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Multiplying and Dividing Integers: 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 Three score entries

A score starts at $0$ . Each of three entries changes the score by $-13$ points. What is the score after all three entries?
1. Hint 1
  Equal changes can be combined by adding the same number repeatedly.
2. Hint 2
  Each entry moves the score to the left of its previous position; combine three moves of 13 points.
Answer

$-39$ points.

Full solution

The three entries contribute three equal negative changes.
$(-13)+(-13)+(-13)=-39$

This is three groups of $-13$ , so it also gives
$3\times(-13)=-39.$
Each change moves left, and the score finishes at $-39$ points.

Answer

$-39$ points.

Key idea

A positive number of equal negative changes gives repeated addition of the negative number.
Problem 1: Mark complete
Problem 2 Maya's check

Maya writes $154\div(-14)=11$ and checks it with $11\times14=154$ . What is wrong with her check? Give the correct quotient and a check that works.
1. Hint 1
  A division check must reproduce the dividend using the divisor exactly as written.
2. Hint 2
  Multiply Maya's quotient by $-14$ and compare the result with $154$ .
3. Hint 3
  Find the size with $154\div14$ , then choose the sign that gives $154$ when multiplied by $-14$ .
Answer

Her check multiplies by $14$ instead of the divisor $-14$ . The correct quotient is $-11$ , checked by $(-11)\times(-14)=154$ .

Full solution

A quotient is the factor that gives the dividend when multiplied by the divisor.
Here the divisor is $-14$ , but Maya's check multiplies by $14$ .

Multiplying her quotient by the divisor as written does not give the dividend.
$11\times(-14)=-154$
So $11$ is not the quotient.

The size of the quotient is found by dividing the sizes.
$154\div14=11$

Only a negative factor times $-14$ gives the positive number $154$ , so the quotient is negative.
$154\div(-14)=-11$

Check with the divisor exactly as written.
$(-11)\times(-14)=154$

Answer

Her check multiplies by $14$ instead of the divisor $-14$ . The correct quotient is $-11$ , checked by $(-11)\times(-14)=154$ .

Key idea

A division check must multiply the quotient by the divisor, sign included.
Problem 2: Mark complete
Problem 3 A calculation to finish

Evaluate $(-3)\times2\times(-4)\times(-2)\times5$ .
1. Hint 1
  For a product with no zero factors, its size and its sign can be determined separately.
2. Hint 2
  Find the product of the sizes, and count all the negative factors before choosing the sign.
3. Hint 3
  The two positive factors can be grouped together to make a round number.
Answer

$-240$ .

Full solution

All five factors are nonzero.
Three are negative: $-3$ , $-4$ and $-2$ .
Two sign changes undo each other, leaving one, so the product is negative.

For its size, group the positive sizes $2$ and $5$ together, and group the other three sizes together.
$\begin{aligned}2\times5&=10\\3\times4&=12\\12\times2&=24\\24\times10&=240\end{aligned}$
The value of the original product is $-240$ .

Check by following the signed products in their original order.
$\begin{aligned}(-3)\times2&=-6\\(-6)\times(-4)&=24\\24\times(-2)&=-48\\(-48)\times5&=-240\end{aligned}$

Answer

$-240$ .

Key idea

For nonzero factors, the product of the sizes gives the size of the answer and the count of negative factors determines its sign.
Problem 3: Mark complete
Problem 4 The scorecard

A player has $-4$ points. Each correct answer earns 12 points, and each penalty removes 7 points. The player then gives three correct answers and receives five penalties. What is the new score?
1. Hint 1
  Find the total change from each kind of event before combining them with the starting score.
2. Hint 2
  The correct answers contribute $3\times12$ points, and the penalties contribute $5\times(-7)$ points.
3. Hint 3
  Add the two changes, then apply that total change to $-4$ .
Answer

$-3$ points.

