Free Math Worksheets

- Which number multiplied by itself gives each product? - Recall the square facts in order: \(1 \times 1\), \(2 \times 2\), \(3 \times 3\), and so on. - Check each result by multiplying the two equal factors.

1. Find the equal factors whose product is each square number. 2. \(4 \times 4 = 16\). 3. \(6 \times 6 = 36\). 4. \(7 \times 7 = 49\). 5. \(8 \times 8 = 64\). 6. \(9 \times 9 = 81\).

\(4 \times 4 = 16\) \(6 \times 6 = 36\) \(7 \times 7 = 49\) \(8 \times 8 = 64\) \(9 \times 9 = 81\)

- Which multiplication facts are usually easiest to recall? - Think about doubling and counting by fives. - What happens when you multiply a whole number by \(1\) or by \(10\)?

1. The fact with \(1\) is \(1 \times 6 = 6\). 2. The fact with \(2\) is \(2 \times 6 = 12\). 3. The fact with \(5\) is \(5 \times 6 = 30\). 4. The fact with \(10\) is \(10 \times 6 = 60\).

\(1 \times 6 = 6\) \(2 \times 6 = 12\) \(5 \times 6 = 30\) \(10 \times 6 = 60\)

- What is special about the factors in a square fact? - Which number multiplied by itself gives each product? - Recall the multiplication facts to match each pair.

1. Evaluate each square fact: \(8 \times 8 = 64\), \(6 \times 6 = 36\), \(10 \times 10 = 100\), and \(7 \times 7 = 49\). 2. Match each product to the fact with that value.

\(64 = 8 \times 8\) \(36 = 6 \times 6\) \(100 = 10 \times 10\) \(49 = 7 \times 7\)

- Say the 4s products in order. - Check each number in the list. - Which numbers appear when you count by \(4\)s?

1. Recall the 4s products through \(4 \times 10\). 2. The listed products in the 4s facts are \(8=2 \times 4\), \(16=4 \times 4\), \(20=5 \times 4\), \(28=7 \times 4\), \(36=9 \times 4\), and \(40=10 \times 4\).

\(8, 16, 20, 28, 36, 40\)

- Evaluate multiplication before addition. - Any number multiplied by \(0\) equals \(0\). - Any number multiplied by \(1\) equals that number.

1. For a), \(0 \times 6 = 0\) and \(0 + 6 = 6\), so \(0 < 6\). 2. For b), \(4 \times 0 + 2 = 2\) and \(2 \times 0 + 4 = 4\), so \(2 < 4\). 3. For c), both sides equal \(8\), so they are equal. 4. For d), both sides equal \(0\), so they are equal.

a) \(<\) b) \(<\) c) \(=\) d) \(=\)

- Which multiplication facts with \(8\) are easiest for you to recall? - Can you break the first factor into two anchor factors? - How can knowing the 2s fact help you find the 4s fact?

1. The anchor facts are \(1 \times 8 = 8\), \(2 \times 8 = 16\), \(5 \times 8 = 40\), and \(10 \times 8 = 80\). 2. Combine \(5 \times 8\) and \(1 \times 8\): \(40 + 8 = 48\), so \(6 \times 8 = 48\). 3. Double \(2 \times 8 = 16\): \(16 + 16 = 32\), so \(4 \times 8 = 32\).

a) \(1 \times 8 = 8\), \(2 \times 8 = 16\), \(5 \times 8 = 40\), \(10 \times 8 = 80\) b) \(5 \times 8 + 1 \times 8 = 40 + 8 = 48\) c) \(2 \times 8 = 16\), and \(16 + 16 = 32\), so \(4 \times 8 = 32\)

- Say the products in the 7s or 9s facts in order. - Which factors make a fact an anchor fact? - What is special about the factors in a square fact?

1. In the 7s facts, \(5 \times 7 = 35\). It is an anchor fact because one factor is \(5\). 2. In the 9s facts, \(2 \times 9 = 18\). It is an anchor fact because one factor is \(2\). 3. The fact is \(7 \times 7 = 49\). It is a square fact because the factors are equal. It is not an anchor fact because neither factor is \(1\), \(2\), \(5\), or \(10\).

a) \(5 \times 7 = 35\); yes, it is an anchor fact. b) \(2 \times 9 = 18\); yes, it is an anchor fact. c) \(7 \times 7 = 49\); it is a square fact, not an anchor fact.

