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Properties of Addition and Multiplication: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Regroup a sum

Rewrite $18 + 6 + 14$ with parentheses so that $6$ and $14$ are added first, keeping the numbers in their original order, and check that the total is unchanged.
1. Hint 1
  A sum keeps its value when you change which pair is combined first without moving the terms.
2. Hint 2
  Parentheses mark the pair that is combined first, and the term outside them is added to that pair's sum.
Answer

$18 + (6 + 14)$ ; both groupings total $38$ .

Full solution

Place the parentheses around the last two terms.
The expression becomes $18 + (6 + 14)$ .
This changes the grouping without changing the order, as the associative property permits.

Check that the grouped expression still has the original total.
$\begin{aligned} 6 + 14 &= 20 \\ 18 + 20 &= 38 \end{aligned}$
Adding the first pair instead gives $24 + 14$ , which is also $38$ .

Answer

$18 + (6 + 14)$ ; both groupings total $38$ .

Key idea

The associative property allows a sum to be grouped differently while its terms stay in the same order.
Problem 1: Mark complete
Problem 2 Tickets at an empty hall

A hall gives each visitor $6$ tickets, but no visitors arrive. How many tickets are handed out?
1. Hint 1
  Multiplication can count equal groups.
2. Hint 2
  Let the number of visitors be the number of groups and the tickets per visitor be the group size.
Answer

$0$ tickets.

Full solution

There are zero visitors, so there are zero groups of six tickets to hand out.
$0 \times 6 = 0$
No group is handed out, so the total is $0$ tickets.

Answer

$0$ tickets.

Key idea

Zero groups of a quantity give a total of zero, which is one meaning of multiplication by zero.
Problem 2: Mark complete
Problem 3 Keep the number unchanged

Which whole number belongs in the box?
$(73 + 0) \times \square = 73$
1. Hint 1
  Look for operations that leave a number unchanged.
2. Hint 2
  Adding nothing preserves the first number, so the missing factor must preserve that same number when multiplying.
Answer

$1$ .

Full solution

Zero is the additive identity, so adding it leaves $73$ unchanged.
$73 + 0 = 73$
One is the multiplicative identity, so a factor of $1$ keeps $73$ unchanged.
$73 \times 1 = 73$
No other whole number works: a factor of $0$ gives $0$ , and a factor of $2$ or more gives at least $146$ .
So the missing factor is $1$ , and the two displayed equations check the completed expression.

Answer

$1$ .

Key idea

The additive identity is zero and the multiplicative identity is one.
Problem 3: Mark complete
Problem 4 Combine the donations

Six groups donate $142$ , $65$ , $58$ , $19$ , $35$ and $81$ books. Find the total by reordering and grouping the counts into helpful pairs, and show your pairs.
1. Hint 1
  A sum keeps its value when its terms are reordered and regrouped.
2. Hint 2
  Look for pairs whose ones digits complete a ten.
3. Hint 3
  Pair $142$ with $58$ , then find pairs among the four remaining counts.
Answer

$400$ books; pairs: $142 + 58$ , $65 + 35$ , and $19 + 81$ .

Full solution

Reorder the counts to bring helpful pairs together, then group each pair.
The commutative property permits the reordering and the associative property permits the grouping.
$\begin{aligned} 142 + 58 &= 200 \\ 65 + 35 &= 100 \\ 19 + 81 &= 100 \end{aligned}$

Add the three subtotals.
$200 + 100 + 100 = 400$
Each donation appears in exactly one pair, so the total includes all $400$ books without adding or losing any.

Answer

$400$ books; pairs: $142 + 58$ , $65 + 35$ , and $19 + 81$ .

Key idea

Reordering and regrouping a long sum can turn several awkward additions into round subtotals.
Problem 4: Mark complete
Problem 5 Count the narration time

A museum has $5$ audio devices. Each device plays a narration $7$ times per day, and each play lasts $2$ minutes. This continues for $3$ days. Find the total minutes of narration by grouping the factors into convenient pairs, and explain why your grouping must give the same total as multiplying the factors in the order the facts are given.
1. Hint 1
  Count the narration minutes by multiplying the number of devices, minutes per play, plays per day and days.
2. Hint 2
  The factors may be reordered and regrouped without changing the total.
3. Hint 3
  Pair the factors $5$ and $2$ , then pair the remaining factors.
Answer

$210$ minutes, for example from $(5 \times 2) \times (7 \times 3)$ .

