Column Subtraction

You hand over a coin for something that costs less than a pound — how much change should you get back? Working out how much is left after you spend, how many days until the holidays, or the gap between two scores all mean taking one number away from a bigger one, and this page shows the tidy way to do it on paper.

Once a number has more than one digit, the easiest way to subtract is to stack the two numbers — one above the other — and line them up by place value: ones above ones, tens above tens. Then we subtract one column at a time, always starting from the ones on the right and working leftwards.

\begin{array}{r}68\\-\;23\\\hline\end{array}

Working right to left keeps everything tidy: each column becomes a tiny one-digit subtraction, and every digit stays under the place it belongs to. For 68 - 23 there is nothing tricky at all — take the bottom digit from the top digit in each column:

Ones: 8 - 3 = 5.
Tens: 6 - 2 = 4.

\begin{array}{r}68\\-\;23\\\hline 45\end{array}

So 68 - 23 = 45. Read the answer straight off the bottom row.

When the top digit is too small: exchanging

Now look at 52 - 27. In the ones column we are asked to take 7 away from 2 — but 2 ones simply isn't enough to give away seven. So we exchange (many people say borrow): we go next door to the tens column, take one ten, and unbundle it into ten ones.

The 5 tens become 4 tens, and the 2 ones become 2 + 10 = 12 ones. The number is rewritten — same value, just carried differently — and now the ones column works:

52 = 5 \text{ tens} + 2 \text{ ones} = 4 \text{ tens} + 12 \text{ ones}

Ones: 12 - 7 = 5.
Tens: 4 - 2 = 2 (remember the 5 is now a 4).

\begin{array}{r}{}^{4}\!\!\not5\;{}^{1}2\\-\;2\;7\\\hline 2\;5\end{array}

So 52 - 27 = 25. Exchanging never changes how big the number is — it only shifts ten from the tens shelf onto the ones shelf so the subtraction becomes possible.

Picture money. You want to pay 7p but you only have 2 loose pennies and a 10p coin. You can't hand over 7 single pennies — so you ask the shopkeeper to change your 10p coin into ten 1p coins. Now you have 2 + 10 = 12 pennies and can easily pay 7. That swap is exactly what borrowing is: one ten traded for ten ones. You never gained or lost money — you just changed which coins you were holding.

One more, with the borrow

Try 73 - 48. The ones column asks for 3 - 8 — too small again, so exchange a ten:

Exchange: 73 = 6 \text{ tens} + 13 \text{ ones}.
Ones: 13 - 8 = 5.
Tens: 6 - 4 = 2.

73 - 48 = 25

The two classic subtraction slips:

You cannot do 3 - 7 in the ones column just because it looks awkward and then write 4. If the top digit is smaller, you must borrow a ten first, then do 13 - 7.
Always take the bottom from the top — never quietly flip a column to "bigger minus smaller". In 52 - 27 the ones column is 2 - 7 (borrow!), not 7 - 2. Flipping it gives the wrong answer every time.

Look only at the ones column before you start. If the top ones digit is as big as or bigger than the bottom one, you're safe — just subtract (68 - 23 needed no borrow because 8 \ge 3). If the top one is smaller (2 < 7 in 52 - 27), you'll need to exchange a ten. Spotting it early stops you trying an impossible 2 - 7 by mistake.

See it worked through

Watch the subtraction step by step. The ones column comes first; when the top digit is too small, one ten is exchanged for ten ones (shown in colour), and then both columns subtract cleanly. Step through it, and press Refresh for a fresh example — sometimes it needs a borrow, sometimes it doesn't.

See it explained

Sal Khan works through a subtraction that needs borrowing, exchanging a ten for ten ones.