- What is special about the factors in a square fact? - Which factors make a fact an anchor fact? - Check the square fact for each of the factors \(1\), \(2\), \(5\), and \(10\). - Make sure each fact uses two equal factors.

1. A square fact has two equal factors. 2. The square fact with factor \(1\) is \(1 \times 1 = 1\). 3. The square fact with factor \(2\) is \(2 \times 2 = 4\). 4. The square fact with factor \(5\) is \(5 \times 5 = 25\). 5. The square fact with factor \(10\) is \(10 \times 10 = 100\).

\(1 \times 1 = 1\), \(2 \times 2 = 4\), \(5 \times 5 = 25\), and \(10 \times 10 = 100\).

- Does the ones digit suggest the 5s or 10s facts? - If the product is even, could a 2s fact work? - A product ending in \(0\) may be a 10s fact.

1. For \(14\), use the factor \(2\): \(7 \times 2 = 14\). 2. For \(45\), use the factor \(5\): \(9 \times 5 = 45\). 3. For \(80\), use the factor \(10\): \(8 \times 10 = 80\). 4. For \(30\), either \(3 \times 10 = 30\) or \(6 \times 5 = 30\) works. 5. For \(25\), use the factor \(5\): \(5 \times 5 = 25\).

a) \(7 \times 2 = 14\) b) \(9 \times 5 = 45\) c) \(8 \times 10 = 80\) d) \(3 \times 10 = 30\) or \(6 \times 5 = 30\) e) \(5 \times 5 = 25\)

- How many groups of \(9\) are needed altogether? - Find the larger anchor-fact part first, then add the smaller part. - Rewrite the multiplication fact as a sum of two easier products.

1. Evaluate \(5 \times 9 = 45\). 2. Evaluate \(1 \times 9 = 9\). 3. Add the partial products: \(45 + 9 = 54\). 4. Therefore, \(6 \times 9 = 54\).

\(5 \times 9 = 45\) \(1 \times 9 = 9\) \(45 + 9 = 54\) \(6 \times 9 = 54\)

- Work through the table one row at a time. - Think about twice, five times, and ten times each row number. - Multiply each number in the first column by each number in the top row.

1. For the row labeled \(4\), compute \(4 \times 2 = 8\), \(4 \times 5 = 20\), and \(4 \times 10 = 40\). 2. For the row labeled \(7\), compute \(7 \times 2 = 14\), \(7 \times 5 = 35\), and \(7 \times 10 = 70\).

Row \(4\): \(8, 20, 40\) Row \(7\): \(14, 35, 70\)

- Find the product on each side before comparing. - Write the two products separately if that helps you compare them. - The open side of the comparison symbol faces the greater value.

1. For part a, \(6 \times 7 = 42\) and \(5 \times 8 = 40\), so \(6 \times 7 > 5 \times 8\). 2. For part b, \(4 \times 9 = 36\) and \(6 \times 6 = 36\), so \(4 \times 9 = 6 \times 6\). 3. For part c, \(3 \times 9 = 27\) and \(4 \times 7 = 28\), so \(3 \times 9 < 4 \times 7\). 4. For part d, \(8 \times 7 = 56\) and \(9 \times 6 = 54\), so \(8 \times 7 > 9 \times 6\).

a) \(>\) b) \(=\) c) \(<\) d) \(>\)

- How does changing a factor from \(10\) to \(9\) change the product? - How are the 10s facts and 9s facts related? - Find the 10s product first, then subtract one group.

