Anka has pretzel rods and Naranbaatar brings more. The ones make , but a place holds only one digit. Where do the extra ones go?

Grade 2 · 1.NBT.C.4 · 2.NBT.B.5

Regrouping in addition

Anka has 47 pretzel rods and Naranbaatar brings 36 more. The ones make 13, but a place holds only one digit. Where do the extra ones go?

You need

Base ten blocks and trading ten for oneGrade 2

Unlocks

The standard addition algorithmGrade 4

Where you are 6 in Multi-digit addition

Number9

Arithmetic5

Addition and subtraction12

Multi-digit addition6

Adding two-digit numbers without regrouping Adding on an open number line

Regrouping in addition

Adding a one-digit number to a two-digit number by making a ten The partial sums method The standard addition algorithm

Why this exists

The child can already add two two-digit numbers when nothing spills over. In Adding two-digit numbers without regrouping, 32 + 45 is 3 + 4 = 7 tens and 2 + 5 = 7 ones, so 77. The child can also add a one-digit number that makes a new ten, as in 38 + 5 from Adding a one-digit number to a two-digit number by making a ten.

Now both numbers have tens and ones, and the ones are big. At Anka’s farm stand, pretzel rods are sold in bags of ten. Anka has 4 bags and 7 loose rods, which is 47. Naranbaatar brings 3 bags and 6 loose rods, which is 36. Bags go with bags: 4 + 3 = 7 bags. Loose rods go with loose rods: 7 + 6 = 13 loose rods.

“7 bags and 13 loose” is a true count, but it is not how the stand keeps its rods. Ten of the loose rods fill a new bag. Then there are 8 bags and 3 loose rods: 83 pretzel rods.

The digits need the same fix. The ones place holds one digit, from 0 to 9. Thirteen ones cannot be written there. Writing the 7 tens next to the 13 ones gives 713, which is seven hundred thirteen, nowhere near the real count. Regrouping is the fix: ten ones become one ten, so no place ever holds more than nine.

Definition

Regrouping in addition means trading ten of one place for one of the next place to the left, so that every place holds at most nine:

ten ones are traded for one ten;
ten tens are traded for one hundred.

The trade itself is the one from Base ten blocks and trading ten for one: ten units swap for one rod, and ten rods swap for one flat. What is new here is when the trade happens. It comes in the middle of a sum, when ones join ones (or tens join tens) and the count reaches ten or more. Some books call it “carrying” or “making a new ten”. It is the same idea.

For Anka’s pretzel rods it reads like this:

\begin{aligned} 47 + 36 &= 7 \text{ tens } 13 \text{ ones} \\ &= 8 \text{ tens } 3 \text{ ones} \\ &= 83 \end{aligned}

The middle line is the trade: 10 of the 13 ones became the eighth ten, and 3 ones are left.

Intuition

The lesson runs over four days. Day 1 makes the trade with blocks. Day 2 draws it. Day 3 writes it with numbers by place. Day 4 lets the tens reach ten too, so a sum passes one hundred.

Part 1: rods and units

Build each number with base ten blocks: 47 is 4 rods and 7 units, and 36 is 3 rods and 6 units. If the blocks come in two colors, build the second number in the other color. Push the two piles together. Rods join rods and units join units: 7 rods and 13 units.

Now line ten of the units up beside a rod. They are exactly as long as the rod, because a rod is ten units joined. Trade them for one rod. The pile is now 8 rods and 3 units, and the number can be read straight off it: 83.

Figure 147 + 36. The ten units in the dashed box are traded for one rod. Both rows are 83.

Have the child say both names aloud: “seven tens and thirteen ones is eight tens and three ones.” Both names describe one pile. The second name is the one that can be written with digits.

The farm stand is the same picture outside the classroom. A bag of ten pretzel rods is a ten and a loose rod is a one. The stand also packs its eggs in trays of ten, so a full tray is a ten and a loose egg is a one. When Lokelani fills a new tray from loose eggs, she is regrouping.

Part 2: a place value drawing

Blocks are slow to count out. A quick drawing does the same job: a stick (a short line) for each ten and a dot for each one. Put the dots in rows of five. Then ten dots are two full rows, easy to see and loop without counting them one by one.

