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Grade 4 · 3.NBT.A.3 · 4.NBT.B.5

Multiplying by tens and hundreds

A bus company has 40 buses, and each one drives 600 miles a month. How can the basic fact 4 \times 6 = 24 give the total miles, and why does the answer end in exactly three zeros?

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Multiplying and dividing by ten, one hundred and one thousandGrade 4
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Dividing by a single digitGrade 4Partial quotientsGrade 4The standard multiplication algorithmGrade 5
Where you are 3 in Multi-digit multiplication
Number9
Arithmetic5
Multiplication and division16
Multi-digit multiplication3
Multiplying by tens and hundreds
Partial productsThe standard multiplication algorithm

When you reach for it

Ishita runs a fleet of school buses. Aroha orders pencils for the schools the buses serve, and Xochi plans the trips. Their questions sound like these. Six buses each drive 400 miles in a week: how many miles is that in all? Forty buses each drive 600 miles in a month: how many miles? Thirty buses have 50 seats each: how many seats?

Each question is a product in which at least one factor is a multiple of ten, a hundred or a thousand, such as 400, 600 or 50. Products like these come up whenever things are counted in tens and hundreds. They are also the pieces of every bigger product: Partial products splits a product like 47 \times 36 into parts such as 40 \times 30, and an estimate such as 500 \times 6 in Estimating by rounding is one of them.

You already have the two tools this needs. One is a basic fact such as 6 \times 4 = 24, a product of two one-digit numbers (Multiplication facts). The other is the shift from Multiplying and dividing by ten, one hundred and one thousand: multiplying by 10, 100 or 1{,}000 moves every digit one, two or three places to the left. This lesson puts the two together over two days. Part 1 multiplies a one-digit number by a multiple of 10, 100 or 1{,}000. Part 2 multiplies two multiples of ten.

The procedure

Part 1: a one-digit number times tens, hundreds or thousands

Take 6 \times 400, the miles for six buses that each drive 400 miles. The number 400 is 4 hundreds, so 6 \times 400 asks for 6 groups of 4 hundreds. Count the hundreds and there are 6 \times 4 = 24 of them.

100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 6 × 4 hundreds = 24 hundreds = 2,400
Figure 1Six groups of 4 hundreds are 24 hundreds in all, and 24 hundreds is 2{,}400.

Twenty-four hundreds is 2{,}400, so

6 \times 400 = 2{,}400.

Here is the same idea written with factors. Since 400 = 4 \times 100,

\begin{aligned} 6 \times 400 &= 6 \times (4 \times 100) \\ &= (6 \times 4) \times 100. \end{aligned}

Moving the parentheses is allowed by the associative property of multiplication: when three numbers are multiplied, you may group them either way and the product is the same. Now 6 \times 4 = 24 is a basic fact, and 24 \times 100 is a shift. Every digit of 24 moves two places to the left, and zeros hold the empty tens and ones places, so 24 \times 100 = 2{,}400.

So the work splits into two easy pieces. The basic fact gives the digits in front. The shift puts them in their places: one place for a factor of 10, two for 100, three for 1{,}000. The zeros at the end of 2{,}400 are there because the 24 moved two places and left the tens and ones empty. They are the result of the move, not a separate rule about writing zeros.

When the bigger number comes first, as in 400 \times 6, the commutative property of multiplication (Properties of multiplication) says that the order of the factors does not change the product. So 400 \times 6 = 6 \times 400 = 2{,}400.

One more case. When the basic fact itself ends in 0, that zero stays. For 5 \times 60 the fact is 5 \times 6 = 30, so the product is 30 tens, which is 300. The answer has two zeros although 60 has only one. One zero belongs to the fact 30, and one holds the ones place after the shift.

Part 2: tens times tens, tens times hundreds

Ishita’s 40 buses each drive 600 miles, so the total is 40 \times 600. Both factors are multiples of ten: 40 = 4 \times 10 and 600 = 6 \times 100. So

40 \times 600 = 4 \times 10 \times 6 \times 100.

