Iluka has buttons, cards of ten and loose, and a customer wants . There are only loose buttons and she needs to hand over . Where do the other loose buttons come from?

Grade 2 · 2.NBT.B.5 · 2.NBT.B.7

Regrouping in subtraction

Iluka has 54 buttons, 5 cards of ten and 4 loose, and a customer wants 27. There are only 4 loose buttons and she needs to hand over 7. Where do the other loose buttons come from?

You need

Subtracting two-digit numbers without regroupingGrade 2 Base ten blocks and trading ten for oneGrade 2

Unlocks

Subtracting three-digit numbers by place value, including across zeroGrade 2 The standard subtraction algorithmGrade 4

Where you are 7 in Multi-digit subtraction

Number9

Arithmetic5

Addition and subtraction12

Multi-digit subtraction7

Subtracting multiples of ten Subtracting two-digit numbers without regrouping

Regrouping in subtraction

Subtracting three-digit numbers by place value, including across zero The standard subtraction algorithm ▾ Show all 7 (2 more)

Why this exists

The child can already subtract two-digit numbers when each place has enough. In Subtracting two-digit numbers without regrouping, tens are taken from tens and ones from ones, and nothing has to be opened up. The child also knows the trade itself from Base ten blocks and trading ten for one: one rod can be swapped for ten units.

Now the ones run short. At the craft store, Iluka keeps buttons sewn on cards of ten. She has 5 cards and 4 loose buttons, which is 54. A customer wants 27 buttons: 2 cards and 7 loose ones. The 2 cards are easy. The 7 loose buttons are not, because Iluka has only 4.

She does not turn the customer away. She has plenty of buttons, just not loose ones. So she snips the buttons off one card. Now she has 4 cards and 14 loose buttons. That is still 54 buttons; nothing was added or lost. But now there are enough loose ones. She hands over 2 cards and 7 loose buttons and keeps 2 cards and 7 loose: 27 buttons.

Regrouping in subtraction is that move: when a place does not have enough to take from, one of the next place up is broken into ten.

Definition

Regrouping in subtraction means breaking one of a place into ten of the place to its right, so that a place has enough to take from:

one ten is broken into ten ones;
one hundred is broken into ten tens.

The number stays the same. Only the way it is split into tens and ones changes. Some books call this “borrowing”, “unbundling” or “breaking a ten”. It is the same idea, and it is regrouping in addition run backward: there ten ones become one ten, here one ten becomes ten ones.

For Iluka’s buttons it reads:

\begin{aligned} 54 &= 5 \text{ tens } 4 \text{ ones} \\ &= 4 \text{ tens } 14 \text{ ones} \end{aligned}

From 4 tens 14 ones, take 2 tens 7 ones. What is left is 2 tens 7 ones, so 54 - 27 = 27.

A trade is needed exactly when the first number has fewer ones than the number being taken away. In 54 - 27 there are 4 ones and 7 to take, so a ten is broken. In 57 - 24 there are 7 ones and only 4 to take, so nothing is broken.

Intuition

The lesson runs over four days. Day 1 breaks a ten with blocks. Day 2 draws it, and decides before starting whether a ten has to be broken. Day 3 writes the numbers by place, takes from a number with no ones at all, and checks by adding back. Day 4 starts from a number over one hundred, where a hundred may be broken into ten tens.

Part 1: rods and units

Build 54 with base ten blocks: 5 rods and 4 units. To take away 27, the child must remove 2 rods and 7 units. The rods are no trouble. The units are: there are only 4.

So take one rod and trade it for ten units, the trade from trading ten for one done in this direction. Lay the ten units beside the rod first if the child is unsure; they are exactly as long, because a rod is ten units joined. The pile is now 4 rods and 14 units. Count it if needed: still 54.

Now take away 2 rods and 7 units. Left are 2 rods and 7 units, which is 27.

Figure 154 - 27. The boxed rod is broken into ten units (shaded). Then 2 rods and 7 units are taken away, and 27 is left.

Have the child say both names of the pile aloud: “five tens and four ones is four tens and fourteen ones.” The first name is how the number is written. The second name is the one that has enough ones.

The two stores tell the same story outside the classroom. A card of ten buttons is a ten, and a loose button is a one; snipping a card open is breaking a ten. At the post office stamps come in strips of ten. A full strip is a ten and a single stamp is a one. Tearing a strip into ten singles is breaking a ten.

Part 2: a quick drawing, and deciding first

A quick drawing does the work of the blocks: a stick for each ten and a dot for each one, the dots in rows of five. To break a ten, cross out one stick and draw ten dots in its place.