For 58 + 25, draw 5 sticks and 8 dots, then 2 sticks and 5 dots. Together that is 7 sticks and 13 dots. Loop ten of the dots. The loop is a ten, so it becomes one more stick: 8 sticks and 3 dots, which is 83.

Figure 258 + 25 as a quick drawing. The ten dots in the loop become one new stick.

The drawing also shows before any adding whether a new ten is coming. Look only at the ones digits. If they add to ten or more, there will be a loop. If they add to nine or less, there will not, and the sum is the kind from adding without regrouping.

Part 3: writing by place

The pictures can now give way to numbers. Write the tens and the ones of each number in their own columns, add each column, then trade. For 49 + 35:

	tens	ones
49	4	9
35	3	5
put together	7	14
after the trade	8	4

Fourteen ones are 1 ten and 4 ones. The new ten joins the 7 tens, making 8 tens. The sum is 8 tens and 4 ones, which is 84.

It does not matter which column is added first; the trade can even wait until both columns are done. Starting with the ones is handy, though, because a new ten can then join the tens while they are being added. Two other ways to write a sum also regroup without naming it: the partial sums method keeps 70 and 14 as two parts and then combines them, and the open number line jumps past a ten.

Part 4: past one hundred

The same trade works one place up. For 76 + 58, the ones are 6 + 8 = 14, which is 1 ten and 4 ones. The tens are 7 + 5 = 12, and the new ten makes 13 tens. Thirteen tens is more than nine, so ten of the rods are traded for a flat, one hundred. What is left is 1 hundred, 3 tens and 4 ones: 134.

Figure 376 + 58. First ten units become a rod, then ten rods become a flat. Every row is 134.

A sum of two two-digit numbers passes one hundred exactly when its tens, with any new ten, reach ten or more. Whatever the place, the trade is always ten for one, and the rest of the lesson asks why that never changes the amount.

Results

Ten for one. Ten ones are one ten, and ten tens are one hundred:

\begin{aligned} 10 \text{ ones} &= 1 \text{ ten} \\ 10 \text{ tens} &= 1 \text{ hundred} \end{aligned}

When the ones make a new ten. If the ones digits add to ten or more, the sum gets one new ten, and its ones digit is what is left after ten are traded. In 47 + 36, 7 + 6 = 13, so there is one new ten and the ones digit is 3.

The trade keeps the amount. Before and after a trade, the pile is the same number:

\begin{aligned} 7 \text{ tens } 13 \text{ ones} &= 70 + 13 = 83 \\ 8 \text{ tens } 3 \text{ ones} &= 80 + 3 = 83 \end{aligned}

Why it is true

Why the trade keeps the amount. A trade takes away ten units, worth 1 each, which is 10. It puts back one rod, which is worth 10. Nothing was added and nothing was lost, so the total stays the same. The blocks show this with no counting at all: ten units laid end to end are as long as one rod. The trade one place up works the same way: ten rods are worth 100, and so is one flat.

Why ones may join ones and tens join tens. The sum 47 + 36 is 40 + 7 + 30 + 6. Adding in any order gives the same total (the commutative property), and adding in any grouping gives the same total (the associative property). So the tens can be gathered as 40 + 30 = 70 and the ones as 7 + 6 = 13, and then 70 + 13 = 83.

Why the trade is needed at all. Each place shows one digit, and there are only ten digits, 0 to 9. A count of ten or more in one place has no single digit. Ten of one place are exactly one of the next place, so trading ten for one always leaves a count from 0 to 9 behind. That is what lets every whole number be written in just one way.

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: Pretzel rods at the stand

Anka has 4 bags of ten pretzel rods and 7 loose rods. Naranbaatar brings 3 more bags and 6 loose rods. How many pretzel rods are there now?

Solution

Anka has 47 rods and Naranbaatar brings 36, so the question is 47 + 36. Model it with blocks: 4 rods and 7 units, then 3 rods and 6 units.

Rods join rods: 4 + 3 = 7 rods. Units join units: 7 + 6 = 13 units. That is 7 tens and 13 ones, which is a true count, but 13 is too many for the ones place.

Ten of the units make a new rod, so trade them. There are now 8 rods and 3 units:

47 + 36 = 83

At the stand, the ten loose rods go into a new bag: 8 bags and 3 loose rods. There are 83 pretzel rods.