The commutative property lets us change the order of these four factors, and the associative property lets us group them as we like. So put the two digits together and the tens and hundreds together:

\begin{aligned} 40 \times 600 &= (4 \times 6) \times (10 \times 100) \\ &= 24 \times 1{,}000 \\ &= 24{,}000. \end{aligned}

The buses drive 24{,}000 miles. The fact 4 \times 6 gives the digits 24. The factor 10 \times 100 = 1{,}000 shifts them three places: one place for the 10 inside 40 and two for the 100 inside 600.

An area model shows the same split. Draw a rectangle 40 by 600 and cut it into strips 10 wide one way and 100 wide the other way.

1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 1,000 100 100 100 100 100 100 10 10 10 10 4 × 6 = 24 pieces of 1,000 = 24,000
Figure 2A 40 by 600 rectangle cut into 10 by 100 pieces. There are 4 \times 6 = 24 pieces, and each piece is 1{,}000. Not drawn to scale, since a real 10 by 100 piece is ten times as long as it is wide.

There are 4 rows and 6 columns, so there are 4 \times 6 = 24 pieces. Each piece is 10 by 100, which is 1{,}000. Twenty-four thousands is 24{,}000.

This gives a quick way to see how far to shift: count the zeros at the right end of both factors. 40 ends in one zero and 600 in two, so the digits of 24 move three places. Each zero at the end of a factor stands for a factor of 10, and each factor of 10 is one place.

A ten times a ten is a hundred, so tens times tens gives hundreds. For example, 30 \times 50 is 3 tens times 5 tens, which is 15 hundreds, or 1{,}500. And as in Part 1, if the basic fact ends in 0, its zero comes along as an extra one; Example 5 shows how.

Worked examples

Examples 1 to 3 go with Part 1. Examples 4 to 7 go with Part 2.

Example 1

Seven times three, three hundred, three thousand

Find 7 \times 30, 7 \times 300 and 7 \times 3{,}000.

Solution

All three start from the same basic fact, 7 \times 3 = 21.

30 is 3 tens, so 7 \times 30 is 7 \times 3 tens, which is 21 tens. Written with factors, 7 \times (3 \times 10) = (7 \times 3) \times 10 by the associative property, and 21 \times 10 moves the digits of 21 one place to the left:

7 \times 30 = 210.

300 is 3 hundreds, so 7 \times 300 is 21 hundreds. The digits move two places: 7 \times 300 = 2{,}100.

3{,}000 is 3 thousands, so 7 \times 3{,}000 is 21 thousands. The digits move three places: 7 \times 3{,}000 = 21{,}000.

Each answer is ten times the one before, because each second factor is ten times the one before. Check the middle one by adding seven groups of 300. Three of them make 900, six make 1{,}800, and one more makes 1{,}800 + 300 = 2{,}100.

Example 2

Boxes of pencils

Aroha orders 6 boxes of pencils. Each box holds 500 pencils. How many pencils is that?

Solution

The total is 6 \times 500. Since 500 is 5 hundreds, 6 \times 500 is 6 \times 5 hundreds, which is 30 hundreds. With factors, using the associative property:

\begin{aligned} 6 \times 500 &= (6 \times 5) \times 100 \\ &= 30 \times 100. \end{aligned}

Now shift 30 two places to the left. Its digits are 3 and 0, and both move: the 3 lands in the thousands place, the 0 in the hundreds place, and two new zeros fill the tens and ones. So 6 \times 500 = 3{,}000 pencils.

The product has three zeros, but 500 has only two. The third zero comes from the basic fact, since 6 \times 5 = 30 already ends in 0. Check: ten hundreds make a thousand, so 30 hundreds make 3 thousands, which is 3{,}000.

