Multi-digit multiplication
The area model and the standard algorithm for multiplying two-digit numbers
About four lessons of 45 to 60 minutes
How do you count the seats in a whole hall without counting one by one?
The school hall is set up for assembly: 13 rows of seats, 14 seats in each row. How many seats is that? Counting them one at a time would take forever and you would lose your place. This is a multiplication, 13 times 14, but it is bigger than any times-table fact you have learned by heart.
The good news is that a big multiplication is just a lot of small ones added up. Today you will break each two-digit number into its tens and ones, multiply the pieces, and add them back together, first by seeing it as the area of a rectangle, then writing it as partial products, and finally as the fast standard algorithm. Same answer, three views, and you will know why each step works.
- 13 rows of 14 seats13 x 14 seats, far too many to count one by one
- A car park, 12 by 15 bays12 x 15 = 180 bays, an array you could walk around
- A theatre of 24 rows of 1324 x 13 = 312 seats, done with the standard algorithm
- A spreadsheet, 24 columns by 30 rows24 x 30 = 720 cells, tens times tens made easy
What students will be able to do
Students will multiply two-digit numbers by breaking each factor into tens and ones, model the product as the area of a rectangle split into four regions, write and add the four partial products, connect those partial products to the lines of the standard algorithm, and estimate to check that an answer is reasonable.
- I can break a two-digit number into tens and ones.
- I can draw an area model and find all four partial products.
- I can add the partial products to get the total.
- I can multiply two-digit numbers with the standard algorithm.
- I can explain why the second line of the algorithm ends in a zero, and estimate to check my answer.
Standards this unit teaches
- 5.NBT.B.5Common Core (US)Multiply multi-digit numbers
Fluently multiply multi-digit whole numbers using the standard algorithm.
- 4.NBT.B.5Common Core (US)Area model and partial products (Grade 4 foundation)
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. The US introduces the area model here, in Grade 4, and Grade 5 turns it into the standard algorithm.
- AC9M6N05Australian Curriculum v9 (ACARA)Multiply larger numbers (Year 6 anchor)
Solve problems that multiply larger numbers by one- or two-digit numbers, choosing efficient strategies and checking the answers. Australia places formal two-digit-by-two-digit multiplication in Year 6, a year later than the US, so the standard-algorithm work here maps to Australian Year 6.
- AC9M4N05Australian Curriculum v9 (ACARA)Efficient multiplication strategies (Year 4 foundation)
Develop efficient strategies and use suitable digital tools for adding, subtracting, multiplying and dividing with no remainder. This Year 4 descriptor is where Australia builds the multiplicative strategies, including partitioning, that the area model here formalises.
Prior knowledge
This unit builds on skills students should already have met. Revisit any that are shaky first.
Words to teach and display
- Factor
- a number being multiplied
- Product
- the result of a multiplication
- Partial product
- one of the smaller products you add up, such as 10 x 4
- Area model
- a rectangle split into regions whose areas are the partial products
- Standard algorithm
- the compact column method for multiplying
- Estimate
- a quick approximate answer from rounded numbers, used to check
Teach it: concrete, pictorial, abstract
The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.
1. Multiplication is still equal rows, the array just grew
ConcreteStart where multiplication began: equal rows. The hall has 13 equal rows of 14 seats, and 13 x 14 counts them all, exactly as 3 x 4 counted three rows of four. The only new thing is the size. Draw the whole array of seats and it is clear that counting them one by one is hopeless, so we need a smarter way to add up equal rows.
The smarter way uses a fact you already have: multiplying by ten and by small numbers is easy. If we can break the big rectangle into pieces whose sides are tens and ones, every piece becomes an easy multiplication. That breaking-up is the whole idea of the next section.
- Why is 13 x 14 harder than a times-table fact like 6 x 7?
- What is easy to multiply by, that might help us break this up?
2. The area model: break each number into tens and ones
PictorialSplit each side of the rectangle at its tens and ones. 13 becomes 10 and 3, and 14 becomes 10 and 4. Those two cuts divide the rectangle into four smaller rectangles, and the area of each is an easy multiplication. Add the four areas and you have the whole thing. This is the distributive property made visible: every ten and one on one side meets every ten and one on the other.
The four regions of 13 x 14 are 10 x 10 = 100, 10 x 4 = 40, 3 x 10 = 30, and 3 x 4 = 12. Add them: 100 + 40 + 30 + 12 = 182. So 13 x 14 = 182 seats. Every seat is counted once, in exactly one of the four boxes.
