You already know your twos and your fives. How do they hand you , and every other fact with a three, a four or a six in it?

Grade 3 · 3.OA.C.7 · 3.OA.B.5

Facts for three, four and six

You already know your twos and your fives. How do they hand you 4 \times 7, 6 \times 7 and every other fact with a three, a four or a six in it?

You need

Facts for two, five and tenGrade 3 The distributive propertyGrade 3

Unlocks

Facts for seven, eight and nineGrade 3

Where you are 4 in Multiplication facts

Number9

Arithmetic5

Multiplication and division16

Multiplication facts4

Facts for two, five and ten Patterns in the multiplication table

Facts for three, four and six

Facts for seven, eight and nine

When you reach for it

By now a child knows the twos, fives and tens from Facts for two, five and ten. Two properties turn those few facts into many more. The commutative property says the order of the factors does not change the product, so 8 \times 3 is the same as 3 \times 8. The The distributive property says a factor can be split into parts, each part multiplied, and the products added, so 3 \times 8 is 2 \times 8 and 1 \times 8 added.

This lesson uses both to fill in the middle of the multiplication table: the facts with a factor of 3, 4 or 6. Each of these three numbers sits right next to a number whose facts are already known:

3 is 2 + 1, so a three fact is a two fact and one more group;
4 is 2 + 2, so a four fact is a two fact doubled;
6 is 5 + 1, so a six fact is a five fact and one more group. It is also 3 + 3, so a six fact is a three fact doubled.

Reach for these moves whenever a fact is not known yet but a fact next to it is. That is the larger skill here, building an unknown fact from a known one, and it keeps working past this lesson. The setting throughout is a flower shop, where flowers are tied in bunches of 3, 4 or 6, and buckets and shelves hold them in rows.

The lesson takes four days, one part each. The aim of the fourth day is fluency: the product comes without rebuilding it, and the building move stays as the way to check it.

The procedure

Each part gives one family of facts, says which property makes the shortcut correct, and shows the array to draw. An array is a picture of equal groups set out in rows, as in Arrays: 3 \times 8 is 3 rows with 8 in each row.

Part 1: threes are twos and one more group

Olumide sets out flower pots in his shop window, 3 rows with 8 pots in each row. The number of pots is 3 \times 8. He does not know that fact yet, but he knows 2 \times 8, since two rows of 8 is the double 8 + 8 = 16.

Draw the pots as an array and cut it after the second row.

Figure 1Three rows of 8 are two rows of 8 and one more row, 16 + 8 = 24.

The cut does not add a pot or take one away. Above the cut are 2 rows of 8; below it is 1 row of 8. So the window holds 16 pots and 8 more:

\begin{aligned} 3 \times 8 &= (2 + 1) \times 8 \\ &= (2 \times 8) + (1 \times 8) \\ &= 16 + 8 \\ &= 24 \end{aligned}

The reason the step is allowed is the distributive property: the 3 was split as 2 + 1, and the 8 multiplied each part. In plain words, a three fact is a two fact and one more group. The extra group is a whole group of the other factor, here a whole row of 8.

The same move gives every three fact from the twos:

Fact	Two fact	One more group	Product
3 \times 3	6	3	9
3 \times 4	8	4	12
3 \times 6	12	6	18
3 \times 7	14	7	21
3 \times 8	16	8	24
3 \times 9	18	9	27

The commutative property covers the other order. Seven bunches of 3 flowers is 7 \times 3, which is the same as 3 \times 7 = 21. So the row of threes is also the column of threes, and one table serves both.

Part 2: fours are double twos

Selam puts 7 tulips in each of 4 vases. That is 4 \times 7 tulips. Four rows split evenly into two rows and two rows, and each half is a two fact she knows: 2 \times 7 = 14.

Figure 2Four rows of 7 are two rows of 7 twice, 14 + 14 = 28.

\begin{aligned} 4 \times 7 &= (2 + 2) \times 7 \\ &= (2 \times 7) + (2 \times 7) \\ &= 14 + 14 \\ &= 28 \end{aligned}

Again the distributive property allows the split, this time of 4 as 2 + 2. Both pieces are the same size, so adding them is doubling. A four fact is a two fact doubled. And a two fact is itself a double (doubles), since 2 \times 7 = 7 + 7. So a four fact can be found by doubling the other factor twice:

7 \;\to\; 14 \;\to\; 28

Fact	Double once	Double again
4 \times 3	6	12
4 \times 4	8	16
4 \times 6	12	24
4 \times 7	14	28
4 \times 8	16	32
4 \times 9	18	36

The second doubling is the harder one, because it can cross a ten: 18 + 18 is 36, found as 10 + 10 = 20 and 8 + 8 = 16, then 20 + 16 = 36.