Example 3

Order and a missing factor

(a) Find 3{,}000 \times 9. (b) Find the missing number: 7 \times \_\_\_ = 6{,}300.

Solution

(a) By the commutative property, 3{,}000 \times 9 = 9 \times 3{,}000. That is 9 groups of 3 thousands, which is 9 \times 3 = 27 thousands:

3{,}000 \times 9 = 27{,}000.

(b) Read 6{,}300 as 63 hundreds. We need 7 times some number to make 63 hundreds. The basic fact 7 \times 9 = 63 says that 7 times 9 hundreds is 63 hundreds. So the missing number is 900. Check: 7 \times 900 = (7 \times 9) \times 100, which is 63 \times 100 = 6{,}300.

Example 4

Seats on the fleet

Ishita’s fleet has 30 buses with 50 seats on each. How many seats are there in all?

Solution

The total is 30 \times 50. Write each factor as a digit times ten: 30 = 3 \times 10 and 50 = 5 \times 10. Reorder with the commutative property and group with the associative property:

\begin{aligned} 30 \times 50 &= (3 \times 5) \times (10 \times 10) \\ &= 15 \times 100 \\ &= 1{,}500. \end{aligned}

So there are 1{,}500 seats. In place value words, 3 tens times 5 tens is 15 hundreds, because a ten times a ten is a hundred. Each factor ends in one zero, so the digits of 15 shift two places.

Check another way: 30 buses is 3 groups of 10 buses. Ten buses have 10 \times 50 = 500 seats, so thirty buses have 3 \times 500 = 1{,}500 seats.

Example 5

When the fact ends in zero

Find 50 \times 80 and 200 \times 50.

Solution

For 50 \times 80, regroup as before:

\begin{aligned} 50 \times 80 &= (5 \times 8) \times (10 \times 10) \\ &= 40 \times 100. \end{aligned}

Here the basic fact is 5 \times 8 = 40, which ends in a zero of its own. Shift 40 two places to the left: 40 \times 100 = 4{,}000. The factors end in two zeros in all, yet the product 4{,}000 ends in three. Two zeros come from the shift and one from the fact.

For 200 \times 50:

\begin{aligned} 200 \times 50 &= (2 \times 5) \times (100 \times 10) \\ &= 10 \times 1{,}000 \\ &= 10{,}000. \end{aligned}

The fact 2 \times 5 = 10 brings an extra zero again, so the product has four zeros although the factors have three in all. Counting the zeros in the factors tells you how far to shift the basic fact. It does not tell you how many zeros the answer has until you know the fact.

Example 6

Enough seats for the trip?

Xochi plans a field trip for 4{,}500 students. The district sends 80 buses, each with 60 seats. Is there a seat for every student? How many seats are left over or missing?

Solution

The number of seats is

\begin{aligned} 80 \times 60 &= (8 \times 6) \times (10 \times 10) \\ &= 48 \times 100 \\ &= 4{,}800. \end{aligned}

Compare: 4{,}800 is more than 4{,}500, so every student has a seat. The seats left over number 4{,}800 - 4{,}500 = 300.

Check the size in hundreds: the seats are 48 hundreds and the students are 45 hundreds, so there are 3 hundreds more seats than students.

Example 7

Miles in twenty days

Each bus in a fleet of 50 buses drives 300 miles a day. How many miles do all the buses drive in 20 days?

Solution

Start with one bus. In 20 days it drives

\begin{aligned} 300 \times 20 &= (3 \times 2) \times (100 \times 10) \\ &= 6 \times 1{,}000 \\ &= 6{,}000 \end{aligned}

miles. Now the fleet: 50 buses each drive 6{,}000 miles, so the total is

\begin{aligned} 50 \times 6{,}000 &= (5 \times 6) \times (10 \times 1{,}000) \\ &= 30 \times 10{,}000 \\ &= 300{,}000. \end{aligned}

All the buses together drive 300{,}000 miles. The fact 5 \times 6 = 30 brought its own zero, so the product has five zeros although 50 and 6{,}000 have four in all.

