M ultiplication is commutative, just like addition. This means you can multiply numbers in any order, so you can rearrange a calculation to make it easier. Division sometimes leaves part of the whole left over. You will learn how to show this in your calculation.
1. Multiply each set of three numbers in any order to simplify the calculation and find the product.
a. $5 \times 4 \times 3$
b. $6 \times 5 \times 3$
c. $6 \times 4 \times 2$
d. $8 \times 3 \times 2$
a) $5 \times 4 \times 3 = 20 \times 3 = 60$
b) $6 \times 5 \times 3 = 30 \times 3 = 90$
c) $6 \times 4 \times 2 = 24 \times 2 = 48$
d) $8 \times 3 \times 2 = 24 \times 2 = 48$
2. Choose one of the sets of numbers from question $1$. Simplify the calculation in a different way to check that the product is the same.
Example using part d:
$8 \times 3 \times 2 = 8 \times 6 = 48$
This matches the previous result, so the product is still $48$.
3. Simplify each multiplication to help you find the product.
a. $13 \times 4 =$
b. $18 \times 5 =$
c. $12 \times 2 =$
d. $15 \times 3 =$
a) $13 \times 4 = 52$
b) $18 \times 5 = 90$
c) $12 \times 2 = 24$
d) $15 \times 3 = 45$
4. Complete each division. Make sure that you include any remainders.
a. $28 \div 4 =$
b. $25 \div 2 =$
c. $53 \div 5 =$
d. $32 \div 4 =$
e. $10 \div 4 =$
f. $46 \div 3 =$
a) $28 \div 4 = 7$
b) $25 \div 2 = 12\text{ r }1$
c) $53 \div 5 = 10\text{ r }3$
d) $32 \div 4 = 8$
e) $10 \div 4 = 2\text{ r }2$
f) $46 \div 3 = 15\text{ r }1$
5. Four children share a bag of $30$ sweets between them equally. How many sweets does each child get? Write the division calculation that shows the result.
$30 \div 4 = 7\text{ r }2$
Each child gets $7$ sweets, with $2$ left over.
Challenge: Find a number between $10$ and $20$ that always leaves a remainder when you divide by $2$, $3$, $4$, and $5$.
Method:
Follow-up Questions: