Grade 2 · 1.NBT.C.4 · 3.NBT.A.2

The partial sums method

One class brings 246 cans and 138 bottles to a school bottle drive, and the numbers are too big to count on. Can each number be broken into hundreds, tens and ones, each place added on its own, and the pieces put back together?

You need

Mental strategiesGrades 2 to 5 Expanded formGrade 2 Adding two-digit numbers without regroupingGrade 1

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

When you reach for it

A school is holding a bottle drive. Every class brings in cans and bottles, and each Friday the counts go up on the hall board. Tumaini’s class brought 246 cans and 138 bottles. How many is that in all?

Counting on 138 from 246, one at a time, would take all afternoon. The child already has two tools that make it quick. From Expanded form they know a number is the sum of what its digits are worth: 246 = 200 + 40 + 6. From Adding two-digit numbers without regrouping they know that tens join tens and ones join ones. The partial sums method puts the two together:

break each number into hundreds, tens and ones;
add the hundreds to the hundreds, the tens to the tens and the ones to the ones;
add those answers, the partial sums, to get the total.

For 246 + 138 the partial sums are 200 + 100 = 300, 40 + 30 = 70 and 6 + 8 = 14. Then 300 + 70 + 14 = 384.

Reach for it whenever two or more whole numbers are put together and no basic fact gives the answer. It is also the method the child will lean on most while bigger sums are new: every step is a fact they know, done in one place at a time, and every partial sum can be seen in the blocks.

The earlier lesson kept the ones at nine or fewer. Here that limit goes away. When the ones make 14, the partial sum is simply 14, and it is added in at the end like any other part. No special step is needed.

The lesson runs over four days. Day 1 adds two-digit numbers within 100. Day 2 moves to three-digit numbers. Day 3 takes sums where more than one place reaches ten or more. Day 4 adds three or four numbers at once, and breaks apart just one of two numbers. Base ten blocks, paper for quick drawings, and a sheet for the written sums are all the lesson needs.

The procedure

Day 1: two-digit numbers, tens with tens and ones with ones

Build both numbers. Pilialoha works at her aunt’s bead shop after school. Beads come on strings of ten, and some are loose. She has 57 red beads and 26 blue beads. Ask the child to build both numbers with base ten blocks: 57 is 5 rods and 7 small squares, and 26 is 2 rods and 6 small squares. Put one number above the other.

Sort by kind. Slide the rods together and the small squares together. Now count each kind on its own. Five rods and two rods are seven rods, which is 70. Seven small squares and six small squares are 13 small squares. Those two counts are the partial sums (Figure 1).

Figure 157 + 26. The rods make 70, the small squares make 13, and 70 + 13 = 83.

Put the parts back together. All the blocks together are 70 + 13. The child knows 13 is 10 + 3, so 70 + 13 is 70 + 10 + 3, which is 80 + 3, or 83. So 57 + 26 = 83.

Write it down. The written record follows the blocks line for line. First break each number apart, as in expanded form. Then write one line for each place, then one line to combine:

\begin{aligned} 57 + 26 &= (50 + 7) + (20 + 6) \\ 50 + 20 &= 70 \\ 7 + 6 &= 13 \\ 70 + 13 &= 83 \end{aligned}

Why it is allowed. Breaking a number into its parts does not change it: 57 and 50 + 7 are the same amount. Once both numbers are broken apart, the sum is four parts added together. The order of the parts can change without changing the total; that is the commutative property of addition. The parts can also be grouped in any way; that is the associative property. So 50 + 7 + 20 + 6 may be added as (50 + 20) + (7 + 6). The method is those two properties, used on purpose.

A partial sum of ten or more is just a number. The ones made 13. Nothing has gone wrong: 13 is a perfectly good part. It is added to 70 the same way any two numbers are added. Seeing that ten of those 13 small squares could be snapped into a rod is the idea of Regrouping in addition, and partial sums leave that trade to the very last line.

Day 2: three-digit numbers

One more place. A three-digit number breaks into three parts, so there are three partial sums. Build 246 and 138 with flats, rods and small squares, one above the other, and sort by kind (Figure 2). Two flats and one flat are 3 flats, 300. Four rods and three rods are 7 rods, 70. Six small squares and eight small squares are 14. Together, 300 + 70 + 14 = 384.