The associative property says the grouping could have gone another way. First find one day for the whole fleet: 50 \times 300 is 5 \times 3 = 15 thousands, so 50 \times 300 = 15{,}000 miles. Then take 20 days:

\begin{aligned} 15{,}000 \times 20 &= (15 \times 2) \times (1{,}000 \times 10) \\ &= 30 \times 10{,}000 \\ &= 300{,}000. \end{aligned}

The same answer comes out, as it must.

Boundaries

It needs a fact you know. The method rests on a basic fact: 40 \times 600 needs only 4 \times 6. A factor such as 47 has two digits that are not zero, so no single basic fact covers it. Split it by place value instead, as in Partial products and the The standard multiplication algorithm.

Only zeros at the right end count. In 305 the zero sits between the 3 and the 5, so 305 is not a one-digit number times 100. Only zeros at the right end of a factor stand for factors of ten.

The count of zeros needs the fact. The zeros at the end of the product are the zeros at the end of the factors, plus any zero at the end of the basic fact. For digits from 1 to 9, the facts that end in 0 are the ones that multiply 5 by an even digit: 2 \times 5 = 10, 4 \times 5 = 20, 6 \times 5 = 30 and 8 \times 5 = 40.

Whole numbers only here. The shift works the same way past the ones place, but a product such as 0.4 \times 600 is not found by counting zeros. That is the work of Multiplying a decimal by ten, one hundred or one thousand.

Where this goes next

Every product in this lesson is one piece of a bigger product. In Partial products, a product like 47 \times 36 is split into 40 \times 30, 40 \times 6, 7 \times 30 and 7 \times 6, and the first three are exactly the kind found here. The The standard multiplication algorithm records the same pieces in a shorter form. In Estimating by rounding, a hard product is rounded to one like 500 \times 6 to get a quick estimate.

The idea also runs backward. Since 6 \times 400 = 2{,}400, the quotient 2{,}400 \div 6 is 400, and that is how Dividing by a single digit and Partial quotients handle large numbers.

Exercises

Part 1: a one-digit number times tens, hundreds or thousands

  1. Find 3 \times 20, 3 \times 200 and 3 \times 2{,}000.
  2. Find 8 \times 60.
  3. Find 9 \times 800.
  4. Find 4 \times 3{,}000.
  5. Find 4 \times 50.
  6. Find 5 \times 4{,}000.
  7. Find 900 \times 4.
  8. Find the missing number: 7 \times \_\_\_ = 3{,}500.
  9. Find the missing number: \_\_\_ \times 600 = 5{,}400.
  10. A bus drives 200 miles a day. How many miles does it drive in 6 days?
  11. Aroha orders 3 boxes of pencils with 900 pencils in each box. How many pencils is that?
  12. Each of 9 buses uses 40 gallons of fuel a week. How many gallons do they use in all?

Part 2: multiples of ten times multiples of ten

  1. Find 20 \times 40.
  2. Find 60 \times 70.
  3. Find 70 \times 800.
  4. Find 500 \times 60.
  5. Find 90 \times 90.
  6. Find 300 \times 300.
  7. Find the missing number: 20 \times \_\_\_ = 1{,}600.
  8. Ishita has 60 buses. Each one drives 300 miles a week. How many miles do they drive in all in a week?
  9. Aroha orders 40 cartons of pencils with 800 pencils in each carton. How many pencils is that?
  10. Xochi’s schedule has 70 buses, each driving 50 miles a day for 20 days. How many miles is that in all?