For 62 - 38, draw 6 sticks and 2 dots. There are 2 dots and 8 to take, so break a stick: now 5 sticks and 12 dots. Cross out 3 sticks and 8 dots. Left are 2 sticks and 4 dots, which is 24.

Figure 262 - 38 as a quick drawing. One stick becomes ten dots, then 3 sticks and 8 dots are crossed out.

Before drawing anything, look at the ones digits. If the first number’s ones digit is smaller than the ones digit being taken, a ten has to be broken. If it is the same or larger, nothing is broken, and the subtraction is the kind from subtracting without regrouping. Deciding first saves breaking a ten that was not needed.

Part 3: writing by place, a zero, and adding back

The pictures now give way to numbers. Write the tens and ones of the first number in columns, break a ten if the ones are short, then take tens from tens and ones from ones. For 60 - 25:

	tens	ones
60	6	0
after breaking a ten	5	10
take away 25	2	5
left	3	5

Sixty has no ones at all. That is no problem: one of its 6 tens becomes ten ones. Then 10 - 5 = 5 ones and 5 - 2 = 3 tens, so 60 - 25 = 35.

Adding back checks the answer. If 25 was taken from 60 and 35 was left, putting the 25 back must give 60:

35 + 25 = 60

This time the adding back regroups too: 5 + 5 = 10 ones make a new ten, and 3 + 2 + 1 = 6 tens. The ten that was broken in the subtraction is made again in the addition. That is no accident, since adding back undoes the subtraction step by step.

Other routes reach the same difference without breaking anything: counting up from the smaller number, or reading the subtraction as an addition with a part missing. They are worth knowing; this lesson stays with taking away by place.

Part 4: a start over one hundred

The same trade works one place up. A flat is ten rods, so a flat can be broken into ten rods when the tens run short. At the craft store a box holds ten cards of buttons, which is 100 buttons; at the post office a sheet holds ten strips, which is 100 stamps.

For 132 - 57, the ones are short first: 2 ones and 7 to take. Break a ten: 1 hundred, 2 tens and 12 ones. Then the tens are short: 2 tens and 5 to take. Break the hundred: 12 tens and 12 ones. Now take away 5 tens and 7 ones. Left are 7 tens and 5 ones: 75.

Figure 3132 - 57. One rod became ten units and the flat became ten rods. Then 5 rods and 7 units are taken away.

Each break is one for ten, so the pile is 132 at every row, until the 57 is taken away. The number taken here is still a two-digit number; only the start has three digits.

Results

Breaking keeps the amount. One ten is ten ones, and one hundred is ten tens:

1 \text{ ten} = 10 \text{ ones} \qquad 1 \text{ hundred} = 10 \text{ tens}

So 5 tens 4 ones and 4 tens 14 ones name the same number:

50 + 4 = 54 \qquad 40 + 14 = 54

When a ten is broken. A ten is broken exactly when the first number’s ones digit is smaller than the ones digit being taken away. One break is always enough for the ones.

Adding back. What is left plus what was taken away is the number at the start:

54 - 27 = 27 \quad \text{because} \quad 27 + 27 = 54

Why it is true

Why breaking a ten keeps the amount. Breaking takes away one rod, worth 10, and puts back ten units, worth 1 each, which is also 10. Nothing is added and nothing is lost. The blocks show it with no counting: the ten units laid end to end are as long as the rod they replaced. Breaking a flat into ten rods works the same way, since both are worth 100.

Why tens may be taken from tens and ones from ones. After the break, 54 is 40 + 14, and 27 is 20 + 7. Taking away 27 means taking away the 20 and taking away the 7. The 20 comes out of the 40, leaving 20; the 7 comes out of the 14, leaving 7. Together that is 20 + 7 = 27. Taking a pile away in two parts leaves the same as taking it in one go.

Why one break is always enough for the ones. The ones digit being taken is at most 9. After one ten is broken, the ones place holds at least 10, which is more than 9. So it never has to be broken twice. The same holds for the tens when a hundred is broken.

Why adding back checks. Subtraction undoes addition. The part taken away and the part left are the two parts of the start, so putting them together must give the start again. If the sum does not come back to the first number, a step went wrong.

Worked examples

The first two examples are for Day 1, the next two for Day 2, then two for Day 3 and two for Day 4.

Example 1

Day 1: Buttons at the craft store

Iluka has 5 cards of ten buttons and 4 loose buttons. A customer buys 27 buttons. How many buttons does Iluka have left?

Solution

Iluka starts with 54 buttons and the customer takes 27, so the question is 54 - 27. Model 54 with blocks: 5 rods and 4 units. The 27 is 2 rods and 7 units.