Figure 2246 + 138. Each kind of block is counted on its own, and the three partial sums make 384.

Hundreds count like tens. The fact 2 + 1 = 3 counts flats here, so it means 200 + 100 = 300. In the same way 4 + 3 = 7 counts rods, so it means 40 + 30 = 70. Saying the unit aloud keeps the place straight: “two hundred and one hundred is three hundred”.

A written record down the page. Once the child can write the lines across, the same work can run down the page. The two numbers go on top, lined up on the right so ones sit under ones. Each partial sum gets its own line under a rule, and a second rule leads to the total:

\begin{array}{rl} 246 & \\ +\;138 & \\ \hline 300 & 200 + 100 \\ 70 & 40 + 30 \\ 14 & 6 + 8 \\ \hline 384 & \end{array}

The notes on the right say where each line came from. With practice they can be left off, but each line still means one place. The partial sums may be written in any order. Starting with the hundreds, as here, matches the way the number is read. Starting with the ones works just as well.

A missing part. Some numbers have a 0 digit, and some have fewer digits than the other number. In 608 + 71, the 608 has no tens and the 71 has no hundreds. A missing part is worth 0, so that place’s partial sum is just the part that is there. Example 4 does this.

Day 3: more than one place reaches ten or more

Two big partial sums. In 357 + 285 the tens give 50 + 80 = 130 and the ones give 7 + 5 = 12. Both are more than one digit’s worth. That is fine: each one is a part, and all three parts are added at the end. The hundreds give 300 + 200 = 500, so the total is 500 + 130 + 12.

From blocks to a quick drawing. Building three-digit numbers with blocks takes a lot of blocks. A quick drawing does the same job with a pencil: a small square for each hundred, a stick for each ten and a dot for each one. Figure 3 shows 357 + 285 this way. The child counts squares, then sticks, then dots, exactly as with the blocks.

Figure 3A quick drawing of 357 + 285. A square is a hundred, a stick is a ten, a dot is a one.

Combining the partial sums. Adding 500 + 130 + 12 is itself a sum of parts. Add the two biggest first: 500 + 130 = 630, because 5 hundreds and 1 hundred are 6 hundreds and the 30 comes along. Then 630 + 12 = 642, because 30 and 12 make 42. So 357 + 285 = 642. No special rule was used. Each step is a sum of hundreds, tens and ones that the child has done before.

Reaching a thousand. A total can land exactly on 1{,}000. If the partial sums are 900, 90 and 10, then 90 + 10 = 100, and 900 + 100 = 1{,}000: ten hundreds make one thousand. Example 6 does this.

Day 4: three or four numbers, and breaking apart just one

More than two numbers. The bottle drive board lists four classes. Partial sums work for any number of addends: break every number apart, add all the tens, add all the ones, and combine. The properties that let two numbers be pulled apart and regrouped work for four numbers just as well. With four two-digit numbers the ones can reach 30 or more, and the tens can reach a hundred or more. Each partial sum is still just a number. Example 7 adds four classes’ counts.

Looking for easy pairs. When there are many ones to add, the child may add them in any order. Pairs that make ten, such as 2 + 8 or 3 + 7, are worth finding first. That is the same make-a-ten idea as in Adding by making a ten, used inside one partial sum.

Breaking apart only one number. Sometimes it is quicker to keep the first number whole and break apart only the second. For 65 + 29: keep 65, break 29 into 20 + 9, add the tens first (65 + 20 = 85), then the ones (85 + 9 = 94). This is the same idea with fewer pieces, and the same two properties allow it. It is also the path of hops on an open number line: one hop of 20, then a hop of 9. Example 8 does a sum both ways.

Worked examples

Examples 1 and 2 go with Day 1, Examples 3 and 4 with Day 2, Examples 5 and 6 with Day 3, and Examples 7 and 8 with Day 4. Blocks or a quick drawing on the table let the child see every partial sum.

Example 1

Day 1: Two colors of beads

At the bead shop, Saltanat strings 43 green beads and then 35 white beads on a necklace. How many beads are on the necklace?

Solution

Break each number into tens and ones: 43 = 40 + 3 and 35 = 30 + 5. With blocks that is 4 rods and 3 small squares, and 3 rods and 5 small squares.

