The standard multiplication algorithm

When you reach for it

The home bleachers at a stadium have 36 sections, and each section has 214 seats. How many seats is that in all? The answer is 214 \times 36, and no times-table fact gives it.

Partial products can do it: split both factors by place value, multiply every pair of parts, and add the pieces. Here that is six products to find and add. It works, but it fills a page.

The standard multiplication algorithm does the same work in a shorter record. You stack the factors in columns by place value. Then you write one row for each digit of the bottom factor, and you add the rows. Reach for it whenever both factors are too big to multiply in your head.

The words: in 214 \times 36 = 7{,}704, the numbers being multiplied, 214 and 36, are the factors, and the answer, 7{,}704, is the product. Each row of the algorithm is itself a partial product.

Two facts carry the whole method. The distributive property lets you split a factor by place value and multiply the parts one at a time: 214 \times 36 = 214 \times 30 + 214 \times 6. And regrouping is a place-value trade: ten ones make one ten, ten tens make one hundred, and so on. The algorithm makes that trade every time a column holds ten or more.

The procedure

Part 1: from partial products to the shortened record

Start with one digit on the bottom: 268 \times 7. Split 268 by place value as 200 + 60 + 8. The distributive property says you may multiply each part by 7 and add:

Each piece uses a basic fact and a shift of place, as in multiplying by tens and hundreds.

The shortened record on the right does the same three multiplications, but it writes only one digit in each column as it goes. It starts at the ones, because whatever the ones make spills into the tens.

• Ones. 8 \times 7 = 56 ones. That is 5 tens and 6 ones. Write 6 in the ones column, and carry the 5 tens: write a small 5 above the tens column. Trading 50 ones for 5 tens is regrouping.
• Tens. 6 \times 7 = 42 tens. Add the 5 tens carried: 42 + 5 = 47 tens. That is 4 hundreds and 7 tens. Write 7, and carry 4 above the hundreds.
• Hundreds. 2 \times 7 = 14 hundreds, and 14 + 4 = 18 hundreds. That is the last place, so write all of 18: 1 thousand and 8 hundreds.

The product is 1{,}876, the same as the partial products give. Nothing was skipped. The 5 in the carry is the 5 tens of 56, and the 4 is the 4 hundreds of 420 + 50.

The carry is added after the multiplying, never before. The carried 5 tens are already part of an answer. They are not more tens of 268 waiting to be multiplied by 7.

Part 2: two digits by two digits, and the zero in the second row

Now take 47 \times 63. Split the bottom factor, 63 = 60 + 3. By the distributive property,

So there are two rows, one for each digit of 63.

First row: 47 \times 3. Ones: 7 \times 3 = 21. Write 1, carry 2. Tens: 4 \times 3 = 12, and 12 + 2 = 14. Write 14. The row is 141.

Second row: 47 \times 60. Since 60 = 6 \times 10, the associative property lets you find 47 \times 6 first and then take ten times it: 47 \times 60 = (47 \times 6) \times 10. So the answer is a number of tens. Show that by writing 0 in the ones column first. Then multiply by 6 and write the digits starting in the tens column. 7 \times 6 = 42: write 2, carry 4. 4 \times 6 = 24, and 24 + 4 = 28: write 28. The row is 2{,}820.

Before starting the second row, cross out the carry from the first row. It belonged to 47 \times 3, not to 47 \times 6.

Add the rows: 141 + 2{,}820 = 2{,}961.

The 0 in the second row is not decoration. The row really is 47 \times 60 = 2{,}820, and the 0 holds the ones place empty so that 282 tens sit in the tens, hundreds and thousands columns. Leave it out, and 141 + 282 = 423 counts those tens as ones. The answer would then be about ten times too small.

Part 3: three digits by two digits

Back to the bleachers: 214 \times 36. The area model below shows all six partial products. Each column of the rectangle is one place of 214, and each row is one place of 36.

The standard algorithm adds each row of the rectangle in one line.

First row: 214 \times 6. 4 \times 6 = 24: write 4, carry 2. 1 \times 6 = 6, and 6 + 2 = 8: write 8. 2 \times 6 = 12: write 12. The row is 1{,}284, the same as 1{,}200 + 60 + 24 in the area model.

Second row: 214 \times 30. Write 0 in the ones column, since this row counts tens. Then 214 \times 3: 4 \times 3 = 12: write 2, carry 1. 1 \times 3 = 3, and 3 + 1 = 4. Then 2 \times 3 = 6. The row is 6{,}420, the same as 6{,}000 + 300 + 120.

Add: 1{,}284 + 6{,}420 = 7{,}704. The bleachers hold 7{,}704 seats.

The number of rows is the number of digits in the bottom factor. By the commutative property, 214 \times 36 = 36 \times 214. So you may put either factor on the bottom. Putting the one with fewer digits there gives fewer rows: two here, not three.

Part 4: four digits by two digits, zeros, and a quick estimate

Try 3{,}476 \times 58. It has longer rows and more carries, but no new idea.