There are only 4 units and 7 are needed, so break one rod into ten units. Now there are 4 rods and 14 units, still 54.

Take away 2 rods: 4 - 2 = 2 rods. Take away 7 units: 14 - 7 = 7 units. Left are 2 rods and 7 units:

54 - 27 = 27

At the store, Iluka snipped one card open, handed over 2 cards and 7 loose buttons, and kept 2 cards and 7 loose buttons. Check by adding back: 27 + 27 = 54.

Example 2

Day 1: Stamps with only one single

Bekzat has 4 strips of ten stamps and 1 single stamp. He uses 15 stamps on packages. How many stamps are left?

Solution

The question is 41 - 15. Model 41: 4 rods and 1 unit. The 15 is 1 rod and 5 units.

One unit is not enough to take 5, so break a rod. Now there are 3 rods and 11 units: the 1 unit that was there plus the ten new ones. Take away 1 rod and 5 units: 3 - 1 = 2 rods and 11 - 5 = 6 units.

41 - 15 = 26

Bekzat tore one strip into singles and has 2 strips and 6 singles left, 26 stamps. Notice the broken ten joined the one single already there. The ones after a break are always ten more than before.

Example 3

Day 2: A drawing for sixty-two take away thirty-eight

Aolani has 62 stamps at the post office counter and sells 38 of them. Use a quick drawing to find how many are left.

Solution

The question is 62 - 38. Draw 6 sticks and 2 dots for 62.

The ones digit of 62 is 2 and the ones digit of 38 is 8. Two dots are not enough to cross out eight, so a ten must be broken. Cross out one stick and draw ten more dots: 5 sticks and 12 dots.

Now cross out 3 sticks and 8 dots. Left are 5 - 3 = 2 sticks and 12 - 8 = 4 dots:

62 - 38 = 24

Aolani has 24 stamps left: 2 strips and 4 singles. Check: 24 + 38 has 4 + 8 = 12 ones, a new ten, and 2 + 3 + 1 = 6 tens, so 62.

Example 4

Day 2: Will a ten be broken?

Without finding the differences first, decide which of 93 - 57 and 97 - 53 needs a ten broken. Then find both differences.

Solution

Only the ones digits decide. In 93 - 57 there are 3 ones and 7 to take, and 3 is smaller than 7, so a ten must be broken. In 97 - 53 there are 7 ones and 3 to take, which is enough, so nothing is broken.

For 93 - 57: break a ten, so 9 tens 3 ones becomes 8 tens 13 ones. Take 5 tens and 7 ones: 8 - 5 = 3 tens, 13 - 7 = 6 ones.

93 - 57 = 36

For 97 - 53: 9 - 5 = 4 tens and 7 - 3 = 4 ones.

97 - 53 = 44

The same four digits appear in both, yet the differences are not the same. Swapping the ones digits decides whether a ten is broken, and the broken ten is why the first answer has fewer tens.

Example 5

Day 3: Buttons from a number with no ones

Iluka has 80 buttons, all on cards: 8 full cards. She sews 46 of them onto a quilt. How many buttons are left? Work by place, then check by adding back.

Solution

Write the tens and ones in columns. There are 0 ones and 6 to take, so break a ten.

	tens	ones
80	8	0
after breaking a ten	7	10
take away 46	4	6
left	3	4

Ones: 10 - 6 = 4. Tens: 7 - 4 = 3. So:

80 - 46 = 34

Iluka snipped one card open, used 4 cards and 6 loose buttons, and has 3 cards and 4 loose buttons left. Check by adding back: 34 + 46 has 4 + 6 = 10 ones, exactly one new ten, and 3 + 4 + 1 = 8 tens. That is 80, the start, so the answer is right.

Example 6

Day 3: How many more

Iluka’s jar holds 72 blue buttons and 29 yellow buttons. How many more blue buttons are there than yellow?

Solution

“How many more” asks for the difference, 72 - 29. Match each yellow button with a blue one; the blue buttons with no partner are the difference, the same number as taking 29 from 72.

There are 2 ones and 9 to take, so break a ten: 7 tens 2 ones becomes 6 tens 12 ones. Then 12 - 9 = 3 ones and 6 - 2 = 4 tens:

72 - 29 = 43

There are 43 more blue buttons. Check: 43 + 29 has 3 + 9 = 12 ones, a new ten and 2 ones, and 4 + 2 + 1 = 7 tens, so 72.