Add the tens: 40 + 30 = 70. Four tens and three tens are seven tens.

Add the ones: 3 + 5 = 8.

Combine the partial sums: 70 + 8 = 78. There are 78 beads on the necklace.

Here the ones stayed below ten, just as in the earlier lesson. What is new is that both numbers have tens and ones, so there are two partial sums to find and one more step to put them together.

Example 2

Day 1: Ones that make more than ten

Pilialoha has 57 red beads and 26 blue beads. How many beads does she have in all?

Solution

Break each number apart: 57 = 50 + 7 and 26 = 20 + 6.

Add the tens: 50 + 20 = 70.

Add the ones: 7 + 6 = 13.

The ones made more than ten, and that changes nothing about the method. Combine the partial sums: 70 + 13 = 83. To see why, think of 13 as 10 + 3: then 70 + 10 = 80, and 80 + 3 = 83.

Pilialoha has 83 beads. Figure 1 shows the blocks. Check by counting on from 57 by tens and then ones: 67, 77, then 78, 79, 80, 81, 82, 83.

Example 3

Day 2: The first bottle count

Tumaini’s class brought 246 cans and 138 bottles to the bottle drive. How many cans and bottles is that?

Solution

Break each number into hundreds, tens and ones:

246 = 200 + 40 + 6, \qquad 138 = 100 + 30 + 8

Add each place on its own. Hundreds: 200 + 100 = 300. Tens: 40 + 30 = 70. Ones: 6 + 8 = 14.

Combine the partial sums. The 300 and 70 make 370, and the 14 is 10 + 4, so 370 + 14 = 384. In one line, 300 + 70 + 14 = 384.

The class brought 384 cans and bottles. Figure 2 shows the same sum with blocks, and the record down the page under Day 2 is this example written out.

Example 4

Day 2: A number with a zero and a shorter number

By Thursday the second grade had collected 608 cans. On Friday 71 more cans came in. How many cans is that now?

Solution

Break each number apart. The 0 in 608 means no tens, and 71 has no hundreds:

608 = 600 + 8, \qquad 71 = 70 + 1

Hundreds: only 608 has any, so the partial sum is 600.

Tens: only 71 has any, so the partial sum is 70.

Ones: 8 + 1 = 9.

Combine: 600 + 70 + 9 = 679. The second grade has 679 cans.

The place-by-place sort is what keeps the 7 tens of 71 away from the hundreds. Lined up on the right, the 7 sits under the 0 of 608, in the tens column, where it belongs.

Example 5

Day 3: Two partial sums of ten or more

On the last Friday of the drive, Room 12 brought 357 bottles and Room 14 brought 285 bottles. How many bottles did the two rooms bring?

Solution

Break each number apart:

357 = 300 + 50 + 7, \qquad 285 = 200 + 80 + 5

Hundreds: 300 + 200 = 500. Tens: 50 + 80 = 130, since 5 tens and 8 tens are 13 tens. Ones: 7 + 5 = 12.

The record down the page is:

\begin{array}{r} 357 \\ +\;285 \\ \hline 500 \\ 130 \\ 12 \\ \hline 642 \end{array}

Combine the partial sums: 500 + 130 = 630, and 630 + 12 = 642. So 500 + 130 + 12 = 642.

The two rooms brought 642 bottles. Figure 3 is the quick drawing. Check by adding the partial sums in another order: 12 + 130 = 142, and 142 + 500 = 642.

Example 6

Day 3: Enough beads for the big jar

The bead shop has a jar that holds 1{,}000 beads. Saltanat pours in a bag of 468 beads and then a bag of 532 beads. Is the jar full?

Solution

Break each number apart: 468 = 400 + 60 + 8 and 532 = 500 + 30 + 2.

Hundreds: 400 + 500 = 900. Tens: 60 + 30 = 90. Ones: 8 + 2 = 10.

Combine. The 90 and the 10 make 100. Then 900 + 100 is ten hundreds, which is one thousand:

900 + 90 + 10 = 900 + 100 = 1{,}000

So 468 + 532 = 1{,}000, and the jar is exactly full. Every place made a ten of its own unit, and each one passed up into the next place when the partial sums were combined.

Example 7

Day 4: Four classes on one morning

On Monday morning four classes counted their bottles: 23, 41, 36 and 18. How many bottles is that?