The biggest box, tens times tens, holds most of the total, and it is the one students most often forget. All four boxes count, always.
Use the area model to find 13 x 14.
- Break the sides: 13 = 10 + 3 and 14 = 10 + 4.
- Find the four box areas: 10 x 10 = 100, 10 x 4 = 40, 3 x 10 = 30, 3 x 4 = 12.
- Add the four partial products: 100 + 40 + 30 + 12 = 182.
Answer: 13 x 14 = 182.
- Which box is the biggest, and why is it easy to forget?
- How do you know every seat is counted exactly once?
3. Partial products written down
PictorialYou do not have to draw the rectangle every time. Once you can see the four boxes in your head, just list the four partial products and add them. Take 24 x 13. Break it into 24 = 20 + 4 and 13 = 10 + 3, and write the four products in a column.
The four partial products of 24 x 13 are 20 x 10 = 200, 20 x 3 = 60, 4 x 10 = 40, and 4 x 3 = 12. Add them: 200 + 60 + 40 + 12 = 312. So 24 x 13 = 312. Writing them from biggest to smallest keeps the place values straight and makes the addition easy.
Find 24 x 13 with partial products.
- Break the factors: 24 = 20 + 4 and 13 = 10 + 3.
- List the four products: 20 x 10 = 200, 20 x 3 = 60, 4 x 10 = 40, 4 x 3 = 12.
- Add them: 200 + 60 + 40 + 12 = 312.
Answer: 24 x 13 = 312.
- Why is it tidy to write the partial products from biggest to smallest?
- What are the four partial products of 16 x 12?
4. The standard algorithm
AbstractThe standard algorithm is the fast, compact version of partial products. Stack 24 over 13. First multiply 24 by the ones digit, 3, to get 72. Then multiply 24 by the tens digit, which is really 10, so write a 0 in the ones place and multiply by 1 to get 240. Add the two lines: 72 + 240 = 312, the same answer the area model gave.
Every line is a group of partial products in disguise. The first line, 72, is 24 x 3, which is the 60 and the 12 from the partial products added together. The second line, 240, is 24 x 10, which is the 200 and the 40. That placeholder zero is not a trick, it is there because you are multiplying by ten, so the answer must land one place to the left.
Always estimate first to catch mistakes. 24 x 13 is roughly 24 x 13 rounded to 20 x 13 = 260, or 24 x 10 = 240 as a floor, so an answer near 300 is reasonable and 312 checks out. An answer like 96 would fail that check at a glance.
Find 24 x 13 with the standard algorithm.
- Multiply 24 by the ones digit 3: 3 x 4 = 12, write 2 and carry 1, then 3 x 2 = 6 plus the carried 1 is 7. First line is 72.
- Multiply 24 by the tens digit. Write a 0 in the ones place, then 1 x 24 = 24. Second line is 240.
- Add the two lines: 72 + 240 = 312.
Answer: 24 x 13 = 312, matching the area model.
- Why does the second line, 24 x 10, end in a zero?
- Estimate 24 x 13 by rounding. Does 312 look reasonable?
Common misconceptions and how to address them
MisconceptionIn the area model, only multiply tens by tens and ones by ones, so 13 x 14 is 100 + 12 = 112.
Why it happens: Students match up the like parts and miss the two cross boxes, tens times ones, that the split also creates.
How to address it: Point at all four boxes on the rectangle. Every ten and one on one side meets every ten and one on the other, so there are always four products, not two. The missing 40 and 30 are real seats.
MisconceptionIn the algorithm, the second line is 24, so 24 x 13 is 72 + 24 = 96.
Why it happens: Students forget the placeholder zero and multiply by the digit 1 as if it were one, not ten.
How to address it: The second line multiplies by the tens digit, which stands for 10. Write the 0 in the ones place first, every time, so the line becomes 240. Say out loud: this line is 24 times ten.
MisconceptionMultiply the tens digits as single digits, so the tens box of 13 x 14 is 1 x 1 = 1.
Why it happens: The digits look like 1 and 1, and place value gets dropped in the rush to multiply.
How to address it: The 1 in 13 is worth 10 and the 1 in 14 is worth 10, so the box is 10 x 10 = 100, not 1. Read each digit as its value before multiplying.
