The standard addition algorithm

When you reach for it

Zahra’s school runs a recycling drive twice a year. In the fall the students collected 48{,}675 cans. In the spring they collected 27{,}548 cans. How many cans is that in all?

The answer is the sum 48{,}675 + 27{,}548. No basic fact gives it, and the numbers are too long to hold in your head. The standard addition algorithm adds whole numbers of any length. You write one number under the other, lined up by place value. Then you add ones to ones, tens to tens, hundreds to hundreds, and so on, starting at the right. When a column adds up to ten or more, you carry: ten of that place become one of the next place, and that one is added into the next column.

Reach for it whenever amounts are put together. “In all”, “altogether” and “total” describe joining. So does “how many more” turned around: if one school collected 6{,}948 more cans than another, you add to find its count. The word “more” alone does not say whether to add or subtract; the story does.

The words: in 48{,}675 + 27{,}548 = 76{,}223, the numbers being added are addends and the answer 76{,}223 is the sum.

You already have the two facts the method rests on. The first is place value: each place is worth ten of the place on its right, the columns of a The place value chart, and you can read numbers up to a million. The second is the addition facts up to 9 + 9 = 18: every column ends as one of them, plus a carry. Two earlier ways of adding lead straight here. The The partial sums method adds each place on its own and then combines the parts. Regrouping in addition trades ten ones for one ten, so that no place holds more than nine.

The procedure

Part 1: from partial sums to the column method

Start with 467 + 258, done first with partial sums. Each addend splits by place: 467 = 400 + 60 + 7 and 258 = 200 + 50 + 8. Add the ones, the tens and the hundreds on their own:

Then combine the parts: 15 + 110 + 600 = 725.

Why is it fair to pull the numbers apart and add them in a new order? Two properties of addition say so. The commutative property says the order of two addends does not change the sum: a + b = b + a. The associative property says the grouping does not change the sum: (a + b) + c = a + (b + c). Together they let you move all six parts around and group them as you like (see properties of addition):

The column method does the same work but writes less. Line the numbers up on the right, so ones sit under ones, and add one column at a time, starting with the ones.

Ones. 7 + 8 = 15 ones. A place can hold only one digit, and 15 ones is more than nine. But fifteen ones are one ten and five ones.

So write 5 in the ones place of the sum, and carry the ten: write a small 1 at the top of the tens column. That small 1 means one ten, because it sits in the tens column.

Tens. Add the carried ten to the tens already there: 1 + 6 + 5 = 12 tens. Twelve tens are 120, which is one hundred and two tens. Write 2 in the tens place and carry 1 to the top of the hundreds column. This carried 1 means one hundred.

Hundreds. 1 + 4 + 2 = 7 hundreds. Seven is less than ten, so nothing is carried.

The sum is 725, the same as by partial sums.

Look at the figure side by side. The partial sum 15 became “write 5, carry 1 ten”. The partial sum 110 is the 6 + 5 = 11 tens of the tens column; with the ten carried from the ones it became 12 tens, so “write 2, carry 1 hundred”. Every part from the left method shows up on the right; the column method just folds each part into the next column as soon as it is found.

Why the carried 1 is worth ten of its place. Each place is worth ten of the place on its right. Ten ones make one ten. Ten tens make one hundred. Ten hundreds make one thousand. So when a column adds up to ten or more, ten of its units can be swapped for one unit of the next place to the left, and the number keeps its value. The carry is that swap, written down. It is the same trade as regrouping in addition, made in whichever column needs it.

Why start at the right. A carry always goes into the column to the left. If you worked from the left, you would write the hundreds digit first and then have to change it when the tens carried. Starting at the right finishes each column’s carry before you reach the column it lands in, so no digit of the answer is ever rewritten.

Part 2: carrying in several places, and adding three or more numbers

Back to the recycling drive: 48{,}675 + 27{,}548. Four of its five columns pass nine.

• Ones. 5 + 8 = 13. Write 3, carry 1 ten.
• Tens. 1 + 7 + 4 = 12 tens. Write 2, carry 1 hundred.
• Hundreds. 1 + 6 + 5 = 12 hundreds. Write 2, carry 1 thousand.
• Thousands. 1 + 8 + 7 = 16 thousands. Write 6, carry 1 ten thousand.
• Ten thousands. 1 + 4 + 2 = 7 ten thousands.

The school collected 76{,}223 cans in all. The carry into the ten thousands column works the same way as the carry into the tens: sixteen thousands are one ten thousand and six thousands. Nothing new happens at the comma. The comma only groups the digits for reading.

How big can a carry be? With two addends, a column holds two digits of at most 9 each, plus a carry. The largest a column can reach is 9 + 9 + 1 = 19. So the carry from a column of two addends is always 0 or 1, never more.

Numbers of different lengths. In 9{,}486 + 785 the second number has no thousands digit. Line up the right ends: the 5 goes under the 6. An empty spot counts as 0.