Solution

Break every number into tens and ones, then add each place across all four numbers.

Tens: 20 + 40 + 30 + 10 = 100. Ten tens make one hundred.

Ones: 3 + 1 + 6 + 8. No two of these make ten, but 3 + 1 = 4, and 4 + 6 = 10. Then add the 8: the ones partial sum is 10 + 8 = 18.

Combine: 100 + 18 = 118. The four classes counted 118 bottles.

Adding in any order and grouping in any way is allowed because of the commutative and associative properties, which is why the ones could be paired up to make ten first.

Example 8

Day 4: Breaking apart one number or both

At the bead shop, Tumaini has 65 beads and buys 29 more. Find the total two ways: by breaking apart only 29, and by breaking apart both numbers.

Solution

First, breaking apart only 29. Keep 65 whole and write 29 = 20 + 9. Add the tens first: 65 + 20 = 85, since six tens and two tens are eight tens and the 5 ones stay. Then add the ones: 85 + 9 = 94.

Now breaking apart both: 65 = 60 + 5 and 29 = 20 + 9. Tens: 60 + 20 = 80. Ones: 5 + 9 = 14. Combine: 80 + 14 = 94.

Both ways give 94, as they must: each one adds the same four parts, 60, 5, 20 and 9, grouped differently. Tumaini has 94 beads.

Boundaries

Like with like. Partial sums only add parts of the same place: hundreds with hundreds, tens with tens, ones with ones. The method rests on each part being worth what its place says, so every number must first be broken apart correctly, as in Expanded form.

Whole numbers. In this lesson every number is a whole number (0, 1, 2, 3 and so on). The method itself works for numbers of any length; here the totals stay within 1{,}000.

The last step is still an addition. Partial sums turn one hard sum into a few easy ones, but combining them is still adding. When a partial sum is ten or more, the combining step crosses a ten or a hundred, as in 70 + 13 or 500 + 130.

Addition only. The method adds. Subtracting place by place, with no trading, works only when each digit being taken away is no bigger than the digit it is taken from, as in Subtracting two-digit numbers without regrouping. Partial sums of ten or more have no match in subtraction.

Where this goes next

Partial sums write every part down. Regrouping in addition looks at the moment a place reaches ten and trades ten ones for one ten, or ten tens for one hundred, so that no place holds more than nine. The standard addition algorithm then folds each partial sum into the next column as soon as it is found and writes less; its first part starts from partial sums like these.

The same breaking apart runs through Mental strategies and the hops of an open number line. Compensation takes a different route: it changes a number to a friendlier one, such as 29 to 30, and then fixes the total.

Exercises

Day 1: two-digit numbers

Add with partial sums: 62 + 35.
Add with partial sums: 36 + 48.
Add with partial sums: 39 + 54.
Tumaini strings 58 yellow beads and 34 orange beads. How many beads is that?
Add with partial sums: 45 + 45. What is special about the ones partial sum?

Day 2: three-digit numbers

Write the two numbers shown by the blocks in Figure 4. Then add them with partial sums.

Figure 4Blocks for Exercise 6.

Add with partial sums: 324 + 152.
Add with partial sums: 517 + 268.
Add with partial sums: 430 + 206.
Add with partial sums: 742 + 56.
Saltanat’s class brought 185 cans in March and 214 cans in April. How many cans is that?

Day 3: partial sums of ten or more

Add with partial sums: 264 + 178.
Add with partial sums: 395 + 238.
Add with partial sums: 486 + 357.
Add with partial sums: 509 + 291.
The bead shop has 375 glass beads and 448 wooden beads. How many beads is that?
Add with partial sums: 199 + 199.

Day 4: more numbers, and breaking apart one

Add with partial sums: 32 + 14 + 21 + 25.
Add with partial sums: 27 + 35 + 18.
Four classes brought 46, 38, 52 and 29 bottles. How many bottles is that?
Find 67 + 29 by keeping 67 whole and breaking apart only 29.
Pilialoha added two three-digit numbers. Her partial sums were 400, 120 and 16. What was the total?

Explain

In Example 3 the hundreds were added first. Would starting with the ones change the answer? Say why or why not.
When two numbers are added, can the ones partial sum ever be 19 or more? What is the largest it can be?
In 41 + 37, the tens digits are 4 and 3. Explain why the tens partial sum is 70 and not 7.