Example 2

Day 1: A full tray from loose eggs

Lokelani has 34 eggs: 3 full trays of ten and 4 loose eggs. Anka gives her 26 more: 2 trays and 6 loose eggs. How many full trays can Lokelani have, and how many eggs is that?

Solution

Cartons join trays: 3 + 2 = 5 trays. Loose eggs join loose eggs: 4 + 6 = 10 loose eggs.

Ten loose eggs fill exactly one more tray, so Lokelani packs them. Now she has 6 trays and no loose eggs. In place value words, 5 tens and 10 ones became 6 tens and 0 ones:

34 + 26 = 60

The ones digit of the sum is 0. That is not “nothing happened in the ones”. The ones made exactly one ten, and the ten moved over.

Example 3

Day 2: A drawing for fifty-eight and twenty-five

Naranbaatar sells 58 pretzel rods on Saturday and 25 on Sunday. Use a place value drawing to find how many he sold.

Solution

The question is 58 + 25. Draw 5 sticks and 8 dots for 58, and 2 sticks and 5 dots for 25, with the dots in rows of five.

Sticks join sticks: 5 + 2 = 7 tens. Dots join dots: 8 + 5 = 13 ones. Two full rows of five dots make ten, so loop them. The loop is a ten, so draw one more stick and cross out the looped dots.

Now the drawing shows 8 sticks and 3 dots, which is 83:

58 + 25 = 83

Naranbaatar sold 83 pretzel rods. This is the same total as Anka’s 47 + 36. Different numbers can give the same sum, and in both the ones made exactly one new ten.

Example 4

Day 2: Will there be a new ten?

Without finding the sums first, decide which of 27 + 45 and 27 + 42 makes a new ten. Then find both sums.

Solution

Only the ones digits decide. In 27 + 45 the ones are 7 + 5 = 12, which is ten or more, so a new ten is coming. In 27 + 42 the ones are 7 + 2 = 9, which is less than ten, so no trade is needed.

For 27 + 45: tens 2 + 4 = 6, ones 12. Trade ten ones for a ten: 7 tens and 2 ones.

27 + 45 = 72

For 27 + 42: tens 2 + 4 = 6, ones 9. Nine ones fit in the ones place, so the sum is 6 tens and 9 ones:

27 + 42 = 69

The two sums differ by only 3 in what was added, yet one sum has 7 tens and the other 6. The new ten is the reason.

Example 5

Day 3: Eggs on two days, by place

Lokelani collects 69 eggs on Monday and 24 eggs on Tuesday. How many eggs is that? Work by place, with no picture.

Solution

Write the tens and the ones of each number in their own columns.

	tens	ones
69	6	9
24	2	4
put together	8	13
after the trade	9	3

The ones add to 9 + 4 = 13, which is 1 ten and 3 ones. The tens add to 6 + 2 = 8, and the new ten makes 9 tens. So:

69 + 24 = 93

Lokelani collected 93 eggs: 9 full trays and 3 loose eggs.

Example 6

Day 3: Three baskets

Anka’s basket holds 18 eggs, Naranbaatar’s holds 27 and Lokelani’s holds 35. How many eggs are there in the three baskets?

Solution

By place, ones first: 8 + 7 + 5 = 20 ones. Twenty ones are two tens and 0 ones, because ten ones make a ten twice over. Tens: 1 + 2 + 3 = 6 tens, and the two new tens make 8 tens.

18 + 27 + 35 = 80

Check by adding two numbers at a time. First 18 + 27: the ones are 15, so 3 tens and 5 ones become 4 tens and 5 ones, which is 45. Then 45 + 35: the ones are 10, so 7 tens and 10 ones become 8 tens, which is 80. The two ways agree, as the associative property says they must. There are 80 eggs, exactly 8 trays.

Example 7

Day 4: Past one hundred

Naranbaatar’s stand sells 76 pretzel rods on Saturday and 58 on Sunday. How many rods is that?

Solution

Ones first: 6 + 8 = 14 ones, which is 1 ten and 4 ones. Tens: 7 + 5 = 12 tens, and the new ten makes 13 tens.

Thirteen tens is too many for the tens place. Ten tens are one hundred, so trade them: 13 tens is 1 hundred and 3 tens. The sum is 1 hundred, 3 tens and 4 ones:

76 + 58 = 134

With blocks, 12 rods and 14 units became 13 rods and 4 units, then 1 flat, 3 rods and 4 units. Each trade was ten for one, so each row was still 134. Naranbaatar sold 134 pretzel rods: 13 bags and 4 loose.