First row: 3{,}476 \times 8. 6 \times 8 = 48: write 8, carry 4. 7 \times 8 = 56, and 56 + 4 = 60: write 0, carry 6. 4 \times 8 = 32, and 32 + 6 = 38: write 8, carry 3. 3 \times 8 = 24, and 24 + 3 = 27: write 27. The row is 27{,}808.

Second row: 3{,}476 \times 50. Write the 0. 6 \times 5 = 30: write 0, carry 3. 7 \times 5 = 35, and 35 + 3 = 38: write 8, carry 3. 4 \times 5 = 20, and 20 + 3 = 23: write 3, carry 2. 3 \times 5 = 15, and 15 + 2 = 17: write 17. The row is 173{,}800.

Add: 27{,}808 + 173{,}800 = 201{,}608.

Zeros inside a factor. A 0 digit in the top factor makes a product of 0 in its column, but a carry from the column before still lands there and is written. A 0 digit in the bottom factor makes a whole row of zeros. You may skip that row, as long as the next row still starts in its own place. Example 7 shows both.

A quick estimate. Round each factor to a number that is easy to multiply: 3{,}476 \approx 3{,}500 and 58 \approx 60. Then 3{,}500 \times 60 = 210{,}000. The product 201{,}608 is close to that, so no row slid into the wrong place. (Choosing numbers that make the estimate easy is the topic of Friendly numbers and deliberate over and under estimates.) An estimate catches a lost zero or a shifted row, since those change the answer about ten times. It does not catch one wrong digit. For that, work the product again another way, such as with the factors swapped.

Worked examples

Boundaries

• The algorithm, as written here, is for whole numbers: 0, 1, 2, 3 and so on. With decimals the digits come out the same, but the decimal point has to be placed, which is a separate step.
• Every digit must stay in its place-value column. Each row’s first digit goes in the place of the bottom digit being used: the ones row starts in the ones column, the tens row in the tens column (after its 0), the hundreds row in the hundreds column (after two zeros). Shift a row and the answer changes by a factor of ten.
• Carrying at ten works because the numbers are written in base ten: ten of one place make one of the next. A carry is added after the column is multiplied, and each row’s carries belong only to that row.
• The method has no size limit. Each extra digit on the bottom adds one more row, so very large factors mean long work, not a different method.
• An estimate checks the size of a product, not each digit. Two products that differ in one middle digit can have the same estimate.

Where this goes next

Long division leans on this algorithm at every step: each digit of a quotient is tested by a multiplication, and the answer is checked by multiplying back, as in The standard long division algorithm. With a two-digit divisor, those multiplications are exactly the two-digit products from this lesson, which is the work of Dividing by a two-digit number.

The same rows, carries and columns multiply decimals, once you know where the decimal point goes: see Multiplying multi-digit decimals with the standard algorithm.

Much later, in algebra, multiplying (x + 4)(x + 7) uses the same distributive property as the area model in Part 3, with letters in place of place values. That is Multiplying two binomials.

Exercises

Part 1: one-digit bottom factors

1. Find 63 \times 7.
2. Find 529 \times 4.
3. Find 3{,}608 \times 5.
4. A bleacher section has 9 rows with 46 seats in each row. How many seats does it have?
5. A hatchery has 7 tanks, each holding 2{,}385 fish. How many fish are in the tanks?
6. Find 538 \times 6 two ways: as three partial products added together, and with the shortened record. Which two carries does the shortened record write?

Part 2: two digits by two digits

7. Find 34 \times 21.
8. Find 58 \times 46.
9. Find 93 \times 78.
10. A stadium end zone has 32 rows of bleachers, with 45 seats in each row. How many seats is that?
11. In 64 \times 27, what number does the second row stand for? What is the product?

Part 3: three digits by two digits

12. Find 312 \times 23.
13. Find 486 \times 35.
14. Find 758 \times 64.
15. A hatchery puts 125 young salmon in each of 48 tanks. How many salmon is that?
16. The visitors’ bleachers have 27 sections with 196 seats in each. How many seats is that?

Part 4: four digits by two digits, zeros and estimates

17. Find 2{,}413 \times 32.
18. Find 5{,}067 \times 24.
19. Find 3{,}190 \times 60.
20. Find 6{,}842 \times 57.
21. Before multiplying 4{,}108 \times 39, estimate the product. Then find it with the standard algorithm and compare the two.
22. A concert at the stadium sold 3{,}645 tickets at 27 dollars each. How much money did the tickets bring in? Give a quick estimate as well.

Reasoning

23. Ryota multiplies 52 \times 37 and writes 156 as his second row, lined up under the first row’s last digit. Explain what 156 is, what the second row should stand for, and find the product.
24. In 48 \times 6, the first step writes 8 and carries 4. Explain why the carried 4 is added after finding 4 \times 6, and not added to the 4 of 48 before multiplying. What answer would the wrong order give?
25. A four-digit whole number is multiplied by a two-digit whole number. How many digits can the product have? Explain using the smallest and the largest possible factors.