Three or more addends. The algorithm does not change. Line all the numbers up on the right and add each column, carry included. The only difference is that a column can now pass nineteen. In 3{,}867 + 2{,}958 + 4{,}775, the ones are 7 + 8 + 5 = 20. Twenty ones are two tens and no ones, so you write 0 and carry 2.

The order you add the numbers in does not matter, by the commutative and associative properties. That gives a free check: add the column from top to bottom, then again from bottom to top. The two totals must agree.

Part 3: estimating first, and checking afterward

Two habits go with the algorithm. One comes before and one comes after.

Estimate first. Round each addend and add in your head, as in estimating sums. For the drive, 48{,}675 is about 49{,}000 and 27{,}548 is about 28{,}000, so the total is about 49{,}000 + 28{,}000 = 77{,}000. The exact sum 76{,}223 is close, so it is in the right range. An estimate catches big slips: numbers lined up on the left, or a lost carry in a big place.

Check afterward. There are two good checks.

• Add in the other order. By the commutative property, 27{,}548 + 48{,}675 must give the same sum. The columns come out the same, but you add each pair the other way round, which is a fresh look at every fact.
• Subtract back. Addition and subtraction come in Fact families. If 48{,}675 + 27{,}548 = 76{,}223, then 76{,}223 - 27{,}548 must give 48{,}675. Do it with the standard subtraction algorithm. If it does not come back, the first place where it goes wrong points to the column to redo.

Reading a word problem. Before adding, ask what is being joined. A sum only makes sense when the addends count the same kind of thing: cans plus cans, or fans plus fans. A problem may need two steps, such as finding one amount and then adding it to another.

Worked examples

Boundaries

• Places lined up on the right. The method adds ones to ones and tens to tens. Lined up on the left, the columns mix places and the sum means nothing.
• Whole numbers, as here. This lesson adds whole numbers. The same column method adds money and decimals once the decimal points are lined up, since that lines up the places; that is adding decimals.
• One digit per place. A column that reaches ten or more must carry. Writing 13 in one place makes a different number.
• Size of the carry. With two addends the carry is never more than 1, because a column is at most 9 + 9 + 1 = 19. With more addends a carry can be 2 or more, and it is still added into the next column in the same way.
• The same kind of thing. The algorithm adds any two whole numbers, but a word problem’s sum only means something when the addends count the same kind of thing.

Where this goes next

Every longer calculation leans on this one. Multiplying multi-digit numbers ends by adding partial products in columns, and long division checks its work by multiplying and adding. The same column method, with the decimal points lined up, adds dollars and cents and other decimals. The habit of estimating first and checking afterward stays useful for every sum you will ever do.

Exercises

Part 1: the column method and one or two carries

1. Find 352 + 417.
2. Find 468 + 325.
3. Find 574 + 263.
4. Find 689 + 247.
5. Find 3{,}476 + 2{,}385.
6. Find 538 + 276 by partial sums and by the column method. Show where each partial sum appears in the column method.
7. Mele’s class collected 1{,}648 cans one week and 2{,}775 cans the next. How many cans did the class collect in the two weeks?
8. In 5{,}829 + 1{,}463, which columns carry? Find the sum.

Part 2: many carries, longer numbers, more addends

9. Find 37{,}856 + 24{,}379.
10. Find 65{,}497 + 18{,}836.
11. Find 8{,}957 + 1{,}068.
12. Find 46{,}738 + 5{,}694.
13. Find 247{,}586 + 135{,}749.
14. Find 2{,}345 + 1{,}876 + 3{,}958.
15. Three classes collected 486, 759 and 897 cans. How many cans is that in all?
16. Find 708 + 1{,}465 + 2{,}089 + 356.

Part 3: estimating, checking, and word problems

17. Estimate 6{,}812 + 2{,}397 by rounding to the nearest thousand, then find the exact sum.
18. Rudo writes 4{,}586 + 3{,}729 = 7{,}215. Is he right? If not, find what went wrong and give the true sum.
19. Is 27{,}468 + 15{,}854 = 43{,}322? Decide by adding in the other order.
20. The ballpark sold 4{,}768 hot dogs on Saturday and 3{,}957 on Sunday. How many hot dogs did it sell over the two days?
21. Zahra’s school collected 12{,}875 cans. Mele’s school collected 6{,}948 more cans than that. How many cans did Mele’s school collect, and how many did the two schools collect together?
22. Find the missing number: \square - 3{,}847 = 5{,}968.

Reasoning

23. In 3{,}476 + 2{,}385 (Exercise 5), the ones carry a 1 into the tens column. Explain how much that 1 is worth, and why it may be added to the tens.
24. Explain why adding two numbers never carries more than 1, but adding three numbers can carry 2. What is the largest carry three addends can make?
25. Zahra finds 375 + 486 + 625 by adding 375 + 625 first. Find the sum her way, and name the properties of addition that make her way give the same sum as adding left to right.