Example 7

Day 4: Opening a box of buttons

The craft store keeps ten cards of buttons in a box, so a box is 100 buttons. Iluka has 1 box, 2 cards and 8 loose buttons. She sells 75 buttons. How many are left?

Solution

Iluka has 128 buttons, so the question is 128 - 75. The 75 is 7 tens and 5 ones.

Ones first: 8 ones and 5 to take is enough, so 8 - 5 = 3 ones and nothing is broken.

Tens: 2 tens and 7 to take is not enough. Break the hundred into ten tens, so 1 hundred 2 tens becomes 12 tens. Then 12 - 7 = 5 tens, and no hundreds are left.

128 - 75 = 53

Iluka opened the box, so she has 5 cards and 3 loose buttons left. Here the ones needed no trade and the tens did. Each place is decided on its own.

Example 8

Day 4: Two breaks

The post office keeps stamps in sheets of ten strips, so a sheet is 100 stamps. Bekzat has 1 sheet, 3 strips and 2 singles. He uses 57 stamps. How many are left?

Solution

The question is 132 - 57. Ones first: 2 ones and 7 to take, so break a ten. That gives 1 hundred, 2 tens and 12 ones, and 12 - 7 = 5 ones.

Tens: only 2 tens are left and 5 are to be taken, so break the hundred. That gives 12 tens, and 12 - 5 = 7 tens. No hundreds are left.

132 - 57 = 75

With blocks, 1 flat, 3 rods and 2 units became 12 rods and 12 units, and 5 rods and 7 units were taken away. Check by adding back: 75 + 57 has 5 + 7 = 12 ones, a new ten and 2 ones, then 7 + 5 + 1 = 13 tens, which is 1 hundred and 3 tens. That is 132.

Boundaries

The first number must be the larger. With whole numbers, 27 - 54 has no answer. Breaking a ten only reshapes what is there; it cannot make 27 hold 54.
A ten must be there to break. In 132 - 57 there were tens to break. When the tens place holds 0, as in 103 - 45, the ten must first come from a hundred; that is subtracting across zeros.
Break only when short. If the ones are enough, breaking a ten anyway still keeps the amount, but leaves ten or more ones in the answer, which must then be traded back to be written.
Ten for one, because each place is ten of the place to its right. Cards of ten buttons and strips of ten stamps match our numbers. If a store sold cards of twelve, snipping one open would give twelve loose buttons, not ten.
One break per place. One broken ten always gives the ones enough, since at most 9 are ever taken from a place.

Where this goes next

The same break carries every subtraction of whole numbers. With three-digit numbers it can happen in the ones, the tens or both, including across a zero, in subtracting within one thousand and subtracting across zeros. Later, the standard subtraction algorithm records each break in the written digits instead of in blocks or columns. Breaking a ten is the reverse of the new ten in regrouping in addition, which is why adding back checks a subtraction.

Exercises

Part 1: rods and units

Iluka has 4 cards of ten buttons and 2 loose buttons. A customer buys 16 buttons. After Iluka snips one card open, how many cards and loose buttons does she have? How many buttons are left after the sale?
Use rods and units to find 35 - 18. How many rods and units are there after breaking a rod, and how many are left?
A pile has 6 rods and 3 units. Break one rod into units. What pile is there now, and what number is it?
Bekzat has 5 strips of ten stamps and 1 single stamp. He uses 26 stamps. How many stamps are left?

Part 2: quick drawings, and deciding first

Look only at the ones digits. Which of these need a ten broken: (a) 46 - 19, (b) 46 - 13, (c) 52 - 17? Then find all three differences.
Make a quick drawing for 74 - 36 and find the difference.
Aolani has 85 stamps and sells 49. How many does she have left?
In 61 - 29, how many tens and ones does 61 become after a ten is broken? Find the difference.

Part 3: writing by place, zeros, adding back

Find 90 - 34 by place. Check by adding back.
Find 70 - 23 by place. Check by adding back.
Find 83 - 45 and check by adding back.
The post office had 92 stamps and Bekzat sold 56. How many are left?
Find the missing digit: 6\square - 27 = 38.
Iluka has 63 red buttons and 37 green buttons. How many more red buttons are there than green?

Part 4: a start over one hundred

Find 116 - 52. Which place needs a break?
Find 143 - 69. Which places need a break?
Iluka has 1 box of ten cards, 4 cards and 6 loose buttons. She sells 29 buttons. How many are left?
Write 152 as tens and ones only, with no hundreds. Then break one of those tens. What do you get?
Aolani has 162 stamps and uses 85. How many are left?

