Multiply by a one-digit number (formal algorithm) - No carrying
Theory
The written algorithm for multiplication sets the larger number on top and the one-digit number underneath, lined up with the ones. Each digit of the top number is multiplied in turn, starting with the ones. When every product is less than ten there is no carrying. This Year 5 maths page shows how to set out and work through these multiplications.
In the written algorithm (short multiplication) the number being multiplied goes on top and the one-digit multiplier goes underneath, under the ones digit.
Multiply the multiplier by each digit of the top number, starting with the ones, then the tens, hundreds and thousands. Write each answer under its own column.
No carrying means every column product is \(9\) or less, so each one is a single 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 algorithm is a short way of writing the distributive property:
How to multiply by a one-digit number
- Write the larger number on top and the one-digit number under its ones digit.
- Multiply the ones and write the answer under the ones.
- Multiply the tens, then the hundreds and thousands, writing each answer in its column.
- Check with an estimate.
| \(\text{O: }4 \times 2\) | \(=\) | \(8\) |
| \(\text{T: }2 \times 2\) | \(=\) | \(4\) |
| \(\text{H: }3 \times 2\) | \(=\) | \(6\) |
The answer is \(648\).
| \(\text{O: }3 \times 3\) | \(=\) | \(9\) |
| \(\text{T: }1 \times 3\) | \(=\) | \(3\) |
| \(\text{H: }2 \times 3\) | \(=\) | \(6\) |
The answer is \(639\).
| \(\text{O: }3 \times 2\) | \(=\) | \(6\) |
| \(\text{T: }2 \times 2\) | \(=\) | \(4\) |
| \(\text{H: }4 \times 2\) | \(=\) | \(8\) |
| \(\text{Th: }1 \times 2\) | \(=\) | \(2\) |
The answer is \(2846\).
| \(\text{O: }1 \times 3\) | \(=\) | \(3\) |
| \(\text{T: }3 \times 3\) | \(=\) | \(9\) |
| \(\text{H: }0 \times 3\) | \(=\) | \(0\) |
| \(\text{Th: }2 \times 3\) | \(=\) | \(6\) |
The zero in the hundreds gives \(0\). The answer is \(6093\).
| \(\text{O: }1 \times 5\) | \(=\) | \(5\) |
| \(\text{T: }1 \times 5\) | \(=\) | \(5\) |
| \(\text{H: }0 \times 5\) | \(=\) | \(0\) |
| \(\text{Th: }1 \times 5\) | \(=\) | \(5\) |
The answer is \(5055\).
| \(\text{O: }1 \times 4\) | \(=\) | \(4\) |
| \(\text{T: }2 \times 4\) | \(=\) | \(8\) |
| \(\text{H: }2 \times 4\) | \(=\) | \(8\) |
There are \(884\) seats.
| \(\text{O: }4 \times 2\) | \(=\) | \(8\) |
| \(\text{T: }2 \times 2\) | \(=\) | \(4\) |
| \(\text{H: }3 \times 2\) | \(=\) | \(6\) |
The multiplier lines up with the ones, so the \(2\) goes under the \(4\). Start with the ones: the answer is \(648\).
Common pitfalls
Frequently asked questions
How do you set out a multiplication in columns?
Write the larger number on top and the one-digit number underneath, lined up with the ones. Draw a line and write the answer below.
Which digit do you multiply first?
The ones. Then the tens, the hundreds and so on, moving left. For \(324 \times 2\): \(4 \times 2\), then \(2 \times 2\), then \(3 \times 2\).
What does no carrying mean?
Each column product is \(9\) or less, so it fits in one column. \(213 \times 3 = 639\) needs no carrying.
What happens to a zero digit?
Zero times any number is zero, so write \(0\) in that column: \(2031 \times 3 = 6093\).
How can I check the answer?
Estimate. \(324 \times 2\) is about \(300 \times 2 = 600\), so \(648\) is sensible.