4th Grade · 4.NBT.B.5

2-digit by 2-digit multiplication in 4th grade

2-digit by 2-digit multiplication means multiplying two numbers that each have two digits, such as $34 \times 27$ . In 4th grade, your child breaks each number into tens and ones, multiplies the pieces and adds the results. The stacked method with carrying that most adults learned is a 5th grade goal.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.

What it means, in plain words

Your child first learns to multiply by tens. $4 \times 70$ is 4 groups of 7 tens, which is 28 tens, or 280. In the same way, $40 \times 70$ is 4 tens times 7 tens, which is 28 hundreds, or 2,800.

For 2-digit by 2-digit multiplication, each number is split into tens and ones, so 58 becomes $50 + 8$ . The area model is a rectangle split into four boxes, one for each pair of pieces. Each box holds a partial product, a smaller multiplication your child can work out mentally. Adding the four partial products gives the answer.

This year your child also multiplies a number with up to four digits by a one-digit number, such as $6 \times 1{,}250$ , in the same way. The goal is that your child can explain each step and say what each box means. Keeping the tens and ones in view now is what makes the 5th grade method make sense later.

Your child should be able to:

Multiply a one-digit number by tens, hundreds or thousands, such as $6 \times 400$
Split two 2-digit numbers into tens and ones and draw an area model
Find the four partial products and add them
Explain what each partial product means, such as 30 times 20

Example

What is $34 \times 27$ ? Use an area model.

Split the numbers: $34 = 30 + 4$ and $27 = 20 + 7$ .
Fill the four boxes: $30 \times 20 = 600$ , $30 \times 7 = 210$ , $4 \times 20 = 80$ and $4 \times 7 = 28$ .
Add the partial products: $600 + 210 + 80 + 28 = 918$ .
$34 \times 27$ is 918.

Example

A theater has 18 rows with 25 seats in each row. How many seats are there?

Find $18 \times 25$ . Split the numbers: $18 = 10 + 8$ and $25 = 20 + 5$ .
$10 \times 20 = 200$ , $10 \times 5 = 50$ , $8 \times 20 = 160$ and $8 \times 5 = 40$ .
$200 + 50 + 160 + 40 = 450$
The theater has 450 seats.

Where children usually get stuck

To find $67 \times 43$ , your child writes the partial product for $67 \times 40$ as 268 instead of 2,680.
The steps were learned as moves with digits, so the child multiplies by 4 and forgets that the 4 means 4 tens. A child who makes this slip tends to make it the same way every time, so it is a habit to fix, not carelessness.
What helps. Ask your child to say each step with its real value: "sixty-seven times forty". Then ask about how much $70 \times 40$ is. It is 2,800, so 268 is far too small.
For $52 \times 31$ , your child writes $50 \times 30 + 2 \times 1 = 1{,}502$ .
It looks like adding: tens with tens and ones with ones. The two cross pieces, $50 \times 1$ and $2 \times 30$ , are missing, and together they are worth 110. The right answer is 1,612.
What helps. Draw the rectangle with four boxes and fill in every box before adding. Four boxes means four partial products.
Your child finds $30 \times 20$ by counting zeros and writes 60 or 6,000.
"Count the zeros" is a trick about how the answer looks. Without knowing that 3 tens times 2 tens is 6 hundreds, it is easy to lose a zero or add one.
What helps. Say it as a sentence together: 3 tens times 2 tens is 6 hundreds, which is 600. Then try $40 \times 50$ : 4 tens times 5 tens is 20 hundreds, which is 2,000.