Part 3: sixes are fives and one more group, or threes doubled

Olumide fills 6 buckets with 7 sunflowers each, 6 \times 7 sunflowers. Six is one more than five, and the fives are known: 5 \times 7 = 35. Cut the array after the fifth row.

Figure 3Six rows of 7 are five rows of 7 and one more row, 35 + 7 = 42.

\begin{aligned} 6 \times 7 &= (5 + 1) \times 7 \\ &= (5 \times 7) + (1 \times 7) \\ &= 35 + 7 \\ &= 42 \end{aligned}

A six fact is a five fact and one more group. It is the same move as the threes, leaning on the fives instead of the twos. The fives are quick (pairs of nickels make dimes), so this is often the fastest way to a six fact.

There is a second way, because 6 is also 3 + 3. Six rows cut in half are two copies of three rows, so a six fact is a three fact doubled:

6 \times 7 = (3 \times 7) + (3 \times 7) = 21 + 21 = 42

Both ways must agree, since both cut the same array. Having two ways is useful: one finds the fact and the other checks it.

Fact	Five fact and one more group	Three fact doubled
6 \times 3	15 + 3 = 18	9 + 9 = 18
6 \times 4	20 + 4 = 24	12 + 12 = 24
6 \times 6	30 + 6 = 36	18 + 18 = 36
6 \times 7	35 + 7 = 42	21 + 21 = 42
6 \times 8	40 + 8 = 48	24 + 24 = 48
6 \times 9	45 + 9 = 54	27 + 27 = 54

Because every six fact is a double, every product of 6 is even.

Part 4: the whole block, and building a fact from one you know

Put the three families together. Facts with a factor of 0, 1, 2, 5 or 10 are already known, and by the commutative property each fact counts once for both orders. That leaves exactly fifteen facts that are new in this lesson:

	\times 3	\times 4	\times 6	\times 7	\times 8	\times 9
3	9	12	18	21	24	27
4		16	24	28	32	36
6			36	42	48	54

The blank cells are not missing: 4 \times 3 is 3 \times 4 = 12, already in the top row, and the same goes for 6 \times 3 and 6 \times 4. After this block only six facts of the table are left, 7 \times 7, 7 \times 8, 7 \times 9, 8 \times 8, 8 \times 9 and 9 \times 9, and Facts for seven, eight and nine builds them.

Three more moves make the block hang together.

One more group, from any known fact. If a fact is known, the fact with one more group is that product plus one group. Olumide knows 3 \times 8 = 24, and by the commutative property that is also 8 \times 3, eight bunches of 3. One more bunch of 3 makes 9 \times 3 = 24 + 3 = 27. This is the distributive property again, with 9 split as 8 + 1. It means a child who knows one fact in a row never has to start the row from zero.

Turn it around when the other order is easier. 8 \times 4 asks for 8 groups of 4. Read it as 4 \times 8 and it is a four fact, double 8 and double again. The commutative property makes the swap correct.

Run it backward. Every fact also answers a missing-factor question. If 24 roses are tied in bunches of 4, the number of bunches is the number in \square \times 4 = 24. Search the fours for the product 24: 6 \times 4 = 24, so there are 6 bunches. This is division, and Division facts takes it up for the whole table.

Fluency means the product comes to mind without rebuilding it. It grows from using the facts, and the building moves are what make that safe: a fact half remembered can always be rebuilt from a neighbor and checked. For 4 \times 7, two sevens doubled give 14 + 14 = 28, and three sevens and one more seven give 21 + 7 = 28. Two routes to the same product are a stronger check than any pattern in the last digit.

Worked examples

Examples 1 and 2 go with Part 1, Examples 3 and 4 with Part 2, Examples 5 and 6 with Part 3, and Examples 7 and 8 with Part 4.

Example 1

Pots in the window

Olumide’s shop window has 3 rows of flower pots with 8 pots in each row. How many pots are in the window?

Solution

The window is 3 rows of 8, so the number of pots is 3 \times 8. This is the array in Part 1.