1. For part a, \(10 \times 4 = 40\), and \(40 - 4 = 36\). Therefore, \(9 \times 4 = 36\). 2. For part b, \(10 \times 8 = 80\), and \(80 - 8 = 72\). Therefore, \(9 \times 8 = 72\). 3. For part c, \(10 \times 6 = 60\), and \(60 - 6 = 54\). Therefore, \(9 \times 6 = 54\).

a) \(10 \times 4 - 1 \times 4 = 40 - 4 = 36\) b) \(10 \times 8 - 1 \times 8 = 80 - 8 = 72\) c) \(10 \times 6 - 1 \times 6 = 60 - 6 = 54\)

- List the square facts from \(1 \times 1\) through \(10 \times 10\). - Then identify which products are between the two numbers in each part. - Write both the multiplication fact and the product.

1. The nearby square products are \(3 \times 3 = 9\), \(4 \times 4 = 16\), and \(5 \times 5 = 25\). Only \(16\) is between \(10\) and \(20\), so part a is \(4 \times 4 = 16\). 2. The nearby square products are \(5 \times 5 = 25\), \(6 \times 6 = 36\), \(7 \times 7 = 49\), and \(8 \times 8 = 64\). The products between \(30\) and \(50\) are \(36\) and \(49\), so part b is \(6 \times 6 = 36\) and \(7 \times 7 = 49\).

a) \(4 \times 4 = 16\) b) \(6 \times 6 = 36\) and \(7 \times 7 = 49\)

- How much less is \(9\) than \(10\)? - What amount should you subtract from the 10s product? - How can ten groups become nine groups?

1. The 10s facts are \(10 \times 5 = 50\), \(10 \times 9 = 90\), and \(10 \times 4 = 40\). 2. Subtract one group to find each 9s fact: a) \(50 - 5 = 45\); b) \(90 - 9 = 81\); c) \(40 - 4 = 36\).

a) \(10 \times 5 = 50\), so \(9 \times 5 = 45\) b) \(10 \times 9 = 90\), so \(9 \times 9 = 81\) c) \(10 \times 4 = 40\), so \(9 \times 4 = 36\)

- Which easy fact with \(2\), \(5\), or \(10\) is close to the target fact? - Do you need to add one group or subtract one group? - Write the anchor fact first.

1. For part a, use \(5 \times 4 = 20\). Add one more group of \(4\): \(20 + 4 = 24\). 2. For part b, use \(10 \times 7 = 70\). Subtract one group of \(7\): \(70 - 7 = 63\). 3. For part c, use \(2 \times 8 = 16\). Add one more group of \(8\): \(16 + 8 = 24\).

a) Anchor fact: \(5 \times 4 = 20\); \(6 \times 4 = 24\) b) Anchor fact: \(10 \times 7 = 70\); \(9 \times 7 = 63\) c) Anchor fact: \(2 \times 8 = 16\); \(3 \times 8 = 24\)

- Which factor leaves the other number unchanged? - Which factor always makes the product \(0\)? - Try both choices mentally and check the equation.

1. Multiplying by \(1\) leaves the other factor unchanged, and multiplying by \(0\) gives a product of \(0\). 2. The completed equations are \(6 \times 1 = 6\), \(9 \times 0 = 0\), \(0 \times 7 = 0\), \(1 \times 8 = 8\), and \(10 \times 0 = 0\).

a) \(1\) b) \(0\) c) \(0\) d) \(1\) e) \(0\)

- Work one row at a time. - In the third row, which number multiplied by \(1\) equals \(6\)? - What happens when the starting number is \(0\)?

1. For \(n=5\), \(5 \times 1 = 5\) and \(5 \times 0 = 0\). 2. For \(n=8\), \(8 \times 1 = 8\). 3. If \(n \times 1 = 6\), then \(n=6\). Therefore, \(6 \times 0 = 0\). 4. For \(n=0\), \(0 \times 1 = 0\) and \(0 \times 0 = 0\).

Row \(n=5\): \(5\), \(0\) Row \(n=8\): \(8\) Row with \(n \times 1=6\): \(n=6\), and \(n \times 0=0\) Row \(n=0\): \(0\), \(0\)

- In which skip-counting sequence do all four numbers appear? - Check the sets of multiplication facts one at a time. - In each matching fact, one factor must be the same number.