Try four practice questions

Real 4th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
The area model below shows $23 \times 41$ . Drag tiles into the blanks to complete the explanation.
40
1
20
20 × 40
20 × 1
3
3 × 40
3 × 1
Tap or click a tile to drop it into the highlighted blank, or tap an empty blank first to choose a different one. Tap a filled blank to send its tile back.
The factor 23 is split into
.
The factor 41 is split into
.
The four rectangles show
.
The total product is found by
.
40 and 1
four partial products
adding those products
adding the side lengths
2 and 3
20 and 3
Tap an item to place it in the next empty slot, or drag it to a specific slot. With a keyboard: select an item with Enter or Space, then select a slot to place it. Press Delete or Backspace on a slot to remove an item.
Show the answer and how to solve it
The factor $23$ splits into $20$ and $3$ . The factor $41$ splits into $40$ and $1$ . The four rectangles show four partial products, and adding them gives $23 \times 41 = 943$ .
1. 1Look at the row and column headers of the area model. The row headers show $23 = 20 + 3$ (two tens and three ones). The column headers show $41 = 40 + 1$ (four tens and one).
2. 2Each rectangle in the area model shows one partial product. There are four rectangles, so there are four partial products: $20 \times 40$ , $20 \times 1$ , $3 \times 40$ , and $3 \times 1$ .
3. 3Compute each partial product:
4. 4To find the total product, add all four partial products together.
Common mistake. A common mistake is using digit-only splits like '2 and 3' instead of place-value splits like '20 and 3'. The 2 in 23 stands for 20 (two tens), not 2 ones.
The idea underneath. An area model breaks two-digit multiplication into four simpler products by place value, then adds them. Every digit of one factor must be multiplied by every digit of the other — each place-value combination needs its own product.
2On level
Drag the partial-product tiles into the slots to complete the equation. Place them in order from the greatest place value to the smallest.
Tap or click a tile to drop it into the highlighted blank, or tap an empty blank first to choose a different one. Tap a filled blank to send its tile back.
3 × 214 =
+
+
(3 × 200)
(3 × 2)
(3 × 4)
(3 × 214)
(3 × 10)
(3 × 40)
Tap an item to place it in the next empty slot, or drag it to a specific slot. With a keyboard: select an item with Enter or Space, then select a slot to place it. Press Delete or Backspace on a slot to remove an item.
Show the answer and how to solve it
The completed equation is $3 \times 214 = (3 \times 200) + (3 \times 10) + (3 \times 4)$ . The tiles go in order from the greatest place value (hundreds) to the smallest (ones).
1. 1Look at the multiplication. The equation breaks $3 \times 214$ into a sum of smaller multiplications — one for each place value in $214$ .
2. 2Write $214$ in expanded form. The hundreds part is $200$ , the tens part is $10$ , and the ones part is $4$ .
3. 3Multiply each place-value part by $3$ . These three products are the partial products.
4. 4Drag the tiles in order from greatest place value to smallest: $(3 \times 200)$ first, then $(3 \times 10)$ , then $(3 \times 4)$ .
Common mistake. A common mistake is using the digit value instead of the place value — for example, dragging $(3 \times 2)$ instead of $(3 \times 200)$ . The $2$ in $214$ stands for $200$ , not $2$ , because it sits in the hundreds place.
The idea underneath. Partial products work because of place value: every multi-digit number is a sum of its place-value parts (like $200 + 10 + 4$ ), so multiplying it by another number is the same as multiplying each part separately and adding the results.
3On level
What number belongs in the missing cell of the area model?
300
70
4
5
?
350
20
A
$\displaystyle 15$
B
$\displaystyle 1{,}500$
C
$\displaystyle 1{,}870$
D
$\displaystyle 15{,}000$
Show the answer and how to solve it
The missing partial product is 1,500. The basic fact is $5 \times 3 = 15$ , and because 300 has two zeros, the partial product also has two zeros: $5 \times 300 = 1{,}500$ .
1. 1Find the missing cell. Its row header is 5 and its column header is 300, so the cell shows the partial product $5 \times 300$ .
2. 2Use the basic fact $5 \times 3 = 15$ . Since $300 = 3 \times 100$ , the partial product has two more zeros than the basic fact:
3. 3The missing partial product is 1,500.
Common mistake. A common mistake is computing only the basic fact $5 \times 3 = 15$ and forgetting the place value, giving 15 instead of 1,500. When multiplying by 300, the answer must end in two zeros.
The idea underneath. An area model breaks multiplication into partial products by place value. To find each partial product, multiply the basic fact and then apply the right number of zeros for the place value.
4Stretch
Use the area model to find $7 \times 1{,}234$ .
1000
200
30
4
7
7000
1400
210
28
A
8,449
B
8,610
C
8,638
D
8,738
Show the answer and how to solve it
$7 \times 1{,}234 = 8{,}638$ . The total of the area model is the sum of all $4$ partial products: $7{,}000 + 1{,}400 + 210 + 28 = 8{,}638$ .
1. 1Look at the partial products inside the area model. The factor $7$ is on the left, and $1{,}234$ is split into $1{,}000 + 200 + 30 + 4$ across the top. Each cell shows one partial product.
2. 2To find the total product, add all $4$ partial products from the cells.
3. 3Add step by step. Combine the cells from left to right until all four partial products are included.
Common mistake. A common mistake is forgetting to add the smallest cell. Stopping after $7{,}000 + 1{,}400 + 210 = 8{,}610$ misses the ones cell, $28$ , so the answer comes out $8{,}610$ instead of $8{,}638$ .
The idea underneath. An area model splits a multi-digit factor into place-value parts. The total product is the sum of every sub-area — none can be skipped, and each cell must use the correct place value.