Multiply by a one-digit number (formal algorithm) - With carrying
Theory
When a column product is ten or more, write its ones digit and carry the tens into the next column. The carried digit is added after multiplying the next digit. This Year 5 maths page shows how to multiply two-, three- and four-digit numbers by a one-digit number with carrying, and how to avoid the common mistakes.
In \(38 \times 7\) the ones give \(8 \times 7 = 56\). Write the \(6\) in the ones and carry the \(5\) tens to the tens column.
Next multiply the tens digit, then add the carry: \(3 \times 7 + 5 = 26\). Multiply first, add second.
The carry can be any digit from \(1\) to \(8\), not just \(1\). In the last column write the whole result.
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 the extra tens (or hundreds) moving left:
How to multiply with carrying
- Multiply the ones. Write the ones digit of the answer and carry the tens digit.
- Multiply the next digit, then add the carry.
- Repeat for every column, carrying when needed.
- Write the whole answer of the last column.
| \(\text{O: }8 \times 7\) | \(=\) | \(56\ (\text{carry } 5)\) |
| \(\text{T: }3 \times 7 + 5\) | \(=\) | \(26\) |
The answer is \(266\).
| \(\text{O: }6 \times 3\) | \(=\) | \(18\ (\text{carry } 1)\) |
| \(\text{T: }5 \times 3 + 1\) | \(=\) | \(16\ (\text{carry } 1)\) |
| \(\text{H: }2 \times 3 + 1\) | \(=\) | \(7\) |
The answer is \(768\).
| \(\text{O: }7 \times 8\) | \(=\) | \(56\ (\text{carry } 5)\) |
| \(\text{T: }6 \times 8 + 5\) | \(=\) | \(53\ (\text{carry } 5)\) |
| \(\text{H: }4 \times 8 + 5\) | \(=\) | \(37\) |
The answer is \(3736\).
| \(\text{O: }9 \times 4\) | \(=\) | \(36\ (\text{carry } 3)\) |
| \(\text{T: }3 \times 4 + 3\) | \(=\) | \(15\ (\text{carry } 1)\) |
| \(\text{H: }6 \times 4 + 1\) | \(=\) | \(25\ (\text{carry } 2)\) |
| \(\text{Th: }2 \times 4 + 2\) | \(=\) | \(10\) |
The last column gives \(10\), so the answer has five digits: \(10\,556\).
| \(\text{O: }6 \times 7\) | \(=\) | \(42\ (\text{carry } 4)\) |
| \(\text{T: }4 \times 7 + 4\) | \(=\) | \(32\) |
\(7\) tickets cost \(\$322\).
| \(\text{O: }5 \times 6\) | \(=\) | \(30\ (\text{carry } 3)\) |
| \(\text{T: }8 \times 6 + 3\) | \(=\) | \(51\ (\text{carry } 5)\) |
| \(\text{H: }1 \times 6 + 5\) | \(=\) | \(11\) |
The mass is \(1110\) g.
| \(\text{O: }8 \times 4\) | \(=\) | \(32\ (\text{carry } 3)\) |
| \(\text{T: }5 \times 4 + 3\) | \(=\) | \(23\) |
Multiply first, then add the carry: \(5 \times 4 + 3 = 23\). The answer is \(232\).
Common pitfalls
Frequently asked questions
What do you do when a column product is more than 9?
Write the ones digit in that column and carry the tens digit to the next column. \(8 \times 7 = 56\): write \(6\), carry \(5\).
When do you add the carried number?
After multiplying the next digit. In \(38 \times 7\) the tens are \(3 \times 7 = 21\), then \(21 + 5 = 26\).
Can the carry be more than 1?
Yes. It can be any digit up to \(8\). \(9 \times 9 = 81\) carries \(8\).
What happens in the last column?
Write the whole result. \(2639 \times 4\) ends with \(2 \times 4 + 2 = 10\), so the answer is \(10\,556\).
How do I check my answer?
Estimate by rounding the large number: \(467 \times 8 \approx 500 \times 8 = 4000\).