1. Each number is a product in the 6s facts. 2. The matching facts are \(2 \times 6 = 12\), \(3 \times 6 = 18\), \(5 \times 6 = 30\), and \(7 \times 6 = 42\).

The 6s facts: \(2 \times 6 = 12\) \(3 \times 6 = 18\) \(5 \times 6 = 30\) \(7 \times 6 = 42\)

- “Twice” means multiply by \(2\). - Use a related multiplication fact to reverse the division statement. - Evaluate the square fact in the last riddle.

1. Twice \(9\) is \(2 \times 9 = 18\). 2. Use the related multiplication fact for \(\square \div 5 = 8\): \(8 \times 5 = 40\). 3. The square fact \(7 \times 7\) has a product of \(49\).

a) \(18\) b) \(40\) c) \(49\)

- Compare the first factor in the given fact with the first factor in the related fact. - Decide whether that factor was doubled or halved. - Make the same change to the product.

1. Part a: \(6\) is double \(3\), so double \(21\): \(42\). 2. Part b: \(5\) is half of \(10\), so halve \(40\): \(20\). 3. Part c: \(8\) is double \(4\), so double \(32\): \(64\). 4. Part d: \(4\) is half of \(8\), so halve \(48\): \(24\).

a) \(42\) b) \(20\) c) \(64\) d) \(24\)

- Look at the second factor on the left side of each equation. - Break that factor into the two addends shown on the right. - Check that the two numbers inside the products add to the original second factor.

1. Decompose the second factor in each multiplication fact. 2. Since \(7 = 5 + 2\), \(6 \times 7 = 6 \times 5 + 6 \times 2\). 3. Since \(8 = 4 + 4\), \(6 \times 8 = 6 \times 4 + 6 \times 4\). 4. Since \(9 = 5 + 4\), \(6 \times 9 = 6 \times 5 + 6 \times 4\). 5. Since \(4 = 2 + 2\), \(6 \times 4 = 6 \times 2 + 6 \times 2\).

\(6 \times 7 = (6 \times 5) + (6 \times 2)\) \(6 \times 8 = (6 \times 4) + (6 \times 4)\) \(6 \times 9 = (6 \times 5) + (6 \times 4)\) \(6 \times 4 = (6 \times 2) + (6 \times 2)\)

- Break the second factor into \(5\) and another number. - Use the familiar multiplication fact with \(5\) first. - Add the two partial products.

1. Part a: Since \(6 = 5 + 1\), \(8 \times 6 = 8 \times 5 + 8 \times 1 = 40 + 8 = 48\). 2. Part b: Since \(8 = 5 + 3\), \(8 \times 8 = 8 \times 5 + 8 \times 3 = 40 + 24 = 64\). 3. Part c: Since \(9 = 5 + 4\), \(8 \times 9 = 8 \times 5 + 8 \times 4 = 40 + 32 = 72\).

a) \(8 \times 6 = 8 \times 5 + 8 \times 1 = 40 + 8 = 48\) b) \(8 \times 8 = 8 \times 5 + 8 \times 3 = 40 + 24 = 64\) c) \(8 \times 9 = 8 \times 5 + 8 \times 4 = 40 + 32 = 72\)

- Break the second factor into two smaller numbers. - Choose multiplication facts with \(9\) that you know well. - Check that the two addends combine to make the original second factor.

1. Part a: Use \(4 = 2 + 2\). Then \(9 \times 4 = 9 \times 2 + 9 \times 2 = 18 + 18 = 36\). 2. Part b: Use \(7 = 5 + 2\). Then \(9 \times 7 = 9 \times 5 + 9 \times 2 = 45 + 18 = 63\). 3. Part c: Use \(8 = 5 + 3\). Then \(9 \times 8 = 9 \times 5 + 9 \times 3 = 45 + 27 = 72\).

a) \(9 \times 4 = (9 \times 2) + (9 \times 2) = 18 + 18 = 36\) b) \(9 \times 7 = (9 \times 5) + (9 \times 2) = 45 + 18 = 63\) c) \(9 \times 8 = (9 \times 5) + (9 \times 3) = 45 + 27 = 72\)

- Identify the factor that is the same in both partial products. - Add the other two factors. - Multiply the common factor by that sum.