Example 8

Day 4: One trade sets off another

Lokelani has 73 eggs and collects 27 more. How many eggs does she have?

Solution

Ones first: 3 + 7 = 10 ones. That is exactly 1 ten and 0 ones. Tens: 7 + 2 = 9 tens, and the new ten makes 10 tens. Ten tens are exactly 1 hundred and 0 tens.

73 + 27 = 100

The first trade caused the second. Before the new ten arrived, the tens were only 9, which fits. The new ten pushed them to 10, and they were traded too. Lokelani has 100 eggs, which is 10 full trays.

Boundaries

The trade is at ten because each place is worth ten of the place to its right. Bags of ten pretzel rods and trays of ten eggs match our numbers. If a stand sold bags of twelve, ten loose rods would not fill a bag, and trading ten for one would change the amount.
Only a full ten is traded. Nine ones stay nine ones. If the ones add to nine or less, there is no trade, and the sum is one without regrouping.
Ten for one, never more. A trade always removes exactly ten from a place and adds exactly one to the next place. Twenty ones are two trades, as in the three baskets.
Addition only. In subtraction the trade runs the other way, one ten broken into ten ones, which is regrouping in subtraction.

Where this goes next

The same trade carries every sum of whole numbers, however long. With three-digit numbers it can happen in the ones, the tens or both, and ten hundreds become a thousand, as in multi-digit addition. Later, the standard addition algorithm records the new ten as a small 1 over the next column. The trade run backward is regrouping in subtraction, used in subtracting within one thousand.

Exercises

Part 1: rods and units

Anka has 5 bags of ten pretzel rods and 7 loose rods. Naranbaatar gives her 1 bag and 6 loose rods. How many bags and loose rods does she have before making a new bag? After? How many rods is that?
Use rods and units to find 28 + 35. How many rods and units are there before the trade, and after?
Lokelani has 44 eggs and collects 18 more. How many full trays of ten and loose eggs does she end with?
A pile has 6 rods and 15 units. Trade until no kind has more than nine. What pile is left, and what number is it?

Part 2: place value drawings

Look at the drawing for Exercise 5. What sum does it show? Loop a ten and find the total.

Figure 4The drawing for Exercise 5.

Look only at the ones digits. Which of these sums make a new ten: (a) 53 + 24, (b) 53 + 28, (c) 36 + 34? Then find all three sums.
Make a place value drawing for 17 + 59 and find the sum.
Lokelani has 45 eggs. Anka brings her 3 trays of ten and 9 loose eggs. How many eggs does Lokelani have now?

Part 3: writing by place

Find 38 + 47 by place, with a tens column and a ones column.
Write each as one number: (a) 4 tens 17 ones, (b) 6 tens 10 ones, (c) 8 tens 19 ones.
Find 56 + 19 + 14 by place.
Find the missing digit: 3\square + 25 = 62.
Anka sells 27 pretzel rods in the morning and 48 in the afternoon. How many does she sell in all?

Part 4: past one hundred

Find 85 + 47. Name the sum in hundreds, tens and ones.
Find 82 + 51. Which place needs a trade, and which does not?
What number is 13 tens and 6 ones? Write it in hundreds, tens and ones.
Find 95 + 27.
The farm stand sells 68 eggs on Monday and 75 on Tuesday. How many eggs is that?

Reasoning

When two two-digit numbers are added, can the ones ever make two new tens? Explain.
Can two two-digit numbers add to more than 200? Explain.
Show that 6 tens 12 ones and 7 tens 2 ones are the same amount.
Find two two-digit numbers whose sum is exactly 100, where the ones make a new ten. Is there more than one answer?
The sum 47 + 36 is 7 tens and 13 ones. Why can it not be written as 713?
A different stand sells pretzel rods in bags of twelve. Anka has 1 bag and 9 loose rods and is given 1 bag and 5 loose rods. Do ten loose rods make a new bag there? How many full bags and loose rods can she end with?