MisconceptionCarrying is optional, or the carried digit gets added before multiplying.
Why it happens: The carry is a small mark that is easy to drop or to fold into the wrong step.
How to address it: Multiply first, then add the carry. For 3 x 4 = 12, write 2 and carry the 1, then do 3 x 2 = 6 and only now add the carried 1 to make 7. Keep the carry written above the next column.
MisconceptionThere is no need to check a multiplication, the algorithm is always right.
Why it happens: Students trust the procedure and never estimate, so a slipped digit sails through.
How to address it: Estimate before or after by rounding the factors. 24 x 13 is about 20 x 13 = 260 or 25 x 13, so the answer should be near 300. Any result far from the estimate signals a mistake to hunt down.
Guided practice (with answers)
1. Break 16 and 12 into tens and ones.
Answer: 16 = 10 + 6 and 12 = 10 + 2.
2. Use the area model to find 16 x 12.
Answer: 10 x 10 = 100, 10 x 2 = 20, 6 x 10 = 60, 6 x 2 = 12, and 100 + 20 + 60 + 12 = 192.
3. Add the partial products 200 + 60 + 40 + 12.
Answer: 312. These are the four products of 24 x 13.
4. What is 45 x 10, and why does it end in a zero?
Answer: 450. Multiplying by ten shifts every digit one place to the left, so a zero fills the ones place.
5. Multiply 23 x 12 with the standard algorithm.
Answer: 23 x 2 = 46, then 23 x 10 = 230, and 46 + 230 = 276.
6. Estimate 19 x 21 to check a friend's answer of 399.
Answer: Round to 20 x 20 = 400, so an answer near 400 is reasonable, and 399 checks out.
Independent practice worksheets
Set the matching ChalkBee worksheets for independent work. The answer keys are computed in code, so they are never wrong. Start with the area model, then move to the standard algorithm once the partial products are secure.
Differentiation
- Keep the area model going longer, drawing the four boxes before every calculation, before moving to the bare algorithm.
- Give a pre-drawn rectangle already split at the tens, so the student only fills in the four products.
- Start with a two-digit number times a one-digit number to lock in carrying, then add the second line.
- Provide a times-table chart so a shaky fact never blocks the method being learned.
- Multiply a three-digit number by a two-digit number, extending the area model to six boxes.
- Compare the area model, partial products and standard algorithm on the same product and explain how the lines correspond.
- Explore what happens with a zero in a factor, such as 20 x 34 or 105 x 12.
- Set a real design task, such as seats for a hall of a given size, and have students choose and justify a method.
Assessment: exit ticket
A three-question exit ticket in the last five minutes. It samples the area model, the standard algorithm, and estimation.
1. Find 14 x 12 with the area model (four products).
Answer: 100 + 20 + 40 + 8 = 168.
2. Multiply 32 x 21 with the standard algorithm.
Answer: 32 x 1 = 32, then 32 x 20 = 640, and 32 + 640 = 672.
3. Estimate 48 x 31 to check reasonableness.
Answer: Round to 50 x 30 = 1,500, so a good answer is near 1,500 (the exact answer is 1,488).
Teacher notes and timings
- Rough timing across four lessons: Lesson 1 arrays grow and the need to break them up (section 1), Lesson 2 the area model (section 2), Lesson 3 partial products (section 3), Lesson 4 the standard algorithm plus the exit ticket (section 4 and assessment).
- Language to keep saying: break it into tens and ones, all four boxes, this line is times ten, estimate to check. These phrases pre-empt most of the misconceptions.
- The area-model arrays are drawn to scale as unit squares, so students can literally count a box to confirm a partial product. Do that at least once so the model is trusted, not just followed.
- Insist the standard algorithm and the area model be linked explicitly: the first line is the ones-digit boxes, the second line is the tens-digit boxes. If students see the algorithm as unrelated magic, the meaning is lost.
- Curriculum note and a US and AU divergence: US Grade 5 (5.NBT.B.5) expects fluency with the standard algorithm, building on the area model introduced in Grade 4 (4.NBT.B.5). Australia places formal two-digit-by-two-digit multiplication a year later, in Year 6 (AC9M6N05), with the strategy groundwork in Year 4 (AC9M4N05). So this unit maps to Australian Year 6 for the algorithm, and the area model belongs to US Grade 4.
- Present mode and print both work: use the Print button for a clean teacher copy or a student handout, and project the page to teach straight from the diagrams.