Full solution

The correct answers add points, while the penalties contribute a negative change.
$\begin{aligned}3\times12&=36\\5\times(-7)&=-35\end{aligned}$

Combine the changes and add them to the starting score.
$\begin{aligned}36+(-35)&=1\\-4+1&=-3\end{aligned}$

The new score is $-3$ points.
The 36 points earned exceed the 35 points removed by 1 point, so a final score 1 point above the starting score is consistent.

Answer

$-3$ points.

Key idea

Multiply each repeated change by its count, then combine those signed totals with the starting amount.
Problem 4: Mark complete
Problem 5 The cooling cycle

A chamber starts at $11$ degrees Celsius. After four cooling cycles, its temperature is $-13$ degrees Celsius. Every cycle changes the temperature by the same integer number of degrees. What will the temperature be after two more cycles?
1. Hint 1
  The change over the first four cycles tells you the change in one cycle.
2. Hint 2
  Subtract the starting temperature from the temperature after four cycles, then divide that change by four.
3. Hint 3
  Apply two more equal changes to the reading of $-13$ degrees Celsius.
Answer

$-25$ degrees Celsius.

Full solution

Subtract the starting reading from the reading after four cycles.
$-13-11=-24$

The four equal changes total $-24$ degrees Celsius, so each cycle changes the temperature by $-6$ degrees Celsius.
$(-24)\div4=-6$

Two more cycles have a combined change of $-12$ degrees Celsius.
$\begin{aligned}2\times(-6)&=-12\\-13+(-12)&=-25\end{aligned}$

The final temperature is $-25$ degrees Celsius.
Check from the original reading over all six cycles.
$11+6\times(-6)=-25$

Answer

$-25$ degrees Celsius.

Key idea

An equal change per cycle is the total signed change divided by the number of cycles.
Problem 5: Mark complete
Problem 6 Above and below the bar

Find the value of $\dfrac{-18+6}{5-8}\times(-4)$ .
1. Hint 1
  The fraction bar groups everything above it and everything below it.
2. Hint 2
  Find the numerator and denominator separately, then check that the denominator is nonzero.
3. Hint 3
  Complete the division before multiplying its result by $-4$ .
Answer

$-16$ .

Full solution

First evaluate the two grouped expressions.
$\begin{aligned}-18+6&=-12\\5-8&=-3\end{aligned}$

The denominator is $-3$ , which is nonzero, so the division is defined.
The quotient is the number that gives $-12$ when multiplied by $-3$ .
$(-12)\div(-3)=4$

Multiply that result by the remaining factor.
$4\times(-4)=-16$

For a check, $4\times(-3)=-12$ , so the quotient used in the last step gives the original numerator.

Answer

$-16$ .

Key idea

Evaluate grouped expressions before dividing, and check that the divisor is nonzero.
Problem 6: Mark complete
Problem 7 The club account

A club account shows a balance of $-30$ dollars after six incorrect charges of 9 dollars each. All six charges are removed. The corrected balance is then shared equally among four members. How many dollars does each member receive?
1. Hint 1
  Removing a charge undoes a negative change to the balance.
2. Hint 2
  The six charges contributed $6\times(-9)$ dollars; subtract that signed total from the displayed balance.
3. Hint 3
  Divide the corrected balance equally among the four members.
Answer

6 dollars.

Full solution

Together the incorrect charges changed the balance by $-54$ dollars.
$6\times(-9)=-54$

Removing those charges means subtracting their negative total.
$\begin{aligned}-30-(-54)&=-30+54\\-30+54&=24\end{aligned}$

The corrected balance is 24 dollars.
Sharing it equally gives each member 6 dollars.
$24\div4=6$

Check both parts: the shares total 24 dollars since $4\times6=24$ , and applying the incorrect charges again gives
$24+(-54)=-30.$

Answer

6 dollars.

