Chapter 1 · Whole number foundations review

Multiply Multi-Digit Numbers by 1-Digit Numbers

Use partial products and the standard algorithm to multiply bigger numbers by a 1-digit number. Then solve word problems and check each product with an estimate.

15 min read

You already know multiplication facts such as $8 \times 3 = 24$ . In this lesson you will use those facts to multiply bigger numbers by a 1-digit number, such as $3$ or $7$ .

Start with what the multiplication means. Think of $23 \times 4$ as $4$ groups of $23$ . The number $23$ is made of $2$ tens and $3$ ones. These are its place-value parts, so $23 = 20 + 3$ . That means $4$ groups of $23$ is the same amount as $4$ groups of $20$ plus $4$ groups of $3$ .

The answer to a multiplication is called the product. When you multiply just one place-value part of a number, you get a partial product. It is one piece of the whole product. Add the partial products and you get the whole product. The table shows each part of $23 \times 4$ in words and as a number sentence.

23 times 4, one place-value part at a time

Part of 23	In words	Number sentence
$3$ ones	$3$ ones times $4$ is $12$ ones	$3 \times 4 = 12$
$2$ tens	$2$ tens times $4$ is $8$ tens, which is $80$	$20 \times 4 = 80$
All of $23$	$12$ ones and $8$ tens make $92$	$23 \times 4 = 12 + 80 = 92$

Worked example

Partial products for a 2-digit number

Find $58 \times 3$ with partial products.

Break 58 into its place-value parts: 5 tens and 8 ones. You can multiply each part by 3 on its own, because 3 groups of 58 is the same as 3 groups of 50 plus 3 groups of 8.
$58 = 50 + 8$
Multiply the ones part: 8 ones times 3 is 24 ones. This is the first partial product.
$8 \times 3 = 24$
Multiply the tens part: 5 tens times 3 is 15 tens, because 5 times 3 is 15. And 15 tens is the same as 150. This is the second partial product.
$50 \times 3 = 150$
Add the partial products. The two parts together make all 3 groups of 58, so their sum is the whole product.
$24 + 150 = 174$

Answer

The product of $58$ and $3$ is $174$ .

Partial products always work, but they take a lot of writing. The standard algorithm is a shorter way to write the same work. An algorithm is a set of steps that you follow in order. In this one you multiply one digit at a time, starting with the ones, and you write only one digit in each place.

Sometimes a place gets $10$ or more. Then you regroup: you trade $10$ of that place for $1$ of the next place to the left. For example, $24$ ones is the same as $2$ tens and $4$ ones. You write the $4$ in the ones place and move the $2$ tens to the tens place. Write a small $2$ above the tens digit so that you remember to add it. This is why you start with the ones: what you regroup always moves left, to a place you have not done yet.

Here is $58 \times 3$ written this way. The small $2$ is the regrouped tens. The table after it matches each step to a partial product.

\begin{array}{cccc} & & \scriptstyle 2 & \\ & & 5 & 8 \\ \times & & & 3 \\ \hline & 1 & 7 & 4 \end{array}

58 times 3: each partial product is one step of the standard algorithm

Place	Partial product	Standard algorithm step
Ones	$8 \times 3 = 24$	$8$ ones times $3$ is $24$ ones. Write $4$ ones. Regroup $2$ tens.
Tens	$50 \times 3 = 150$	$5$ tens times $3$ is $15$ tens. Add the $2$ regrouped tens: $17$ tens. Write $7$ tens and $1$ hundred.
Whole product	$24 + 150 = 174$	The digits you wrote read $174$ . Adding the regrouped tens joined the two partial products.

Key concept

The standard algorithm is partial products, written short

To multiply by a 1-digit number, multiply each place-value part and put the results together. Each result is a partial product. The standard algorithm finds the same partial products one place at a time, starting with the ones and moving left. In each place:

Multiply the digit by the 1-digit number.
Add any amount that you regrouped into that place.
Write the last digit of your result in that place. Regroup the digit in front of it, if there is one, to the next place on the left.

At the last place, write the whole result. Always multiply first and add the regrouped amount second.

