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Division fact families

Learning goals

  • Connects related addition/subtraction or multiplication/division facts.
  • Generates complete fact families from given numbers.
  • Uses inverse operations to find missing values.

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Division fact families 1
A fact family includes \(3\times4=12\) and \(12\div3=4\). Write the other division equation in this family. Check it with multiplication.

Hints

- Use the same three numbers as in the two given facts. - Think about another way to share the same total into equal groups.

Solution

1. The other division fact divides the same total by the other factor: \(12\div4=3\). 2. Check by multiplying: \(4\times3=12\).

Answer

\(12\div4=3\); check: \(4\times3=12\).
Division fact families 2
For \(4\times3=12\), write the multiplication fact with the factors switched and one matching division fact.

Hints

- How else could you describe the same \(12\) objects in equal groups? - If the total is already known, what might division find about those groups?

Solution

1. Switching the factors gives \(3\times4=12\). 2. Undoing four groups of three gives \(12\div4=3\).

Answer

\(3\times4=12\) and \(12\div4=3\).
Division fact families 3
Use each set of three numbers to write a fact family with two multiplication equations and two division equations. a) \(7, 9, 63\) b) \(4, 8, 32\)

Hints

- A fact family uses the same three numbers in every equation. - The greatest number is the product in the multiplication equations. - Each division equation starts with the greatest number. - Switch the two factors to write the second multiplication equation.

Solution

1. For \(7, 9, 63\), the equations are \(7 \times 9 = 63\), \(9 \times 7 = 63\), \(63 \div 9 = 7\), and \(63 \div 7 = 9\). 2. For \(4, 8, 32\), the equations are \(4 \times 8 = 32\), \(8 \times 4 = 32\), \(32 \div 8 = 4\), and \(32 \div 4 = 8\).

Answer

a) \(7 \times 9 = 63\), \(9 \times 7 = 63\), \(63 \div 9 = 7\), \(63 \div 7 = 9\) b) \(4 \times 8 = 32\), \(8 \times 4 = 32\), \(32 \div 8 = 4\), \(32 \div 4 = 8\)
Division fact families 4
Use the array to write the complete multiplication-and-division fact family for the three numbers shown by its rows, columns, and total.
Figure for problem 550354

Hints

- Read the number of rows and columns from the array. - The same three numbers appear in all four equations. - Two equations multiply to the total, and two divide the total.

Solution

1. The array has \(4\) rows and \(6\) columns, so it contains \(24\) squares. 2. The multiplication facts are \(4 \times 6 = 24\) and \(6 \times 4 = 24\). 3. The related division facts are \(24 \div 4 = 6\) and \(24 \div 6 = 4\).

Answer

\(4 \times 6 = 24\) \(6 \times 4 = 24\) \(24 \div 4 = 6\) \(24 \div 6 = 4\)
Division fact families 5
Complete the fact family for \(5,8,40\). Write two multiplication equations and two division equations. How do you know none is missing?

Hints

- Begin with the two ways to order the factors. - Make two division facts using the product as the number being divided.

Solution

1. Multiplication: \(5\times8=40\) and \(8\times5=40\). 2. Division: \(40\div5=8\) and \(40\div8=5\). 3. Both factor orders and both ways to undo the multiplication have been used.

Answer

\(5\times8=40\), \(8\times5=40\), \(40\div5=8\), and \(40\div8=5\). Both factor orders and both division facts are included.
Division fact families 6
The numbers in a fact family are \(6,6,36\). Does switching the two factors give a new multiplication equation? Write all the different multiplication and division equations you can make from this family.

Hints

- Try changing the order of the factors. - Decide whether that makes different numbers or the very same equation.

Solution

1. Switching the equal factors does not change the written equation \(6\times6=36\). 2. The two division equations are also identical: \(36\div6=6\). 3. There are only two different equations.

Answer

No. The two different equations are \(6\times6=36\) and \(36\div6=6\).
Division fact families 7
Complete each set of three numbers so they make a multiplication-and-division fact family. Then write all four equations. a) \(6,8,\square\) b) \(\square,7,63\) c) \(36,\square,4\) d) \(4,\square,32\)

Hints

- In a fact family, two numbers are factors and the third is their product. - Use multiplication or division to find the missing number. - Then write two multiplication equations and two division equations with the same three numbers.

Solution

1. a) The missing number is \(48\): \(6 \times 8=48\), \(8 \times 6=48\), \(48 \div 6=8\), \(48 \div 8=6\). 2. b) The missing number is \(9\): \(9 \times 7=63\), \(7 \times 9=63\), \(63 \div 9=7\), \(63 \div 7=9\). 3. c) The missing number is \(9\): \(9 \times 4=36\), \(4 \times 9=36\), \(36 \div 9=4\), \(36 \div 4=9\). 4. d) The missing number is \(8\): \(4 \times 8=32\), \(8 \times 4=32\), \(32 \div 4=8\), \(32 \div 8=4\).

