Properties of Addition and Multiplication: Practice

Reorder so the round pair $4 \times 25 = 100$ multiplies first, then finish.

$4 \times 17 \times 25 = (4 \times 25) \times 17 = 100 \times 17 = 1700$

The commutative and associative properties guarantee this equals the original product.

Move the $5$ past the $47$ (commutative), then group it with the $2$ (associative).

$2 \times 47 \times 5 = (2 \times 5) \times 47 = 10 \times 47 = 470$

Evaluate each expression and compare.

$6 + 9 = 15, \quad 9 + 6 = 15, \quad (3 + 3) + 9 = 15, \quad 9 - 6 = 3$

The first two are the same sum with the order swapped, which is the commutative property. The third is not a reorder or a regroup: the $6$ has been split into $3 + 3$. Since $3 + 3$ is $6$ again, that expression still adds to $15$. The odd one out is $9 - 6$, which subtracts instead of adding and comes to $3$.

The last pair makes a round $40$, so combine that pair first.

$58 + 36 + 4 = 58 + (36 + 4) = 58 + 40 = 98$

Nothing had to change order here, so this step is the associative property on its own. Adding in the printed order gives $94 + 4$, the same $98$ by a harder route.

Moving the $1$ to sit next to the $19$ changes only the order of the terms.

$19 + 46 + 1 = 19 + 1 + 46$

Reordering is the commutative property. (Grouping $19 + 1$ afterward is the associative step.)

Subtraction is not commutative, so swapping the numbers changes the result.

$12 - 4 = 8, \qquad 4 - 12 \ne 8$

Reversing $12 - 4$ asks you to take $12$ from just $4$, which drops below zero and cannot give $8$. So that equation is false, while the other three are true.

Only $+$ and $\times$ let you choose the first pair freely.

$(20 + 6) + 4 = 30 = 20 + (6 + 4)$

Each of the others changes value with the grouping: $(20 - 6) - 4 = 10$ but $20 - (6 - 4) = 18$; $(36 \div 6) \div 3 = 2$ but $36 \div (6 \div 3) = 18$; and $(20 - 5) + 2 = 17$ but $20 - (5 + 2) = 13$.

Move the $2$ next to the $18$ and the $3$ next to the $27$ (commutative), then group each pair (associative).

$18 + 27 + 2 + 3 = (18 + 2) + (27 + 3) = 20 + 30 = 50$

Two round tens are far easier to add than the four numbers taken in the printed order.

One is the multiplicative identity, so the product is just the other factor.

$1 \times 64 = 64$

Group the round pair $250 \times 4 = 1000$ first.

$250 \times 4 \times 3 = (250 \times 4) \times 3 = 1000 \times 3 = 3000$

A counterexample needs two unequal results from swapping the order.

$10 \div 2 = 5, \qquad 2 \div 10 \ne 5$

Reversing $10 \div 2$ asks how many $10$s fit in $2$, which is less than one whole, so it cannot be $5$. Since the two results differ, division is not commutative. The other options each involve a single quotient and prove nothing about order.

Move the $5$ past the $38$ (commutative), then group it with the $2$ (associative).

$2 \times 38 \times 5 = 2 \times 5 \times 38 = 10 \times 38 = 380$

Every pairing reaches $380$, which is what the two properties guarantee. This one turns the work into multiplying by $10$.