 3 and you have 5 with a remainder of 1. Write down the 5 in your answer after the 4. Now there are no numbers left to divide. Your answer to the overall question is 045 (or 45), with a remainder of 1 as the number did not divide equally into sets of 3 and there was 1 left over.
- 5
Leave the problem there if this answer will do for you, but if you want an exact answer to your problem you can take the number to decimal places. To do this, place a decimal after the 136 then add a 0. So it looks like this 136.0. The remaining 1 that you were left with, you can now carry over in front of the 0 to make the number 10.
- 1
Carry the 1 over again and put it in front of the next 0. Again you have 10 to divide by 3 which is 3, remainder 1. Write the 3 down next in your answer. Now you have 136.33, remainder 1. As you can see, if you carry the 1 over again in front of the next 0, you have a 10 again. So you will continue to end up with the same numbers. You will end up with a string of 3s and there will be no end to the problem. When this happens your answer is 136.33 "recurring." This means there are an infinite number of 3s.