Answer

a) \(48\): \(6 \times 8=48\), \(8 \times 6=48\), \(48 \div 6=8\), \(48 \div 8=6\) b) \(9\): \(9 \times 7=63\), \(7 \times 9=63\), \(63 \div 9=7\), \(63 \div 7=9\) c) \(9\): \(9 \times 4=36\), \(4 \times 9=36\), \(36 \div 9=4\), \(36 \div 4=9\) d) \(8\): \(4 \times 8=32\), \(8 \times 4=32\), \(32 \div 4=8\), \(32 \div 8=4\)
Division fact families 8
Find the missing number so each set makes a multiplication-and-division fact family. Then write all four equations. a) \(5,45,\square\) b) \(\square,4,28\)

Hints

- In each part, the greatest number is the product. - Which number multiplied by \(5\) gives \(45\)? - Which number multiplied by \(4\) gives \(28\)? - Use the two multiplication equations and the two related division equations.

Solution

1. For part a, \(45 \div 5 = 9\), so the missing number is \(9\). The fact family is \(5 \times 9 = 45\), \(9 \times 5 = 45\), \(45 \div 9 = 5\), and \(45 \div 5 = 9\). 2. For part b, \(28 \div 4 = 7\), so the missing number is \(7\). The fact family is \(7 \times 4 = 28\), \(4 \times 7 = 28\), \(28 \div 4 = 7\), and \(28 \div 7 = 4\).

Answer

a) Missing number: \(9\) \(5 \times 9 = 45\), \(9 \times 5 = 45\), \(45 \div 9 = 5\), \(45 \div 5 = 9\) b) Missing number: \(7\) \(7 \times 4 = 28\), \(4 \times 7 = 28\), \(28 \div 4 = 7\), \(28 \div 7 = 4\)
Division fact families 9
The numbers \(6\), \(8\), and \(48\) make a multiplication-and-division fact family. Write all four equations in the family. Then explain why each division equation must begin with \(48\).

Hints

- Start with the two multiplication equations made from \(6\), \(8\), and \(48\). - Each related division equation starts with the product, \(48\). - Explain why dividing \(48\) by one factor gives the other factor.

Solution

1. The multiplication facts are \(6\times8=48\) and \(8\times6=48\). 2. The related division facts are \(48\div6=8\) and \(48\div8=6\). 3. Each division equation begins with \(48\) because \(48\) is the product. Dividing it by one factor gives the other factor.

Answer

\(6 \times 8=48\) \(8 \times 6=48\) \(48 \div 6=8\) \(48 \div 8=6\) Both division equations begin with \(48\) because \(48\) is the whole product. Dividing it by one factor gives the other factor.
Division fact families 10
Elena writes this fact family: \(8\times7=56\) \(7\times8=56\) \(56\div7=8\) \(8\div56=7\) One equation is wrong. Find it and correct it. Then explain why both division equations in this fact family must start with \(56\).

Hints

- Check that every equation uses the same three numbers correctly. - In the multiplication equations, \(56\) is the whole product. - In the related division equations, start with that whole product.

Solution

1. The incorrect equation is \(8 \div 56 = 7\). 2. The correct fourth equation is \(56 \div 8 = 7\). 3. In both division facts, the product \(56\) is the number being divided because division starts with the whole product and separates it using one factor.

Answer

The incorrect equation is \(8 \div 56 = 7\). It should be \(56 \div 8 = 7\). The number being divided is \(56\) in both division facts.
Division fact families 11
The same two numbers can sometimes be used in two different fact families. Example with \(10\) and \(5\): - If both are factors, \(10\times5=50\), so the third number is \(50\). - If \(10\) is the product, \(2\times5=10\), so the third number is \(2\). For each pair below, find both possible third numbers. For each one, write one multiplication equation and one matching division equation. a) \(8\) and \(4\) b) \(9\) and \(3\) c) \(10\) and \(2\)

Hints

- First use the two given numbers as factors. What product completes that fact family? - Then treat the larger given number as the product. What missing factor completes that fact family? - Check each third number with one multiplication equation and one related division equation.

Solution

1. For \(8\) and \(4\), one third number is \(32\): \(8 \times 4 = 32\) and \(32 \div 4 = 8\). Another third number is \(2\): \(2 \times 4 = 8\) and \(8 \div 4 = 2\). 2. For \(9\) and \(3\), one third number is \(27\): \(9 \times 3 = 27\) and \(27 \div 3 = 9\). Another third number is \(3\): \(3 \times 3 = 9\) and \(9 \div 3 = 3\). 3. For \(10\) and \(2\), one third number is \(20\): \(10 \times 2 = 20\) and \(20 \div 2 = 10\). Another third number is \(5\): \(5 \times 2 = 10\) and \(10 \div 2 = 5\).

Answer

a) \(32\): \(8 \times 4 = 32\), \(32 \div 4 = 8\); \(2\): \(2 \times 4 = 8\), \(8 \div 4 = 2\) b) \(27\): \(9 \times 3 = 27\), \(27 \div 3 = 9\); \(3\): \(3 \times 3 = 9\), \(9 \div 3 = 3\) c) \(20\): \(10 \times 2 = 20\), \(20 \div 2 = 10\); \(5\): \(5 \times 2 = 10\), \(10 \div 2 = 5\)

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