Answers

6 bags and 13 loose rods before; 7 bags and 3 loose rods after; 73 rods. Bags: 5 + 1 = 6; loose: 7 + 6 = 13, and ten of them fill a bag. So 57 + 16 = 73.
5 rods and 13 units before; 6 rods and 3 units after, so 28 + 35 = 63. Rods: 2 + 3 = 5; units: 8 + 5 = 13, and ten units trade for a rod.
6 trays and 2 loose eggs, which is 62. Cartons: 4 + 1 = 5; loose: 4 + 8 = 12, and ten of them fill one more tray. So 44 + 18 = 62.
7 rods and 5 units, which is 75. Ten of the 15 units trade for a rod, leaving 5 units.
The sum is 39 + 46 = 85. Sticks: 3 + 4 = 7 tens; dots: 9 + 6 = 15 ones. Loop ten dots to make an eighth stick, leaving 5 dots: 8 tens and 5 ones.
(b) and (c) make a new ten; (a) does not. (a) 3 + 4 = 7 ones, so 53 + 24 = 77. (b) 3 + 8 = 11 ones, so 7 tens and 11 ones become 53 + 28 = 81. (c) 6 + 4 = 10 ones, so 6 tens and 10 ones become 36 + 34 = 70.
17 + 59 = 76. The drawing has 1 + 5 = 6 sticks and 7 + 9 = 16 dots; loop ten dots for a seventh stick, leaving 6 dots.
84 eggs. She is brought 39, and 45 + 39 is 7 tens and 14 ones, which is 8 tens and 4 ones.
38 + 47 = 85. Tens: 3 + 4 = 7; ones: 8 + 7 = 15, which is 1 ten and 5 ones. The new ten makes 8 tens.
(a) 57, since 17 ones is 1 ten and 7 ones, so 5 tens and 7 ones. (b) 70, since 10 ones is 1 ten. (c) 99, since 19 ones is 1 ten and 9 ones, so 9 tens and 9 ones.
56 + 19 + 14 = 89. Ones: 6 + 9 + 4 = 19, which is 1 ten and 9 ones. Tens: 5 + 1 + 1 = 7, and the new ten makes 8.
The digit is 7, so 37 + 25 = 62. The tens 3 + 2 = 5 became 6, so the ones made a new ten and \square + 5 = 12, which gives 7.
75 rods. 27 + 48 is 6 tens and 15 ones, which is 7 tens and 5 ones.
85 + 47 = 132, which is 1 hundred, 3 tens and 2 ones. Ones: 5 + 7 = 12, so one new ten and 2 ones. Tens: 8 + 4 + 1 = 13, so ten tens trade for a hundred, leaving 3 tens.
82 + 51 = 133. The ones need no trade: 2 + 1 = 3. The tens do: 8 + 5 = 13 tens is 1 hundred and 3 tens.
136, which is 1 hundred, 3 tens and 6 ones. Ten of the 13 tens make the hundred.
95 + 27 = 122. Ones: 5 + 7 = 12, one new ten and 2 ones. Tens: 9 + 2 + 1 = 12, which is 1 hundred and 2 tens.
143 eggs. 68 + 75: ones 8 + 5 = 13, one new ten and 3 ones; tens 6 + 7 + 1 = 14, which is 1 hundred and 4 tens.
No. The largest ones digits are 9 and 9, and 9 + 9 = 18, which is one ten and 8 ones. Two new tens would need at least 20 ones.
No. The largest two-digit number is 99, and 99 + 99 = 198, which is less than 200. Any other pair is smaller still.
6 tens 12 ones is 60 + 12 = 72, and 7 tens 2 ones is 70 + 2 = 72. Both are 72; ten of the 12 ones were traded for the seventh ten.
One answer is 45 + 55 = 100: ones 5 + 5 = 10 make a new ten, and tens 4 + 5 + 1 = 10 make a hundred. There are many answers, such as 63 + 37 or 18 + 82; any pair adding to 100 whose ones digits are not both 0 works.
Because 713 means 7 hundreds, 1 ten and 3 ones, far more than the rods there are. Each place holds one digit, so the 13 ones must first become 1 ten and 3 ones, giving 8 tens and 3 ones, which is 83.
No. A bag there holds twelve, so ten loose rods do not fill one. She has 2 bags and 9 + 5 = 14 loose rods, which is 38 rods. Twelve of the loose rods fill a third bag, so she ends with 3 bags and 2 loose rods. The trade at ten works only when a bag holds ten.