Multiplication and division

Multiplication - partition method

In this method the smaller number is partitioned (broken down into tens and units).

Example

Use the partition method to work out \({352}\times{27}\).

Method 1

\({27}\) is broken down into \({20}\) and \({7}\).

Method 2

\({27}\) is broken down into \({10}\), \({10}\) and \({7}\).

The method of partitioning into tens, simplifies the multiplication further because to multiply successively by \({10}\) is to simply move the digits one place to the left.

This method uses a grid to multiply numbers.

Multiplying a two-digit number with another two-digit number would require a \({2}\) by \({2}\) grid, for example \({43}\times{26}\).

Multiplying \({264}\times{53}\) would require a \({3}\) by \({2}\) grid.

Follow the steps below to see how Napier’s method is used to calculate \({43}\times{26}\).

• The first step is to draw a \({2}\) by \({2}\) grid.
• The second step is to draw a diagonal in each box. The diagonal line separates the tens and the units. Always write the tens above the diagonal line in each box.
• Start by multiplying \({4}\) and \({2}\) to fill the left box on the top row.
• \({4}\times{2}={8}\)
• Write \({0}\) and \({8}\) to show that there are no tens.

• Now multiply \({3}\) and \({2}\) to fill the right box on the top row.
• \({3}\times{2}={6}\)

• Next, multiply \({4}\) and \({6}\) to fill the left box on the bottom row.
• \({4}\times{6}={24}\)
• Remember to put the \({2}\) above the diagonal.

• Complete the grid by multiplying \({3}\) and \({6}\) to fill the right box on the bottom row.
• \({3}\times{6}={18}\)

After completing the grid, add the columns along the diagonals, starting at the bottom-right.

We sometimes need to carry over from one diagonal to the next.

To get the answer, read the totals down the left and to the right.

\({43}\times{26}={1,118}\)

The area method breaks down the numbers to be multiplied at the sides of a rectangle.

The area of the rectangle will be the answer to the multiplication.

Example

Calculate:

\({56}\times{34}\)

Therefore:

\(({50}+{6})\times({30}+{4})\)

To calculate the answer, the individual areas need to be added together:

\({56}\times{34}={1,500}+{180}+{200}+{24}={1,904}\)

To multiply \(237\) by \(4\) without using a calculator, you can set it out like this:

1. Start with \(4 \times 7\), which is \(28\), so write the \(8\) and carry the \(2\) to the tens column.
2. \(4 \times 3 = 12\), but remember to add the carried \(2\) to get \(14\). Write the \(4\) and carry the \(1\) to the hundreds column.
3. \(4 \times 2 = 8\), and we add the carried \(1\) to get \(9\).

Therefore:

\(237 \times 4 = 948\)

This method is called short multiplication.

To divide a large number by a \({1}\)-digit number you can set it out like this:

1. \(6\) goes into \(9\) once with \(3\) remaining (remainder \(3\)), so put \(1\) above the hundreds column and carry the \(3\) to the 1.

2. \(6\) goes into \(31\) five times, remainder \(1\), so put \(5\) above the tens column and carry the \(1\) to the units column.

3. \(6\) goes into \(18\) exactly three times, so put \(3\) above the units column.

4. Read the answer from the top line, \({153}\).

Sometimes, a number does not divide into another number exactly.

In this case, we can either give the answer as a whole number along with a remainder, or we can give the answer as a decimal.

The decimal answer may be a terminating decimal or a recurring decimal.

Often, decimal answers may be rounded to a required number of decimal places.

\({964}\div{7}\)

1. \(7\) goes into \(9\) once with \(2\) remaining (remainder \(2\)), so put \(1\) above the hundreds column and carry the \(2\) to the tens column.
2. \(7\) goes into \(26\) three times, remainder \(5\), so put \(3\) above the tens column and carry the \(5\) to the units column.
3. \(7\) goes into \(54\) seven times, remainder \(5\), so put \(7\) above the units column and have a remainder of \(5\).

So, \(964 \div 7 = 137\) remainder \(5\).

We write this with an \(r\) for 'remainder', so it looks like this:

\({964}\div{7}={137~r5}\)

1. \(7\) goes into \(9\) once with \(2\) remaining (remainder \(2\)), so put \(1\) above the hundreds column and carry the \(2\) to the tens column.
2. \(7\) goes into \(26\) three times, remainder \(5\), so put \(3\) above the tens column and carry the \(5\) to the units column.
3. \(7\) goes into \(54\) seven times, remainder \(5\), so put \(7\) above the units column and carry the \(5\) to the tenths column (writing the \({964}\) as \({964.000}\) as \({7}\) does not divide into it exactly).
4. \({7}\) goes into \({50}\) seven times, remainder \({1}\), so put \({7}\) above the tenths column and carry the \({1}\) to the hundredths column.
5. \({7}\) goes into \({10}\) once, remainder \({3}\), etc.

So, \({964}\div{7}={137.714…}\)\({=137.7}\) (correct to \({1}\) decimal place).

This method is called short division.

Multiplying by a number between 0 and 1

The multiplication sign can be replaced by the phrase 'lots of'.

For example:

\(2 \times 3\) means \(2\) lots of \(3\) (\({=6}\))

\(6 \times 8\) means \(6\) lots of \(8\) (\({=48}\))

So, \({10}\times\frac{1}{2}\) means \({10}\) lots of \(\frac{1}{2}\) (\({=5}\))

And \({12}\times\frac{1}{3}\) means \({12}\) lots of \(\frac{1}{3}\) (\({=4}\))

When you multiply by a number greater than 1, the answer is greater than the original number.

But when you multiply by a number between 0 and 1, the answer is smaller than the original number.

We have seen that: \({10}\times\frac{1}{2}={5}\)

We also know that \({10}\div{2}={5}\)

Therefore: \({10}\times\frac{1}{2}={10}\div{2}\)

This shows that '\(\times~\frac{1}{2}\)' is the same as '\(\div~{2}\)'.

In general:

\(m \times \frac{1}{n} = m \div n\)

Examples

\(8 \times \frac{1}{4} = 8 \div 4 = 2\) (as '\(\times~\frac{1}{4}\)' is the same as '\(\div~{4}\)')

\(20 \times \frac{1}{5} = 20 \div 5 = 4\) (as '\(\times~\frac{1}{5}\)' is the same as '\(\div~{5}\)')

Dividing by a number between 0 and 1

Imagine that you had \(10\) bars of chocolate that you wanted to share amongst some children.

If you gave the children \(2\) bars each, you would have enough for \(5\) children.

\(10 \div 2 = 5\)

If you gave the children \(\frac{1}{2}\) bar each, you would have enough for \(20\) children.

\(10 \div \frac{1}{2} = 20\)

The pattern

Can you see what's happening?

\(10 \div 2 = 5\)

\(10 \div \frac{1}{2} = 20\)

When you divide by a whole number the answer is smaller than the original number.

When you divide by \(\frac{1}{2}\) the answer (\({20}\)) is greater than the original number (\({10}\)).

When you divide by a number greater than 1, the answer is smaller than the original number. But when you divide by a number between 0 and 1, the answer is greater than the original number.

So: \(10 \div \frac{1}{2} = 20\)

Similarly: \(10 \div \frac{1}{3} = 30\) and \(10 \div \frac{1}{4} = 40\)

In general: \(m \div \frac{1}{n} = mn\)