Taking a single digit from a two-digit number

Tak­ing away is some­thing you do all the time. You have 13 mar­bles and give 7 to a friend. A bus has 23 peo­ple and 5 get off. A shop has 35 man­goes and sells 7. Each time you ask the same ques­tion: how many are left? This les­son shows you quick and safe ways to find out.

Tak­ing a digit from a two-digit num­ber is easy when the units are big enough — $47 - 5$ is just $40 + (7 - 5)$, which is $42$. This les­son is about what to do when they are not, as in $13 - 7$, where seven is more than the three you have.

Ten take away a digit

Start here, because it is the tens com­ple­ments once more, read back­wards.

If $9 + 1 = 10$, then $10 - 1 = 9$. Every addi­tion fact you know is a sub­trac­tion fact as well.

Try it with ten beads on a string. Push $6$ beads to one side. Count the beads left on the other side. There are $4$. So $10 - 6 = 4$, and $6 + 4 = 10$. The two num­bers that make ten are called part­ners. Learn the part­ners of ten well. You will use them in every sum below.

A whole ten take away a digit

The answer drops into the ten below, and the units are the part­ner of what you took.

Look at $60 - 7 = 53$. Sixty becomes fifty, and the units are 3 — the part­ner of 7. That is the pat­tern in every row.

Why does this work? Think of $60$ as six bags of ten sweets. To take away $7$, you open one bag. Five bags stay shut, which is $50$. From the open bag of ten you take $7$, and $3$ are left. So you have $50 + 3 = 53$.

When the units are big enough

Noth­ing has to be bor­rowed. Take the digit off the units and leave the tens alone.

For exam­ple, $47 - 5$. The units digit is $7$, and $7$ is more than $5$. So take $5$ from the $7$ to get $2$. The four tens do not change. The answer is $42$.

When the units are not big enough

Here is the idea every­thing below depends on. A sub­trac­tion asks how far apart two num­bers are. If you move both num­bers up by the same amount, or down by the same amount, the gap between them stays exactly the same — so the answer does not change.

Imag­ine two chil­dren stand­ing on the num­ber lad­der. If they both climb three rungs, they are still the same dis­tance apart.

The pic­ture shows this for $13 - 7$. The gap between $7$ and $13$ is $6$ steps. Slide both num­bers down by $3$ and you get $4$ and $10$. The gap is still $6$. Slide both up by $3$ and you get $10$ and $16$. The gap is still $6$.

First way: move both down, so the top becomes a ten

13 - 7

$$\begin{aligned}13 - 7 &= (13 - 3) - (7 - 3) \\ &= 10 - 4 \\ &= 6\end{aligned}$$

Three came off both num­bers. The top became a round ten, and $10 - 4$ is a fact from the first table in this les­son.

Sec­ond way: move both up, so the num­ber being taken becomes a ten

13 - 7

$$\begin{aligned}13 - 7 &= (13 + 3) - (7 + 3) \\ &= 16 - 10 \\ &= 6\end{aligned}$$

Three went on to both num­bers this time. Tak­ing away a whole ten is easy: $16 - 10 = 6$.

Third way: know it

After enough prac­tice $13 - 7 = 6$ comes straight to mind with no work­ing at all. That is the goal — the two ways above are the scaf­fold­ing you use until it does.

Two more, step by step

32 - 9, mov­ing both down

The units digit of $32$ is $2$. Take $2$ off both num­bers. Then the top is a round $30$.

$$\begin{aligned}32 - 9 &= (32 - 2) - (9 - 2) \\ &= 30 - 7 \\ &= 23\end{aligned}$$

And $30 - 7$ is a whole ten take away a digit. The answer drops into the twen­ties, and $3$ is the part­ner of $7$. So it is $23$.

71 - 6, mov­ing both up

The part­ner of $6$ is $4$. Add $4$ to both num­bers. Then you take away a round $10$.

$$\begin{aligned}71 - 6 &= (71 + 4) - (6 + 4) \\ &= 75 - 10 \\ &= 65\end{aligned}$$

Tak­ing away ten is easy. The tens digit goes down by one and the units stay the same.

Which way to choose

Both ways always give the same answer, so pick the one that feels eas­ier. If the units digit of the top num­ber is small, like the $2$ in $32$, mov­ing down is quick. If the num­ber you take away is close to ten, like $8$ or $9$, mov­ing up is quick, because you only add a lit­tle. With prac­tice you will choose with­out think­ing, and one day you will just know the answer.

Check by adding back

You can always check a sub­trac­tion with an addi­tion. If $71 - 6 = 65$, then $65 + 6$ must give $71$. It does. If the addi­tion does not give the num­ber you started with, look at your work­ing again.

More worked exam­ples

Each one is done both ways, so you can see that they always agree.

Remem­ber. Move both num­bers the same way and the answer stays the same. Move them until one of them is a round ten.

Your turn

Com­mon mis­takes

• Tak­ing the smaller units digit from the big­ger one. In $13 - 7$, some chil­dren do $7 - 3 = 4$ and write $14$. The right answer is $6$.
• Mov­ing only one num­ber. If you add $3$ to the top, you must add $3$ to the num­ber you take away too.
• Mov­ing the num­bers in oppo­site direc­tions. Both go up, or both go down.
• For­get­ting that the tens drop by one when the units are not big enough, as in $60 - 7 = 53$, not $63$.
• Mix­ing up the part­ners of ten. Say them aloud until they come quickly.

Key terms

Sub­tract

To take one num­ber away from another.

Dif­fer­ence

The answer to a sub­trac­tion: how far apart two num­bers are.

Units digit

The last digit of a num­ber. In $47$ it is $7$.

Tens digit

The digit that counts tens. In $47$ it is $4$.

Part­ners of ten

Two num­bers that add to make $10$, such as $7$ and $3$.

Round ten

A num­ber that ends in zero, such as $10$, $30$ or $80$.

Answers

1) $5$, 2) $5$, 3) $9$, 4) $8$, 5) $19$, 6) $28$, 7) $38$, 8) $45$, 9) $58$, 10) $66$, 11) $77$, 12) $85$, 13) $9$, 14) $9$, 15) $5$, 16) $15$, 17) $48$, 18) $35$, 19) $28$, 20) $34$, 21) $44$, 22) $57$, 23) $69$, 24) $75$, 25) $69$, 26) $86$, 27) $38$, 28) $49$, 29) $27$, 30) $19$

Check any one by adding back. For ques­tion 1, $5 + 6 = 11$. For ques­tion 30, $19 + 8 = 27$.