Key idea

Undo a collection of equal charges by subtracting their signed total before sharing the corrected balance.
Problem 7: Mark complete
Problem 8 A learner's question

A learner asks whether either sign could be used for $(-11)\times(-7)$ . Give its value and explain why the other sign would break an arithmetic rule.
1. Hint 1
  A calculation must keep the same value when an arithmetic property rewrites it.
2. Hint 2
  Pair $7$ with $-7$ , then multiply their sum by $-11$ .
3. Hint 3
  The two separate products must add to zero; one of them can be found by adding $-11$ seven times.
Answer

$77$ ; a negative value would break the distributive property.

Full solution

The integers $7$ and $-7$ add to zero.
Multiplying their sum by $-11$ must therefore give zero.
$(-11)\times(7+(-7))=0$

The distributive property says the same calculation is the sum of the products $(-11)\times7$ and $(-11)\times(-7)$ .
The first is seven groups of $-11$ .
$(-11)\times7=-77$

The two products must add to zero.
$-77+(-11)\times(-7)=0$

The opposite of $-77$ is $77$ , so the second product is forced to be positive.
$(-11)\times(-7)=77$

Check the two choices: $-77+77=0$ , but
$-77+(-77)=-154.$
The negative choice would make the distributed calculation disagree with multiplying the original zero sum.

Answer

$77$ ; a negative value would break the distributive property.

Key idea

The product of two negative integers must be positive for multiplication to stay consistent with distribution over a zero pair.
Problem 8: Mark complete
Problem 9 Two replacements

A product of five nonzero integers is negative. Two of its factors are replaced by their opposites. Is the new product positive, negative or zero? Explain.
1. Hint 1
  Consider what replacing one nonzero factor by its opposite does to the whole product.
2. Hint 2
  Each replacement has the same effect as multiplying the product by $-1$ ; track both replacements.
Answer

Negative, and equal to the original product.

Full solution

Replacing a nonzero factor by its opposite keeps its size and changes its sign.
The number of negative factors therefore goes up by one or down by one.
That turns an odd count even, or an even count odd, so the sign of the product reverses.
The original product is negative, so after the first replacement it is positive.

The second replacement reverses the sign again, back to negative.
A reversal that keeps the size is the same as multiplying by $-1$ , so the two replacements together multiply the original product by $1$ .
$(-1)\times(-1)=1$

The sizes of the factors are unchanged, so the size of the product is unchanged too.
The new product is negative, with the same value as the original product.

Answer

Negative, and equal to the original product.

Key idea

Replacing two nonzero factors by their opposites reverses the product sign twice and leaves the product unchanged.
Problem 9: Mark complete
Problem 10 Recovering an input

A machine accepts any integer and multiplies it by $0$ . For an output of $-12$ and for an output of $0$ , identify all possible inputs. Can either output determine exactly one input? Explain.
1. Hint 1
  To recover an input, ask which numbers reproduce the specified output under the machine operation.
2. Hint 2
  Try a positive input, a negative input and zero, then consider what those trials have in common.
Answer

Output $-12$ : no integer inputs. Output $0$ : every integer input. Neither output determines exactly one input.

Full solution

Every input is multiplied by zero, so every output is zero.
For example,
$\begin{aligned}8\times0&=0\\(-5)\times0&=0\\0\times0&=0\end{aligned}$

An output of $-12$ has no possible input.
Trying to recover it as $(-12)\div0$ fails because no number multiplied by zero gives $-12$ .

An output of $0$ could come from every integer input, so it does not select one input.
Trying to recover it as $0\div0$ also fails to give one definite value.

Neither output determines exactly one input.
These are the two reasons division by zero is undefined: a nonzero dividend has no matching factor, and a zero dividend does not single out a factor.

Answer

Output $-12$ : no integer inputs. Output $0$ : every integer input. Neither output determines exactly one input.

Key idea

Division by zero fails to specify one quotient, while multiplication by zero sends every integer to zero.
Problem 10: Mark complete

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