Split the 3 rows as 2 + 1. The first two rows are a two fact, 2 \times 8 = 8 + 8 = 16. The last row is one more group of 8. The distributive property says the two pieces together make the whole:

3 \times 8 = (2 \times 8) + (1 \times 8) = 16 + 8 = 24

The window holds 24 pots. Check by adding the rows: 8 + 8 + 8 = 24.

Example 2

Seven bunches of three

Faustine ties daisies in bunches of 3. She makes 7 bunches. How many daisies does she use?

Solution

Seven bunches of 3 is 7 groups of 3, so the count is 7 \times 3.

One way turns it around. By the commutative property, 7 \times 3 = 3 \times 7, and 3 \times 7 is a three fact: a two fact and one more group.

3 \times 7 = (2 \times 7) + 7 = 14 + 7 = 21

Another way keeps the order and splits the 7 as 5 + 2, both of which are known groups:

7 \times 3 = (5 \times 3) + (2 \times 3) = 15 + 6 = 21

Both give 21. Faustine uses 21 daisies.

Example 3

Four vases of tulips

Selam puts 7 tulips in each of 4 vases. How many tulips does she use?

Solution

Four vases of 7 is 4 \times 7. Split the 4 as 2 + 2, as in the picture in Part 2. Each half is the two fact 2 \times 7 = 14:

\begin{aligned} 4 \times 7 &= (2 \times 7) + (2 \times 7) \\ &= 14 + 14 = 28 \end{aligned}

The step is the distributive property with 4 = 2 + 2. Said as doubling: double 7 is 14, and double 14 is 28. Selam uses 28 tulips. Check: 7 + 7 + 7 + 7 is 14 + 14 = 28.

Example 4

Eight bunches of carnations

Faustine sells 8 bunches of carnations, with 4 carnations in each bunch. How many carnations is that?

Solution

Eight bunches of 4 is 8 \times 4. Counting by fours eight times is slow, so turn it around: by the commutative property, 8 \times 4 = 4 \times 8, which is a four fact.

Double the 8, then double again:

8 \;\to\; 16 \;\to\; 32

The first doubling is 2 \times 8 = 16. The second is 16 + 16: the tens make 10 + 10 = 20 and the ones make 6 + 6 = 12, so 20 + 12 = 32. In one line,

8 \times 4 = 4 \times 8 = (2 \times 8) + (2 \times 8) = 16 + 16 = 32

Faustine sells 32 carnations. Check with another split of the 8 bunches, 5 + 3: (5 \times 4) + (3 \times 4) = 20 + 12 = 32.

Example 5

Buckets of sunflowers

Olumide fills 6 buckets with 7 sunflowers each. How many sunflowers does he use?

Solution

Six buckets of 7 is 6 \times 7. Six is 5 + 1, so split off five buckets, as in the picture in Part 3. Five buckets is the five fact 5 \times 7 = 35, and the sixth bucket is one more group of 7:

\begin{aligned} 6 \times 7 &= (5 \times 7) + (1 \times 7) \\ &= 35 + 7 = 42 \end{aligned}

Olumide uses 42 sunflowers. Check by doubling a three fact: three buckets hold 3 \times 7 = 21, and six buckets hold twice that, 21 + 21 = 42.

Example 6

Six bunches of six

Selam ties lilies in bunches of 6 and makes 6 bunches. Find the number of lilies two ways.

Solution

The count is 6 \times 6.

Five sixes and one more six. Split the 6 bunches as 5 + 1:

6 \times 6 = (5 \times 6) + (1 \times 6) = 30 + 6 = 36

Three sixes doubled. Split the 6 bunches as 3 + 3 instead. Three bunches of 6 is 3 \times 6, a three fact: (2 \times 6) + 6 = 12 + 6 = 18. The other three bunches hold the same, so

6 \times 6 = (3 \times 6) + (3 \times 6) = 18 + 18 = 36

Figure 4Six rows of 6 cut in half are two copies of 3 \times 6, 18 + 18 = 36.

Both splits cut the same array, so they agree. Selam uses 36 lilies.

Example 7

How many bunches?

Faustine has 27 daisies and ties them in bunches of 3. How many bunches does she make?

Solution

The total is known and the number of bunches is missing:

\square \times 3 = 27

Start from a fact that is close. Example 1 showed 3 \times 8 = 24, so by the commutative property 8 bunches of 3 use 24 daisies. That leaves 27 - 24 = 3 daisies, exactly one more bunch. So \square = 8 + 1 = 9.

Check with a three fact: 3 \times 9 = (2 \times 9) + 9 = 18 + 9 = 27. Faustine makes 9 bunches.

