Add whole numbers (written strategies) - With regrouping
Theory
When a column adds to ten or more, you regroup: write the ones digit of the column total and carry the ten into the next column to the left. This Year 5 maths page shows how to add three-, four- and five-digit numbers with carrying, including chains of carries, three-number sums, missing numbers and word problems.
Regrouping (carrying) happens when a column adds to ten or more. Write the ones digit of the total and carry the ten into the next column as a small \(1\).
Ten ones make one ten, ten tens make one hundred and ten hundreds make one thousand, so the value is kept, just moved into a higher place.
A carry can make the next column reach ten as well, so carries can chain across several columns. If the last column reaches ten, write the whole total, which gives the answer an extra digit.
In the working, O, T, H, Th, TTh, HTh and M stand for ones, tens, hundreds, thousands, ten thousands, hundred thousands and millions.
The carry is just a ten moving into the next place:
How to add with regrouping
- Line up the numbers on the right and add the ones.
- If the total is ten or more, write the ones digit and carry \(1\) to the next column.
- Add the next column, including any carry, and repeat.
- Write the whole total of the last column.
How to check
- Estimate: round each number and add.
- Inverse: subtract one number from the total; you should get the other.
| \(\text{O: }3 + 8\) | \(=\) | \(11\ (\text{carry } 1)\) |
| \(\text{T: }5 + 1 + 1\) | \(=\) | \(7\) |
| \(\text{H: }2 + 4\) | \(=\) | \(6\) |
The answer is \(671\).
| \(\text{O: }6 + 8\) | \(=\) | \(14\ (\text{carry } 1)\) |
| \(\text{T: }7 + 5 + 1\) | \(=\) | \(13\ (\text{carry } 1)\) |
| \(\text{H: }4 + 3 + 1\) | \(=\) | \(8\) |
The answer is \(834\).
| \(\text{O: }7 + 5\) | \(=\) | \(12\ (\text{carry } 1)\) |
| \(\text{T: }8 + 4 + 1\) | \(=\) | \(13\ (\text{carry } 1)\) |
| \(\text{H: }6 + 5 + 1\) | \(=\) | \(12\ (\text{carry } 1)\) |
| \(\text{Th: }3 + 2 + 1\) | \(=\) | \(6\) |
The answer is \(6232\).
| \(\text{O: }9 + 7\) | \(=\) | \(16\ (\text{carry } 1)\) |
| \(\text{T: }5 + 8 + 1\) | \(=\) | \(14\) |
The hundreds column is new, so the answer has three digits: \(146\).
| \(\text{O: }6 + 9\) | \(=\) | \(15\ (\text{carry } 1)\) |
| \(\text{T: }8 + 4 + 1\) | \(=\) | \(13\ (\text{carry } 1)\) |
| \(\text{H: }7 + 5 + 1\) | \(=\) | \(13\ (\text{carry } 1)\) |
| \(\text{Th: }5 + 6 + 1\) | \(=\) | \(12\ (\text{carry } 1)\) |
| \(\text{TTh: }1 + 1\) | \(=\) | \(2\) |
The answer is \(22\,335\).
| \(\text{O: }5 + 9 + 6\) | \(=\) | \(20\ (\text{carry } 2)\) |
| \(\text{T: }8 + 3 + 7 + 2\) | \(=\) | \(20\ (\text{carry } 2)\) |
| \(\text{H: }2 + 1 + 4 + 2\) | \(=\) | \(9\) |
The ones add to \(20\), so write \(0\) and carry \(2\). The total is \(900\).
| \(6003 - 2687\) | \(=\) | \(3316\) |
| \(2687 + 3316\) | \(=\) | \(6003\) |
The missing number is \(3316\).
| \(\text{O: }6 + 8\) | \(=\) | \(14\ (\text{carry } 1)\) |
| \(\text{T: }7 + 5 + 1\) | \(=\) | \(13\ (\text{carry } 1)\) |
| \(\text{H: }4 + 3 + 1\) | \(=\) | \(8\) |
Leo did not add the carried \(1\)s. The correct answer is \(834\).
Common pitfalls
Frequently asked questions
What is regrouping in addition?
Regrouping, or carrying, is what you do when a column adds to ten or more: write the ones digit and carry the ten into the next column. \(7 + 8 = 15\) means write \(5\), carry \(1\).
Why do you carry the 1?
Ten ones make one ten, so the extra ten moves into the tens column as a single 1. Its value is not lost.
Can a carry happen more than once?
Yes. A carry can push the next column to ten as well, so it carries again. \(3687 + 2545\) carries three times.
When a column adds to 13, what do you write?
Write \(3\) in that column and carry \(1\) to the next column.
What happens when the last column reaches ten?
Write the whole total. \(59 + 87 = 146\) has one more digit than either number.
How do you check an addition?
Estimate by rounding, or subtract one of the numbers from your answer to see if you get the other one.