1. Part a: Add the numbers of groups: \(3 + 5 = 8\). Then \((4 \times 3) + (4 \times 5) = 4 \times 8 = 32\). 2. Part b: Add the numbers of groups: \(2 + 6 = 8\). Then \((7 \times 2) + (7 \times 6) = 7 \times 8 = 56\). 3. Part c: Add the numbers of groups: \(8 + 2 = 10\). Then \((3 \times 8) + (3 \times 2) = 3 \times 10 = 30\). 4. Part d: Add the numbers of groups: \(4 + 4 = 8\). Then \((6 \times 4) + (6 \times 4) = 6 \times 8 = 48\).

a) \(4 \times 8 = 32\) b) \(7 \times 8 = 56\) c) \(3 \times 10 = 30\) d) \(6 \times 8 = 48\)

- Identify the factor that stays the same in each expression. - Add or subtract the numbers of groups. - Multiply the common factor by the resulting number of groups.

1. Part a: Combine the groups: \(3 + 7 = 10\). Then \((8 \times 3) + (8 \times 7) = 8 \times 10 = 80\). 2. Part b: Subtract the groups: \(9 - 4 = 5\). Then \((5 \times 9) - (5 \times 4) = 5 \times 5 = 25\). 3. Part c: Combine all three sets of groups: \(2 + 2 + 2 = 6\). Then the expression equals \(4 \times 6 = 24\). 4. Part d: Subtract the groups: \(10 - 1 = 9\). Then \((9 \times 10) - (9 \times 1) = 9 \times 9 = 81\).

a) \(8 \times 10 = 80\) b) \(5 \times 5 = 25\) c) \(4 \times 6 = 24\) d) \(9 \times 9 = 81\)

- Break \(7\) into the two numbers Mia uses. - Multiply both parts by the other factor. - Add the two partial products.

1. Part a: Break \(7\) into \(5 + 2\). Then \(7 \times 4 = 5 \times 4 + 2 \times 4 = 20 + 8 = 28\). 2. Part b: Break \(7\) into \(5 + 2\). Then \(7 \times 7 = 5 \times 7 + 2 \times 7 = 35 + 14 = 49\). 3. Part c: Break \(7\) into \(5 + 2\). Then \(7 \times 9 = 5 \times 9 + 2 \times 9 = 45 + 18 = 63\).

a) \(7 \times 4 = (5 \times 4) + (2 \times 4) = 20 + 8 = 28\) b) \(7 \times 7 = (5 \times 7) + (2 \times 7) = 35 + 14 = 49\) c) \(7 \times 9 = (5 \times 9) + (2 \times 9) = 45 + 18 = 63\)

- Identify the factor that appears in both products. - Combine the other two factors before multiplying. - Pay attention to whether the expression uses addition or subtraction.

1. Part a: \((4 + 6) \times 9 = 10 \times 9 = 90\). 2. Part b: \((8 - 3) \times 7 = 5 \times 7 = 35\). 3. Part c: \(5 \times (8 + 2) = 5 \times 10 = 50\). 4. Part d: \((10 - 2) \times 6 = 8 \times 6 = 48\).

a) \(90\) b) \(35\) c) \(50\) d) \(48\)

- Which number is a factor of every number in a group? - Check the sets of multiplication facts one at a time. - The last number in each group is a square number. Which square fact makes it?

1. The numbers in part a are \(2 \times 7\), \(3 \times 7\), \(5 \times 7\), and \(7 \times 7\), so they are in the 7s facts. 2. The numbers in part b are \(2 \times 8\), \(3 \times 8\), \(5 \times 8\), and \(8 \times 8\), so they are in the 8s facts. 3. The numbers in part c are \(2 \times 9\), \(3 \times 9\), \(5 \times 9\), and \(9 \times 9\), so they are in the 9s facts.

a) 7s facts b) 8s facts c) 9s facts

- Recall the names of the numbers and result in a multiplication equation. - Evaluate both products before comparing them. - Use the related division fact to find the missing factor.