Example 8

One more bunch of six

A customer orders 7 bunches of 6 roses. Faustine already knows from Example 6 that 6 bunches of 6 is 36 roses. How many roses does the order need?

Solution

Seven bunches is six bunches and one more bunch, so the distributive property, with 7 split as 6 + 1, gives

7 \times 6 = (6 \times 6) + (1 \times 6) = 36 + 6 = 42

The order needs 42 roses.

Check against a different fact. By the commutative property 7 \times 6 = 6 \times 7, and Example 5 found 6 \times 7 = 42 from five sevens and one more seven. Two different routes reach the same product, which is what makes a built fact trustworthy.

Boundaries

The parts must add up to the factor. 3 = 2 + 1, 4 = 2 + 2 and 6 = 5 + 1 = 3 + 3. That is what lets the pieces of the array fill it exactly. Pieces that add to 5 count only five rows.
The extra group is a whole group of the other factor. In 3 \times 8, the one more group is a row of 8, not a single pot, so 3 \times 8 = 16 + 8.
Doubling needs an even factor. 4 and 6 split into two equal whole parts, so their facts are doubles. 3 does not split into two equal whole parts, so splitting the 3 never gives a doubling; the threes use 2 + 1 instead.
Whole groups and factors from 0 to 10. Every fact here counts whole bunches of whole flowers. The moves still hold at zero (3 \times 0 = (2 \times 0) + 0 = 0) and past ten, but the facts to know by heart stop at 10 \times 10.
A backward question needs an exact fit. \square \times 4 = 30 has no whole number answer, because 4 \times 7 = 28 and 4 \times 8 = 32 jump over 30. Thirty flowers cannot all go into bunches of 4.
This lesson splits a factor into a sum. Splitting 9 as 10 - 1 also works, and that is taken up in Facts for seven, eight and nine.

Where this goes next

Facts for seven, eight and nine finishes the table with the same idea: a nine fact is a ten fact take away one group, and an eight fact is a two fact doubled and doubled again. With the commutative property, the block in this lesson and the earlier twos, fives and tens, only six facts are left for it. Every fact here also answers a division question, \square \times 4 = 24 being 24 \div 4, and Division facts builds division on that link. The doubling between the twos and the fours, and between the threes and the sixes, shows up as whole rows of the table in Patterns in the multiplication table. Later, doubling and halving to find a product becomes a strategy of its own in Doubling, halving and quick products, and a fact like 4 \times 7 grows into 4 \times 70 in Multiplying by tens and hundreds. All of these sit under Multiplication facts.

Exercises

Exercises 1 to 4 go with Part 1, 5 to 9 with Part 2, 10 to 14 with Part 3, and 15 to 21 with Part 4. The reasoning group draws on all four.

Part 1: threes

Find 3 \times 6 by finding 2 \times 6 and adding one more group of 6.
Faustine makes 4 bunches of 3 daisies. Selam makes 3 bunches of 4 daisies. How many daisies does each use?
Fill in the boxes: 3 \times 3 = (2 \times 3) + \square, which is 6 + \square, which is \square.
Find 3 \times 5 two ways: as a two fact and one more group, and as a five fact.

Part 2: fours

Find 4 \times 6 by doubling 2 \times 6.
Selam sells 9 bunches of 4 tulips. How many tulips does she sell?
Find 4 \times 4 by doubling 4 and doubling again.
Fill in the boxes: 4 \times 3 = (2 \times 3) + (2 \times \square), which is \square + \square, which is \square.
Olumide puts 5 roses in each of 4 buckets. Find the number of roses by doubling, then check with a five fact.

Part 3: sixes

Find 6 \times 8 as a five fact and one more group.
Find 6 \times 9 by doubling 3 \times 9.
Find 6 \times 4 two ways: as a five fact and one more group, and as a three fact doubled.
Selam has 3 bunches of 6 lilies. Faustine has 6 bunches of 3 lilies. Who has more lilies?
Fill in the boxes: 6 \times 10 = (5 \times 10) + \square, which is \square.