Reasoning

  1. The factors in 40 \times 50 end in two zeros in all, but the product 2{,}000 ends in three. Explain where the third zero comes from.
  2. Use the commutative and associative properties to show that 20 \times 400 equals (2 \times 4) \times (10 \times 100), and then find the product.
  3. Xochi says that 7 \times 400 and 70 \times 40 have the same product. Is she right? Explain using place value.
Answers
  1. 60, 600 and 6{,}000. The fact 3 \times 2 = 6 gives 6 tens, 6 hundreds and 6 thousands.
  2. 480. 8 \times 6 = 48, so 8 \times 60 is 48 tens, which is 480.
  3. 7{,}200. 9 \times 8 = 72, so the product is 72 hundreds: 72 \times 100 = 7{,}200.
  4. 12{,}000. 4 \times 3 = 12, so the product is 12 thousands.
  5. 200. 4 \times 5 = 20, so the product is 20 tens, which is 200. One zero comes from the fact and one from the shift.
  6. 20{,}000. 5 \times 4 = 20, so the product is 20 thousands. The fact brings one zero and the shift three.
  7. 3{,}600. By the commutative property 900 \times 4 = 4 \times 900, which is 4 \times 9 = 36 hundreds.
  8. 500. 3{,}500 is 35 hundreds and 7 \times 5 = 35, so 7 \times 500 = 3{,}500.
  9. 9. 5{,}400 is 54 hundreds and 600 is 6 hundreds; 9 \times 6 = 54, so 9 \times 600 = 5{,}400.
  10. 1{,}200 miles. 6 \times 200 is 6 \times 2 = 12 hundreds.
  11. 2{,}700 pencils. 3 \times 900 is 3 \times 9 = 27 hundreds.
  12. 360 gallons. 9 \times 40 is 9 \times 4 = 36 tens.
  13. 800. The fact is 2 \times 4 = 8, and 10 \times 10 = 100, so the product is 8 hundreds.
  14. 4{,}200. The fact is 6 \times 7 = 42, and 10 \times 10 = 100, so the product is 42 \times 100.
  15. 56{,}000. The fact is 7 \times 8 = 56, and 10 \times 100 = 1{,}000, so the product is 56 \times 1{,}000.
  16. 30{,}000. The fact is 5 \times 6 = 30, and 100 \times 10 = 1{,}000, so the product is 30 \times 1{,}000. The fact 30 brings the fourth zero.
  17. 8{,}100. The fact is 9 \times 9 = 81, and 10 \times 10 = 100, so the product is 81 \times 100.
  18. 90{,}000. The fact is 3 \times 3 = 9, and 100 \times 100 = 10{,}000, so the product is 9 \times 10{,}000.
  19. 80. 1{,}600 is 16 hundreds, and tens times tens gives hundreds, so the missing number is a number of tens: 2 \times 8 = 16, so 20 \times 80 = 1{,}600.
  20. 18{,}000 miles. 60 \times 300 = (6 \times 3) \times (10 \times 100), which is 18 \times 1{,}000 = 18{,}000.
  21. 32{,}000 pencils. 40 \times 800 = (4 \times 8) \times (10 \times 100), which is 32 \times 1{,}000 = 32{,}000.
  22. 70{,}000 miles. One bus drives 50 \times 20 = 10 \times 100 = 1{,}000 miles, so 70 buses drive 70 \times 1{,}000 = 70{,}000 miles.
  23. 40 \times 50 = (4 \times 5) \times (10 \times 10), which is 20 \times 100. The two zeros in the factors make the shift of two places. The third zero is the one at the end of the basic fact 4 \times 5 = 20: the digits 2 and 0 both move two places, giving 2{,}000.
  24. 20 \times 400 = 2 \times 10 \times 4 \times 100. The commutative property lets the 4 move next to the 2, and the associative property lets us group the digits and the tens and hundreds: (2 \times 4) \times (10 \times 100), which is 8 \times 1{,}000 = 8{,}000.
  25. Yes, both are 2{,}800. 7 \times 400 is 7 times 4 hundreds, which is 28 hundreds. 70 \times 40 is 7 tens times 4 tens, and a ten times a ten is a hundred, so it is also 28 hundreds.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0