Full solution

The count is $5 \times 7 \times 2 \times 3$ : devices, plays per day, minutes per play and days.
Reorder the factors to bring $5$ and $2$ together, then group the factors into two convenient pairs.
$\begin{aligned} 5 \times 2 &= 10 \\ 7 \times 3 &= 21 \\ 10 \times 21 &= 210 \end{aligned}$
The grouped product is $(5 \times 2) \times (7 \times 3)$ , giving $210$ minutes.

Reordering and regrouping change only which partial count is made first.
Every play of every device on every day is still counted exactly once, so no grouping can reach a different total.
Check by counting one device over all three days: $2 \times 7$ gives $14$ minutes per day, and $14 \times 3$ gives $42$ minutes.
All $5$ devices then play $5 \times 42$ , or $210$ minutes.

Answer

$210$ minutes, for example from $(5 \times 2) \times (7 \times 3)$ .

Key idea

Regrouping the factors of a repeated count changes the subtotals used without changing what is counted.
Problem 5: Mark complete
Problem 6 Check the art order

Each art kit contains $1$ brush, $0$ paint tubes and $6$ paper sheets. A studio orders $15$ kits and $9$ separate paint tubes. How many brushes, paint tubes and paper sheets will the studio receive?
1. Hint 1
  Count each kind of item separately, using the number of kits and the contents of one kit.
2. Hint 2
  Multiplying the number of kits by one preserves that number, while multiplying it by zero gives no items of that kind.
3. Hint 3
  After counting the kit contents, include the paint tubes ordered separately.
Answer

$15$ brushes, $9$ paint tubes and $90$ paper sheets.

Full solution

There is one brush per kit, so the multiplicative identity gives $15$ brushes.
$15 \times 1 = 15$
Each kit has six paper sheets, giving $90$ paper sheets.
$15 \times 6 = 90$

Fifteen kits with no paint tubes in any kit contribute no paint tubes.
The nine separate paint tubes must still be included.
$\begin{aligned} 15 \times 0 &= 0 \\ 0 + 9 &= 9 \end{aligned}$
The zero property counts the paint tubes in the kits, and the additive identity leaves the nine separate paint tubes unchanged.
Every item in the order has been counted in its own category.

Answer

$15$ brushes, $9$ paint tubes and $90$ paper sheets.

Key idea

Multiplication by one preserves a count, while empty groups contribute zero before other items are added.
Problem 6: Mark complete
Problem 7 Repair a product

A game score is the product $16 \times 0 \times 1 \times 7$ . Replace exactly one of its four factors with a whole number so that the score becomes $112$ . Which factor should be replaced, and what should replace it?
1. Hint 1
  A zero factor makes the entire product zero.
2. Hint 2
  If you replace a factor other than zero, consider which factor will still force the score to zero.
3. Hint 3
  Find the product of the three factors that stay, then choose a replacement that preserves it.
Answer

Replace the factor $0$ with $1$ .

Full solution

Changing $16$ , $1$ or $7$ leaves the zero factor in place, so each of those choices would still give a score of $0$ .
The factor that must change is $0$ .

The three unchanged factors already give the desired score.
$\begin{aligned} 16 \times 1 &= 16 \\ 16 \times 7 &= 112 \end{aligned}$
The replacement must keep this product at $112$ .
A replacement of $1$ does, while $0$ gives $0$ and any whole number from $2$ up gives at least $224$ .
Check the repaired product.
$16 \times 1 \times 1 \times 7 = 112$

Answer

Replace the factor $0$ with $1$ .

Key idea

A zero factor makes a product zero, while a factor of one preserves the product of the other factors.
Problem 7: Mark complete
Problem 8 Check Nora's label

Nora rewrites $(11 + 9) + 26$ as $26 + (11 + 9)$ . She calls this the associative property since the parentheses appear in a different place on the page. Is her label correct? Name the property used and explain your decision.
1. Hint 1
  Distinguish moving a group from changing which terms belong to it.
2. Hint 2
  Look at the pair inside the parentheses before and after the rewrite.
3. Hint 3
  Treat the whole parenthesized sum as one number and compare its position with that of $26$ .
Answer

No; the commutative property of addition.

Full solution

The terms $11$ and $9$ remain grouped together in both expressions, so the grouping has not changed.
The two things being added are the sum $(11 + 9)$ and the number $26$ .
Those two things trade places, which is the commutative property of addition.