1. Part a: The numbers being multiplied are factors, and the result is the product. 2. Part b: \(8 \times 7 = 56\) and \(9 \times 6 = 54\). Therefore, \(8 \times 7\) is greater. 3. Part c: Find the missing factor with \(72 \div 8 = 9\). The other factor is \(9\).

a) factors; product b) \(8 \times 7\) is greater because \(56 > 54\). c) The other factor is \(9\).

- Multiply each neighboring pair to find the brick above. - Use multiplication facts through \(10 \times 10\).

1. The second row is \(2 \times 2 = 4\), \(2 \times 1 = 2\), and \(1 \times 3 = 3\). 2. The third row is \(4 \times 2 = 8\) and \(2 \times 3 = 6\). 3. The top brick is \(8 \times 6 = 48\).

Second row: \(4\), \(2\), \(3\) Third row: \(8\), \(6\) Top: \(48\)

- Array b) has exactly one more row. - Use a neighboring multiplication fact.

1. Array a) shows \(8 \times 6 = 48\) plants. 2. One additional bed adds \(6\) plants: \(48 + 6 = 54\). 3. The new situation is \(9 \times 6 = 54\).

\(8 \times 6 = 48\). One more group of \(6\) gives \(48 + 6 = 54\), so \(9 \times 6 = 54\).

- Count rows and columns starting with \(1\). - Multiply to find the total number of squares, then subtract the uncolored square.

1. The uncolored square is in row \(3\), column \(3\). 2. A \(6 \times 6\) array has \(36\) squares. Since one is uncolored, \(36 - 1 = 35\) squares are blue.

The uncolored square is in row \(3\), column \(3\). There are \(35\) blue squares.

- How can you split \(7\) into two anchor factors? - Evaluate the two easier multiplication facts first. - What should you do with the two partial products?

1. Decompose \(7\) as \(5 + 2\). 2. Then \(7 \times 8 = 5 \times 8 + 2 \times 8\). 3. Evaluate the anchor facts: \(5 \times 8 = 40\) and \(2 \times 8 = 16\). 4. Add the partial products: \(40 + 16 = 56\).

\(7 \times 8 = 5 \times 8 + 2 \times 8 = 40 + 16 = 56\).

- First list every multiplication fact you know with a product of \(24\). - Then list the multiplication facts with a product of \(25\). - Count the equations in each list and compare the totals.

1. The multiplication equations with a product of \(24\) are \(3 \times 8 = 24\), \(8 \times 3 = 24\), \(4 \times 6 = 24\), and \(6 \times 4 = 24\). There are \(4\) equations. 2. The only multiplication equation with a product of \(25\) is \(5 \times 5 = 25\). There is \(1\) equation. 3. Since \(4 > 1\), Max is correct.

Max is correct. Product of \(24\): \(3 \times 8\), \(8 \times 3\), \(4 \times 6\), \(6 \times 4\) Product of \(25\): \(5 \times 5\)

- What happens to the product when the number of equal groups doubles? - How are the factors \(2\), \(4\), and \(8\) related? - Can repeated addition of a known product help you find the next one?

1. For part a, \(2 \times 8 = 16\). Doubling \(16\) gives \(32\), so \(4 \times 8 = 32\). Doubling \(32\) gives \(64\), so \(8 \times 8 = 64\). 2. For part b, \(3 \times 6 = 18\). Since \(6\) is twice \(3\), double \(18\) to get \(36\). Therefore, \(6 \times 6 = 36\).

a) \(2 \times 8 = 16\); \(4 \times 8 = 32\); \(8 \times 8 = 64\) b) \(6 \times 6 = 36\). Double the product \(18\) from \(3 \times 6\).

- How are \(5\) and \(10\) related? - What happens to a product when one factor is halved and the other stays the same? - Which easy multiplication fact is close to \(6 \times 7\)?

