Mathematics

Written arithmetic

Add and subtract big numbers in columns, one column at a time.

Adding $347 + 285$ in your head is hard work, and a shopping receipt often carries even bigger numbers. That is what written arithmetic is for: it breaks every calculation into tiny steps. You write the numbers under each other and then only ever add single digits, one column at a time. Once you can do this cleanly, even five-digit numbers hold no fear.

Lined up by place value

Everything depends on how you write the numbers down. Ones go under ones, tens under tens, hundreds under hundreds. For whole numbers that simply means: right aligned, even when the numbers have different lengths. In $347 + 85$ the 5 sits under the 7 and the 8 sits under the 4, and above the 8 the hundreds column just stays empty. If a digit slips by a single column, you add ones to tens and end up hundreds away from the right answer.

\begin{array}{cccc} & 3 & 4 & 7 \\ + & & 8 & 5 \\ \hline \end{array}

Each column is one place: ones on the right, then tens, then hundreds. For 85 the hundreds column stays empty.

The carry

Now you work column by column, from right to left. For $347 + 285$ you start with the ones: $7 + 5 = 12$ . Twelve ones are one ten and two ones, so you write the 2 in the ones column and note the single ten as a small mark above the tens column. That is the carry. There you work out $4 + 8 + 1 = 13$ , write the 3 and pass a 1 on to the left again, this time a hundred. The hundreds column gives $3 + 2 + 1 = 6$ , so the total is 632. That is exactly why you start on the right: the carry always travels to the left, so that column must not be finished yet.

Place-value board

Hundreds3

Tens4

Ones7

3Hundreds

4Tens

7Ones

3 × 100 + 4 × 10 + 7 × 1 = 347

Try it: the ones column stops at 9, because a single column holds no more than that. That is exactly why ten ones become one ten in the column to the left, and that is what the carry records.

When the digit is too small

When subtracting, sooner or later you hit a column where the top digit is too small. In $52 - 27$ the ones column asks for $2 - 7$ , and 2 ones are simply not enough. So you borrow a ten from the neighbouring column and split it into 10 ones: 2 ones become 12, and $12 - 7 = 5$ . In return the tens column gave a ten away, 5 tens have become 4, and $4 - 2 = 2$ . The result is 25. This is called ungrouping: a bundle of ten is opened up again, while the number itself stays exactly as big as before. If the neighbouring column holds a 0, it has nothing to give away either. Then you first fetch something one column further left, and the borrowing travels across two columns.

\begin{array}{cccc} & 6 & 2 & 3 \\ - & 2 & 7 & 5 \\ \hline & 3 & 4 & 8 \end{array}

Here you borrow twice: first for

13 - 5 = 8

, then for

11 - 7 = 4

. The hundreds column leaves

5 - 2 = 3

Exercises

0 of 6 solved

Time to try it yourself. You can't break anything, every attempt counts.

You write $46 + 27$ in columns, lined up by place value. Which two digits end up in the ones column?

Work out in columns: $34 + 25$

Adding $27 + 15$ in columns, the ones column gives 12. What do you do with it?

The ones column says $3 - 8$ . Before you can calculate, you borrow one … from the neighbouring column.

Work out in columns: $347 + 285$

Put the steps in the right order to work out $623 - 275$ in columns.

1Write the numbers so that ones sit under ones
2Finally do the hundreds column: $5 - 2 = 3$ , and the result is 348
3Borrow a ten and work out $13 - 5 = 8$
4Set up $3 - 5$ in the ones column and notice that it does not work
5The tens column now shows 1 instead of 2, so borrow again: $11 - 7 = 4$

Where this leads

Polynomial division

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