The same steps work for bigger numbers. A 3-digit number has a hundreds place, and a 4-digit number has a thousands place too. You just keep moving left, one place at a time.

Some numbers have a $0$ in one of their places. The number $1{,}308$ has $0$ tens. Do not skip that place. Zero tens times any number is still $0$ tens, but tens that you regrouped from the ones may be waiting there.

Here is the finished work for $1{,}308 \times 7$ . The small digits are the amounts that were regrouped. The example after it goes through the work one place at a time.

\begin{array}{ccccc} & \scriptstyle 2 & & \scriptstyle 5 & \\ & 1 & 3 & 0 & 8 \\ \times & & & & 7 \\ \hline & 9 & 1 & 5 & 6 \end{array}

Worked example

Standard algorithm with a 0 in the number

Find $1{,}308 \times 7$ with the standard algorithm.

Start with the ones: 8 ones times 7 is 56 ones. That is 5 tens and 6 ones, so write 6 in the ones place and regroup the 5 tens.
$8 \times 7 = 56$
Move to the tens: 0 tens times 7 is 0 tens. Now add the 5 tens you regrouped. That makes 5 tens, so write 5 in the tens place. Since 5 is less than 10, there is nothing to regroup.
$(0 \times 7) + 5 = 5$
Move to the hundreds: 3 hundreds times 7 is 21 hundreds. That is 2 thousands and 1 hundred, so write 1 in the hundreds place and regroup the 2 thousands.
$3 \times 7 = 21$
Move to the thousands: 1 thousand times 7 is 7 thousands. Add the 2 thousands you regrouped to get 9 thousands. Write 9 in the thousands place.
$(1 \times 7) + 2 = 9$
Read the digits you wrote: 9,156. Then check with partial products. The four parts add up to the same number, so the regrouping is right.
$7{,}000 + 2{,}100 + 0 + 56 = 9{,}156$

Answer

The product of $1{,}308$ and $7$ is $9{,}156$ .

Many word problems are about equal groups, which means the same amount again and again. When a problem tells you the size of one group and the number of groups, multiply to find the total.

It is easy to make a small mistake in a long multiplication, so check your product with an estimate. An estimate is a number that is close to the exact answer and quick to find. To make one, round the big number. That means you swap it for a nearby number that is easy to multiply, such as the nearest hundred or thousand. Then multiply the rounded number. If your exact product is far from your estimate, look for a mistake.

Worked example

A word problem with an estimate check

A bakery bakes $289$ bread rolls every day. How many rolls does it bake in $6$ days?

Decide what to do. The bakery bakes the same number of rolls each day, so 6 days make 6 equal groups of 289 rolls. To find the total of equal groups, multiply.
$289 \times 6$
Estimate first. Round 289 to the nearest hundred. It is between 200 and 300, and it is closer to 300. Then 3 hundreds times 6 is 18 hundreds, which is 1,800. The exact product should be a little less, because 289 is less than 300.
$300 \times 6 = 1{,}800$
Now use the standard algorithm. In the ones place, 9 ones times 6 is 54 ones, so write 4 and regroup 5 tens. In the tens place, 8 tens times 6 is 48 tens. Add the 5 regrouped tens to get 53 tens, so write 3 and regroup 5 hundreds.
$9 \times 6 = 54 \qquad (8 \times 6) + 5 = 53$
Multiply the hundreds: 2 hundreds times 6 is 12 hundreds. Add the 5 regrouped hundreds to get 17 hundreds. This is the last place, so write 17. The digits now read 1,734.
$(2 \times 6) + 5 = 17$
Check with the estimate. The product 1,734 is close to 1,800, and it is a little less, just as you expected. The product makes sense.
$289 \times 6 = 1{,}734$

Answer

The bakery bakes $1{,}734$ rolls in $6$ days.

The big idea: multiplying by a 1-digit number is place-value work. You multiply each place, and the partial products together make the whole product. The standard algorithm is the same work in a shorter form. In every place you multiply first and then add what you regrouped, even when the digit is $0$ . When a problem has equal groups, multiply, and then use a quick estimate to see that your product makes sense.

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