The grouped sum is $20$ .
Both orders count that same $20$ together with the same $26$ , with nothing added or removed.
$\begin{aligned} 11 + 9 &= 20 \\ 20 + 26 &= 46 \\ 26 + 20 &= 46 \end{aligned}$
The equal totals check the rewrite, while the unchanged pair identifies which property was used.

Answer

No; the commutative property of addition.

Key idea

Moving a whole group past another term changes their order, even when the parentheses move on the page.
Problem 8: Mark complete
Problem 9 Test the subtraction claims

A student says that subtraction is commutative and associative, so its numbers can be swapped or regrouped freely.

Give a numerical counterexample to each conclusion, and explain why your examples settle the claims. Use positive whole numbers. You may describe a reversed subtraction as below zero without calculating a negative answer. For your grouping example, choose numbers that keep every subtraction above zero.
1. Hint 1
  A property must hold for every allowed choice, so one failed example can disprove it.
2. Hint 2
  To test order, choose two different positive numbers and compare both orders.
3. Hint 3
  To test grouping, keep three numbers in the same order and compare the two choices of first pair.
4. Hint 4
  A large first number and two smaller numbers can keep both grouped calculations above zero.
Answer

One choice: $9 - 2 = 7$ , but $2 - 9 < 0$ ; $(18 - 7) - 3 = 8$ , but $18 - (7 - 3) = 14$ .

Full solution

For order, choose $9$ and $2$ .
$9 - 2 = 7$
Reversing them asks for $9$ to be taken from $2$ .
That goes below zero, so it does not give $7$ .
Thus $9 - 2$ and $2 - 9$ disagree, disproving commutativity.

For grouping, keep $18$ , $7$ and $3$ in that order.
First combine the first pair.
$\begin{aligned} 18 - 7 &= 11 \\ 11 - 3 &= 8 \end{aligned}$
Then combine the last pair instead.
$\begin{aligned} 7 - 3 &= 4 \\ 18 - 4 &= 14 \end{aligned}$
Every subtraction stays above zero, but the final results differ, disproving associativity.

Each claimed property would have to hold for every choice of numbers.
One counterexample to each claim is enough to show that the claimed rule fails.

Answer

One choice: $9 - 2 = 7$ , but $2 - 9 < 0$ ; $(18 - 7) - 3 = 8$ , but $18 - (7 - 3) = 14$ .

Key idea

A single counterexample disproves a claimed property of subtraction.
Problem 9: Mark complete
Problem 10 Correct the division instructions

An instruction card says that division allows both swapping the two numbers and regrouping three numbers freely. Show that each instruction can change the result by giving one numerical counterexample for swapping and one for regrouping, and explain your comparisons.

Use positive whole numbers. For a reversed division, you may describe the answer as less than one without writing a fraction. In your regrouping example, every division must have a whole number result.
1. Hint 1
  An instruction that is meant to work freely must preserve the result for every allowed choice.
2. Hint 2
  For swapping, choose a larger number that contains several equal groups of a smaller number.
3. Hint 3
  For regrouping, compare dividing twice with dividing once by the result of the other division.
4. Hint 4
  Choose a large first number and two small divisors, with the middle number divisible by the last number.
Answer

One choice: $24 \div 8 = 3$ , but $8 \div 24 < 1$ ; $(72 \div 6) \div 3 = 4$ , but $72 \div (6 \div 3) = 36$ .

Full solution

For swapping, choose $24$ and $8$ .
$24 \div 8 = 3$
In the reverse order, $8$ contains less than one whole group of $24$ , so $8 \div 24$ is less than $1$ and cannot equal $3$ .
Swapping changes the result.

For regrouping, keep $72$ , $6$ and $3$ in that order.
Grouping the first pair gives these steps.
$\begin{aligned} 72 \div 6 &= 12 \\ 12 \div 3 &= 4 \end{aligned}$
Grouping the last pair instead gives these steps.
$\begin{aligned} 6 \div 3 &= 2 \\ 72 \div 2 &= 36 \end{aligned}$
The results are $4$ and $36$ .
Every divisor is nonzero and every division has a whole number result, yet regrouping changes the value.

The first counterexample shows division is not commutative, and the second shows it is not associative.
Each instruction therefore fails even though it may happen to work for some choices.

Answer

One choice: $24 \div 8 = 3$ , but $8 \div 24 < 1$ ; $(72 \div 6) \div 3 = 4$ , but $72 \div (6 \div 3) = 36$ .

Key idea

Division depends on both the order of its numbers and the grouping of successive divisions.
Problem 10: Mark complete

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