Answers

97. Tens: 60 + 30 = 90. Ones: 2 + 5 = 7. Then 90 + 7 = 97.
84. Tens: 30 + 40 = 70. Ones: 6 + 8 = 14. Then 70 + 14 = 84.
93. Tens: 30 + 50 = 80. Ones: 9 + 4 = 13. Then 80 + 13 = 93.
92 beads. Tens: 50 + 30 = 80. Ones: 8 + 4 = 12. Then 80 + 12 = 92.
90. Tens: 40 + 40 = 80. Ones: 5 + 5 = 10. Then 80 + 10 = 90. The ones make exactly ten, so the total has 0 ones.
The numbers are 163 and 228. Hundreds: 100 + 200 = 300. Tens: 60 + 20 = 80. Ones: 3 + 8 = 11. Total: 300 + 80 + 11 = 391.
476. Hundreds: 300 + 100 = 400. Tens: 20 + 50 = 70. Ones: 4 + 2 = 6. Then 400 + 70 + 6 = 476.
785. Hundreds: 500 + 200 = 700. Tens: 10 + 60 = 70. Ones: 7 + 8 = 15. Then 700 + 70 + 15 = 785.
636. Hundreds: 400 + 200 = 600. Tens: only 430 has any, so 30. Ones: only 206 has any, so 6. Then 600 + 30 + 6 = 636.
798. Hundreds: only 742 has any, so 700. Tens: 40 + 50 = 90. Ones: 2 + 6 = 8. Then 700 + 90 + 8 = 798.
399 cans. Hundreds: 100 + 200 = 300. Tens: 80 + 10 = 90. Ones: 5 + 4 = 9. Then 300 + 90 + 9 = 399.
442. Hundreds: 200 + 100 = 300. Tens: 60 + 70 = 130. Ones: 4 + 8 = 12. Then 300 + 130 = 430 and 430 + 12 = 442.
633. Hundreds: 300 + 200 = 500. Tens: 90 + 30 = 120. Ones: 5 + 8 = 13. Then 500 + 120 = 620 and 620 + 13 = 633.
843. Hundreds: 400 + 300 = 700. Tens: 80 + 50 = 130. Ones: 6 + 7 = 13. Then 700 + 130 = 830 and 830 + 13 = 843.
800. Hundreds: 500 + 200 = 700. Tens: 0 + 90 = 90. Ones: 9 + 1 = 10. Then 90 + 10 = 100 and 700 + 100 = 800.
823 beads. Hundreds: 300 + 400 = 700. Tens: 70 + 40 = 110. Ones: 5 + 8 = 13. Then 700 + 110 = 810 and 810 + 13 = 823.
398. Hundreds: 100 + 100 = 200. Tens: 90 + 90 = 180. Ones: 9 + 9 = 18. Then 200 + 180 = 380 and 380 + 18 = 398.
92. Tens: 30 + 10 + 20 + 20 = 80. Ones: 2 + 4 + 1 + 5 = 12. Then 80 + 12 = 92.
80. Tens: 20 + 30 + 10 = 60. Ones: 7 + 5 + 8 = 20. Then 60 + 20 = 80.
165 bottles. Tens: 40 + 30 + 50 + 20 = 140. Ones: 6 + 8 + 2 + 9 = 25 (pair 2 + 8 first). Then 140 + 25 = 165.
96. Write 29 = 20 + 9. Then 67 + 20 = 87 and 87 + 9 = 96.
536. Combine the partial sums: 400 + 120 = 520 and 520 + 16 = 536.
No. The three partial sums are the same three numbers, 300, 70 and 14, whichever is found first, and adding them in any order gives the same total by the commutative and associative properties. Ones first: 14 + 70 = 84 and 84 + 300 = 384.
No. A ones digit is at most 9, so the ones partial sum of two numbers is at most 9 + 9 = 18. It can be 18, as in 199 + 199.
The 4 in 41 is in the tens place, so it is worth 40, and the 3 in 37 is worth 30. The tens partial sum adds what the digits are worth: 40 + 30 = 70. The fact 4 + 3 = 7 counts tens, so it means seven tens, which is 70. Writing 7 would count those seven rods as seven small squares.