Part 4: the whole block

Faustine ties 28 roses in bunches of 4. How many bunches does she make?
Olumide ties 30 daisies in bunches of 6. How many bunches does he make?
A customer buys 2 bunches of 6 tulips and 3 bunches of 4 tulips. How many tulips is that in all?
Selam knows 4 \times 8 = 32. Use it to find 4 \times 9, then 4 \times 10.
Fill in the missing factors: 3 \times \square = 18, 4 \times \square = 36 and \square \times 6 = 54.
Which is larger, 6 \times 4 or 3 \times 9, and by how much?
A customer wants 40 flowers tied in bunches that are all the same size. Can the bunches be of 4? Of 6? Give the number of bunches where it works.

Reasoning

Explain, with an array, why 3 \times 7 is the same as (2 \times 7) + 7. Name the property that makes it true.
Explain why doubling 2 \times 9 gives 4 \times 9.
Explain why 6 \times 8 can be found both as (5 \times 8) + 8 and as (3 \times 8) + (3 \times 8), and why the two ways must give the same answer.
Is every product with a factor of 6 even? Explain.

Answers

18. 2 \times 6 = 12, and one more group of 6 makes 12 + 6 = 18.
Each uses 12 daisies. 4 \times 3 = 12 and 3 \times 4 = (2 \times 4) + 4 = 8 + 4 = 12; the commutative property says the two orders give the same product.
The boxes are 3, 3 and 9: (2 \times 3) + 3 = 6 + 3 = 9.
15 both ways. (2 \times 5) + 5 = 10 + 5 = 15, and the five fact 3 \times 5 = 15.
24. 2 \times 6 = 12, and doubling gives 12 + 12 = 24.
36 tulips. 9 \times 4 = 4 \times 9; double 9 is 18 and double 18 is 36.
16. Double 4 is 8, and double 8 is 16.
The boxes are 3, then 6 + 6, then 12: (2 \times 3) + (2 \times 3) = 6 + 6 = 12.
20 roses. Double 5 is 10 and double 10 is 20. Check: 4 \times 5 = 20, two pairs of fives making two tens.
48. (5 \times 8) + 8 = 40 + 8 = 48.
54. 3 \times 9 = (2 \times 9) + 9 = 18 + 9 = 27, and 27 + 27 = 54.
24 both ways. (5 \times 4) + 4 = 20 + 4 = 24, and 3 \times 4 = 12 doubled is 12 + 12 = 24.
Neither; both have 18. 3 \times 6 = 18 and 6 \times 3 = 18, the same factors in the other order.
The boxes are 10 and 60: 50 + 10 = 60.
7 bunches. \square \times 4 = 28, and 7 \times 4 = 4 \times 7 = 28.
5 bunches. \square \times 6 = 30, and 5 \times 6 = 30.
24 tulips. 2 \times 6 = 12 and 3 \times 4 = 12, so 12 + 12 = 24.
4 \times 9 = 32 + 4 = 36, one more group of 4. Then 4 \times 10 = 36 + 4 = 40, which agrees with the ten fact.
6, 9 and 9. 3 \times 6 = 18, 4 \times 9 = 36 and 9 \times 6 = 54.
3 \times 9, by 3. 6 \times 4 = 24 and 3 \times 9 = 27, and 27 - 24 = 3.
Bunches of 4: yes, 10 bunches, since 10 \times 4 = 40. Bunches of 6: no, since 6 \times 6 = 36 and 7 \times 6 = 42 jump over 40, so 4 flowers would be left over.
Draw 3 rows of 7 and cut after the second row. The top piece is 2 rows of 7, which is 2 \times 7, and the bottom piece is 1 row of 7, which is 7. The cut moves no dot, so the two pieces hold all 3 \times 7 dots. That is the distributive property with 3 split as 2 + 1: 3 \times 7 = (2 \times 7) + (1 \times 7) = 14 + 7 = 21.
4 \times 9 is 4 rows of 9. Cut it after the second row and each half is 2 rows of 9, so 4 \times 9 = (2 \times 9) + (2 \times 9) by the distributive property with 4 = 2 + 2. Adding a number to itself is doubling, so 4 \times 9 is double 2 \times 9: 18 + 18 = 36.
Both are cuts of the same array of 6 rows of 8. The first cuts after the fifth row, splitting 6 as 5 + 1: 40 + 8 = 48. The second cuts after the third row, splitting 6 as 3 + 3: 24 + 24 = 48. A cut never adds or removes a dot, so every cut of the array counts the same 48 dots.
Yes. A product with a factor of 6 can be written with the 6 first, by the commutative property, and 6 \times n = (3 \times n) + (3 \times n), which is a double. A number that is a double splits into two equal whole parts, so it is even. For example, 6 \times 7 = 21 + 21 = 42.