1. Since \(5\) is half of \(10\), halve the product \(40\): \(40 \div 2 = 20\). Therefore, \(5 \times 4 = 20\). 2. Since \(4\) is half of \(8\), halve the product \(24\): \(24 \div 2 = 12\). Therefore, \(4 \times 3 = 12\). 3. One strategy is to use \(5 \times 7 = 35\). One more group of \(7\) gives \(35 + 7 = 42\), so \(6 \times 7 = 42\).

a) \(20\); halve \(40\). b) \(12\); halve \(24\). c) \(42\); for example, \(5 \times 7 = 35\) and \(35 + 7 = 42\).

- How does the product change when the first factor increases by \(1\)? - How does it change when the first factor decreases by \(1\)? - Identify which factor stays the same.

1. The square products are \(6 \times 6 = 36\), \(9 \times 9 = 81\), and \(5 \times 5 = 25\). 2. For part a, add one group of \(6\): \(36 + 6 = 42\). 3. For part b, subtract one group of \(9\): \(81 - 9 = 72\). 4. For part c, subtract one group of \(5\): \(25 - 5 = 20\).

a) \(6 \times 6 = 36\), so \(7 \times 6 = 42\) b) \(9 \times 9 = 81\), so \(8 \times 9 = 72\) c) \(5 \times 5 = 25\), so \(4 \times 5 = 20\)

- Determine how \(8\) is being broken apart in each line. - In part b, compare \(8\) with \(10\). - In part c, split \(8\) into two equal addends.

1. Part a: Use \(8 = 5 + 3\). Then \(7 \times 8 = 7 \times 5 + 7 \times 3 = 35 + 21 = 56\). 2. Part b: Use \(8 = 10 - 2\). Then \(7 \times 8 = 7 \times 10 - 7 \times 2 = 70 - 14 = 56\). 3. Part c: Use \(8 = 4 + 4\). Then \(7 \times 8 = 7 \times 4 + 7 \times 4 = 28 + 28 = 56\).

a) \(7 \times 8 = 7 \times 5 + 7 \times 3 = 35 + 21 = 56\) b) \(7 \times 8 = 7 \times 10 - 7 \times 2 = 70 - 14 = 56\) c) \(7 \times 8 = 7 \times 4 + 7 \times 4 = 28 + 28 = 56\)

- Pay attention to whether the groups of \(8\) are being added or subtracted. - Combine or subtract the numbers of groups before multiplying by \(8\). - In part d, find the number that must be added to \(2\) to make \(10\).

1. Part a: \(7 - 2 = 5\), so \((8 \times 7) - (8 \times 2) = 8 \times 5 = 40\). 2. Part b: \(3 + 4 = 7\), so \((8 \times 3) + (8 \times 4) = 8 \times 7 = 56\). 3. Part c: \(9 - 5 = 4\), so \((8 \times 9) - (8 \times 5) = 8 \times 4 = 32\). 4. Part d: \(2 + 8 = 10\), so the missing factor is \(8\), and \(8 \times 10 = 80\).

a) \(8 \times 5 = 40\) b) \(8 \times 7 = 56\) c) \(8 \times 4 = 32\) d) The missing factor is \(8\), and the result is \(80\).

- The two parts of the second factor must combine to make the original second factor. - Evaluate each smaller multiplication fact. - Add the two partial products.

1. Part a: Since \(7 = 5 + 2\), \(6 \times 7 = 6 \times 5 + 6 \times 2 = 30 + 12 = 42\). 2. Part b: Since \(9 = 5 + 4\), \(8 \times 9 = 8 \times 5 + 8 \times 4 = 40 + 32 = 72\). 3. Part c: Since \(8 = 4 + 4\), \(7 \times 8 = 7 \times 4 + 7 \times 4 = 28 + 28 = 56\).

a) \(6 \times 7 = (6 \times 5) + (6 \times 2) = 30 + 12 = 42\) b) \(8 \times 9 = (8 \times 5) + (8 \times 4) = 40 + 32 = 72\) c) \(7 \times 8 = (7 \times 4) + (7 \times 4) = 28 + 28 = 56\)

- Identify the factor that appears in both partial products. - Add the other factors, or find the missing addend needed to reach the factor on the right. - Substitute your number into the equation to check it.