Reasoning

When a two-digit number is taken from a two-digit number, can the ones ever need two tens broken? Explain.
Show that 5 tens 12 ones and 6 tens 2 ones are the same amount.
Is this always true: when the first number has 0 ones and the number taken away does not, a ten must be broken? Explain, and give an example.
Find two subtractions with a difference of 18, one that needs a ten broken and one that does not.
Both numbers are two-digit and the ones need a ten broken. What is the largest the difference can be? Explain.

Answers

3 cards and 12 loose buttons; 26 buttons are left. Taking 1 card and 6 loose leaves 3 - 1 = 2 cards and 12 - 6 = 6 loose, so 42 - 16 = 26.
2 rods and 15 units after the break; 1 rod and 7 units are left, so 35 - 18 = 17. There are 5 units and 8 to take, so a rod is broken; then 15 - 8 = 7 and 2 - 1 = 1.
5 rods and 13 units, which is still 63, because 50 + 13 = 63.
25 stamps. 51 - 26: 1 single is not enough for 6, so tear a strip: 4 strips and 11 singles. Then 11 - 6 = 5 and 4 - 2 = 2, so 51 - 26 = 25.
(a) and (c) need a ten broken; (b) does not. (a) 6 ones and 9 to take: 3 tens 16 ones, so 46 - 19 = 27. (b) 6 ones and 3 to take is enough, so 46 - 13 = 33. (c) 2 ones and 7 to take: 4 tens 12 ones, so 52 - 17 = 35.
74 - 36 = 38. Draw 7 sticks and 4 dots; break a stick to get 6 sticks and 14 dots; cross out 3 sticks and 6 dots, leaving 3 sticks and 8 dots.
36 stamps. 85 - 49: 7 tens 15 ones take away 4 tens 9 ones leaves 3 tens 6 ones.
5 tens and 11 ones; 61 - 29 = 32, since 11 - 9 = 2 ones and 5 - 2 = 3 tens.
90 - 34 = 56. Break a ten: 8 tens 10 ones. Then 10 - 4 = 6 ones and 8 - 3 = 5 tens. Check: 56 + 34 = 90.
70 - 23 = 47. Break a ten: 6 tens 10 ones. Then 10 - 3 = 7 ones and 6 - 2 = 4 tens. Check: 47 + 23 = 70.
83 - 45 = 38. Break a ten: 7 tens 13 ones. Then 13 - 5 = 8 and 7 - 4 = 3. Check: 38 + 45 = 83.
36 stamps. 92 - 56: 8 tens 12 ones take away 5 tens 6 ones leaves 3 tens 6 ones. Check: 36 + 56 = 92.
The digit is 5. Adding back, 38 + 27 = 65, so the start is 65; and 65 - 27 = 38.
26 more. 63 - 37: 5 tens 13 ones take away 3 tens 7 ones leaves 2 tens 6 ones, so 63 - 37 = 26.
116 - 52 = 64. The ones need no break: 6 - 2 = 4. The tens do: 1 ten is not enough for 5, so the hundred is broken into 11 tens, and 11 - 5 = 6.
143 - 69 = 74. Both places. Ones: break a ten, 13 - 9 = 4. Tens: 3 are left, not enough for 6, so break the hundred: 13 - 6 = 7 tens.
117 buttons. She has 146. Ones: 6 is not enough for 9, so snip a card: 16 - 9 = 7. Tens: 3 - 2 = 1, and the box is not opened. So 146 - 29 = 117.
15 tens and 2 ones; breaking one ten gives 14 tens and 12 ones. Both are 152, since 150 + 2 = 152 and 140 + 12 = 152.
77 stamps. 162 - 85: ones, break a ten, 12 - 5 = 7. Tens: 5 left, not enough for 8, so break the hundred, 15 - 8 = 7. So 162 - 85 = 77.
No. At most 9 ones are taken. One broken ten gives the ones place at least 10, which is always enough.
5 tens 12 ones is 50 + 12 = 62, and 6 tens 2 ones is 60 + 2 = 62. Both are 62; one of the 6 tens was broken into ten ones.
Yes. The ones digit taken away is at least 1, and 0 ones cannot give 1 or more, so a ten must be broken. Example: 50 - 23 becomes 4 tens 10 ones, and 50 - 23 = 27.
One answer: 41 - 23 = 18 needs a break (1 one, 3 to take), and 39 - 21 = 18 does not. There are many others, such as 50 - 32 and 58 - 40.
79, for example 98 - 19 = 79. The first number has at most 9 tens, and one is broken, leaving at most 8. The number taken has at least 1 ten. So at most 8 - 1 = 7 tens are left, and the difference is less than 80.