1. Part a: The common factor is \(8\), and \(3 + 4 = 7\). The missing number is \(7\). 2. Part b: The two numbers of groups must add to \(9\). Since \(7 + 2 = 9\), the missing number is \(2\). 3. Part c: The common factor is \(9\), and \(6 + 4 = 10\). The missing number is \(10\). 4. Part d: The two numbers of groups must add to \(8\). Since \(5 + 3 = 8\), the missing number is \(5\).

a) \(7\) b) \(2\) c) \(10\) d) \(5\)

- How are a multiplication fact and its commutative fact related? - What should be added when one factor increases by \(1\)? - What should be subtracted when one factor decreases by \(1\)?

1. Switch the factors to get \(4 \times 8 = 32\). 2. For \(9 \times 4\), add one more group of \(4\): \(32 + 4 = 36\). 3. For \(8 \times 3\), subtract one group of \(8\): \(32 - 8 = 24\).

a) \(4 \times 8 = 32\) b) \(9 \times 4 = 32 + 4 = 36\) c) \(8 \times 3 = 32 - 8 = 24\)

- For the first method, write \(6\) as a sum of two positive whole numbers. - For the second method, write \(8\) as a sum of two positive whole numbers. - Check that both methods give the same product.

1. For Method 1, one possible decomposition is \(6 = 3 + 3\). Then \(6 \times 8 = 3 \times 8 + 3 \times 8 = 24 + 24 = 48\). 2. For Method 2, one possible decomposition is \(8 = 4 + 4\). Then \(6 \times 8 = 6 \times 4 + 6 \times 4 = 24 + 24 = 48\). 3. Other correct decompositions of the requested factor are also possible.

Method 1: \((3 \times 8) + (3 \times 8) = 24 + 24 = 48\) Method 2: \((6 \times 4) + (6 \times 4) = 24 + 24 = 48\) Other valid decompositions are possible.

- Determine how many groups of \(4\) there must be altogether. - The two numbers multiplied by \(4\) must add to \(9\). - For part c, choose a decomposition of \(9\) that is different from parts a and b.

1. The two numbers of groups must add to \(9\). 2. Part a: Since \(5 + 4 = 9\), the missing number is \(4\). 3. Part b: Since \(7 + 2 = 9\), the missing number is \(7\). 4. Part c: One different decomposition is \(9 = 1 + 8\), so \((4 \times 1) + (4 \times 8) = 4 \times 9\). Other valid decompositions are possible.

a) \(4\) b) \(7\) c) For example, \((4 \times 1) + (4 \times 8) = 4 \times 9\).

- What changed in part a? - How many groups of \(8\) are in the given fact and in part b? - Can you add one more group instead of recalculating the whole product?

1. Part a is the commutative fact, so \(8 \times 6 = 48\). 2. Part b has one more group of \(8\) than \(6 \times 8\), so \(48 + 8 = 56\). Therefore, \(7 \times 8 = 56\).

a) \(8 \times 6 = 48\) b) \(7 \times 8 = 56\), because \(48 + 8 = 56\).

- Start with the common related fact \(7 \times 6 = 42\). - Determine what must be added to \(42\) for each bus.

1. Use \(7 \times 6 = 42\) as the common related fact. 2. Bus A has one additional row of \(6\) seats: \(8 \times 6 = 42 + 6 = 48\). 3. Bus B has one additional seat in each of \(7\) rows: \(7 \times 7 = 42 + 7 = 49\). 4. Since \(49 > 48\), Bus B has one more seat.

Bus B has \(49\) seats, which is one more than Bus A’s \(48\) seats.

- Use a multiplication fact that is closely related to both products. - Compare what must be added to that common fact.

1. Use \(8 \times 6 = 48\) as a common related fact. 2. Then \(8 \times 7 = 8 \times 6 + 8 = 48 + 8 = 56\). 3. Also, \(9 \times 6 = 8 \times 6 + 6 = 48 + 6 = 54\). 4. Therefore, \(8 \times 7\) is greater by \(56 - 54 = 2\).

\(8 \times 7 = 56\) and \(9 \times 6 = 54\), so \(8 \times 7\) is greater by \(2\).